TEST BANK FOR An Introduction to Management Science Quantitative Approaches to Decision Making, 15 E

An Introduction to Management Science Quantitative Approaches to Decision Making, 15e David Anderson, Dennis Sweeney, Thomas Williams et al (Test Bank All Chapters, 100% Original Verified, A+ Grade) CH 01 - Introduction True / False 1. The process of decision making is more limited than that of problem solving. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.01 - 1.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.1 Problem Solving and Decision Making KEYWORDS: Bloom's: Understand 2. The breakeven point is the point at which the volume of output produced is the result of total revenue equaling total cost. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.04 - 1.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.4 Models of Cost, Revenue, and Profit KEYWORDS: Bloom's: Understand 3. Problem solving encompasses both the identification of a problem and the action to resolve it. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.01 - 1.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.1 Problem Solving and Decision Making KEYWORDS: Bloom's: Remember 4. The decision-making process includes implementation and subsequent evaluation of the decision. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.01 - 1.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.1 Problem Solving and Decision Making © Cengage. Testing Powered by Cognero.

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CH 01 - Introduction KEYWORDS:

Bloom's: Understand

5. Most successful quantitative analysis models will advise separating the management analyst from the managerial team until after the problem has been fully structured. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.3 Quantitative Analysis KEYWORDS: Bloom's: Understand 6. The value of making a decision based on models is dependent on how closely the model represents the real situation. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.3 Quantitative Analysis KEYWORDS: Bloom's: Understand 7. Uncontrollable inputs are the decision variables for a model. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.3 Quantitative Analysis KEYWORDS: Bloom's: Remember 8. The feasible solution is the best solution possible for a mathematical model. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking © Cengage. Testing Powered by Cognero.

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CH 01 - Introduction TOPICS: KEYWORDS:

1.3 Quantitative Analysis Bloom's: Understand

9. Frederic W. Taylor is credited with providing the foundation for quantitative methodology in the early part of the 20th century. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.01 - 1.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.1 Problem Solving and Decision Making KEYWORDS: Bloom's: Remember 10. To identify the choice that provides the highest profit and also uses the fewest employees, we apply a single-criterion decision process. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.01.01 - 1.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.1 Problem Solving and Decision Making KEYWORDS: Bloom's: Understand 11. The most critical component in determining the success or failure of any quantitative approach to decision making is problem definition. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.3 Quantitative Analysis KEYWORDS: Bloom's: Remember 12. The first step in the decision-making process is to identify the problem. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy © Cengage. Testing Powered by Cognero.

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CH 01 - Introduction LEARNING OBJECTIVES: IMS.ASWC.19.01.01 - 1.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.1 Problem Solving and Decision Making KEYWORDS: Bloom's: Remember 13. All uncontrollable inputs or data must be specified before we can analyze the model and recommend a decision or solution for the problem. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.3 Quantitative Analysis KEYWORDS: Bloom's: Understand 14. If you are deciding to buy either machine A, B, or C with the objective of minimizing the sum of labor, material and utility costs, you are dealing with a single-criterion decision. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.01.01 - 1.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.1 Problem Solving and Decision Making KEYWORDS: Bloom's: Understand 15. Model development should be left to quantitative analysts; the model user's involvement should begin at the implementation stage. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.01.01 - 1.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.1 Problem Solving and Decision Making KEYWORDS: Bloom's: Understand 16. A feasible solution is one that satisfies at least one of the constraints in the problem. a. True b. False ANSWER: False © Cengage. Testing Powered by Cognero.

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CH 01 - Introduction POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.3 Quantitative Analysis KEYWORDS: Bloom's: Understand 17. A toy train layout designed to represent an actual railyard is an example of an analog model. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.3 Quantitative Analysis KEYWORDS: Bloom's: Remember 18. The last step in any problem-solving process is to choose the correct alternative among those available. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.01 - 1.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.1 Problem Solving and Decision Making KEYWORDS: Bloom's: Understand 19. Decision variables in a production process are those that cannot be controlled by the manager. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.3 Quantitative Analysis KEYWORDS: Bloom's: Understand 20. The optimal solution to a model is one in which known, specific values provide the best output. a. True b. False ANSWER: True © Cengage. Testing Powered by Cognero.

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CH 01 - Introduction POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.3 Quantitative Analysis KEYWORDS: Bloom's: Understand 21. If all the uncontrollable inputs into the decision-making process are known to the decision maker, the model of decision making is known as "stochastic." a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.3 Quantitative Analysis KEYWORDS: Bloom's: Remember Multiple Choice 22. The field of management science a. concentrates on the use of quantitative methods to assist in decision making. b. approaches decision making rationally, with techniques based on the scientific method. c. is another name for decision science and for operations research. d. All of these are correct.

ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.01 - 1.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.1 Problem Solving and Decision Making KEYWORDS: Bloom's: Remember 23. By identifying and defining a problem, we have a. taken the final step in the decision-making process. b. proposed all viable alternatives. c. considered multiple criteria. d. taken the first step of decision making.

ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.01 - 1.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking © Cengage. Testing Powered by Cognero.

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CH 01 - Introduction TOPICS: KEYWORDS:

1.1 Problem Solving and Decision Making Bloom's: Remember

24. The set of decision alternatives a. should be identified before the decision criteria are established. b. are limited to quantitative solutions. c. are evaluated as a part of the problem definition stage. d. are best generated by brainstorming.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.01 - 1.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.1 Problem Solving and Decision Making KEYWORDS: Bloom's: Remember 25. Decision criteria a. are the choices faced by the decision maker. b. are the problems faced by the decision maker. c. are the ways to evaluate the choices faced by the decision maker. d. are unique for any problem.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.01.01 - 1.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.1 Problem Solving and Decision Making KEYWORDS: Bloom's: Understand 26. In a multicriteria decision problem a. it is impossible to select a single decision alternative. b. the decision maker must evaluate each alternative with respect to each criterion. c. successive decisions must be made over time. d. All of these are correct.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.01.01 - 1.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.1 Problem Solving and Decision Making KEYWORDS: Bloom's: Understand 27. The quantitative analysis approach requires a. the manager's prior experience with a similar problem. © Cengage. Testing Powered by Cognero.

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CH 01 - Introduction b. a relatively uncomplicated problem. c. mathematical expressions for the relationships. d. All of these are correct.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.02 - 1.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.2 Quantitative Analysis and Decision Making KEYWORDS: Bloom's: Remember 28. A thermometer is an example of a model that does not have the same physical appearance as that which is being modeled; thus, it is a(n) a. analog model. b. iconic model. c. mathematical model. d. qualitative model.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.3 Quantitative Analysis KEYWORDS: Bloom's: Understand 29. Inputs to a quantitative model a. are a trivial part of the problem solving process. b. are uncertain for a stochastic model. c. are uncontrollable for the decision variables. d. must all be deterministic if the problem is to have a solution.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.3 Quantitative Analysis KEYWORDS: Bloom's: Understand 30. When the value of the output cannot be determined even if the value of the controllable input is known, the model is a. analog. b. digital. c. stochastic. d. deterministic.

ANSWER: c © Cengage. Testing Powered by Cognero.

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CH 01 - Introduction POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.3 Quantitative Analysis KEYWORDS: Bloom's: Understand 31. The volume that results in total revenue being equal to total cost is known as the a. breakeven point. b. marginal volume. c. marginal cost. d. profit mix.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.04 - 1.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.4 Models of Cost, Revenue, and Profit KEYWORDS: Bloom's: Remember 32. Management science and operations research both involve a. qualitative managerial skills. b. quantitative approaches to decision making. c. operational management skills. d. scientific research as opposed to applications.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.01 - 1.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.1 Problem Solving and Decision Making KEYWORDS: Bloom's: Remember 33. George Dantzig is important in the history of management science because he developed a. the scientific management revolution. b. World War II operations research teams. c. the simplex method for linear programming. d. powerful digital computers.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.01 - 1.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.1 Problem Solving and Decision Making © Cengage. Testing Powered by Cognero.

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CH 01 - Introduction KEYWORDS:

Bloom's: Remember

34. In order to undertake problem solving, the first step must be a. the determination of the correct analytical solution procedure. b. the definition of decision variables. c. the identification of a difference between the actual and desired state of affairs. d. None of these are correct.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.01.01 - 1.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.1 Problem Solving and Decision Making KEYWORDS: Bloom's: Understand 35. The process of problem definition must a. include specific objectives and operating constraints. b. occur prior to the quantitative analysis process. c. involve both the analyst and the user of the results. d. All of these are correct.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.3 Quantitative Analysis KEYWORDS: Bloom's: Understand 36. A model that uses a system of symbols or expressions to represent a problem is called a(n) ______ model. a. mathematical b. iconic c. analog d. constrained ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.3 Quantitative Analysis KEYWORDS: Bloom's: Understand 37. Which of the following is NOT one of the commonly used names for the body of knowledge involving quantitative approaches to decision making? a. management science © Cengage. Testing Powered by Cognero.

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CH 01 - Introduction b. business analytics c. operations research d. efficiency studies ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.01.01 - 1.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.1 Problem Solving and Decision Making KEYWORDS: Bloom's: Remember 38. A valid reason for using a quantitative approach to the decision-making process is when a. the problem is repetitive, necessitating routine decision making. b. the problem is unique and the manager has no prior experience solving this sort of problem. c. the problem is particularly complex. d. All of these are correct.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.01.02 - 1.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 1.2 Quantitative Analysis and Decision Making KEYWORDS: Bloom's: Understand Subjective Short Answer 39. A snack food manufacturer buys corn for tortilla chips from two cooperatives, one in Iowa and one in Illinois. The price per unit of the Iowa corn is $6.00 and the price per unit of the Illinois corn is $5.50. a. Define variables that would tell how many units to purchase from each source. b. Develop an objective function that would minimize the total cost.

The manufacturer needs at least 12,000 units of corn. The Iowa cooperative can supply up to c. 8000 units, and the Illinois cooperative can supply at least 6000 units. Develop constraints for these conditions.

ANSWER: a. b. c.

Let x1 = the number of units from Iowa Let x2 = the number of units from Illinois Min 6x1 + 5.5x2 x1 + x 2 ≥ 12,000 x1 ≥ 8000 x1 ≥ 6000

POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic © Cengage. Testing Powered by Cognero.

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CH 01 - Introduction TOPICS: KEYWORDS:

1.3 Quantitative Analysis Bloom's: Apply

40. The relationship d = 5000 − 25p describes what happens to demand (d) as price (p) varies. Here, price can vary between $10 and $50. a. How many units can be sold at the $10 price? How many can be sold at the $50 price? b. Model the expression for total revenue.

Consider prices of $20, $30, and $40. Which of these three price alternatives will maximize c. total revenue? What are the values for demand and revenue at this price?

ANSWER:

a. b. c.

For p = 10, d = 4750 For p = 50, d = 3750 TR = p(5000 − 25p) For p = 20, d = 4500, TR = $90,000 For p = 30, d = 4250, TR = $127,500 For p = 40, d = 4000, TR = $160,000 (maximum total revenue)

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 1.3 Quantitative Analysis KEYWORDS: Bloom's: Apply 41. There is a fixed cost of $50,000 to start a production process. Once the process has begun, the variable cost per unit is $25. The revenue per unit is projected to be $45. a. Write an expression for total cost. b. Write an expression for total revenue. c. Write an expression for total profit. d. Find the breakeven point.

ANSWER:

a. C(x) = 50000 + 25x b. R(x) = 45x c. P(x) = 45x − (50000 + 25x) d. x = 2500 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.01.04 - 1.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 1.4 Models of Cost, Revenue, and Profit KEYWORDS: Bloom's: Apply 42. An author has received an advance against royalties of $10,000. The royalty rate is $1.00 for every book sold in the United States, and $1.35 for every book sold outside the United States. Define variables for this problem and write an expression that could be used to calculate the number of books to be sold to cover the advance.

ANSWER: Let x1 = the number of books sold in the U.S. Let x2 = the number of books sold outside the U.S. 10000 = 1x1 + 1.35x2 © Cengage. Testing Powered by Cognero.

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CH 01 - Introduction POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.01.04 - 1.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 1.4 Models of Cost, Revenue, and Profit KEYWORDS: Bloom's: Apply 43. A university schedules summer school courses based on anticipated enrollment. The cost for faculty compensation, laboratories, student services, and allocated overhead for a computer class is $8500. If students pay $920 to enroll in the course, how large would enrollment have to be for the university to break even?

ANSWER: Enrollment would need to be 10 students.

POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.01.04 - 1.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 1.4 Models of Cost, Revenue, and Profit KEYWORDS: Bloom's: Evaluate 44. As part of their application for a loan to buy Lakeside Farm, a property they hope to develop as a bed-and-breakfast operation, the prospective owners have projected: Monthly fixed cost (loan payment, taxes, insurance, maintenance) Variable cost per occupied room per night Revenue per occupied room per night a. b. c.

$6000 20 75

Write the expression for total cost per month. Assume 30 days per month.

Write the expression for total revenue per month.

If there are 12 guest rooms available, can they break even? What percentage of rooms would need to be occupied, on average, to break even?

ANSWER:

a. b. c.

C(x) = 6000 + 20(30)x (monthly) R(x) = 75(30)x (monthly) Breakeven occupancy = 3.64 or 4 occupied rooms per night, so they have enough rooms to breakeven. This would be a 33% occupancy rate.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.01.04 - 1.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 1.4 Models of Cost, Revenue, and Profit KEYWORDS: Blooms': Apply 45. Organizers of an Internet training session will charge participants $150 to attend. It costs $3000 to reserve the room, hire the instructor, bring in the equipment, and advertise. Assume it costs $25 per student for the organizers to provide the course materials. a. How many students would have to attend for the company to break even?

If the trainers think, realistically, that 20 people will attend, then what price should be b. charged per person for the organization to break even? © Cengage. Testing Powered by Cognero.

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CH 01 - Introduction ANSWER:

a.

b.

C(x) = 3000 + 25x R(x) = 150x Breakeven students = 24 Cost = 3000 + 25(20) Revenue = 20p Breakeven price = $175

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.01.04 - 1.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 1.4 Models of Cost, Revenue, and Profit KEYWORDS: Bloom's: Apply 46. In this portion of an Excel spreadsheet, the user has given values for selling price, the costs, and a sample volume.

Give the cell formula for a. cell E12, breakeven volume. b. cell E16, total revenue. c. cell E17, total cost. d. cell E19, profit (loss).

A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

B

C

D

E

Breakeven calculation Selling price per unit

10

Costs Fix cost Variable cost per unit

8400 4.5

Breakeven volume Sample calculation Volume Total revenue Total cost

2000

Profit (loss)

ANSWER:

a. =E9/(E6-E10) b. =E15*E6 c. =E9+E10*E15 d. =E16-E17 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.01.05 - 1.5 © Cengage. Testing Powered by Cognero.

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CH 01 - Introduction NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 1.5 Management Science Techniques KEYWORDS: Bloom's: Apply 47. A furniture store has set aside 800 square feet to display its sofas and chairs. Each sofa utilizes 50 sq. ft. and each chair utilizes 30 sq. ft. At least five sofas and at least five chairs are to be displayed. a. Write a mathematical model representing the store's constraints.

Suppose the profit on sofas is $200 and on chairs is $100. On a given day, the probability b. that a displayed sofa will be sold is .03 and that a displayed chair will be sold is .05.

Mathematically model each of the following objectives: 1. Maximize the total pieces of furniture displayed.

2. Maximize the total expected number of daily sales.

3. Maximize the total expected daily profit.

ANSWER: a. 50s + 30c ≤ 800 s≥5 c≥5 b. (1) Max s + c (2) Max .03s + .05c (3) Max 6s + 5c POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 1.3 Quantitative Analysis KEYWORDS: Bloom's: Apply 48. A manufacturer makes two products, doors and windows. Each must be processed through two work areas. Work area #1 has 60 hours of available production time per week. Work area #2 has 48 hours of available production time per week.

Manufacturing of a door requires 4 hours in work area #1 and 2 hours in work area #2. Manufacturing of a window requires 2 hours in work area #1 and 4 hours in work area #2. Profit is $8 per door and $6 per window. a. Define decision variables that will tell how many units to build (doors and windows) per week. b. Develop an objective function that will maximize total profit per week. c. Develop production constraints for work area #1 and #2.

ANSWER: a. Let D = the number of doors to build per week Let N = the number of windows to build per week b. Weekly Profit = 8D + 6W c. 4D + 2W ≤ 60 2D + 4W ≤ 48 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 1.3 Quantitative Analysis KEYWORDS: Bloom's: Apply 49. A firm builds and sells small, ergonomic conference tables. The investment in plant and equipment is $165,000. The variable cost per table is $1,500. The selling price of each table is $3,000. How many tables would have to be sold for the © Cengage. Testing Powered by Cognero.

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CH 01 - Introduction firm to break even?

ANSWER: 110 tables POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.01.04 - 1.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 1.4 Models of Cost, Revenue, and Profit KEYWORDS: Bloom's: Evaluate 50. A computer rework center has the capacity to rework 300 computers per day. The expected number of computers needing to be reworked per day is 225. The center is paid $26 for each computer reworked. The fixed cost of renting the reworking equipment is $250 per day. Work space rents for $150 per day. The cost of material is $18 per computer and labor costs $3 per computer. What is the breakeven number of computers reworked per day?

ANSWER: 80 computers POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.01.04 - 1.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 1.4 Models of Cost, Revenue, and Profit KEYWORDS: Bloom's: Evaluate 51. To establish a driver education school, organizers must decide how many cars, instructors, and students to have. Costs are estimated as follows. Annual fixed costs to operate the school are $30,000. The annual cost per car is $3000. The annual cost per instructor is $11,000 and one instructor is needed for each car. Tuition for each student is $350. Let x be the number of cars and y be the number of students. a. Write an expression for total cost. b. Write an expression for total revenue. c. Write an expression for total profit. d. The school offers the course eight times each year. Each time the course is offered, there are two sessions. If they decide to operate five cars, and if four students can be assigned to each car, will they break even?

ANSWER: a. C(x) = 30000 + 14000x b. R(y) = 350y c. P(x,y) = 350y − (30000 + 14000x) d. Each car/instructor can serve up to (4 students/session)(2 sessions/course)(8 courses/year) = 64 students annually. Five cars can serve 320 students. If the classes are filled, then profit for five cars is 350(320) − (30000 + 14000(5)) = 12000. So, the school can reach the breakeven point.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.01.04 - 1.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 1.4 Models of Cost, Revenue, and Profit KEYWORDS: Bloom's: Apply 52. Zipco Printing operates a shop that has five printing machines. The machines differ in their capacities to perform © Cengage. Testing Powered by Cognero.

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CH 01 - Introduction various printing operations due to differences in the machines' designs and operator skill levels. At the start of the workday there are five printing jobs to schedule. The manager must decide what the job-machine assignments should be. a. How could a quantitative approach to decision making be used to solve this problem? b. What would be the uncontrollable inputs for which data must be collected?

Define the decision variables, objective function, and constraints to appear in the c. mathematical model. d. Is the model deterministic or stochastic? e. Suggest some simplifying assumptions for this problem.

ANSWER: a. A quantitative approach to decision making can provide a systematic way for deciding the job-machine pairings so that total job processing time is minimized. b. How long it takes to process each job on each machine, and any job-machine pairings that are unacceptable. c. Decision variables: one for each job-machine pairing, taking on a value of 1 if the pairing is used and 0 otherwise.

Objective function: minimize total job processing time.

Constraints: each job is assigned to exactly one machine, and each machine be assigned no more than one job. d. Stochastic: job processing times vary due to varying machine set-up times, variable operator performance, and more. e. Assume that processing times are deterministic (known/fixed).

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 1.3 Quantitative Analysis KEYWORDS: Bloom's: Evaluate 53. Consider a department store that must make weekly shipments of a certain product from two different warehouses to four different stores. a. How could a quantitative approach to decision making be used to solve this problem? b. What would be the uncontrollable inputs for which data must be gathered?

What would be the decision variables of the mathematical model? the objective function? the c. constraints? d. Is the model deterministic or stochastic? e. Suggest assumptions that could be made to simplify the model.

ANSWER: A quantitative approach to decision making can provide a systematic way to determine a a. minimum shipping cost from the warehouses to the stores.

Fixed costs and variable shipping costs; the demand each week at each store; the b. supplies each week at each warehouse.

Decision variables--how much to ship from each warehouse to each store; objective c. function--minimize total shipping costs; constraints--meet the demand at the stores without exceeding the supplies at the warehouses.

Stochastic--weekly demands fluctuate as do weekly supplies; transportation costs could d. vary depending upon the amount shipped, other goods sent with a shipment, etc.

Make the model deterministic by assuming fixed shipping costs per item, demand is e. constant at each store each week, and weekly supplies in the warehouses are constant.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.01.03 - 1.3 © Cengage. Testing Powered by Cognero.

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CH 01 - Introduction NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 1.3 Quantitative Analysis KEYWORDS: Bloom's: Evaluate 54. Three production processes—A, B, and C—have the following cost structure: Process

Fixed Cost per Year

Variable Cost per Unit

A B C

$120,000 90,000 80,000

$3.00 4.00 4.50

a. What is the most economical process for a volume of 8,000 units?

How many units per year must be sold with each process to have annual profits of $50,000 if the selling price is $6.95 b. per unit? c. What is the breakeven volume for each process?

ANSWER:

a. C(x) = FC + VC(x) Process A: C(x) = $120,000 + $3.00(8,000) = $144,000 per year Process B: C(x) = $ 90,000 + $4.00(8,000) = $122,000 per year Process C: C(x) = $ 80,000 + $4.50(8,000) = $116,000 per year Process C has the lowest annual cost for a production volume of 8,000 units. b. Q = (profit + FC)/(price - VC) Process A: Q = ($50,000 + $120,000)/($6.95 - $3.00) = 43,038 units Process B: Q = ($50,000 + $ 90,000)/($6.95 - $4.00) = 47,458 units Process C: Q = ($50,000 + $ 80,000)/($6.95 - $4.50) = 53,062 units Process A requires the lowest production volume for an annual profit of $50,000. c. At breakeven, profit (the pretax profits per period) is equal to zero.

Q = FC/(price - VC) Process A: Q = $120,000/ ($6.95 - $3.00) = 30,380 units Process B: Q = $ 90,000/ ($6.95 - $4.00) = 30,509 units Process C: Q = $ 80,000/ ($6.95 - $4.50) = 32,654 units Process A has the lowest breakeven quantity, while Process B’s is almost as low.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.01.04 - 1.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 1.4 Models of Cost, Revenue, and Profit KEYWORDS: Bloom's: Apply 55. Jane Persico, facility engineer at the El Paso plant of Computer Products Corporation (CPC), is studying a process selection decision at the plant. A new printer is to be manufactured and she must decide whether the printer will be autoassembled or manually assembled. The decision is complicated by the fact that annual production volume is expected to increase by almost 50% over three years. Jane has developed these estimates for two alternatives for the printer assembly process: © Cengage. Testing Powered by Cognero.

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CH 01 - Introduction

Annual fixed cost Variable cost per product Estimated annual production (in number of products):

Year 1 Year 2 Year 3

AutoAssembly Process

Manual Assembly Process

$690,000 $29.56

$269,000 $31.69

152,000 190,000 225,000

152,000 190,000 225,000

a. Which production process would be the least-cost alternative in Years 1, 2, and 3? b. How much would the variable cost per unit have to be in Year 2 for the auto-assembly process to justify the additional annual fixed cost for the auto-assembly process over the manual assembly process?

ANSWER:

a. C(x) = fixed cost + variable cost(x) Year 1: CA = 690,000 + 29.56(152,000) = $5,183,120 CM = 269,000 + 31.69(152,000) = $5,085,880 (least-cost alternative) Year 2: CA = 690,000 + 29.56(190,000) = $6,306,400 CM = 269,000 + 31.69(190,000) = $6,290,100 (least-cost alternative) Year 3: CA = 690,000 + 29.56(225,000) = $7,341,000 (least-cost alternative) CM = 269,000 + 31.69(225,000) = $7,399,250 b. CA = CM FCA + vA(190,000) = FCM + vM(190,000) 690,000 + v(190,000) = 269,000 + 31.69(190,000) vA = (269,000 + 6,021,100 - 690,000)/190,000 vA = $29.47 (roughly a 0.3% reduction)

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.01.04 - 1.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 1.4 Models of Cost, Revenue, and Profit KEYWORDS: Bloom's: Analyze

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CH 02 - An Introduction to Linear Programming True / False 1. In a linear programming problem, the objective function and the constraints must be linear functions of the decision variables. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.01 - 2.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.1 A Simple Maximization Problem KEYWORDS: Bloom's: Remember 2. Only binding constraints form the shape (boundaries) of the feasible region. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Remember 3. It is not possible to have more than one optimal solution to a linear programming problem. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Understand 4. A linear programming problem can be both unbounded and infeasible. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.06 - 2.6 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.6 Special Cases © Cengage. Testing Powered by Cognero.

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CH 02 - An Introduction to Linear Programming KEYWORDS:

Bloom's: Understand

5. An infeasible problem is one in which the objective function can be increased to infinity. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.06 - 2.6 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.6 Special Cases KEYWORDS: Bloom's: Understand 6. An unbounded feasible region might not result in an unbounded solution for a minimization or maximization problem. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.06 - 2.6 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.6 Special Cases KEYWORDS: Bloom's: Understand 7. An optimal solution to a linear programming problem can be found at an extreme point of the feasible region for the problem. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.03 - 2.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.3 Extreme Points and the Optimal Solution KEYWORDS: Bloom's: Understand 8. The optimal solution to any linear programming problem is the same as the optimal solution to the standard form of the problem. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.01 - 2.1 © Cengage. Testing Powered by Cognero.

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CH 02 - An Introduction to Linear Programming NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.1 A Simple Maximization Problem KEYWORDS: Bloom's: Understand 9. The constraint 2x1 − x2 = 0 passes through the point (200, 100). a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Understand 10. The point (3, 2) is feasible for the constraint 2x1 + 6x2 ≤ 30. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Understand 11. No matter what value it has, each objective function line is parallel to every other objective function line in a problem. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Understand 12. Constraints limit the degree to which the objective in a linear programming problem is satisfied. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.01 - 2.1 © Cengage. Testing Powered by Cognero.

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CH 02 - An Introduction to Linear Programming NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.1 A Simple Maximization Problem KEYWORDS: Bloom's: Remember 13. Alternative optimal solutions occur when there is no feasible solution to the problem. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.06 - 2.6 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.6 Special Cases KEYWORDS: Bloom's: Understand 14. Because surplus variables represent the amount by which the solution exceeds a minimum target, they are given positive coefficients in the objective function. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.02.01 - 2.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.1 A Simple Maximization Problem KEYWORDS: Bloom's: Understand 15. A redundant constraint cannot be removed from the problem without affecting the feasible region. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Understand 16. The constraint 5x1 − 2x2 ≤ 0 passes through the point (20, 50). a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate © Cengage. Testing Powered by Cognero.

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CH 02 - An Introduction to Linear Programming LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Understand 17. At a problem's optimal solution, a redundant constraint will have zero slack. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Blooms: Understand 18. If a constraint is redundant, it can be removed from the problem without affecting the feasible region. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Understand 19. For a minimization problem, the solution is considered to be unbounded if the value may be made infinitely small. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.06 - 2.6 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.6 Special Cases KEYWORDS: Bloom's: Remember Multiple Choice 20. The maximization or minimization of a desired quantity is the a. goal of management science. b. decision for decision analysis. c. constraint of operations research. d. objective of linear programming. © Cengage. Testing Powered by Cognero.

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CH 02 - An Introduction to Linear Programming ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.01 - 2.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.1 A Simple Maximization Problem KEYWORDS: Bloom's: Remember 21. Decision variables a. are values that are used to determine how much or how many of something to produce, invest, etc. b. represent the values of the constraints. c. are values that measure the objective function. d. must be unique for each constraint.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.01 - 2.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.1 A Simple Maximization Problem KEYWORDS: Bloom's: Understand 22. Which of the following is a valid objective function for a linear programming problem? a. Min 8xy b. Min 4x + 3y + (1/2)z c. Min 5x2 + 6y2 d. Max (x1 + x2)/x3 ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.01 - 2.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.1 A Simple Maximization Problem KEYWORDS: Bloom's: Understand 23. Which of the following statements is NOT true? a. A feasible solution satisfies all constraints. b. An optimal solution satisfies all constraints. c. An infeasible solution violates all constraints. d. A feasible solution point does not have to lie on the boundary of the feasible region.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 © Cengage. Testing Powered by Cognero.

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CH 02 - An Introduction to Linear Programming NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Understand 24. When no solution to the linear programming problem satisfies all the constraints, including the nonnegativity conditions, it is considered a. optimal. b. feasible. c. infeasible. d. semifeasible.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.02.06 - 2.6 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.6 Special Cases KEYWORDS: Bloom's: Understand 25. The amount by which the left side of a less-than-or-equal-to constraint is smaller than the right side a. is known as a surplus. b. is known as slack. c. is optimized for the linear programming problem. d. exists for each variable in a linear programming problem.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Understand 26. To find the optimal solution to a linear programming problem using the graphical method, a. find the feasible point that is the farthest away from the origin. b. find the feasible point that is at the highest location. c. find the feasible point that is closest to the origin. d. None of these are correct.

ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.03 - 2.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.3 Extreme Points and the Optimal Solution KEYWORDS: Blooms: Understand

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CH 02 - An Introduction to Linear Programming 27. Which of the following special cases does NOT require reformulation of the problem in order to obtain a solution? a. alternative optimality b. infeasibility c. unboundedness d. Each case requires a reformulation.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.02.06 - 2.6 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.6 Special Cases KEYWORDS: Bloom's: Understand 28. Infeasibility means that the number of solutions to the linear programming models that satisfies all constraints is a. at least 1. b. 0. c. an infinite number. d. at least 2.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.06 - 2.6 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.6 Special Cases KEYWORDS: Bloom's: Remember 29. A constraint that does NOT affect the feasible region of the solution is a a. nonnegativity constraint. b. redundant constraint. c. standard constraint. d. slack constraint.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Remember 30. Whenever all the constraints in a linear program are expressed as equalities, the linear program is said to be written in a. standard form. b. bounded form. c. feasible form. d. alternative form. © Cengage. Testing Powered by Cognero.

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CH 02 - An Introduction to Linear Programming ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Remember 31. All of the following statements about a redundant constraint are correct EXCEPT a. a redundant constraint does not affect the optimal solution. b. a redundant constraint does not affect the feasible region. c. recognizing a redundant constraint is easy with the graphical solution method. d. at the optimal solution, a redundant constraint will have zero slack.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Understand 32. All linear programming problems have all of the following properties EXCEPT a. a linear objective function that is to be maximized or minimized. b. a set of linear constraints. c. alternative optimal solutions. d. variables that are all restricted to nonnegative values.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.02.01 - 2.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.1 A Simple Maximization Problem KEYWORDS: Bloom's: Understand 33. If there is a maximum of 4,000 hours of labor available per month and 300 ping-pong balls (x1) or 125 wiffle balls (x2) can be produced per hour of labor, which of the following constraints reflects this situation? a. 300x1 + 125x2 > 4,000 b. 300x1 + 125x2 < 4,000 c. 425(x1 + x2) < 4,000 d. 300x1 + 125x2 = 4,000 ANSWER: POINTS: DIFFICULTY:

b 1 Moderate

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CH 02 - An Introduction to Linear Programming LEARNING OBJECTIVES: IMS.ASWC.19.02.01 - 2.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.1 A Simple Maximization Problem KEYWORDS: Bloom's: Apply 34. In which part(s) of a linear programming formulation would the decision variables be stated? a. objective function and the left-hand side of each constraint b. objective function and the right-hand side of each constraint c. the left-hand side of each constraint only d. the objective function only ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.01 - 2.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.1 A Simple Maximization Problem KEYWORDS: Bloom's: Understand 35. The three assumptions necessary for a linear programming model to be appropriate include all of the following EXCEPT a. proportionality. b. additivity. c. divisibility. d. normality.

ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.01 - 2.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.1 A Simple Maximization Problem KEYWORDS: Bloom's: Remember 36. A redundant constraint results in a. no change in the optimal solution(s). b. an unbounded solution. c. no feasible solution. d. alternative optimal solutions.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Remember © Cengage. Testing Powered by Cognero.

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CH 02 - An Introduction to Linear Programming 37. A variable added to the left-hand side of a less-than-or-equal-to constraint to convert the constraint into an equality is a a. standard variable. b. slack variable. c. surplus variable. d. nonnegative variable.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Remember Subjective Short Answer 38. Solve the following system of simultaneous equations.

6X + 2Y = 50 2X + 4Y = 20 ANSWER: X = 8, Y =1 POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Apply 39. Solve the following system of simultaneous equations.

6X + 4Y = 40 2X + 3Y = 20 ANSWER: X = 4, Y = 4 POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Apply 40. Consider the following linear programming problem: Max s.t.

8X + 7Y 15X + 5Y ≤ 75 10X + 6Y ≤ 60 X+ Y≤8

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CH 02 - An Introduction to Linear Programming X, Y ≥ 0 a.

Use a graph to show each constraint and the feasible region.

Identify the optimal solution point on your graph. What are the values of X and Y at the b. optimal solution? c. What is the optimal value of the objective function?

ANSWER:

a.

b.

The optimal solution occurs at the intersection of constraints 2 and 3. The point is X = 3, Y = 5.

The value of the objective function is 59.

c.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Apply

41. For the following linear programming problem, determine the optimal solution using the graphical solution method. −X + 2Y 6X − 2Y ≤ 3 −2X + 3Y ≤ 6 X+ Y≤3 X, Y ≥ 0 ANSWER: Max s.t.

X = 0.6 and Y = 2.4

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CH 02 - An Introduction to Linear Programming

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Apply 42. Use this graph to answer the questions.

Max s.t.

20X + 10Y 12X + 15Y ≤ 180 15X + 10Y ≤ 150 3X − 8Y ≤ 0 X, Y ≥ 0

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CH 02 - An Introduction to Linear Programming a. b. c. d.

Which area (I, II, III, IV, or V) forms the feasible region?

Which point (A, B, C, D, or E) is optimal?

Which constraints are binding?

Which slack variables equal zero?

ANSWER:

a. b. c. d.

Area III is the feasible region.

Point D is optimal.

Constraints 2 and 3 are binding.

S2 and S3 are equal to 0.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Analyze 43. Find the complete optimal solution to this linear programming problem.

Min s.t.

5X + 6Y 3X + Y ≥ 15 X + 2Y ≥ 12 3X + 2Y ≥ 24 X, Y ≥ 0

ANSWER:

The complete optimal solution is X = 6, Y = 3, Z = 48, S1 = 6, S2 = 0, S3 = 0 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic © Cengage. Testing Powered by Cognero.

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CH 02 - An Introduction to Linear Programming TOPICS: KEYWORDS:

2.2 Graphical Solution Procedure Bloom's: Apply

44. Find the complete optimal solution to this linear programming problem.

Max s.t.

5X + 3Y 2X + 3Y ≤ 30 2X + 5Y ≤ 40 6X − 5Y ≤ 0 X, Y ≥ 0

ANSWER:

The complete optimal solution is X = 15, Y = 0, Z = 75, S1 = 0, S2 = 10, S3 = 90 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Analyze 45. Find the complete optimal solution to this linear programming problem.

Max s.t.

2X + 3Y 4X + 9Y ≤ 72 10X + 11Y ≤ 110 17X + 9Y ≤ 153 X, Y ≥ 0

ANSWER:

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CH 02 - An Introduction to Linear Programming

The complete optimal solution is X = 4.304, Y = 6.087, Z = 26.87, S1 = 0, S2 = 0, S3 = 25.043 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Analyze 46. Find the complete optimal solution to this linear programming problem.

Min s.t.

3X + 3Y 12X + 4Y ≥ 48 10X + 5Y ≥ 50 4X + 8Y ≥ 32 X, Y ≥ 0

ANSWER:

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CH 02 - An Introduction to Linear Programming

The complete optimal solution is X = 4, Y = 2, Z = 18, S1 = 8, S2 = 0, S3 = 0 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Analyze 47. For the following linear programming problem, determine the optimal solution using the graphical solution method.

Are any of the constraints redundant? If yes, identify the constraint that is redundant.

Max s.t.

X + 2Y X+ Y≤3 X − 2Y ≥ 0 Y≤1 X, Y ≥ 0 ANSWER:

X = 2 and Y = 1 Yes, there is a redundant constraint; Y ≤ 1

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CH 02 - An Introduction to Linear Programming

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Analyze 48. Maxwell Manufacturing makes two models of felt-tip marking pens. Requirements for each lot of pens are given below.

Plastic Ink assembly Molding time

Fliptop Model 3 5 5

Tiptop Model 4 4 2

Available 36 40 30

The profit for either model is $1000 per lot. a. What is the linear programming model for this problem? b. Find the optimal solution. c. Will there be excess capacity in any resource?

ANSWER:

a.

Let F = number of lots of Fliptop pens to produce T = number of lots of Tiptop pens to produce Max s.t.

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1000F + 1000T 3F + 4T ≤ 36 5F + 4T ≤ 40 5F + 2T ≤ 30 F, T ≥ 0

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CH 02 - An Introduction to Linear Programming

b.

The complete optimal solution is F = 2, T = 7.5, Z = 9500, S1 = 0, S2 = 0, S3 = 5 c. There is an excess of 5 units of molding time available.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.02.01 - 2.1 IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.1 A Simple Maximization Problem 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Analyze 49. The Sanders Garden Shop mixes two types of grass seed into a blend. Each type of grass has been rated (per pound) according to its shade tolerance, ability to stand up to traffic, and drought resistance, as shown in the table. Type A seed costs $1 and Type B seed costs $2.

Shade tolerance Traffic resistance Drought resistance

Type A 1 2 2

Type B 1 1 5

a. If the blend needs to score at least 300 points for shade tolerance, 400 points for traffic resistance, and 750 points for drought resistance, how many pounds of each seed should be in the blend? b. Which targets will be exceeded? c. How much will the blend cost?

ANSWER:

a. Let A = pounds of Type A seed in the blend B = pounds of Type B seed in the blend Min s.t.

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1A + 2B 1A + 1B ≥ 300 2A + 1B ≥ 400 2A + 5B ≥ 750 A, B ≥ 0

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CH 02 - An Introduction to Linear Programming

The optimal solution is at A = 250, B = 50. b. Constraint 2 has a surplus value of 150. c. The cost is 350.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.02.01 - 2.1 IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.2 Graphical Solution Procedure 2.1 A Simple Maximization Problem KEYWORDS: Bloom's: Analyze 50. Muir Manufacturing produces two popular grades of commercial carpeting among its many other products. In the coming production period, Muir needs to decide how many rolls of each grade should be produced in order to maximize profit. Each roll of Grade X carpet uses 50 units of synthetic fiber, requires 25 hours of production time, and needs 20 units of foam backing. Each roll of Grade Y carpet uses 40 units of synthetic fiber, requires 28 hours of production time, and needs 15 units of foam backing.

The profit per roll of Grade X carpet is $200, and the profit per roll of Grade Y carpet is $160. In the coming production period, Muir has 3000 units of synthetic fiber available for use. Workers have been scheduled to provide at least 1800 hours of production time (overtime is a possibility). The company has 1500 units of foam backing available for use.

Develop and solve a linear programming model for this problem.

ANSWER: Let X = number of rolls of Grade X carpet to make Y = number of rolls of Grade Y carpet to make Max

200X + 160Y

s.t.

50X + 40Y ≤ 3000 25X + 28Y ≥ 1800 20X + 15Y ≤ 1500 X, Y ≥ 0

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CH 02 - An Introduction to Linear Programming

The complete optimal solution is X = 30, Y = 37.5, Z = 12,000, S1 = 0, S2 = 0, S3 = 337.5 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.02.01 - 2.1 IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.2 Graphical Solution Procedure 2.1 A Simple Maximization Problem KEYWORDS: Bloom's: Analyze 51. Does the following linear programming problem exhibit infeasibility, unboundedness, or alternative optimal solutions? Explain.

Min s.t.

3X + 3Y 1X + 2Y ≤ 16 1X + 1Y ≤ 10 5X + 3Y ≤ 45 X, Y ≥ 0

ANSWER:

The problem has alternative optimal solutions.

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POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.02.06 - 2.6 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.6 Special Cases KEYWORDS: Bloom's: Analyze 52. Does the following linear programming problem exhibit infeasibility, unboundedness, or alternative optimal solutions? Explain.

Min s.t.

1X + 1Y 5X + 3Y ≤ 30 3X + 4Y ≥ 36 Y≤7 X, Y ≥ 0 ANSWER:

The problem is infeasible.

POINTS: DIFFICULTY:

1 Challenging

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CH 02 - An Introduction to Linear Programming LEARNING OBJECTIVES: IMS.ASWC.19.02.06 - 2.6 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.6 Special Cases KEYWORDS: Bloom's: Analyze 53. A businessman is considering opening a small specialized trucking firm. To make the firm profitable, it must have a daily trucking capacity of at least 84,000 cubic feet. Two types of trucks are appropriate for the specialized operation.

Their characteristics and costs are summarized in the table below. Note that truck two requires three drivers for long haul trips. There are 41 potential drivers available, and there are facilities for at most 40 trucks. The businessman's objective is to minimize the total cost outlay for trucks.

Truck Small Large

Cost $18,000 $45,000

Capacity (cu. ft.)

2,400 6,000

Drivers Needed 1 3

Solve the problem graphically and note that there are alternative optimal solutions. a. Which optimal solution uses only one type of truck? b. Which optimal solution utilizes the minimum total number of trucks? c. Which optimal solution uses the same number of small and large trucks?

ANSWER:

a. 35 small, 0 large b. 5 small, 12 large c. 10 small, 10 large POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.02.06 - 2.6 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.6 Special Cases KEYWORDS: Bloom's: Analyze 54. Consider the following linear program: Max s.t.

60X + 43Y X + 3Y ≥ 9 6X − 2Y = 12 X + 2Y ≤ 10 X, Y ≥ 0

a. b.

Write the problem in standard form.

What is the feasible region for the problem?

Show that regardless of the values of the actual objective function coefficients, the optimal c. solution will occur at one of two points. Solve for these points and then determine which one maximizes the current objective function.

ANSWER: a. Max 60X + 43Y s.t.

X + 3Y − S1 = 9 6X − 2Y = 12 X + 2Y + S3 = 10 X, Y, S1, S3 ≥ 0 © Cengage. Testing Powered by Cognero.

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CH 02 - An Introduction to Linear Programming b. Line segment of 6X − 2Y = 12 between (22/7, 24/7) and (27/10, 21/10). c. Extreme points: (22/7, 24/7) and (27/10, 21/10). First one is optimal, giving Z = 336.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.02.03 - 2.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.3 Extreme Points and the Optimal Solution KEYWORDS: Bloom's: Analyze 55. Solve the following linear program graphically.

Max s.t.

5X + 7Y X ≤6 2X + 3Y ≤ 19 X+ Y≤8 X, Y ≥ 0 ANSWER:

From the graph below, we see that the optimal solution occurs at X = 5, Y = 3, and Z = 46.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.2 Graphical Solution Procedure KEYWORDS: Bloom's: Analyze 56. Solve the following linear program graphically. How many extreme points exist for this problem?

Min s.t.

150X + 210Y 3.8X + 1.2Y ≥ 22.8 Y≥6 Y ≤ 15

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CH 02 - An Introduction to Linear Programming 45X + 30Y = 630 X, Y ≥ 0 ANSWER:

Two extreme points exist (points A and B below). The optimal solution is X = 10, Y = 6, and Z = 2760 (point B).

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 IMS.ASWC.19.02.03 - 2.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.2 Graphical Solution Procedure 2.3 Extreme Points and the Optimal Solution KEYWORDS: Bloom's: Analyze 57. Solve the following linear program graphically.

Max s.t.

4X + 5Y X + 3Y ≤ 22 −X + Y ≤ 4 Y≤6 2X − 5Y ≤ 0 X, Y ≥ 0 ANSWER:

Two extreme points exist (points A and B below). The optimal solution is X = 10, Y = 6, and Z = 2760 (point B).

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CH 02 - An Introduction to Linear Programming

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.02.02 - 2.2 IMS.ASWC.19.02.03 - 2.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 2.2 Graphical Solution Procedure 2.3 Extreme Points and the Optimal Solution KEYWORDS: Bloom's: Analyze

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CH 03 - Linear Programming: Sensitivity Analysis and Interpretation of Solution True / False 1. Classical sensitivity analysis provides no information about changes resulting from a change in the coefficient of a variable in a constraint. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.03.04 - 3.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.4 Limitations of Classical Sensitivity Analysis KEYWORDS: Bloom's: Understand 2. The reduced cost of a variable is the dual value of the corresponding nonnegativity constraint. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.03.03 - 3.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.3 Sensitivity Analysis: Computer Solution KEYWORDS: Bloom's: Understand 3. When the right-hand sides of two constraints are each increased by one unit, the objective function value will be adjusted by the sum of the constraints' dual prices. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.03.02 - 3.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.2 Graphical Sensitivity Analysis KEYWORDS: Bloom's: Understand 4. If the range of feasibility indicates that the original amount of a resource, which was 20, can increase by 5, then the amount of the resource can increase to 25. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.03.03 - 3.3 © Cengage. Testing Powered by Cognero.

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CH 03 - Linear Programming: Sensitivity Analysis and Interpretation of Solution NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.3 Sensitivity Analysis: Computer Solution KEYWORDS: Bloom's: Apply 5. When two or more objective function coefficients are changed simultaneously, further analysis is necessary to determine whether the optimal solution will change. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.03.02 - 3.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 3.2 Graphical Sensitivity Analysis KEYWORDS: Bloom's: Understand 6. The dual value and dual price are identical for a minimization problem. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.A3.02 - A3.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: Appendix 3.2 Sensitivity Analysis with Lingo KEYWORDS: Bloom's: Understand 7. A negative dual price indicates that increasing the right-hand side of the associated constraint would be detrimental to the objective. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.03.02 - 3.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.2 Graphical Sensitivity Analysis KEYWORDS: Bloom's: Understand 8. In order to tell the impact of a change in a constraint coefficient, the change must be made and then the model resolved. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy © Cengage. Testing Powered by Cognero.

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CH 03 - Linear Programming: Sensitivity Analysis and Interpretation of Solution LEARNING OBJECTIVES: IMS.ASWC.19.03.04 - 3.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.4 Limitations of Classical Sensitivity Analysis KEYWORDS: Bloom's: Understand 9. A small change in the objective function coefficient can necessitate modifying the optimal solution. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.03.02 - 3.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.2 Graphical Sensitivity Analysis KEYWORDS: Bloom's: Understand 10. The dual price associated with a constraint is the change in the value of the solution per unit decrease in the right-hand side of the constraint. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.03.02 - 3.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 3.2 Graphical Sensitivity Analysis KEYWORDS: Bloom's: Remember 11. For a minimization problem, a positive dual price indicates the value of the objective function will increase. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.03.02 - 3.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.2 Graphical Sensitivity Analysis KEYWORDS: Bloom's: Understand 12. There is a dual price for every decision variable in a model. a. True b. False ANSWER: False POINTS: 1 © Cengage. Testing Powered by Cognero.

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CH 03 - Linear Programming: Sensitivity Analysis and Interpretation of Solution DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.03.03 - 3.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.3 Sensitivity Analysis: Computer Solution 3.2 Graphical Sensitivity Analysis KEYWORDS: Bloom's: Understand 13. The amount of a sunk cost will vary depending on the values of the decision variables. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.03.03 - 3.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.3 Sensitivity Analysis: Computer Solution KEYWORDS: Bloom's: Understand 14. If the optimal value of a decision variable is zero and its reduced cost is zero, this indicates that alternative optimal solutions exist. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.A3.01 - A3.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: Appendix 3.1 Sensitivity Analysis with Excel Solver KEYWORDS: Bloom's: Understand 15. Relevant costs should be reflected in the objective function, but sunk costs should not. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.03.03 - 3.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.3 Sensitivity Analysis: Computer Solution KEYWORDS: Bloom's: Understand 16. If the range of feasibility for b1 is between 16 and 37, then if b1 = 22, the optimal solution will not change from the original optimal solution. a. True b. False © Cengage. Testing Powered by Cognero.

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CH 03 - Linear Programming: Sensitivity Analysis and Interpretation of Solution ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.03.03 - 3.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.3 Sensitivity Analysis: Computer Solution KEYWORDS: Bloom's: Apply 17. If the dual price for the right-hand side of a ≤ constraint is zero, there is no upper limit on its range of feasibility. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.03.02 - 3.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.2 Graphical Sensitivity Analysis KEYWORDS: Bloom's: Understand 18. Increasing the right-hand side of a nonbinding constraint will not cause a change in the optimal solution. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.03.02 - 3.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.2 Graphical Sensitivity Analysis KEYWORDS: Bloom's: Understand 19. If two or more objective function coefficients are changed simultaneously, further analysis is necessary to determine whether the optimal solution will change. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.03.02 - 3.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.2 Graphical Sensitivity Analysis KEYWORDS: Bloom's: Understand Multiple Choice 20. To solve a linear programming problem with thousands of variables and constraints, © Cengage. Testing Powered by Cognero.

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CH 03 - Linear Programming: Sensitivity Analysis and Interpretation of Solution a. a personal computer can be used. b. a mainframe computer is required. c. the problem must be partitioned into subparts. d. unique software would need to be developed.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.03.03 - 3.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.3 Sensitivity Analysis: Computer Solution 3.1 Introduction to Sensitivity Analysis KEYWORDS: Bloom's: Remember 21. A negative dual price for a constraint in a minimization problem means a. as the right-hand side increases, the objective function value will increase. b. as the right-hand side decreases, the objective function value will increase. c. as the right-hand side increases, the objective function value will decrease. d. as the right-hand side decreases, the objective function value will decrease.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.A3.02 - A3.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: Appendix 3.2 Sensitivity Analysis with Lingo KEYWORDS: Bloom's: Understand 22. If a decision variable is not positive in the optimal solution, its reduced cost is a. what its objective function value would need to be before it could become positive. b. the amount its objective function value would need to improve before it could become positive. c. zero. d. its dual price.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.03.03 - 3.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.3 Sensitivity Analysis: Computer Solution KEYWORDS: Bloom's: Understand 23. A constraint with a positive slack value a. will have a positive dual price. b. will have a negative dual price. c. will have a dual price of zero. d. has no restrictions for its dual price. © Cengage. Testing Powered by Cognero.

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CH 03 - Linear Programming: Sensitivity Analysis and Interpretation of Solution ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.03.03 - 3.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.3 Sensitivity Analysis: Computer Solution KEYWORDS: Bloom's: Understand 24. The amount by which an objective function coefficient can change before a different set of values for the decision variables becomes optimal is the a. optimal solution. b. dual solution. c. range of optimality. d. range of feasibility.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.03.02 - 3.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.2 Graphical Sensitivity Analysis KEYWORDS: Bloom's: Remember 25. The range of feasibility measures a. the right-hand-side values for which the objective function value will not change. b. the right-hand-side values for which the values of the decision variables will not change. c. the right-hand-side values for which the dual prices will not change. d. All of these are correct.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.03.03 - 3.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.3 Sensitivity Analysis: Computer Solution KEYWORDS: Bloom's: Remember 26. An objective function reflects the relevant cost of labor hours used in production rather than treating them as a sunk cost. The correct interpretation of the dual price associated with the labor hours constraint is the a. maximum premium (say for overtime) over the normal price that the company would be willing to pay. b. upper limit on the total hourly wage the company would pay. c. reduction in hours that could be sustained before the solution would change. d. number of hours by which the right-hand side can change before there is a change in the solution point.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate © Cengage. Testing Powered by Cognero.

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CH 03 - Linear Programming: Sensitivity Analysis and Interpretation of Solution LEARNING OBJECTIVES: IMS.ASWC.19.03.03 - 3.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.3 Sensitivity Analysis: Computer Solution KEYWORDS: Bloom's: Understand 27. The graphical solution procedure is useful only for linear programs involving a. two decision variables. b. more than two decision variables. c. a single constraint. d. None of these are correct.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.03.03 - 3.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.3 Sensitivity Analysis: Computer Solution KEYWORDS: Bloom's: Understand 28. An improvement in the value of the objective function per unit increase in a right-hand side is the a. sensitivity value. b. constraint coefficient. c. slack value. d. None of these are correct.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.03.02 - 3.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.2 Graphical Sensitivity Analysis KEYWORDS: Bloom's: Understand 29. The amount the objective function coefficient of a decision variable would have to improve before that variable would have a positive value in the solution is the a. dual price. b. surplus variable. c. reduced cost. d. upper limit.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.03.03 - 3.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.3 Sensitivity Analysis: Computer Solution KEYWORDS: Bloom's: Remember © Cengage. Testing Powered by Cognero.

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CH 03 - Linear Programming: Sensitivity Analysis and Interpretation of Solution 30. Based on the per-unit increase in the right-hand side of the constraint, the dual price measures the a. increase in the value of the optimal solution. b. decrease in the value of the optimal solution. c. improvement in the value of the optimal solution. d. change in the value of the optimal solution.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.A3.02 - A3.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: Appendix 3.2 Sensitivity Analysis with Lingo KEYWORDS: Bloom's: Remember 31. Sensitivity analysis information in computer output is based on the assumption that a. no coefficient changes. b. one coefficient changes. c. two coefficients change. d. all coefficients change.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.03.04 - 3.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.4 Limitations of Classical Sensitivity Analysis KEYWORDS: Bloom's: Remember 32. When the cost of a resource is sunk, then the dual price can be interpreted as the a. minimum amount the firm should be willing to pay for one additional unit of the resource. b. maximum amount the firm should be willing to pay for one additional unit of the resource. c. minimum amount the firm should be willing to pay for multiple additional units of the resource. d. maximum amount the firm should be willing to pay for multiple additional units of the resource.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.03.03 - 3.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.3 Sensitivity Analysis: Computer Solution KEYWORDS: Bloom's: Understand 33. Which of the following is NOT a question answered by standard sensitivity analysis information? a. If the right-hand-side value of a constraint changes, will the objective function value change? b. Over what range can a constraint's right-hand-side value change without the constraint's dual price possibly changing? © Cengage. Testing Powered by Cognero.

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CH 03 - Linear Programming: Sensitivity Analysis and Interpretation of Solution c. By how much will the objective function value change if the right-hand-side value of a constraint changes beyond the range of feasibility? d. By how much will the objective function value change if a decision variable's coefficient in the objective function changes within the range of optimality?

ANSWER: c POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.03.03 - 3.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.3 Sensitivity Analysis: Computer Solution KEYWORDS: Bloom's: Understand 34. The cost that varies depending on the values of the decision variables is a a. reduced cost. b. relevant cost. c. sunk cost. d. dual cost.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.03.03 - 3.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.3 Sensitivity Analysis: Computer Solution KEYWORDS: Bloom's: Remember 35. A cost that is incurred no matter what values the decision variables assume is a(n) a. reduced cost. b. optimal cost. c. sunk cost. d. dual cost.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.03.03 - 3.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.3 Sensitivity Analysis: Computer Solution KEYWORDS: Bloom's: Remember 36. Sensitivity analysis is sometimes referred to as a. feasibility testing. b. duality analysis. c. alternative analysis. d. postoptimality analysis.

ANSWER: d © Cengage. Testing Powered by Cognero.

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CH 03 - Linear Programming: Sensitivity Analysis and Interpretation of Solution POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.03.01 - 3.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.1 Introduction to Sensitivity Analysis KEYWORDS: Bloom's: Remember 37. Sensitivity analysis is concerned with how certain changes affect the a. feasible solution. b. unconstrained solution. c. optimal solution. d. degenerative solution.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.03.01 - 3.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.1 Introduction to Sensitivity Analysis KEYWORDS: Bloom's: Remember 38. The dual price for a < constraint will a. always be < 0. b. always be > 0. c. be < 0 in a minimization problem and > 0 in a maximization problem. d. always equal 0.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.03.02 - 3.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 3.2 Graphical Sensitivity Analysis KEYWORDS: Bloom's: Understand Subjective Short Answer 39. In a linear programming problem, the binding constraints for the optimal solution are: 5X + 3Y ≤ 30 2X + 5Y ≤ 20 a.

b.

Fill in the blanks in the following sentence: As long as the slope of the objective function stays between _______ and _______, the current optimal solution point will remain optimal.

Which of these objective functions will lead to the same optimal solution?

(1) 2X + 1Y (2) 7X + 8Y (3) 80X + 60Y (4) 25X + 35Y

ANSWER: © Cengage. Testing Powered by Cognero.

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CH 03 - Linear Programming: Sensitivity Analysis and Interpretation of Solution a. −5/3 and −2/5 b. objective functions (2), (3), and (4) POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.03.02 - 3.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 3.2 Graphical Sensitivity Analysis KEYWORDS: Bloom's: Analyze 40. The optimal solution of this linear programming problem is at the intersection of constraints 1 and 2.

Max s.t.

a. b. c.

2x1 + x2 4x1 + 1x2 ≤ 400 4x1 + 3x2 ≤ 600 1x1 + 2x2 ≤ 300 x1, x2 ≥ 0 Over what range can the coefficient of x1 vary before the current solution is no longer optimal?

Over what range can the coefficient of x2 vary before the current solution is no longer optimal?

Compute the dual prices for the three constraints.

ANSWER:

a. 1.33 ≤ c1 ≤ 4 b. 0.5 ≤ c2 ≤ 1.5 c. Dual prices are 0.25, 0.25, and 0.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.03.02 - 3.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 3.2 Graphical Sensitivity Analysis KEYWORDS: Bloom's: Analyze 41. The binding constraints for this problem are the first and second.

Min

x1 + 2x2

s.t.

x1 + x2 ≥ 300 2x1 + x2 ≥ 400 2x1 + 5x2 ≤ 750 x1, x2 ≥ 0

a. b.

Keeping c2 fixed at 2, over what range can c1 vary before there is a change in the optimal solution point?

Keeping c1 fixed at 1, over what range can c2 vary before there is a change in the optimal solution point?

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CH 03 - Linear Programming: Sensitivity Analysis and Interpretation of Solution c. d. e.

If the objective function becomes Min 1.5x1 + 2x2, what will be the optimal values of x1, x2, and the objective function?

If the objective function becomes Min 7x1 + 6x2, what constraints will be binding?

Find the dual price for each constraint in the original problem.

ANSWER:

a. 0.8 ≤ c1 ≤ 2 b. 1 ≤ c2 ≤ 2.5 c. x1 = 250, x2 = 50, z = 475 d. Constraints 1 and 2 will be binding. e. Dual prices are 0.33, 0, and 0.33. (The first and third values are negative.)

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.03.02 - 3.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 3.2 Graphical Sensitivity Analysis KEYWORDS: Bloom's: Analyze 42. Excel's Solver tool has been used in the spreadsheet below to solve a linear programming problem with a maximization objective function and all ≤ constraints.

Input Section Objective Function Coefficients X Y 4 6 Constraints #1 #2 #3

3 3 1

5 2 1

Avail.

60 48 20

Variables Profit

13.333333 53.333333

4 24

77.333333

Constraint #1 #2 #3

Usage 60 48 17.333333

Slack 1.789E-11 -2.69E-11 2.6666667

Output Section

a. Give the original linear programming problem. b. Give the complete optimal solution.

ANSWER: a. Max 4X + 6Y s.t. © Cengage. Testing Powered by Cognero.

3X + 5Y ≤ 60 Page 13

CH 03 - Linear Programming: Sensitivity Analysis and Interpretation of Solution

b.

3X + 2Y ≤ 48 1X + 1Y ≤ 20 X, Y ≥ 0 The complete optimal solution is X = 13.333, Y = 4, Z = 73.333, S1 = 0, S2 = 0, S3 = 2.667

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.A3.01 - A3.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: Appendix 3.1 Sensitivity Analysis with Excel Solver KEYWORDS: Bloom's: Analyze 43. Excel's Solver tool has been used in the spreadsheet below to solve a linear programming problem with a minimization objective function and all ≥ constraints.

Input Section Objective Function Coefficients X 5

Y 4

Constraints #1 #2 #3

4 2 9

3 5 8

Req'd 60 50 144

Variables Profit

9.6 48

7.2 28.8

76.8

Constraint #1 #2 #3

Usage 60 55.2 144

Slack 1.35E-11 -5.2 -2.62E-11

Output Section

a. Give the original linear programming problem. b. Give the complete optimal solution.

ANSWER: a. Min 5X + 4Y 4X + 3Y ≥ 60 2X + 5Y ≥ 50 9X + 8Y ≥ 144 X, Y ≥ 0 The complete optimal solution is X = 9.6, Y = 7.2, Z = 76.8, S1 = 0, S2 = 5.2, S3 = 0 s.t.

b.

POINTS: DIFFICULTY:

1 Challenging

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CH 03 - Linear Programming: Sensitivity Analysis and Interpretation of Solution LEARNING OBJECTIVES: IMS.ASWC.19.A3.01 - A3.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: Appendix 3.1 Sensitivity Analysis with Excel Solver KEYWORDS: Bloom's: Analyze 44. Use the spreadsheet and Excel Solver sensitivity report to answer these questions. a. What is the cell formula for B12? b. What is the cell formula for C12? c. What is the cell formula for D12? d. What is the cell formula for B15? e. What is the cell formula for B16? f. What is the cell formula for B17? g. What is the optimal value for x1? h. What is the optimal value for x2? i. Would you pay $0.50 each for up to 60 more units of resource 1?

Is it possible to figure the new objective function value if the profit on product 1 increases by j. a dollar, or do you have to rerun Solver?

A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

B

C

D

E

Var. 1 2 3 1

Var. 2 5 1 1

(type) < < >

Avail.

40 30 12

5

4

Input Information Constraint 1 Constraint 2 Constraint 3 Profit Output Information Variables Profit Resources Constraint 1 Constraint 2 Constraint 3

= Total Used

Slack/Surplus

Name Variable 1 Variable 2

Final Reduced Value Cost 8.461538462 0 4.615384615 0

Objective Coefficient 5 4

Final

Constraint Allowable

Sensitivity Report Changing Cells Cell $B$12 $C$12

Allowable Increase 7 8.5

Allowable Decrease 3.4 2.333333333

Constraints © Cengage. Testing Powered by Cognero.

Shadow

Allowable Page 15

CH 03 - Linear Programming: Sensitivity Analysis and Interpretation of Solution Cell

Name constraint 1 Used constraint 2 Used constraint 3 Used

$B$15 $B$16 $B$17

Value

Price

40 30

Increase

Decrease

0.538461538 40

110

7

1.307692308 30

30

4.666666667

13.07692308 0

R.H. Side

12

1.076923077 1E+30

ANSWER:

a. =B8*B11 b. =C8*C11 c. =B12+C12 d. =B4*B11+C4*C11 e. =B5*B11+C5*C11 f. =B6*B11+C6*C11 g. 8.46 h. 4.61 i. yes j. no POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.A3.01 - A3.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: Appendix 3.1 Sensitivity Analysis with Excel Solver KEYWORDS: Bloom's: Evaluate 45. A large sporting goods store is placing an order for bicycles with its supplier. Four models can be ordered: the adult Open Trail, the adult Cityscape, the girl's Sea Sprite, and the boy's Trail Blazer. It is assumed that every bike ordered will be sold, and their profits, respectively, are 30, 25, 22, and 20. The LP model should maximize profit. The store needs to worry about several conditions. One of these is space to hold the inventory. An adult's bike needs two feet, but a child's bike needs only one foot. The store has 500 feet of space. There are 1200 hours of assembly time available. The child's bikes need 4 hours of assembly time, the Open Trail needs 5 hours, and the Cityscape needs 6 hours. The store would like to place an order for at least 275 bikes. a. Formulate a model for this problem. b. Solve your model with any computer package available to you. c. How many of each kind of bike should be ordered, and what will the profit be? d. What would the profit be if the store had 100 more feet of storage space? e. If the profit on the Cityscape increases to 35, will any of the Cityscape bikes be ordered? f. Over what range of assembly hours is the dual price applicable?

If we require 5 more bikes in inventory, what will happen to the value of the optimal g. solution? h. Which resource should the company work to increase, inventory space or assembly time?

ANSWER:

NOTE TO INSTRUCTOR: This problem is suitable for a take-home or lab exam. The student must formulate the model, solve the problem with a computer package, and then interpret the solution to answer the questions. a.

MAX 30 X1 + 25 X2 + 22 X3 + 20 X4 SUBJECT TO 2) 2 X1 + 2 X2 + X3 + X4 <= 500

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CH 03 - Linear Programming: Sensitivity Analysis and Interpretation of Solution 3) 5 X1 + 6 X2 + 4 X3 + 4 X4 <= 1200 4) X1 + X2 + X3 + X4 >= 275 b.

OBJECTIVE FUNCTION VALUE 1) 6850.0000 VARIABLE X1 X2 X3 X4 ROW 2) 3) 4)

VALUE 100.000000 0.000000 175.000000 0.000000

REDUCED COST 0.000000 13.000000 0.000000 2.000000

SLACK OR SURPLUS 125.000000 0.000000 0.000000

DUAL PRICE 0.000000 8.000000 −10.000000

NO. ITERATIONS= 2 RANGES IN WHICH THE BASIS IS UNCHANGED: OBJ. COEFFICIENT RANGES CURRENT ALLOWABLE ALLOWABLE VARIABLE COEFFICIENT INCREASE DECREASE X1 30.000000 INFINITY 2.500000 X2 25.000000 13.000000 INFINITY X3 22.000000 2.000000 2.000000 X4 20.000000 2.000000 INFINITY

ROW 2 3 4 c. d. e. f. g. h.

CURRENT RHS 500.000000 1200.000000 275.000000

RIGHTHAND SIDE RANGES ALLOWABLE INCREASE INFINITY 125.000000 25.000000

ALLOWABLE DECREASE 125.000000 100.000000 35.000000

Order 100 Open Trails, 0 Cityscapes, 175 Sea Sprites, and 0 Trail Blazers. Profit will be 6850.

6850 No. The $10 increase is below the reduced cost.

1100 to 1325 It will decrease by 50. assembly time

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.03.05 - 3.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 3.5 The Electronic Communications Problem © Cengage. Testing Powered by Cognero.

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CH 03 - Linear Programming: Sensitivity Analysis and Interpretation of Solution KEYWORDS:

Bloom's: Analyze

46. Consider the following linear program: Max

5x1 + 7x2

x1 ≤ 6 2x1 + 3x2 ≤ 19 x1 + x2 ≤ 8 x1 , x2 ≥ 0 The graphical solution to the problem is shown below. From the graph, we see that the optimal solution occurs at x1 = 5, x2 = 3, and z = 46. s.t.

a. Calculate the range of optimality for each objective function coefficient. b. Calculate the dual price for each resource.

ANSWER: a. Ranges of optimality: 14/3 ≤ c1 ≤ 7 and 5 ≤ c2 ≤ 15/2 The dual price for the first resource is 0, for the second resource is 2, and for the third is b.

1.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.03.01 - 3.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 3.1 Introduction to Sensitivity Analysis KEYWORDS: Bloom’s: Analyze

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM True / False 1. Media selection problems can maximize exposure quality and use number of customers reached as a constraint, or maximize the number of customers reached and use exposure quality as a constraint. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.01 - 4.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.1 Marketing Applications KEYWORDS: Bloom's: Remember 2. Linear programming is appropriate for financial problem situations involving capital budgeting, asset allocation, financial planning, and portfolio selection. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.02 - 4.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.2 Financial Applications KEYWORDS: Bloom's: Remember 3. Portfolio selection problems should acknowledge both risk and return. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.02 - 4.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.2 Financial Applications KEYWORDS: Bloom's: Remember 4. It is improper to combine manufacturing costs and overtime costs in the same objective function. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking © Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM TOPICS: KEYWORDS:

4.3 Operations Management Applications Bloom's: Understand

5. Production constraints frequently take the form: Beginning inventory + Sales − Production = Ending inventory a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Understand 6. If a real-world problem is correctly formulated, it is NOT possible to have alternative optimal solutions. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.02 - 4.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.2 Financial Applications KEYWORDS: Bloom's: Understand 7. To properly interpret dual prices, one must know how costs were allocated in the objective function. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Understand 8. A company produces two different products from steel. One requires 3 tons of steel and the other requires 5 tons. There are 100 tons of steel available daily. A constraint on daily production could be written as: 3x1 + 5x2 ≥ 100. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy © Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Understand 9. Double-subscript notation for decision variables should be avoided unless the number of decision variables exceeds nine. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Understand 10. Using minutes as the unit of measurement on the left-hand side of a constraint and using hours on the right-hand side is acceptable since both are a measure of time. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Understand 11. Compared to the problems in the textbook, real-world problems generally require more variables and constraints. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Understand 12. For the multiperiod production scheduling problem in the textbook, period n − 1's ending inventory variable was also used as period n's beginning inventory variable. a. True b. False ANSWER: True © Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Understand 13. A company makes two products, A and B. A sells for $100 and B sells for $90. The variable production costs are $30 per unit for A and $25 for B. The company's objective could be written as: Max 190x1 − 55x2. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Understand 14. The primary limitation of linear programming's applicability is the requirement that all decision variables be nonnegative. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Understand 15. A decision maker would be wise to not deviate from the optimal solution found by an LP model because it is the best solution. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Understand 16. It can take longer to collect the data for a large-scale linear programming model than it does for either the formulation of the model or the development of the computer solution. © Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Understand 17. The purpose of media selection applications of linear programming is to help marketing managers allocate a fixed advertising budget to various advertising media. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.01 - 4.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.1 Marketing Applications KEYWORDS: Bloom's: Remember 18. The marketing research model presented in the textbook involves minimizing total interview cost subject to interview quota guidelines. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.01 - 4.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.1 Marketing Applications KEYWORDS: Bloom's: Understand 19. A marketing research firm must determine how many daytime interviews (D) and evening interviews (E) to conduct.

At least 40% of the interviews must be in the evening. A correct modeling of this constraint is: –0.4D + 0.6E > 0. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.04.01 - 4.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.1 Marketing Applications © Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM KEYWORDS:

Bloom's: Understand

20. An important part of the LP process is to take the time to ensure the linear programming model accurately reflects the real problem. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.01 - 4.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.1 Marketing Applications KEYWORDS: Bloom's: Understand Multiple Choice 21. Media selection problems usually determine a. how many times to use each media source. b. the coverage provided by each media source. c. the cost of each advertising exposure. d. the relative value of each medium.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.01 - 4.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.1 Marketing Applications KEYWORDS: Bloom's: Understand 22. To study consumer characteristics, attitudes, and preferences, a company would engage in a. client satisfaction processing. b. marketing research. c. capital budgeting. d. production planning.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.01 - 4.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.1 Marketing Applications KEYWORDS: Bloom's: Remember 23. An LP model for a marketing research application uses the variable HD to represent the number of homeowners interviewed during the day. The objective function minimizes the cost of interviewing this and other categories. There is a constraint that HD ≥ 100. The solution indicates that interviewing another homeowner during the day will increase © Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM costs by 10.00. What does this tell you about the HD variable? a. The objective function coefficient of HD is 10. b. The dual price for the HD constraint is 10. c. The objective function coefficient of HD is −10. d. The dual price for the HD constraint is −10.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.04.01 - 4.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.1 Marketing Applications KEYWORDS: Bloom's: Understand 24. The dual price for a constraint that compares funds used with funds available is 0.058. This means that a. the cost of additional funds is 5.8%. b. if more funds can be obtained at a rate of 5.5%, some should be. c. no more funds are needed. d. the objective was to minimize.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.04.02 - 4.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.2 Financial Applications KEYWORDS: Bloom's: Understand 25. Let M be the number of units to make and let B be the number of units to buy. If it costs $2 to make a unit and $3 to buy a unit, and 4000 units are needed, the objective function is a. Max 2M + 3B. b. Min 4000 (M + B). c. Max 8000M + 12,000B. d. Min 2M + 3B. ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Apply 26. If Pij = the production of product i in period j, then to indicate that the limit on production of the company's three products in period 2 is 400, we would express it as a. P21 + P22 + P23 ≤ 400. © Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM b. P12 + P22 + P32 ≤ 400. c. P32 ≤ 400. d. P23 ≤ 400.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Apply 27. Let Pij = the production of product i in period j. To specify that production of product 1 in period 3 and in period 4 differs by no more than 100 units, we would write it as a. P13 − P14 ≤ 100; P14 − P13 ≤ 100. b. P13 − P14 ≤ 100; P13 − P14 ≥ 100. c. P13 − P14 ≤ 100; P14 − P13 ≥ 100. d. P13 − P14 ≥ 100; P14 − P13 ≥ 100.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Apply 28. Let A, B, and C be the amounts invested in companies A, B, and C. If no more than 50% of the total investment can be in company B, then a. B ≤ 5. b. A − 0.5B + C ≤ 0. c. 0.5A − B − 0.5C ≤ 0. d. −0.5A + 0.5B − 0.5C ≤ 0.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.04.02 - 4.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.2 Financial Applications KEYWORDS: Bloom's: Apply 29. Department 3 has 2500 hours. Transfers are allowed to departments 2 and 4, and from departments 1 and 2. If Ai measures the labor hours allocated to department i and Tij the hours transferred from department i to department j, then © Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM a. T13 + T23 − T32 − T34 − A3 = 2500. b. T31 + T32 − T23 − T43 + A3 = 2500. c. A3 + T13 + T23 − T32 − T34 = 2500. d. A3 − T13 − T23 + T32 + T34 = 2500.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Apply 30. Blending problems arise whenever a manager must decide how to a. mix several different asset types in one investment strategy. b. mix two or more resources to produce one or more products. c. combine the results of two or more research studies into one. d. allocate workers with different skill levels to various work shifts.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Understand 31. In a production scheduling LP, the demand requirement constraint for a time period takes the form a. Beginning inventory + Production + Ending inventory > Demand. b. Beginning inventory − Production + Ending inventory = Demand. c. Beginning inventory + Production − Ending inventory = Demand. d. Beginning inventory − Production − Ending inventory > Demand.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Remember 32. A mutual fund manager must decide how much money to invest in Atlantic Oil (A) and how much to invest in Pacific Oil (P). At least 60% of the money invested in the two oil companies must be in Pacific Oil. A correct modeling of this constraint is a. 0.4A + 0.6P > 0. b. –0.4A + 0.6P > 0. © Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM c. 0.6A + 0.4P > 0. d. –0.6A + 0.4P > 0.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.04.02 - 4.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.2 Financial Applications KEYWORDS: Bloom's: Apply 33. The production scheduling problem modeled in the textbook involves capacity constraints on all of the following types of resources EXCEPT a. material. b. labor. c. machine. d. storage.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Remember 34. Which operations management application model is most likely to have an objective function that minimizes the sum of manufacturing costs, purchasing costs, and overtime costs? a. blending b. make-or-buy decision c. workforce assignment d. production scheduling ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Understand 35. Blending problems occur frequently in which of the following industries? a. chemical b. petroleum c. food services d. All of these are correct. © Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Remember Subjective Short Answer 36. A&C Distributors represents many outdoor products companies and schedules deliveries to discount stores, garden centers, and hardware stores. Currently, scheduling needs to be done for two lawn sprinklers, the Water Wave and Spring Shower models. Requirements for shipment to a warehouse for a national chain of garden centers are shown below.

Minimum Unit Cost Per-Unit Requirement to Ship Inventory Cost March Water Wave 3000 0.30 0.06 Spring Shower 1800 0.25 0.05 April 7000 Water Wave 4000 0.40 0.09 Spring Shower 4000 0.30 0.06 May 6000 Water Wave 5000 0.50 0.12 Spring Shower 2000 0.35 0.07 Let Sij be the number of units of sprinkler i shipped in month j, where i = 1 or 2, and j = 1, 2, or 3. Let Wij be the number of sprinklers that are at the warehouse at the end of a month, in excess of the minimum requirement. a Write the portion of the objective function that minimizes shipping costs.

An inventory cost is assessed against this ending inventory. Give the portion of the objective b. function that represents inventory cost.

There will be three constraints that guarantee, for each month, that the total number of c. sprinklers shipped will not exceed the shipping capacity. Write these three constraints.

There are six constraints that work with inventory and the number of units shipped, making d. sure that enough sprinklers are shipped to meet the minimum requirements. Write these six constraints.

Month

Shipping Capacity 8000

Product

ANSWER: a. b. c.

d.

Min 0.3S11 + 0.25S21 + 0.40S12 + 0.30S22 + 0.50S13 + 0.35S23 Min 0.06W11 + 0.05W21 + 0.09W12 + 0.06W22 + 0.12W13 + 0.07W23 S11 + S21 ≤ 8000 S12 + S22 ≤ 7000 S13 + S23 ≤ 6000 S11 − W11 = 3000 S21 − W21 = 1800 W11 + S12 − W12 = 4000 W21 + S22 − W22 = 4000 W12 + S13 − W13 = 5000 W22 + S23 − W23 = 2000

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 © Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Apply 37. An ad campaign for a new snack chip will be conducted in a limited geographical area and can use TV time, radio time, and newspaper ads. Information about each medium is shown below.

Medium Cost per Ad # Reached Exposure Quality TV 500 10,000 30 Radio 200 3000 40 Newspaper 400 5000 25 If the number of TV ads cannot exceed the number of radio ads by more than 4, and if the advertising budget is $10,000, develop the model that will maximize the number reached and achieve an exposure quality of at least 1000.

ANSWER:

Let T = number of TV ads R = number of radio ads N = number of newspaper ads Max

10,000T + 3000R + 5000N

s.t.

500T + 200R + 400N ≤ 10,000 30T + 40R + 25N ≥ 1000 T−R≤4 T, R, N ≥ 0

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.04.01 - 4.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.1 Marketing Applications KEYWORDS: Bloom's: Analyze 38. Information on a prospective investment for Wells Financial Services is given below.

Loan funds available Investment income (% of previous period's investment) Maximum investment Payroll payment

1 3000

4500 100

Period 2 3 7000 4000

4 5000

110%

112%

113%

8000 120

6000 150

7500 100

In each period, funds available for investment come from two sources: loan funds and income from the previous period's investment. Expenses, or cash outflows, in each period must include repayment of the previous period's loan plus 8.5% interest, and the current payroll payment. In addition, to end the planning horizon, investment income from period 4 (at 110% of the investment) must be sufficient to cover the loan plus interest from period 4. The difference in these two quantities represents net income and is to be maximized. How much should be borrowed, and how much should be invested each period?

ANSWER: Let © Cengage. Testing Powered by Cognero.

Lt = loan in period t, t = 1, ... ,4 Page 12

CH 04 - Linear Programming Applications in Marketing, Finance, and OM It = investment in period t, t = 1, ... ,4 Max

1.1I4 − 1.085L4

s.t.

L1 ≤ 3000 I1 ≤ 4500 L1 − I1 = 100 L2 ≤ 7000 I2 ≤ 8000 L2 + 1.1I1 − 1.085L1 − I2 = 120 L3 ≤ 4000 I3 ≤ 6000 L3 + 1.12I2 − 1.085L2 − I3 = 150 L4 ≤ 5000 I4 ≤ 7500 L4 + 1.13I3 −1.085L3 −I4 = 100 1.10I4 − 1.085L4 ≥ 0 Lt, It ≥ 0

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.04.02 - 4.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.2 Financial Applications KEYWORDS: Bloom's: Analyze 39. Tots Toys makes a plastic tricycle that is composed of three major components: a handlebar-front wheel-pedal assembly, a seat and frame unit, and rear wheels. The company has orders for 12,000 of these tricycles. Current schedules yield the following information: Requirements Cost to Cost to Component Plastic Time Space Manufacture Purchase Front 3 10 2 8 12 Seat/frame 4 6 2 6 9 Rear wheel (each) 0.5 2 0.1 1 3 Available 50,000 160,000 30,000 The company obviously does not have the resources available to manufacture everything needed for the completion of 12,000 tricycles so it has gathered purchase information for each component. Develop a linear programming model to tell the company how many of each component should be manufactured and how many should be purchased in order to provide 12,000 fully completed tricycles at the minimum cost.

ANSWER: Let FM = number of fronts made SM = number of seats made WM = number of wheels made FP = number of fronts purchased SP = number of seats purchased WP = number of wheels purchased © Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM Min

8FM + 6SM + 1WM + 12FP + 9SP + 3WP

s.t.

3FM + 4SM + 0.5WM ≤ 50,000 10FM + 6SM + 2WM ≤ 160,000 2FM + 2SM + 0.1WM ≤ 30,000 FM + FP ≥ 12,000 SM + SP ≥ 12,000 WM + WP ≥ 24,000 FM, SM, WM, FP, SP, WP ≥ 0

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Analyze 40. Tots Toys Company is trying to schedule production of two very popular toys for the next three months: a rocking horse and a scooter. Information about both toys is given below.

Toy Rocking horse Scooter

Beg. Inv. June 1 25 55

Required Plastic 5 4

Required Time 2 3

Production Cost 12 14

Summer Schedule June July August

Plastic Available 3500 5000 4800

Time Available 2100 3000 2500

Monthly Demand Horse 220 350 600

Production Cost 1 1.2

Monthly Demand Scooter 450 700 520

Develop a model that would tell the company how many of each toy to produce during each month. You are to minimize total cost. Inventory cost will be levied on any items in inventory on June 30, July 31, or August 31 after demand for the month has been satisfied. Your model should make use of the relationship: Beginning inventory + Production − Demand = Ending inventory for each month. The company wants to end the summer with 150 rocking horses and 60 scooters as beginning inventory for September 1. Don't forget to define your decision variables.

ANSWER: Let Pij = number of toy i to produce in month j Sij = surplus (inventory) of toy i at end of month j Min s.t.

© Cengage. Testing Powered by Cognero.

12P11 + 12P12 + 12P13 + 14P21 + 14P22 + 14P23 + 1S11 + 1S12 + 1S13 + 1.2S21 + 1.2S22 + 1.2S23 P11 − S11 = 195 S11 + P12 − S12 = 350 S12 + P13 − S13 = 600 S13 ≥ 150 P21 − S21 = 395 Page 14

CH 04 - Linear Programming Applications in Marketing, Finance, and OM S21 + P22 − S22 = 700 S22 + P23 − S23 = 520 S23 ≥ 60 5P11 + 4P21 ≤ 3500 5P12 + 4P22 ≤ 5000 5P13 + 4P23 ≤ 4800 2P11 + 3P21 ≤ 2100 2P12 + 3P22 ≤ 3000 2P13 + 3P23 ≤ 2500 Pij, Sij ≥ 0 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Analyze 41. Larkin Industries manufactures several lines of decorative and functional metal items. The most recent order was for 1200 door lock units for an apartment complex developer. The sales and production departments must work together to determine delivery schedules. Each lock unit consists of three components: the knob and face plate, the actual lock itself, and a set of two keys. Although the processes used in the manufacture of the three components vary, there are three areas where the production manager is concerned about the availability of resources. These three areas, their usage by the three components, and their availability are detailed in the table below.

Resource Brass alloy Machining Finishing

Knob and Plate 12 18 15

Lock 5 20 5

Key (each) 1 10 1

Available 15,000 units 36,000 minutes 12,000 minutes

A quick look at the amounts available confirms that Larkin does not have the resources to fill this contract. A subcontractor, who can make an unlimited number of each of the three components, quotes the prices below.

Component Knob and plate Lock Keys (set of 2)

Subcontractor Cost 10.00 9.00 1.00

Larkin Cost 6.00 4.00 0.50

Develop a linear programming model that would tell Larkin how to fill the order for 1200 lock sets at the minimum cost.

ANSWER:

Let

PM = number of knob and plate units to make PB = number of knob and plate units to buy LM = number of lock units to make LB = number of lock units to buy KM = number of key sets to make KB = number of key sets to buy

Min

6PM + 10PB + 4LM + 9LB + 0.5KM + 1KB

s.t.

12PM + 5LM + 2KM ≤ 15,000

© Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM 18PM + 20LM + 20KM ≤ 36,000 15PM + 5LM + 2KM ≤ 12,000 PM + PB ≥ 1200 LM + LB ≥ 1200 KM + KB ≥ 1200 PM, PB, LM, LB, KM, KB ≥ 0 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Analyze 42. G&P Manufacturing would like to minimize the labor cost of producing dishwasher motors for a major appliance manufacturer. Although two models of motors exist, the finished models are indistinguishable from one another; their cost difference is due to a different production sequence. The time in hours required for each model in each production area, along with the labor cost, is shown in the table below.

Area A Area B Area C Cost

Model 1 15 4 4 80

Model 2 3 10 8 65

Currently, labor assignments provide for 10,000 hours in both areas A and B and 18,000 hours in area C. If 2000 hours are available to be transferred from area B to area A and 3000 hours are available to be transferred from area C to either area A or area B, develop the linear programming model whose solution would tell G&P how many of each model to produce and how to allocate the workforce.

ANSWER: Let P1 = number of model 1 motors to produce P2 = number of model 2 motors to produce AA = number of hours allocated to area A AB = number of hours allocated to area B AC = number of hours allocated to area C TBA = number of hours transferred from B to A TCA = number of hours transferred from C to A TCB = number of hours transferred from C to B Min

80P1 + 65P2

s.t.

15P1 + 3P2 − AA ≤ 0 4P1 + 10P2 − AB ≤ 0 4P1 + 8P2 − AC ≤ 0 AA − TBA − TCA = 10,000 AB − TCB + TBA = 10,000 AC + TCA + TCB = 18,000 TBA ≤ 2000

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM TCA + TCB ≤ 3000 all variables ≥ 0 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Apply 43. FarmFresh Foods manufactures a snack mix called TrailTime by blending three ingredients: a dried fruit mixture, a nut mixture, and a cereal mixture. Information about the three ingredients (per ounce) is shown below.

Ingredient Dried fruit Nut mix Cereal mix

Cost 0.35 0.50 0.20

Volume 1/4 cup 3/8 cup 1 cup

Fat Grams 0 10 1

Calories 150 400 50

The company needs to develop a linear programming model whose solution would tell them how many ounces of each mix to put into the TrailTime blend. TrailTime is packaged in boxes that will hold between three and four cups. The blend should contain no more than 1000 calories and no more than 25 grams of fat. Dried fruit must be at least 20% of the volume of the mixture, and nuts must be no more than 15% of the weight of the mixture. Develop a model that meets these restrictions and minimizes the cost of the blend.

ANSWER:

Let

D = number of ounces of dried fruit mix in the blend N = number of ounces of nut mix in the blend C = number of ounces of cereal mix in the blend

Min

0.35D + 0.50N + 0.20C

s.t.

0.25D + 0.375N + C ≥ 3 0.25D + 0.375N + C ≤ 4 150D + 400N + 50C ≤ 1000 10N + C ≤ 25 0.2D − 0.075N − 0.2C ≥ 0 −0.15D + 0.85N − 0.15C ≤ 0 D, N, C ≥ 0

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Apply 44. Meredith Ribbon Company produces paper and fabric decorative ribbon that it sells to paper products companies and craft stores. The demand for ribbon is seasonal. Information about projected demand and production for a particular type of ribbon is given. © Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM Quarter 1 Quarter 2 Quarter 3 Quarter 4

Demand (yards) 10,000 18,000 16,000 30,000

Production Cost per Yard 0.03 0.04 0.06 0.08

Production Capacity (yards) 30,000 20,000 20,000 15,000

An inventory holding cost of $0.005 is levied on every yard of ribbon carried over from one quarter to the next. a. Define the decision variables needed to model this problem.

The objective is to minimize total cost, the sum of production and inventory holding cost. b.

Give the objective function. c. Write the production capacity constraints.

Write the constraints that balance inventory, production, and demand for each quarter. d.

Assume there is no beginning inventory in quarter 1.

To attempt to balance the production and avoid large changes in the workforce, production in e. period 1 must be within 5000 yards of production in period 2. Write this constraint.

ANSWER: a. b. c.

d.

Let Pi = production in yards in quarter i Si = ending surplus (inventory) in quarter i Min 0.03P1 + 0.04P2 + 0.06P3 + 0.08P4 + 0.005(S1 + S2 + S3 + S4) P1 ≤ 30,000 P2 ≤ 20,000 P3 ≤ 20,000 P4 ≤ 15,000 P1 − S1 = 10,000 S1 + P2 − S2 = 18,000 S2 + P3 − S3 = 16,000 S3 + P4 − S4 = 30,000 P1 ≤ 30,000 P2 ≤ 20,000 P3 ≤ 20,000 P4 ≤ 15,000

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Apply 45. Island Water Sports provides rental equipment and instruction for a variety of water sports in a resort town. On one particular morning, a decision must be made on how many Wildlife Raft Trips and how many Group Sailing Lessons should be scheduled. Each Wildlife Raft Trip requires one captain and one crew person and can accommodate six passengers. The revenue per raft trip is $120. Ten rafts are available, and at least 30 people are on the list for reservations this morning. Each Group Sailing Lesson requires one captain and two crew people for instruction. Two boats are needed for each group. Four students form each group. There are 12 sailboats available, and at least 20 people are on the list for sailing instruction this morning. The revenue per group sailing lesson is $160. The company has 12 captains and 18 crew available this morning. The company would like to maximize the number of customers served while generating at least © Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM $1800 in revenue and honoring all reservations.

ANSWER: Let R = number of Wildlife Raft Trips to schedule S = number of Group Sailing Lessons to schedule Max

6R + 4S

s.t.

R + S ≤ 12 R + 2S ≤ 18 6R ≥ 30 4S ≥ 20 120R + 160S ≥ 1800 R ≤ 10 2S ≤ 12 R, S ≥ 0

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective - BUSPROG: Analytic TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Analyze 46. Evans Enterprises bought a prime parcel of beachfront property and plans to build a luxury hotel on it. After meeting with the architectural team, the Evans family has drawn up some information to make preliminary plans for construction.

Excluding the suites, which are not part of this decision, the hotel will have four kinds of rooms: beachfront non-smoking, beachfront smoking, lagoon view non-smoking, and lagoon view smoking. To decide how many of the four kinds of rooms to plan for, the Evans family will consider the following information: a.

b.

c. d. e.

After adjusting for expected occupancy, the average nightly revenue for a beachfront nonsmoking room is $175. The average nightly revenue for a lagoon view non-smoking room is $130. Smokers will be charged an extra $15.

Construction costs vary. The cost estimate is $12,000 for a lagoon view room and $15,000 for a beachfront room. Air purifying systems and additional smoke detectors and sprinklers ad $3000 to the cost of any smoking room. Evans Enterprises has raised $6.3 million in construction guarantees for this portion of the building.

There will be at least 100 but no more than 180 beachfront rooms.

Design considerations require that the number of lagoon view rooms be at least 1.5 times the number of beachfront rooms, and no more than 2.5 times that number.

Industry trends recommend that the number of smoking rooms be no more than 50% of the number of non-smoking rooms.

Develop the linear programming model to maximize revenue.

ANSWER:

Let

BN = number of beachfront non-smoking rooms BS = number of beachfront smoking rooms LN = number of lagoon view non-smoking rooms LS = number of lagoon view smoking rooms

Max

175BN + 190BS + 130LN + 145LS

s.t.

15,000BN + 18,000BS + 12,000LN + 15,000LS ≤ 6,300,000 BN + BS ≥ 100

© Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM BN + BS ≤ 180 −1.5BN − 1.5BS + LN + LS ≥ 0 −2.5BN − 2.5BS + LN + LS ≤ 0 −0.5BN + BS − 0.5LN + LS ≤ 0 BN, BS, LN, LS ≥ 0 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Analyze 47. Super City Discount Department Store is open 24 hours a day. The number of cashiers need in each four-hour period of a day is listed below.

Period 10 p.m. to 2 a.m.

2 a.m. to 6 a.m.

6 a.m. to 10 a.m.

10 a.m. to 2 p.m.

2 p.m. to 6 p.m.

6 p.m. to 10 p.m.

Cashiers Needed 8 4 7 12 10 15

If cashiers work for eight consecutive hours, how many should be scheduled to begin working in each period in order to minimize the number of cashiers needed?

ANSWER: Let TNP = number of cashiers who begin working at 10 p.m.

TWA = number of cashiers who begin working at 2 a.m.

SXA = number of cashiers who begin working at 6 a.m.

TNA = number of cashiers who begin working at 10 a.m.

TWP = number of cashiers who begin working at 2 p.m.

SXP = number of cashiers who begin working at 6 p.m.

Min

TNP + TWA + SXA + TNA + TWP + SXP

s.t.

TNP + TWA ≥ 4 TWA + SXA ≥ 7 SXA + TNA ≥ 12 TNA + TWP ≥ 10 TWP + SXP ≥ 15 SXP + TNP ≥ 8 all variables ≥ 0

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Analyze 48. National Wing Company (NWC) is gearing up for the new B-48 contract. Currently, NWC has 100 equally qualified © Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM workers. Over the next three months, NWC has made the following commitments for wing production: Month May June July

Wing Production 20 24 30

Each worker can either be placed in production or train new recruits. A new recruit can be trained to be an apprentice in one month. The next month, he, himself, becomes a qualified worker (after two months from the start of training). Each trainer can train two recruits. The production rate and salary per employee are estimated below.

Employee Production Trainer Apprentice Recruit

Production Rate (Wings/Month) 0.6 0.3 0.4 0.05

Salary per Month $3,000 3,300 2,600 2,200

At the end of July, NWC wishes to have no recruits or apprentices but at least 140 full-time workers. Formulate and solve a linear program for NWC to accomplish this at minimum total cost.

ANSWER: Pi = number of producers in month i (where i = 1, 2, or 3) Ti = number of trainers in month i (where i = 1 or 2) Ai = number of apprentices in month i (where i = 2 or 3) Ri = number of recruits in month i (where i = 1 or 2) Min s.t.

3000P1 + 3300T1 + 2200R1 + 3000P2 + 3300T2 + 2600A2 + 2200R2 + 3000P3 + 2600A3 0.6P1 + 0.3T1 + 0.05R1 ≥ 20 0.6P1 + 0.3T1 + 0.05R1 + 0.6P2 + 0.3T2 + 0.4A2 + 0.05R2 ≥ 44 0.6P1 + 0.3T1 + 0.05R1 + 0.6P2 + 0.3T2 + 0.4A2 + 0.05R2 + 0.6P3 + 0.4A3 ≥ 74 P1 − P2 + T1 − T2 = 0 P2 − P3 + T2 + A2 = 0 A2 − R1 = 0 A3 − R2 = 0 2T1 − R1 ≥ 0 2T2 − R2 ≥ 0 P1 + T1 = 100 P3 + A3 ≥ 140 Pj, Tj, Aj, Rj ≥ 0 for all j P1 = 100, T1 = 0, R1 = 0, P2 = 80, T2 = 20, A2 = 0, R2 = 40, P3 = 100, A3 = 40 Total cost = $1,098,000

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic © Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM TOPICS: KEYWORDS:

4.3 Operations Management Applications Bloom's: Analyze

49. John Sweeney is an investment advisor who is attempting to construct an "optimal portfolio" for a client who has $400,000 cash to invest. There are ten different investments, falling into four broad categories that John and his client have identified as potential candidates for this portfolio.

The following table lists the investments and their important characteristics. Note that Unidyne equities (stocks) and Unidyne debt (bonds) are two separate investments, whereas First General REIT is a single investment that is considered both an equities and a real estate investment.

Category Equities

Debt

Real Estate Money

Investment Unidyne Corp.

Col. Mustard Restaurant First General REIT Metropolitan Electric Unidyne Corp.

Lemonville Transit Fairview Apartment Partnership First General REIT T-Bill Account Money Market Fund All Saver's Certificates

Expected Annual After-Tax Return 15.0% 17.0% 17.5% 11.8% 12.2% 12.0% 22.0% (see above) 9.6% 10.5% 12.6%

Liquidity Factor 100 100 100 95 92 79 0 (see above) 80 100 0

Risk Factor 60 70 75 20 30 22 50 (see above) 0 10 0

Formulate and solve a linear program to accomplish John's objective as an investment advisor, which is to construct a portfolio that maximizes his client's total expected after-tax return over the next year, subject to the following constraints placed upon him by the client for the portfolio: 1. Its (weighted) average liquidity factor must be at least 65.

2. The (weighted) average risk factor must be no greater than 55.

3. At most, $60,000 is to be invested in Unidyne stocks or bonds.

4. No more than 40% of the investment can be in any one category except the money category.

No more than 20% of the investment can be in any one investment except the money market 5. fund.

6. At least $1,000 must be invested in the money market fund.

7. The maximum investment in All Saver's Certificates is $15,000.

8. The minimum investment desired for debt is $90,000.

9. At least $10,000 must be placed in a T-bill account.

ANSWER:

Xj = $ invested in investment j; where j = 1(Uni Eq.), 2(Col. Must.), 3(1st Gen REIT), 4(Met. Elec.), 5(Uni Debt), 6(Lem. Trans.), 7(Fair. Apt.), 8(T-Bill), 9(Money Market), 10(All Saver's) Max

0.15X1 + 0.17X2 + 0.175X3 + 0.118X4 + 0.122X5 + 0.12X6 + 0.22X7 + 0.096X8 + 0.105X9 + 0.126X10

s.t.

X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10 = 400,000 100X1 + 100X2 + 100X3 + 95X4 + 92X5 + 79X6 + 80X8 + 100X9 ≥ 65(X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10) 60X1 + 70X2 + 75X3 + 20X4 + 30X5 + 22X6 + 50X7 + 10X9 ≤ 55(X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10)

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM X1 + X5 ≤ 60,000 X1 + X2 + X3 ≤ 160,000 X4 + X5 + X6 ≤ 160,000 X3 + X7 ≤ 160,000 X1 ≤ 80,000 X2 ≤ 80,000 X3 ≤ 80,000 X4 ≤ 80,000 X5 ≤ 80,000 X6 ≤ 80,000 X7 ≤ 80,000 X8 ≤ 80,000 X9 ≥ 1,000 X10 ≤ 15,000 X4 + X5 + X6 ≥ 90,000 X8 ≥ 10,000 Xj ≥ 0 j = 1, ... , 10 Solution:

X1 = 0; X2 = 80,000; X3 = 80,000; X4 = 0; X5 = 60,000; X6 = 74,000; X7 = 80,000; X8 = 10,000; X9 = 1000; X10 = 15,000; Total return = $64,355

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.04.02 - 4.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.2 Financial Applications KEYWORDS: Bloom's: Analyze 50. BP Cola must decide how much money to allocate for new soda and traditional soda advertising over the coming year.

The advertising budget is $10,000,000. Because BP wants to push its new sodas, at least one-half of the advertising budget is to be devoted to new soda advertising. However, at least $2,000,000 is to be spent on its traditional sodas. BP estimates that each dollar spent on traditional sodas will translate into 100 cans sold, whereas, because of the harder sell needed for new products, each dollar spent on new sodas will translate into 50 cans sold.

To attract new customers, BP has lowered its profit margin on new sodas to 2 cents per can as compared to 4 cents per can for traditional sodas. How should BP allocate its advertising budget if it wants to maximize its profits while selling at least 750 million cans?

ANSWER: X1 = amount invested in new soda advertising X2 = amount invested in traditional soda advertising Max s.t. © Cengage. Testing Powered by Cognero.

X1 + 4X2 X1 + X2 ≤ 10,000,000 X1 ≥ 5,000,000 Page 23

CH 04 - Linear Programming Applications in Marketing, Finance, and OM X2 ≥ 2,000,000 50X1 + 100X2 ≥ 750,000,000 X1, X2 ≥ 0 Spend $5,000,000 on new soda ad, spend $5,000,000 on traditional ad, profit = $25 million POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Analyze 51. Wes Wheeler is the production manager of Wheeler Wheels, Inc. Wes has just received orders for 1,000 standard wheels and 1,250 deluxe wheels next month and for 800 standard and 1,500 deluxe wheels the following month. All orders are to be filled.

The cost of producing standard wheels is $10 and the cost for deluxe wheels is $16. Overtime rates are 50% higher.

There are 1,000 hours of regular time and 500 hours of overtime available each month. It takes 0.5 hour to make a standard wheel and 0.6 hour to make a deluxe wheel. The cost of storing one wheel from one month to the next is $2.

Wes wants to develop a two-month production schedule for standard and deluxe wheels. Formulate this production planning problem as a linear program.

ANSWER: Define the decision variables We want to determine the production levels, xj, as follows: Month 1 Month 2 Reg. Time Overtime Reg. Time Overtime Standard x1 x2 x5 x6 Deluxe x3 x4 x7 x8 Let y1 = number of standard wheels stored from month 1 to month 2 y2 = number of deluxe wheels stored from month 1 to month 2 Define the objective function Minimize total production and storage costs: Min (cost per wheel)(number of wheels produced) + 2y1 + 2y2 Min 10x1 + 15x2 + 16x3 + 24x4 + 10x5 + 15x6 + 16x7 + 24x8 + 2y1 + 2y2 Define the constraints Standard Wheel Production Month 1 = (Requirements) + (Amount Stored) (1) x1 + x2 = 1000 + y1 or x1 + x2 – y1 = 1000 Deluxe Wheel Production Month 1 = (Requirements) + (Amount Stored) (2) x3 + x4 = 1250 + y2 or x3 + x4 – y2 = 1250 Standard Wheel Production Month 2 = (Requirements) – (Amount Stored) (3) x5 + x6 = 800 – y1 or x5 + x6 + y1 = 800 Deluxe Wheel Production Month 2 = (Requirements) – (Amount Stored) (4) x7 + x8 = 1500 – y2 or x7 + x8 + y2 = 1500 Regular Hours Used Month 1 < Regular Hours Available Month 1: © Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM (5) 0.5x1 + 0.6x3 < 1000 Overtime Hours Used Month 1 < Overtime Hours Available Month 1: (6) 0.5x2 + 0.6x4 < 500 Regular Hours Used Month 2 < Regular Hours Available Month 2: (7) 0.5x5 + 0.6x7 < 1000 Overtime Hours Used Month 2 < Overtime Hours Available Month 2: (8) 0.5x6 + 0.6x8 < 500 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Apply 52. Comfort Plus Inc. (CPI) manufactures a standard dining chair used in restaurants. The demand forecasts for quarter 1 (January–March) and quarter 2 (April–June) are 3700 chairs and 4200 chairs, respectively. CPI has a policy of satisfying all demand in the quarter in which it occurs.

The chair contains an upholstered seat that can be produced by CPI or purchased from DAP, a subcontractor. DAP currently charges $12.50 per seat, but has announced a new price of $13.75, effective April 1. CPI can produce the seat at a cost of $10.25. CPI can produce up to 3800 seats per quarter.

Seats that are produced or purchased in quarter 1 and used to satisfy demand in quarter 2 cost CPI $1.50 each to hold in inventory, but maximum inventory cannot exceed 300 seats.

Formulate this problem as a linear programming problem.

ANSWER: x1 = number of seats produced by CPI in quarter 1 x2 = number of seats purchased from DAP in quarter 1 x3 = number of seats carried in inventory from quarters 1 to 2 x4 = number of seats produced by CPI in quarter 2 x5 = number of seats purchased from DAP in quarter 2 Min 10.25x1 + 12.5x2 + 1.5x3 + 10.25x4 + 13.75x5 (costs) s.t. x1 + x2 – x3 > 3700 (quarter 1 demand) x3 + x4 + x5 > 4200 (quarter 2 demand) x1 < 3800 (CPI's production capacity in quarter 1) x4 < 3800 (CPI’s production capacity in quarter 2) x3 < 300 (inventory capacity) POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Analyze 53. Target Shirt Company makes three varieties of shirts: Collegiate, Traditional, and European. These shirts are made © Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM from different combinations of cotton and polyester.

The cost per yard of unblended cotton is $5 and for unblended polyester is $4. Target can receive up to 4000 yards of raw cotton and 3000 yards of raw polyester fabric weekly. The table below gives pertinent data concerning the manufacture of the shirts.

Shirt Collegiate

Total Yards 1.00

Traditional

1.20

European

0.90

Fabric Requirement At least 50% cotton No more than 20% polyester As much as 80% polyester

Weekly Contracts 500

Weekly Demand 600

Selling Price $14.00

650

850

$15.00

280

675

$18.00

Formulate and solve this blending problem as a linear program.

ANSWER:

Formulation Decision variables Not only must we decide how many shirts to make and how much fabric to purchase, we also need to decide how much of each fabric is blended into each shirt.

Let

sj = total number of shirt style j produced fi = number of yards of material i purchased xij = yards of fabric i blended into shirt style j where i = 1 (cotton) or 2 (polyester) and j = 1 (Collegiate), 2 (Traditional), or 3 (European)

Objective function Maximize the overall profit.

To determine the profit function, subtract the cost of purchasing the fabric from the shirt sales revenue. Thus, the objective function is: Max 14s1 + 15s2 + 18s3 − 5f1 − 4f2 Constraints Definition of total number of shirts of each style Total number of each style = Total yardage used in making the style/Yardage per shirt) (1) Collegiate: s1 = (x11 + x21)/1 >>> s1 − x11 − x21 = 0 (2) Traditional: s2 = (x12 + x22)/1.2 >>> 1.2s2 − x12 − x22 = 0 (3) European: s3 = (x13 + x23)/0.9 >>> 0.95s3 − x13 − x23 = 0 Definition of total yardage of materials (4) Cotton: f1 = x11 + x12 + x13 (5) Polyester: f2 = x21 + x22 + x23

>>> >>>

f1 − x11 − x12 − x13 = 0 f2 − x21 − x22 − x23 = 0

Weekly availability of the resources (6) Cotton: f1 < 4000 (7) Polyester: f2 < 3000 Meet weekly contracts © Cengage. Testing Powered by Cognero.

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CH 04 - Linear Programming Applications in Marketing, Finance, and OM (8) Collegiate: s1 > 500 (9) Traditional: s2 > 650 (10) European: s3 > 280 Do not exceed weekly demand (11) Collegiate: s1 < 600 (12) Traditional: s2 < 850 (13) European: s3 < 675 Fabric requirements Collegiate at least 50% cotton: (Total yds. of cotton in Colleg. shirts) > [(0.5(1.00) yds./shirt)(number of Colleg. shirts)] (14) x11 > 0.5s1 >>> x11 − 0.5s1 > 0 Traditional at most 20% polyester: (Total yds. polyester in Tradit. shirts) < [0.2(1.20) yds./shirt)(number of Tradit. shirts)] (15) x22 < 0.24s2 >>> x22 − 0.24s2 < 0 European at most 80% polyester: (Total yds. polyester in Europ. shirts) < [0.8(0.90) yds./shirt)(number of Europ. shirts)] (16) x23 < 0.72s3 >>> x23 − 0.72s3 < 0 Nonnegativity of variables sj, fi, xij > 0 for i = 1 or 2 and j = 1, 2, or 3 Total profit = $23,152.50 Cotton Collegiate yds. = 300.0, cotton Traditional yds. = 816.0, cotton European yds. = 121.5, polyester Collegiate yds. = 300.0, polyester Traditional yds. = 204.0, polyester European yds. = 486.0, Collegiate shirts = 600, Traditional shirts = 850, European shirts = 675, cotton yards = 1237.5, polyester yards = 990.0 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.04.03 - 4.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 4.3 Operations Management Applications KEYWORDS: Bloom's: Analyze

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CH 05 - Advanced Linear Programming Applications True / False 1. Revenue management methodology was originally developed for the banking industry. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.02 - 5.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.2 Revenue Management KEYWORDS: Bloom's: Remember 2. The goal of portfolio models is to create a portfolio that provides the best balance between risk and return. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.03 - 5.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.3 Portfolio Models and Asset Allocation KEYWORDS: Bloom's: Remember 3. DEA is used to measure the relative efficiency of two (or more) units in different industries, such as a bank and a school. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.01 - 5.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.1 Data Envelopment Analysis KEYWORDS: Bloom's: Remember 4. If the inputs of the composite unit in DEA are greater than the inputs for an individual unit, then the composite is more efficient. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.01 - 5.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking © Cengage. Testing Powered by Cognero.

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CH 05 - Advanced Linear Programming Applications TOPICS: KEYWORDS:

5.1 Data Envelopment Analysis Bloom's: Understand

5. It is possible for a single DEA programming model to identify all operating units that are inefficient. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.01 - 5.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.1 Data Envelopment Analysis KEYWORDS: Bloom's: Understand 6. Upon completion of a DEA model, all operating units that are relatively efficient are not necessarily identified. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.01 - 5.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.1 Data Envelopment Analysis KEYWORDS: Bloom's: Understand 7. DEA will show all but one operating unit to be relatively inefficient in the case where an operating unit produces the most of every output and also consumes the least of every input. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.01 - 5.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.1 Data Envelopment Analysis KEYWORDS: Bloom's: Understand 8. For a two-person, zero-sum, mixed-strategy game, it is necessary to solve the LP for each player in order to learn both players' optimal strategies. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.04 - 5.4 © Cengage. Testing Powered by Cognero.

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CH 05 - Advanced Linear Programming Applications NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.4 Game Theory KEYWORDS: Bloom's: Understand 9. If a pure strategy solution exists for a two-person, zero-sum game, it is the optimal solution to the game. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.04 - 5.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.4 Game Theory KEYWORDS: Bloom's: Understand 10. In a two-person, zero-sum, pure-strategy game, it can be advantageous for a player to switch from their strategy when the player discovers the other player's strategy in advance. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.05.04 - 5.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.4 Game Theory KEYWORDS: Bloom's: Understand 11. In portfolio models, risk is minimized by diversification. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.03 - 5.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.3 Portfolio Models and Asset Allocation KEYWORDS: Bloom's: Remember 12. Revenue management methodology can be applied in the case of nonperishable assets. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy © Cengage. Testing Powered by Cognero.

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CH 05 - Advanced Linear Programming Applications LEARNING OBJECTIVES: IMS.ASWC.19.05.02 - 5.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.2 Revenue Management KEYWORDS: Bloom's: Remember 13. In a data envelopment model, dual prices inform reservation agents of the revenue loss associated with overbooking each origin-destination-itinerary fare (ODIF). a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.02 - 5.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.2 Revenue Management KEYWORDS: Bloom's: Understand 14. In game theory, the player seeking to maximize the value of the game selects a maximax strategy. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.04 - 5.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.4 Game Theory KEYWORDS: Bloom's: Understand 15. The analysis of a two-person, zero-sum game begins with checking to see whether a pure strategy exists. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.04 - 5.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.4 Game Theory KEYWORDS: Bloom's: Remember 16. Revenue management methodology enables an airline to maximize the number of full-fare seats it sells on each flight. a. True b. False ANSWER: False POINTS: 1 © Cengage. Testing Powered by Cognero.

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CH 05 - Advanced Linear Programming Applications DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.02 - 5.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.2 Revenue Management KEYWORDS: Bloom's: Understand 17. Zero-sum means that what one player wins, the other player loses, and the gain and loss balance each other out. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.04 - 5.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.4 Game Theory KEYWORDS: Bloom's: Understand Multiple Choice 18. Revenue management methodology was originally developed for a(n) a. cruise line. b. airline. c. car rental company. d. hotel chain.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.02 - 5.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.2 Revenue Management KEYWORDS: Bloom's: Remember 19. The overall goal of portfolio models is to create a portfolio that provides the best balance between a. short- and long-term investments. b. gains and losses. c. risk and return. d. liquidity and stability.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.03 - 5.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.3 Portfolio Models and Asset Allocation KEYWORDS: Bloom's: Understand © Cengage. Testing Powered by Cognero.

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CH 05 - Advanced Linear Programming Applications 20. The composite unit in DEA a. has as output a weighted average of the outputs of the individual units. b. has as input a weighted average of the inputs of the individual units. c. has outputs greater than or equal to the outputs of any individual unit. d. All of these are correct.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.05.01 - 5.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.1 Data Envelopment Analysis KEYWORDS: Bloom's: Understand 21. Modern revenue management systems maximize revenue potential for an organization by helping to manage a. pricing strategies. b. reservation policies. c. short-term supply decisions. d. All of these are correct.

ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.02 - 5.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.2 Revenue Management KEYWORDS: Bloom's: Understand 22. A DEA linear programming model involving four input measures and three output measures will have a. 7 constraints. b. 8 constraints. c. 12 constraints. d. 13 constraints.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.05.01 - 5.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.1 Data Envelopment Analysis KEYWORDS: Bloom's: Understand 23. In general, every DEA linear programming model will include a constraint that requires the weights for the operating units to sum to a. 1. b. the number of operating units. c. 100. © Cengage. Testing Powered by Cognero.

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CH 05 - Advanced Linear Programming Applications d. None of these are correct.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.01 - 5.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.1 Data Envelopment Analysis KEYWORDS: Bloom's: Remember 24. For a two-person, zero-sum, mixed-strategy game, each player selects its strategy according to a. what strategy the other player used last. b. a fixed rotation of strategies. c. a probability distribution. d. the outcome of the previous game.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.05.04 - 5.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.4 Game Theory KEYWORDS: Bloom's: Understand 25. If it is an optimal solution for both players in a two-person, zero-sum game to select one strategy and stay with that strategy—regardless of what the other player does—then the game a. has a mixed strategy solution. b. will have alternating winners. c. will have no winner. d. has a pure strategy solution.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.05.04 - 5.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.4 Game Theory KEYWORDS: Bloom's: Understand 26. A linear programming application used to measure the relative efficiency of operating units with the same goals and objectives is a. game theory. b. asset allocation. c. data envelopment analysis. d. revenue management.

ANSWER: c POINTS: 1 © Cengage. Testing Powered by Cognero.

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CH 05 - Advanced Linear Programming Applications DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.01 - 5.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.1 Data Envelopment Analysis KEYWORDS: Bloom's: Remember 27. In data envelopment analysis, the percentage of an individual operating unit's resources that are available to the composite operating unit is the a. efficiency index. b. saddle point. c. maximin. d. minimax.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.05.01 - 5.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.1 Data Envelopment Analysis KEYWORDS: Bloom's: Remember 28. To develop a portfolio that provides the best return possible with a minimum risk, the linear programming model will have an objective function that a. minimizes the maximum risk. b. minimizes total risk. c. maximizes return and minimizes risk. d. maximizes the minimum return.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.05.03 - 5.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 5.3 Portfolio Models and Asset Allocation KEYWORDS: Bloom's: Understand 29. Revenue management methodology enables an airline to a. maximize the number of full-fare seats it sells on each flight. b. minimize the number of discount-fare seats it sells on each flight. c. increase the average number of passengers per flight. d. decrease the number of overbooked flights.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.05.02 - 5.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking © Cengage. Testing Powered by Cognero.

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CH 05 - Advanced Linear Programming Applications TOPICS: KEYWORDS:

5.2 Revenue Management Bloom's: Understand

Subjective Short Answer 30. Eastern Washington County School Corporation is interested in comparing educational performance at four elementary schools and has hired you to prepare a DEA model to do so. After detailed conversations with the corporation administrative staff and the building principals, you have isolated the following input and output measures: Input Measures Average classroom size Percent of students on reduced price lunch Average number of parent volunteer hours per week

Output Measures Percent of children in remedial classes Average first grade score on standard test Average fifth grade score on standard test

Data are collected for each school on each measure.

Archer Inputs Classroom size % reduced lunch Avg. vol. hours Outputs % remediation 1st grade score 5th grade score

School Hayes Ralston

Creekside

21 10 30

28 8 46

32 25 15

20 2 64

15 83 76

12 84 72

19 62 53

3 85 79

Develop a DEA model that would evaluate the efficiency of Ralston Elementary School.

ANSWER: Let A = weight applied to inputs and outputs for Archer Elementary H = weight applied to inputs and outputs for Hayes Elementary R = weight applied to inputs and outputs for Ralston Elementary C = weight applied to inputs and outputs for Creekside Elementary Min s.t.

E A+H+R+C=1 0.15A + 0.12H + 0.19R + 0.03C ≥ 0.19 83A + 84H + 62R + 85C ≥ 62 76A + 72H + 53R + 79C ≥ 53 21A + 28H + 32R + 20C ≤ 32E 0.10A + 0.08H + 0.25R + 0.02C ≤ 0.25E 30A + 46H + 15R + 64C ≤ 15

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.05.01 - 5.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 5.1 Data Envelopment Analysis KEYWORDS: Bloom's: Create 31. Mountainside State Park has four visitor centers. To study the operation of these centers, a DEA model has been developed that compares inputs (size, number of staff, weekly hours of operation) and outputs (% of visitors attending educational program, daily sales in gift shop). The computer solution is shown below. What can you conclude about the © Cengage. Testing Powered by Cognero.

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CH 05 - Advanced Linear Programming Applications efficiency of the North center?

LINEAR PROGRAMMING PROBLEM Min s.t.

1E+0wn+0ws+0we+0ww 1) −400E+400wn+1200ws+2400we+1500ww<0 2) −3E+3wn+6ws+10we+7ww<0 3) −56E+56wn+108ws+92we+108ww<0 4) −49E+49wn+83ws+56we+72ww>0 5) −38E+38wn+425ws+1200we+630ww>0 6) +1wn+1ws+1we+1ww=1

OPTIMAL SOLUTION Objective Function Value =

1.000

Variable E wn ws we ww

Reduced Cost 0.000 0.000 0.000 0.000 0.000

Value 1.000 1.000 0.000 0.000 0.000

Constraint 1 2 3 4 5 6 ANSWER:

Slack/Surplus Dual Price 0.000 0.008 0.000 0.000 0.000 0.001 0.000 −0.034 0.000 −0.013 0.000 −1.000 Because E = 1, the composite visitor center requires as much input as the North center does.

There is no evidence that the North center is inefficient.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.05.01 - 5.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 5.1 Data Envelopment Analysis KEYWORDS: Bloom's: Analyze 32. The output shows the solution to a DEA model where facilities in Seaview (S), Farmington (F), Lewiston (L), and San Domingo (D) are compared. The inputs, in order, are number of machines, size of workforce, and goodness of location.

The outputs, in order, are production, quality rating, and on-time completion percentage. The model examines the efficiency of Lewiston.

Min

0S+0F+0L+0D+1E

s.t.

1) 1S+1F+1L+1D=1 2) 5S+22F+36L+15D−36E<0 3) 400S+1500F+3150L+1060D−3150E<0 4) 24S+13F+32L+17D−32E<0 5) 800S+2900F+1860L+1700D+0E>1860 6) 95S+92F+83L+94D+0E>83 7) 83S+85F+90L+91D+0E>90 OPTIMAL SOLUTION © Cengage. Testing Powered by Cognero.

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CH 05 - Advanced Linear Programming Applications Objective Function Value =

0.510

Variable S F L D E

Value 0.000 0.167 0.000 0.833 0.510

Reduced Cost 0.385 0.000 0.490 0.000 0.000

Constraint 1 2 3 4 5 6 7

Slack/Surplus 0.000 2.208 474.479 0.000 40.000 10.667 0.000

a. b. c.

Dual Price 1.365 0.000 0.000 0.031 0.000 0.000 −0.021

Is the Lewiston plant efficient? Why or why not? If not, which plants should it emulate in order to improve?

How much more production does the composite facility provide than the Lewiston site?

What is the quality rating for the composite facility?

ANSWER:

a.

No. The composite plant requires less input than Lewiston to obtain the output. Lewiston should emulate Farmington and San Domingo.

40 83 + 10.667 = 93.667

b. c.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.05.01 - 5.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 5.1 Data Envelopment Analysis KEYWORDS: Bloom's: Evaluate

33. Consider a two-person, zero-sum game where the payoffs listed below are the winnings for Player A. Identify the pure strategy solution. What is the value of the game?

Player A Strategies a1 a2 a3

b1 5 1 7

Player B Strategies b2 5 6 2

b3 4 2 3

ANSWER:

Optimal pure strategies: Player A uses strategy a1; Player B uses strategy b3.

Value of game: Gain of 4 for Player A; loss of 4 for Player B. POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.05.04 - 5.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 5.4 Game Theory © Cengage. Testing Powered by Cognero.

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CH 05 - Advanced Linear Programming Applications KEYWORDS:

Bloom's: Evaluate

34. Consider a two-person, zero-sum game where the payoffs listed below are the winnings for Company X. Identify the pure strategy solution. What is the value of the game?

Company Y Strategies y1 y2 y3 3 5 9 8 4 3 7 6 7

Company X Strategies x1 x2 x3 ANSWER:

Optimal pure strategies: Company X uses strategy x3; Company Y uses strategy y2.

Value of game: Gain of 6 for Company X; loss of 6 for Company Y. POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.05.04 - 5.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 5.4 Game Theory KEYWORDS: Bloom's: Evaluate 35. Consider the following two-person, zero-sum game. Payoffs are the winnings for Company X. Formulate the linear program that determines the optimal mixed strategy for Company X. Company Y Strategies y1 y2 y3 4 3 9 2 5 1 6 1 7

Company X Strategies x1 x2 x3 ANSWER:

Max

GAINA

s.t.

4PA1 + 2PA2 + 6PA3 − GAINA ≥ 0 3PA1 + 5PA2 + 1PA3 − GAINA ≥ 0 9PA1 + 1PA2 + 7PA3 − GAINA ≥ 0 PA1 + PA2 + PA3 = 1 PA1, PA2, PA3, GAINA ≥ 0

(Strategy B1) (Strategy B2) (Strategy B3) (Probabilities must sum to 1) (Nonnegativity)

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.05.04 - 5.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 5.4 Game Theory KEYWORDS: Bloom's: Create 36. Shown below is the solution to the linear program for finding Player A's optimal mixed strategy in a two-person, zerosum game.

OBJECTIVE FUNCTION VALUE = VARIABLE

VALUE

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3.500 REDUCED COSTS Page 12

CH 05 - Advanced Linear Programming Applications PA1 PA2 PA3 GAINA

0.050 0.600 0350 3.500

0.000 0.000 0.000 0.000

CONSTRAINT 1 2 3 4

SLACK/SURPLUS 0.000 0.000 0.000 0.000

DUAL PRICES −0.500 −0.500 0.000 3.500

a. What is Player A's optimal mixed strategy? b. What is Player B's optimal mixed strategy? c. What is Player A's expected gain? d. What is Player B's expected loss?

ANSWER: a. Player A's optimal mixed strategy:

b.

Player B's optimal mixed strategy:

c.

Player A's expected gain: 3.500

Use strategy A1 with 0.05 probability.

Use strategy A2 with 0.60 probability.

Use strategy A3 with 0.35 probability.

Use strategy B1 with 0.50 probability.

Use strategy B2 with 0.50 probability.

Do not use strategy B3.

d. Player B's expected loss: 3.500 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.05.04 - 5.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 5.4 Game Theory KEYWORDS: Bloom's: Evaluate 37. Portfolio manager Max Gaines needs to develop an investment portfolio for his conservative clients. His task is to determine the proportion of the portfolio to invest in each of the five mutual funds listed below so that the portfolio provides the best return possible with a minimum risk. Formulate the maximin linear program.

Mutual Fund International Stock Large-Cap Blend Mid-Cap Blend Small-Cap Blend Intermediate Bond ANSWER:

Year 1 22.37 14.88 19.45 13.79 7.29

Annual Returns (Planning Scenarios) Year 2 Year 3 26.73 6.46 18.61 10.52 18.04 5.91 11.33 −2.07 8.05 9.18

Year 4 −3.19 5.25 −1.94 6.85 3.92

Max M −M + 22.37IS + 14.88LC + 19.45MC + 13.79SC + 7.29IB (Scenario 1) ≥0 −M + 26.73IS + 18.61LC + 18.04MC + 11.33SC + (Scenario 2)

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CH 05 - Advanced Linear Programming Applications 8.05IB ≥ 0 −M + 6.46IS + 10.52LC + 5.91MC − 2.07SC + 9.18IB ≥ 0(Scenario 3) −M − 3.19IS + 5.25LC − 1.94MC + 6.85SC + 3.92IB ≥ 0 (Scenario 4) IS + LC + MC + SC + IB = 1 (Total proportion must equal 1) M, IS, LC, MC, SC, IB ≥ 0

(Nonnegativity)

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.05.03 - 5.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 5.3 Portfolio Models and Asset Allocation KEYWORDS: Bloom's: Create 38. Portfolio manager Max Gaines needs to develop an investment portfolio for his clients who are willing to accept a moderate amount of risk. His task is to determine the proportion of the portfolio to invest in each of the five mutual funds listed below so that the portfolio provides an annual return of no less than 3%. Formulate the appropriate linear program.

Mutual Fund International Stock Large-Cap Blend Mid-Cap Blend Small-Cap Blend Intermediate Bond ANSWER:

Year 1 22.37 14.88 19.45 13.79 7.29 Max

Annual Returns (Planning Scenarios) Year 2 Year 3 26.73 6.46 18.61 10.52 18.04 5.91 11.33 −2.07 8.05 9.18

Year 4 −3.19 5.25 −1.94 6.85 3.92

13.093IS + 12.315LC + 10.365MC + 7.475SC + 7.110IB

22.37IS + 14.88LC + 19.45MC + 13.79SC + 7.29IB ≥ 3 (Scenario 1) 26.73IS + 18.61LC + 18.04MC + 11.33SC + 8.05IB ≥ 3 (Scenario 2) 6.46IS + 10.52LC + 5.91MC − 2.07SC + 9.18IB ≥ 3 (Scenario 3) −3.19IS + 5.25LC − 1.94MC + 6.85SC + 3.92IB ≥ 3 (Scenario 4) IS + LC + MC + SC + IB = 1 (Total proportion must equal 1) IS, LC, MC, SC, IB ≥ 0

(Nonnegativity)

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.05.03 - 5.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 5.3 Portfolio Models and Asset Allocation KEYWORDS: Bloom's: Create 39. Hervis Car Rental in Austin, Texas, has 50 high-performance Shelby-H Mustangs in its rental fleet. These cars will be in greater demand than usual during the last weekend in July when the Central Texas Mustang Club holds its annual rally in Austin. At times like this, Hervis uses a revenue management system to determine the optimal number of reservations to have available for the Shelby-H cars.

Hervis has agreed to have at least 60% of its Shelby-H Mustangs available for rally attendees at a special rate. Although many of the rally attendees will request a Saturday and Sunday two-day package, some attendees may select a Saturdayonly or a Sunday-only reservation. Customers not attending the rally may also request a Saturday and Sunday two-day package, or make a Saturday-only or Sunday-only reservation. Thus, six types of reservations are possible. The cost for © Cengage. Testing Powered by Cognero.

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CH 05 - Advanced Linear Programming Applications each type of reservation is shown here.

Two-Day SaturdaySundayPackage Only Only Rally $125 $75 $65 Regular 150 85 75 The anticipated demand for each type of reservation is as follows: Two-Day SaturdaySundayPackage Only Only Rally 20 10 15 Regular 10 20 25 Hervis Car Rental would like to determine how many Shelby-H Mustangs to make available for each type of reservation in order to maximize total revenue. a. Define the decision variables. b. Formulate a linear programming model for this revenue management application.

ANSWER: a. RATW = number of reservations available for rally attendees for two days RGTW = number of reservations available for regular customers for two days RASA = number of reservations available for rally attendees for Saturday only RGSA = number of reservations available for regular customers for Saturday only RASU = number of reservations available for rally attendees for Sunday only RGSU = number of reservations available for regular customers for Sunday only b. Max

125RATW + 155RGTW + 75RASA + 90RGSA + 65RASU + 80RGSU

s.t.

RATW + RETW + RASA + RGSA ≤ 50 RATW + RETW + RASU + RGSU ≤ 50

RASA ≤ 10 RGSA ≤ 20 RASU ≤ 15 RGSU ≤ 15

(Shelby-H Saturday capacity) (Shelby-H Sunday capacity) (60% Saturday capacity for rally attendees) (60% Sunday capacity for rally attendees) (two-day package rally demand) (two-day package regular demand) (Saturday-only rally demand) (Saturday-only regular demand) (Sunday-only rally demand) (Sunday-only regular demand)

RATW,RGTW,RASA,RGSA,RASU,RGSU ≥ 0

(Nonnegativity)

RATW + RASA ≥ 30 RATW + RASU ≥ 30 RATW ≤ 20 RGTW ≤ 10

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.05.02 - 5.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 5.2 Revenue Management KEYWORDS: Bloom's: Create 40. LeapFrog Airways provides passenger service for Indianapolis, Baltimore, Memphis, Austin, and Tampa. LeapFrog has two WB828 airplanes, one based in Indianapolis and the other in Baltimore. Each morning, the Indianapolis-based © Cengage. Testing Powered by Cognero.

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CH 05 - Advanced Linear Programming Applications plane flies to Austin with a stopover in Memphis, and the Baltimore-based plane flies to Tampa with a stopover in Memphis. Both planes have a coach section with a 120-seat capacity.

LeapFrog uses two fare classes: a discount-fare D class and a full-fare F class. LeapFrog's products, each referred to as an origin-destination-itinerary fare (ODIF), are listed below with their fares and forecasted demand.

ODIF 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Origin Indianapolis Indianapolis Indianapolis Indianapolis Indianapolis Indianapolis Baltimore Baltimore Baltimore Baltimore Baltimore Baltimore Memphis Memphis Memphis Memphis

Destination Memphis Austin Tampa Memphis Austin Tampa Memphis Austin Tampa Memphis Austin Tampa Austin Tampa Austin Tampa

Fare Class D D D F F F D D D F F F D D F F

ODIF Code IMD IAD ITD IMF IAF ITF BMD BAD BTD BMF BAF BTF MAD MTD MAF MTF

Fare 175 275 285 395 425 475 185 315 290 385 525 490 190 180 310 295

Demand 44 25 40 15 10 8 26 50 42 12 16 9 58 48 14 11

Develop a linear programming model for LeapFrog's problem situation and determine how many seats LeapFrog should allocate to each ODIF. ANSWER:

Define the decision variables There are 16 variables, one for each ODIF. IMD = number of seats allocated to Indianapolis-Memphis discount class IAD = number of seats allocated to Indianapolis-Austin discount class ITD = number of seats allocated to Indianapolis-Tampa discount class IMF = number of seats allocated to Indianapolis-Memphis full-fare class IAF = number of seats allocated to Indianapolis-Austin full-fare class ITF = number of seats allocated to Indianapolis-Tampa full-fare class BMD = number of seats allocated to Baltimore-Memphis discount class BAD = number of seats allocated to Baltimore-Austin discount class BTD = number of seats allocated to Baltimore-Tampa discount class BMF = number of seats allocated to Baltimore-Memphis full-fare class BAF = number of seats allocated to Baltimore-Austin full-fare class BTF = number of seats allocated to Baltimore-Tampa full-fare class MAD = number of seats allocated to Memphis-Austin discount class MTD = number of seats allocated to Memphis-Tampa discount class MAF = number of seats allocated to Memphis-Austin full-fare class MTF = number of seats allocated to Memphis-Tampa full-fare class Define the objective function Maximize total revenue: Max (Fare per seat for each ODIF) × (Number of seats allocated to the ODIF) Max 175IMD + 275IAD + 285ITD + 395IMF + 425IAF + 475ITF + 185BMD + 315BAD + 290BTD + 385BMF + 525BAF + 490BTF + 190MAD + 180MTD + 310MAF + 295MTF

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CH 05 - Advanced Linear Programming Applications Define the constraints There are four capacity constraints, one for each flight leg: (1) IMD + IAD + ITD + IMF + IAF + ITF  120 (Indianapolis-Memphis leg) (2) BMD + BAD + BTD + BMF + BAF + BTF  120 (Baltimore-Memphis leg) (3) IAD + IAF + BAD + BAF + MAD + MAF  120 (Memphis-Austin leg) (4) ITD + ITF + BTD + BTF + MTD + MTF  120 (Memphis-Tampa leg) There are 16 demand constraints, one for each ODIF: (5) IMD  44 (11) BMD  26 (17) MAD  58 (6) IAD  25 (12) BAD  50 (18) MTD  48 (7) ITD  40 (13) BTD  42 (19) MAF  14 (8) IMF  15 (14) BMF  12 (20) MTF  11 (15) BAF < 16 (9) IAF  10 (10) ITF  8 (16) BTF  9 Nonnegativity constraints: IMD, IAD, ITD, … , MTF  0 OBJECTIVE FUNCTION VALUE = 94735.000 VARIABLE IMD IAD ITD IMF IAF ITF BMD BAD BTD BMF BAF BTF MAD MTD MAF MTF

VALUE 44.000 3.000 40.000 15.000 10.000 8.000 26.000 50.000 7.000 12.000 16.000 9.000 27.000 45.000 14.000 11.000

CONSTRAINT 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

SLACK/SURPLUS 0.000 0.000 0.000 0.000 0.000 22.000 0.000 0.000 0.000 0.000 0.000 0.000 35.000 0.000 0.000 0.000 31.000

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REDUCED COSTS 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 DUAL PRICE 85.000 110.000 190.000 180.000 90.000 0.000 20.000 310.000 150.000 210.000 75.000 15.000 0.000 275.000 225.000 200.000 0.000 Page 17

CH 05 - Advanced Linear Programming Applications 18 19 20

3.000 0.000 0.000

0.000 120.000 115.000

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.05.01 - 5.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 5.1 Data Envelopment Analysis 5.2 Revenue Management KEYWORDS: Bloom's: Create

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CH 06 - Distribution and Network Models True / False 1. Whenever total supply is less than total demand in a transportation problem, the LP model does not determine how the unsatisfied demand is handled. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Remember 2. Converting a transportation problem LP from cost minimization to profit maximization requires only changing the objective function; the conversion does not affect the constraints. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Remember 3. A transportation problem with three sources and four destinations will have seven decision variables. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Understand 4. If a transportation problem has four origins and five destinations, the LP formulation of the problem will have nine constraints. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 © Cengage. Testing Powered by Cognero.

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CH 06 - Distribution and Network Models NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Understand 5. When a transportation problem has capacity limitations on one or more of its routes, it is known as a capacitated transportation problem. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Remember 6. When the number of agents exceeds the number of tasks in an assignment problem, one or more dummy tasks must be introduced in the LP formulation or else the LP will not have a feasible solution. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Understand 7. A transshipment constraint must contain a variable for every arc entering or leaving the node. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Understand 8. The shortest-route problem is a special case of the transshipment problem. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy © Cengage. Testing Powered by Cognero.

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CH 06 - Distribution and Network Models LEARNING OBJECTIVES: IMS.ASWC.19.06.03 - 6.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.3 Shortest-Route Problem KEYWORDS: Bloom's: Remember 9. Transshipment problems allow shipments both in and out of some nodes, while transportation problems do not. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Understand 10. A dummy origin in a transportation problem is used when supply exceeds demand. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Understand 11. When a route in a transportation problem is unacceptable, the corresponding variable can be removed from the LP formulation. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Understand 12. In the LP formulation of a maximal flow problem, a conservation of flow constraint ensures that an arc's flow capacity is not exceeded. a. True b. False ANSWER: False POINTS: 1 © Cengage. Testing Powered by Cognero.

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CH 06 - Distribution and Network Models DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.06.04 - 6.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.4 Maximal Flow Problem KEYWORDS: Bloom's: Understand 13. The maximal flow problem can be formulated as a capacitated transshipment problem and determines the maximum amount of flow (such as messages, vehicles, fluid, etc.) that can enter and exit a network system in a given period of time. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.04 - 6.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.4 Maximal Flow Problem KEYWORDS: Bloom's: Understand 14. The direction of flow in the shortest-route problem is always out of the origin node and into the destination node. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.03 - 6.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.3 Shortest-Route Problem KEYWORDS: Bloom's: Remember 15. A transshipment problem is a generalization of the transportation problem in which certain nodes are neither supply nodes nor destination nodes. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Understand 16. The assignment problem is a special case of the transportation problem in which one agent is assigned to one, and only one, task. a. True b. False © Cengage. Testing Powered by Cognero.

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CH 06 - Distribution and Network Models ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.02 - 6.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.2 Assignment Problem KEYWORDS: Bloom's: Remember 17. A transportation problem with three sources and four destinations will have seven variables in the objective function. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Understand 18. Flow in a transportation network is limited to one direction. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Remember 19. In a transportation problem with total supply equal to total demand, if there are four origins and seven destinations, and there is a unique optimal solution, the optimal solution will utilize 11 shipping routes. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Understand 20. In an assignment problem, one agent can be assigned to several tasks. a. True © Cengage. Testing Powered by Cognero.

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CH 06 - Distribution and Network Models b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.02 - 6.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.2 Assignment Problem KEYWORDS: Bloom's: Remember 21. In a capacitated transshipment problem, some or all of the transfer points are subject to capacity restrictions. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Understand 22. In a transportation problem, excess supply will appear as slack in the linear programming solution. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Remember 23. There are two specific types of problems common in supply chain models that can be solved using linear programing: transportation problems and transshipment problems. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Remember Multiple Choice © Cengage. Testing Powered by Cognero.

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CH 06 - Distribution and Network Models 24. The problem that deals with the distribution of goods from several sources to several destinations is a(n) a. maximal flow problem. b. transportation problem. c. assignment problem. d. shortest-route problem.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Remember 25. The parts of a network that represent the origins are called a. capacities. b. flows. c. nodes. d. arcs.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Remember 26. The objective of the transportation problem is to a. identify one origin that can satisfy total demand at the destinations and at the same time minimize total shipping cost. b. minimize the number of origins used to satisfy total demand at the destinations. c. minimize the number of shipments necessary to satisfy total demand at the destinations. d. minimize the cost of shipping products from several origins to several destinations.

ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Understand 27. We represent the number of units shipped from origin i to destination j by a. xij . b. xji . c. oij . © Cengage. Testing Powered by Cognero.

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CH 06 - Distribution and Network Models d. oji .

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Remember 28. Which of the following is NOT true regarding the linear programming formulation of a transportation problem? a. Costs appear only in the objective function. b. The number of variables is calculated as number of origins times number of destinations. c. The number of constraints is calculated as number of origins times number of destinations. d. The constraints' left-hand-side coefficients are either 0 or 1.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Understand 29. The difference between the transportation and assignment problems is that a. total supply must equal total demand in the transportation problem. b. the number of origins must equal the number of destinations in the transportation problem. c. each supply and demand value is 1 in the assignment problem. d. there are many differences between the transportation and assignment problems.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.02 - 6.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.2 Assignment Problem KEYWORDS: Bloom's: Understand 30. In the general linear programming model of the assignment problem, a. one agent can do parts of several tasks. b. one task can be done by several agents. c. each agent is assigned to its own best task. d. one agent is assigned to one and only one task.

ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.02 - 6.2 © Cengage. Testing Powered by Cognero.

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CH 06 - Distribution and Network Models NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.2 Assignment Problem KEYWORDS: Bloom's: Understand 31. The assignment problem is a special case of the a. transportation problem. b. transshipment problem. c. maximal flow problem. d. shortest-route problem.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.02 - 6.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.2 Assignment Problem KEYWORDS: Bloom's: Remember 32. Which of the following is NOT true regarding an LP model of the assignment problem? a. Costs appear in the objective function only. b. All constraints are of the ≥ form. c. All constraint left-hand-side coefficient values are 1. d. All decision variable values are either 0 or 1.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.06.02 - 6.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.2 Assignment Problem KEYWORDS: Bloom's: Understand 33. The assignment problem constraint x31 + x32 + x33 + x34 ≤ 2 means a. agent 3 can be assigned to two tasks. b. agent 2 can be assigned to three tasks. c. a mixture of agents 1, 2, 3, and 4 will be assigned to tasks. d. there is no feasible solution.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.06.02 - 6.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 6.2 Assignment Problem KEYWORDS: Bloom's: Understand 34. Arcs in a transshipment problem © Cengage. Testing Powered by Cognero.

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CH 06 - Distribution and Network Models a. must connect every node to a transshipment node. b. represent the cost of shipments. c. indicate the direction of the flow. d. All of these are correct.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Remember 35. Constraints in a transshipment problem a. correspond to arcs. b. include a variable for every arc. c. require the sum of the shipments out of an origin node to equal supply. d. All of these are correct.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Understand 36. In a transshipment problem, shipments a. cannot occur between two origin nodes. b. cannot occur between an origin node and a destination node. c. cannot occur between a transshipment node and a destination node. d. can occur between any two nodes.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Understand 37. Consider a shortest-route problem in which a bank courier must travel between branches and the main operations center. When represented with a network, a. the branches are the arcs and the operations center is the node. b. the branches are the nodes and the operations center is the source. c. the branches and the operations center are all nodes and the streets are the arcs. d. the branches are the network and the operations center is the node. © Cengage. Testing Powered by Cognero.

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CH 06 - Distribution and Network Models ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.06.03 - 6.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.3 Shortest-Route Problem KEYWORDS: Bloom's: Understand 38. The shortest-route problem finds the shortest route a. from the source to any other node. b. from any node to any other node. c. from any node to the destination. d. None of these are correct.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.03 - 6.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.3 Shortest-Route Problem KEYWORDS: Bloom's: Remember 39. Consider a maximal flow problem in which vehicle traffic entering a city is routed among several routes before eventually leaving the city. When represented with a network, a. the nodes represent stoplights. b. the arcs represent one-way streets. c. the nodes represent locations where speed limits change. d. None of these are correct.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.06.04 - 6.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.4 Maximal Flow Problem KEYWORDS: Bloom's: Understand 40. In a maximal flow problem, a. the flow out of a node is less than the flow into the node. b. the objective is to determine the maximum amount of flow that can enter and exit a network system in a given period of time. c. the number of arcs entering a node is equal to the number of arcs exiting the node. d. None of these are correct.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate © Cengage. Testing Powered by Cognero.

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CH 06 - Distribution and Network Models LEARNING OBJECTIVES: IMS.ASWC.19.06.04 - 6.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.4 Maximal Flow Problem KEYWORDS: Bloom's: Understand 41. If a transportation problem has four origins and five destinations, the LP formulation of the problem will have a. 5 constraints. b. 9 constraints. c. 18 constraints. d. 20 constraints.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Understand 42. Which of the following is NOT a characteristic of assignment problems? a. Costs appear in the objective function only. b. The RHS of all constraints is 1. c. The value of all decision variables is either 0 or 1. d. The signs of constraints are always <.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.06.02 - 6.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.2 Assignment Problem KEYWORDS: Bloom's: Understand 43. The network flows into and out of demand nodes are what makes the production and inventory application modeled in the textbook a a. shortest-route model. b. maximal flow model. c. transportation model. d. transshipment model.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.06.05 - 6.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 6.5 A Production and Inventory Application KEYWORDS: Bloom's: Understand © Cengage. Testing Powered by Cognero.

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CH 06 - Distribution and Network Models Subjective Short Answer 44. Write the LP formulation for this transportation problem.

ANSWER: Min

5X1A + 6X1B + 4X2A + 2X2B + 3X3A + 6X3B + 9X 4A + 7X4B

s.t.

X1A + X1B ≤ 100 X2A + X2B ≤ 200 X3A + X3B ≤ 150 X4A + X4B ≤ 50 X1A + X2A + X3A + X4A = 250 X1B + X 2B + X3B + X4B = 250 all Xij ≥ 0

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Create 45. Draw the network for this transportation problem.

Min

2XAX + 3XAY + 5XAZ+ 9XBX + 12XBY + 10XBZ

XAX + XAY + XAZ ≤ 500 X BX + XBY + XBZ ≤ 400 XAX + XBX = 300 XAY + XBY = 300 XAZ + XBZ = 300 Xij ≥ 0 ANSWER: s.t.

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CH 06 - Distribution and Network Models

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Create 46. Canning Transport is to move goods from three factories to three distribution centers. Information about the move is given below. Give the network model and the linear programming model for this problem.

Source A B C

Supply 200 100 150

Destination X Y Z

Demand 50 125 125

Shipping costs are: Source A B C

Destination X Y Z 3 2 5 9 10 — 5 6 4 (Source B cannot ship to destination Z)

ANSWER:

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CH 06 - Distribution and Network Models Min

3XAX + 2XAY + 5XAZ + 9XBX + 10XBY + 5XCX + 6XCY + 4XCZ

s.t.

XAX + XAY + XAZ ≤ 200 XBX + XBY ≤ 100 XCX + XCY + XCZ ≤ 150 XDX + XDY + XDZ ≤ 50 XAX + XBX + XCX + XDX = 250 X AY + XBY + XCY + XDY = 125 XAZ + XBZ + XCZ + XDZ = 125 Xij ≥ 0

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Create 47. After some special presentations, the employees of AV Center have to move projectors back to classrooms. The table below indicates the buildings where the projectors are now (the sources), where they need to go (the destinations), and a measure of the distance between sites.

Source Baker Hall Tirey Hall Arena Demand

Business 10 12 15 12

Education 9 11 14 20

Destination Parsons Hall 5 1 7 10

Holmstedt Hall 2 6 6 10

Supply 35 10 20

a. If you were going to write this as a linear programming model, how many decision variables would there be, and how many constraints would there be?

The solution to this problem is shown below. Use it to answer questions b through e.

TRANSPORTATION PROBLEM ***************************** OPTIMAL TRANSPORTATION SCHEDULE **************************************** FROM TO DESTINATION FROM ORIGIN 1 2 ------------------- ----------1 12 20 2 0 0 3 0 0

3 -----0 10 0

4 -----3 0 7

TOTAL TRANSPORTATION COST OR REVENUE IS 358 NOTE: THE TOTAL SUPPLY EXCEEDS THE TOTAL DEMAND BY 13 ORIGIN ----------

EXCESS SUPPLY -----------------------

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CH 06 - Distribution and Network Models 3

13

b. How many projectors are moved from Baker to Business? c. How many projectors are moved from Tirey to Parsons? d. How many projectors are moved from Arena to Education? e. Which site(s) has (have) projectors left?

ANSWER:

a. 12 decision variables, 7 constraints b. 12 c. 10 d. 0 e. Arena POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Evaluate 48. Show both the network and the linear programming formulation for this assignment problem.

Task Person 1 2 3

A 9 12 11

B 5 6 6

C 4 3 5

D 2 5 7

ANSWER:

Let

Xij = 1 if person i is assigned to job j = 0 otherwise

Min

9X1A + 5X1B + 4X1C + 2X1D + 12X2A + 6X2B + 3X2C + 5X2D + 11X3A + 6X3B + 5X3C + 7X3D

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CH 06 - Distribution and Network Models s.t.

X1A + X1B + X1C + X1D ≤ 1 X2A + X2B + X2C + X2D ≤ 1 X 3A + X3B + X3C + X3D ≤ 1 X4A + X4B + X4C + X4D ≤ 1 X1A + X2A + X3A + X4A = 1 X1B + X2B + X3B + X4B = 1 X1C + X2C + X3C + X4C = 1 X1D + X2D + X3D + X4D = 1

POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.06.02 - 6.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 6.2 Assignment Problem KEYWORDS: Bloom's: Analyze 49. Draw the network for this assignment problem.

Min

10x1A + 12x1B + 15x1C + 25x1D + 11x2A + 14x2B + 19x2C + 32x2D + 18x3A + 21x3B + 23x3C + 29x3D + 15x4A + 20x4B + 26x4C + 28x4D

s.t.

x1A + x1B + x1C + x1D = 1 x2A + x2B + x2C + x2D = 1 x3A + x3B + x3C + x3D = 1 x4A + x4B + x4C + x4D = 1 x1A + x2A + x3A + x4A = 1 x1B + x 2B + x3B + x4B = 1 x1C + x 2C + x3C + x4C = 1 x1D + x2D + x3D + x4D = 1 ANSWER:

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CH 06 - Distribution and Network Models POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.06.02 - 6.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 6.2 Assignment Problem KEYWORDS: Bloom's: Create 50. A professor has been contacted by four not-for-profit agencies that are willing to work with student consulting teams.

The agencies need help with such things as budgeting, information systems, coordinating volunteers, and forecasting.

Although each of the four student teams could work with any of the agencies, the professor feels that there is a difference in the amount of time it would take each group to solve each problem. The professor's estimate of the time, in days, is given in the table below. Use the computer solution to see which team works with which project.

Projects Team Budgeting Information Volunteers A 32 35 15 B 38 40 18 C 41 42 25 D 45 45 30 ASSIGNMENT PROBLEM ************************ OBJECTIVE: MINIMIZATION

Forecasting 27 35 38 42

SUMMARY OF UNIT COST OR REVENUE DATA ********************************************* TASK AGENT 1 2 3 4 -------------------------1 32 35 15 27 2 38 40 18 35 3 41 42 25 38 4 45 45 30 42 OPTIMAL ASSIGNMENTS COST/REVENUE ************************ *************** ASSIGN AGENT 3 TO TASK 1 41 ASSIGN AGENT 4 TO TASK 2 45 ASSIGN AGENT 2 TO TASK 3 18 ASSIGN AGENT 1 TO TASK 4 27 ----------------------------------------------TOTAL COST/REVENUE 131 ANSWER: Team A works with the forecast, Team B works with volunteers, Team C works with budgeting, and Team D works with information. The total time is 131.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.06.02 - 6.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 6.2 Assignment Problem KEYWORDS: Bloom's: Evaluate 51. Write the linear program for this transshipment problem. © Cengage. Testing Powered by Cognero.

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CH 06 - Distribution and Network Models

ANSWER: 3x16 + 2x14 + 3x15 + 5x24 + 6x25 + 2x32 + 8x 34 + 10x35 + 5x46 + 9x47 + 12x56 + 15x57

Min

x16 + x14 + x35 ≤ 500 x24 + x25 − x23 ≤ 400 x32 + x34 + x35 ≤ 300 x46 + x47 − (x14 + x24 + x34) = 0 x56 + x57 − (x15 + x25 + x35) = 0 x16 + x46 + x56 = 600 x56 + x57 = 600

s.t.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Create 52. Peaches are to be transported from three orchard regions to two canneries. Intermediate stops at a consolidation station are possible.

Orchard Riverside Sunny Slope Old Farm

Supply 1200 1500 2000

Station Waterford Northside

Cannery Sanderson Millville

Capacity 2500 3000

Shipment costs are shown in the table below. Where no cost is given, shipments are not possible. Where costs are shown, shipments are possible in either direction. Draw the network model for this problem.

R Riverside Sunny Slope Old Farm Waterford Northside

SS 1

OF

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W 5 4 6

N 5 3 2

S 3

M

2 5

4 9 Page 19

CH 06 - Distribution and Network Models Sanderson Millville

2

ANSWER:

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Create 53. RVW (Restored Volkswagens) buys 15 used VW's at each of two car auctions each week held at different locations. It then transports the cars to repair shops it contracts with. When they are restored to RVW's specifications, RVW sells 10 each to three different used car lots. Various costs are associated with the average purchase and transportation prices from each auction to each repair shop. There are also transportation costs from the repair shops to the used car lots. RVW is concerned with minimizing its total cost given the costs in the table below. a. Draw a network representation for this problem.

Repair Shops S1 S2 Auction 1

550

500

Auction 2

600

450

S1 S2

Used Car Lots L1

L2

L3

250

300

500

350

650

450

b. Formulate this problem as a transshipment linear programming model.

ANSWER:

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CH 06 - Distribution and Network Models

a.

b.

Denote A1 as node 1, A2 as node 2, S1 as node 3, S2 as node 4, L1 as node 5, L2 as node 6, and L3 as node 7.

Min

50X13 + 500X14 + 600X23 + 450X24 + 250X35 + 300X36 + 500X37 + 350X45 + 650X46 + 450X47

s.t.

X13 + X14 ≤ 15 X23 + X24 ≤ 15 X13 + X23 − X35 − X36 − X37 = 0 X14 + X24 − X45 − X46 − X47 = 0 X35 + X45 = 10 X36 + X46 = 10 X37 + X4 = 10 Xij ≥ 0 for all i and j

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Create 54. Consider the network below. Formulate the LP for finding the shortest-route path from node 1 to node 7.

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CH 06 - Distribution and Network Models

ANSWER: Min

10X12 + 12X13 + 4X24 + 8X25 + 7X35 + 9X36 + 4X 42 + 3X45 + 6X47 + 8X52 + 7X53 + 3X54 + 4X57 + 9X63 + 3X67

s.t.

X12 + X13 =1 −X12 + X24 + X25 − X42 − X52 =0 −X13 + X35 + X36 − X53 − X63 =0 −X24 + X42 + X45 + X47 − X54 =0 −X25 − X35 − X45 + X52 + X53 + X57 = 0 −X36 + X63 + X67 =0 X47 + X57 + X67 =1 Xij ≥ 0 for all i and j

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.06.03 - 6.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 6.3 Shortest-Route Problem KEYWORDS: Bloom's: Create 55. Consider the following shortest-route problem involving six cities with the distances given. Draw the network for this problem and formulate the LP for finding the shortest distance from City 1 to City 6.

Path 1 to 2 1 to 3 2 to 4 2 to 5 3 to 4 3 to 5 4 to 6 5 to 6 ANSWER:

Distance 3 2 4 5 3 7 6 2

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CH 06 - Distribution and Network Models

Min

3X12 + 2X13 + 4X24 + 5X25 + 3X34 + 7X35 + 4X42 + 3X43 + 6X46 + 5X 52 + 7X53 + 2X56

s.t.

X12 + X13 =1 −X12 + X24 + X25 − X42 − X52 = 0 −X13 + X34 + X35 − X 43 − X53 = 0 −X24 − X34 + X42 + X43 + X46 = 0 −X25 − X35 + X52 + X53 + X56 = 0 X46 + X56 =1 Xij ≥ 0 for all i and j

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.06.03 - 6.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 6.3 Shortest-Route Problem KEYWORDS: Bloom's: Create 56. A beer distributor needs to plan how to make deliveries from its warehouse (node 1) to a supermarket (node 7), as shown in the network below. Develop the LP formulation for finding the shortest route from the warehouse to the supermarket.

ANSWER: Min

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3X12 + 3X15 + 12X16 + 5X23 + 5X32 + 6X34 + 6X 43 + 4X46 + 5X47 + 8X56 + 4X64 + 8X65 + 3X67

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CH 06 - Distribution and Network Models s.t.

X12 + X15 + X16 =1 −X 12 + X23 =0 −X23 + X32 + X34 =0 −X34 + X43 + X46 + X47 − 4X64 =0 −X15 + X56 =0 −X16 − X46 − X56 + X64 + X65 + X67 = 0 X47 + X67 =1 Xij ≥ 0 for all i and j

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.06.03 - 6.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 6.3 Shortest-Route Problem KEYWORDS: Bloom's: Create 57. Consider the following shortest-route problem involving seven cities. The distances between the cities are given below. Draw the network model for this problem and formulate the LP for finding the shortest route from City 1 to City 7.

Path 1 to 2 1 to 3 1 to 4 2 to 3 2 to 5 3 to 4 3 to 5 3 to 6 4 to 6 5 to 7 6 to 7 ANSWER:

Distance 6 10 7 4 5 5 2 4 8 7 5

Min

6X12 + 10X13 + 7X14 + 4X23 + 5X25 + 4X32 + 5X34 + 2X35 + 4X36 + 5X43 + 8X46 + 5X52 + 2X53 + 7X57 + 4X63 + 8X64 + 5X67

s.t.

X12 + X13 + X14 −X12 + X23 + X25 − X32 − X52

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=1 =0 Page 24

CH 06 - Distribution and Network Models −X13 − X23 + X32 + X34 + X35 + X36 − X43 − X53 − X63 = 0 −X14 − X34 + X43 + X46 − X64 =0 −X25 − X35 + X52 + X53 + X57 =0 −X36 − X46 + X63 + X64 + X67 =0 X57 + X67 =1 Xij ≥ 0 for all i and j POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.06.03 - 6.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 6.3 Shortest-Route Problem KEYWORDS: Bloom's: Create 58. The network below shows the flows possible between pairs of six locations. Formulate an LP to find the maximal flow possible from node 1 to node 6.

ANSWER:

POINTS: DIFFICULTY:

Min X61 s.t.

X12 + X13 + X15 − X61 =0 X23 + X 24 − X12 − X32 =0 X32 + X34 + X35 − X13 − X23 − X53 = 0 X43 + X46 − X24 − X34 =0 X53 + X56 − X15 − X35 =0 X61 − X36 − X46 − X56 =0 X12 ≤ 18 X13 ≤ 20 X23 ≤ 9 X24 ≤ 15 X32 ≤ 4 X34 ≤ 10 X43 ≤ 10 X46 ≤ 14 X53 ≤ 8 X56 ≤ 10 Xij ≥ 0 for all i and j 1 Challenging

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X15 ≤ 10 X35 ≤ 8

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CH 06 - Distribution and Network Models LEARNING OBJECTIVE IMS.ASWC.19.06.04 - 6.4 S: NATIONAL STANDARDS United States - BUSPROG: Analytic : TOPICS: 6.4 Maximal Flow Problem KEYWORDS: Bloom's: Create 59. A network of railway lines connects the main lines entering and leaving a city. Speed limits, track reconstruction, and train length restrictions lead to the flow diagram below, where the numbers represent how many cars can pass per hour.

Formulate an LP to find the maximal flow in cars per hour from node 1 to node F.

ANSWER: Min XF1 s.t.

X12 + X15 + X16 − XF1 =0 X23 + X24 − X 12 =0 X34 − X23 =0 X48 + X4F − X24 − X34 − X84 = 0 X57 − X15 =0 X67 + X69 − X16 − X76 =0 X76 + X79 − X57 − X67 =0 X84 + X89 + X8F − X48 − X98 = 0 X98 + X9F − X69 − X79 − X89 = 0 XF1 − X4F − X8F − X9F =0 X12 ≤ 500 X15 ≤ 300 X23 ≤ 300 X24 ≤ 400 X34 ≤ 150 X48 ≤ 400 X4F ≤ 600 X57 ≤ 400 X67 ≤ 300 X69 ≤ 500 X76 ≤ 200 X79 ≤ 350 © Cengage. Testing Powered by Cognero.

X16 ≤ 600

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CH 06 - Distribution and Network Models X84 ≤ 200 X98 ≤ 300

X89 ≤ 300 X8F ≤ 450 X9F ≤ 500 Xij ≥ 0 for all i and j

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.06.04 - 6.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 6.4 Maximal Flow Problem KEYWORDS: Bloom's: Create 60. A foreman is trying to assign crews to produce the maximum number of parts per hour of a certain product. He has three crews and four possible work centers. The estimated number of parts per hour for each crew at each work center is summarized below. Solve for the optimal assignment of crews to work centers.

Crew A Crew B Crew C ANSWER:

WC1 15 20 25

Work Center WC2 WC3 WC4 20 18 30 22 26 30 26 27 30 OBJECTIVE FUNCTION VALUE = 82.000 VARIABLE AA1 AA2 AA3 AA4 AB1 AB2 AB3 AB4 AC1 AC2 AC3 AC4 AD1 AD2 AD3 AD4 CONSTRAINT 1 2 3 4 5 6 7

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VALUE 0.000 0.000 0.000 1.000 0.000 0.000 1.000 0.000 0.000 1.000 0.000 0.000 1.000 0.000 0.000 0.000

REDUCED COST 12.000 0.000 0.000 1.000 2.000 4.000 0.000 0.000 12.000 0.000 0.000 1.000 2.000 4.000 0.000 0.000

SLACK/SURPLUS 0.000 0.000 0.000 0.000 0.000 0.000 0.000

DUAL PRICE 18.000 23.000 24.000 −1.000 1.000 2.000 3.000 Page 27

CH 06 - Distribution and Network Models 8 0.000 An optimal solution is: Crew Crew A Crew B Crew C ----------

Work Center WC4 WC3 WC2 WC1 Total Parts Per Hour

12.000 Parts/Hour 30 26 26 --82

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.06.02 - 6.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 6.2 Assignment Problem KEYWORDS: Bloom's: Evaluate 61. A plant manager for a sporting goods manufacturer is in charge of assigning the manufacture of four new aluminum products to four different departments. Because of varying expertise and workloads, the different departments can produce the new products at various rates. If only one product is to be produced by each department and the daily output rates are given in the table below, which department should manufacture which product to maximize total daily product output? (Note: Department 1 does not have the facilities to produce golf clubs.)

Department 1 2 3 4

Baseball Bats 100 100 110 85

Tennis Rackets 60 80 75 50

Golf Clubs X 140 150 100

Racquetball Rackets 80 100 120 75

Formulate this assignment problem as a linear program.

ANSWER: xi represent all possible combinations of departments and products.

For example: x1 = 1 if Department 1 is assigned baseball bats; = 0 otherwise x2 = 1 if Department 1 is assigned tennis rackets; = 0 otherwise x5 = 1 if Department 2 is assigned baseball bats; = 0 otherwise x15 = 1 if Department 4 is assigned golf clubs; = 0 otherwise (Note: x3 is not used because Dept.1/golf clubs is not a feasible assignment.)

Min Z = 100x1+60x2+80x4+100x5+80x6+140x7+100x8+110x9 +75x10+150x11+120x12+85x13+50x14+100x15+75x16 s.t. x1 + x2 + x4 = 1 x5 + x6 + x7 + x8 = 1 x9 + x10 + x11 + x12 = 1 x13 + x14 + x15 + x16 = 1 x1 + x5 + x9 + x13 = 1 x2 + x6 + x10 + x14 = 1 x7 + x11 + x15 = 1 x4 + x8 + x12 + x16 = 1 © Cengage. Testing Powered by Cognero.

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CH 06 - Distribution and Network Models POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.06.02 - 6.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 6.2 Assignment Problem KEYWORDS: Bloom's: Evaluate 62. A clothing distributor has four warehouses that serve four large cities. Each warehouse has a monthly capacity of 5,000 blue jeans. It is considering using a transportation LP approach to match demand and capacity. The following table provides data on shipping cost, capacity, and demand constraints on a per-month basis. Develop a linear programming model for this problem.

Warehouse A B C D City Demand

City E City F City G City H 0.53 0.21 0.52 0.41 0.31 0.38 0.41 0.29 0.56 0.32 0.54 0.33 0.42 0.55 0.34 0.52 2,000 3,000 3,500 5,500

ANSWER: Xij = each combination of warehouse i and city j Min 0.53XAE + 0.21XAF + 0.52XAG + 0.41XAH + 0.31XBE + 0.38XBF + 0.41XBG + 0.29XBH + 0.56XCE + 0.32XCF + 0.54XCG + 0.33XCH + 0.42XDE + 0.55XDF + 0.34XDG + 0.52XDH s.t.

XAE + XAF + XAG + XAH  5,000 XAE + XBE + XCE + XDE = 2,000 XBE + XBF + XBG + XBH  5,000 XAF + XBF + XCF + XDF = 3,000 XCE + XCF + XCG + XCH  5,000 XAG + XBG + XCG + XDG = 3,500 XDE + XDF + XDG + XDH  5,000 XAH + XBH + XCH + XDH = 5,500 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Create 63. A computer manufacturing company wants to develop a monthly plan for shipping finished products from three of its manufacturing facilities to three regional warehouses. It is thinking about using a transportation LP formulation to exactly match capacities and requirements. Data on transportation costs (in dollars per unit), capacities, and requirements are given below.

Plant A B C Requirement

Warehouse 1 2 3 2.41 1.63 2.09 3.18 5.62 1.74 4.12 3.16 3.09 8,000 2,000 3,000

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Capacities 4,000 6,000 3,000

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CH 06 - Distribution and Network Models a. How many variables are involved in the LP formulation? b. How many constraints are there in this problem? c. What is the constraint corresponding to Plant B? d. What is the constraint corresponding to Warehouse 3?

ANSWER: The problem formulation is shown below. Use it to answer questions a through d.

Xij = each combination of plant i and warehouse j Min 2.41XA1 + 1.63XA2 + 2.09XA3 + 3.18XB1 + 5.62XB2 + 1.74XB3 + 4.12XC1 + 3.16XC2 + 3.09XC3 s.t. XA1 + XA2 + XA3 = 4,000 (capacities) XA1 + XB1 + XC1 = 8,000 (requirements) XB1 + XB2 + XB3 = 6,000 XA2 + XB2 + XC2 = 2,000 XC1 +XC2 + XC3 = 3,000 XA3 + XB3 + XC3 = 3,000 a. 9 variables b. 6 constraints c. XB1 + XB2 + XB3 = 6,000 d. XA3 + XB3 +XC3 = 3,000 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.06.01 - 6.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 6.1 Supply Chain Models KEYWORDS: Bloom's: Evaluate

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CH 07 - Integer Linear Programming True / False 1. The LP Relaxation contains the objective function and constraints of the IP problem, but drops all integer restrictions. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.07.02 - 7.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.2 Graphical and Computer Solutions for an All-Integer Linear Program KEYWORDS: Bloom's: Understand 2. In general, rounding large values of decision variables to the nearest integer value causes fewer problems than rounding small values. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.07.02 - 7.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.2 Graphical and Computer Solutions for an All-Integer Linear Program KEYWORDS: Bloom's: Understand 3. If the optimal solution to the LP Relaxation problem is an integer, it is the optimal solution to the integer linear program. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.07.02 - 7.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.2 Graphical and Computer Solutions for an All-Integer Linear Program KEYWORDS: Bloom's: Understand 4. Slack and surplus variables are not useful in integer linear programs. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.07.03 - 7.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking © Cengage. Testing Powered by Cognero.

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CH 07 - Integer Linear Programming TOPICS: KEYWORDS:

7.3 Applications Involving 0-1 Variables Bloom's: Understand

5. A multiple-choice constraint involves selecting k out of n alternatives, where k ≥ 2. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.04 - 7.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.4 Modeling Flexibility Provided by 0-1 Integer Variables KEYWORDS: Bloom's: Understand 6. In a model involving fixed costs, the 0-1 variable guarantees that the capacity is not available unless the cost has been incurred. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.03 - 7.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.3 Applications Involving 0-1 Variables KEYWORDS: Bloom's: Understand 7. If x1 + x2 ≤ 500y1 and y1 is 0-1, then if y1 is 0, x1 and x2 will be 0. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.03 - 7.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 7.3 Applications Involving 0-1 Variables KEYWORDS: Bloom's: Apply 8. The constraint x1 + x2 + x3 + x4 ≤ 2 means that two out of the first four projects must be selected. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.07.04 - 7.4 © Cengage. Testing Powered by Cognero.

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CH 07 - Integer Linear Programming NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 7.4 Modeling Flexibility Provided by 0-1 Integer Variables KEYWORDS: Bloom's: Apply 9. The constraint x1 − x2 = 0 implies that if project 1 is selected, project 2 cannot be. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.07.04 - 7.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 7.4 Modeling Flexibility Provided by 0-1 Integer Variables KEYWORDS: Bloom's: Understand 10. A product design and market share optimization problem involves choosing a product design that maximizes the number of consumers preferring it. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.07.03 - 7.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.3 Applications Involving 0-1 Variables KEYWORDS: Bloom's: Remember 11. If a problem has only less-than-or-equal-to constraints with positive coefficients for the variables, rounding down will always provide a feasible integer solution. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.07.02 - 7.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.2 Graphical and Computer Solutions for an All-Integer Linear Program KEYWORDS: Bloom's: Understand 12. The computer output for integer programs in the textbook does not include reduced costs, dual values, or sensitivity ranges because these variables are not meaningful for integer programs. a. True b. False © Cengage. Testing Powered by Cognero.

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CH 07 - Integer Linear Programming ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.07.02 - 7.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.2 Graphical and Computer Solutions for an All-Integer Linear Program KEYWORDS: Bloom's: Understand 13. Some linear programming problems have a special structure that guarantees the variables will have integer values. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.07.01 - 7.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.1 Types of Integer Linear Programming Models KEYWORDS: Bloom's: Remember 14. Generally, the optimal solution to an integer linear program is less sensitive to the constraint coefficients than is a linear program. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.07.04 - 7.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.4 Modeling Flexibility Provided by 0-1 Integer Variables KEYWORDS: Bloom's: Understand 15. For the classic assignment problem, the optimal linear programming solution will consist of 0s and 1s. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.07.01 - 7.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.1 Types of Integer Linear Programming Models KEYWORDS: Bloom's: Understand 16. If project 5 must be completed before project 6, the constraint would be x5 − x6 ≤ 0. a. True © Cengage. Testing Powered by Cognero.

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CH 07 - Integer Linear Programming b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.04 - 7.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.4 Modeling Flexibility Provided by 0-1 Integer Variables KEYWORDS: Bloom's: Understand 17. If the LP Relaxation of an integer program has a feasible solution, then the integer program has a feasible solution. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.02 - 7.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.2 Graphical and Computer Solutions for an All-Integer Linear Program KEYWORDS: Bloom's: Understand 18. Binary variables can be used to model multiple-choice and mutually exclusive constraints. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.07.04 - 7.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.4 Modeling Flexibility Provided by 0-1 Integer Variables KEYWORDS: Bloom's: Remember 19. Most practical applications of integer linear programming involve only 0-1 integer variables. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.07.03 - 7.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.3 Applications Involving 0-1 Variables KEYWORDS: Bloom's: Remember 20. Integer linear programs provide substantial modeling flexibility and thus are harder to solve than linear programs. a. True © Cengage. Testing Powered by Cognero.

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CH 07 - Integer Linear Programming b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.07.01 - 7.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.1 Types of Integer Linear Programming Models KEYWORDS: Bloom's: Understand 21. In general, whenever rounding has a minimal impact on the objective function and constraints, most managers find it acceptable. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.07.02 - 7.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.2 Graphical and Computer Solutions for an All-Integer Linear Program KEYWORDS: Bloom's: Remember Multiple Choice 22. Which of the following is the most useful contribution of integer programming? a. finding whole number solutions where fractional solutions would not be appropriate b. using 0-1 variables for modeling flexibility c. increased ease of solution d. provision for solution procedures for transportation and assignment problems ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.01 - 7.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.1 Types of Integer Linear Programming Models KEYWORDS: Bloom's: Understand 23. In a model, x1 ≥ 0 and integer, x2 ≥ 0, and x3 = 0, 1. Which of the following solutions would NOT be feasible? a. x1 = 5, x2 = 3, x3 = 0 b. x1 = 4, x2 = 0.389, x3 = 1 c. x1 = 2, x2 = 3, x3 = 0.578 d. x1 = 0, x2 = 8, x3 = 0 ANSWER: POINTS:

c 1

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CH 07 - Integer Linear Programming DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.01 - 7.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 7.1 Types of Integer Linear Programming Models KEYWORDS: Bloom's: Analyze 24. Rounded solutions to linear programs must be evaluated for a. feasibility and optimality. b. sensitivity and duality. c. relaxation and boundedness. d. All of these are correct.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.02 - 7.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.2 Graphical and Computer Solutions for an All-Integer Linear Program KEYWORDS: Bloom's: Understand 25. Rounding the solution of an LP Relaxation to the nearest integer values provides a(n) a. feasible but not necessarily optimal integer solution. b. integer solution that is optimal. c. integer solution that might be neither feasible nor optimal. d. infeasible solution.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.02 - 7.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.2 Graphical and Computer Solutions for an All-Integer Linear Program KEYWORDS: Bloom's: Understand 26. The solution to the LP Relaxation of a maximization integer linear program provides a(n) a. upper bound for the value of the objective function. b. lower bound for the value of the objective function. c. upper bound for the value of the decision variables. d. lower bound for the value of the decision variables.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.02 - 7.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.2 Graphical and Computer Solutions for an All-Integer Linear Program KEYWORDS: Bloom's: Understand © Cengage. Testing Powered by Cognero.

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CH 07 - Integer Linear Programming 27. The graph of a problem that requires x1 and x2 to be integer has a feasible region a. the same as its LP Relaxation. b. consisting of dots. c. consisting of horizontal stripes. d. consisting of vertical stripes.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.02 - 7.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.2 Graphical and Computer Solutions for an All-Integer Linear Program KEYWORDS: Bloom's: Understand 28. The 0-1 variables in the fixed cost models correspond to a. a process for which a fixed cost occurs. b. the number of products produced. c. the number of units produced. d. the actual value of the fixed cost.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.03 - 7.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.3 Applications Involving 0-1 Variables KEYWORDS: Bloom's: Understand 29. Sensitivity analysis for integer linear programming a. can be provided only by computer. b. has precisely the same interpretation as that from linear programming. c. does not have the same interpretation as that from linear programming and should be disregarded. d. is most useful for 0-1 models.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.04 - 7.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.4 Modeling Flexibility Provided by 0-1 Integer Variables KEYWORDS: Bloom's: Understand 30. Let x1 and x2 be 0-1 variables whose values indicate whether projects 1 and 2 are/are not done. Which of the following answers indicates that project 2 can be done only if project 1 is done? a. x1 + x2 = 1 b. x1 + x2 = 2 © Cengage. Testing Powered by Cognero.

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CH 07 - Integer Linear Programming c. x1 − x2 ≤ 0 d. x1 − x2 ≥ 0 ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.04 - 7.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 7.4 Modeling Flexibility Provided by 0-1 Integer Variables KEYWORDS: Bloom's: Analyze 31. Let x1, x2, and x3 be 0-1 variables whose values indicate whether the projects are not done (0) or are done (1). Which of the following answers indicates that at least two of the projects must be done? a. x1 + x2 + x3 ≥ 2 b. x1 + x2 + x3 ≤ 2 c. x1 + x2 + x3 = 2 d. x1 − x2 = 0 ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.04 - 7.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 7.4 Modeling Flexibility Provided by 0-1 Integer Variables KEYWORDS: Bloom's: Analyze 32. If the acceptance of project A is conditional on the acceptance of project B, and vice versa, the appropriate constraint to use is a a. multiple-choice constraint. b. k out of n alternatives constraint. c. mutually exclusive constraint. d. corequisite constraint.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.04 - 7.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.4 Modeling Flexibility Provided by 0-1 Integer Variables KEYWORDS: Bloom's: Understand 33. In an all-integer linear program, a. all objective function coefficients must be integer. b. all right-hand-side values must be integer. c. all variables must be integer. d. all objective function coefficients and right-hand-side values must be integer. © Cengage. Testing Powered by Cognero.

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CH 07 - Integer Linear Programming ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.01 - 7.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.1 Types of Integer Linear Programming Models KEYWORDS: Bloom's: Understand 34. To perform sensitivity analysis involving an integer linear program, it is best to a. use the dual prices very cautiously. b. make multiple computer runs. c. use the same approach as you would for a linear program. d. use LP Relaxation.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.04 - 7.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.4 Modeling Flexibility Provided by 0-1 Integer Variables KEYWORDS: Bloom's: Understand 35. Modeling a fixed cost problem as an integer linear program requires a. adding the fixed costs to the corresponding variable costs in the objective function. b. using 0-1 variables. c. using multiple-choice constraints. d. using LP Relaxation.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.03 - 7.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.3 Applications Involving 0-1 Variables KEYWORDS: Bloom's: Understand 36. Most practical applications of integer linear programming involve a. only 0-1 integer variables and not ordinary integer variables. b. mostly ordinary integer variables and a small number of 0-1 integer variables. c. only ordinary integer variables. d. a near equal number of ordinary integer variables and 0-1 integer variables.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.07.03 - 7.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking © Cengage. Testing Powered by Cognero.

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CH 07 - Integer Linear Programming TOPICS: KEYWORDS:

7.3 Applications Involving 0-1 Variables Bloom's: Understand

37. Assuming W1, W2, and W3 are 0-1 integer variables, the constraint W1 + W2 + W3 < 1 is often called a a. multiple-choice constraint. b. mutually exclusive constraint. c. k out of n alternatives constraint. d. corequisite constraint.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.04 - 7.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.4 Modeling Flexibility Provided by 0-1 Integer Variables KEYWORDS: Bloom's: Understand 38. Which of the following applications modeled in the textbook does NOT involve only 0-1 integer variables? a. supply chain design b. bank location c. capital budgeting d. product design and market share optimization ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.07.03 - 7.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 7.3 Applications Involving 0-1 Variables KEYWORDS: Bloom's: Remember Subjective Short Answer 39. Solve the following problem graphically.

Max s.t.

a. b. c.

5X + 6Y 17X + 8Y ≤ 136 3X + 4Y ≤ 36 X, Y ≥ 0 and integer Graph the constraints for this problem. Indicate all feasible solutions.

Find the optimal solution to the LP Relaxation. Round down to find a feasible integer solution. Is this solution optimal?

Find the optimal solution.

ANSWER:

a.

The feasible region is those integer values in the space labeled Feasible Region.

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CH 07 - Integer Linear Programming

b.

The optimal LP relaxed solution occurs at X = 5.818, Y = 4.636, with Z = 56.909. The rounded down solution occurs at X = 5, Y = 4, Z= 49.

The optimal solution is at X = 4, Y = 6, and Z = 56.

c.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.07.02 - 7.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 7.2 Graphical and Computer Solutions for an All-Integer Linear Program KEYWORDS: Bloom's: Analyze 40. Solve the following problem graphically.

Max s.t.

a. b. c.

X + 2Y 6X + 8Y ≤ 48 7X + 5Y ≥ 35 X, Y ≥ 0 Y is integer Graph the constraints for this problem. Indicate all feasible solutions.

Find the optimal solution to the LP Relaxation. Round down to find a feasible integer solution. Is this solution optimal?

Find the optimal solution.

ANSWER:

a.

The feasible region consists of the portions of the horizontal lines that lie within the area labeled F. R.

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CH 07 - Integer Linear Programming

b. c.

The optimal LP relaxed solution occurs at X = 1.538, Y = 4.846, with Z = 11.231. The rounded down solution occurs at X = 1.538, Y = 4.

The optimal solution is at X = 2.667, and Y = 4, and Z = 10.667.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.07.02 - 7.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 7.2 Graphical and Computer Solutions for an All-Integer Linear Program KEYWORDS: Bloom's: Analyze 41. Solve the following problem graphically.

Min s.t.

6X + 11Y 9X + 3Y ≥ 27 7X + 6Y ≥ 42 4X + 8Y ≥ 32 X, Y ≥ 0 and integer

a.

Graph the constraints for this problem. Indicate all feasible solutions.

Find the optimal solution to the LP Relaxation. Round up to find a feasible integer solution. b.

Is this solution optimal? c. Find the optimal solution.

ANSWER: a. The feasible region is the set of integer points in the area labeled Feasible Region.

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CH 07 - Integer Linear Programming

b.

The optimal relaxed solution is at X = 4.5, Y = 1.75, and Z = 46.25.

The rounded solution is X = 5, Y = 2.

The optimal solution is at X = 6, Y = 1, and Z = 47.

c.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.07.02 - 7.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 7.2 Graphical and Computer Solutions for an All-Integer Linear Program KEYWORDS: Bloom's: Evaluate 42. Consider a capital budgeting example with five projects from which to select. Let xi = 1 if project i is selected, 0 if not, for i = 1, … , 5. Write the appropriate constraint(s) for each condition. Conditions are independent. a. Choose no fewer than three projects. b. If project 3 is chosen, project 4 must be chosen. c. If project 1 is chosen, project 5 must not be chosen. d. Projects cost 100, 200, 150, 75, and 300, respectively. The budget is 450. e. No more than two of projects 1, 2, and 3 can be chosen.

ANSWER: a. b. c. d. e.

x1 + x2 + x3 + x4 + x5 ≥ 3 x3 − x4 ≤ 0 x1 + x5 ≤ 1 100x1 + 200x2 + 150x3 + 75x4 + 300x5 ≤ 450 x1 + x2 + x3 ≤ 2

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.07.03 - 7.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 7.3 Applications Involving 0-1 Variables KEYWORDS: Bloom's: Create

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CH 07 - Integer Linear Programming 43. Grush Consulting has five projects to consider. Each will require time in the next two quarters according to the table below.

Project A B C D E

Time in First Quarter 5 3 7 2 15

Time in Second Quarter 8 12 5 3 1

Revenue 12,000 10,000 15,000 5,000 20,000

Revenue from each project is also shown. Develop a model whose solution would maximize revenue, meet the time budget of 25 in the first quarter and 20 in the second quarter, and not do both projects C and D. ANSWER: Let A = 1 if project A is selected, 0 otherwise; same for B, C, D, and E Max s.t.

12,000A + 10,000B + 15,000C + 5,000D + 20,000E 5A + 3B + 7C + 2D + 15E ≤ 25 8A + 12B + 5C + 3D + 1E ≤ 20 C+D≤1

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.07.03 - 7.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 7.3 Applications Involving 0-1 Variables KEYWORDS: Bloom's: Create 44. Westfall Company has a contract to produce 10,000 garden hoses for a large discount chain. Westfall has four different machines that can produce this kind of hose. Because these machines are from different manufacturers and use differing technologies, their specifications are not the same.

Machine 1 2 3 4 a. b. c. d.

Fixed Cost to Set Up Production Run 750 500 1000 300

Variable Cost per Hose 1.25 1.50 1.00 2.00

Capacity 6000 7500 4000 5000

This problem requires two different kinds of decision variables. Clearly define each kind.

The company wants to minimize total cost. Give the objective function.

Give the constraints for the problem.

Write a constraint to ensure that if machine 4 is used, machine 1 cannot be.

ANSWER: a. b. c.

Let

Pi = number of hoses produced on machine i Ui = 1 if machine i is used; 0 otherwise Min 750U1 + 500U2 + 1000U3 + 300U4 + 1.25P 1 + 1.5P2 + P3 + 2P4 P1 ≤ 6000U1 P2 ≤ 7500U2 P3 ≤ 4000U3 P4 ≤ 5000U4 P1 + P2 + P3 + P4 ≥ 10,000

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CH 07 - Integer Linear Programming d.

U1 + U4 ≤ 1

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.07.03 - 7.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 7.3 Applications Involving 0-1 Variables KEYWORDS: Bloom's: Analyze 45. Hansen Controls has been awarded a contract for a large number of control panels. To meet this demand, it will use its existing plants in San Diego and Houston and consider new plants in Tulsa, St. Louis, and Portland. Finished control panels are to be shipped to Seattle, Denver, and Kansas City. Pertinent information is given in the table below.

Shipping Cost to Destination Sources San Diego Houston Tulsa St. Louis Portland

Construction Cost ------350,000 200,000 480,000 Demand

Seattle

Denver

Kansas City

Capacity

5 10 9 12 4 3000

7 8 4 6 10 8000

8 6 3 2 11 9000

2500 2500 10,000 10,000 10,000

Develop a model whose solution would reveal which plants to build and the optimal shipping schedule.

ANSWER: Let Pij = number of panels shipped from source i to destination j Bi = 1 if plant i is built; 0 otherwise (i = 3, 4, 5) Min

350,000B3 + 200,000B4 + 480,000B5 + 5P11 + 7P12 + 8P13 + 10P21 + 8P22 + 6P23 + 9P31 + 4P32 + 3P33 + 12P41 + 6P42 + 2P43 + 4P51 + 10P52 + 11P53

s.t.

P11 + P12 + P13 ≤ 2500 P21 + P22 + P23 ≤ 2500 P31 + P32 + P33 ≤ 10,000B3 P41 + P42 + P43 ≤ 10,000B4 P51 + P52 + P53 ≤ 10,000B5 P11 + P21 + P31 + P41 + P51 = 3000 P12 + P22 + P32 + P42 + P52 = 8000 P13 + P23 + P33 + P43 + P53 = 9000

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.07.03 - 7.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 7.3 Applications Involving 0-1 Variables KEYWORDS: Bloom's: Create 46. Your express package courier company is drawing up new zones for the location of drop boxes for customers. The © Cengage. Testing Powered by Cognero.

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CH 07 - Integer Linear Programming city has been divided into seven zones. You have targeted six possible locations for drop boxes. The list of the seven zones and which drop boxes could be reached easily from each zone is listed below.

Zone Downtown Financial Downtown Legal Retail South Retail East Manufacturing North Manufacturing East Corporate West

Can Be Served by Locations 1, 2, 5, 6 2, 4, 5 1, 2, 4, 6 3, 4, 5 1, 2, 5 3, 4 1, 2, 6

Let xi = 1 if drop box location i is used, 0 otherwise. Develop a model to provide the smallest number of locations, but make sure that each zone is covered by at least two boxes.

ANSWER: Min s.t.

Σxi x1 + x2 + x5 + x6 ≥ 2 x2 + x4 + x5 ≥ 2 x1 + x2 + x4 + x6 ≥ 2 x3 + x4 + x5 ≥ 2 x1 + x 2 + x5 ≥ 2 x3 + x4 ≥ 2 x1 + x2 + x6 ≥ 2

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.07.01 - 7.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 7.1 Types of Integer Linear Programming Models KEYWORDS: Bloom's: Create 47. Consider the problem faced by a summer camp recreation director who is trying to choose activities for a rainy day.

Information about possible choices is given in the table below.

Category Art

Music Sports Computer

a. b.

Activity 1 - Painting 2 - Drawing 3 - Nature craft 4 - Rhythm band 5 - Relay races 6 - Basketball 7 - Internet 8 - Creative writing 9 - Games

Time (minutes) 30 20 30 20 45 60 45 30 40

Popularity with Campers 4 5 3 5 2 1 1 4 1

Popularity with Counselors 2 2 1 5 1 3 1 3 2

Give a general definition of the variables necessary in this problem so that each activity can be considered for inclusion in the day's schedule.

The popularity ratings are defined so that 1 is the most popular. If the objective is to keep the campers happy, what should the objective function be?

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CH 07 - Integer Linear Programming Write constraints for these restrictions: c. At most one art activity can be done. d. No more than two computer activities can be done. e. If basketball is chosen, then rhythm band must be chosen. f. At least 120 minutes of activities must be selected. g. No more than 165 minutes of activities may be selected. h. To keep the staff happy, the counselor rating should be no higher than 10.

ANSWER: a. b. c. d. e. f. g. h.

Let xi = 1 if activity i is chosen, 0 if not, for i = 1, … , 9 Max 4x1 + 5x2 + 3x3 + 5x4 + 2x5 + 1x6 + 1x7 + 4x8 + 1x9 x1 + x2 + x3 ≤ 1 x7 + x8 + x9 ≤ 2 x6 ≤ x4 30x1 + 20x2 + 30x3 + 20x4 + 45x5 + 60x6 + 45x7 + 30x8 + 40x9 ≥ 120 30x1 + 20x 2 + 30x3 + 20x4 + 45x5 + 60x6 + 45x7 + 30x8 + 40x9 ≤ 165 2x1 + 2x2 + 1x3 + 5x4 + 1x5 + 3x6 + 1x7 + 3x8 + 2x9 ≤ 10

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.07.02 - 7.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 7.2 Graphical and Computer Solutions for an All-Integer Linear Program KEYWORDS: Bloom's: Create 48. Tower Engineering Corporation is considering undertaking several proposed projects for the next fiscal year. The projects, the number of engineers and support personnel required for each project, and the expected profits for each project are summarized in the following table: Project 1 20 15 1.0

Engineers required Support personnel required Profit ($1,000,000s)

2 55 45 1.8

3 47 50 2.0

4 38 40 1.5

5 90 70 3.6

6 63 70 2.2

Formulate an integer program that maximizes Tower's profit subject to the following management constraints: a. Use no more than 175 engineers. b. Use no more than 150 support personnel. c. If either project 6 or project 4 is done, both must be done. d. Project 2 can be done only if project 1 is done. e. If project 5 is done, project 3 must not be done and vice versa. f. No more than three projects are to be done.

ANSWER: Max s.t.

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P1 + 1.8P2 + 2P3 + 1.5P4 + 3.6P5 + 2.2P6 20P1 + 55P2 + 47P 3 + 38P4 + 90P5 + 63P6 ≤ 175 15P1 + 45P2 + 50P3 + 40P4 + 70P5 + 70P6 ≤ 150 P4 − P6 = 0 P1 − P2 ≥ 0 P3 + P5 ≤ 1 Page 18

CH 07 - Integer Linear Programming P1 + P2 + P3 + P4 + P5 + P6 ≤ 3 Pi = 0 or 1 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.07.02 - 7.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 7.2 Graphical and Computer Solutions for an All-Integer Linear Program KEYWORDS: Bloom's: Create 49. Given the following all-integer linear program: Max s.t.

a. b. c. d.

3x1 + 2x2 3x1 + x2 ≤ 9 x1 + 3x2 ≤ 7 −x1 + x2 ≤ 1 x1, x2 ≥ 0 and integer Solve the problem as a linear program ignoring the integer constraints. Show that the optimal solution to the linear program gives fractional values for both x1 and x2.

What solution is obtained by rounding fractions greater than or equal to 1/2 to the next larger number? Show that this solution is not a feasible solution.

What is the solution obtained by rounding down all fractions? Is it feasible?

Enumerate all points in the linear programming feasible region in which both x1 and x2 are integers, and show that the feasible solution obtained in part (c) is not optimal and that in fact the optimal integer is not obtained by any form of rounding.

ANSWER:

a. From the graph that follows, the optimal solution to the linear program is x1 = 2.5, x2 = 1.5, z = 10.5. b. By rounding the optimal solution of x1 = 2.5, x2 = 1.5 to x1 = 3, x2 = 2, this point lies outside the feasible c. By rounding the optimal solution down to x1 = 2, x2 = 1, we see that this solution indeed is an integer sol the feasible region, and substituting in the objective function, it gives z = 8. d. There are eight feasible integer solutions in the linear programming feasible region with z values as follo

1.

2.

3.

4.

5.

6.

7.

8.

x1 0 1 2 3 0 1 2 1

x2 0 0 0 0 1 1 1 2

z 0 3 6 9 2 5 8 7

optimal

part (c) solution

x1 = 3, x2 = 0 is the optimal solution. Rounding the LP solution (x1 = 2.5, x2 = 1.5) would not have bee

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CH 07 - Integer Linear Programming

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTI IMS.ASWC.19.07.02 - 7.2 VES: NATIONAL STANDARUnited States - BUSPROG: Analytic DS: TOPICS: 7.2 Graphical and Computer Solutions for an All-Integer Linear Program KEYWORDS: Bloom's: Evaluate 50. Market Pulse Research has conducted a study for Lucas Furniture on some designs for a new commercial office desk.

Three attributes were found to be most influential in determining which desk is most desirable: number of file drawers, the presence or absence of pullout writing boards, and simulated wood or solid color finish. Listed below are the partworths for each level of each attribute provided by a sample of seven potential Lucas customers.

Consumer 1 2 3 4 5 6 7

0 5 18 4 12 19 6 9

File Drawers 1 2 26 20 11 5 16 22 8 4 9 3 15 21 6 3

Pullout Writing Boards Present Absent 18 11 12 16 7 13 18 9 4 14 8 17 13 5

Finish Simulated Wood Solid Color 17 10 15 26 11 19 22 14 30 19 20 11 16 28

Suppose the overall utility (sum of part-worths) of the current favorite commercial office desk is 50 for each customer.

What product design will maximize the share of choices for the seven sample participants? Using Excel, formulate and solve this 0-1 integer programming problem.

ANSWER: Define the decision variables: There are seven lij decision variables, one for each level of attribute. lij = 1 if Lucas chooses level i for attribute j; 0 otherwise There are seven Yk decision variables, one for each consumer in the sample.

Yk = 1 if consumer k chooses the Lucas brand; 0 otherwise Define the objective function: © Cengage. Testing Powered by Cognero.

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CH 07 - Integer Linear Programming Maximize the number of consumers preferring the Lucas brand desk.

Max Y1 + Y2 + Y3 + Y4 + Y5 + Y6 + Y7 Define the constraints: There is one constraint for each consumer in the sample.

5l11 + 26l21 + 20l31 + 18l12 + 11l22 + 17l13 + 10l23 − 50Y1 ≥ 1 18l11 + 11l21 + 5l31 + 12l12 + 16l22 + 15l13 + 26l23 − 50Y2 ≥ 1 4l11 + 16l21 + 22l31 + 7l12 + 13l22 + 11l13 + 19l23 − 50Y3 ≥ 1 12l11 + 8l21 + 4l31 + 18l12 + 9l22 + 22l13 + 14l23 − 50Y4 ≥ 1 19l11 + 9l21 + 3l31 + 4l12 + 14l22 + 30l13 + 19l23 − 50Y5 ≥ 1 6l11 + 15l21 + 21l31 + 8l12 + 17l22 + 20l13 + 11l23 − 50Y6 ≥ 1 9l11 + 6l21 + 3l31 + 13l12 + 5l22 + 16l13 + 28l23 − 50Y7 ≥ 1 There is one constraint for each attribute. l11 + l21 + l31 = 1 l12 + l22 = 1 l13 + l23 = 1 Optimal solution: Lucas should choose these product features: 1 file drawer (I21 = 1) No pullout writing boards (I22 = 1) Simulated wood finish (I13 = 1) Three sample participants would choose the Lucas design: Participant 1 (Y1 = 1) Participant 5 (Y5 = 1) Participant 6 (Y6 = 1) POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.A7.01 - A7.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: Appendix 7.1 Excel Solution of Integer Linear Programs KEYWORDS: Bloom's: Create 51. Kloos Industries has projected the availability of capital over each of the next three years to be $850,000, $1,000,000, and $1,200,000, respectively. It is considering four options for the disposition of the capital: a.

Research and development of a promising new product b.

Plant expansion c.

Modernization of its current facilities d.

Investment in a valuable piece of nearby real estate Monies not invested in these projects in a given year will NOT be available for the following year's investment in the projects. The expected benefits three years hence from each of the four projects and the yearly capital outlays of the four options are summarized in the table below in millions of dollars.

In addition, Kloos has decided to undertake exactly two of the projects. If plant expansion is selected, it will also modernize its current facilities. © Cengage. Testing Powered by Cognero.

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CH 07 - Integer Linear Programming Capital Outlay Options

Year 1

Year 2

Year 3

New product R&D Plant expansion Modernization Real estate

0.35 0.50 0.35 0.50

0.55 0.50 0.40 0

0.75 0 0.45 0

Projected Benefits 5.2 3.6 3.2 2.8

Formulate and solve this problem as a binary programming problem.

ANSWER: Max s.t.

5.2X1 + 3.6X2 + 3.2X3 + 2.8X4 0.35X1 + 0.50X2 + 0.35X3 + 0.50X4 ≤ 0.85 (Year 1) 0.55X1 + 0.50X2 + 0.40X3 ≤ 1.00 ( Year 2) 0.75X 1 + 0.45X3 ≤ 1.20 (Year 3) X1 + X2 + X3 + X4 = 2 X2 + X3 ≥0 Xi = 0 or 1 X1 = 1, X2 = 0, X3 = 1, X4 = 0, total projected benefits = $8.4 million

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.07.03 - 7.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 7.3 Applications Involving 0-1 Variables KEYWORDS: Bloom's: Evaluate 52. Given the following all-integer linear program: Max s. t.

15x1 + 2x2 7x1 + x2 < 23 3x1 - x2 < 5 x1, x2 > 0 and integer

a. Solve the problem as an LP, ignoring the integer constraints. b. What solution is obtained by rounding up fractions greater than or equal to 1/2? Is this the optimal integer solution? c. What solution is obtained by rounding down all fractions? Is this the optimal integer solution? Explain. d. Show that the optimal objective function value for the ILP (integer linear programming) is lower than that for the optimal LP. e. Why is the optimal objective function value for the ILP problem always less than or equal to the corresponding LP's optimal objective function value? When would they be equal? Comment on the optimal objective function of the MILP (mixed-integer linear programming) compared to the corresponding LP and ILP. ANSWER: a. x1 = 2.8, x2 = 3.4, Obj. function = 48.8 b. x1 = 3, x2 = 3 → infeasible c. x1 = 2, x2 = 3 → feasible but not optimal d. optimal x1 = 2, x2 = 9, objective function = 48 < 48.8 e. Additional integer constraints restrict the feasible region further; © Cengage. Testing Powered by Cognero.

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CH 07 - Integer Linear Programming optimal solution values: ILP < mixed ILP < LP. POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.07.02 - 7.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 7.2 Graphical and Computer Solutions for an All-Integer Linear Program KEYWORDS: Bloom's: Evaluate 53. A business manager for a grain distributor is asked to decide how many containers of each of two grains to purchase to fill its 1,600-pound capacity warehouse. The table below summarizes the container size, availability, and expected profit per container upon distribution.

Grain A B

Container Size 500 lbs.

600 lbs.

Containers Available 3 2

Container Profit $1,200 $1,500

a. Formulate as a linear program with the decision variables representing the number of containers purchased of each grain. Solve for the optimal solution. b. What would be the optimal solution if you were not allowed to purchase fractional containers? c. There are three possible results from rounding an LP solution to obtain an integer solution: (1) The rounded optimal LP solution will be the optimal IP solution.

(2) The rounded optimal LP solution gives a feasible, but not optimal IP solution.

(3) The rounded optimal LP solution is an infeasible IP solution.

For this problem, (i) round down all fractions; (ii) round up all fractions; and (iii) round off (to the nearest integer) all fractions (Note: Two of these are equivalent.) Which result above (1, 2, or 3) occurred under each rounding method?

ANSWER: a. 4/5 container of Grain A, 2 containers of Grain B, $3960 b. 2 containers of Grain A, 1 container of Grain B, $3900 c. (i) Round down all fractions: For (0,2), the result is feasible but not optimal. (ii) Round up all fractions: For (1,2), the result is infeasible. (iii) Round off all fractions: For (1,2), the result is infeasible.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.07.02 - 7.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 7.2 Graphical and Computer Solutions for an All-Integer Linear Program KEYWORDS: Bloom's: Create 54. Given the following all-integer linear programming problem: Max 3x1 + 10x2 s.t.

2x1 + x2 < 5 x1 + 6x2 < 9 x1 – x2 > 2 x1, x2 > 0 and integer

a. Solve the problem graphically as a linear program. © Cengage. Testing Powered by Cognero.

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CH 07 - Integer Linear Programming b. Show that there is only one integer point and that it is optimal. c. Suppose the third constraint was changed to x1 – x2 > 2.1. What is the new optimal solution to the LP? To the ILP? ANSWER: a. x1 = 7/3, x2 = 1/3, objective function = 31/3 b. x1 = 2, x2 = 0, objective function = 6 c. LP—optimal solution changes slightly; ILP—infeasible POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.07.02 - 7.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 7.2 Graphical and Computer Solutions for an All-Integer Linear Program KEYWORDS: Bloom's: Analyze

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CH 08 - Nonlinear Optimization Models True / False 1. A nonlinear optimization problem is any optimization problem in which at least one term in the objective function or a constraint is nonlinear. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Remember 2. A function is considered a quadratic function if its nonlinear terms have a power of 2. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Remember 3. Nonlinear programming algorithms are more complex than linear programming algorithms. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Remember 4. Many linear programming algorithms such as the simplex method optimize by examining only the extreme points and selecting the extreme point that gives the best solution value. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 © Cengage. Testing Powered by Cognero.

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CH 08 - Nonlinear Optimization Models NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Understand 5. Nonlinear optimization problems can have only one local optimal solution. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Remember 6. A feasible solution is a global optimum if no other feasible solutions with a better objective function value are found in the immediate neighborhood. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Remember 7. A feasible solution is a global optimum if no other feasible points with a better objective function value are found in the feasible region. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Remember 8. For a typical nonlinear problem, if you change the right-hand side by even a small amount, the dual value changes. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy © Cengage. Testing Powered by Cognero.

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CH 08 - Nonlinear Optimization Models LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Understand 9. The interpretation of the dual price for nonlinear models is different than the interpretation of the dual price for linear models. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Remember 10. In the case of functions with multiple local optima, most nonlinear optimization software methods can get stuck and terminate at a local optimum. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Remember 11. For a minimization problem, a point is a global minimum if there are no other feasible points with a smaller objective function value. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Remember 12. It is possible that a nonlinear application in which there is a single local optimal solution also shows that solution as the global optimal solution. a. True b. False © Cengage. Testing Powered by Cognero.

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CH 08 - Nonlinear Optimization Models ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Understand 13. Functions that are convex have a single local maximum that is also the global maximum. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Understand 14. The function f (X, Y) = X 2 + Y 2 has a single global minimum and is relatively easy to minimize. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Understand 15. The problem of maximizing a concave quadratic function over a linear constraint set is relatively difficult to solve. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Understand 16. Each point on the efficient frontier is the maximum possible risk, measured by portfolio variance, for the given return. a. True b. False © Cengage. Testing Powered by Cognero.

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CH 08 - Nonlinear Optimization Models ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.03 - 8.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.3 Markowitz Portfolio Model KEYWORDS: Bloom's: Understand 17. Any feasible solution to a blending problem with pooled components is feasible to the problem with no pooling. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.04 - 8.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.4 Blending: The Pooling Problem KEYWORDS: Bloom's: Understand 18. Any feasible solution to a blending problem without pooled components is feasible to the problem with pooled components. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.04 - 8.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.4 Blending: The Pooling Problem KEYWORDS: Bloom's: Understand 19. When components (or ingredients) in a blending problem must be pooled, the number of feasible solutions is reduced. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.04 - 8.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.4 Blending: The Pooling Problem KEYWORDS: Bloom's: Understand 20. The value of the coefficient of imitation, q, in the Bass model for forecasting adoption of a new product cannot be negative. a. True © Cengage. Testing Powered by Cognero.

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CH 08 - Nonlinear Optimization Models b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.08.05 - 8.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.5 Forecasting Adoption of a New Product KEYWORDS: Bloom's: Understand 21. The Markowitz mean-variance portfolio model presented in the text is a convex optimization problem. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.03 - 8.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.3 Markowitz Portfolio Model KEYWORDS: Bloom's: Understand 22. Because most nonlinear optimization codes will terminate with a local optimum, the solution returned by the codes will be the best solution. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Understand 23. It is possible for the optimal solution to a nonlinear optimization problem to lie in the interior of the feasible region. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Understand 24. The minimum value for a convex function is 0, and the point (0, 0) gives the minimum value of 0. © Cengage. Testing Powered by Cognero.

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CH 08 - Nonlinear Optimization Models a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Understand Multiple Choice 25. Which of the following statements is incorrect? a. A global optimum is a local optimum in a nonlinear optimization problem. b. A local maximum is a global maximum in a concave nonlinear optimization problem. c. A global minimum is a local minimum in a convex nonlinear optimization problem. d. A local optimum is a global optimum in a nonlinear optimization problem.

ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Understand 26. The measure of risk most often associated with the Markowitz portfolio model is the portfolio a. average return. b. minimum return. c. variance. d. standard deviation.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.08.03 - 8.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.3 Markowitz Portfolio Model KEYWORDS: Bloom's: Understand 27. An investor can pick the mean-variance tradeoff that he or she is most comfortable with by looking at a graph of the a. feasible region. b. pooled components. c. rolling horizon. d. efficient frontier.

ANSWER: d © Cengage. Testing Powered by Cognero.

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CH 08 - Nonlinear Optimization Models POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.08.03 - 8.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.3 Markowitz Portfolio Model KEYWORDS: Bloom's: Understand 28. The Bass model is used to forecast the adoption of a new product. Which of the following is NOT a parameter of this Bass model? a. coefficient of innovation b. coefficient of interaction c. coefficient of imitation d. estimated pool of current adopters ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.08.05 - 8.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.5 Forecasting Adoption of a New Product KEYWORDS: Bloom's: Understand 29. When the number of blending components exceeds the number of storage facilities, the number of feasible solutions to the blending problem is a. reduced. b. increased. c. unchanged. d. zero.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.08.04 - 8.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.4 Blending: The Pooling Problem KEYWORDS: Bloom's: Understand 30. In the Bass model for forecasting the adoption of a new product, the objective function a. minimizes the sum of forecast errors. b. minimizes the sum of squared forecast errors. c. maximizes the number of adoptions. d. maximizes the number of adoptions and imitations.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.08.05 - 8.5 © Cengage. Testing Powered by Cognero.

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CH 08 - Nonlinear Optimization Models NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.5 Forecasting Adoption of a New Product KEYWORDS: Bloom's: Understand 31. Which of the following is NOT true regarding a concave function? a. It is bowl-shaped down. b. It is relatively easy to maximize. c. It has multiple local maxima. d. It has a single global maximum.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Understand 32. A convex function is a. bowl-shaped up. b. bowl-shaped down. c. elliptical in shape. d. sinusoidal in shape.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Remember 33. If the coefficient of each squared term in a quadratic function is positive, the function is a. concave. b. convex. c. elliptical. d. sinusoidal.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Understand 34. When components share a storage facility, they are called © Cengage. Testing Powered by Cognero.

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CH 08 - Nonlinear Optimization Models a. constrained components. b. correlated components. c. blended components. d. None of these are correct.

ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.04 - 8.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.4 Blending: The Pooling Problem KEYWORDS: Bloom's: Remember 35. The key idea behind constructing an index fund is to choose a portfolio of securities that a. is a mix of growth- and income-oriented stocks. b. minimizes risk without sacrificing liquidity. c. mimics the performance of a broad market index. d. balances short- and long-term investments.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.08.02 - 8.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.2 Constructing an Index Fund KEYWORDS: Bloom's: Understand 36. It is common that components that are referred to as pooled a. are shared by two or more customers. b. have common ingredients. c. share a storage facility or storage tank. d. are interchangeable.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.08.04 - 8.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 8.4 Blending: The Pooling Problem KEYWORDS: Bloom's: Understand Subjective Short Answer 37. Investment manager Max Gaines has several clients who wish to own a mutual fund portfolio that matches, as a whole, the performance of the S&P 500 stock index. His task is to determine what proportion of the portfolio should be invested in each of the five mutual funds listed below so that the portfolio most closely mimics the performance of the S&P 500 index. Formulate the appropriate nonlinear program. © Cengage. Testing Powered by Cognero.

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CH 08 - Nonlinear Optimization Models Mutual Fund International Stock Large-Cap Blend Mid-Cap Blend Small-Cap Blend Intermediate Bond S&P 500 Index ANSWER:

Year 1 26.73 18.61 18.04 11.33 8.05 21.00

Annual Returns (Planning Scenarios) Year 2 Year 3 22.37 6.46 14.88 10.52 19.45 15.91 13.79 −2.07 7.29 9.18 19.00 12.00

Year 4 −3.19 5.25 −1.94 6.85 3.92 4.00

IS = proportion of portfolio invested in international stock LC = proportion of portfolio invested in large-cap blend MC = proportion of portfolio invested in mid-cap blend SC = proportion of portfolio invested in small-cap blend IB = proportion of portfolio invested in intermediate bond R1 = portfolio return for scenario 1 (Year 1) R2 = portfolio return for scenario 2 (Year 2) R3 = portfolio return for scenario 3 (Year 3) R4 = portfolio return for scenario 4 (Year 4) (R1 − 21)2 + (R2 − 19)2 + (R3 − 12)2 + (R4 − 4)2

Min

26.73IS + 18.61LC + 18.04MC + 11.33SC + 8.05IB = R1 22.37IS + 14.88LC + 19.45MC + 13.79SC + 7.29IB = R2 6.46IS + 10.52LC + 15.91MC − 2.07SC + 9.18IB = R3 −3.19IS + 5.25LC − 1.94MC + 6.85SC + 3.92IB = R4 IS + LC + MC + SC + IB = 1 IS, LC, MC, SC, IB ≥ 0 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.08.02 - 8.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 8.2 Constructing an Index Fund KEYWORDS: Bloom's: Create 38. Financial planner Minnie Margin has a substantial number of clients who wish to own a mutual fund portfolio that matches, as a whole, the performance of the Russell 2000 index. Her task is to determine what proportion of the portfolio should be invested in each of the five mutual funds listed below so that the portfolio most closely mimics the performance of the Russell 2000 index. Formulate the appropriate nonlinear program.

Mutual Fund International Stock Large-Cap Value Mid-Cap Value Small-Cap Growth Short-Term Bond Russell 2000 Index ANSWER:

Year 1 22.37 15.48 17.42 23.18 9.26 20.00

Annual Returns (Planning Scenarios) Year 2 Year 3 26.73 4.86 19.64 11.50 20.07 −4.97 12.36 3.25 8.81 6.15 22.00 8.00

Year 4 2.17 −5.25 −1.69 3.81 4.04 −2.00

IS = proportion of portfolio invested in international stock LC = proportion of portfolio invested in large-cap value MC = proportion of portfolio invested in mid-cap value

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CH 08 - Nonlinear Optimization Models SC = proportion of portfolio invested in small-cap growth SB = proportion of portfolio invested in short-term bond R1 = portfolio return for scenario 1 (Year 1) R2 = portfolio return for scenario 2 (Year 2) R3 = portfolio return for scenario 3 (Year 3) R4 = portfolio return for scenario 4 (Year 4) Min

(R1 − 20)2 + (R2 − 22)2 + (R3 − 8)2 + (R4 + 2)2 22.37IS + 15.48LC + 17.42MC + 23.18SC + 9.26SB = R1 26.73IS + 19.64LC + 20.07MC + 12.36SC + 8.81SB = R2 4.86IS + 11.50LC − 4.97MC + 3.25SC + 6.15SB = R3 2.17IS − 5.25LC − 1.69MC + 3.81SC + 4.04SB = R4 IS + LC + MC + SC + SB = 1 IS, LC, MC, SC, SB ≥ 0

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.08.02 - 8.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 8.2 Constructing an Index Fund KEYWORDS: Bloom's: Create 39. Investment manager Max Gaines wishes to develop a mutual fund portfolio based on the Markowitz portfolio model.

He needs to determine the proportion of the portfolio to invest in each of the five mutual funds listed below so that the variance of the portfolio is minimized subject to the constraint that the expected return of the portfolio be at least 4%.

Formulate the appropriate nonlinear program.

Mutual Fund International Stock Large-Cap Blend Mid-Cap Blend Small-Cap Blend Intermediate Bond ANSWER:

Annual Returns (Planning Scenarios) Year 1 Year 2 Year 3 Year 4 26.73 22.37 6.46 −3.19 18.61 14.88 10.52 5.25 18.04 19.45 15.91 −1.94 11.33 13.79 6.85 −2.07 8.05 7.29 9.18 3.92 IS = proportion of portfolio invested in international stock LC = proportion of portfolio invested in large-cap blend MC = proportion of portfolio invested in mid-cap blend SC = proportion of portfolio invested in small-cap blend IB = proportion of portfolio invested in intermediate bond R1 = portfolio return for scenario 1 (Year 1) R2 = portfolio return for scenario 2 (Year 2) R3 = portfolio return for scenario 3 (Year 3) R4 = portfolio return for scenario 4 (Year 4) AR = expected (average) portfolio return Min

0.25(R1 − AR)2 + 0.25(R2 − AR)2 + 0.25(R3 − AR)2 + 0.25(R4 − AR)2 26.73IS + 8.61LC + 18.04MC + 11.33SC + 8.05IB = R1 22.37IS + 4.88LC + 19.45MC + 13. 79SC + 7.29IB = R2 6.46IS + 10.52LC + 15.91MC − 2.07SC + 9.18IB = R3

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CH 08 - Nonlinear Optimization Models −3.19IS + 5.25LC − 1.69MC + 3.81SC + 4.04IB = R4 IS + LC + MC + SC + IB = 1 0.25R1 + 0.25R2 + 0.25R3 + 0.25R4 = AR AR ≥ 4 IS, LC, MC, SC, IB ≥ 0 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.08.03 - 8.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 8.3 Markowitz Portfolio Model KEYWORDS: Bloom's: Create 40. Financial planner Minnie Margin wishes to develop a mutual fund portfolio based on the Markowitz portfolio model.

She needs to determine the proportion of the portfolio to invest in each of the five mutual funds listed below so that the variance of the portfolio is minimized subject to the constraint that the expected return of the portfolio be at least 5%.

Formulate the appropriate nonlinear program.

Mutual Fund International Stock Large-Cap Value Mid-Cap Value Small-Cap Growth Short-Term Bond ANSWER:

Annual Returns (Planning Scenarios) Year 1 Year 2 Year 3 Year 4 22.37 26.73 4.86 2.17 15.48 19.64 11.50 −5.25 17.42 20.07 −4.97 −1.69 23.18 12.36 3.25 3.81 9.26 8.81 6.15 4.04 IS = proportion of portfolio invested in international stock LC = proportion of portfolio invested in large-cap value MC = proportion of portfolio invested in mid-cap value SC = proportion of portfolio invested in small-cap growth SB = proportion of portfolio invested in short-term bond R1 = portfolio return for scenario 1 (Year 1) R2 = portfolio return for scenario 2 (Year 2) R3 = portfolio return for scenario 3 (Year 3) R4 = portfolio return for scenario 4 (Year 4) AR = expected (average) portfolio return Min

0.25(R1 − AR)2 + 0.25(R2 − AR)2 + 0.25(R3 − AR)2 + 0.25(R4 − AR)2 22.37IS + 15.48LC + 17.42MC + 23.18SC + 9.26SB = R1 26.73IS + 19.64LC + 20.07MC + 12.36SC + 8.81SB = R2 4.86IS + 11.50LC − 4.97MC + 3.25SC + 6.15SB = R3 2.17IS − 5.25LC − 1.69MC + 3.81SC + 4.04SB = R4 IS + LC + MC + SC + SB = 1 0.25R1 + 0.25R2 + 0.25R3 + 0.25R4 = AR AR ≥ 5 IS, LC, MC, SC, SB ≥ 0

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.08.03 - 8.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 8.3 Markowitz Portfolio Model © Cengage. Testing Powered by Cognero.

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CH 08 - Nonlinear Optimization Models KEYWORDS:

Bloom's: Create

41. Cutting Edge Yard Care is a residential and commercial lawn service company that has been in business in the Atlanta metropolitan area for almost one year. Cutting Edge would like to use its Atlanta service subscription data below to develop a model for forecasting service subscriptions in other metropolitan areas where it might expand. The first step is to estimate values for p (coefficient of innovation) and q (coefficient of imitation). Formulate the appropriate nonlinear program.

Month 1 2 3 4 5 6 7 8 9 10 ANSWER:

Subscribers St 0.53 2.04 3.37 4.85 5.14 3.90 2.26 1.43 0.98 0.42 Min

Cum. Subscribers Ct 0.53 2.57 5.94 10.79 15.93 19.83 22.09 23.52 24.50 24.92 E12 + E22 + E32 + E42 + E52 + E62 + E72 + E82 + E92 + E102 F1 = pm F2 = [p + q(0.53/m)] (m − 0.53) F3 = [p + q(2.57/m)] (m − 2.57) F4 = [p + q(5.94/m)] (m − 5.94) F5 = [p + q(10.79/m)] (m − 10.79) F6 = [p + q(15.93/m)] (m − 15.93) F7 = [p + q(19.83/m)] (m − 19.83) F8 = [p + q(22.09/m)] (m − 22.09) F9 = [p + q(23.52/m)] (m − 23.52) F10 = [p + q(24.50/m)] (m − 24.50) E1 = F1 − 0.53 E2 = F2 − 2.04 E3 = F3 − 3.37 E4 = F4 − 4.85 E5 = F 5 − 5.14 E6 = F6 − 3.90 E7 = F7 − 2.26 E8 = F8 − 1.43 E9 = F9 − 0.98 E10 = F10 − 0.42

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.08.05 - 8.5 © Cengage. Testing Powered by Cognero.

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CH 08 - Nonlinear Optimization Models NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 8.5 Forecasting Adoption of a New Product KEYWORDS: Bloom's: Create 42. MegaSports, Inc. produces two high-priced metal baseball bats, the Slugger and the Launcher, that are made from special aluminum and steel alloys. The cost to produce a Slugger bat is $100, and the cost to produce a Launcher bat is $120. We cannot assume that MegaSports will sell all the bats it can produce. As the selling price of each bat model— Slugger and Launcher—increases, the quantity demanded for each model goes down.

Assume that the demand, S, for Slugger bats is given by S = 640 − 4PS and the demand, L, for Launcher bats is given by L = 450 − 3PL where PS is the price of a Slugger bat and PL is the price of a Launcher bat. The profit contributions are PS S − 100S for Slugger bats and PL L − 120L for Launcher bats. Develop the total profit contribution function for this problem.

ANSWER: Solving S = 640 − 4PS for PS, we get: PS = 160 − 1/4S Substituting 160 − 1/4S for PS in PS S − 100S, we get: 60S − 1/4 S 2 Solving L = 450 − 3PL for PL, we get: PL = 150 − 1/3L Substituting 150 − 1/3L for PL in PL L − 120L, we get: 30L − 1/3L2 Total profit contribution = 60S − 1/4S2 + 30L − 1/3L2 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Create 43. Pacific-Gulf Oil Company is faced with the problem of refining three petroleum components into regular and premium gasoline in order to maximize profit. Components 1 and 2 are pooled in a single storage tank, and component 3 has its own storage tank. Regular and premium gasolines are made from blending the pooled components and component 3. Prices per gallon for the two products and three components, as well as product specifications, are listed below.

Regular gasoline Premium gasoline Component 1 Component 2 Component 3 Product Regular gasoline

Premium gasoline

Price per Gallon $2.80 3.10 2.40 2.50 2.75 Specifications At most 25% component 1 At least 40% component 2 At most 30% component 3 At least 30% component 1 At most 50% component 2 At least 25% component 3

The maximum number of gallons available for each of the three components is 4000, 8000, and 8000, respectively.

Formulate a nonlinear program to determine: (1) what percentages of components 1 and 2 should be used in the pooled mixture and (2) how to make regular and premium gasoline by blending the mixture of components 1 and 2 from the pooling tank with component 3. © Cengage. Testing Powered by Cognero.

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CH 08 - Nonlinear Optimization Models ANSWER:

y1 = gallons of component 1 in the pooling tank y2 = gallons of component 2 in the pooling tank xpr = gallons of pooled components 1 and 2 in regular gasoline xpp = gallons of pooled components 1 and 2 in premium gasoline x3r = gallons of component 3 in regular gasoline x3p = gallons of component 2 in premium gasoline Max

2.8(x pr + x3 r) + 3.1(x pp + x3 p) − 2.4y1 − 2.5y2 − 2.75(x3 r + x3 p) y1 + y2 = x pr + x pp [y1 /(y1 + y2)] x pr ≤ 0.25(x pr + x3 r) [y2 /(y1 + y2)] x pr ≥ 0.4(x pr + x3 r) x3 r ≤ 0.3(x pr + x3 r) [y1 /(y1 + y2)] x pp ≥ 0.3(x pp + x3 p) [y2 /(y1 + y2)] x pp ≤ 0.5(x pp + x3 p) x3 p ≥ 0.25(x pp + x3 p) y1 ≤ 4000 y2 ≤ 8000 x3 r + x3 p ≤ 8000 x pr + x3 r ≤ 8000 x pr , x pp , x3 r , x3 p , y1 , y2 ≥ 0

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.08.04 - 8.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 8.4 Blending: The Pooling Problem KEYWORDS: Bloom's: Create 44. Skooter's Skateboards produces two models of skateboards, the FX and the ZX. Skateboard revenue (in $l000s) for the firm is nonlinear and is stated as (number of FXs)(5 − 0.2 number of FXs) + (number of ZXs)(7 − 0.3 number of ZXs).

Skooter's has 80 labor-hours available per week in its paint shop. Each FX requires two labor-hours to paint, and each ZX requires three labor-hours. Formulate this nonlinear production planning problem to determine how many FX and ZX skateboards should be produced per week at Scooter's.

ANSWER: FX = number of FX skateboards to produce per week ZX = number of ZX skateboards to produce per week Max

5FX − 0.2FX 2 + 7ZX − 0.3ZX 2

s.t.

2FX + 3ZX ≤ 80 FX, ZX ≥ 0

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic © Cengage. Testing Powered by Cognero.

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CH 08 - Nonlinear Optimization Models TOPICS: KEYWORDS:

8.1 A Production Application—Par, Inc., Revisited Bloom's: Create

45. Native Customs sells two popular styles of hand-sewn footwear: a sandal and a moccasin. The cost to make a pair of sandals is $18, and the cost to make a pair of moccasins is $24. The demand for these two items is sensitive to the price, and historical data indicate that the monthly demands are given by S = 400 − 10P1 and M = 450 − 15P2, where S = demand for sandals (in pairs), M = demand for moccasins (in pairs), P1 = price for a pair of sandals, and P2 = price for a pair of moccasins. To remain competitive, Native Customs must limit the price (per pair) to no more than $60 and $75 for its sandals and moccasins, respectively. Formulate this nonlinear programming problem to find the optimal production quantities and prices for sandals and moccasins that maximize total monthly profit.

ANSWER: Max 22S − 1 /10 S 2 + 6M − 1/15 M 2 s.t.

S + 10P1 = 400 M − 15P2 = 450 P1 ≤ 60 P2 ≤ 75 S, M, P1, P2 ≥ 0

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.08.01 - 8.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 8.1 A Production Application—Par, Inc., Revisited KEYWORDS: Bloom's: Create

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CH 09 - Project Scheduling: PERT/CPM True / False 1. It is normal for activities on the critical path to be delayed without delaying the entire project. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Remember 2. PERT and CPM are applicable only when there is no dependence among activities. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Remember 3. A path through a project network must reach every node. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Understand 4. A critical activity can be part of a noncritical path. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times © Cengage. Testing Powered by Cognero.

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CH 09 - Project Scheduling: PERT/CPM KEYWORDS:

Bloom's: Understand

5. When activity times are uncertain, an activity's most likely time is the same as its expected time. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.09.02 - 9.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.2 Project Scheduling Considering Uncertain Activity Times KEYWORDS: Bloom's: Understand 6. The final activity's earliest finish time is the anticipated project duration. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Understand 7. The length of time an activity can be delayed without affecting the project completion time is the slack. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Remember 8. When activity times are uncertain, total project time is normally distributed by considering the overall mean being equal to the sum of the means of all of the critical activities. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.09.02 - 9.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking © Cengage. Testing Powered by Cognero.

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CH 09 - Project Scheduling: PERT/CPM TOPICS: KEYWORDS:

9.2 Project Scheduling Considering Uncertain Activity Times Bloom's: Understand

9. Crashing refers to an unanticipated delay in a critical path activity that causes the total time to exceed its limit. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.09.03 - 9.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.3 Considering Time–Cost Trade-Offs KEYWORDS: Bloom's: Remember 10. Constraints in the LP models for crashing decisions are required to compare the activity's earliest finish time with the earliest finish time of each predecessor. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.03 - 9.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.3 Considering Time–Cost Trade-Offs KEYWORDS: Bloom's: Understand 11. The project manager should monitor the progress of any activity with a large time variance even if the expected time does not identify the activity as a critical activity. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.09.02 - 9.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.2 Project Scheduling Considering Uncertain Activity Times KEYWORDS: Bloom's: Understand 12. The variance in the project completion time is the sum of the variances of all activities in the project. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.09.02 - 9.2 © Cengage. Testing Powered by Cognero.

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CH 09 - Project Scheduling: PERT/CPM NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.2 Project Scheduling Considering Uncertain Activity Times KEYWORDS: Bloom's: Understand 13. The latest finish time for an activity is the largest of the latest start times for all activities that immediately follow the activity. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Understand 14. The earliest start time for an activity is equal to the smallest of the earliest finish times for all its immediate predecessors. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Understand 15. Shortening activity times, which is usually accomplished by adding resources to the project, is called crashing. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.09.03 - 9.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.3 Considering Time–Cost Trade-Offs KEYWORDS: Bloom's: Remember 16. A key distinction of the critical path is that all activities on the critical path have zero slack time. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy © Cengage. Testing Powered by Cognero.

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CH 09 - Project Scheduling: PERT/CPM LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Understand 17. It is possible to have more than one critical path at a time. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Remember 18. Precedence relationships among activities are critical in CPM analysis but not in PERT. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.02 - 9.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.2 Project Scheduling Considering Uncertain Activity Times KEYWORDS: Bloom's: Understand 19. The normal distribution tends to be a better approximation of the distribution of total time for shorter projects where the critical path has relatively few activities. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.02 - 9.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.2 Project Scheduling Considering Uncertain Activity Times KEYWORDS: Bloom's: Understand 20. The earliest start time for an activity is equivalent to the latest of the earliest finish times for all its immediate predecessors. a. True b. False ANSWER:

True

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CH 09 - Project Scheduling: PERT/CPM POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Understand Multiple Choice 21. PERT and CPM a. are most valuable when a small number of activities must be scheduled. b. have different features and are not applied to the same situation. c. do not require a chronological relationship among activities. d. have been combined to develop a procedure that uses the best of each.

ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Remember 22. Which of the following is NOT a significant challenge of project scheduling? a. Deadlines exist. b. Activities are independent. c. Many employees could be required. d. Delays are costly.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Understand 23. Arcs in a project network indicate a. completion times. b. precedence relationships. c. activities. d. the critical path.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 © Cengage. Testing Powered by Cognero.

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CH 09 - Project Scheduling: PERT/CPM NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Remember 24. We identify the critical path of a project as a. any path that goes from the starting node to the completion node. b. the combined time of all paths of the project. c. the shortest path through the project. d. the longest path through the project.

ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Understand 25. The earliest start time rule a. compares the starting times of all activities for successors of an activity. b. compares the finish times for all immediate predecessors of an activity. c. determines when the project can begin. d. determines when the project must begin.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Understand 26. Activities following a node a. can begin as soon as any activity preceding the node has been completed. b. have an earliest start time equal to the largest of the earliest finish times for all activities entering the node. c. have a latest start time equal to the largest of the earliest finish times for all activities entering the node. d. None of these are correct.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Understand 27. Activities G, P, and R are the immediate predecessors for activity W. If the earliest finish times for the three are 12, © Cengage. Testing Powered by Cognero.

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CH 09 - Project Scheduling: PERT/CPM 15, and 10, then the earliest start time for W a. is 10. b. is 12. c. is 15. d. cannot be determined.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Apply 28. Activities K, M, and S immediately follow activity H, and their latest start times are 14, 18, and 11. The latest finish time for activity H a. is 11. b. is 14. c. is 18. d. cannot be determined.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Apply 29. When activity times are uncertain, a. we assume they are normally distributed. b. we start by calculating the optimistic, the most probable, and the pessimistic time estimates. c. we use the most probable time. d. we need to obtain at least four time estimates for each activity.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.02 - 9.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.2 Project Scheduling Considering Uncertain Activity Times KEYWORDS: Bloom's: Understand 30. To determine how to crash activity times, a. normal activity costs and costs under maximum crashing must be known. b. shortest times with crashing must be known. c. realize that new paths may become critical. © Cengage. Testing Powered by Cognero.

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CH 09 - Project Scheduling: PERT/CPM d. All of these are correct.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.03 - 9.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.3 Considering Time–Cost Trade-Offs KEYWORDS: Bloom's: Understand 31. Slack equals a. LF – EF. b. EF – LF. c. EF – LS. d. LF – ES. ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Remember 32. Activities with zero slack a. can be delayed. b. must be completed first. c. lie on a critical path. d. have no predecessors.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Understand 33. In deciding which activities to crash, one must crash a. all critical activities. b. largest-duration activities. c. lowest-cost activities. d. activities on the critical path(s) only.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.03 - 9.3 © Cengage. Testing Powered by Cognero.

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CH 09 - Project Scheduling: PERT/CPM NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.3 Considering Time–Cost Trade-Offs KEYWORDS: Bloom's: Understand 34. For an activity with more than one immediate predecessor activity, which of the following is used to compute its earliest finish (EF) time? a. the largest EF among the immediate predecessors b. the average EF among the immediate predecessors c. the largest LF among the immediate predecessors d. the difference in EF among the immediate predecessors ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Understand 35. Which of the following is always true about a critical activity? a. LS = EF b. LF = LS c. ES = LS d. EF = ES ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Understand 36. For an activity with more than one immediate successor activity, its latest finish time is equal to the a. largest latest finish time among its immediate successors. b. smallest latest finish time among its immediate successors. c. largest latest start time among its immediate successors. d. smallest latest start time among its immediate successors.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Understand

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CH 09 - Project Scheduling: PERT/CPM 37. Which of the following is a general rule for crashing activities? a. Crash only noncritical activities. b. Crash activities with zero slack. c. Crash activities with the greatest number of predecessors. d. Crash the path with the fewest activities.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.03 - 9.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.3 Considering Time–Cost Trade-Offs KEYWORDS: Bloom's: Understand 38. To calculate an activity’s latest finish time, you should consider its a. predecessors’ latest finish times. b. predecessors’ latest start times. c. successors’ earliest start times. d. successors’ latest start times.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Understand 39. A critical activity is a. an activity that consumes no time but shows precedence between events. b. a milestone accomplishment within the project. c. an activity with zero slack. d. the beginning of an event.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Remember 40. The main difference between CPM and PERT is a. CPM and PERT use different activity time estimates. b. PERT analysis is less expensive to conduct. c. PERT lends itself to computerization while CPM networks must be constructed manually. d. CPM integrates time and cost performance while PERT is based solely on time performance. © Cengage. Testing Powered by Cognero.

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CH 09 - Project Scheduling: PERT/CPM ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.02 - 9.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.2 Project Scheduling Considering Uncertain Activity Times KEYWORDS: Bloom's: Understand 41. In PERT, the activity duration time is equal to the a. pessimistic time. b. optimistic time. c. most likely time. d. mean duration.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.02 - 9.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 9.2 Project Scheduling Considering Uncertain Activity Times KEYWORDS: Bloom's: Remember Subjective Short Answer 42. From this schedule of activities, draw the PERT/CPM network.

Activity A B C D E F G

Immediate Predecessor — A B B A C, D E, F

ANSWER:

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 © Cengage. Testing Powered by Cognero.

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CH 09 - Project Scheduling: PERT/CPM NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Create 43. From this PERT/CPM network, create a list of activities and their predecessors.

ANSWER: Activity A B C D E F G H

Immediate Predecessor — — A A, B C, D D E F, G

POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Create 44. A cookie recipe gives the following numbered steps.

1. Preheat oven.

2. Grease cookie sheets.

3. Cream shortening and sugar.

4. Add eggs and flavoring.

5. Measure and sift dry ingredients.

6. Add dry ingredients to mixture.

7. Drop by spoonfuls onto sheets and bake for 10 minutes.

Although the steps are numbered, they do not always reflect immediate precedence relationships. Develop a table that lists the immediate predecessors for each activity.

ANSWER: Immediate Activity Predecessor 1 — 2 — 3 — 4 3 5 — 6 4, 5 © Cengage. Testing Powered by Cognero.

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CH 09 - Project Scheduling: PERT/CPM 7

1, 2, 6

POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Analyze 45. A senior MIS design class project team has developed the following schedule of activities for their project, using their best estimate of completion times. Both written and oral reports are required. Draw the project network. Can they complete the project in the 38 class days remaining until the end of the semester?

A. B. C. D. E. F. G.

Activity Find client.

Write prospectus.

Obtain approval from client and professor.

Complete programming.

Do industry background research.

Write final paper.

Write oral report.

Time 4 2 3 12 10 6 5

Immediate Predecessor — A B C — D, E D, E

ANSWER:

Yes, they can complete it in time. The critical path is A-B-C-D-F. POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Create 46. A project network is shown below. Use a forward and a backward pass to determine the critical path, and then fill out the table below. Activity times are in weeks.

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CH 09 - Project Scheduling: PERT/CPM

Activity A B C D E F G H I

Precedence Activities

Activity Time (weeks)

ES

LS

EF

LF

Slack

Critical Path?

Now assume that the times listed are only the expected times instead of being fixed times. Is the probability of being finished in fewer than 25 weeks more or less than 50%?

ANSWER: Activity A B C D E F G H I

Precedence Activities — — — A, B B, C D A, F E, F G, H

Activity Time (weeks) 5 4 8 4 12 10 3 2 5

ES

LS

EF

LF

Slack

0 0 0 5 8 9 19 20 22

0 1 0 5 8 9 19 20 22

5 4 8 9 20 19 22 22 27

5 5 8 9 20 19 22 22 27

0 1 0 0 0 0 0 0 0

Critical Path?

Yes Yes Yes Yes Yes Yes Yes Yes

The probability is less than 50% because 25 weeks is less than the mean time of 27 weeks.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.09.02 - 9.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic © Cengage. Testing Powered by Cognero.

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CH 09 - Project Scheduling: PERT/CPM TOPICS: KEYWORDS:

9.2 Project Scheduling Considering Uncertain Activity Times Bloom's: Evaluate

47. A project network is shown below. Use a forward and a backward pass to determine the critical path, and then fill out the table below. Activity times are in weeks.

Activity A B C D E F G H I

Precedence Activities

Activity Time (weeks)

ES

LS

EF

LF

Slack

Critical Path?

Now assume that the times listed are only the expected times instead of being fixed times. Is the probability of being finished in more than 28 weeks more or less than 50%?

ANSWER: Precedence Activity Critical Activity ES LS EF LF Slack Activities Time (weeks) Path?

A — 5 0 13 5 18 13 B — 4 0 14 4 18 14 C — 8 0 0 8 8 0 Yes D A, B 4 5 18 9 22 13 E C 12 8 8 20 20 0 Yes F C 10 8 10 18 20 2 G D, E 3 20 22 23 25 2 H E 2 20 23 22 25 3 I E, F 5 20 20 25 25 0 Yes The probability is less than 50% because 28 weeks is more than the mean time of 25 weeks.

POINTS:

1

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CH 09 - Project Scheduling: PERT/CPM DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.09.02 - 9.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 9.2 Project Scheduling Considering Uncertain Activity Times KEYWORDS: Bloom's: Evaluate 48. Use the following network of related activities with their duration times (weeks) to complete a row for each activity under the column headings below.

Activity A B C D E F G H

Immediate Predecessors

Activity Time (weeks)

ES

LS

EF

LF

Slack

ES 0 0 2 8 2 2 11 8

LS 0 1 3 8 4 7 11 13

Critical Path?

ANSWER: Activity A B C D E F G H

Precedence Activities — — B A, C B B D, E F

Activity Time (weeks) 8 2 5 3 7 5 12 10

EF 8 2 7 11 9 8 23 18

LF 8 3 8 11 11 13 23 23

Slack 0 1 1 0 2 5 0 5

Critical Path?

Yes

Yes

Yes

CRITICAL PATH: A-D-G PROJECT COMPLETION TIME = 23 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 © Cengage. Testing Powered by Cognero.

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CH 09 - Project Scheduling: PERT/CPM NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Create 49. Use the following network of related activities with their duration times (weeks) to complete a row for each activity under the column headings below.

Activity A B C D E F G H I J

Immediate Predecessors

Activity Time (weeks)

ANSWER: Activity A B C D E F G H I J

ES

Precedence Activities — A A A B B C D E F, G, H

EF

LF

Slack

Activity Time (weeks) 2 4 3 7 6 5 8 5 4 6

ES 0 2 2 2 6 6 5 9 12 14

LS 0 5 3 2 10 9 6 9 16 14

LS

EF 2 6 5 9 12 11 13 14 16 20

Critical Path?

LF 2 9 6 9 16 14 14 14 20 20

Slack 0 3 1 0 4 3 1 0 4 0

Critical Path?

Yes

Yes

Yes Yes

CRITICAL PATH: A-D-H-J PROJECT COMPLETION TIME = 20 © Cengage. Testing Powered by Cognero.

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CH 09 - Project Scheduling: PERT/CPM POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Create 50. Use the following network with activities and times estimated in days to answer questions a through c.

Activity

Optimistic

A B C D E F G H I J K

2 1 6 5 3 8 4 3 5 12 1

a. b. c.

Most Probable 5 3 7 12 4 9 6 6 7 13 3

Pessimistic 6 7 10 14 5 12 8 8 12 14 4

What are the critical path activities?

What is the expected time to complete the project?

What is the probability the project will take more than 28 days to complete?

ANSWER:

a. and b.

Activity A B C D

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Expected Time 4.67 3.33 7.33 11.17

Variance 0.44 1.00 0.44 2.25 Page 19

CH 09 - Project Scheduling: PERT/CPM E F G H I J K Activity A B C D E F G H I J K

c.

4.00 9.33 6.00 5.83 7.50 13.00 2.83 Precedence Activities

— — — A A A, B C C E, F, G H D, J

Time (days) 4.67 3.33 7.33 11.17 4.00 9.33 6.00 5.83 7.50 13.00 2.83

0.11 0.44 0.44 0.69 1.36 0.11 0.25 ES

LS

EF

LF

Slack

0.00 4.67 4.67 9.33 4.67 0.00 6.00 3.33 9.33 6.00 0.00 0.00 7.33 7.33 0.00 4.67 14.67 8.67 18.67 10.33 4.67 14.67 8.67 18.67 10.00 4.67 9.33 14.00 18.67 4.67 7.33 12.67 13.33 18.67 5.33 7.33 7.33 13.17 13.17 0.00 14.00 18.67 21.50 26.17 4.67 13.17 13.17 26.17 26.17 0.00 26.17 26.17 29.00 29.00 0.00

Critical Path?

Yes

Yes Yes Yes

CRITICAL PATH: C-H-J-K EXPECTED PROJECT COMPLETION TIME = 29 VARIANCE OF PROJECT COMPLETION TIME = 1.5 The probability that the project will take more than 28 days is P(z > (28 − 29)/1.22) = P(z > −0.82) = 0.7939.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.09.02 - 9.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 9.2 Project Scheduling Considering Uncertain Activity Times KEYWORDS: Bloom's: Evaluate 51. The critical path for this network is A - E - F, and the project completion time is 22 weeks.

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CH 09 - Project Scheduling: PERT/CPM

Activity A B C D E F G

Normal Time 12 14 8 5 4 6 10

Crash Time 8 10 8 3 3 5 8

Normal Cost 8,000 5,000 10,000 6,000 5,000 9,000 5,000

Crash Cost 12,000 7,500 10,000 8,000 7,000 12,000 8,000

If a deadline of 17 weeks is imposed, give the linear programming model for the crashing decision.

ANSWER: Let Ei = earliest finish time for activity i Ci = amount to crash activity i Min s.t.

1000CA + 625CB + 1000CD + 2000CE + 3000CF + 1500CG EA ≥ 0 + 12 − CA EB ≥ 0 + 14 − CB EC ≥ 0 + 8 − CC ED ≥ EA + 5 − CD EE ≥ EA + 4 − CE EF ≥ EE + 6 − CF EF ≥ EC + 6 − CF EG ≥ EC + 10 − CG EA ≤ 4 EB ≤ 4 ED ≤ 2 EE ≤ 1 EF ≤ 1 EG ≤ 2

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.09.03 - 9.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic © Cengage. Testing Powered by Cognero.

Page 21

CH 09 - Project Scheduling: PERT/CPM TOPICS: KEYWORDS:

9.3 Considering Time–Cost Trade-Offs Bloom's: Create

52. For the project represented below, determine the earliest and latest start and finish times for each activity as well as the expected overall completion time.

Activity A B C D E F G H

Duration 4 3 4 2 5 2 5 6

ES

EF

LS

LF

Slack

ANSWER:

[Activity, ES, EF, LS, LF, Slack] [A, 0, 4, 0, 4, 0]; [B, 4, 7, 4, 7, 0]; [C, 4, 8, 6, 10, 2]; [D, 7, 9, 8, 10, 1]; [E, 7, 12, 7, 12, 0] [F, 9, 11, 10, 12, 1]; [G, 12, 17, 12, 17, 0]; [H, 9, 15, 11, 17, 2] POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Evaluate 53. Consider the following PERT/CPM network with estimated times in weeks. The project is scheduled to begin on May 1.

The three-time estimate approach was used to calculate the expected times and the following table gives the variance for © Cengage. Testing Powered by Cognero.

Page 22

CH 09 - Project Scheduling: PERT/CPM each activity: Activity A B C D a. b.

Variance 1.1 0.5 1.2 0.8

Activity E F G H

Variance 0.3 0.6 0.6 1.0

Give the expected project completion date and the critical path.

By what date are you 99% sure the project will be completed?

ANSWER:

a. 16 weeks = August 21 b. About September 22 (20.66 weeks from May 1) POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.09.02 - 9.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 9.2 Project Scheduling Considering Uncertain Activity Times KEYWORDS: Bloom's: Evaluate 54. A project consists of five activities. Naturally, the paint mixing precedes the painting activities. Also, both ceiling painting and floor sanding must be done prior to floor buffing.

Activity Floor sanding Floor buffing Paint mixing Wall painting Ceiling painting a. b. c.

Optimistic Time (hr.)

3 1 0.5 1 1

Most Probable Time (hr.)

4 2 1 2 5.5

Pessimistic Time (hr.)

5 3 1.5 9 7

Construct the PERT/CPM network for this problem.

What is the expected completion time of this project?

What is the probability that the project can be completed within nine hours?

ANSWER:

a.

b. c.

8 hours 0.8264

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.09.02 - 9.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 9.2 Project Scheduling Considering Uncertain Activity Times © Cengage. Testing Powered by Cognero.

Page 23

CH 09 - Project Scheduling: PERT/CPM KEYWORDS:

Bloom's: Evaluate

55. Consider a project that has been modeled as follows: Activity A B C D E F G H I J K L a. b. c. d. e.

Immediate Predecessors — — A A A B, C B, C E, F E, F E, F D, H G, J

Duration (hr.)

7 10 4 30 7 12 15 11 25 6 21 25

Draw the PERT/CPM network for this project and determine the project's expected completion time and its critical path.

Can activities E and G be performed simultaneously without delaying the minimum project completion time?

Can one person perform A, G, and I without delaying the project?

By how much can activities G and L be delayed without delaying the entire project?

How much would the project be delayed if activity G was delayed by seven hours and activity L was delayed by four hours? Explain.

ANSWER:

a.

Expected completion time = 58; Critical path = A - D - K b. Yes c. Yes d. Slack on G = 7 hours; Slack on L = 4 hours e. 4 hours; the slack times are not independent as G and L are on the same path.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.09.01 - 9.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 9.1 Project Scheduling Based on Expected Activity Times KEYWORDS: Bloom's: Evaluate © Cengage. Testing Powered by Cognero.

Page 24

CH 09 - Project Scheduling: PERT/CPM 56. Joe King has ambitions to be mayor of Williston, North Dakota. Joe has determined the breakdown of the steps to the nomination and has estimated normal and crash costs and times for the campaign as follows (times are in weeks).

A. B. C. D. E. F. G. H.

Activity Solicit volunteers Initial "free" exposure Raise money Organize schedule Hire advertising firm Arrange TV interview Advertising campaign Personal campaigning

Time 6 3 9 4 2 3 5 7

Normal Cost $5,000 4,000 4,000 1,000 1,500 4,000 7,000 8,000

Time 4 3 6 2 1 1 4 5

Crash Cost $ 9,000 4,000 10,000 2,000 2,000 8,000 12,000 20,000

Immediate Predecessors — — A A B B C, E D, F

Joe is not a wealthy man and would like to organize a 16-week campaign at minimum cost. Write and solve a linear program to accomplish this task.

ANSWER:

Let Xi = earliest finish time for activity i Yi = amount of time activity i is crashed Min

2000YA + 2000YC + 500YD + 500YE + 2000YF + 5000YG + 6000Y H

s.t.

XA ≥ 0 + (6 − YA) XB ≥ 0 + 3 XC ≥ XA + (9 − YC) XD ≥ XA + (4 − YD) XE ≥ XB + (2 − YE) XF ≥ XB + (3 − YF) XG ≤ 16 XG ≥ XC + (5 − YG) XG ≥ XE + (5 − YG) XH ≥ XD + (7 − YH) XH ≥ XF + (7 − YH) XH ≤ 16 YA ≤ 2 YC ≤ 3 YD ≤ 2 YE ≤ 1 YF≤2

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CH 09 - Project Scheduling: PERT/CPM YG ≤ 1 YH ≤ 2 Xi, Yj ≥ 0 for all i Solution:

XA = 4, XB = 6, XC = 11, XD = 9, XE = 11, XF = 9, XG = 16, XH = 16, YA = 2, YC = 2, YD = 0, YE = 0, YF = 0, YG = 0, YH = 0, Total crash cost = $8,000

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.09.03 - 9.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 9.3 Considering Time–Cost Trade-Offs KEYWORDS: Bloom's: Create 57. Marcy Fetter, a staff analyst at the Los Angeles plant of Computer Products Corporation, is assigned to the team that is developing the process design for producing an RFID sensor. The corporate planning group in San Jose, California, has contacted her and asked how confident the design group is about completing the project in 60 days. She has developed the following estimated time durations in days for the project:

Activity A B C D E F G H I J

Immediate Predecessor Activities — A A A B C, E B B, D B, D F, G, H

Optimistic Time (to)

Most Likely Time (tm)

Pessimistic Time (tp)

10 6 10 9 5 10 12 18 10 8

12 10 15 9 6 12 14 21 15 10

15 14 20 18 8 13 16 24 20 14

a. Compute the expected time and variance for each activity. b. Determine the critical path and expected duration of the project. c. What is the probability that the project will take longer than 58 days to complete? d. Which path in the project network offers the greatest risk of overrunning a new deadline of 56 days?

ANSWER:

a.

Activity

Optimistic Time (to)

Most Likely Time (tm)

Pessimistic Time (tp)

Activity Expected Time (te)

Activity Variance (Vt)

A

10

12

15

12.17

0.694

B

6

10

14

10.00

1.778

C

10

15

20

15.00

2.778

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CH 09 - Project Scheduling: PERT/CPM D

9

10

14

10.50

0.694

E

5

6

8

6.17

0.250

F

10

12

13

11.83

0.250

G

12

14

16

14.00

0.444

H

18

21

24

21.00

1.000

I

10

15

20

15.00

2.778

J

8

10

14

10.33

1.000

b.

Path

Length of Path (days)

A-C-F-J A-B-E-F-J A-B-G-J A-B-H-J A-B-I

12.17 + 15 + 11.83 + 10.33 12.17+ 10 + 6.17 + 11.83 + 10.33 12.17 + 10 + 14 + 10.33 12.17 + 10 + 21 + 10.33 12.17 + 10 + 15

= 49.33 = 50.50 = 46.50 = 53.50 = 37.17

A-D-H-J

12.17 + 10.5 + 21 + 10.33

= 54.00

A-D-I

12.17 + 10.5 + 15

= 37.67

The critical path is A-D-H-J; its duration is 54 days. c. For path A-D-H-J: SVt = 0.694 + 0.694 + 1.000 + 1.000 = 3.388 days z = (x – m)/st = (58 – 54)/1.84 = 2.17 P(D > 58) = 1 – 0.9850 = 0.0150 or about 1.5% (There is a 98.5% chance of the project being completed within 58 days.) d. The longest paths (the ones offering the greatest risk of exceeding 56 days) are A-B-E-FJ, A-B-H-J, and A-D-H-J. The variances for these paths are: Path A-C-F-J A-B-E-F-J A-B-G-J A-B-H-J A-B-I A-D-H-J A-D-I

Length of Path (days) 49.33 50.50 46.50 53.50 37.17 54.00 37.67

Variance of Path (days)

0.694 + 1.778 + 1.0 + 1.0 = 4.472 0.694 + 0.694 + 1.0 + 1.0 = 3.388

Path A-B-H-J offers a risk roughly as great as path A-D-H-J. Path A-B-H-J: z = (56 – 53.5) = 2.5/2.115 = 1.18 P(D > 56) = 0.1190 or about a 12% chance Path A-D-H-J: z = (56 – 54.0) = 2.0/1.841 = 1.09 P(D > 56) = 0.1380 or about a 14% chance Both paths need to be managed closely in order to complete the project in no more than 56 days. © Cengage. Testing Powered by Cognero.

Page 27

CH 09 - Project Scheduling: PERT/CPM POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.09.02 - 9.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 9.2 Project Scheduling Considering Uncertain Activity Times KEYWORDS: Bloom's: Evaluate 58. A project has the following activities, precedence relationships, and time estimates in weeks:

Activity

Immediate Predecessor Activities

Optimistic Time (to)

Most Likely Time (tm)

Pessimistic Time (tp)

A

—

15

20

25

B

—

8

10

12

C

A

25

30

40

D

B

15

15

15

E

B

22

25

27

F

E

15

20

22

G

D

20

20

22

a. Compute the expected time and variance for each activity. b. Determine the critical path and the expected duration of the project. c. What is the probability that the project will take longer than 56 weeks to complete?

ANSWER: a.

Activity

Optimistic Time (to)

Most Likely Time (tm)

Pessimistic Time (tp)

Activity Expected Time (te)

Activity Variance (Vt)

A

15

20

25

20

2.778

B

8

10

12

10

0.444

C

25

30

40

30.83

6.250

D

15

15

15

15

0

E

22

25

27

24.83

0.694

F

15

20

22

19.50

1.361

G

20

20

22

20.33

0.111

b.

Path

Length of Path (weeks)

B-D-G B-E-F A-C

10 + 15 + 20.33 10 + 24.83 + 19.5 20 + 30.83

= 45.33 = 54.33* = 50.83

c. SVt = 2.499 weeks; z = 1.06; P(D > 56) = 0.1446 © Cengage. Testing Powered by Cognero.

Page 28

CH 09 - Project Scheduling: PERT/CPM POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.09.02 - 9.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 9.2 Project Scheduling Considering Uncertain Activity Times KEYWORDS: Bloom's: Evaluate 59. Three paths of a PERT network have these mean durations and variances in weeks:

Path

Mean Duration

Variance

1

45

2.75

2

44

5.50

3

46

1.20

Which path offers the greatest risk of overrunning a contract deadline of 48 weeks?

ANSWER: Path 1: st = 1.658; z = 1.809; P(D > 48) = 3.5% Path 2: st = 2.345; z = 1.706; P(D > 48) = 4.4% Path 3: st = 1.095; z = 1.826; P(D > 48) = 3.4% Path 2 offers the greatest risk (but only by a small margin).

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.09.02 - 9.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 9.2 Project Scheduling Considering Uncertain Activity Times KEYWORDS: Bloom's: Evaluate 60. A project has the following activities, durations, costs, and precedence relationships:

Activity

Present Duration (Weeks)

Accelerated Duration (Weeks)

Immediate Predecessor Activities

Present Cost

Accelerated Cost

A

10

9

—

$11,000

$15,000

B

15

13

—

20,000

25,000

C

10

6

A

9,000

20,000

D

20

18

A

25,000

30,000

E

15

10

C

20,000

35,000

F

17

15

B

20,000

30,000

G

12

10

B

15,000

25,000

H

9

8

D, F

12,000

18,000

I

7

6

G, H

10,000

15,000

Develop a time-cost trade-off analysis. Detail the steps that you would use to accelerate or crash the project to its © Cengage. Testing Powered by Cognero.

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CH 09 - Project Scheduling: PERT/CPM minimum duration at the lowest cost. Determine each step's cost and the duration of the project.

ANSWER:

Activity

Present Duration (Weeks)

Accelerated Duration (Weeks)

Present Cost

Accelerated Cost

Crashing Cost per Week

A

10

9

$11,000

$15,000

$4,000

B

15

13

20,000

25,000

2,500

C

10

6

9,000

20,000

2,750

D

20

18

25,000

30,000

2,500

E

15

10

20,000

35,000

3,000

F

17

15

20,000

30,000

5,000

G

12

10

15,000

25,000

5,000

H

9

8

12,000

18,000

6,000

I

7

6

10,000

15,000

5,000

Length of Path (weeks) 10 + 10 + 15 A-C-E 35 = A-D-H- 10 + 20 + 9 46 I +7= 15 + 17 + 9 B-F-H-I 48 +7= 15 + 12 + 7 B-G-I 34 = Activity Crashed Additional Cost for Crashing Path

35

35

35

35

35

35

46

46

45

44

43

42

47

46

45

44

43

42

33

32

31

31

31

31

B 2,500

B 2,500

I 5,000

H 6,000

D, F 7,500

D, F 7,500

Project is reduced from 48 to 42 weeks at an additional cost of $31,000.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.09.03 - 9.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 9.3 Considering Time–Cost Trade-Offs KEYWORDS: Bloom's: Evaluate 61. National Oil Company (NATOCO) must plan the shutdown of its Houston refinery for routine preventive maintenance. Each hour of downtime is lost production time and is very costly, so NATOCO wants the maintenance project completed in 22 hours. The PERT network below shows the precedence relationships of the activities involved in the project. The table gives the activity times and costs under normal operations and maximum crashing.

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CH 09 - Project Scheduling: PERT/CPM

NATOCO wants to know the minimum cost of completing the maintenance project within the 22-hour period. Formulate and solve a linear program that will yield this information.

Activity A B C D E F G H I J K L M ANSWER:

Normal Time (hr.)

2 4 1 4 6 10 8 2 5 12 7 11 4

Normal Cost $2,000 3,000 1,500 5,300 5,400 6,000 4,800 2,800 4,500 6,000 7,000 8,800 1,000

Crash Time (hr.)

1.5 3 1 2.5 5 8 5 1 4 6 4 9 1

Crash Cost $3,000 3,500 1,500 8,000 7,000 9,000 9,900 2,900 5,000 9,600 9,700 9,200 7,000

m = Normal time – Time under maximum crashing k = (Cost under maximum crashing – Normal cost)/m Activity A B C D E F G H I J K

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m 0.5 1 0 1.5 1 2 3 1 1 6 3

k 2000 500 0 1800 1600 1500 1700 100 500 600 900 Page 31

CH 09 - Project Scheduling: PERT/CPM L M

2 3

200 2000

LP Formulation Let xA= earliest finish time for activity A yA = amount of time activity A is crashed Min 2000yA + 500yB + 1800yD + 1600yE + 1500yF + 1700yG + 100yH + 500yI + 600yJ + 900yK + 200yL + 2000yM Subject to: (1) The project must be completed within 22 hours: xK < 22, xL < 22, xM < 22 (2) The amount an activity is crashed cannot exceed its maximum crashing: yA < 0.5 yB < 1 yD < 1.5 yE < 1 yF < 2 yG < 3 yH < 1 yI < 1 yJ < 6 yK < 3 yL < 2 yM < 3 (3) For each activity: Earliest finish time > Earliest finish time of preceding activity + (Normal activity time) − Amount of time the activity is crashed) xA > 0 + (2 – yA) or xB > 0 + (4 – yB) or xC > 0 + (1 – 0) or

xA + yA > 2 xB + yB > 4 xC > 1

xD > xA + (4 – yD) or xD + yD – xA > 4 xE > xB + (6 – yE) or xE + yE – xB > 6 xG > xB + (8 – yG) or xG + yG – xB > 8 xH > xB + (2 – yH) or xH + yH – xB > 2 xI > xC + (5 – yI) or xI + yI – xC > 5 xJ > xC + (12 – yJ) or xJ + yJ – xC > 12 xF > xD + (10 – yF) or xF > xE + (10 – yF) or xL > xG + (11 – yL) or

xF + yF – xD > 10 xF + yF – xE > 10 xL + yL – xG > 11

xK > xF + (7 – yK) or xK > xI + (7 – yK) or xM > xH + (4 – yM) or

xK + yK – xF > 7 xK + yK – xI > 7 xM + y M – xH > 4

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CH 09 - Project Scheduling: PERT/CPM xM > xJ + (4 – yM) or

xM + y M – xJ > 4

(4) Nonnegativity of the variables: xi > 0 for i = A, B, C, … , M yj > 0 for j = A, B, D, … , M Solution OBJECTIVE FUNCTION VALUE = VARIABLE XA XB XC XD XE XF XG XH XI XJ XK XL XM YA YB YD YE YF YG YH YI YJ YK YL YM

VALUE 5.000 3.000 1.000 9.000 9.000 18.000 11.000 13.000 18.000 13.000 22.000 22.000 17.000 0.000 1.000 0.000 0.000 1.000 0.000 0.000 0.000 0.000 3.000 0.000 0.000

4700.000

REDUCED COSTS 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 2000.000 0.000 1800.000 100.000 0.000 1500.000 100.000 500.000 600.000 0.000 0.000 2000.000

Summary: Crash activity B one hour, activity F one hour, and activity K three hours. The preventive maintenance project will be completed in 22 hours at a crashing cost of $4700.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.09.03 - 9.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 9.3 Considering Time–Cost Trade-Offs KEYWORDS: Bloom's: Evaluate

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Page 33

CH 10 - Inventory Models True / False 1. To be considered inventory, goods must be finished and waiting for delivery. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Remember 2. When demand is independent, it is not related to demand for other components or items produced by the firm. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Remember 3. Constant demand is a key assumption of the EOQ model. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Understand 4. In the EOQ model, if the average inventory during each cycle is 1/2Q, the average inventory over any number of cycles is Q/2. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.1 Economic Order Quantity (EOQ) Model © Cengage. Testing Powered by Cognero.

Page 1

CH 10 - Inventory Models KEYWORDS:

Bloom's: Understand

5. The single-period inventory model is most applicable to items that are perishable or have seasonal demand. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.05 - 10.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.5 Single-Period Inventory Model with Probabilistic Demand KEYWORDS: Bloom's: Remember 6. The time between placing orders is the lead time. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Remember 7. If the optimal production lot size decreases, average inventory increases. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.02 - 10.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.2 Economic Production Lot Size Model KEYWORDS: Bloom's: Remember 8. If an item's per-unit backorder cost is greater than its per-unit holding cost, no intentional shortage should be planned. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.10.03 - 10.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.3 Inventory Model with Planned Shortages © Cengage. Testing Powered by Cognero.

Page 2

CH 10 - Inventory Models KEYWORDS:

Bloom's: Understand

9. It is logical to order an amount from the highest discount category when quantity discounts are available. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.04 - 10.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.4 Quantity Discounts for The EOQ Model KEYWORDS: Bloom's: Remember 10. When there is probabilistic demand in a multiperiod model, the inventory level will not decrease smoothly and can fall below 0. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.10.06 - 10.6 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.6 Order-Quantity, Reorder Point Model with Probabilistic Demand KEYWORDS: Bloom's: Understand 11. In the periodic review model, the order quantity at each review period must be sufficient to cover demand for the review period plus the demand for the following lead time. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.07 - 10.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.7 Periodic Review Model with Probabilistic Demand KEYWORDS: Bloom's: Understand 12. Periodic review systems require smaller safety stock levels than corresponding continuous review systems. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.07 - 10.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking © Cengage. Testing Powered by Cognero.

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CH 10 - Inventory Models TOPICS: KEYWORDS:

10.7 Periodic Review Model with Probabilistic Demand Bloom's: Understand

13. The cost of overestimating demand is usually harder to determine than the cost of underestimating demand. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.05 - 10.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.5 Single-Period Inventory Model with Probabilistic Demand KEYWORDS: Bloom's: Understand 14. The terms "inventory on hand" and "inventory position" have the same meaning. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Remember 15. The EOQ model is insensitive to small variations or errors in the cost estimates. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Blooms: Understand 16. At the optimal order quantity for the quantity discount model, the sum of the annual holding and ordering costs is minimized. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.10.04 - 10.4 © Cengage. Testing Powered by Cognero.

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CH 10 - Inventory Models NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.4 Quantity Discounts for The EOQ Model KEYWORDS: Bloom's: Understand 17. As lead time for an item increases, the cycle time increases. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Understand 18. An assumption in the economic production lot size model is that there is storage capacity to hold the entire production lot. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.02 - 10.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.2 Economic Production Lot Size Model KEYWORDS: Bloom's: Understand 19. If lead time is longer than the review period, the order quantity at any review point is the amount needed for the inventory on hand plus all outstanding orders to reach the replenishment level. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.07 - 10.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.7 Periodic Review Model with Probabilistic Demand KEYWORDS: Bloom's: Understand 20. For the periodic review inventory model presented in the textbook, it is assumed that a special replenishment order will be placed in the event of a stockout between review points. a. True b. False ANSWER: False POINTS: 1 © Cengage. Testing Powered by Cognero.

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CH 10 - Inventory Models DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.10.07 - 10.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.7 Periodic Review Model with Probabilistic Demand KEYWORDS: Bloom's: Understand 21. For the production lot size, Q, the average inventory is one-half the maximum inventory, or 1/2Q. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.02 - 10.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.2 Economic Production Lot Size Model KEYWORDS: Bloom's: Remember 22. In a periodic review model, inventory is checked and reordering is done only at specified points in time, such as on a weekly, biweekly, monthly, or some other periodic basis. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.10.07 - 10.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.7 Periodic Review Model with Probabilistic Demand KEYWORDS: Bloom's: Understand Multiple Choice 23. Inventory a. is held against uncertain usage so that a supply of items is available if needed. b. constitutes a small part of the cost of doing business. c. is not something that can be managed effectively. d. All of these are correct.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Remember 24. Inventory models for which the rate of demand is constant or nearly constant are called © Cengage. Testing Powered by Cognero.

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CH 10 - Inventory Models a. fixed models. b. deterministic models. c. absolute models. d. requirement models.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Remember 25. The EOQ model a. determines only how frequently to order. b. considers total cost. c. minimizes both ordering and holding costs. d. All of these are correct.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Remember 26. Which of the following would NOT be considered part of a holding cost? a. cost of capital b. shipping cost c. insurance cost d. warehouse overhead ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Understand 27. For inventory systems with constant demand and a fixed lead time, a. the reorder point = lead-time demand. b. the reorder point > lead-time demand. c. the reorder point < lead-time demand. d. the reorder point is unrelated to lead-time demand.

ANSWER: a © Cengage. Testing Powered by Cognero.

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CH 10 - Inventory Models POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Understand 28. Safety stock a. can be determined by the EOQ formula. b. depends on the inventory position. c. depends on the variability of demand during lead time. d. is not needed if Q* is the actual order quantity.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.10.06 - 10.6 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.6 Order-Quantity, Reorder Point Model with Probabilistic Demand KEYWORDS: Bloom's: Understand 29. The economic production lot size model is appropriate when a. demand exceeds the production rate. b. there is a constant supply rate for every period, without pause. c. ordering cost is equivalent to the production setup cost. d. All of these are correct.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.10.02 - 10.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.2 Economic Production Lot Size Model KEYWORDS: Bloom's: Understand 30. If S indicates the number of backorders that have accumulated by the time a new shipment of size Q is received, and we ship existing backorders to our customers and place the remaining units in inventory, then __________ is the maximum inventory we subsequently have on hand. a. Q b. Q − S c. S d. (Q − S)/2 ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.10.03 - 10.3 © Cengage. Testing Powered by Cognero.

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CH 10 - Inventory Models NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.3 Inventory Model with Planned Shortages KEYWORDS: Bloom's: Understand 31. Annual purchase cost is included in the total cost in a. the EOQ model. b. the economic production lot size model. c. the quantity discount model. d. all inventory models.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.10.04 - 10.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.4 Quantity Discounts for the EOQ Model KEYWORDS: Bloom's: Understand 32. In the single-period inventory model with probabilistic demand, a. surplus items are not allowed to be carried in future inventory. b. co = cu. c. probabilities are used to calculate expected losses. d. All of these are correct.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.10.05 - 10.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.5 Single-Period Inventory Model with Probabilistic Demand KEYWORDS: Bloom's: Understand 33. For the inventory model with planned shortages, the optimal order quantity results in a. annual holding cost = annual ordering cost. b. annual holding cost = annual backordering cost. c. annual ordering cost = annual holding cost + annual backordering cost. d. annual ordering cost = annual holding cost − annual backordering cost.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.10.03 - 10.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.3 Inventory Model with Planned Shortages KEYWORDS: Bloom's: Understand 34. Periodic review inventory systems © Cengage. Testing Powered by Cognero.

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CH 10 - Inventory Models a. are less subject to stockouts than corresponding continuous review systems. b. require larger safety stock levels than corresponding continuous review systems. c. have constant order quantities. d. make the coordination of orders for multiple products more difficult.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.10.07 - 10.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.7 Periodic Review Model with Probabilistic Demand KEYWORDS: Bloom's: Understand 35. Inventory position is a. the amount of inventory on hand in excess of expected demand. b. the amount of inventory on hand. c. the amount of inventory on hand plus the amount of inventory on order. d. None of these are correct.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Remember 36. A firm that is presently using the economic order quantity model and is planning to switch to the economic production lot size model can expect a. the Q* to increase. b. the maximum inventory level to increase. c. the order cycle to decrease. d. annual holding cost to be less than annual setup cost.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.10.02 - 10.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.2 Economic Production Lot Size Model KEYWORDS: Bloom's: Understand 37. Which of the following is NOT implied when average inventory is Q/2, where Q is the order quantity? a. An entire order quantity arrives at one time. b. The previous order quantity is entirely depleted when the next order arrives. c. An order quantity is depleted at a uniform rate over time. d. Backorders are permitted. © Cengage. Testing Powered by Cognero.

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CH 10 - Inventory Models ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Understand 38. For the EOQ model, which of the following relationships is NOT correct? a. As the order quantity increases, the number of orders placed annually decreases. b. As the order quantity increases, annual holding cost increases. c. As the order quantity increases, annual ordering cost increases. d. As the order quantity increases, average inventory increases.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Understand 39. The objective of the EOQ with quantity discounts model is to a. determine the minimum order quantity required for the maximum discount. b. balance annual ordering and holding costs. c. minimize annual purchase cost. d. minimize the sum of annual carrying, holding, and purchase costs.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.10.04 - 10.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.4 Quantity Discounts for the EOQ Model KEYWORDS: Bloom's: Understand 40. When the reorder point r exceeds Q*, the difference is a. safety stock. b. one or more outstanding orders. c. surplus inventory. d. backorders.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking © Cengage. Testing Powered by Cognero.

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CH 10 - Inventory Models TOPICS: KEYWORDS:

10.1 Economic Order Quantity (EOQ) Model Bloom's: Understand

41. Inventory position is the amount of inventory on hand plus the amount a. on order. b. promised to customers. c. on reserve. d. to be returned to suppliers.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Remember 42. For the EOQ model, cycle time is the time between a. placing successive orders. b. placing and receiving an order. c. stocking out and receiving an order. d. receiving and storing an order.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Understand 43. The expression Q* refers to the a. maximum inventory on hand. b. optimal inventory for a stated period. c. optimal order quantity. d. None of these are correct.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.A10.01 - A10.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: Appendix 10.1 Development of the Optimal Order Quantity (Q*) Formula for the EOQ Model KEYWORDS: Bloom's: Understand Subjective Short Answer © Cengage. Testing Powered by Cognero.

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CH 10 - Inventory Models 44. Show the total cost expression and calculate the EOQ for an item with holding cost rate 18%, unit cost $8.00, annual demand of 40,000, and ordering cost of $48.

ANSWER: TC = 1/2Q(0.18)(8) + 40,000(48)/Q Q* = 1633 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Apply 45. Demand for a popular athletic shoe is nearly constant at 800 pairs per week for a regional division of a national retailer. The cost per pair is $54. It costs $72 to place an order, and annual holding costs are charged at 22% of the cost per unit. The lead time is two weeks. a. What is the EOQ? b. What is the reorder point? c. What is the cycle time? d. What is the total annual cost?

ANSWER: a. Q* = 710.1 b. r = 1600 c. T = 710.1/800 = 0.8876 weeks or 0.8876(7) = 6.2 days d. TC = 0.5(710.1)(0.22)(54) + (41,600/710.1)(72) = 8435.99 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Evaluate 46. The Super Discount store (open 24 hours a day, every day) sells 8-packs of paper towels, at the rate of approximately 420 packs per week. Because the towels are so bulky, the annual cost to carry them in inventory is estimated at $0.50. The cost to place an order for more is $20, and it takes four days for an order to arrive. a. Find the optimal order quantity. b. What is the reorder point? c. How often should an order be placed?

ANSWER: a. Q* = 1322 b. r = (60)(4) = 240 c. T = 365(1322)/21,840 = 22 days POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Evaluate 47. An office supply store open five days a week must determine the best inventory policy for boxes of copier paper.

Weekly demand is nearly constant at 250 boxes. When orders are placed, the entire shipment arrives at once. The cost per © Cengage. Testing Powered by Cognero.

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CH 10 - Inventory Models box is $22, and the inventory holding cost is 30%. Orders are placed at a cost of $40 each, including preparation time and communication charges, and the lead time is two days. a. Find the optimal order quantity. b. What is the reorder point? c. How often should an order be placed? d. What is the cycle time?

ANSWER: a. Q* = 397 b. TC = 0.5(397)(6.6) + (13,000/397)(40) = 2620 c. r = 100 d. Cycle = 7.94 days POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Analyze 48. Zip Games purchases blank DVD disks onto which it copies its software for sale through its mail-order operation. A disk costs Zip $0.20. Processing an order for more disks cost $15. Zip uses 60,000 disks annually, and the company has a 25% cost of capital. a. Find the optimal order quantity. b. How many orders are placed annually? c. How frequently will orders be placed?

ANSWER: a. Q* = 6000 b. 10 orders placed annually c. every 36.5 days POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Analyze 49. Kellam Images prints snack food bags on long rolls of plastic film. The plant operates 250 days a year. The daily production rate is 6000 bags, and the daily demand is 3500 bags. The cost to set up the design for printing is $300. The holding cost is estimated at 2 cents per bag. a. What is the recommended production lot size? b. If there is a five-day lead time to set up the line, what is the recommended reorder point?

ANSWER: a. Q* = 250,998 b. 17,500 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.10.02 - 10.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 10.2 Economic Production Lot Size Model KEYWORDS: Bloom's: Analyze © Cengage. Testing Powered by Cognero.

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CH 10 - Inventory Models 50. Henderson Furniture sells reproductions of 18th century furniture. For a particular table, the assumptions of the inventory model with backorders are valid.

D = 200 tables per year I = 25% per year C = $800 per table Co = $80 per order Cb = $50 per table per year The store is open 250 days a year. a. b. c. d.

What are the values for order quantity and number of planned backorders that will minimize total cost?

What is the maximum inventory?

What is the cycle time?

What is total cost?

ANSWER:

a.

Q* = 28.28 S* = 28.28(200/250) = 22.62 28 − 23 = 5 tables Cycle time = (28/200)250 = 35 days TC = 89.29 + 571.43 + 473.32 = 1134.04 (using whole numbers of tables for Q and S)

b. c. d.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.10.03 - 10.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 10.3 Inventory Model with Planned Shortages KEYWORDS: Bloom's: Analyze

51. The Tiernan Gallery and Art Museum distributes to its visitors a printed guide to its collections. The museum has about 18,000 visitors per year. Holding costs for the brochures are 20%, and it costs $30 to place an order with the printer.

The printer has offered the following discount schedule: Category 1 2 3

Order Size 0–1499 1500–2999 3000 and over

Unit Cost $2.50 $2.20 $1.80

How many brochures should be printed at a time?

ANSWER:

EOQ3 = 1732.1 (raise to 3000) EOQ2 = 1566.7 (feasible, so no reason to compute EOQ1) TC(1566.7) = 689.35 TC(3000) = 720.00 So the gallery should order 1566.7 or 1567 brochures at a time.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.10.04 - 10.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic © Cengage. Testing Powered by Cognero.

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CH 10 - Inventory Models TOPICS: KEYWORDS:

10.4 Quantity Discounts for the EOQ Model Bloom's: Evaluate

52. A weekly sports magazine publishes a special edition for the World Series. The sales forecast is for the number of copies to be normally distributed with mean 800,000 copies and standard deviation 60,000 copies. It costs $0.35 to print a copy, and the newsstand price is $1.95. Unsold copies will be scrapped. How many copies should be printed?

ANSWER: P(demand < Q*) = 1.6/1.95 = 0.8205 Q* = 800,000 + 0.92(60,000) = 855,200 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.10.05 - 10.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 10.5 Single-Period Inventory Model with Probabilistic Demand KEYWORDS: Bloom's: Evaluate 53. The Fitness Shop is considering ordering a special model exercise machine. Each unit will cost the shop $410, and it will sell for $750. Any units not sold at the regular price will be sold at the year-end model clearance for $340. Assume that demand follows a normal probability distribution with μ = 20 and σ = 6. What is the recommended order quantity?

ANSWER: cu = 340, co = 70 P(demand ≤ Q*) = 0.8293 Q* = 20 + 0.95(6) = 25.7 so order 26 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.10.05 - 10.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 10.5 Single-Period Inventory Model with Probabilistic Demand KEYWORDS: Bloom's: Evaluate 54. Daily demand for packages of five videotapes at a warehouse store is found to be normally distributed with mean 50 and standard deviation 5. When the store orders more tapes, the ordering cost is $42 and the orders take four days to arrive. Each pack of tapes costs $7.20, and there is a 24% annual holding cost for inventory. Assume the store is open 360 days a year. a. What is the EOQ? b. If the store wants the probability of stocking out to be no more than 5%, and demand each day is independent of the day before, what reorder point should be set? c. How much of your reorder point in part b) is safety stock?

ANSWER:

a. b.

Q* = 866.025 DDLT is N(200,10) r = 200 + 1.645(10) = 216.45 16.45

c.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.10.06 - 10.6 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 10.6 Order-Quantity, Reorder Point Model with Probabilistic Demand KEYWORDS: Bloom's: Analyze © Cengage. Testing Powered by Cognero.

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CH 10 - Inventory Models 55. Chez Paul Restaurant orders special Styrofoam "doggy bags" for its customers once a month, and lead time is one week. Weekly demand for doggy bags is approximately normally distributed with an average of 120 bags and a standard deviation of 25. Chez Paul wants at most a 3% chance of running out of doggy bags during the replenishment period. If he has 150 bags in stock when he places an order, how many additional bags should he order? What is the safety stock in this case?

ANSWER: Order 555 doggy bags. Safety stock is 105 bags.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.10.07 - 10.7 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 10.7 Periodic Review Model with Probabilistic Demand KEYWORDS: Bloom's: Evaluate 56. Bank Drugs sells Jami Michelle lipstick. Jami Michelle Company offers a 6% discount on orders of at least 500 tubes, a 10% discount on orders of at least 1000 tubes, a 12% discount on orders of at least 1800 tubes, and a 15% discount on orders at least 2500 tubes.

Bank sells an average of 40 tubes of Jami Michelle lipstick weekly. The normal price paid by Bank Drugs is $1 per tube.

If it costs Bank $30 to place an order and Bank's annual holding cost rate is 27%, determine the optimal order policy for Bank Drugs.

ANSWER: Bank Drugs should order 1000 tubes at a time.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.10.04 - 10.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 10.4 Quantity Discounts for the EOQ Model KEYWORDS: Bloom's: Analyze 57. Amazing Bakers sells bread to 40 supermarkets. It costs Amazing $1250 per day to operate its plant. The profit per loaf of bread sold in the supermarket is $0.025. Any unsold bread is returned to Amazing to be sold at a loss of $0.015. a. If sales follow a normal distribution with μ = 70,000 and σ = 5000 per day, how many loaves should Amazing bake daily? b. Amazing is considering a different sales plan for which the profit per loaf of bread sold in the supermarket is $0.03 and the loss per loaf of bread returned is $0.018. If μ = 60,000 and σ = 4000 per day, how many loaves should Amazing bake daily?

ANSWER:

a. 71,600 loaves b. 61,280 loaves POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.10.05 - 10.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 10.5 Single-Period Inventory Model with Probabilistic Demand KEYWORDS: Bloom's: Evaluate 58. A lawn and garden shop that is open for business seven days a week orders bags of grass seed every other Monday.

Lead time for seed orders is five days. On Monday, at ordering time, a clerk found 112 bags of seed in stock so he ordered © Cengage. Testing Powered by Cognero.

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CH 10 - Inventory Models 198 bags. Daily demand for grass seed is normally distributed with a mean of 15 bags and a standard deviation of four bags.

The manager would like to know the probability that a grass seed stockout will occur before the next order arrives.

ANSWER: z = 1.43, so P(stockout) = 0.0764 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.10.07 - 10.7 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 10.7 Periodic Review Model with Probabilistic Demand KEYWORDS: Bloom's: Evaluate 59. Kelly's Service Station does a large business in tune-ups. Demand has been averaging 210 spark plugs per week.

Holding costs are $0.01 per plug per week, and reorder costs are estimated at $10 per order.

Kelly does not want to be out of stock on more than 1% of his orders. There is a one-day delivery time. The standard deviation of demand is five plugs per day. Assume a normal distribution of demand during lead time and a seven-day work week. a. What inventory policy do you suggest for Kelly's station? b. What is the average amount of safety stock for the reorder point in part (a)? c. What is the total variable weekly cost including safety stock cost?

ANSWER:

a. order 648 every 3.08 weeks when supply reaches 42 b. 12 spark plugs c. $6.60 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Analyze 60. Non-Slip Tile Company (NST) has been using production runs of 100,000 tiles, 10 times per year to meet the demand of 1,000,000 tiles annually. The setup cost is $5000 per run, and holding cost is estimated at 10% of the manufacturing cost of $1 per tile. The production capacity of the machine is 500,000 tiles per month. The factory is open 365 days per year. a. What production schedule do you recommend? b. How much is NST losing annually with its present production schedule? c. What is the maximum number of tiles in inventory under the current policy? Under the optimal policy? d. What fraction of time is the machine idle (not producing tiles) under the current policy? Under the optimal policy?

ANSWER: D = 1,000,000, P = 6,000,000, Ch = 0.10, Co = 5,000 a. Q = 346,410 D/Q = 2.89 times per year The recommended policy is produce 346,410 tiles at a time, 2.89 times per year. b. Optimal TC = $28,867.51 Current TC = $54,166.67 © Cengage. Testing Powered by Cognero.

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CH 10 - Inventory Models Difference = $25,299.16 c. Current maximum inventory = 83,333 Optimal maximum inventory = 288,675 d. The machine is producing tiles d/p = 1/6 of the time.

The machine is idle 5/6 of the time.

Optimal Policy: There are 2.89 cycles per year.

Each cycle lasts 126.3 days.

Each run is 346,410 units.

Each run takes 21.1 days.

Machine is idle 105.2 days between runs.

Machine is idle 0.833 (or 5/6) of time.

Current Policy: There are 10 cycles per year.

Each cycle lasts 36.5 days.

Each run is 100,000 units.

Each run takes 6.083 days.

Machine is idle 30.417 days between runs.

The machine is idle 0.833 (or 5/6) of time.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.10.02 - 10.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 10.2 Economic Production Lot Size Model KEYWORDS: Bloom's: Evaluate 61. Ken Ells, owner and operator of Kennels, Inc., is concerned that the person in charge of ordering dog food is often incurring an order expediting expense because he is waiting too long (letting the inventory level drop too low) before ordering.

Past data indicate that demand during lead time (when expediting does not occur) is normally distributed with a mean of 340 pounds and a standard deviation of 45 pounds.

Ken wants the probability of running out of dog food to be 0.03. a. If the current order point is 400 pounds, what is the resulting service level? b. If Ken wants a service level of 97% for dog food, what should the order point be?

ANSWER: a. 90.82% b. r = 340 + 1.88(45) = 424.6 or 425 pounds POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.10.01 - 10.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 10.1 Economic Order Quantity (EOQ) Model KEYWORDS: Bloom's: Evaluate © Cengage. Testing Powered by Cognero.

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CH 10 - Inventory Models

© Cengage. Testing Powered by Cognero.

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CH 11 - Waiting Line Models True / False 1. For an M/M/1 queuing system, if the service rate, µ, is doubled, the average wait in the system, W, is cut in half. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.11.09 - 11.9 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.9 Waiting Line Models with Finite Calling Populations KEYWORDS: Bloom's: Understand 2. A waiting line situation with a single server is referred to as an M/M/1 model with a finite calling population. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.09 - 11.9 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.9 Waiting Line Models with Finite Calling Populations KEYWORDS: Bloom's: Remember 3. Use of the Poisson probability distribution assumes that arrivals are not random. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.1 Structure of a Waiting Line System KEYWORDS: Bloom's: Understand 4. Queue discipline refers to the assumption that a customer has the patience to remain in a slow moving queue. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.1 Structure of a Waiting Line System © Cengage. Testing Powered by Cognero.

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CH 11 - Waiting Line Models KEYWORDS:

Bloom's: Remember

5. Before waiting lines can be analyzed economically, the arrivals' cost of waiting must be estimated. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.05 - 11.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.5 Economic Analysis of Waiting Lines KEYWORDS: Bloom's: Understand 6. A variation of the waiting line models involves a system in which no waiting is allowed, and arriving units are denied access to the system. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.11.08 - 11.8 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.8 Multiple-Server Model with Poisson Arrivals, Arbitrary Service Times, and No Waiting Line KEYWORDS: Bloom's: Understand 7. Little's flow equations indicate that the relationship of L to Lq is the same as that of W to Wq. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.11.04 - 11.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.4 Some General Relationships for Waiting Line Models KEYWORDS: Bloom's: Understand 8. Kendall's notation is helpful when classifying the wide variety of different waiting line models and can indicate that the waiting line system is assumed to have infinite capacity. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.06 - 11.6 © Cengage. Testing Powered by Cognero.

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CH 11 - Waiting Line Models NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.6 Other Waiting Line Models KEYWORDS: Bloom's: Understand 9. When blocked customers are cleared, an important decision is how many servers to provide. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.08 - 11.8 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.8 Multiple-Server Model with Poisson Arrivals, Arbitrary Service Times, and No Waiting Line KEYWORDS: Bloom's: Understand 10. If service time follows an exponential probability distribution, approximately 63% of the service times are less than the mean service time. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.1 Structure of a Waiting Line System KEYWORDS: Bloom's: Remember 11. Queue discipline refers to the manner in which waiting units are arranged for service. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.1 Structure of a Waiting Line System KEYWORDS: Bloom's: Remember 12. Waiting line models describe the transient-period operating characteristics of a waiting line. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate © Cengage. Testing Powered by Cognero.

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CH 11 - Waiting Line Models LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.1 Structure of a Waiting Line System KEYWORDS: Bloom's: Understand 13. For a single-server waiting line, the utilization factor is the probability that an arriving unit must wait for service. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.02 - 11.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.2 Single-Server Waiting Line Model with Poisson Arrivals and Exponential Service Times KEYWORDS: Bloom's: Understand 14. After the startup or transient period, a waiting system is in steady-state operation and considered to be the normal operation of the waiting line. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.1 Structure of a Waiting Line System KEYWORDS: Bloom's: Understand 15. Adding more servers always improves the operating characteristics of the waiting line and reduces the waiting cost. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.05 - 11.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.5 Economic Analysis of Waiting Lines KEYWORDS: Bloom's: Remember 16. In developing the total cost for a waiting line, waiting cost takes into consideration both the time spent waiting in line and the time spent being served. a. True b. False © Cengage. Testing Powered by Cognero.

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CH 11 - Waiting Line Models ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.05 - 11.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.5 Economic Analysis of Waiting Lines KEYWORDS: Bloom's: Remember 17. In waiting line systems where the length of the waiting line is limited, the mean number of units entering the system might be less than the arrival rate. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.04 - 11.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.4 Some General Relationships for Waiting Line Models KEYWORDS: Bloom's: Understand 18. A multiple-server waiting line is one that has two or more parallel service facilities. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.03 - 11.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.3 Multiple-Server Waiting Line Model with Poisson Arrivals and Exponential Service Times KEYWORDS: Bloom's: Remember 19. For a single-server queuing system, the average number of customers in the waiting line is one less than the average number in the system. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.11.02 - 11.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.2 Single-Server Waiting Line Model with Poisson Arrivals and Exponential Service Times KEYWORDS: Bloom's: Understand © Cengage. Testing Powered by Cognero.

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CH 11 - Waiting Line Models 20. In waiting line applications, the exponential probability distribution indicates that approximately 63% of the service times are less than the mean service time. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.1 Structure of a Waiting Line System KEYWORDS: Bloom's: Understand 21. Waiting line models consist of mathematical formulas and relationships that can be used to determine the operating characteristics (also known as performance measures) for a waiting line. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.1 Structure of a Waiting Line System KEYWORDS: Bloom's: Remember 22. Little’s flow equations can apply to a single-server as well as multiple-server waiting line model. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.11.04 - 11.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.4 Some General Relationships for Waiting Line Models KEYWORDS: Bloom's: Understand 23. The body of knowledge dealing with waiting lines is known as queueing theory. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.1 Structure of a Waiting Line System © Cengage. Testing Powered by Cognero.

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CH 11 - Waiting Line Models KEYWORDS:

Bloom's: Remember

Multiple Choice 24. Decision makers in queuing situations attempt to balance a. operating characteristics against the arrival rate. b. service levels against service cost. c. the number of units in the system against the time in the system. d. the service rate against the arrival rate.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.1 Structure of a Waiting Line System KEYWORDS: Bloom's: Remember 25. Performance measures dealing with the number of units in line and the time spent waiting are called a. queuing facts. b. performance queues. c. system measures. d. operating characteristics.

ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.1 Structure of a Waiting Line System KEYWORDS: Bloom's: Remember 26. If arrivals occur according to the Poisson distribution every 20 minutes, then which of the following is NOT true? a. λ = 20 arrivals per hour b. λ = 3 arrivals per hour c. λ = 1/20 arrivals per minute d. λ = 72 arrivals per day ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 11.1 Structure of a Waiting Line System KEYWORDS: Bloom's: Apply 27. The manner in which units receive their service, such as FCFS, is the a. queue discipline. © Cengage. Testing Powered by Cognero.

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CH 11 - Waiting Line Models b. server. c. steady state. d. operating characteristic.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.1 Structure of a Waiting Line System KEYWORDS: Bloom's: Remember 28. In a waiting line situation, arrivals occur, on average, every 10 minutes, and 10 units can be received every hour.

Which of the following represents λ and μ in this situation? a. λ = 10, μ = 10 b. λ = 6, μ = 6 c. λ = 6, μ = 10 d. λ = 10, μ = 6 ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 11.1 Structure of a Waiting Line System KEYWORDS: Bloom's: Apply 29. Operating characteristics formulas for the single-server queue do NOT require a. λ ≥ μ. b. Poisson distribution of arrivals. c. an exponential distribution of service times. d. an FCFS queue discipline.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.11.02 - 11.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.2 Single-Server Waiting Line Model with Poisson Arrivals and Exponential Service Times KEYWORDS: Bloom's: Understand 30. The mean number of units that can be served per time period is called a. the service rate and is denoted by λ. b. the service rate and is denoted by μ. c. the steady state. d. None of these are correct. © Cengage. Testing Powered by Cognero.

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CH 11 - Waiting Line Models ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.1 Structure of a Waiting Line System KEYWORDS: Bloom's: Understand 31. Little's flow equations a. require Poisson and exponential assumptions. b. are applicable to any waiting line model. c. require independent calculation of W, L, Wq, and Lq. d. All of these are correct.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.11.04 - 11.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.4 Some General Relationships for Waiting Line Models KEYWORDS: Bloom's: Understand 32. The total cost for a waiting line does NOT specifically depend on the a. cost of waiting. b. cost of service. c. number of units in the system. d. cost of a lost customer.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.11.05 - 11.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.5 Economic Analysis of Waiting Lines KEYWORDS: Bloom's: Understand 33. Models with a finite calling population a. have an arrival rate independent of the number of units in the system. b. have a service rate dependent on the number of units in the system. c. have no limit placed on how many units may seek service. d. All of these are correct.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.09 - 11.9 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking © Cengage. Testing Powered by Cognero.

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CH 11 - Waiting Line Models TOPICS: KEYWORDS:

11.9 Waiting Line Models with Finite Calling Populations Bloom's: Understand

34. When no limit is placed on how many units may seek service, the waiting line model a. can assume that no units are in the system. b. is said to have an infinite calling population. c. maintains a constant arrival rate. d. None of these are correct.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.11.09 - 11.9 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.9 Waiting Line Models with Finite Calling Populations KEYWORDS: Bloom's: Understand 35. The arrival rate in queuing formulas is expressed as the a. mean time between arrivals. b. minimum number of arrivals per time period. c. mean number of arrivals per server. d. mean number of arrivals per time period.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.1 Structure of a Waiting Line System KEYWORDS: Bloom's: Remember 36. Which of the following queue disciplines is assumed by the waiting line models presented in the textbook? a. first-come, first-served b. last-in, first-out c. shortest processing, time first d. first-in, last-out ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.1 Structure of a Waiting Line System KEYWORDS: Bloom's: Understand 37. For many waiting line situations, the arrivals occur randomly and independently of other arrivals and it has been found that a good description of the arrival pattern is provided by a(n) © Cengage. Testing Powered by Cognero.

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CH 11 - Waiting Line Models a. normal probability distribution. b. exponential probability distribution. c. uniform probability distribution. d. Poisson probability distribution.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.1 Structure of a Waiting Line System KEYWORDS: Bloom's: Understand 38. The assumption of exponentially distributed service times indicates that a. 37% of the service times are less than the mean service time. b. 50% of the service times are less than the mean service time. c. 63% of the service times are less than the mean service time. d. service time increase at an exponential rate as the waiting line grows.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.1 Structure of a Waiting Line System KEYWORDS: Bloom's: Remember 39. Single-booth ticket sales at a theater are an example of which of the following queuing models? a. single-server, Poisson service rate distribution, unlimited queue length b. single-server, Poisson service rate distribution, limited queue length c. single-server, constant service rate distribution, unlimited queue length d. single-server, normal service rate distribution, unlimited queue length ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.11.02 - 11.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.2 Single-Server Waiting Line Model with Poisson Arrivals and Exponential Service Times KEYWORDS: Bloom's: Understand 40. The equations provided in the textbook for computing operating characteristics apply to a waiting line operating a. at start-up. b. at steady state. c. at peak demand times. d. in transition. © Cengage. Testing Powered by Cognero.

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CH 11 - Waiting Line Models ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.11.02 - 11.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 11.2 Single-Server Waiting Line Model with Poisson Arrivals and Exponential Service Times KEYWORDS: Bloom's: Understand Subjective Short Answer 41. During summer weekdays, boats arrive at the inlet drawbridge according to the Poisson distribution at a rate of three per hour. In a two-hour period, a. what is the probability that no boats arrive? b. what is the probability that two boats arrive? c. what is the probability that eight boats arrive?

ANSWER:

a. P(0) = 0.0025 b. P(2) = 0.0446 c. P(8) = 0.1033 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 11.1 Structure of a Waiting Line System KEYWORDS: Bloom's: Analyze 42. The time to process a registration at the Sea View Resort follows the exponential distribution and has a mean of six minutes. a. What is the probability of a registration time shorter than three minutes? b. What is the probability of a registration time shorter than six minutes? c. What is the probability of a registration time between three and six minutes?

ANSWER:

a. P(time ≤ 3) = 1 − 0.6065 = 0.3935 b. P(time ≤ 6) = 1 − 0.3679 = 0.6321 c. P(3 ≤ time ≤ 6) = 0.6321 − 0.3935 = 0.2386 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.11.01 - 11.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 11.1 Structure of a Waiting Line System KEYWORDS: Bloom's: Analyze 43. The Grand Movie Theater has one box office clerk. On average, each customer that comes to see a movie can be sold a ticket at the rate of six per minute. For the theater's normal offerings of older movies, customers arrive at the rate of three per minute. Assume arrivals follow the Poisson distribution and service times follow the exponential distribution. a. What is the average number of customers waiting in line? © Cengage. Testing Powered by Cognero.

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CH 11 - Waiting Line Models b. What is the average time a customer spends in the waiting line? c. What is the average number of customers in the system? d. What is a customer's average time in the system? e. What is the probability that someone will be buying tickets when an arrival occurs?

The Grand has booked the Stars Wars Trilogy and expects more customers. From conversations with other theater owners, it estimates that the arrival rate will increase to 10 per minute. Output is supplied for two- and three-cashier systems.

Number of servers Arrival rate Service rate Probability of no units in system Average waiting time Average time in system Average number waiting Average number in system Probability of waiting Probability of 11 in system f. g.

2 10 6 0.0909 0.3788 0.5455 3.78790 5.4545 0.7576 0.0245

3 10 6 0.1727 0.0375 0.2041 0.3747 2.0414 0.2998 less than 0.0088

The Grand has space for 10 customers to wait indoors to buy tickets. Which system will be better?

Do you think it is more sensible for the Grand to continue the one-cashier system?

ANSWER: a. b. c. d. e. f. g.

Lq = .5 Wq = 0.1667 L=1 W = 0.3333 Pw = 0.5 The three-cashier system is probably too good and not cost effective.

The one-cashier system won't work because now the arrival rate is faster than the service rate.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.11.02 - 11.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 11.2 Single-Server Waiting Line Model with Poisson Arrivals and Exponential Service Times KEYWORDS: Bloom's: Evaluate 44. The Arctic Flyers minor league hockey team has one box office clerk. On average, each customer that comes to see a game can be sold a ticket at the rate of eight per minute. For normal games, customers arrive at the rate of five per minute.

Assume arrivals follow the Poisson distribution and service times follow the exponential distribution. a. What is the average number of customers waiting in line? b. What is the average time a customer spends in the waiting line? c. What is the average number of customers in the system? d. What is a customer's average time in the system? e. What is the probability that someone will be buying tickets when an arrival occurs?

The Flyers are playing in the league playoffs and anticipate more fans, estimating that the arrival rate will increase to 12 per minute. Output is supplied for two- and three-cashier systems. © Cengage. Testing Powered by Cognero.

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CH 11 - Waiting Line Models Number of servers Arrival rate Service rate Probability of no units in system Average waiting time Average time in system Average number waiting Average number in system Probability of waiting Probability of 7 in system f. g.

2 12 8 0.1429 0.1607 0.2857 1.9286 3.4286 0.6429 0.0381

3 12 8 0.2105 0.0197 0.1447 0.2368 1.7368 0.2368 0.0074

The rink has space for six customers to wait indoors to buy tickets. Which system will be better?

Do you think it is more sensible for the Flyers to continue the one-cashier system?

ANSWER: a. b. c. d. e. f. g.

Lq = 1.04 Wq = 0.2083 L = 1.665 W = 0.3333 Pw = 0.625 The three-cashier system is probably too good and not cost effective.

The one-cashier system won't work because now the arrival rate is faster than the service rate.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.11.02 - 11.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 11.2 Single-Server Waiting Line Model with Poisson Arrivals and Exponential Service Times KEYWORDS: Bloom's: Evaluate 45. In a waiting line situation, arrivals occur at a rate of two per minute, and the service times average 18 seconds.

Assume the Poisson and exponential distributions. a. What is λ? b. What is μ? c. Find the probability of no units in the system. d. Find the average number of units in the system. e. Find the average time in the waiting line. f. Find the average time in the system. g. Find the probability that there is one person waiting. h. Find the probability that an arrival will have to wait.

ANSWER:

a. b. c. d. e.

λ = 3/min. μ = 4/min.

P0 = 0.25 L=3 Wq = 0.75 min.

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CH 11 - Waiting Line Models f. g. h.

W = 1.00 min.

P(2) = 0.140625 Pw = 0.75

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.11.09 - 11.9 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 11.9 Waiting Line Models with Finite Calling Populations KEYWORDS: Bloom's: Evaluate 46. In a waiting line situation, arrivals occur around the clock at a rate of six per day, and the service occurs at one every three hours. Assume the Poisson and exponential distributions. a. What is λ? b. What is μ? c. Find the probability of no units in the system. d. Find the average number of units in the system. e. Find the average time in the waiting line. f. Find the average time in the system. g. Find the probability that there is one person waiting. h. Find the probability that an arrival will have to wait.

ANSWER:

a. b. c. d. e. f. g. h.

λ=6 μ=8 P0 = 0.25 L=3 Wq = 0.375 W = 0.5 P(2) = 0.1406 Pw = 0.75

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.11.09 - 11.9 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 11.9 Waiting Line Models with Finite Calling Populations KEYWORDS: Bloom's: Evaluate 47. Two new checkout scanning systems are under consideration by a retail store. Arrivals to the checkout stand follow the Poisson distribution with λ = 2 per minute. The cost for waiting is $18 per hour. The first system has an exponential service rate of five per minute and costs $10 per hour to operate. The second system has an exponential service rate of eight per minute and costs $20 per hour to operate. Which system should be chosen?

ANSWER: The first system costs 0.37 per minute, and the second system costs 0.43 per minute. Choose the first system.

POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.11.05 - 11.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 11.5 Economic Analysis of Waiting Lines © Cengage. Testing Powered by Cognero.

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CH 11 - Waiting Line Models KEYWORDS:

Bloom's: Analyze

48. Circle Electric Supply is considering opening a second service counter to better serve the electrical contractor customers. The arrival rate is 10 per hour. The service rate is 14 per hour. If the cost of waiting is $30 and the cost of each service counter is $22 per hour, should the second counter be opened?

ANSWER: With one window, the cost per hour is 97. With two windows, the cost per hour is 68.56.

Choose two windows.

POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.11.05 - 11.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 11.5 Economic Analysis of Waiting Lines KEYWORDS: Bloom's: Analyze 49. For an M/G/1 system with λ = 6, μ = 9, and σ = 0.03, find a. the probability the system is idle. b. the average length of the queue. c. the average number in the system.

ANSWER: a. P0 = 0.67 b. Lq = 0.7153 c. L = 1.3819 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.11.07 - 11.7 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 11.7 Single-Server Waiting Line Model with Poisson Arrivals and Arbitrary Service Times KEYWORDS: Bloom's: Analyze 50. For an M/G/1 system with λ = 20, μ = 35, and σ = 0.005, find a. the probability the system is idle. b. the average length of the queue. c. the average number in the system.

ANSWER: a. P0 = 0.4286 b. Lq = 0.3926 c. L = 0.964 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.11.07 - 11.7 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 11.7 Single-Server Waiting Line Model with Poisson Arrivals and Arbitrary Service Times KEYWORDS: Bloom's: Analyze 51. Arrivals at a box office in the hour before the show follow the Poisson distribution with λ = 7 per minute. Service times are constant at 7.5 seconds. Find the average length of the waiting line.

ANSWER: Lq = 3.0625 © Cengage. Testing Powered by Cognero.

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CH 11 - Waiting Line Models POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.11.07 - 11.7 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 11.7 Single-Server Waiting Line Model with Poisson Arrivals and Arbitrary Service Times KEYWORDS: Bloom's: Analyze 52. The eight students in a seminar class must come to the professor's office to turn in a paper and give a five-minute oral summary. Assume there is a service rate of 10 per hour and adequate time is available for all. The arrival rate for each unit is five per hour. What is the probability there is no one in the office or waiting when you come?

ANSWER: λ = 5, μ = 10, N = 8 P0 = 0.0009 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.11.09 - 11.9 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 11.9 Waiting Line Models with Finite Calling Populations KEYWORDS: Bloom's: Analyze 53. Andy Archer, Ph.D., is a training consultant for six mid-sized manufacturing firms. On average, each of his six clients calls him for consulting assistance once every 25 days. Andy typically spends an average of five days at the client's firm during each consultation.

Assuming the time between client calls follows an exponential distribution, determine a. the average number of clients Andy has on backlog. b. the average time a client must wait before Andy arrives. c. the proportion of time Andy is busy.

ANSWER: a. b. c.

Lq = 0.7094 Wq = 4.96 days 1 − P0 = 0.7151

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.11.09 - 11.9 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 11.9 Waiting Line Models with Finite Calling Populations KEYWORDS: Bloom's: Analyze 54. The Quick Snap photo machine at the Lemon County bus station takes four snapshots in exactly 75 seconds.

Customers arrive at the machine according to a Poisson distribution at the mean rate of 20 per hour. On the basis of this information, determine a. the average number of customers waiting to use the photo machine. b. the average time a customer spends using the system. c. the probability an arriving customer must wait for service.

ANSWER: a.

Lq = 0.15

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CH 11 - Waiting Line Models b. c.

W = 0.028 hour = 1.7 minutes PW = 0.417

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.11.07 - 11.7 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 11.7 Single-Server Waiting Line Model with Poisson Arrivals and Arbitrary Service Times KEYWORDS: Bloom's: Analyze 55. Quick Clean Rooter cleans out clogged drains. Due to the competitive nature of the drain cleaning business, if a customer calls Quick Clean and finds the line busy, they immediately try another company and Quick Clean loses the business.

Quick Clean management estimates that, on average, a customer tries to call Quick Clean every three minutes and the average time to take a service order is 200 seconds. The company wishes to hire enough operators so that at most 4% of its potential customers get the busy signal. a. How many operators should be hired to meet this objective? b. Given your answer to part (a), what is the probability that all the operators are idle?

ANSWER: a. b.

k = 4; PW = 0.021 P0 = 0.331

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.11.08 - 11.8 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 11.8 Multiple-Server Model with Poisson Arrivals, Arbitrary Service Times, and No Waiting Line KEYWORDS: Bloom's: Analyze 56. A company has tool cribs where workmen draw parts. Two men have applied for the position of distributing parts to the workmen. George Fuller is fresh out of trade school and expects a $6 per hour salary. His average service time is four minutes. John Cox is a veteran who expects $12 per hour. His average service time is two minutes. A workman's time is figured at $10 per hour. Workmen arrive to draw parts at an average rate of 12 per hour. a. What is the average waiting time a workman would spend in the system under each applicant? b. Which applicant should be hired?

ANSWER:

a. George: W = 20 minutes; John: W = 3 1/3 minutes b. Total cost: George = $46.00, John = $18.67. Hire John.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.11.05 - 11.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 11.5 Economic Analysis of Waiting Lines KEYWORDS: Bloom's: Analyze © Cengage. Testing Powered by Cognero.

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CH 11 - Waiting Line Models 57. The insurance department at Shear's has two agents, each working at a mean speed of eight customers per hour.

Customers arrive at the insurance desk at a mean rate of one every six minutes and form a single queue. Management feels that some customers are going to find the wait at the desk too long and take their business to Word's, Shear's competitor.

In order to reduce the time required by an agent to serve a customer, Shear's is contemplating installing one of two minicomputer systems: System A that leases for $18 per day and will increase an agent's efficiency by 25% or System B that leases for $23 per day and will increase an agent's efficiency by 50%. Agents work eight-hour days.

If Shear's estimates its cost of having a customer in the system at $3 per hour, determine if it should install a new minicomputer system, and if so, which one?

ANSWER: No system: $49.20; System A: $50.00; System B: $47.50. Install System B. POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.11.05 - 11.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 11.5 Economic Analysis of Waiting Lines KEYWORDS: Bloom's: Analyze 58. The postmaster at the Oak Hill Post Office expects that the mean arrival rate of people to her customer counter will soon increase by 50% due to a large apartment complex being built. Currently, the mean arrival rate is 15 people per hour.

The postmaster can serve an average of 25 people per hour. By what percentage must the postmaster's mean service rate increase when the apartment complex is completed in order that the average time spent at the post office remains at its current value?

ANSWER: 30% POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.11.09 - 11.9 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 11.9 Waiting Line Models with Finite Calling Populations KEYWORDS: Bloom's: Analyze

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CH 12 - Simulation True / False 1. Simulation is an excellent technique to use when a situation is too complicated to use standard analytical procedures. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.01 - 12.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.1 What-If Analysis KEYWORDS: Bloom's: Remember 2. Simulation is a trial-and-error approach to problem solving. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.01 - 12.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.1 What-If Analysis KEYWORDS: Bloom's: Understand 3. The degree of risk is associated with the probability or magnitude of loss. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.01 - 12.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.1 What-If Analysis KEYWORDS: Bloom's: Understand 4. To use Excel to generate a normally distributed random variable, you must know the mean and standard deviation of the distribution and have a random number between 0 and 1. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.02 - 12.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.2 Simulation of Sanotronics Problem © Cengage. Testing Powered by Cognero.

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CH 12 - Simulation KEYWORDS:

Bloom's: Understand

5. Trials of a simulation show what would happen when values of the probabilistic input change. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.02 - 12.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.2 Simulation of Sanotronics Problem KEYWORDS: Bloom's: Understand 6. Verification is the process of ensuring that the simulation model provides an accurate representation of the real system. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.05 - 12.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.5 Simulation Considerations KEYWORDS: Bloom's: Remember 7. Simulation is a method that uses repeated random sampling of values in order to represent uncertainty in a model that represents a real system and computes the values of model outputs. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.12.05 - 12.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.5 Simulation Considerations KEYWORDS: Bloom's: Remember 8. Validation determines that the computer procedure is operating as it is intended to operate. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.05 - 12.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking © Cengage. Testing Powered by Cognero.

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CH 12 - Simulation TOPICS: KEYWORDS:

12.5 Simulation Considerations Bloom's: Remember

9. A discrete-event simulation reviews the status of the system periodically, whether or not an event occurs. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.04 - 12.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.4 Waiting Line Simulation KEYWORDS: Bloom's: Understand 10. A static simulation model is used in situations where the state of the system affects how the system changes or evolves over time. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.04 - 12.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.4 Waiting Line Simulation KEYWORDS: Bloom's: Remember 11. The following relationship is valid for any waiting line system: the average number in the waiting line is equal to the total waiting time divided by the total time of simulation. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.12.04 - 12.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.4 Waiting Line Simulation KEYWORDS: Bloom's: Understand 12. Varying the values of controllable inputs enables simulation models to identify good operating policies and decisions. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.12.03 - 12.3 © Cengage. Testing Powered by Cognero.

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CH 12 - Simulation NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.3 Inventory Simulation KEYWORDS: Bloom's: Understand 13. Using simulation to perform risk analysis is like playing out many what-if scenarios by randomly generating values for the probabilistic inputs. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.12.01 - 12.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.1 What-If Analysis KEYWORDS: Bloom's: Understand 14. All values of the computer-generated random numbers are equally likely and are uniformly distributed over the interval from 0 to 1. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.12.02 - 12.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.2 Simulation of Sanotronics Problem KEYWORDS: Bloom's: Understand 15. Computer-generated random numbers are not technically random, because they are generated through the use of mathematical formulas. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.02 - 12.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.2 Simulation of Sanotronics Problem KEYWORDS: Bloom's: Understand 16. Simulation is an optimization technique. a. True b. False ANSWER: False POINTS: 1 © Cengage. Testing Powered by Cognero.

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CH 12 - Simulation DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.03 - 12.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.3 Inventory Simulation KEYWORDS: Bloom's: Remember 17. Simulation models that must take into account how the system changes or evolves over time are referred to as dynamic simulation models. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.04 - 12.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.4 Waiting Line Simulation KEYWORDS: Bloom's: Remember 18. Each simulation run provides only a sample of how the real system will operate. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.05 - 12.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.5 Simulation Considerations KEYWORDS: Bloom's: Understand 19. One disadvantage of simulation is that it is limited in the variety of probability distributions that can be used in modeling a system. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.12.05 - 12.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.5 Simulation Considerations KEYWORDS: Bloom's: Understand 20. A simulation model provides a convenient experimental laboratory for the real system. a. True b. False ANSWER: True © Cengage. Testing Powered by Cognero.

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CH 12 - Simulation POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.05 - 12.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.5 Simulation Considerations KEYWORDS: Bloom's: Understand 21. Simulation allows the user to specify certain desired results (e.g., profit or service level values), and then the necessary model parameters and operating policies are determined. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.12.03 - 12.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.3 Inventory Simulation KEYWORDS: Bloom's: Understand 22. Simulation models manually generate specific values for the random variables so that the values used reflect what we might observe in practice. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.02 - 12.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.2 Simulation of Sanotronics Problem KEYWORDS: Bloom's: Understand 23. Independent trials are those in which the results for one trial have no effect on what happens in subsequent trials. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.04 - 12.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.4 Waiting Line Simulation KEYWORDS: Bloom's: Remember 24. When making a decision in the presence of uncertainty, decision makers should only be interested in the average, or expected, outcome. a. True © Cengage. Testing Powered by Cognero.

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CH 12 - Simulation b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.01 - 12.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.1 What-If Analysis KEYWORDS: Bloom's: Understand 25. Simulation analysis requires a model foundation of logical formulas that correctly express the relationships between parameters and decisions to generate outputs of interest. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.01 - 12.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.1 What-If Analysis KEYWORDS: Bloom's: Understand Multiple Choice 26. A simulation model uses the mathematical expressions and logical relationships of the a. real system. b. computer model. c. performance measures. d. estimated inferences.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.01 - 12.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.1 What-If Analysis KEYWORDS: Bloom's: Understand 27. Values for the random variables or uncertain variable inputs to a simulation are a. selected by the decision maker. b. controlled by the decision maker. c. randomly generated from the specified probability distributions. d. calculated by fixed mathematical formulas.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate © Cengage. Testing Powered by Cognero.

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CH 12 - Simulation LEARNING OBJECTIVES: IMS.ASWC.19.12.01 - 12.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.1 What-If Analysis KEYWORDS: Bloom's: Understand 28. A quantity that is difficult to measure with certainty is called a a. risk analysis. b. project determinant. c. probabilistic input. d. profit/loss process.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.12.02 - 12.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.2 Simulation of Sanotronics Problem KEYWORDS: Bloom's: Remember 29. A value for probabilistic input from a discrete probability distribution a. is the value given by the RAND() function. b. is given by matching the probabilistic input with an interval of random numbers. c. is between 0 and 1. d. must be nonnegative.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.12.02 - 12.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.2 Simulation of Sanotronics Problem KEYWORDS: Bloom's: Understand 30. The number of units expected to be sold is uniformly distributed between 300 and 500. If r is a random number between 0 and 1, then the proper expression for sales is a. 200(r). b. r + 300. c. 300 + 500(r). d. 300 + r(200).

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.12.02 - 12.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 12.2 Simulation of Sanotronics Problem KEYWORDS: Bloom's: Apply © Cengage. Testing Powered by Cognero.

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CH 12 - Simulation 31. When events occur at discrete points in time, a. a simulation clock is required. b. the simulation advances to the next event. c. the model is a discrete-event simulation. d. All of these are correct.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.12.02 - 12.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.2 Simulation of Sanotronics Problem KEYWORDS: Bloom's: Understand 32. If customer 2 has a service time of 1.6, and if customer 3 has an interarrival time of 1.1 and a service time of 2.3, when will customer 3's service be completed? a. 5.0 b. 3.9 c. 3.4 d. There is not enough information to answer.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.12.04 - 12.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 12.4 Waiting Line Simulation KEYWORDS: Bloom's: Apply 33. In order to verify a simulation model, a. compare results from several simulation languages. b. be sure that the procedures for calculations are logically correct. c. confirm that the model accurately represents the real system. d. run the model long enough to overcome initial start-up results.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.12.05 - 12.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.5 Simulation Considerations KEYWORDS: Bloom's: Understand 34. Simulation a. does not guarantee optimality. b. is flexible and does not require the assumptions of theoretical models. © Cengage. Testing Powered by Cognero.

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CH 12 - Simulation c. allows testing of the system without affecting the real system. d. All of these are correct.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.12.05 - 12.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.5 Simulation Considerations KEYWORDS: Bloom's: Understand 35. A simulation model in which each trial is used in situations where the state of the system at one point in time does not affect the state of the system at future points in time is called a a. dynamic simulation model. b. static simulation model. c. steady-state simulation model. d. discrete-event simulation model.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.12.04 - 12.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.4 Waiting Line Simulation KEYWORDS: Bloom's: Remember 36. The process of determining that the computer procedure that performs the simulation calculations is logically correct is called a. implementation. b. validation. c. verification. d. repetition.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.05 - 12.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.5 Simulation Considerations KEYWORDS: Bloom's: Remember 37. Using simulation models, we are able to explore the sensitivity of various decisions to some of the model's a. parameters. b. optimizations. c. uncontrollable input. d. None of these are correct.

ANSWER: a © Cengage. Testing Powered by Cognero.

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CH 12 - Simulation POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.12.03 - 12.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.3 Inventory Simulation KEYWORDS: Bloom's: Understand 38. Which of the following statements is NOT true about the advantages of simulations? a. Simulation models are flexible. b. Simulation models can be used to describe systems without requiring the assumptions that are often required by mathematical models. c. Generally, the larger the number of random variables a system has, the less likely that a simulation model will provide the most suitable approach for studying the system. d. Simulation models provide a convenient experimental "laboratory" for a real system.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.12.05 - 12.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.5 Simulation Considerations KEYWORDS: Bloom's: Understand 39. Which of the following statements is true regarding the advantages of using simulations? a. Its methodology can be used to model and learn about the behavior of complex systems. b. Simulations do not guarantee an optimal solution. c. Simulation models can be used to describe systems without requiring the assumptions often required by mathematical models. d. All of these are correct.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.12.05 - 12.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.5 Simulation Considerations KEYWORDS: Bloom's: Understand 40. The word "uniform" in the term "uniform random numbers" means a. all the numbers have the same number of digits. b. if one number is, say, 10 units above the mean, the next number will be 10 units below the mean. c. all the numbers are odd or all are even. d. each number has an equal probability of being drawn.

ANSWER: d POINTS: 1 DIFFICULTY: Easy © Cengage. Testing Powered by Cognero.

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CH 12 - Simulation LEARNING OBJECTIVES: IMS.ASWC.19.A12.01 - A12.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: Appendix 12.1 Probability Distributions for Random Variables KEYWORDS: Bloom's: Understand 41. The process of generating probabilistic inputs and computing the value of the output is called a. simulation. b. verification. c. validation. d. implementation.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.02 - 12.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.2 Simulation of Sanotronics Problem KEYWORDS: Bloom's: Remember 42. A graphical tool that helps describe the logic of the simulation model is a a. Gantt chart. b. histogram. c. flowchart. d. stem-and-leaf display.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.03 - 12.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.3 Inventory Simulation KEYWORDS: Bloom's: Remember 43. A set of values for the random variables is called a a. simulation. b. risk analysis. c. trial. d. None of these are correct.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.12.02 - 12.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.2 Simulation of Sanotronics Problem KEYWORDS: Bloom's: Remember © Cengage. Testing Powered by Cognero.

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CH 12 - Simulation 44. A trial-and-error approach to learning about the range of possible outputs for a model is known as a. what-if analysis. b. worst-case scenario analysis. c. static simulation modeling. d. base-case scenario.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.12.01 - 12.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 12.1 What-If Analysis KEYWORDS: Bloom's: Remember Subjective Short Answer 45. For the past 50 days, daily sales of laundry detergent in a large grocery store have been recorded (to the nearest 10).

Units Sold 30 40 50 60 70

Number of Times 8 12 15 10 5

a. Determine the relative frequency for each number of units sold. b. Suppose that the following random numbers were obtained using Excel: 0.12 0.96 0.53 0.80 0.95 0.10 0.40 0.45 0.77 0.29 Use these random numbers to simulate 10 days of sales.

ANSWER:

a. Units Sold 30 40 50 60 70 b.

Random Number 0.12 0.96 0.53 0.80 0.95 0.10 0.40 0.45 0.77 0.29

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Relative Frequency 0.12 0.24 0.32 0.26 0.06

Cumulative Frequency 0.12 0.36 0.68 0.94 1.00

Interval of Random Numbers 0.00 but less than 0.12 0.12 but less than 0.36 0.36 but less than 0.68 0.68 but less than 0.94 0.94 but less than 1.00

Units Sold 40 70 50 60 70 30 50 50 60 40 Page 13

CH 12 - Simulation

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.12.02 - 12.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 12.2 Simulation of Sanotronics Problem KEYWORDS: Bloom's: Analyze 46. The time required to set up lighting for a portrait studio is uniformly distributed between 12 and 20 minutes. Use the following random numbers to generate the setup time for 10 customers.

0.27

0.53

0.06

0.92

0.16

0.74

0.06

0.29

0.82

0.23

ANSWER:

Random Number Time 0.27 14.16 0.53 16.24 0.06 12.48 0.92 19.36 0.16 13.28 0.74 17.92 0.06 12.48 0.29 14.32 0.82 18.56 0.23 13.84 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.12.02 - 12.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 12.2 Simulation of Sanotronics Problem KEYWORDS: Bloom's: Analyze 47. Estimates of the financial information for a new product include the following information: Units Sold 600 800 1000

Probability 0.35 0.45 0.20

Fixed cost Variable cost Revenue

$8,000 $6/unit $22/unit

Use the random numbers 0.51, 0.97, 0.58, 0.22, and 0.16 to simulate five trials. What is the net profit for each trial?

ANSWER: Trial 1 2 3 4 5 Random number 0.51 0.97 0.58 0.22 0.16 Units sold 800 1000 800 600 600 Revenue 17,600 22,000 17,600 13,200 13,200 Variable cost 4800 6000 4800 3600 3600 Fixed cost 8000 8000 8000 8000 8000 Net profit 4800 8000 4800 1600 1600 POINTS:

1

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CH 12 - Simulation DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.12.02 - 12.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 12.2 Simulation of Sanotronics Problem KEYWORDS: Bloom's: Analyze 48. Seventy-five percent of calls arriving at a help line can be handled by the person who answers the phone, but the remaining 25% of them will need to be referred to someone else. Assume that every call requires one minute of attention by the person who answers the phone (either to answer the question or to figure out how the referral should be handled).

Calls that are referred need an additional amount of time, as given in the table below.

Time Required Probability 3 minutes 0.25 5 minutes 0.35 10 minutes 0.40 Callers are served on a first-come, first-served basis and are put on hold until the line is free. Use random numbers to simulate what happens to 10 callers. (Use the random numbers in order, from left to right, first row first, as you need them.) What percentage of your callers needs to be referred? Of those who had to be referred, what is the average referral time?

0.82 0.85

0.39 0.32

ANSWER:

0.16 0.64

0.79 0.90

0.56 0.73

Call 1 2 3 4 5 6 7 8 9 10

0.62 0.02 RN 0.82 0.16 0.79 0.62 0.13 0.04 0.42 0.81 0.85 0.64

0.13 0.76

0.04 0.03

Referred? yes no yes no no no no yes yes no

0.42 0.86

0.81 0.67

RN 0.39

Time 5

0.56

5

0.85 0.32

10 5

40% of the callers were referred. The average referral time was 6.25 minutes.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.12.04 - 12.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 12.4 Waiting Line Simulation KEYWORDS: Bloom's: Evaluate 49. On a visit to an amusement park, you pass someone who has just ridden a roller coaster and asks you for directions to the First Aid Station. Realizing that traffic at the First Aid Station would be something to study with simulation, you gather some information. Two EMTs staff the station, and patients wait and go to the first one available. People coming there can be divided into two groups: those who need something minor (e.g., Tylenol, a Band-Aid) or those who need more help.

Assume those in the first group constitute 25% of the patients and take five minutes to have their problem solved. Those in the second group need an uncertain amount of time, as given by a probability distribution. Develop a flowchart for this © Cengage. Testing Powered by Cognero.

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CH 12 - Simulation simulation problem.

ANSWER:

Let

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i = patient counter AT(i) = arrival time IAT(i) = interarrival time Start(i) = time patient i begins service ST(i) = service time Finish(i) = time patient leaves CT1 = time that EMT 1 completes current service CT2 = time that EMT 2 completes current service

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CH 12 - Simulation POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.12.04 - 12.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: KEYWORDS:

12.4 Waiting Line Simulation Bloom's: Create

50. Using the spreadsheet below, give the cell address that would have the formula shown.

Cell Formula =VLOOKUP(B18,$B$10:$C$12,2) =VLOOKUP(D23,$F$11:$G$14,2) =K19*($I$16-I19) =VLOOKUP(H27,$B$10:$C$12,2) =AVERAGE(L18:L27)

Belongs in Cell

ANSWER: Cell Formula =VLOOKUP(B18,$B$10:$C$12,2) =VLOOKUP(D23,$F$11:$G$14,2) =K19*($I$16-I19) =VLOOKUP(H27,$B$10:$C$12,2) =AVERAGE(L18:L27)

Belongs in Cell C18 E23 L19 I27 L28

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.12.02 - 12.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic © Cengage. Testing Powered by Cognero.

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CH 12 - Simulation TOPICS: KEYWORDS:

12.2 Simulation of Sanotronics Problem Bloom's: Evaluate

51. As the owner of a rent-a-car agency, you have determined the following statistics: Potential Rental Probability Probability Rentals Daily Duration 0 0.10 1 day 0.50 1 0.15 2 day 0.30 2 0.20 3 days 0.15 3 0.30 4 days 0.05 4 0.25 The gross profit is $40 per car per day rented. When there is demand for a car but none is available, there is a goodwill loss of $80 and the rental is lost. Each day a car is unused costs you $5 per car. Your firm initially has four cars. a. Conduct a 10-day simulation of this business using Row #1 below for demand and Row #2 for rental length.

Row #1: Row #2:

63 59

88 09

55 57

46 87

55 07

69 92

13 29

17 28

36 64

81 36

b. If your firm can obtain another car for $200 for 10 days, should you take the extra car?

ANSWER:

a. The 10-day profit is $885. b. The 10-day profit is $970. Take the extra car.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.12.05 - 12.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 12.5 Simulation Considerations KEYWORDS: Bloom's: Evaluate 52. Susan Winslow has two alternative routes to travel from her home in Olport to her office in Lewisburg. She can travel on Freeway 5 to Freeway 57 or on Freeway 55 to Freeway 91. The time distributions are as follows: Freeway 5 Relative Time Frequency 5 0.30 6 0.20 7 0.40 8 0.10

Freeway 57 Relative Time Frequency 4 0.10 5 0.20 6 0.35 7 0.20 8 0.15

Freeway 55 Relative Time Frequency 6 0.20 7 0.20 8 0.40 9 0.20

Freeway 91 Relative Time Frequency 3 0.30 4 0.35 5 0.20 6 0.15

Do a five-day simulation of each of the two combinations of routes using the random numbers below. Based on this simulation, which routes should Susan take if her objective is to minimize her total travel time?

Freeway 5 Freeway 57 Freeway 55 Freeway 91 ANSWER:

63 59 71 51

88 09 95 79

55 57 83 09

46 87 44 67

55 07 34 15

Freeways 5 and 57 have 62 minutes; Freeways 55 and 91 have 61 minutes; select 55 and 91.

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CH 12 - Simulation POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.12.05 - 12.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 12.5 Simulation Considerations KEYWORDS: Bloom's: Evaluate 53. Three airlines compete on the route between New York and Los Angeles. Stanton Marketing has performed an analysis of first class business travelers to determine their airline choice. Stanton has modeled this choice as a Markov process and has determined the following transition probabilities: Last Airline A B C

A 0.50 0.30 0.10

Next Airline B 0.30 0.45 0.35

C 0.20 0.25 0.55

a. Show the random number assignments that can be used to simulate the first class business traveler's next airline when her last airline is A, B, and C. b. Assume the traveler used airline C last. Simulate which airline the traveler will be using over her next 25 flights. What percent o her flights are on each of the three airlines? Use the following random numbers, going from left to right, top to bottom.

71 49 75 77 37 ANSWER:

95 88 12 76 46

83 56 03 57 85 a.

44 05 59 15 24 A Last

A Next B Next C Next b.

34 39 29 53 53 B Last

RNs 00–49 50–79 80–99

A Next B Next C Next

C Last RNs 00–29 30–74 75–99

A Next B Next C Next

RNs 00–09 10–44 45–99

20% of flights on airline A 48% of flights on airline B 32% of flights on airline C

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.12.04 - 12.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 12.4 Waiting Line Simulation KEYWORDS: Bloom's: Evaluate

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CH 13 - Decision Analysis True / False 1. Sample information with an efficiency rating of 100% is perfect information. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.05 - 13.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.5 Decision Analysis with Sample Information KEYWORDS: Bloom's: Remember 2. States of nature should be defined so that one and only one will actually occur. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.01 - 13.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.1 Problem Formulation KEYWORDS: Bloom's: Remember 3. Decision alternatives are structured so that several could occur simultaneously. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.01 - 13.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.1 Problem Formulation KEYWORDS: Bloom's: Understand 4. Square nodes in a decision tree indicate that a decision must be made. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.01 - 13.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.1 Problem Formulation © Cengage. Testing Powered by Cognero.

Page 1

CH 13 - Decision Analysis KEYWORDS:

Bloom's: Remember

5. Circular nodes in a decision tree are considered chance nodes. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.01 - 13.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.1 Problem Formulation KEYWORDS: Bloom's: Remember 6. Risk analysis helps the decision maker recognize the difference between the expected value of a decision alternative and the payoff that may actually occur. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.04 - 13.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.4 Risk Analysis and Sensitivity Analysis KEYWORDS: Bloom's: Remember 7. The expected value of an alternative can never be negative. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.03 - 13.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.3 Decision Making with Probabilities KEYWORDS: Bloom's: Understand 8. For a chance node, the expected value is the weighted average of the payoffs, where the weights are the state-of-nature probabilities. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.03 - 13.3 © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.3 Decision Making with Probabilities KEYWORDS: Bloom's: Understand 9. EVPI is always greater than or equal to EVSI. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.05 - 13.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.5 Decision Analysis with Sample Information KEYWORDS: Bloom's: Understand 10. After all probabilities and payoffs are placed on a decision tree, the decision maker calculates expected values at stateof-nature nodes and makes selections at decision nodes. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.05 - 13.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.5 Decision Analysis with Sample Information KEYWORDS: Bloom's: Understand 11. A decision strategy is a sequence of decisions and chance outcomes where the decisions chosen depend on the yet-tobe-determined outcomes of chance events. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.05 - 13.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.5 Decision Analysis with Sample Information KEYWORDS: Bloom's: Remember 12. Under the minimax regret approach to decision making, EVPI equals the expected regret that is associated with the minimax decision. a. True b. False ANSWER: True POINTS: 1 © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.02 - 13.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.2 Decision Making Without Probabilities KEYWORDS: Bloom's: Remember 13. The expected value approach is more appropriate for a one-time decision than a repetitive decision. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.03 - 13.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.3 Decision Making with Probabilities KEYWORDS: Bloom's: Understand 14. Maximizing the expected payoff and minimizing the expected opportunity loss result in the same recommended decision. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.02 - 13.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.2 Decision Making Without Probabilities KEYWORDS: Bloom's: Understand 15. The expected value of sample information can never be less than the expected value of perfect information. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.05 - 13.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.5 Decision Analysis with Sample Information KEYWORDS: Bloom's: Understand 16. Regret is the difference between the payoff associated with a particular decision alternative and the payoff associated with the decision that would yield the most desirable payoff for a given state of nature. a. True b. False ANSWER: True © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.02 - 13.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.2 Decision Making Without Probabilities KEYWORDS: Bloom's: Remember 17. The primary value of decision trees is that they provide a useful way to organize how operations managers think about complex multiphase decisions. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.01 - 13.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.1 Problem Formulation KEYWORDS: Bloom's: Understand 18. A high efficiency rating indicates that the sample information is almost as good as perfect information. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.05 - 13.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.5 Decision Analysis with Sample Information KEYWORDS: Bloom's: Understand 19. When the expected value approach is used to select a decision alternative, the payoff that actually occurs will usually have a value different from the expected value. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.03 - 13.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.3 Decision Making with Probabilities KEYWORDS: Bloom's: Understand 20. The decision alternative with the best expected monetary value will always be the most desirable decision. a. True b. False © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Understand 21. When monetary value is not the sole measure of the true worth of the outcome to the decision maker, monetary value should be replaced by utility. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Understand 22. Utility is the term for a measure of the total worth or relative desirability of a particular outcome. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Remember 23. As long as the monetary value of payoffs stays within a range that the decision maker considers reasonable, selecting the decision alternative with the best expected value usually leads to selection of the most preferred decision. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Understand 24. The expected utility is the utility of the expected monetary value. a. True © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Understand 25. A risk avoider is a decision maker who would choose a guaranteed payoff over a lottery with a superior expected payoff. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Remember 26. The utility function for a risk avoider typically shows a diminishing marginal return for money. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Understand 27. Given two decision makers, one a risk taker and the other a risk avoider, the risk avoider will show a diminishing marginal return for money. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Understand © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis 28. When the expected utility approach and the expected value approach applied to monetary payoffs result in the same action, these are characteristics generally associated with a risk-neutral decision maker. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Understand Multiple Choice 29. The options from which a decision maker chooses a course of action are a. called the decision alternatives. b. not under the control of the decision maker. c. the same as the states of nature. d. None of these are correct.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.01 - 13.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.1 Problem Formulation KEYWORDS: Bloom's: Remember 30. States of nature a. are the possible outcomes for a chance event. b. can be selected by the decision maker. c. cannot be enumerated by the decision maker. d. None of these are correct.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.01 - 13.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.1 Problem Formulation KEYWORDS: Bloom's: Understand 31. A payoff a. is always measured in profit. b. is always measured in cost. c. exists for each pair of a decision alternative and a state of nature. © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis d. exists for each state of nature.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.01 - 13.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.1 Problem Formulation KEYWORDS: Bloom's: Remember 32. Making a good decision a. requires probabilities for all states of nature. b. requires a clear understanding of decision alternatives, states of nature, and payoffs. c. implies that a desirable outcome will occur. d. All of these are correct.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.02 - 13.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.2 Decision Making Without Probabilities KEYWORDS: Bloom's: Understand 33. A decision tree a. presents all decision alternatives first and follows them with all states of nature. b. presents all states of nature first and follows them with all decision alternatives. c. alternates the decision alternatives and states of nature. d. arranges decision alternatives and states of nature in their natural chronological order.

ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.01 - 13.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.1 Problem Formulation KEYWORDS: Bloom's: Remember 34. Which of the following methods for decision making best protects the decision maker from undesirable results? a. the optimistic approach b. the conservative approach c. minimum regret d. minimax regret ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.02 - 13.2 © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.2 Decision Making Without Probabilities KEYWORDS: Bloom's: Understand 35. Sensitivity analysis a. considers how sensitive the decision maker is to risk. b. considers how changes in the number of states of nature can impact the recommended decision. c. considers how changes in various aspects of the problem affect the recommended decision alternative. d. None of these are correct.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.01 - 13.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.1 Problem Formulation KEYWORDS: Bloom's: Remember 36. To find the expected value of sample information (EVSI), a. use prior and sample information probabilities to calculate revised probabilities. b. use indicator probabilities to calculate prior probabilities. c. use sample information to revise the sample information probabilities. d. None of these are correct.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.13.05 - 13.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.5 Decision Analysis with Sample Information KEYWORDS: Bloom's: Understand 37. If P(high) = 0.3, P(low) = 0.7, P(favorable | high) = 0.9, and P(unfavorable | low) = 0.6, then P(favorable) = a. 0.10. b. 0.27. c. 0.30. d. 0.55.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.13.06 - 13.6 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.6 Computing Branch Probabilities with Bayes’ Theorem KEYWORDS: Bloom's: Apply 38. The efficiency of sample information is © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis a. EVSI*(100%). b. EVSI/EVPI*(100%). c. EVwoSI/EVwoPI*(100%). d. EVwSI/EVwoSI*(100%).

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.13.05 - 13.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.5 Decision Analysis with Sample Information KEYWORDS: Bloom's: Apply 39. Decision tree probabilities refer to the probability of a. finding the optimal strategy. b. the decision being made. c. overlooked choices. d. an uncertain event occurring.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.13.05 - 13.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.5 Decision Analysis with Sample Information KEYWORDS: Bloom's: Understand 40. For a maximization problem, the conservative approach is often referred to as the a. minimax approach. b. maximin approach. c. maximax approach. d. minimin approach.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.02 - 13.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.2 Decision Making Without Probabilities KEYWORDS: Bloom's: Remember 41. For a minimization problem, the optimistic approach is often referred to as the a. minimax approach. b. maximin approach. c. maximax approach. d. minimin approach.

ANSWER: d © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.02 - 13.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.2 Decision Making Without Probabilities KEYWORDS: Bloom's: Remember 42. For a maximization problem, the optimistic approach is often referred to as the a. minimax approach. b. maximin approach. c. maximax approach. d. minimin approach.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.02 - 13.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.2 Decision Making Without Probabilities KEYWORDS: Bloom's: Remember 43. For a minimization problem, the conservative approach is often referred to as the a. minimax approach. b. maximin approach. c. maximax approach. d. minimin approach.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.02 - 13.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.2 Decision Making Without Probabilities KEYWORDS: Bloom's: Remember 44. In an influence diagram, decision nodes are represented by a. circles or ovals. b. squares or rectangles. c. diamonds. d. triangles.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.01 - 13.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.1 Problem Formulation © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis KEYWORDS:

Bloom's: Remember

45. Which of the following approaches to decision making requires knowledge of the probabilities of the states of nature? a. minimax regret b. maximin c. expected value d. conservative ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.03 - 13.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.3 Decision Making with Probabilities KEYWORDS: Bloom's: Understand 46. Decision tree probabilities are primarily used to a. analyze more complex problems and to identify an optimal sequence of decisions. b. analyze less complex problems while identifying the optimal sequence of decisions. c. find overlooked choices to the problem. d. None of these are correct.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.13.01 - 13.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.1 Problem Formulation KEYWORDS: Bloom's: Understand 47. Which of the following is NOT an advantage of using decision tree analysis? a. the ability to see clearly what decisions must be made b. the ability to see clearly in what sequence the decisions must occur c. the ability to see clearly the interdependence of decisions d. the ability to see clearly the future outcome of a decision ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.13.03 - 13.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.3 Decision Making with Probabilities KEYWORDS: Bloom's: Understand 48. A decision tree provides a. the only method for analyzing decisions. b. a deterministic approach to decision analysis. © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis c. the absolute value of the decision. d. an objective way of determining the relative value of each decision alternative.

ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.05 - 13.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.5 Decision Analysis with Sample Information KEYWORDS: Bloom's: Understand 49. The difference between the expected value of an optimal strategy based on sample information and the "best" expected value without any sample information is called the a. information sensitivity. b. expected value of sample information. c. expected value of perfect information. d. efficiency of sample information.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.13.05 - 13.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.5 Decision Analysis with Sample Information KEYWORDS: Bloom's: Remember 50. The purchase of insurance and lottery tickets shows that people make decisions based on a. expected value. b. sample information. c. utility. d. maximum likelihood.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Understand 51. The expected utility approach a. does not require probabilities. b. leads to the same decision as the expected value approach. c. is most useful when excessively large or small payoffs are possible. d. requires a decision tree.

ANSWER: c POINTS: 1 © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Understand 52. Utility reflects the decision maker's attitude toward a. probability and profit. b. profit, loss, and risk. c. risk and regret. d. probability and regret.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Understand 53. The probability for which a decision maker cannot choose between a certain amount and a lottery based on that probability is the a. indifference probability. b. lottery probability. c. uncertain probability. d. utility probability.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Remember 54. A decision maker has chosen 0.4 as the probability for which he cannot choose between a certain loss of 10,000 and the lottery p(−25,000) + (1 − p)(5000). If the utility of −25,000 is 0 and of 5000 is 1, then the utility of −10,000 is a. 0.5. b. 0.6. c. 0.4. d. 4.

ANSWER: b POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Analytic © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis TOPICS: KEYWORDS:

13.7 Utility Theory Bloom's: Apply

55. When the decision maker prefers a guaranteed payoff value that is smaller than the expected value of the lottery, the decision maker is a(n) a. risk avoider. b. risk taker. c. optimist. d. optimizer.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Remember 56. A decision maker whose utility function graphs as a straight line is a. conservative. b. a risk taker. c. risk neutral. d. a risk avoider.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Remember 57. When the utility function for a risk-neutral decision maker is graphed (with monetary value on the horizontal axis and utility on the vertical axis), the function appears as a(n) a. convex curve. b. concave curve. c. 'S' curve. d. straight line.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Understand Subjective Short Answer © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis 58. East West Distributing is in the process of trying to determine where it should schedule next year's production of a popular line of kitchen utensils that it distributes. Manufacturers in four different countries have submitted bids to East West. However, a pending trade bill in Congress will greatly affect the cost to East West due to proposed tariffs, favorable trading status, etc.

After careful analysis, East West has determined the following cost breakdown for the four manufacturers (in $1000s) based on whether or not the trade bill passes: Country A Country B Country C Country D a. b.

Bill Passes 260 320 240 275

Bill Fails 210 160 240 210

If East West estimates that there is a 40% chance of the bill passing, which country should it choose for manufacturing?

Over what range of values for the "bill passing" will the solution in part (a) remain optimal?

ANSWER:

a. Using the EV approach: EV(A) = 230, EV(B) = 224, EV(C) = 240; choose Country B (lowest EV) b. As long as the probability of the bill passing is less than 0.455, East West should choose Country B.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.03 - 13.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.3 Decision Making with Probabilities KEYWORDS: Bloom's: Evaluate © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis 59. Super Cola is also considering the introduction of a root beer drink. The company thinks the probability that the product will be a success is 0.6. The payoff table is as follows:

Produce (d1) Do not produce (d2)

Success (s1) $250,000 − 50,000

Failure (s2) −$300,000 − 20,000

The company has a choice of two research firms to obtain information for this product. Stanton Marketing has market indicators I1 and I2 for which P(I1 | s1) = 0.7 and P(I1 | s2) = 0.4. New World Marketing has indicators J1 and J2 for which P(J1 | s1) = 0.6 and P(J1 | s2) = 0.3. a. What is the optimal decision if neither firm is used? Over what probability of success range is this decision optimal? b. What is the EVPI? c. Find the EVSIs and efficiencies for Stanton and New World. d. If both firms charge $5,000, which firm should be hired? e. If Stanton charges $10,000 and New World charges $4000, which firm should Super Cola hire?

ANSWER:

a.

Introduce root beer; p ≤ 0.483 b.

EVPI = $112,000 NOTE: The answers to parts (c) through (e) are very sensitive to rounding error.

Figures in parentheses are for two decimal places only. c.

Stanton: EVSI = $13,200 ($11,862) Efficiency = 0.118 (0.106) New World: EVSI = $6400 ($6424) Efficiency = 0.057 (0.057) d.

Hire Stanton (Stanton) e.

Hire New World (Stanton) POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.06 - 13.6 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.6 Computing Branch Probabilities with Bayes’ Theorem KEYWORDS: Bloom's: Evaluate 60. Dollar Department Stores has just acquired the chain of Wenthrope and Sons Custom Jewelers. Dollar has received an offer from Harris Diamonds to purchase the Wenthrope store on Grove Street for $120,000. Dollar has determined probability estimates of the store's future profitability, based on economic outcomes, as: P($80,000) = 0.2, P($100,000) = 0.3, P($120,000) = 0.1, and P($140,000) = 0.4. a. Should Dollar sell the store on Grove Street? b. What is the EVPI? c. Dollar can have an economic forecast performed, costing $10,000, that produces indicators I1 and I2, for which P(I1 | 80,000) = 0.1; P(I1 | 100,000) = 0.2; P(I1 | 120,000) = 0.6; P(I1 | 140,000) = 0.3. Should Dollar purchase the forecast?

ANSWER:

a. b. c.

Yes, Dollar should sell the store.

EVPI = $8000 No; the survey cost exceeds EVPI.

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CH 13 - Decision Analysis POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.06 - 13.6 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.6 Computing Branch Probabilities with Bayes’ Theorem KEYWORDS: Bloom's: Evaluate 61. An appliance dealer must decide how many (if any) new microwave ovens to order for next month. The ovens cost $220 and sell for $300. Because the oven company is coming out with a new product line in two months, any ovens not sold next month will have to be sold at the dealer's half price clearance sale. Additionally, the appliance dealer feels he suffers a loss of $25 for every oven demanded when he is out of stock. On the basis of past months' sales data, the dealer estimates the probabilities of monthly demand (D) for 0, 1, 2, or 3 ovens to be 0.3, 0.4, 0.2, and 0.1, respectively.

The dealer is considering conducting a telephone survey on customers' attitudes towards microwave ovens. The results of the survey will either be favorable (F), unfavorable (U), or no opinion (N). The dealer's probability estimates for the survey results based on the number of units demanded are: P(F | D = 0) = 0.1 P(F | D = 1) = 0.2 a. b. c. d.

P(F | D = 2) = 0.3 P(F | D = 3) = 0.9

P(U | D = 0) = 0.8 P(U | D = 1) = 0.3

P(U | D = 2) = 0.1 P(U | D = 3) = 0.1

What is the dealer's optimal decision without conducting the survey?

What is the EVPI? Based on the survey results, what is the optimal decision strategy for the dealer?

What is the maximum amount he should pay for this survey?

ANSWER: Ovens Ordered 0 1 2 3

0 0 −70 −140 −210

Demand for Ovens 1 2 −25 −50 80 55 10 160 −60 90

3 −75 30 135 240

a. Order one oven: EV = $25.00 b. EVPI = $63.00 c. Favorable: order 2; Unfavorable: order 0; No opinion: order 1 d. EVSI = $9.10 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.06 - 13.6 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.6 Computing Branch Probabilities with Bayes’ Theorem KEYWORDS: Bloom's: Evaluate 62. Lakewood Fashions must decide how many lots of assorted ski wear to order for its three stores. Information on pricing, sales, and inventory costs has led to the following payoff table, in thousands. Show a regret table.

Order Size 1 lot 2 lots

Low 12 9

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Demand Medium 15 25

High 15 35 Page 19

CH 13 - Decision Analysis 3 lots

6

35

60

a. What decision should be made by the optimist? b. What decision should be made by the conservative? c. What decision should be made using minimax regret?

ANSWER: a.

3 lots b.

1 lot c.

3 lots Regret Table Order Size 1 lot 2 lots 3 lots

Low 0 3 6

Demand Medium 20 10 0

High 45 25 0

Maximum Regret 45 25 6

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.02 - 13.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.2 Decision Making Without Probabilities KEYWORDS: Bloom's: Evaluate 63. The table shows both prospective profits and losses for a company, depending on what decision is made and what state of nature occurs. Use the information to determine what the company should do. Show your work (regret table).

Decision d1 d2 d3 d4 a. b. c.

s1 30 100 −80 20

State of Nature s2 80 30 −10 20

s3 −30 −40 120 20

If an optimistic strategy is used.

If a conservative strategy is used.

If minimax regret is the strategy.

ANSWER: a. d3 b. d1 c. d4 Regret Table: Decision d1 d2 d3 d4 © Cengage. Testing Powered by Cognero.

State of Nature s1 s2 s3 70 0 150 0 50 160 180 90 0 80 60 100

Maximum Regret 150 160 180 100 Page 20

CH 13 - Decision Analysis POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.02 - 13.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.2 Decision Making Without Probabilities KEYWORDS: Bloom's: Evaluate 64. A decision maker has developed the following decision tree. How sensitive is the choice between N and P to the probabilities of states of nature U and V, and which would you choose?

ANSWER: Choose N if p ≤ 0.78.

POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.13.04 - 13.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.4 Risk Analysis and Sensitivity Analysis KEYWORDS: Bloom's: Evaluate 65. If p is the probability of Event 1 and (1 − p) is the probability of Event 2, for what values of p would you choose A? B? C? Values in the table are payoffs.

Choice/Event Event 1 Event 2 A 0 20 B 4 16 C 8 0 ANSWER: Choose A if p ≤ 0.5, choose B if 0.5 ≤ p ≤ 0.8, and choose C if p ≥ 0.8.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.04 - 13.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.4 Risk Analysis and Sensitivity Analysis KEYWORDS: Bloom's: Evaluate 66. Fold back this decision tree. Clearly state the decision strategy you determine.

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CH 13 - Decision Analysis

ANSWER:

Choose A. If F happens, choose K. POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.05 - 13.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.5 Decision Analysis with Sample Information KEYWORDS: Bloom's: Evaluate 67. Use graphical sensitivity analysis to determine the range of values of the probability of state of nature s1 over which each of the decision alternatives has its largest expected value.

Decision d1 d2

State of Nature s1 s2 8 10 4 16

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CH 13 - Decision Analysis 10

d3 ANSWER:

0 EV(d1) and EV(d2) intersect at p = 0.6. EV(d1) and EV(d3) intersect at p = 0.8333. Therefore, when 0 ≤ p ≤ 0.6, choose d2. When 0.6 ≤ p ≤0.8333, choose d1. When p ≥ 0.8333, choose d3.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.04 - 13.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.4 Risk Analysis and Sensitivity Analysis KEYWORDS: Bloom's: Evaluate 68. A paint company has three sources for buying bright red pigment for its paints: Vietnam, Taiwan, or Thailand.

Unfortunately, the pigment is made from a bush whose annual growth is heavily dependent upon the amount of rainfall during the growing season. The tables below show probabilities and prices for wet, dry, and normal growing seasons.

Probabilities Dry Normal 0.2 0.3 0.3 0.1 0.4 0.2

Vietnam Taiwan Thailand

Wet 0.5 0.6 0.4

Vietnam Taiwan Thailand

Price/Pound ($) Wet Dry Normal 0.95 1.10 1.00 0.85 1.20 0.98 0.90 1.15 1.05

What country should the company select, and what is the expected value (price) associated with it?

ANSWER: EV(Vietnam) = 0.5(0.95) + 0.2(l.10) + 0.3(1.00) = $0.995 EV(Taiwan) = 0.6(0.85) + 0.3(l.20) + 0.1(0.98) = $0.968 EV(Thailand) = 0.4(0.90) + 0.4(1.15) + 0.2(l.05) = $1.03 Select Taiwan, because it has the lowest EV. POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.05 - 13.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.5 Decision Analysis with Sample Information © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis KEYWORDS:

Bloom's: Evaluate

69. A regional fast‑food restaurant is considering an expansion program. The major factor influencing the success of such a program is the future level of interest rates. It is estimated that there is a 20% chance that interest rates will increase by 2 percentage points, a 50% chance that they will remain the same, and a 30% chance that they will decrease by 2 percentage points. The alternatives they are considering and possible payoffs are shown in the table below. Which alternative is best, based on expected value?

Build 50 restaurants Build 25 restaurants Do nothing ANSWER:

Rates up 2% –$200,000 –$115,000 –$70,000

Rates unchanged $50,000 $26,000 0

Rates down 2% $150,000 $80,000 $5,000

EV(A—50 restaurants) = 0.2(–200,000) + 0.5(50,000) + 0.3(150,000) = $30,000 EV(B—25 restaurants) = 0.2(–115,000) + 0.5(26,000) + 0.3(80,000) = $14,000 EV(C—do nothing) = 0.2(–70,000) + 0.5(0) + 0.3(5000) = –$12,500 The company should build 50 restaurants; EV of $30,000.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.05 - 13.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.5 Decision Analysis with Sample Information KEYWORDS: Bloom's: Evaluate 70. A chemical company is trying to decide whether to build a pilot plant now for a new chemical process or to build the full plant now. If it builds a pilot plant now, it could expand later to a full plant or license the plant to another company. It would cost $2 million to build the pilot plant and another $2 million later to expand it. If the company builds the full plant now, it would cost $3.5 million to construct.

The returns the company expects to get from the full production plant depend on the market. There is a 60% chance the market will be robust, a 30% chance it will remain stable, and a 10% chance it will become stagnate. The returns are estimated to be $5 million if it is robust, $3 million if it is stable, and $1 million if it is stagnate.

Before the company expands the pilot plant, it plans to conduct a comprehensive study. Based on past experience, it expects the study to report a 60% chance of favorable outcome for expansion and a 40% unfavorable chance. In either case, it will have to decide whether to expand to a full plant or license the pilot plant. If the report is favorable and the company licenses it, the company expects to get $3 million. However, if the report is unfavorable and the company licenses it, the company will only get $1 million.

Develop a decision tree for this problem and determine the optimal decision strategy.

ANSWER:

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CH 13 - Decision Analysis

The company should build the pilot plant. If the report is unfavorable, the company should expand. If the report is favorable, the company should license. EV of $600,000.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIV IMS.ASWC.19.13.05 - 13.5 ES: NATIONAL STANDARD United States - BUSPROG: Analytic S: TOPICS: 13.5 Decision Analysis with Sample Information KEYWORDS: Bloom's: Create © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis 71. A manufacturing company is considering expanding its production capacity to meet a growing demand for its product line of air fresheners. The alternatives are to build a new plant, expand the old plant, or do nothing. The marketing department estimates a 35% probability of a market upturn, a 40% probability of a stable market, and a 25% probability of a market downturn. Georgia Swain, the firm's capital appropriations analyst, estimates the following annual returns for these alternatives: Market Upturn

Stable Market

Market Downturn

Build new plant

$690,000

$(130,000)

$(150,000)

Expand old plant

490,000

(45,000)

(65,000)

Do nothing

50,000

0

(20,000)

a. Use a decision tree to analyze these decision alternatives. b. What should the company do? c. What returns will accrue to the company if your recommendation is followed?

ANSWER: a.

b. Decision: Build the new plant. c. Returns to accrue: $690,000; ($130,000); or ($150,000) POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.05 - 13.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.5 Decision Analysis with Sample Information KEYWORDS: Bloom's: Create 72. Sunshine Manufacturing Company has developed a unique new product and must now decide between two facility plans. The first alternative is to build a large new facility immediately. The second alternative is to build a small plant © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis initially and to consider expanding it to a larger facility three years later if the market has proven favorable.

Marketing has provided the following probability estimates for a 10-year plan: First 3-Year Demand

Next 7-Year Demand

Probability

Unfavorable Unfavorable Favorable Favorable

Unfavorable Favorable Favorable Unfavorable

0.2 0.0 0.7 0.1

If the small plant is expanded, the probability of demands over the remaining seven years is 7/8 for favorable and 1/8 for unfavorable. The accounting department has provided the payoff for each outcome.

Demand Favorable, favorable Favorable, unfavorable Unfavorable, unfavorable Favorable, favorable Favorable, unfavorable Favorable, favorable Favorable, unfavorable Unfavorable, unfavorable

Facility Plan

Payoff

1 1 1 2—expanded 2—expanded 2—not expanded 2—not expanded 2—not expanded

$5,000,000 2,500,000 1,000,000 4,000,000 100,000 1,500,000 500,000 300,000

Use these estimates to analyze Sunshine's facility decision. a. Perform a complete decision tree analysis. b. Recommend a strategy to Sunshine. c. Determine what payoffs will result from your recommendation.

ANSWER:

a.

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CH 13 - Decision Analysis

b. Recommended strategy: Build the large plant. c. Possible payoffs that will result: $5,000,000; $2,500,000; or $1,000,000 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.05 - 13.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.5 Decision Analysis with Sample Information KEYWORDS: Bloom's: Create 73. A Pacific Northwest lumber company is considering the expansion of one of its mills. The question is whether to do it now or wait one year and reconsider. If it expands now, the major factors are the state of the economy and the level of interest rates. The combination of these two factors results in five possible situations. If it does not expand now, only the state of the economy is important and three conditions characterize the possibilities. The following table summarizes the situation: Expand Very favorable Favorable Neutral Unfavorable Very unfavorable Don't expand Expansion Steady Contraction

Probabilities

Revenues

0.2 0.2 0.1 0.3 0.2

$80,000 $60,000 $20,000 –$20,000 –$30,000

0.2 0.5 0.3

$50,000 $30,000 $10,000

a. Draw the decision tree for this problem. b. What is the expected value for expanding? c. What is the expected value for not expanding? © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis d. Based on expected value, what should the company’s decision(s) be?

ANSWER:

a.

b. EV(Expanding) = 0.2(80,000) + 0.2(60,000) + 0.1(20,000) + 0.3(–20,000) + 0.2(–30,000) = $18,000 c. EV(Not Expanding) = 0.2(50,000) + 0.5(30,000) + 0.3(10,000) = $28,000 d. Do not expand.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.05 - 13.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.5 Decision Analysis with Sample Information KEYWORDS: Bloom's: Create 74. For the payoff table below, the decision maker will use P(s1) = 0.15, P(s2) = 0.5, and P(s3) = 0.35.

Decision d1 d2 a. b.

State of Nature s2 1000 −2000

s3 10,000 40,000

What alternative would be chosen according to expected value?

For a lottery having a payoff of 40,000 with probability p and −15,000 with probability (1 − p), the decision maker expressed the following indifference probabilities: Payoff 10,000 1000 −2000 −5000

c.

s1 −5000 −15,000

Probability 0.85 0.60 0.53 0.50

Let U(40,000) = 10 and U(−15,000) = 0 and find the utility value for each payoff.

What alternative would be chosen according to expected utility?

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CH 13 - Decision Analysis ANSWER:

a. b.

EV(d1) = 3250 and EV(d2) = 10,750, so choose d2.

Payoff Probability Utility 10,000 0.85 8.5 1000 0.60 6.0 0.53 5.3 −2000 0.50 5.0 −5000

c.

EU(d1) = 6.725 and EU(d2) = 6.15, so choose d1.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Evaluate 75. A decision maker who is considered to be a risk taker is faced with this set of probabilities and payoffs.

Decision d1 d2 d3 Probability

s1 5 −25 −50 0.30

State of Nature s2 10 0 −10 0.35

s3 20 50 80 0.35

For the lottery p(80) + (1 − p)(−50), this decision maker has assessed the following indifference probabilities: Payoff 50 20 10 5 0 −10 −25

Probability 0.60 0.35 0.25 0.22 0.20 0.18 0.10

Rank the decision alternatives on the basis of expected value and on the basis of expected utility.

ANSWER: EV(d1) = 12 EV(d2) = 10 EV(d3) = 9.5 EU(d1) = 2.76 EU(d3) = 3.1 EU(d3) = 4.13 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Evaluate © Cengage. Testing Powered by Cognero.

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CH 13 - Decision Analysis 76. Three decision makers have assessed utilities for the problem whose payoff table appears below.

Decision d1 d2 d3 Probability Payoff 300 200 150 100 −100 a. b. c.

State of Nature s2 100 150 200 0.6

s1 500 200 −100 0.2

s3 −400 100 300 0.2

Indifference Probability for Person A B C 0.95 0.68 0.45 0.94 0.64 0.32 0.91 0.62 0.28 0.89 0.60 0.22 0.75 0.45 0.10

Plot the utility function for each decision maker.

Characterize each decision maker's attitude toward risk.

Which decision will each person prefer?

ANSWER:

a.

b. c.

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Person A is a risk avoider, Person B is fairly risk neutral, and Person C is a risk avoider.

For person A, EU(d1) = 0.734 EU(d2) = 0.912 EU(d3) = 0.904 For person B, EU(d1) = 0.56 EU(d2) = 0.62 EU(d3) = 0.61 For person C, EU(d1) = 0.332 EU(d2) = 0.276 EU(d3) = 0.302 Page 31

CH 13 - Decision Analysis Decision 1 would be chosen by person C. Decision 2 would be chosen by persons A and B.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Evaluate 77. A decision maker has the following utility function: Payoff 200 150 50 0 −50

Indifference Probability 1.00 0.95 0.75 0.60 0

What is the risk premium for the payoff of 50?

ANSWER: EV = 0.75(200) + 0.25(−50) = 137.50 Risk premium is 137.50 − 50 = 87.50.

POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Analyze 78. Determine decision strategies based on expected value and on expected utility for this decision tree. Use the following utility function: Payoff 500 350 300 180 100 40 20 0

Indifference Probability 1.00 0.89 0.84 0.60 0.43 0.20 0.13 0

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CH 13 - Decision Analysis

ANSWER:

Let U(500) = 1 and U(0) = 0. Then After Branch A J K B C

Expected Value 120 316 150 127.2 100

Expected Utility 0.336 0.680 0.522 0.381 0.430

Based on expected value, the decision strategy is to select B. If G happens, select J. Based on expected utility, it is best to choose C. POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Analyze 79. Burger Prince Restaurant is considering the purchase of a $100,000 fire insurance policy. The fire statistics indicate that in a given year the probability of property damage in a fire is as follows: Fire Damage Probability a.

$100,000 0.006

$75,000 0.002

$50,000 0.004

$25,000 0.003

$10,000 0.005

$0 0.980

If Burger Prince was risk neutral, how much would it be willing to pay for fire insurance?

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CH 13 - Decision Analysis b.

If Burger Prince has the utility values given below, approximately how much would it be willing to pay for fire insurance?

Loss Utility

$100,000 0

ANSWER:

$75,000 30

a. b.

$50,000 60

$25,000 85

$10,000 95

$5,000 99

$0 100

$1,075 $5,000

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Evaluate 80. Super Cola is considering the introduction of a new eight-oz. root beer. The probability that the root beer will be a success is believed to equal 0.6. The payoff table is as follows:

Produce Do not produce

Success (s1) $250,000 −$50,000

Failure (s2) −$300,000 −$20,000

Company management has determined the following utility values: Amount Utility a. b.

−$20,000 60

$250,000 100

−$50,000 55

−$300,000 0

Is the company a risk taker, risk averse, or risk neutral?

What is Super Cola's optimal decision?

ANSWER:

a. b.

Risk averse Produce root beer as long as p  60/105 = 0.571

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Evaluate 81. Chez Paul is contemplating either opening another restaurant or expanding its existing location. The payoff table for these two decisions is as follows: Decision New restaurant Expand

s1 −$80,000 −40,000

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State of Nature s2 $20,000 20,000

s3 $160,000 100,000 Page 34

CH 13 - Decision Analysis Paul has calculated the indifference probability for the lottery having a payoff of $160,000 with probability p and −$80,000 with probability (1 −p) as follows: Amount −$ 40,000 20,000 100,000 a. b.

c.

Indifference Probability (p) 0.4 0.7 0.9

Is Paul a risk avoider, a risk taker, or risk neutral?

Suppose Paul has defined the utility of −$80,000 to be 0 and the utility of $160,000 to be 80.

What would be the utility values for −$40,000, $20,000, and $100,000 based on the indifference probabilities?

Suppose P(s1) = 0.4, P(s2) = 0.3, and P(s3) = 0.3. Which decision should Paul make?

Compare with the decision using the expected value approach.

ANSWER:

a. b.

A risk avoider Amount −$ 40,000 20,000 100,000

c.

Decision is d2; EV criterion decision would be d1.

Utility 32 56 72

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Evaluate 82. The Dollar Department Store chain has the opportunity of acquiring 3, 5, or 10 leases from the bankrupt Granite Variety Store chain. Dollar estimates the profit potential of the leases depends on the state of the economy over the next five years. There are four possible states of the economy as modeled by Dollar Department Stores, and its president estimates P(s1) = 0.4, P(s2) = 0.3, P(s3) = 0.1, and P(s4) = 0.2. The utility has also been estimated. Given the payoffs (in $1,000,000s) and utility values below, which decision should Dollar make?

Payoff Table Decision d1 — buy 10 leases d2 — buy 5 leases d3 — buy 3 leases d4 — do not buy

s1 10 5 2 0

State of the Economy over the Next 5 Years s2 s3 5 0 0 −1 1 0 0 0

+10 +10

+5 +5

s4 −20 −10 −1 0

Utility Table Payoff (in $1,000,000s) Utility

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+2 +2

0 0

−1 −1

−10 −20

−20 −50

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CH 13 - Decision Analysis ANSWER:

Buy three leases POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Evaluate 83. Consider the following problem with four states of nature, three decision alternatives, and the following payoff table (in $s):

The indifference probabilities for three individuals are as follows: Payoff $ 2600 400 200 0 – 200 –1400

Person 1 1.00 0.40 0.35 0.30 0.25 0

Person 2 1.00 0.45 0.40 0.35 0.30 0

Person 3 1.00 0.55 0.50 0.45 0.40 0

a. Classify each person as a risk avoider, risk taker, or risk neutral. b. For the payoff of $400, what premium will the risk avoider pay to avoid risk? What premium will the risk taker pay to have the opportunity of the high payoff? c. Suppose each state is equally likely. What are the optimal decisions for each of these three people? s1 s2 s3 s4 d1 200 2600 –1400 200 d2 0 200 – 200 200 d3 –200 400 0 200 ANSWER:

a. Person 1 is a risk taker, Person 2 is risk neutral, and Person 3 is a risk avoider. b. The risk avoider would pay $400. The risk taker would pay $200. c. Person 1, d1; Person 2, d1; Person 3, d1

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Evaluate 84. Metropolitan Cablevision has the choice of using one of three DVR systems. Profits are believed to be a function of customer acceptance. The payoff to Metropolitan for the three systems is as follows: Acceptance Level High

System I $150,000

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II $200,000

III $200,000 Page 36

CH 13 - Decision Analysis Medium Low

80,000 20,000

20,000 – 50,000

80,000 –100,000

The probabilities of customer acceptance for each system are as follows: System Acceptance Level High Medium Low

I 0.4 0.3 0.3

II 0.3 0.4 0.3

III 0.3 0.5 0.2

The first vice president believes that the indifference probabilities for Metropolitan should be as follows: Amount $150,000 80,000 20,000 – 50,000

Probability 0.90 0.70 0.50 0.25

The second vice president believes Metropolitan should assign the following utility values: Amount $200,000 150,000 80,000 20,000 – 50,000 –100,000

Utility 125 95 55 30 10 0

a. Which vice president is a risk taker? Which one is risk averse? b. Which system will each vice president recommend? c. Which system would a risk-neutral vice president recommend?

ANSWER: a. Risk taker, second vice president Risk avoider, first vice president b. First vice president, System I Second vice president, System III c. Risk-neutral vice president, System I POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.13.07 - 13.7 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 13.7 Utility Theory KEYWORDS: Bloom's: Evaluate

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CH 14 - Multicriteria Decisions True / False 1. Objectives in multicriteria problems seldom conflict. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.1 Goal Programming: Formulation and Graphical Solution KEYWORDS: Bloom's: Remember 2. Target values will never be met precisely in a goal programming problem. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.1 Goal Programming: Formulation and Graphical Solution KEYWORDS: Bloom's: Understand 3. Goal equations consist of a function that defines goal achievement and deviation variables that measure the distance from the target. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.1 Goal Programming: Formulation and Graphical Solution KEYWORDS: Bloom's: Understand 4. In a goal programming model, there can only be one goal at each priority level. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.1 Goal Programming: Formulation and Graphical Solution © Cengage. Testing Powered by Cognero.

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CH 14 - Multicriteria Decisions KEYWORDS:

Bloom's: Understand

5. To solve a goal programming problem with preemptive priorities, successive linear programming programs, with an adjustment to the objective function and an additional constraint, must be solved. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.1 Goal Programming: Formulation and Graphical Solution KEYWORDS: Bloom's: Understand 6. If a problem has multiple goals at different priority levels, then it is rare that all the goals will be achieved with existing resources. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.14.02 - 14.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.2 Goal Programming: Solving More Complex Problems KEYWORDS: Bloom's: Understand 7. For a scoring model, the decision maker evaluates each decision alternative by considering the weight or importance of each criterion. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.14.03 - 14.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.3 Scoring Models KEYWORDS: Bloom's: Understand 8. In goal programming, deviation variables allow for the possibility of not meeting the target value exactly. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 © Cengage. Testing Powered by Cognero.

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CH 14 - Multicriteria Decisions NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.1 Goal Programming: Formulation and Graphical Solution KEYWORDS: Bloom's: Understand 9. An item's priority reveals how it compares to its competitors on a specific criterion. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.14.06 - 14.6 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.6 Using AHP to Develop an Overall Priority Ranking KEYWORDS: Bloom's: Understand 10. The priority matrix shows the priority for each item on each criterion. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.14.05 - 14.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.5 Establishing Priorities Using AHP KEYWORDS: Bloom's: Understand 11. One limitation of a scoring model is that it uses arbitrary weights that do not necessarily reflect the preferences of the individual decision maker. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.14.03 - 14.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.3 Scoring Models KEYWORDS: Bloom's: Understand 12. A consistency ratio greater than 0.10 indicates inconsistency in the pairwise comparisons. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy © Cengage. Testing Powered by Cognero.

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CH 14 - Multicriteria Decisions LEARNING OBJECTIVES: IMS.ASWC.19.14.05 - 14.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.5 Establishing Priorities Using AHP KEYWORDS: Bloom's: Understand 13. Requiring a decision maker to provide judgments about the relative importance of each criterion and then specifying a preference for each decision alternative using each criterion is the purpose of the analytic hierarchy process. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.14.04 - 14.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.4 Analytic Hierarchy Process KEYWORDS: Bloom's: Understand 14. The goal programming approach can be used when an analyst is confronted with an infeasible solution to an ordinary linear program. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.14.02 - 14.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.2 Goal Programming: Solving More Complex Problems KEYWORDS: Bloom's: Understand 15. A problem involving only one priority level is not considered a goal programming problem. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.14.02 - 14.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.2 Goal Programming: Solving More Complex Problems KEYWORDS: Bloom's: Understand 16. AHP allows a decision maker to express personal preferences about the various aspects of a multicriteria problem. a. True b. False ANSWER: True POINTS: 1 © Cengage. Testing Powered by Cognero.

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CH 14 - Multicriteria Decisions DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.14.04 - 14.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.4 Analytic Hierarchy Process KEYWORDS: Bloom's: Understand 17. Goal priorities are referred to as preemptive priorities because the satisfaction of a higher level goal cannot be traded for the satisfaction of a lower level goal. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.1 Goal Programming: Formulation and Graphical Solution KEYWORDS: Bloom's: Remember Multiple Choice 18. A decision with more than one objective a. cannot have an optimal solution. b. requires the decision maker to place the objectives in some order of importance. c. depends on the probability of satisfying each objective. d. should be decomposed into a separate model for each objective.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.1 Goal Programming: Formulation and Graphical Solution KEYWORDS: Bloom's: Understand 19. The objective function in a goal programming model calls for minimizing a function of the a. goal variables. b. target variables. c. deviation variables. d. preemptive variables.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.1 Goal Programming: Formulation and Graphical Solution © Cengage. Testing Powered by Cognero.

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CH 14 - Multicriteria Decisions KEYWORDS:

Bloom's: Understand

20. Preemptive priorities in goal programming a. show the target values for the problem. b. prevent sacrifice of a priority level goal to satisfy a lower level one. c. force the problem to be a standard linear program. d. limit deviations to d− only.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.1 Goal Programming: Formulation and Graphical Solution KEYWORDS: Bloom's: Understand 21. Deviation variables that occur in the objective function a. indicate the targets. b. indicate the priorities. c. allow for the possibility of not meeting the target value exactly. d. identify the difference between all actual and target values.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.1 Goal Programming: Formulation and Graphical Solution KEYWORDS: Bloom's: Understand 22. The variable d− measures the amount a. over the target and is similar to a slack. b. less than the target and is similar to a slack. c. less than the target and is similar to a surplus. d. under the target and is similar to a surplus.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.1 Goal Programming: Formulation and Graphical Solution KEYWORDS: Bloom's: Understand 23. The constraint 5x1 + 3x2 ≤ 150 is modified to become a goal equation, and priority one is to avoid overutilization.

Which of the following is appropriate? a. Min P1d1− ; 5x1 + 3x2 + d1− − d1+ = 150 © Cengage. Testing Powered by Cognero.

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CH 14 - Multicriteria Decisions b. Min P1d1+ ;

5x1 + 3x2 + d1− − d1+ = 150

c. Min P1d1+ ;

5x1 + 3x2 + d1+ = 150

d. Min P1d1+ ;

5x1 + 3x2 − d1+ = 150 ANSWER: b POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 14.1 Goal Programming: Formulation and Graphical Solution KEYWORDS: Bloom's: Analyze 24. The goal programming problem with the objective function Min P1(d1+) + P2(d2−) is initially solved by the computer and the objective function value is 0. Which of the following constraints should be added for the second problem? a. d1+ = 0 b. d1+ + d2− = 0 c. −d1+ + d2− = 0 d. d1+ ≤ 0 ANSWER: a POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 14.1 Goal Programming: Formulation and Graphical Solution KEYWORDS: Bloom's: Analyze 25. A required step in the analytic hierarchy process is to determine a. the goals to be satisfied. b. the expected value of the criteria. c. the relative importance of the criteria themselves. d. how many hierarchies to use.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.14.04 - 14.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.4 Analytic Hierarchy Process KEYWORDS: Bloom's: Understand 26. Pairwise comparisons are used to a. compare criteria in terms of the overall goal. b. compare choices on each criterion. c. compare criteria in terms of the overall goal and compare choices on each criterion. © Cengage. Testing Powered by Cognero.

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CH 14 - Multicriteria Decisions d. None of these are correct.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.14.05 - 14.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.5 Establishing Priorities Using AHP KEYWORDS: Bloom's: Understand 27. The overall priorities for decision alternatives a. are the sum of the products of the criterion priority times the priority of the decision alternative with respect to that criterion. b. sum to 1. c. indicate what choice is preferred, but do not force that choice to be made. d. All of these are correct.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.14.06 - 14.6 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.6 Using AHP to Develop an Overall Priority Ranking KEYWORDS: Bloom's: Understand 28. The steps of the scoring model include all of the following EXCEPT a. list the decision-making criteria and assign a weight to each. b. develop a pairwise comparison matrix for each criterion. c. rate how well each decision alternative satisfies each criterion. d. compute the total score for each decision alternative.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.14.03 - 14.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.3 Scoring Models KEYWORDS: Bloom's: Understand 29. Goal programming with preemptive priorities never permits trade-offs between a. goals with the same priority level and the same weights. b. goals with different priority levels. c. goals with the same priority level and different weights. d. any goals.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate © Cengage. Testing Powered by Cognero.

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CH 14 - Multicriteria Decisions LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.1 Goal Programming: Formulation and Graphical Solution KEYWORDS: Bloom's: Understand 30. Inconsistency in pairwise judgments is indicated by a consistency ratio that is a. less than 0. b. greater than 0.10. c. greater than 0.50. d. greater than 1.00.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.14.05 - 14.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.5 Establishing Priorities Using AHP KEYWORDS: Bloom's: Remember 31. When using a linear programming approach to solving a goal programming problem, a linear program must be solved for each a. goal. b. pair of deviation variables. c. priority level. d. pairwise comparison.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.1 Goal Programming: Formulation and Graphical Solution KEYWORDS: Bloom's: Understand 32. Computing the consistency ratio for a criterion's pairwise comparison matrix is the next step after a. developing the criterion's pairwise comparison matrix. b. converting the criterion's pairwise comparison matrix to a normalized matrix. c. developing the criterion's priority vector. d. developing the overall priority vector.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.14.05 - 14.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 14.5 Establishing Priorities Using AHP KEYWORDS: Bloom's: Understand © Cengage. Testing Powered by Cognero.

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CH 14 - Multicriteria Decisions Subjective Short Answer 33. Solve the following problem graphically: Min s.t.

P1(d1+) + P2(d2−) 3x1 + 5x2 ≤ 45 3x1 + 2x2 − d1+ + d1− = 24 x1 + x2 − d2+ + d2− = 10 x1, x2, d1−, d1+, d2−, d2+ ≥ 0

ANSWER:

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 14.1 Goal Programming: Formulation and Graphical Solution KEYWORDS: Bloom's: Evaluate 34. Lofton Company has developed the following linear programming problem: Max s.t.

x1 + x2 2x1 + x2 ≤ 10 2x1 + 3x2 ≤ 24 3x1 + 4x2 ≥ 36

Lofton finds this problem is infeasible. In revision, Lofton drops the original objective and establishes three goals.

Goal 1: Don't exceed 10 in constraint 1.

Goal 2: Don't fall short of 36 in constraint 3.

Goal 3: Don't exceed 24 in constraint 2.

Give the goal programming model and solve it graphically.

ANSWER: Min s.t. © Cengage. Testing Powered by Cognero.

P1d1+, P2d3−, P3d2+ 2x1 + x2 + d1− − d1+ = 10 Page 10

CH 14 - Multicriteria Decisions 2x1 + 3x2 + d2− − d2+ = 24 3x1 + 4x2 + d3− − d3+ = 36

The small triangle designated by the arrow satisfies the first two goals. After examining the corner points, the vertex (0.8, 8.4) minimizes d2+.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 14.1 Goal Programming: Formulation And Graphical Solution KEYWORDS: Bloom's: Create 35. An ATM is to be located in a campus union building so that it minimizes the distance from the food court, the gift shop, and the theater. They are located at coordinates (2, 2), (0, 6), and (8, 0). Develop a goal programming model to locate the best place for the ATM. ANSWER: Let x1 = first coordinate of the ATM x2 = second coordinate of the ATM Min

Σ (di+ + di− )

s.t.

x1 + d1− − d1+ = 2 x1 + d2− − d2+ = 2 x1 + d3− − d3+ = 0 x2 + d4− − d4+ = 6 x2 + d5− − d5+ = 8 x2 + d6− − d6+ = 0

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 14.1 Goal Programming: Formulation and Graphical Solution © Cengage. Testing Powered by Cognero.

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CH 14 - Multicriteria Decisions KEYWORDS:

Bloom's: Evaluate

36. As treasurer of the school PTA, you chair the committee that decides how the $20,000 raised by candy sales will be spent. Four kinds of projects have been proposed, and facts on each are shown below.

Project Basketball goals Encyclopedia sets Field trips Computer stations

Number Requested 12 5 6 20

Unit Cost $400 750 300 800

Volunteers Needed (Person-Days, Each) 2 0 3 0.5

Develop a goal programming model that would represent these goals and priorities.

Priority 1 Goal 1 Goal 2 Priority 2 Goal 3 Goal 4 Goal 5 ANSWER:

Spend the entire $20,000.

Do not use more than 50 person-days of volunteer time.

Provide at least as many encyclopedias and computers as requested.

Provide at least as many field trips as requested.

Do not provide any more basketball goals than requested.

Let

B = number of basketball goals E = number of encyclopedia sets F = number of field trips C = number of computer stations

Min

P1[d1− + d2+] + P2[d3− + d4− + d5− +d6+]

400B + 750E + 300F + 800C + d1− − d1+ = 20,000 2B + 3F + 0.5C + d2− − d2+ = 50 E + d3− − d3+ = 5 C + d4− − d4+ = 20 F + d5− − d5+ = 6 B + d6− − d6+ = 12 B, E, F, C, and all di ≥ 0 Note: This formulation assumes expenditures could exceed $20,000. Delete d1+ from the problem if this is not so. s.t.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 14.1 Goal Programming: Formulation and Graphical Solution KEYWORDS: Bloom's: Create 37. Rosie's Ribs is in need of an office management software package. After considerable research, Rosie has narrowed her choice to one of three packages: N-able, VersaSuite, and SoftTrack. She has determined her decision-making criteria, assigned a weight to each criterion, and rated how well each alternative satisfies each criterion. © Cengage. Testing Powered by Cognero.

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CH 14 - Multicriteria Decisions Criterion Ease of use Report generation Functional integration Online help Entry error checking Price Support cost

Weight 4 3 5 3 2 4 3

N-Able 3 8 5 8 8 4 6

Decision Alternatives VersaSuite 5 7 8 6 3 7 5

SoftTrack 8 6 6 4 4 5 7

Using a scoring model, determine the recommended software package for Rosie's.

ANSWER: N-Able = 135, VersaSuite = 148, SoftTrack = 141 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.14.03 - 14.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 14.3 Scoring Models KEYWORDS: Bloom's: Evaluate 38. A consumer group is using AHP to compare four used car models. Part of the pairwise comparison matrix for "repair frequency" is shown below. a. b.

Complete the matrix.

Does it seem to be consistent?

Repair Frequency Model A Model B Model C Model D Model A 1 5 1/6 Model B 1/5 1 1/2 Model C 1/4 1/3 1 Model D 1/4 1 ANSWER: The matrix is not consistent. A to B is 5, B to C is 3, C to D is 4, yet D to A is 6.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.14.05 - 14.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 14.5 Establishing Priorities Using AHP KEYWORDS: Bloom's: Evaluate 39. A computer company looking for a new location for a plant has determined three criteria to use to rate cities. Pairwise comparisons are given.

Recreation opportunities Proximity to university Cost of living

Recreation Opportunities 1 3 5

Proximity to University 1/3 1 4

Cost of Living 1/5 1/4 1

Determine priorities for the three relative to the overall location goal.

ANSWER: © Cengage. Testing Powered by Cognero.

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CH 14 - Multicriteria Decisions Normalized Pairwise Comparison Matrix 0.1111 0.0625 0.1379 0.3333 0.1875 0.1724 0.5556 0.7500 0.6897

Priorities 0.1038 0.2311 0.6651

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.14.05 - 14.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 14.5 Establishing Priorities Using AHP KEYWORDS: Bloom's: Evaluate 40. In an AHP problem, the priorities for three criteria are as follows: Criterion 1 Criterion 2 Criterion 3

0.1722 0.1901 0.6377

The priority matrix is as follows Choice 1 Choice 2 Choice 3

Criterion 1 0.425 0.330 0.245

Criterion 2 0.292 0.251 0.457

Criterion 3 0.850 0.105 0.045

Compute the overall priority for each choice.

ANSWER: a. States 1 and 2 (recover and die) are the absorbing states. b.

The probability that a person with symptom 2 will recover is 0.633.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.14.05 - 14.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 14.5 Establishing Priorities Using AHP KEYWORDS: Bloom's: Analyze 41. Like many high school seniors, Anne has several universities to consider when making her final college choice. To assist in her decision, she has decided to use AHP to develop a ranking for school R, school P, and school M. The schools will be evaluated on five criteria, and Anne's pairwise comparison matrix for the criteria is shown below.

Distance Program Size Climate

Distance 1 4 1/2 1/3

Program 1/4 1 1/4 1/5

© Cengage. Testing Powered by Cognero.

Size 2 4 1 1/3

Campus Climate 3 5 3 1

Cost 4 6 3 2 Page 14

CH 14 - Multicriteria Decisions Cost

1/4

1/6

1/3

1/2

1

The universities' pairwise comparisons on the criteria are shown below.

Distance R P M

R 1 1/2 1/3

P 2 1 2/3

M 3 3/2 1

Programs R P M

R 1 2 1

P 1/2 1 1/2

M 1 2 1

Size R P M

R 1 1/4 1/2

P 4 1 2

M 2 1/2 1

Climate R P M

R 1 1 1/3

P 1 1 1/3

M 3 3 1

Cost R P M

R 1 5 1

P 1/5 1 1/5

M 1 5 1

a. What is the overall ranking of the five criteria? b. What is the overall ranking of the three universities?

ANSWER: The spreadsheet shows all calculations. a.

Distance Program Size Climate Cost

The ranking of the criteria is: 0.2098 0.4980 0.1548 0.1548 0.0553

R P M

The ranking of the schools is: 0.3705 0.4030 0.2265

b.

AHP COLLEGE

Distance Program Size Climate Cost sums

Distance Program Size 1.0000 0.2500 2.0000 4.0000 1.0000 4.0000 0.5000 0.2500 1.0000 0.3333 0.2000 0.3333 0.2500 0.1667 0.3333 6.0833 1.8667 7.6666

Distance

R

© Cengage. Testing Powered by Cognero.

P

M

Climate Cost 3.0000 4.0000 5.0000 6.0000 3.0000 3.0000 1.0000 2.0000 0.5000 1.0000 12.5000 16.0000

NPCM 0.1644 0.6575 0.0822 0.0548 0.0411

NPCM

0.1339 0.5357 0.1339 0.1071 0.0893

0.2609 0.5217 0.1304 0.0435 0.0435

0.2400 0.4000 0.2400 0.0800 0.0400

0.2500 0.3750 0.1875 0.1250 0.0625

Avg. 0.2098 0.4980 0.1548 0.0821 0.0553

Avg. Page 15

CH 14 - Multicriteria Decisions R

1.0000

2.0000

3.0000

0.5455

0.5454

0.5455

0.5455

P

0.5000

1.0000

1.5000

0.2727

0.2727

0.2727

0.2727

M

0.3333

0.6667

1.0000

0.1818

0.1818

0.1818

0.1818

sums

1.8333

3.6667

5.5000

Program

R

P

M

NPCM

R

1.0000

0.5000

1.0000

0.2500

0.2500

0.2500

0.2500

P

2.0000

1.0000

2.0000

0.5000

0.5000

0.5000

0.5000

M

1.0000

0.5000

1.0000

0.2500

0.2500

0.2500

0.2500

sums

4.0000

2.0000

4.0000

Size

R

P

M

NPCM

R

1.0000

4.0000

2.0000

0.5714

0.5714

0.5714

0.5714

P

0.2500

1.0000

0.5000

0.1429

0.1429

0.1429

0.1429

M

0.5000

2.0000

1.0000

0.2857

0.2857

0.2857

0.2857

sums

1.7500

7.0000

3.5000

Climate

R

P

M

NPCM

R

1.0000

1.0000

3.0000

0.4286

0.4286

0.4286

0.4286

P

1.0000

1.0000

3.0000

0.4286

0.4286

0.4286

0.4286

M

0.3333

0.3333

1.0000

0.1428

0.1428

0.1429

0.1428

sums

2.3333

2.3333

7.0000

Cost

R

P

M

NPCM

R

1.0000

0.2000

1.0000

0.1429

0.1429

0.1429

0.1429

P

5.0000

1.0000

5.0000

0.7143

0.7143

0.7143

0.7143

M

1.0000

0.2000

1.0000

0.1429

0.1429

0.1429

0.1429

sums

7.0000

1.4000

7.0000

Priority Matrix Distance Program Size

Avg.

Avg.

Avg.

Avg.

Criterion Priority Vector Climate Cost

R

0.5455

0.2500

0.5714 0.4286

0.1429

Distance Program Size

P

0.2727

0.5000

0.1429 0.4286

0.7143

0.2098

M

0.1818

0.2500

0.2857 0.1428

0.1429

Overall Rankings R

0.3705

P

0.4030

M

0.2265

0.4980

Climate Cost

0.1548 0.0821 0.0553

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.14.06 - 14.6 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 14.6 Using AHP to Develop an Overall Priority Ranking KEYWORDS: Bloom's: Create 42. John Harris is interested in purchasing a new Harley-Davidson motorcycle. He has narrowed his choice to

one of three models: Sportster Classic, Heritage Softtail, and Electra Glide. After much consideration, John has determined his decision-making criteria, assigned a weight to each criterion, and rated how well each decision © Cengage. Testing Powered by Cognero.

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CH 14 - Multicriteria Decisions alternative satisfies each criterion.

Criterion Wind protection Fuel tank capacity Passenger comfort Seat height Acceleration Vehicle weight Storage capacity

Weight 5 3 2 3 4 3 3

Decision Alternative Sportster Heritage Classic Softtail 3 6 5 7 5 6 8 5 8 5 8 6 4 5

Electra Glide 8 6 8 6 3 3 8

Using a scoring model, determine the recommended motorcycle model for John.

ANSWER:

S1 = 5(3) + 3(5) + 2(5) + 3(8) + 4(8) + 3(8) + 3(4) = 132 S2 = 5(6) + 3(7) + 2(6) + 3(5) + 4(5) + 3(6) + 3(5) = 131 S3 = 5(8) + 3(6) + 2(8) + 3(6) + 4(3) + 3(3) + 3(8) = 137 (Electra Glide recommended.)

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.14.03 - 14.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 14.3 Scoring Models KEYWORDS: Bloom's: Evaluate 43. The campaign headquarters of Jerry Black, a candidate for the board of supervisors, has 100 volunteers. With one week to go in the election, three major strategies are remaining: media advertising, door-to-door canvassing, and telephone campaigning. Each phone call is estimated to take approximately four minutes and each door-to-door personal contact will average seven minutes. These times include time between contacts for breaks, transportation, dialing, etc.

Volunteers who work on advertising will not be able to handle any other duties. Each ad will utilize the talents of three workers for the entire week.

Volunteers are expected to work 12 hours per day during the final seven days of the campaign. At a minimum, Jerry Black feels he needs 30,000 phone contacts, 20,000 personal contacts, and three advertisements during the last week.

However, he would like to see 50,000 phone contacts and 50,000 personal contacts made and five advertisements developed. Advertising is believed to be 50 times as important as personal contacts, which in turn are twice as important as phone contacts.

Formulate this goal programming problem with a single weighted priority to determine how the work should be distributed during the final week of the campaign.

ANSWER: Variables x1 = number of volunteers doing phone work during the week x2 = number of volunteers making personal contacts during the week x3 = number of volunteers preparing advertising during the week Goals 1. 50,000 phone contacts: d1+ and d1– = amount this quantity is overachieved and underachieved, respectively 2. 50,000 personal contacts: © Cengage. Testing Powered by Cognero.

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CH 14 - Multicriteria Decisions d2+ and d2– = amount this quantity is overachieved and underachieved, respectively 3. 5 advertisements: d3+ and d3– = amount this quantity is overachieved and underachieved, respectively Objective Function It would not hurt Jerry Black if his goals were exceeded; however, since the importance of his goals are in the ratio 1:2:100, the goal programming objective function would be: Min d1– + 2d2– + 100d3– LP Constraints 1) At least 30,000 phone contacts: 1260x1 > 30,000 2) At least 20,000 personal contacts: 720x2 > 20,000 3) At least 3 advertisements: (1/3)x3 > 3 4) 100 volunteers: x1 + x2 + x3 = 100 Goal Constraints 5) Make 50,000 phone contacts: 1260x1 – d1+ + d1– = 50,000 6) Make 50,000 personal contacts: 720x2 – d2+ + d2– = 50,000 7) Design 5 advertisements: (1/3)x3 – d3+ + d3– = 5 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.14.01 - 14.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 14.1 Goal Programming: Formulation and Graphical Solution KEYWORDS: Bloom's: Evaluate

© Cengage. Testing Powered by Cognero.

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CH 15 - Time Series Analysis and Forecasting True / False 1. If we focus upon the historical data, or past values of the variable to be forecast, we refer to this as a time series method of forecasting. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.01 - 15.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.1 Time Series Patterns KEYWORDS: Bloom's: Remember 2. Quantitative forecasting methods can be used when past information about the variable being forecast is unavailable. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.01 - 15.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.1 Time Series Patterns KEYWORDS: Bloom's: Remember 3. Trend in a time series must always be linear. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.01 - 15.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.1 Time Series Patterns KEYWORDS: Bloom's: Remember 4. All quarterly time series contain seasonality. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.01 - 15.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.1 Time Series Patterns © Cengage. Testing Powered by Cognero.

Page 1

CH 15 - Time Series Analysis and Forecasting KEYWORDS:

Bloom's: Understand

5. A four-period moving average forecast for period 10 would be found by averaging the values from periods 10, 9, 8, and 7. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Understand 6. If the random variability in a time series is great, a small value of the smoothing constant is preferred so that we do not overreact and adjust our forecasts too quickly. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Understand 7. With fewer periods in a moving average, it will take longer to adjust to a new level of data values. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Understand 8. Qualitative forecasting methods are appropriate when historical data on the variable being forecast are either unavailable or not applicable. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy © Cengage. Testing Powered by Cognero.

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CH 15 - Time Series Analysis and Forecasting LEARNING OBJECTIVES: IMS.ASWC.19.15.01 - 15.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.1 Time Series Patterns KEYWORDS: Bloom's: Remember 9. A sequence of observations on a variable measured at successive points in time or over successive periods of time is known as a time series. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.01 - 15.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.1 Time Series Patterns KEYWORDS: Bloom's: Remember 10. Any recurring sequence of points above and below the trend line lasting less than one year can be attributed to the cyclical component of the time series. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.01 - 15.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.1 Time Series Patterns KEYWORDS: Bloom's: Understand 11. Smoothing methods are more appropriate for a stable time series than when significant trend or seasonal patterns are present. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Understand 12. The exponential smoothing forecast for any period is a weighted average of all the previous actual values for the time series. a. True b. False © Cengage. Testing Powered by Cognero.

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CH 15 - Time Series Analysis and Forecasting ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Understand 13. The mean squared error is obtained by computing the average of the squared forecast errors. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.02 - 15.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.2 Forecast Accuracy KEYWORDS: Bloom's: Remember 14. If a time series has a significant trend pattern, then one should not use a moving average to forecast. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.01 - 15.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.1 Time Series Patterns KEYWORDS: Bloom's: Understand 15. For a time series with relatively little random variability, we should use larger values of the smoothing constant to provide the advantage of allowing the forecasts to react more quickly to changing conditions. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Understand 16. Time series data can exhibit seasonal patterns of less than one month in duration. a. True © Cengage. Testing Powered by Cognero.

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CH 15 - Time Series Analysis and Forecasting b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.01 - 15.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.1 Time Series Patterns KEYWORDS: Bloom's: Understand 17. When using a moving average of order k to forecast, a small value for k is preferred if only the most recent values of the time series are considered relevant. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Understand 18. In situations where you need to compare forecasting methods for different time periods, relative measures such as mean absolute error (MAE) are preferred. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.02 - 15.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.2 Forecast Accuracy KEYWORDS: Bloom's: Understand 19. In order to use moving averages to forecast a time series, the first step is to select the order k, the number of time series values to be included in the moving average. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Understand © Cengage. Testing Powered by Cognero.

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CH 15 - Time Series Analysis and Forecasting 20. When forecasting, if a greater number of past values are considered relevant, then we generally opt for a larger value of k. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Understand Multiple Choice 21. All of the following are true about time series methods EXCEPT a. they discover a pattern in historical data and project it into the future. b. they involve the use of expert judgment to develop forecasts. c. they assume that the pattern of the past will continue into the future. d. their forecasts are based solely on past values of the variable or past forecast errors.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.01 - 15.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.1 Time Series Patterns KEYWORDS: Bloom's: Understand 22. Gradual shifting of a time series to relatively higher or lower values over a long period of time is called a. periodicity. b. a cycle. c. seasonality. d. a trend pattern.

ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.01 - 15.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.1 Time Series Patterns KEYWORDS: Bloom's: Remember 23. Seasonal patterns a. cannot be predicted. b. are regular repeated patterns. c. are multiyear runs of observations above or below the trend line. d. reflect a shift in the time series over time. © Cengage. Testing Powered by Cognero.

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CH 15 - Time Series Analysis and Forecasting ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.01 - 15.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.1 Time Series Patterns KEYWORDS: Bloom's: Remember 24. The focus of smoothing methods is to smooth out a. random fluctuations. b. wide seasonal variations. c. significant trend effects. d. long-range forecasts.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Remember 25. Forecast errors a. are the difference in successive values of a time series. b. are the differences between actual and forecast values. c. should all be nonnegative. d. should be summed to judge the goodness of a forecasting model.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.02 - 15.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.2 Forecast Accuracy KEYWORDS: Bloom's: Understand 26. Linear trend is calculated as . The trend projection for period 15 is a. 11.25. b. 28.50. c. 39.75. d. 44.25.

ANSWER: c POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.04 - 15.4 © Cengage. Testing Powered by Cognero.

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CH 15 - Time Series Analysis and Forecasting NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.4 Linear Trend Projection KEYWORDS: Bloom's: Apply 27. The trend pattern is easy to identify by using a. a moving average. b. exponential smoothing. c. regression analysis. d. a weighted moving average.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.01 - 15.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.1 Time Series Patterns KEYWORDS: Bloom's: Understand 28. The forecasting method that is appropriate when the time series has no significant trend, cyclical, or seasonal pattern is a. moving average. b. mean squared error. c. mean average error. d. qualitative forecasting.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Understand 29. We can model a time series with a seasonal pattern by treating the season itself as a(n) a. categorical variable. b. qualitative variable. c. annual variable d. None of these are correct.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.15.05 - 15.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.5 Seasonality KEYWORDS: Bloom's: Understand 30. One measure of the accuracy of a forecasting model is the © Cengage. Testing Powered by Cognero.

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CH 15 - Time Series Analysis and Forecasting a. smoothing constant. b. linear trend. c. mean absolute error. d. seasonal index.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.15.05 - 15.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.5 Seasonality KEYWORDS: Bloom's: Understand 31. Using a naive forecasting method, the forecast for next week’s sales volume equals a. the most recent week’s sales volume. b. the most recent week’s forecast. c. the average of the last four weeks’ sales volumes. d. next week’s production volume.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.15.02 - 15.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.2 Forecast Accuracy KEYWORDS: Bloom's: Understand 32. All of the following are true about a cyclical pattern EXCEPT a. it is often due to multiyear business cycles. b. it is often combined with long-term trend patterns and called trend-cycle patterns. c. it usually is easier to forecast than a seasonal pattern due to less variability. d. it is an alternating sequence of data points above and below the trend line.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.15.01 - 15.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.1 Time Series Patterns KEYWORDS: Bloom's: Understand 33. All of the following are true about a stationary time series EXCEPT a. its statistical properties are independent of time. b. a plot of the series will always exhibit a horizontal pattern. c. the process generating the data has a constant mean. d. there is no variability in the time series over time.

ANSWER: d © Cengage. Testing Powered by Cognero.

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CH 15 - Time Series Analysis and Forecasting POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.01 - 15.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.1 Time Series Patterns KEYWORDS: Bloom's: Understand 34. In situations where you need to compare forecasting methods for different time periods, the most appropriate accuracy measure is a. MSE. b. MAPE. c. MAE. d. ME. ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.15.02 - 15.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.2 Forecast Accuracy KEYWORDS: Bloom's: Understand 35. Whenever a categorical variable such as season has k levels, the number of dummy variables required is a. k − 1. b. k. c. k + 1. d. 2k.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.15.05 - 15.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.5 Seasonality KEYWORDS: Bloom's: Apply 36. To select a value for α for exponential smoothing a. use a small α when the series varies substantially. b. use a large α when the series has little random variability. c. use a value between 0 and 1. d. All of these are correct.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking © Cengage. Testing Powered by Cognero.

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CH 15 - Time Series Analysis and Forecasting TOPICS: KEYWORDS:

15.3 Moving Averages and Exponential Smoothing Bloom's: Understand

37. Which of the following forecasting methods puts the least weight on the most recent time series value? a. exponential smoothing with α = 0.3 b. exponential smoothing with α = 0.2 c. moving average using the most recent four periods d. moving average using the most recent three periods ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Understand 38. Using exponential smoothing, the demand forecast for time period 10 equals the demand forecast for time period 9 plus a. α times the demand forecast for time period 8. b. α times the error in the demand forecast for time period 9. c. α times the observed demand in time period 9. d. α times the demand forecast for time period 9.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Apply 39. Which of the following exponential smoothing constant values puts the same weight on the most recent time series value as does a five-period moving average? a. α = 0.2 b. α = 0.25 c. α = 0.75 d. α = 0.8 ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Apply Subjective Short Answer © Cengage. Testing Powered by Cognero.

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CH 15 - Time Series Analysis and Forecasting 40. The number of cans of soft drinks sold in a machine each week is recorded below. Develop forecasts using a threeperiod moving average.

338, 219, 278, 265, 314, 323, 299, 259, 287, 302 ANSWER: FORECASTING WITH MOVING AVERAGES ***************************************** THE MOVING AVERAGE USES 3 TIME PERIODS Time Period 1 2 3 4 5 6 7 8 9 10

Actual Value

Forecast

Forecast Error

265 314 323 299 259 287 302

278.33 254.00 285.67 300.67 312.00 293.67 281.67

−13.33 60.00 37.33 −1.67 −53.00 −6.67 20.33

THE FORECAST FOR PERIOD 11 282.67 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Analyze 41. Use a four-period moving average to forecast attendance at baseball games. Historical records show the following: 5346, 7812, 6513, 5783, 5982, 6519, 6283, 5577, 6712, 7345 ANSWER: FORECASTING WITH MOVING AVERAGES ***************************************** THE MOVING AVERAGE USES 4 TIME PERIODS Time Period 1 2 3 4 5 6 7 8 9 10

Actual Value

Forecast

Forecast Error

5,982 6,519 6,283 5,577 6,712 7,345

6,363.50 6,522.50 6,199.25 6,141.75 6,090.25 6,272.75

−381.50 −3.50 83.75 −564.75 621.75 1,072.25

THE FORECAST FOR PERIOD 11 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 © Cengage. Testing Powered by Cognero.

6479.25

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CH 15 - Time Series Analysis and Forecasting NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Analyze 42. A hospital records the number of floral deliveries its patients receive each day. For a two-week period, the records show the following: 15, 27, 26, 24, 18, 21, 26, 19, 15, 28, 25, 26, 17, 23 Use exponential smoothing with a smoothing constant of 0.4 to forecast the number of deliveries.

ANSWER: FORECASTING WITH EXPONENTIAL SMOOTHING ************************************************ THE SMOOTHING CONSTANT IS 0.4 Time Period 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Actual Value

Forecast

Forecast Error

27 26 24 18 21 26 19 15 28 25 26 17 23

15.00 19.80 22.28 22.97 20.98 20.99 22.99 21.40 18.84 22.50 23.50 24.50 21.50

12.00 6.20 1.72 −4.97 0.02 5.01 −3.99 −6.40 9.16 2.50 2.50 −7.50 1.50

THE FORECAST FOR PERIOD 15 22.10 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Analyze 43. The number of girls who attend a summer basketball camp has been recorded for the seven years the camp has been offered. Use exponential smoothing with a smoothing constant of 0.8 to forecast attendance for the eighth year.

47, 68, 65, 92, 98, 121, 146 ANSWER: FORECASTING WITH EXPONENTIAL SMOOTHING ************************************************ THE SMOOTHING CONSTANT IS 0.8 Time Period 1 2 3 4 5 © Cengage. Testing Powered by Cognero.

Actual Value

Forecast

Forecast Error

68 65 92 98

47.00 63.80 64.76 86.55

21.00 1.20 27.24 11.45 Page 13

CH 15 - Time Series Analysis and Forecasting 6 7

121 146

95.71 115.94

25.29 30.06

THE FORECAST FOR PERIOD 8 139.99 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Analyze 44. A trend line for the weekly attendance at a restaurant's Sunday brunch is given by the following:

How many guests would you expect in Week 20?

ANSWER: POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.04 - 15.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.4 Linear Trend Projection KEYWORDS: Bloom's: Analyze 45. The number of new contributors to a public radio station's annual fund drive over the last 10 years is as follows: 63, 58, 61, 72, 98, 103, 121, 147, 163, 198 Develop a trend equation for this information, and use it to predict next year's number of new contributors.

ANSWER: FORECASTING WITH LINEAR TREND ************************************ THE LINEAR TREND EQUATION:

where

trend value of the time series in period t

Time Period Actual Value Forecast 1 63 39.35 2 58 54.69 3 61 70.04 4 72 85.38 5 98 100.73 6 103 116.07 7 121 131.42 8 147 146.76 9 163 162.11 10 198 177.45 MEAN SQUARE ERROR 154.11 FORECAST FOR PERIOD 11 192.80 © Cengage. Testing Powered by Cognero.

Forecast Error 23.66 3.31 −9.03 −13.38 −2.73 −13.07 −10.42 0.24 0.89 20.55

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CH 15 - Time Series Analysis and Forecasting POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.04 - 15.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.4 Linear Trend Projection KEYWORDS: Bloom's: Analyze 46. The average SAT verbal score for students from one high school over the last 10 exams is as follows: 508, 490, 502, 505, 493, 506, 492, 490, 503, 501 Do the scores support an increasing or a decreasing trend?

ANSWER: FORECASTING WITH LINEAR TREND ************************************ THE LINEAR TREND EQUATION:

where

trend value of the time series in period t

The negative slope for the trend indicates that scores are decreasing. (However, we have not tested for the significance of this coefficient.)

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.04 - 15.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.4 Linear Trend Projection KEYWORDS: Bloom's: Analyze 47. The number of properties newly listed with a real estate agency in each quarter over the last four years is given below.

Assume the time series has seasonality without trend.

Year Quarter 1 2 3 4

1 73 89 123 92

2 81 87 115 95

3 76 91 108 87

4 77 88 120 97

a. Develop the optimization model that finds the estimated regression equation that minimizes the sum of squared error. b. Solve for the estimated regression equation. c. Forecast the four quarters of Year 5.

ANSWER: a.

Min [(73 – Ȳ1)2 + (89 – Ȳ2)2 + (123 – Ȳ3)2 + … + (97 – Ȳ16)2] s.t. Ȳ1 = b0 + 1b1 + 0b2 + 0b3 Ȳ2 = b0 + 0b1 + 1b2 + 0b3 Ȳ3 = b0 + 0b1 + 0b2 + 1b3 Ȳ4 = b0 + 0b1 + 0b2 + 0b3 . © Cengage. Testing Powered by Cognero.

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CH 15 - Time Series Analysis and Forecasting . . Ȳ13 = b0 + 1b1 + 0b2 + 0b3 Ȳ14 = b0 + 0b1 + 1b2 + 0b3 Ȳ15 = b0 + 0b1 + 0b2 + 1b3 Ȳ16 = b0 + 0b1 + 0b2 + 0b3 b. Ȳt = 92.75 − 16Qtr1 − 4Qtr2 + 23.75Qtr3 c. Ȳ17 = 76.75, Ȳ18 = 88.75, Ȳ19 = 116.50, Ȳ20 = 92.75 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.04 - 15.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.4 Linear Trend Projection KEYWORDS: Bloom's: Create 48. Quarterly billing for water usage is shown below.

Year Quarter Winter Spring Summer Fall a. b.

1 64 103 152 73

2 66 103 160 72

3 68 104 162 78

4 73 120 176 88

Solve for the forecast equation that minimizes the sum of squared error.

Forecast the summer of Year 5 and spring of Year 6.

ANSWER:

a. Ȳt = 64.875 − 6.1375Qtr1t + 32.325Qtr2t + 86.0375Qtr3t + 1.2875t b. Summer, Year 5 = 175.375; Spring, Year 6 = 125.525 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.05 - 15.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.5 Seasonality KEYWORDS: Bloom's: Analyze 49. A customer comment phone line is staffed from 8:00 a.m. to 4:30 p.m. five days a week. Records are available that show the number of calls received every day for the last five weeks.

Week Day 1 M T W TH F 2 M T

Number 28 12 16 15 23 25 10

Week Day 4 M T W TH F 5 M T

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Number 27 13 16 18 24 26 11 Page 16

CH 15 - Time Series Analysis and Forecasting

3

a. b. c.

W TH F M T W TH F

14 14 26 32 15 15 13 21

W TH F

18 17 25

Develop the optimization model that finds the estimated regression equation that minimizes the sum of squared error.

Solve for the estimated regression equation.

Forecast the five days of Week 6.

ANSWER:

a.

Min [(28 – Ȳ1)2 + (12 – Ȳ2)2 + (16 – Ȳ3)2 + … + (25 – Ȳ25)2] s.t. Ȳ1 = b0 + 1b1 + 0b2 + 0b3 + 0b4 Ȳ2 = b0 + 0b1 + 1b2 + 0b3 + 0b4 Ȳ3 = b0 + 0b1 + 0b2 + 1b3 + 0b4 Ȳ4 = b0 + 0b1 + 0b2 + 0b3 + 1b4 Ȳ5 = b0 + 0b1 + 0b2 + 0b3 + 0b4 . . . Ȳ21 = b0 + 1b1 + 0b2 + 0b3 + 0b4 Ȳ22 = b0 + 0b1 + 1b2 + 0b3 + 0b4 Ȳ23 = b0 + 0b1 + 0b2 + 1b3 + 0b4 Ȳ24 = b0 + 0b1 + 0b2 + 0b3 + 1b4 Ȳ25 = b0 + 0b1 + 0b2 + 0b3 + 0b4

b. Ȳt = 23.8 + 3.8M − 11.6T − 8.0W − 8.4TH c. Ȳ26 = 27.6, Ȳ27 = 12.2, Ȳ28 = 15.8, Ȳ29 = 15.4, Ȳ30 = 23.8 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.01 - 15.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.1 Time Series Patterns KEYWORDS: Bloom's: Create 50. Monthly sales at a coffee shop have been analyzed. The seasonal index values are as follows: Month January February March April May June

Index 1.38 1.42 1.35 1.03 0.99 0.62

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CH 15 - Time Series Analysis and Forecasting July 0.51 August 0.58 September 0.82 October 0.82 November 0.92 December 1.56 The trend line is 74,123 + 26.9t. Assume there is no cyclical component and forecast sales for Year 8 (months 97–108).

ANSWER: SI Month Trend Forecast 1.38 97 76,994.2 106,252.0 1.42 98 77,023.8 109,373.8 1.35 99 77,053.4 104,022.1 1.03 100 77,083.0 79,395.5 0.99 101 77,112.6 76,341.5 0.62 102 77,142.2 47,828.2 0.51 103 77,171.8 39,357.6 0.58 104 77,201.4 44,776.8 0.82 105 77,231.0 63,329.4 0.82 106 77,260.6 63,353.7 0.92 107 77,290.2 71,107.0 1.56 108 77,319.8 120,618.9 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.05 - 15.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.5 Seasonality KEYWORDS: Bloom's: Create 51. A 24-hour coffee/donut shop makes donuts every eight hours. The manager must forecast donut demand so that the bakers have the fresh ingredients they need. Listed below is the actual number of glazed donuts (in dozens) sold in each of the preceding 13 eight-hour shifts.

Date June 3

June 4

June 5

June 6

June 7

Shift Day Evening Night Day Evening Night Day Evening Night Day Evening Night Day

Demand (dozens) 59 47 40 64 43 39 62 46 42 60 45 40 58

a. Develop the optimization model that finds the estimated regression equation that minimizes the sum of squared error. b. Solve for the estimated regression equation. c. Forecast the demand for glazed donuts for the day, evening, and night shifts of June 8.

ANSWER: a.

Min [(59 – Ȳ1)2 + (47 – Ȳ2)2 + (40 – Ȳ3)2 + … + (58 – Ȳ13)2] © Cengage. Testing Powered by Cognero.

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CH 15 - Time Series Analysis and Forecasting s.t. Ȳ1 = b0 + 1b1 + 0b2 Ȳ2 = b0 + 0b1 + 1b2 Ȳ3 = b0 + 0b1 + 0b2 . . . Ȳ10 = b0 + 1b1 + 0b2 Ȳ11 = b0 + 0b1 + 1b2 Ȳ12 = b0 + 0b1 + 0b2 Ȳ13 = b0 + 1b1 + 0b2 b. Ȳt = 40.25 + 20.35Dt + 5.00Et c. Day = 60.60, Evening = 45.25, Night = 40.25 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.01 - 15.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.1 Time Series Patterns KEYWORDS: Bloom's: Evaluate 52. The number of plumbing repair jobs performed by Auger's Plumbing Service in each of the last nine months is listed below.

Month March April May a. b.

Jobs 353 387 342

Month June July August

Jobs 374 396 409

Month September October November

Jobs 399 412 408

Assuming a linear trend function, forecast the number of repair jobs Auger's will perform in December using the least squares method.

What is your forecast for December using a three-period weighted moving average with weights of 0.6, 0.3, and 0.1? How does it compare with your forecast from part (a)?

ANSWER: a. b.

Ȳ10 = 349.667 + (10)(7.4) = 423.667 Ȳ10 = 0.1(399) + 0.3(412) + 0.6(408) = 408.3

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.01 - 15.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.1 Time Series Patterns KEYWORDS: Bloom's: Create 53. Quarterly revenues (in $1,000,000s) for a national restaurant chain for a five-year period were as follows: Quarter 1

Year 1 33

Year 2 42

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Year 3 54

Year 4 70

Year 5 85 Page 19

CH 15 - Time Series Analysis and Forecasting 2 3 4 a. b.

36 35 38

40 42 47

53 54 62

67 70 77

82 87 99

Solve for the forecast equation that minimizes the sum of squared error.

Forecast the four quarters of Year 6.

ANSWER:

a. Ȳt = 24.475 + 2.23125Qtr1t − 2.3125Qtr2t − 3.65625Qtr3t + 3.34375t b. Year 6: Q1 = 96.925, Q2 = 95.725, Q3 = 97.725, Q4 = 104.725 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.01 - 15.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.1 Time Series Patterns KEYWORDS: Bloom's: Create 54. Business at Terry's Tie Shop can be viewed as falling into three distinct seasons: (1) Christmas (November– December); (2) Father's Day (late May–mid-June); and (3) all other times. Average weekly sales (in $s) for each of these three seasons during the past four years have been as follows: Season 1 2 3

Year 1 1856 2012 985

Year 2 1995 2168 1072

Year 3 2241 2306 1105

Year 4 2280 2408 1120

Determine a forecast for the average weekly sales in Years 5 and 6 for each of the three seasons.

ANSWER: Ȳt = 797 + 1095.433S1t + 1189.467S2t + 36.467t Forecasts Year 5: Seas.1 = 2366.5, Seas.2 = 2497.0, Seas.3 = 1344.0 Forecasts Year 6: Seas.1 = 2475.9, Seas.2 = 2606.4, Seas.3 = 1453.4 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.01 - 15.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.1 Time Series Patterns KEYWORDS: Bloom's: Evaluate 55. Coyote Cable has been experiencing an increase in cable service subscribers over the last few years due to increased advertising and an influx of new residents to the region. The number of subscribers (in 1000s) for the last 16 months is as follows: Month

Sales

Month

Sales

Month

Sales

1

12.8

7

20.6

12

23.8

2

14.6

8

18.5

13

25.1

3

15.2

9

19.9

14

24.7

4

16.1

10

23.6

15

26.5

5

15.8

11

24.2

16

28.9

6

17.2

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CH 15 - Time Series Analysis and Forecasting Forecast the number of subscribers for Months 17, 18, 19, and 20.

ANSWER: Month 17: 29.0; Month 18: 30.0; Month 19: 31.0; Month 20: 32.0 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.04 - 15.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.4 Linear Trend Projection KEYWORDS: Bloom's: Evaluate 56. Weekly sales of the Weber food processor for the past 10 weeks have been as follows: Week 1 2 3 4 5

Sales 980 1040 1120 1050 960

Week 6 7 8 9 10

Sales 990 1030 1260 1240 1100

a. Determine, on the basis of minimizing the mean square error, whether a three- or four-period simple moving average model gives a better forecast for this problem. b. For each model, forecast sales for Week 11.

ANSWER: a. The three-week moving average gives the better forecast (MSE = 16,337) (compared to a MSE = 17,911 for the four-week moving average). b. Three-week moving average forecast for Week 11 = 1200.

Four-week moving average forecast for Week 11 = 1158.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Analyze 57. Below you are given information on John's Hair Salon's profit for the past seven years.

Year 1 2 3 4 5 6 7

Profit (in 1000s) 15.0 16.2 17.1 18.1 18.8 19.2 20.5

a. Use regression analysis to obtain an expression for the linear trend projection. b. Forecast John's Hair Salon's profit for the next five years.

ANSWER: © Cengage. Testing Powered by Cognero.

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CH 15 - Time Series Analysis and Forecasting a. b. 21.3, 22.2, 23.0, 23.9, 24.8 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.04 - 15.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.4 Linear Trend Projection KEYWORDS: Bloom's: Analyze 58. The number of pizzas ordered on Friday evenings between 5:30 and 6:30 at a pizza delivery location for the last 10 weeks is shown below. Use exponential smoothing with smoothing constants of 0.2 and 0.8 to forecast a value for Week 11. Compare your forecasts using MSE. Which smoothing constant would you prefer?

58, 46, 55, 39, 42, 63, 54, 55, 61, 52 ANSWER: FORECASTING WITH EXPONENTIAL SMOOTHING ************************************************ THE SMOOTHING CONSTANT IS 0.2 Time Period Actual Value Forecast 1 2 46 58.00 3 55 55.60 4 39 55.48 5 42 52.18 6 63 50.15 7 54 52.72 8 55 52.97 9 61 53.38 10 52 54.90 THE MEAN SQUARE ERROR 84.12 THE FORECAST FOR PERIOD 11 54.32

Forecast Error −12.00 −0.60 −16.48 −10.18 12.85 1.28 2.03 7.62 −2.90

FORECASTING WITH EXPONENTIAL SMOOTHING ************************************************ THE SMOOTHING CONSTANT IS 0.8 Time Period Actual Value Forecast 1 2 46 58.00 3 55 48.40 4 39 53.68 5 42 41.94 6 63 41.99 7 54 58.80 8 55 54.96 9 61 54.99 10 52 59.80 THE MEAN SQUARE ERROR 107.17 THE FORECAST FOR PERIOD 11 53.56

Forecast Error −12.00 6.60 −14.68 0.06 21.01 −4.80 0.04 6.01 −7.80

Based on MSE, smoothing with α = 0.2 provides a better model. © Cengage. Testing Powered by Cognero.

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CH 15 - Time Series Analysis and Forecasting POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Evaluate 59. Sales (in 1000s) of the new Thorton Model 506 convection oven over the eight-week period since its introduction have been as follows: Week 1 2 3 4 5 6 7 8 a. b.

Sales 18.6 21.4 25.2 22.4 24.6 19.2 21.7 23.8

Which exponential smoothing model provides better forecasts, one using α = 0.6 or α = 0.2?

Compare them using mean squared error.

Using the two forecast models in part (a), what are the forecasts for Week 9?

ANSWER:

a. b.

For α = 0.6, MSE = 9.12; for α = 0.2, MSE = 10.82 For α = 0.6, Ȳ9 = 22.86; for α = 0.2, Ȳ9 = 21.72

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.15.03 - 15.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 15.3 Moving Averages and Exponential Smoothing KEYWORDS: Bloom's: Evaluate

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CH 16 - Markov Processes True / False 1. Markov processes use historical probabilities. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Remember 2. All entries in a matrix of transition probabilities sum to 1. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Understand 3. All Markov chain transition matrices have the same number of rows as columns. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.16.02 - 16.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.2 Accounts Receivable Analysis KEYWORDS: Bloom's: Understand 4. A unique matrix of transition probabilities should be developed for each customer. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.16.02 - 16.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.2 Accounts Receivable Analysis © Cengage. Testing Powered by Cognero.

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CH 16 - Markov Processes KEYWORDS:

Bloom's: Understand

5. The probability that a system is in state 2 in the fifth period is π5(2). a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Understand 6. The fundamental matrix is derived from the matrix of transition probabilities and is relatively easy to compute for Markov processes with a small number of states. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.16.02 - 16.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.2 Accounts Receivable Analysis KEYWORDS: Bloom's: Understand 7. Steady-state probabilities are independent of initial state. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Understand 8. A Markov chain cannot consist of all absorbing states. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.16.02 - 16.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking © Cengage. Testing Powered by Cognero.

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CH 16 - Markov Processes TOPICS: KEYWORDS:

16.2 Accounts Receivable Analysis Bloom's: Understand

9. If an absorbing state exists, then the probability that a unit will ultimately move into the absorbing state is given by the steady-state probability. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.16.02 - 16.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.2 Accounts Receivable Analysis KEYWORDS: Bloom's: Understand 10. All Markov chains have steady-state probabilities. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.16.02 - 16.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.2 Accounts Receivable Analysis KEYWORDS: Bloom's: Understand 11. All entries in a row of a matrix of transition probabilities sum to 1. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Understand 12. A state i is a transient state if there exists a state j that is reachable from i, but the state i is not reachable from state j. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 © Cengage. Testing Powered by Cognero.

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CH 16 - Markov Processes NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Understand 13. When absorbing states are present, each row of the transition matrix corresponding to an absorbing state will have a single 1 and all other probabilities will be 0. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.16.02 - 16.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.2 Accounts Receivable Analysis KEYWORDS: Bloom's: Understand 14. For Markov processes having the memoryless property, the prior states of the system must be considered in order to predict the future behavior of the system. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Understand 15. When absorbing states are present, each row of the transition matrix corresponding to an absorbing state will have a single 1 and all other probabilities will be 0. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.16.02 - 16.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.2 Accounts Receivable Analysis KEYWORDS: Bloom's: Remember 16. Transition probabilities are conditional probabilities. a. True b. False ANSWER: True © Cengage. Testing Powered by Cognero.

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CH 16 - Markov Processes POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Understand 17. A state, i, is an absorbing state if, when i = j, pij = 1. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.16.02 - 16.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.2 Accounts Receivable Analysis KEYWORDS: Bloom's: Understand 18. If a Markov chain has at least one absorbing state, steady-state probabilities cannot be calculated. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Understand 19. State j is an absorbing state if pij = 1. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.16.02 - 16.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.2 Accounts Receivable Analysis KEYWORDS: Bloom's: Understand 20. A state is said to be absorbing if the probability of making a transition out of that state is zero. a. True b. False ANSWER: True © Cengage. Testing Powered by Cognero.

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CH 16 - Markov Processes POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.16.02 - 16.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.2 Accounts Receivable Analysis KEYWORDS: Bloom's: Understand 21. Transition probabilities indicate that a customer moves, or makes a transition, from a state in a given period to each state in the following period. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Understand Multiple Choice 22. In Markov analysis, we are concerned with the probability that the a. state is part of a system. b. system is in a particular state at a given time. c. time has reached a steady state. d. transition will occur.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Understand 23. For a situation with weekly dining at either an Italian or Mexican restaurant, a. the weekly visit is the trial and the restaurant is the state. b. the weekly visit is the state and the restaurant is the trial. c. the weekly visit is the trend and the restaurant is the transition. d. the weekly visit is the transition and the restaurant is the trend.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking © Cengage. Testing Powered by Cognero.

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CH 16 - Markov Processes TOPICS: KEYWORDS:

16.1 Market Share Analysis Bloom's: Understand

24. A transition probability describes a. the probability of a success in repeated, independent trials. b. the probability a system in a particular state now will be in a specific state next period. c. the probability of reaching an absorbing state. d. None of these are correct.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Understand 25. The probability of going from state 1 in period 2 to state 4 in period 3 is a. p12. b. p23. c. p14. d. p43.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Apply 26. The probability that a system is in a particular state after a large number of periods is a. independent of the beginning state of the system. b. dependent on the beginning state of the system. c. equal to one half. d. the same for every ending system.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Remember 27. At steady state, © Cengage. Testing Powered by Cognero.

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CH 16 - Markov Processes a. π1(n + 1) > π1(n). b. π1 = π2. c. π1 + π2 ≥ 1. d. π1(n + 1) = π1.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Apply 28. Markov process trials a. are used to describe future behavior of the system. b. are used to optimize the system. c. lead to higher-order decision making. d. All of these are correct.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Understand 29. If the probability of making a transition from a state is 0, then that state is called a(n) a. steady state. b. final state. c. origin state. d. absorbing state.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.16.02 - 16.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.2 Accounts Receivable Analysis KEYWORDS: Bloom's: Remember 30. Absorbing state probabilities are the same as a. steady-state probabilities. b. transition probabilities. c. fundamental probabilities. d. None of these are correct. © Cengage. Testing Powered by Cognero.

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CH 16 - Markov Processes ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.16.02 - 16.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.2 Accounts Receivable Analysis KEYWORDS: Bloom's: Remember 31. The probability of reaching an absorbing state is given by the a. R matrix. b. NR matrix. c. Q matrix. d. (I − Q)−1 matrix.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.16.02 - 16.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.2 Accounts Receivable Analysis KEYWORDS: Bloom's: Understand 32. Markov process models a. study system evolution over repeated trials. b. often study successive time periods. c. are often used when the state of the system in any particular period cannot be determined with certainty. d. All of these are correct.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Understand 33. The probability of making a transition from state i in a given period to state j in the next period is denoted as a. Pij. b. P = [i – j]. c. P( )ij. d. None of these are correct.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 © Cengage. Testing Powered by Cognero.

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CH 16 - Markov Processes NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Apply Subjective Short Answer 34. Calculate the steady-state probabilities for this transition matrix.

ANSWER:

π1 = 3/7, π2 = 4/7 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Analyze 35. Two airlines offer conveniently scheduled flights to the airport nearest your corporate headquarters. Historically, flights have been scheduled as reflected in this transition matrix.

Current Flight Airline A Airline B a. b. c.

Next Flight Airline A Airline B 0.6 0.4 0.2 0.8

If your last flight was on B, what is the probability your next flight will be on A? If your last flight was on B, what is the probability your second next flight will be on A? What are the steady-state probabilities?

ANSWER:

a. 0.2 b. 0.28 c. 1/3, 2/3 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Analyze 36. This matrix of transition probabilities deals with brand loyalty to Bark Bits and Canine Chow dog food.

Current Purchase Bark Bits Canine Chow

Next Purchase Bark Bits Canine Chow 0.75 0.25 0.20 0.80

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CH 16 - Markov Processes a. b.

What are the steady-state probabilities?

What is the probability that a customer will switch brands on the next purchase after a large number of periods?

ANSWER: a.

(π1 π2) = (π1 π2)

b.

π1 = 4/9, π2 = 5/9 P(switching) = (4/9)(1/4) + (5/9)(1/5) = 2/9

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Analyze 37. Bark Bits Company is planning an advertising campaign to raise the brand loyalty of its customers to 0.80. a. The former transition matrix is as follows:

b. c.

What is the new one?

What are the new steady-state probabilities?

If each point of market share increases profit by $15,000, what is the most you would pay for the advertising?

ANSWER:

a.

b. c.

π1 = 0.5, π2 = 0.5 The increase in market share is 0.5 − 0.444 = 0.056.

(5.6 points)($15,000/point) = $84,000 value for the campaign.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Analyze 38. A city is served by three cable TV companies: Xcellent Cable, Your Cable, and Zephyr Cable. A survey of 1000 cable subscribers shows this breakdown of customers from the beginning to the end of August.

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CH 16 - Markov Processes Company on August 1 Xcellent Your Zephyr a. b. c.

Company on August 31 Xcellent Your Zephyr 300 50 50 10 200 40 40 80 230

Construct the transition matrix.

What was each company's market share at the beginning and end of the month?

If the current trend continues, what will the market shares be?

ANSWER: a.

P=

b.

Market shares, August 1: X: 0.40, Y: 0.25, Z: 0.35 Market shares, August 31: X: 0.35, Y: 0.33, Z: 0.32 Steady states: 0.2156, 0.48125, 0.30315

c.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Evaluate 39. A television ratings company surveys 100 viewers on March 1 and April 1 to find out what is being watched at 6:00 p.m.—the local NBC affiliate's local news, the CBS affiliate's local news, or "Other" which includes all other channels and not watching TV. The results show the following: March 1 Choice NBC CBS Other a. b. c.

Record of Switches During March to NBC CBS Other – 5 10 15 – 5 5 5 –

Number 30 40 30

What are the numbers in each choice for April 1?

What is the transition matrix?

What ratings percentages do you predict for May 1?

ANSWER:

a.

NBC 35, CBS 30, Other 35

b.

P=

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CH 16 - Markov Processes c. π3 =

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Evaluate 40. Accounts receivable have been grouped into the following states: State 1: Paid State 2: Bad debt State 3: 0–30 days old State 4: 31–60 days old Sixty percent of all new bills are paid before they are 30 days old. The remainder go to state 4. Seventy percent of all 30day-old bills are paid before they become 60 days old. If not paid, they are permanently classified as bad debts. a. Set up the one-month Markov transition matrix. b. What is the probability that an account in state 3 will be paid?

ANSWER:

a.

Paid Bad 0–30 31–60

Paid 1 0 0.6 0.7

b.

Bad 0 1 0 0.3

0–30 0 0 0 0

31–60 0 0 0.4 0

NR = So, the probability that an account in state 3 will be paid is 0.88.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.16.02 - 16.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 16.2 Accounts Receivable Analysis KEYWORDS: Bloom's: Evaluate 41. Rent-To-Keep rents household furnishings by the month. At the end of a rental month, a customer can (a) rent the item for another month, (b) buy the item, or (c) return the item. The matrix below describes the month-to-month transition probabilities for the 32-inch stereo televisions the shop stocks.

This

Next Month

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CH 16 - Markov Processes Month Rent Buy Return

Rent 0.72 0 0

Buy 0.10 1 0

Return 0.18 0 1

What is the probability that a customer who rented a TV this month will eventually buy it?

ANSWER: P(buy) = 0.357, P(return) = 0.643 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.16.02 - 16.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 16.2 Accounts Receivable Analysis KEYWORDS: Bloom's: Evaluate 42. On any particular day, an individual can take one of two routes to work. Route A has a 25% chance of being congested, whereas route B has a 40% chance of being congested.

The probability of the individual taking a particular route depends on his previous day's experience. If one day he takes route A and it is not congested, he will take route A again the next day with probability 0.8. If it is congested, he will take route B the next day with probability 0.7.

On the other hand, if he takes route B one day and it is not congested, he will take route B again the next day with probability 0.9. Similarly, if route B is congested, he will take route A the next day with probability 0.6. a. Construct the transition matrix for this problem. (Hint: There are four states corresponding to the route taken and the congestion. The transition probabilities are products of the independent probabilities of congestion and next-day choice.) b. What is the long-run proportion of time that route A is taken?

ANSWER:

a.

This Day Route A – Not Congested (RA-NC) Route A – Congested (RAC) Route B – Not Congested (RB-NC) Route B – Congested (RBC)

RA-NC 0.600

Next Day RA-C RB-NC 0.200 0.120

RB-C 0.080

0.225

0.075

0.420

0.280

0.075

0.025

0.540

0.360

0.450

0.150

0.240

0.160

b. 0.36 + 0.12 = 0.48 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Evaluate 43. Henry, a persistent salesman, calls North's Hardware Store once a week hoping to speak with the store's buying agent, Shirley. If Shirley does not accept Henry's call this week, the probability she will do the same next week is 0.35. On the other hand, if she accepts Henry's call this week, the probability she will not do so next week is 0.20. © Cengage. Testing Powered by Cognero.

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CH 16 - Markov Processes a. b. c. d.

Construct the transition matrix for this problem.

How many times per year can Henry expect to talk to Shirley?

What is the probability Shirley will accept Henry's next two calls if she does not accept his call this week?

What is the probability of Shirley accepting exactly one of Henry's next two calls if she accepts his call this week?

ANSWER:

a. The transition matrix is: This Week's Call Refuses Accepts

Refuses 0.35 0.20

Next Week's Call Accepts 0.65 0.80

b. 39.76 or about 40 c. 0.52 d. 0.13 + 0.16 = 0.29 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Create 44. The evening television news broadcast that individuals view on one evening is influenced by which broadcast they viewed previously. An executive at network C has determined the following transition probability matrix describing this phenomenon: Current Network News Watched A C N

A 0.80 0.08 0.08

Next Network News Watched C N 0.12 0.08 0.85 0.07 0.09 0.83

a. Which network has the most loyal viewers? b. What are the three networks' long-run market shares? c. Suppose each of the three networks earns $1,250 in daily profit from advertising revenue for each 1,000,000 viewers it has. If, on average, 40,000,000 people watch the evening television news, compute the long-run average daily profit each network generates from its evening news broadcast.

ANSWER: a. network C b. (pA, pC, pN) = (0.29, 0.41, 0.30) c. network A: $14,300; network C: $20,550; network N: $15,200 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Evaluate 45. Precision Craft, Inc., manufactures ornate pedestal sinks. On any day, the status of a given sink is either: (a) © Cengage. Testing Powered by Cognero.

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CH 16 - Markov Processes somewhere in the normal manufacturing process, (b) being reworked because of a detected flaw, (c) finished successfully, or (d) scrapped because a flaw could not be corrected. The transition matrix is as follows: Today's Status In-process Rework Finished Scrapped

In-Process 0.30 0.40 0 0

Tomorrow's Status Rework Finished 0.15 0.50 0.10 0.30 0 1 0 0

Scrapped 0.05 0.20 0 1

a. What is the probability of a sink eventually being finished if it is currently in process? b. What is the probability of a sink eventually being scrapped if it is currently in rework? c. What is the probability that a sink currently in rework will have a "finished" status either tomorrow or the next day?

(Hint: There are three ways this can happen.)

ANSWER:

a. 0.868 b. 0.281 c. 0.3 + 0.03 + 0.2 = 0.53 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.16.01 - 16.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 16.1 Market Share Analysis KEYWORDS: Bloom's: Evaluate

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CH 17 - Linear Programming: Simplex Method True / False 1. When a system of simultaneous equations has more variables than equations, there is a unique solution. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.17.01 - 17.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.1 An Algebraic Overview of the Simplex Method KEYWORDS: Bloom's: Remember 2. In order to determine a basic solution, set n − m of the variables equal to zero, and solve the m linear constraint equations for the remaining m variables. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.17.01 - 17.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.1 An Algebraic Overview of the Simplex Method KEYWORDS: Bloom's: Understand 3. A basic feasible solution also satisfies the nonnegativity restriction. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.17.01 - 17.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.1 An Algebraic Overview of the Simplex Method KEYWORDS: Bloom's: Understand 4. The simplex method is an iterative procedure for moving from one basic feasible solution (an extreme point) to another until the optimal solution is reached. a. True b. False ANSWER: POINTS:

True 1

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CH 17 - Linear Programming: Simplex Method DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.17.01 - 17.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.1 An Algebraic Overview of the Simplex Method KEYWORDS: Bloom's: Remember 5. In a simplex tableau, a variable is associated with each column and both a constraint and a basic variable are associated with each row. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.17.03 - 17.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.3 Setting up the Initial Simplex Tableau KEYWORDS: Bloom's: Remember 6. At each iteration of the simplex procedure, a new variable becomes basic and a currently basic variable becomes nonbasic, preserving the same number of basic variables and improving the value of the objective function. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.17.04 - 17.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.4 Improving the Solution KEYWORDS: Bloom's: Understand 7. Coefficients in a nonbasic column in a simplex tableau indicate the amount of decrease in the current basic variables when the value of the nonbasic variable is increased from 0 to 1. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.17.03 - 17.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.3 Setting up the Initial Simplex Tableau KEYWORDS: Bloom's: Understand 8. If a variable is not in the basis, its value is 0. a. True b. False © Cengage. Testing Powered by Cognero.

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CH 17 - Linear Programming: Simplex Method ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.17.03 - 17.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.3 Setting up the Initial Simplex Tableau KEYWORDS: Bloom's: Understand 9. The purpose of row operations is to create a unit column for the entering variable while maintaining unit columns for the remaining basic variables. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.17.05 - 17.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.5 Calculating the Next Tableau KEYWORDS: Bloom's: Understand 10. Infeasibility can be recognized when the optimality criterion indicates that an optimal solution has been obtained, and one or more of the artificial variables remain in the solution at a positive value. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.17.08 - 17.8 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.8 Special Cases KEYWORDS: Bloom's: Understand 11. Artificial variables are added for the purpose of obtaining an initial basic feasible solution. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.17.06 - 17.6 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.6 Tableau Form: The General Case KEYWORDS: Bloom's: Understand 12. A solution is optimal when all values in the cj − zj row of the simplex tableau are either zero or positive. a. True © Cengage. Testing Powered by Cognero.

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CH 17 - Linear Programming: Simplex Method b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.17.05 - 17.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.5 Calculating the Next Tableau KEYWORDS: Bloom's: Understand 13. The variable to enter into the basis is the variable with the largest positive cj − zj value. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.17.07 - 17.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.7 Solving a Minimization Problem KEYWORDS: Bloom's: Understand 14. The variable to remove from the current basis is the variable with the smallest positive cj − zj value. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.17.07 - 17.7 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.7 Solving a Minimization Problem KEYWORDS: Bloom's: Understand 15. The coefficient of an artificial variable in the objective function is zero. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.17.06 - 17.6 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.6 Tableau Form: The General Case KEYWORDS: Bloom's: Understand Multiple Choice © Cengage. Testing Powered by Cognero.

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CH 17 - Linear Programming: Simplex Method 16. Algebraic methods such as the simplex method are used to solve a. nonlinear programming problems. b. any size linear programming problem. c. programming problems under uncertainty. d. graphical models.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.17.01 - 17.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.1 An Algebraic Overview of the Simplex Method KEYWORDS: Bloom's: Understand 17. A basic feasible solution is a basic solution that a. also satisfies the nonnegativity conditions. b. contains 0 variables. c. corresponds to no extreme points. d. None of these are correct.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.17.01 - 17.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.1 An Algebraic Overview of the Simplex Method KEYWORDS: Bloom's: Understand 18. A basic solution and a basic feasible solution a. are the same thing. b. differ in the number of variables allowed to be zero. c. describe interior points and exterior points, respectively. d. differ in their inclusion of nonnegativity restrictions.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.17.01 - 17.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.1 An Algebraic Overview of the Simplex Method KEYWORDS: Bloom's: Understand 19. Whenever a system of simultaneous linear equations has more variables than equations, a. it is a basic set. b. it is a feasible set. c. there is a unique solution. d. we can expect an infinite number of solutions. © Cengage. Testing Powered by Cognero.

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CH 17 - Linear Programming: Simplex Method ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.17.01 - 17.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.1 An Algebraic Overview of the Simplex Method KEYWORDS: Bloom's: Understand 20. Unit columns are used to identify the a. tableau. b. c row. c. b column. d. basic variables.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.17.03 - 17.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.3 Setting up the Initial Simplex Tableau KEYWORDS: Bloom's: Understand 21. The values in the cj − zj , or net evaluation, row indicate the a. value of the objective function. b. decrease in value of the objective function that will result if one unit of the variable corresponding to the jth column of the A matrix is brought into the basis. c. net change in the value of the objective function that will result if one unit of the variable corresponding to the jth column of the A matrix is brought into the basis. d. values of the decision variables.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.17.03 - 17.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.3 Setting up the Initial Simplex Tableau KEYWORDS: Bloom's: Understand 22. The purpose of the tableau form is to provide a(n) a. infeasible solution. b. optimal infeasible solution. c. initial basic feasible solution. d. degenerate solution.

ANSWER: c POINTS: 1 DIFFICULTY: Easy © Cengage. Testing Powered by Cognero.

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CH 17 - Linear Programming: Simplex Method LEARNING OBJECTIVES: IMS.ASWC.19.17.03 - 17.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.3 Setting up the Initial Simplex Tableau KEYWORDS: Bloom's: Understand 23. Which of the following is NOT a step that is necessary to prepare a linear programming problem for solution using the simplex method? a. Formulate the problem. b. Set up the standard form by adding slack and/or subtracting surplus variables. c. Perform elementary row and column operations. d. Set up the tableau form.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.17.02 - 17.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.2 Tableau Form KEYWORDS: Bloom's: Understand 24. In the simplex method, a tableau is optimal only if all the cj − zj values are a. zero or negative. b. zero. c. negative and nonzero. d. positive and nonzero.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.17.05 - 17.5 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.5 Calculating the Next Tableau KEYWORDS: Bloom's: Understand 25. The way we guarantee that artificial variables will be eliminated before the optimal solution is reached is to assign each artificial variable the coefficient value of a. 0. b. 1. c. a very large negative number. d. a very large positive number.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.17.06 - 17.6 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.6 Tableau Form: The General Case © Cengage. Testing Powered by Cognero.

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CH 17 - Linear Programming: Simplex Method KEYWORDS:

Bloom's: Understand

26. If one or more of the basic variables in a linear program have a value of zero, a. post-optimality analysis is required. b. their dual prices will be equal. c. converting the pivot element will break the tie. d. a condition of degeneracy is present.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.17.08 - 17.8 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.8 Special Cases KEYWORDS: Bloom's: Understand 27. An alternative optimal solution is indicated when, in the simplex tableau, a a. nonbasic variable has a value of zero in the cj − zj row. b. basic variable has a positive value in the cj − zj row. c. basic variable has a value of zero in the cj − zj row. d. nonbasic variable has a positive value in the cj − zj row.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.17.08 - 17.8 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.8 Special Cases KEYWORDS: Bloom's: Understand 28. Infeasibility exists when one or more of the artificial variables a. remain in the final solution as a negative value. b. remain in the final solution as a positive value. c. have been removed from the basis. d. remain in the basis.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.17.08 - 17.8 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.8 Special Cases KEYWORDS: Bloom's: Understand 29. Which of the following steps is included in the preparation of a linear programming problem for a solution using the simplex method? a. Formulate the problem. © Cengage. Testing Powered by Cognero.

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CH 17 - Linear Programming: Simplex Method b. Set up the standard form by adding slack and/or subtracting surplus variables. c. Set up the tableau form. d. All of these are correct.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.17.02 - 17.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 17.2 Tableau Form KEYWORDS: Bloom's: Understand Subjective Short Answer 30. Write the following problem in tableau form. Which variables would be in the initial basic solution?

Min Z = 3x1 + 8x2 s.t. x1 + x2 ≤ 200 x1 ≤ 80 x2 ≤ 60 ANSWER: Min Z s.t.

= 3x1 + 8x2 + Ma1 + Ma2 x1 + x2 + a1 = 200 x1 + s1 = 80 x2 + s2 + a2 = 60

The initial basis includes a1, a2, and s2.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.17.06 - 17.6 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 17.6 Tableau Form: The General Case KEYWORDS: Bloom's: Create 31. An initial simplex tableau is shown below.

Basis s1 s2 s3

cB 0 0 0 zj cj − zj

a. b. c.

x1 5

x2 8

x3 12

s1 0

s2 0

s3 0

3 9 1

4 15 −1

5 20 2

1 0 0

0 1 0

0 0 0

80 250 20

0 5

0 8

0 12

0 0

0 0

0 0

0

What variables form the basis?

What are the current values of the decision variables?

What is the current value of the objective function?

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CH 17 - Linear Programming: Simplex Method d. Which variable will be made positive next, and what will its value be? e. Which variable that is currently positive will become 0? f. What value will the objective function have next?

ANSWER: a. s1, s2 , s3 b. x1 = 0, x2 = 0, x3 = 0, s1 = 80, s2 = 250, s3 = 20 c. 0 d. x3, 10 e. s3 f. z = 120 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.17.05 - 17.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 17.5 Calculating the Next Tableau KEYWORDS: Bloom's: Evaluate 32. A simplex tableau is shown below.

Basis

cB

x1 3 1/2 0

x2 4 1 0

x3 5 0 1

s1 0 1/2 −1/4

s2 0 −1/2 1

6 3

cj cj − zj a. What variables form the basis? b. What are the current values of the decision variables? c. What is the current value of the objective function? d. Which variable will be made positive next, and what will its value be? e. Which variable that is currently positive will become 0? f. What value will the objective function have next?

ANSWER: a. x2, x3 b. x1 = 0, x2 = 6, x3 = 3 c. 39 d. x1, 12 e. x2 f. 51 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.17.05 - 17.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 17.5 Calculating the Next Tableau KEYWORDS: Bloom's: Evaluate 33. A simplex tableau is shown below. © Cengage. Testing Powered by Cognero.

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CH 17 - Linear Programming: Simplex Method

Basis s1 s2 x3

a. b. c.

cB 0 0 8

x1 3

x2 5

x3 8

s1 0

s2 0

s3 0

3 −5/2 1/2

6 −1/2 1/2

0 0 1

1 0 0

0 1 0

−9 −9/2 1/2

126 45 18

zj cj − zj

4 −1

4 1

8 0

0 0

0 0

4 −4

144

Perform one more iteration of the simplex procedure.

What is the current complete solution?

Is this solution optimal? Why or why not?

ANSWER:

a.

Basis x2 s2 x3

x1 3

x2 5

x3 8

s1 0

s2 0

1/2 −9/4 1/4

1 0 0

0 0 1

1/6 1/12 −1/12

0 1 0

−3/2 21 −21/4 111/2 −5/4 15/2

9/2 zj cj − zj −3/2

5 0

8 0

1/6 −1/6

0 0

5/2 −5/2

cB 5 0 8

s3 0

165

b. x1 = 0, x2 = 21, x3 = 7.5, s1 = 0, s2 = 55.5, s3 = 0, Z = 165 c. The solution is optimal because all cj − zj ≤ 0.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.17.05 - 17.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 17.5 Calculating the Next Tableau KEYWORDS: Bloom's: Analyze 34. A simplex tableau is shown below.

Basis s1 x3 s3

cB 0 8 0

x1 5

x2 4

x3 8

s1 0

s2 0

s3 0

2/5 4/5 4/5

−3/5 4/5 9/5

0 1 0

1 0 0

−2/5 1/5 1/5

0 0 1

4 8 10

zj cj − zj

32/5 −7/5

32/5 −12/5

8 0

0 0

8/5 −8/5

0 0

64

a. What is the current complete solution? b. The 32/5 for z1 is composed of 0 + 8(4/5) + 0. Explain the meaning of this number. c. Explain the meaning of the −12/5 value for c 2 − z2.

ANSWER: © Cengage. Testing Powered by Cognero.

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CH 17 - Linear Programming: Simplex Method a.

x1 = 0, x2 = 0, x3 = 8, s1 = 4, s2 = 0, s3 = 10, Z = 64

b.

If we make one unit of x1, it will require 0.4 unit of s1, at no cost; 0.8 unit of x3 production will have to be foregone at a cost of 8 each; and 0.8 unit of s3 will be needed at no cost. In order to make a unit of x1, profit of 32/5 will be given up.

c.

The profit for a unit of x2 is 4. However, in order to make a unit of x2, profit of 32/5 would be given up as resources are diverted. The net improvement is a loss, shown by the −12/5. Therefore, it makes no sense to produce x2.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.17.05 - 17.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 17.5 Calculating the Next Tableau KEYWORDS: Bloom's: Evaluate 35. Solve the following problem by the simplex method.

Max s.t.

14x1 + 14.5x2 + 18x3 x1 + 2x2 + 2.5x3 ≤ 50 x1 + x2 + 1.5x3 ≤ 30 x1 , x2 , x3 ≥ 0 ANSWER: Basis s1 s2

x1 14

x2 14.5

x3 18

s1 0

s2 0

1 1

2 1

2.5 1.5

1 0

0 1

50 30

0 14

0 14.5

0 18

0 0

0 0

0

cB 18 0

x1 14

x2 14.5

x3 18

s1 0

s2 0

0.4 0.4

0.8 −0.2

1 0

0.4 −0.6

0 1

20 0

zj cj − zj

7.20 6.8

14.4 0.1

18 0

7.2 −7.2

0 0

360

x1 14

x2 14.5

x3 18

s1 0

s2 0

0 1

1 −0.5

1 0

1 −1.5

−1 2.5

20 0

14 0

11 3.5

18 0

−3 −3

17 −17

360

cB 0 0 zj cj − zj

Basis x3 s2

Basis x3 x1

cB 18 14 zj cj − zj

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Page 12

CH 17 - Linear Programming: Simplex Method

Basis x2 x1

cB 14.5 14 zj cj − zj

x1 14

x2 14.5

x3 18

s1 0

s2 0

0 1

1 0

1 0.5

1 −1

−1 2

20 10

14 0

14.5 0

21.5 −3.5

0.5 −0.5

13.5 −13.5

430

POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.17.05 - 17.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 17.5 Calculating the Next Tableau KEYWORDS: Bloom's: Evaluate 36. Solve the following problem by the simplex method.

Max s.t.

100x1 + 120x2 + 85x3 3x1 + 1x2 + 6x3 ≤ 120 5x1 + 8x2 + 2x3 ≤ 160 x1 , x2 , x3 ≥ 0 ANSWER: Basis s1 s2

x1 100

x2 120

x3 85

s1 0

s2 0

3 5

1 8

6 2

1 0

0 1

120 160

0 100

0 120

0 85

0 0

0 0

0

cB 0 120

x1 100

x2 120

x3 85

s1 0

s2 0

2.375 0.625

0 1

5.75 0.25

1 0

−0.125 0.125

100 20

zj cj − zj

75 25

120 0

30 55

0 0

15 −15

2400

cB 85 120

x1 100

x2 120

x3 85

s1 0

s2 0

0.413 0.522

0 1

1 0

0.174 −0.043

−0.0217 0.1304

17.391 15.652

zj cj − zj

97.745 2.283

120 0

85 0

9.63 −9.565

13.8035 −13.8035

3356.52

cB

x1 100

x2 120

x3 85

s1 0

s2 0

cB 0 0 zj cj − zj

Basis s1 x2

Basis x3 x2

Basis © Cengage. Testing Powered by Cognero.

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CH 17 - Linear Programming: Simplex Method x3 x1

85 100

0 1

−0.792 1.917

1 0

0.2083 −0.0833

−0.125 0.250

5 30

zj cj − zj

100 0

124.38 −4.375

85 0

9.3755 −9.375

14.375 −14.375

3425

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.17.05 - 17.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 17.5 Calculating the Next Tableau KEYWORDS: Bloom's: Evaluate 37. Determine from a review of the following tableau whether the linear programming problem has multiple optimal solutions.

Basis s3 x2 x1

cB 0 2 3

x1 0 0 1

x2 0 1 0

s1 1 0 0

s2 −1/5 1/5 1/5

s3 8/6 −3/5 2/5

zj cj −zj

3 0

2 0

0 0

0 −1

0 0

6 1 4 14

ANSWER:

Since nonbasic variable s3 has a relative profit of zero, any increase in s3 will produce no change in the objective function value. Thus, since s3 can be made a basic variable, the resulting basic feasible solution will also have an optimum value of 14. An alternative optimal solution is indicated whenever there exists a nonbasic variable whose relative profit (cj − z) row coefficient is zero in the optimal solution.

POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.17.08 - 17.8 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 17.8 Special Cases KEYWORDS: Bloom's: Analyze 38. Write the following problem in tableau form. Which variables would be in the initial basis?

Max s.t.

x1 + 2x2 3x1 + 4x2 ≤ 100 2x1 + 3.5x2 ≥ 60 2x1 − 1x2 = 4 x1 , x2 ≥ 0 ANSWER: Max s.t.

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x1 + 2x2 − Ma1 − Ma2 3x1 + 4x2 + s1 = 100 2x1 + 3.5x2 − s2 + a1 = 60 2x1 − 1x2 + a2 = 50 Page 14

CH 17 - Linear Programming: Simplex Method x1 , x2 , s1 , s2 , a1 , a2 ≥ 0 The initial basis would include s1, a1, and a2.

POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.17.06 - 17.6 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 17.6 Tableau Form: The General Case KEYWORDS: Bloom's: Analyze 39. Write the following problem in tableau form. Which variables would be in the initial basic solution?

Min Z s.t.

= −3x1 + x2 + x3 x1 − 2x2 + x3 ≤ 11 −4 x1 + x2 + 2x3 ≥ 3 2x1 − x3 ≥ −1

ANSWER:

= −3x1 + x2 + x3 + Ma1 + Ma2 x1 − 2x2 + x3 + s1 = 11 −4x1 + x2 + 2 x3 − s2 + a1 = 3 −2x1 + x3 + a2 = 1 The initial basis includes s1, a1, and a2.

2x1 + 3.5x2 − s2 + a1 = 60 2x1 − 1x2 + a2 = 50 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.17.06 - 17.6 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 17.6 Tableau Form: The General Case KEYWORDS: Bloom's: Analyze Min Z s.t.

40. Identify the type of solution shown in this simplex tableau.

Basis a1 x3

cB −M 5

x1 1

x2 2

x3 5

s1 0

s2 0

a1 −M

−3 1

−1 1/2

0 1

−1 0

−2 1/2

1 0

4 4

5 0

M −M

2.5 + 2M −2.5 − 2M

−M 0

−4M + 20

5 + 3M 2.5 + M zj cj − zj −4 − 3M −0.5 − M

ANSWER: The tableau indicates an infeasible solution.

POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.17.08 - 17.8 NATIONAL STANDARDS: United States - BUSPROG: Analytic © Cengage. Testing Powered by Cognero.

Page 15

CH 17 - Linear Programming: Simplex Method TOPICS: KEYWORDS:

17.8 Special Cases Bloom's: Analyze

41. What type of solution is shown in this simplex tableau?

Basis s1 x2

cB 0 6

x1 4

x2 6

x3 5

s1 0

s2 0

3 1

0 1

−1 1

1 0

−1 −1

26 10

6 6 0 −6 60 0 −1 0 6 ANSWER: The tableau indicates an unbounded solution.

POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.17.08 - 17.8 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 17.8 Special Cases KEYWORDS: Bloom's: Analyze zj cj − zj

6 −2

42. Joe Forrester, an operations analyst for a manufacturing company, developed the following LP formulation. From it, create an initial simplex tableau.

Max 40xl + 30x2 + 50x3 s.t. 2xl + 3x2 + 4x3 < 200 x1 + 2x2 + 2x3 < 300 3xl + x2 + 5x3 < 500 ANSWER: Basis s1 s2 s3

cB 0 0 0 zj cj – zj

x1 40

x2 30

x3 50

s1 0

s2 0

s3 0

3 9 1

4 15 –1

5 20 2

1 0 0

0 1 0

0 0 1

200 300 500

0 40

0 30

0 50

0 0

0 0

0 0

0

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.17.03 - 17.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 17.3 Setting up the Initial Simplex Tableau KEYWORDS: Bloom's: Create 43. Jackie Quinn developed the following LP formulation for a problem she is working on and now needs to create an initial simplex tableau. Create the initial simplex tableau. © Cengage. Testing Powered by Cognero.

Page 16

CH 17 - Linear Programming: Simplex Method Min 75xl + 45x2 s.t. 3xl + 2x2 > 10 xl + 6x2 > 15 ANSWER: Basis

cB 0 0

a1 a2

zj cj – zj

x1 –75

x2 –45

a1 –M

a2 –M

s1 0

s2 0

3 1

2 6

1 0

0 1

–1 0

0 –1

10 15

–4M –75 + 4M

–8M –45 + 8M

–M 50

–M 0

M –M

M –M

–25M

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.17.03 - 17.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 17.3 Setting up the Initial Simplex Tableau KEYWORDS: Bloom's: Create 44. A student in a Management Science class developed this initial tableau for a maximization problem and now wants to perform row operations to create the next tableau and check for an optimal solution. Create the next tableau. Is there an optimal solution?

Basis s1 s2 s3

cB 0 0 0 zj cj – zj

ANSWER:

x1 550

x2 350

s1 0

s2 0

s3 0

2 2 1

2 1 2

1 0 0

0 1 0

0 0 1

2000 1500 3000

0 550

0 350

0 0

0 0

0 0

0

x1 550

x2 350

s1 0

s2 0

s3 0

0 1 0

1 1/2 3/2

1 0 0

–1 1/2 –1/2

0 0 1

500 750 2250

550 0

275 75

0 0

275 –275

0 0

412,500

Second tableau Basis s1 x1 s3

cB 0 550 0 zj cj – zj

POINTS:

Solution is not optimal. There is a positive cj – zj value.

1

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Page 17

CH 17 - Linear Programming: Simplex Method DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.17.05 - 17.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 17.5 Calculating the Next Tableau KEYWORDS: Bloom's: Create 45. A student in a Management Science class wants to perform row operations on this second tableau and complete the third tableau to see if it is optimal. If it is optimal, what is the optimal answer?

Basis s1 x1 s3

cB 0 550 0 zj cj – zj

ANSWER:

x1 550

x2 350

s1 0

s2 0

s3 0

0 1 0

1 1/2 3/2

1 0 0

–1 1/2 –1/2

0 0 1

500 750 2250

550 0

275 75

0 0

275 –275

0 0

412,500

x1 550

x2 350

s1 0

s2 0

s3 0

0 1 0

1 0 0

1 –1/2 –3/2

–1 1 1

0 0 1

500 500 1500

550 0

350 0

75 –75

200 –200

0 0

450,000

Third tableau Basis x2 x1 s3

cB 350 550 0 zj cj – zj

Solution is optimal. There are no positive cj – zj values. x1 = 550, x2 = 350, and objective function value is 450,000.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.17.05 - 17.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 17.5 Calculating the Next Tableau KEYWORDS: Bloom's: Create 46. An operations research analyst for a communications company has the following LP problem and wants to solve it using the simplex method.

Max 50x1 + 20x2 s.t. 2x1 + x2 < 200 x1 + x2 < 350 xl + 2x2 < 275 ANSWER:

Initial tableau

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CH 17 - Linear Programming: Simplex Method

Basis s1 s2 s3

cB 0 0 0 zj cj – zj

x1 50

x2 20

s1 0

s2 0

s3 0

2 1 1

1 1 2

1 0 0

0 1 0

0 0 1

200 350 275

0 50

0 20

0 0

0 0

0 0

0

x1 50

x2 20

s1 0

s2 0

s3 0

1 0 0

1/2 1/2 3/2

1/2 –1/2 –1/2

0 1 0

0 0 1

100 250 175

50 0

25 –5

25 –25

0 0

0 0

5000

Second tableau

Basis x1 s2 s3

cB 50 0 0 zj cj – zj

Tableau is optimal. Optimal solution is x1 = 100, and the objective function = 5000.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.17.05 - 17.5 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 17.5 Calculating the Next Tableau KEYWORDS: Bloom's: Create 47. The operations research analyst for a big manufacturing firm in Oregon developed the following variable definitions for an LP maximization problem she was working on. The company was trying to determine how many consoles of each model to produce next week given each console had to go through three production departments. She obtained the following optimal simplex tableau for the problem and wanted to interpret its meaning.

Variable definitions: xl = number of model 1 consoles produced x2 = number of model 2 consoles produced s1 = unused personnel hours in department 1 s2 = unused personnel hours in department 2 s3 = unused personnel hours in department 3 objective function = total profit on model 1 and model 2 consoles produced in the coming week Optimal tableau Basis x1 s2 x2

cB 180 0 350

x1

x2

s1

s2

s3

180

350

0

0

0

1 0 0

0 0 1

1 –1 0

0 1 0

–1 1 1

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15 10 20 Page 19

CH 17 - Linear Programming: Simplex Method 180 0

zj cj – zj ANSWER:

350 0

Optimal solution:

180 –180

0 0

170 –170

9700

Produce 15 model 1 consoles, produce 20 model 2 consoles, all personnel hours will be used in departments 1 and 3, 10 personnel hours will be unused in department 2, and total profit next week will be $9,700.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.17.08 - 17.8 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 17.8 Special Cases KEYWORDS: Bloom's: Evaluate 48. A manager for a food company is putting together a buffet, and she is trying to determine the best mix of crab and steak to be served. Below are variable definitions she developed including vitamin, mineral, and protein requirements and an optimal simplex tableau she obtained from her computations. She is interested in interpreting what it means.

Variable definitions: xl = amount of crab (oz.) to be served per buffet batch x2 = amount of steak (oz.) to be served per buffet batch s1 = vitamin A units provided in excess of requirements s2 = mineral units provided in excess of requirements s3 = protein units provided in excess of requirements Optimal tableau Basis s2 x2 x1

cB 0 –4 –7 zj cj – zj

ANSWER:

x1 –7

x2 –4

a1 –M

a2 –M

a3 –M

s1 0

s2 0

s3 0

0 0 1

0 1 0

4 1/50 –1/100

–2 0 0

–1 –1/100 1/100

–4 –1/50 1/100

1 0 0

2 1/100 –1/100

2500 20 15

–7 0

–4 0

1/100 –M

0 –M

–3/100 –M

1/100 –1/100

0 0

170 –3/100

–185

Optimal solution:

15 ounces of crab in each batch, 20 ounces of steak in each batch, 0 excess units of vitamin A provided per batch, 2500 excess units of minerals provided per batch, 0 excess units of protein provided per batch, and total cost per batch = $185.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.17.08 - 17.8 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 17.8 Special Cases © Cengage. Testing Powered by Cognero.

Page 20

CH 17 - Linear Programming: Simplex Method KEYWORDS:

Bloom's: Evaluate

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Page 21

CH 18 - Simplex-Based Sensitivity Analysis and Duality True / False 1. The range of optimality is useful only for basic variables. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Remember 2. The range of optimality is calculated by considering changes in the cj − zj value of the variable in question. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Understand 3. As long as the objective function coefficient remains within the range of optimality, the variable values will not change although the value of the objective function could. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Understand 4. If the simplex tableau is from a maximization converted from a minimization, the signs and directions of the inequalities that give the objective function ranges will need to be adjusted to apply to the original coefficients. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking © Cengage. Testing Powered by Cognero.

Page 1

CH 18 - Simplex-Based Sensitivity Analysis and Duality TOPICS: KEYWORDS:

18.1 Sensitivity Analysis with the Simplex Tableau Bloom's: Understand

5. The ranges for which the right-hand-side values are valid are the same as the ranges over which the dual prices are valid. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Understand 6. A dual price is associated with each decision variable. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Understand 7. The dual price for an equality constraint is the zj value for its artificial variable. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Understand 8. The entries in the associated slack column of the final tableau can also be interpreted as the changes in the values of the current basic variables corresponding to a one-unit increase in the right-hand side. a. True b. False ANSWER: POINTS: DIFFICULTY:

True 1 Easy

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CH 18 - Simplex-Based Sensitivity Analysis and Duality LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Understand 9. The range of optimality for a basic variable defines the objective function coefficient values for which that variable will remain part of the current optimal basic feasible solution. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Remember 10. The dual price is the improvement in value of the optimal solution per unit increase in a constraint's right-hand-side value. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.18.02 - 18.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 18.2 Duality KEYWORDS: Bloom's: Remember 11. As long as the actual value of the objective function coefficient is within the range of optimality, the current basic feasible solution will remain optimal. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Understand 12. We can often avoid the process of formulating and solving a modified linear programming problem by using the range of optimality to determine whether a change in an objective function coefficient is large enough to cause a change in the optimal solution. a. True © Cengage. Testing Powered by Cognero.

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CH 18 - Simplex-Based Sensitivity Analysis and Duality b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Understand 13. Within the concept of duality is the original formulation of a linear programming problem known as the primal problem. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.18.02 - 18.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 18.2 Duality KEYWORDS: Bloom's: Remember Multiple Choice 14. Dual prices and ranges for objective function coefficients and right-hand-side values are found by considering a. dual analysis. b. optimality analysis. c. ranging analysis. d. sensitivity analysis.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Understand 15. For the basic feasible solution to remain optimal, a. all cj − zj values must remain ≤ 0. b. no objective function coefficients are allowed to change. c. the value of the objective function must not change. d. All of these are correct.

ANSWER: a POINTS: 1 DIFFICULTY: Easy © Cengage. Testing Powered by Cognero.

Page 4

CH 18 - Simplex-Based Sensitivity Analysis and Duality LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Understand 16. A one-sided range of optimality a. always occurs for nonbasic variables. b. always occurs for basic variables. c. indicates changes in more than one coefficient. d. indicates changes in a slack variable's coefficient.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Understand 17. A linear programming problem with the objective function 3x1 + 8x2 has the optimal solution x1 = 5, x2 = 6. If c2 decreases by 2 and the range of optimality shows 5 ≤ c2 ≤ 12, the value of Z a. will decrease by 12. b. will decrease by 2. c. will not change. d. cannot be determined from this information.

ANSWER: a POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Apply 18. The improvement in the value of the optimal solution per unit increase in a constraint's right-hand side is a. the slack value. b. the dual price. c. never negative. d. the 100% rule.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Remember © Cengage. Testing Powered by Cognero.

Page 5

CH 18 - Simplex-Based Sensitivity Analysis and Duality 19. The dual variable represents the a. marginal value of the constraint. b. right-hand-side value of the constraint. c. artificial variable. d. technical coefficient of the constraint.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.18.02 - 18.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 18.2 Duality KEYWORDS: Bloom's: Understand 20. The range of feasibility indicates right-hand-side values for which a. the value of the objective function will not change. b. the values of the decision variables will not change. c. those variables that are in the basis will not change. d. more simplex iterations must be performed.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Understand 21. If the dual price for b1 is 2.7, the range of feasibility is 20 ≤ b1 ≤ 50, and the original value of b1 was 30, which of the following is true? a. There currently is no slack in the first constraint. b. We would be willing to pay up to $2.70 per unit for up to 20 more units of resource 1. c. If only 25 units of resource 1 were available, profit would drop by $13.50. d. All of these are correct.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Analyze 22. The number of constraints to the dual of the following problem is Max Z s.t.

= 3x1 + 2x2 + 6x3 4x1 + 2x2 + 3x3 ≥ 100

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CH 18 - Simplex-Based Sensitivity Analysis and Duality 2x1 + x2 − 2x3 ≤ 200 4x2 + x3 ≥ 200 a. 1. b. 2. c. 3. d. 4.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.18.02 - 18.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 18.2 Duality KEYWORDS: Bloom's: Apply 23. Given the simplex tableau for the optimal primal solution, a. the values of the dual variables can be found from the cj − zj values of the slack/surplus variable columns. b. the values of the dual surplus variables can be found from the cj − zj values of the primal decision variable columns. c. the value of the dual objective function will be the same as the objective function value for the primal problem. d. All of these are correct.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.18.02 - 18.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 18.2 Duality KEYWORDS: Bloom's: Understand 24. An LP maximization problem with all less-than-or-equal-to constraints and nonnegativity requirements for the decision variables is known as a. a canonical form for a minimization problem. b. a canonical form for a maximization problem. c. always unbounded. d. None of these are correct.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.18.02 - 18.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 18.2 Duality KEYWORDS: Bloom's: Remember Subjective Short Answer © Cengage. Testing Powered by Cognero.

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CH 18 - Simplex-Based Sensitivity Analysis and Duality 25. For the following linear programming problem Max Z s.t.

0.5x1 + 6x2 + 5x3 4x1 + 6x2 + 3x3 ≤ 24 1x1 + 1.5x2 + 3x3 ≤ 12 3x1 + x2 ≤ 12

the final tableau is

Basis x2 x3 s3

a. b.

cB 6 5 0

x1 0.5

x2 6

x3 5

s1 0

s2 0

s3 0

1 0 2.33

1 0 0

0 1 0

0.22 0.11 −0.22

−0.22 −0.44 0.22

0 0 1

2.67 2.67 9.33

zj cj − zj

4 0.5

6 0

5 0

0.77 −0.77

0.88 −0.88

0 0

29.33

Find the range of optimality for c1, c2, c3, c4, c5, and c6.

Find the range of feasibility for b1, b2, and b3.

ANSWER: a.

c1 ≤ 4 2.5 < c2 ≤ 10 3 < c3 ≤ 12 c4 ≤ 0.77 c5 ≤ 0.88 −1.5 ≤ c6 ≤ 3.5

b.

12 ≤ b1 ≤ 48 6 ≤ b2 ≤ 24 b3 ≥ 2.67

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Analyze 26. For the following linear programming problem Max Z s.t.

−2x1 + x2 − x3 2x1 + x2 ≤ 7 1x1 + x2 + x3 ≥ 4

the final tableau is © Cengage. Testing Powered by Cognero.

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CH 18 - Simplex-Based Sensitivity Analysis and Duality

Basis s2 x2

cB 0 1 zj cj − zj

a. b.

x1 −2

x2 1

x3 −1

s1 0

s2 0

a2 −M

1 2

0 1

−1 0

1 1

1 0

−1 0

3 7

2 −4

1 0

0 1

1 −1

0 0

0 −M

7

Find the range of optimality for c1, c2, c3, c4, c5, and c6.

Find the range of feasibility for b1 and b2.

ANSWER: a.

c1 ≤ 2 0 ≤ c2 c3 ≤ 0 c4 ≤ 1 −1 ≤ c5 ≤ 1 c6 ≤ 0

b.

4 ≤ b1 b2 ≤ 7

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Analyze 27. Write the dual of the following problem: Min Z s.t.

= 2x1 − 3x2 + 5x3 −3x1 + 2x2 + 5x3 ≥ 7 2x1 − x3 ≥ 5 4x2 + 3x3 ≥ 8

ANSWER: Max Z2 s.t.

= 7y1 + 5y2 + 8y3 −3y1 + 2y2 ≤ 2 2y1 + 4y3 ≤ −3 5y1 − y2 + 3y3 ≤ 5

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.18.02 - 18.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 18.2 Duality © Cengage. Testing Powered by Cognero.

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CH 18 - Simplex-Based Sensitivity Analysis and Duality KEYWORDS:

Bloom's: Create

28. For the following linear programming problem Max s.t.

10x1 + 12x2 1x1 + 2x2 ≥ 40 5x1 + 8x2 ≤ 160 1x1 + 1x2 ≤ 40 x1, x2 ≥ 0

the final tableau is

Basis x2 x1 s3

cB 12 10 0 zj cj − zj

x1 10

x2 12

s1 0

s2 0

s3 0

0 1 0

1 0 0

−2.5 4 −1.5

−0.5 1 −0.5

0 0 1

20 0 20

10 0

12 0

10 −10

4 −4

0 0

240

a. Find the range of optimality for c1 and c2. b. Find the range of feasibility for b1, b2, and b3. c. Find the dual prices.

ANSWER: a. 7.5 ≤ c1 < ∞ −∞ < c2 ≤ 16 b.

32 ≤ b1 < 0 160 ≤ b2 ≤ 200 20 ≤ b3 < ∞

c.

u1 = 10 u2 = 4 u3 = 0

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Create 29. For this optimal simplex tableau, the original right-hand sides were 100 and 90. The problem was a maximization.

Basis x3

cB 8

x1 2

x2 4

x3 8

s1 0

s2 0

0

0.48

1

0.12

−0.04

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CH 18 - Simplex-Based Sensitivity Analysis and Duality x1

a. b.

2

1

0.2

0

−0.2

0.4

16

zj cj − zj

2 0

4.24 −0.24

8 0

0.56 −0.56

0.48 −0.48

99.2

What would the new solution be if there had been 150 units available in the first constraint?

What would the new solution be if there had been 70 units available in the second constraint?

ANSWER: a. b.

x3 = 14.4, x1 = 6, Z = 127.2 x3 = 9.2, x1 = 8, Z = 89.6

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Evaluate 30. For this optimal simplex tableau, the right-hand sides for the two original ≥ constraints were 300 and 250. The problem was a minimization.

Basis x3 x1

cB −95 −100 zj cj − zj

a. b.

x1 −100

x2 −110

x3 −95

s1 0

s2 0

0 1

−1.5 2

1 0

1 −1

−1.5 1

75 50

−100 0

−57.5 −52.5

−95 0

5 −5

42.5 −42.5

−12,125

What would the new solution be if the right-hand-side value in the first constraint had been 325?

What would the new solution be if the right-hand-side value for the second constraint had been 220?

ANSWER: a. b.

x3 = 50, x1 = 75, Z = 12,250 x3 = 30, x1 = 80, Z = 10,850

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Evaluate 31. Creative Kitchen Tools manufactures a wide line of gourmet cooking tools from stainless steel. For the coming production period, there is demand of 1200 for eight-quart stock pots and unlimited demand for three-quart mixing bowls and large slotted spoons. In the following model, the three variables measure the number of pots, bowls, and spoons to make. The objective function measures profit. Constraint 1 measures steel, constraint 2 measures manufacturing time, © Cengage. Testing Powered by Cognero.

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CH 18 - Simplex-Based Sensitivity Analysis and Duality constraint 3 measures finishing time, and constraint 4 measures the stock pot demand.

Max s.t.

5x1 + 3x2 + 6x3 3x1 + 1x2 + 2x3 ≤ 15,000 4x1 + 4x2 + 5x3 ≤ 18,000 2x1 + 1x2 + 2x3 ≤ 10,000 x1 ≤ 1200 x1, x2, x3 ≥ 0

The final tableau is as follows:

Basis s1 s4 s3 x1

cB 0 0 0 5 zj cj − zj

a. b. c. d. e.

x1 5

x2 3

x3 6

s1 0

s2 0

s3 0

s4 0

0 0 0 1

−2 1 −1 1

−1.75 1.25 −0.5 1.25

1 0 0 0

−0.75 0.25 −0.5 0.25

0 0 1 0

0 1 0 0

1500 3300 1000 4500

5 0

5 −2

6.25 −0.25

0 0

1.25 −1.25

0 0

0 0

22,500

Calculate the range of optimality for c1, c2, and c3.

Calculate the range of feasibility for b1, b2, b3, and b4.

Suppose that the inventory records were incorrect and the company really has only 14,000 units of steel. What effect will this have on your solution?

Suppose that a cost increase will change the profit on the pots to $4.62. What effect will this have on your solution?

Assume that the cost of time in production and finishing is relevant. Would you be willing to pay a $1.00 premium over the normal cost for 1000 more hours in the production department? What would this do to your solution?

ANSWER: a.

4.80 ≤ c1 < ∞ −∞ < c2 ≤ 5 −∞ < c3 ≤ 6.25

b.

13,500 ≤ b1 < ∞ 480 ≤ b2 ≤ 20,000 9000 ≤ b3 < ∞ −∞ < b4 ≤ 4500

c.

This would affect only the amount of slack, decreasing it from 1500 to 500.

d.

This change is out of the range of optimality so the basis would change.

e.

An increase of 1000 hours is within the range of feasibility and the cost is less than the dual price, so it makes sense to do this. The new solution would be as follows: s1 = 1500 − 0.75(1000) = 750

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CH 18 - Simplex-Based Sensitivity Analysis and Duality s4 = 3300 + 0.25(1000) = 3550 s3 = 1000 − 0.5(1000) = 500 x1 = 4500 + 0.25(1000) = 4750 Z = 22,500 + (1.25 − 1)(1000) = 22,750 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Evaluate 32. Stelle Office Supplies (SOS) must fill an order for 2000 modular office dividers. Each divider consists of a frame, a set of legs, and a panel. SOS has limited production and finishing time available and is considering the purchase of some of the components. Let x1, x2, and x3 be the number of frames, leg sets, and panels to make, respectively, and let x4, x5, and x6 be the number of each to buy, respectively. The model reflects the costs to be minimized, the amount of production time, the amount of assembly time, and the need for 2000 of each component.

Min s.t.

20x1 + 14x2 + 15x3 + 28x4 + 20x5 + 25x6 30x1 + 40x2 + 25x3 ≤ 180,000 15x1 + 10x2 + 30x3 ≤ 90,000 x1 + x4 = 2000 x2 + x5 = 2000 x3 + x6 = 2000 all xi ≥ 0

The final tableau is as follows:

Basis x6 s1 x1 x2 x3

cB −25 0 −20 −14 −15 zj cj − zj

a. b. c. d. e. f. g.

x1 −20

x2 −14

x3 −15

x4 −28

x5 −20

x6 −25

s1 0

s2 0

0 0 1 0 0

0 0 0 1 0

0 0 0 0 1

0.5 −17.5 1 0 −0.5

0.333 −31.67 0 1 −0.333

1 0 0 0 0

0 1 0 0 0

−0.033 −0.833 0 0 0.033

666.67 6666.67 2000 2000 1333.33

−20 −14 0 0

−15 0

−25 −3

−17.33 −2.67

−25 0

0 0

0.33 −0.33

68,667

Calculate the range of optimality for all of the objective function coefficients.

Calculate the range of feasibility for the first two right-hand sides.

How much less expensive would it have to be to buy frames before you would consider it?

How much more expensive would legs have to be to make before you would change your solution?

What would the total cost be if the cost to make a panel increased by $3.00?

What would you be willing to pay for more production time?

What would happen to the total cost if the amount of assembly time decreased by 2000 hours?

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CH 18 - Simplex-Based Sensitivity Analysis and Duality ANSWER: a.

−∞ < c1 ≤ 23 −∞ < c2 ≤ 16.67 9 ≤ c3 ≤ 25 25 ≤ c4 < ∞ 17.33 ≤ c5 < ∞ 15 ≤ c6 ≤ 31

b.

173,333.33 ≤ b1 < ∞ 50,000 ≤ b2 ≤ 98,000 6666.67 ≤ b3 ≤ 2380.95 0 ≤ b4 ≤ 2210.53 1333.33 ≤ b5 < ∞

c.

$3.00 per frame

d.

$2.67 additional

e.

Cost would increase by 1333.33(3) = 4000.

f.

nothing

g.

0.33(2000) = 666.67 more cost

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.18.01 - 18.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 18.1 Sensitivity Analysis with the Simplex Tableau KEYWORDS: Bloom's: Evaluate 33. Write the dual to the following problem.

Min s.t.

12x1 + 15x2 + 20x3 + 18x4 x1 + x2 + x3 + x4 ≥ 50 3x1 + 4x3 ≥ 60 2x2 + x3 − 2x4 ≤ 10 x1, x2, x3, x4 ≥ 0 ANSWER: Max 50u1 + 60u2 − 10u3 s.t. u1 + 3u2 ≤ 12 u1 − 2u3 ≤ 15 u1 + 4u2 − u3 ≤ 20 u1 + 2u3 ≤ 18 u1, u2, u3 ≥ 0 © Cengage. Testing Powered by Cognero.

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CH 18 - Simplex-Based Sensitivity Analysis and Duality POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.18.02 - 18.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 18.2 Duality KEYWORDS: Bloom's: Create 34. The primal problem is as follows: Min s.t.

2x1 + 5x2 + 4x3 x1 + 3x2 + 3x3 ≥ 30 3x1 + 7x2 + 5x3 ≥ 70 x1, x2, x3 ≥ 0

The final tableau for its dual problem is as follows:

Basis u2 s2 u1

cB 70 0 30 zj cj − zj

u1 30

u2 70

s1 0

s2 0

s3 0

0 0 1

1 0 0

3/4 −3/2 −5/4

0 1 0

−1/4 −1/2 3/4

1/2 0 1/2

30 0

70 0

15 −15

0 0

5 −5

50

Give the complete solution to the primal problem.

ANSWER: The complete primal solution is x1 = 15, x2 = 0, x3 = 5, s1 = 0, s2 = 0, Z = 50.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.18.02 - 18.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 18.2 Duality KEYWORDS: Bloom's: Evaluate

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CH 19 - Solution Procedures for Transportation and Assignment Problems True / False 1. The transportation simplex method can be used to solve the assignment problem. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.19.02 - 19.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.2 Assignment Problem: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Remember 2. The transportation simplex method is limited to minimization problems. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Remember 3. For an assignment problem with three agents and four tasks, the assignment matrix will have three rows and four columns. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.19.02 - 19.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.2 Assignment Problem: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Understand 4. The transportation simplex method can be applied only to a balanced problem; if a problem is not balanced, a dummy origin or dummy destination must be added. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking © Cengage. Testing Powered by Cognero.

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CH 19 - Solution Procedures for Transportation and Assignment Problems TOPICS: KEYWORDS:

19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure Bloom's: Understand

5. In assignment problems, dummy agents or tasks are created when the number of agents and tasks is not equal. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Understand 6. The transportation simplex method is more efficient than general-purpose linear programming for solving large-sized transportation problems. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Understand 7. A dummy origin in a transportation problem is used when supply exceeds demand. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Understand 8. The net evaluation index for occupied cells in the transportation simplex method is 0. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking © Cengage. Testing Powered by Cognero.

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CH 19 - Solution Procedures for Transportation and Assignment Problems TOPICS: KEYWORDS:

19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure Bloom's: Understand

9. Optimal assignments are made in the Hungarian method to cells in the reduced matrix that contain a 0. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.19.02 - 19.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.2 Assignment Problem: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Understand 10. Using the Hungarian method, the optimal solution to an assignment problem is found when the minimum number of lines required to cover the zero cells in the reduced matrix equals the number of agents. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.19.02 - 19.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.2 Assignment Problem: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Understand 11. To handle unacceptable routes in a transportation problem where cost is to be minimized, infeasible arcs must be assigned negative cost values. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Understand 12. The transportation simplex method is a two-phase method that first finds an initial feasible solution and then makes iterative improvements in the solution until an optimal solution is reached. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy © Cengage. Testing Powered by Cognero.

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CH 19 - Solution Procedures for Transportation and Assignment Problems LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Understand 13. In the transportation simplex method, when total supply is greater than total demand, we introduce a dummy destination with demand less than the excess of supply over demand. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Understand 14. Finding an initial feasible solution can be done by using a heuristic procedure. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Remember Multiple Choice 15. A solution to a transportation problem that has less than m + n − 1 cells with positive allocations in the transportation tableau is a(n) a. optimal solution. b. initial feasible solution. c. minimum cost solution. d. degenerate solution.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Understand 16. The optimal solution is found in an assignment matrix when the minimum number of straight lines needed to cover all © Cengage. Testing Powered by Cognero.

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CH 19 - Solution Procedures for Transportation and Assignment Problems the zeros equals a. number of agents − 1. b. number of agents. c. number of agents + 1. d. number of agents + number of tasks.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.19.02 - 19.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.2 Assignment Problem: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Understand 17. The stepping-stone method requires that one or more artificially occupied cells with a flow of zero be created in the transportation tableau when the number of occupied cells is fewer than a. m + n − 2. b. m + n − 1. c. m + n. d. m + n + 1.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.19.02 - 19.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.2 Assignment Problem: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Understand 18. The per-unit change in the objective function associated with assigning flow to an unused arc in the transportation simplex method is called the a. net evaluation index. b. degenerate value. c. opportunity loss. d. simplex multiplier.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.19.02 - 19.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.2 Assignment Problem: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Remember 19. The difference between the transportation and assignment problems is that a. total supply must equal total demand in the transportation problem. b. the number of origins must equal the number of destinations in the transportation problem. c. each supply and demand value is 1 in the assignment problem. © Cengage. Testing Powered by Cognero.

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CH 19 - Solution Procedures for Transportation and Assignment Problems d. There are many differences between the transportation and assignment problems.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.19.02 - 19.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.2 Assignment Problem: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Understand 20. An example of a heuristic is the a. minimum cost method. b. stepping-stone method. c. Hungarian method. d. MODI method.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.19.02 - 19.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.2 Assignment Problem: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Remember 21. Using the transportation simplex method, the optimal solution to the transportation problem has been found when a. there is a shipment in every cell. b. more than one stepping-stone path is available. c. there is a tie for the outgoing cell. d. the net evaluation index for each unoccupied cell is ≥ 0.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Understand 22. Identifying the outgoing arc in phase II of the transportation simplex method is performed using the a. minimum cost method. b. MODI method. c. stepping-stone method. d. matrix reduction method.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 © Cengage. Testing Powered by Cognero.

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CH 19 - Solution Procedures for Transportation and Assignment Problems NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Understand 23. In order to identify the arc that will cause the largest reduction per unit in the total cost of the current solution, we use the MODI method. The MODI method is used to identify a. an outgoing arc. b. an incoming arc. c. unoccupied cells. d. an initial feasible solution.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Understand 24. To use the transportation simplex method, a transportation problem that is unbalanced requires the use of a. artificial variables. b. one or more transshipment nodes. c. a dummy origin or destination. d. matrix reduction.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Understand 25. To use the Hungarian method, a profit-maximization assignment problem requires a. converting all profits to opportunity losses. b. a dummy agent or task. c. matrix expansion. d. finding the maximum number of lines to cover all the zeros in the reduced matrix.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.19.02 - 19.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.2 Assignment Problem: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Understand 26. To use the transportation simplex method, © Cengage. Testing Powered by Cognero.

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CH 19 - Solution Procedures for Transportation and Assignment Problems a. there can be no unacceptable routes. b. the initial feasible solution cannot be degenerate. c. a minimization objective function must be the case. d. total supply must equal total demand.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Understand 27. The Hungarian method involves a. matrix reduction. b. subtracting and adding appropriate values to yield an optimal solution. c. continuing iteratively until the value of one of the solutions is zero. d. All of these are correct.

ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.19.02 - 19.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 19.2 Assignment Problem: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Understand Subjective Short Answer 28. Develop the transportation tableau for this transportation problem.

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CH 19 - Solution Procedures for Transportation and Assignment Problems

ANSWER: Destination Origin

A

1 2 3 4 Demand

250

B

Supply

5

6

4

2

3

6

9

7

100 200 150 50

250

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Create © Cengage. Testing Powered by Cognero.

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CH 19 - Solution Procedures for Transportation and Assignment Problems 29. Solve the following transportation problem using the transportation simplex method. State the minimum total shipping cost.

Origin A B

Supply 500 400

Destination X Y Z

Demand 300 300 300

Shipping costs are as follows: Source A B

X 2 9

Destination Y 3 12

Z 5 10

ANSWER: Origin

1

A

200

B

100

Demand

2

Destination 2 3 300

9

300

12 300

3

Supply 5

300

10

500 400

300

Total shipping cost is $5200.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Analyze 30. Canning Transport is to move goods from three factories (origins) to three distribution centers (destinations).

Information about the move is given below. Solve the problem using the transportation simplex method and compute the total shipping cost.

Origin A B C

Supply 200 100 150

Destination X Y Z

Demand 50 125 125

Shipping costs are as follows: Origin A B C

Destination X Y Z 3 2 5 9 10 – 5 6 4 (Source B cannot ship to destination Z.)

ANSWER: © Cengage. Testing Powered by Cognero.

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CH 19 - Solution Procedures for Transportation and Assignment Problems

Origin

Destination Y 2 125

X 3

A

75

B

100

C

25

D

50

Demand

Z

9

10

5

6

0

0

250

Supply 5

200

999

100

4

125

150

0

125

50

125

Total shipping cost is $2000.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Analyze 31. The following table shows the unit shipping cost between cities, the supply at each origin city, and the demand at each destination city. Solve this minimization problem using the transportation simplex method and compute the optimal total cost.

Origin St. Louis Evansville Bloomington Demand

Terre Haute 8 5 3 150

Destination Indianapolis Ft. Wayne 6 12 5 10 2 9 60 45

South Bend 9 8 10 45

Supply 100 100 100

ANSWER: Origin St. Louis Evansville

100

Bloomington

50

Demand

POINTS:

Terre Haute 8 5 3 150

Destination Indianapolis Ft. Wayne 6 12 10 45

50 60

South Bend 9 45

5

10

8

2

9

10

45

Supply 100 100 100

45

Ship 10 from St. Louis to Indianapolis, 45 from St. Louis to Ft. Wayne, 45 from St. Louis to South Bend, 100 from Evansville to Terre Haute, 50 from Bloomington to Terre Haute, and 50 from Bloomington to Indianapolis. The total cost is $1755.

1

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CH 19 - Solution Procedures for Transportation and Assignment Problems DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Analyze 32. After some special presentations, the employees of the AV Center have to move overhead projectors back to classrooms. The table below indicates the buildings where the projectors are now (the origins), where they need to go (the destinations), and a measure of the distance between sites. Determine the transport arrangement that minimizes the total transport distance.

Origin Baker Hall Tirey Hall Arena Demand

Business 10 12 15 12

ANSWER:

Education 9 11 14 20

Destination Parsons Hall 5 1 7 10

Origin Baker Hall Baker Hall Baker Hall Tirey Hall Arena Arena

Destination Business Education Holmstedt Hall Parsons Hall Holmstedt Hall (none)

Holmstedt Hall 2 6 6 10 Units 12 20 3 10 7 13 Total

Supply 35 10 20

Distance 120 180 6 10 42 0 358

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Evaluate 33. Solve the following assignment problem using the Hungarian method. No agent can be assigned to more than one task. Total cost is to be minimized.

Task Agent 1 2 3

A 9 12 11

B 5 6 6

ANSWER:

POINTS:

Agent 1 2 3 Unassigned

C 4 3 5

D 2 5 7 Task D C B A Total cost

Cost 2 3 6 0 11

1

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CH 19 - Solution Procedures for Transportation and Assignment Problems DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.19.02 - 19.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 19.2 Assignment Problem: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Evaluate 34. Use the Hungarian method to obtain the optimal solution to the following assignment problem in which total cost is to be minimized. All tasks must be assigned, and no agent can be assigned to more than one task.

Task Agent 1 2 3 4

A 10 11 18 15

ANSWER:

B 12 14 21 20

C 15 19 23 26

D 25 32 29 28

Agent 1 2 3 4

Task C B D A Total cost

Cost 15 14 29 15 73

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.19.02 - 19.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 19.2 Assignment Problem: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Evaluate 35. A professor has been contacted by four not-for-profit agencies that are willing to work with student consulting teams.

The agencies need help with such things as budgeting, information systems, coordinating volunteers, and forecasting.

Although each of the four student teams could work with any of the agencies, the professor feels that there is a difference in the amount of time it would take each group to solve each problem. The professor's estimate of the time, in days, is given in the table below. Use the Hungarian method to determine which team works with which project. All projects must be assigned, and no team can be assigned to more than one project.

Team A B C D

Budgeting 32 38 41 45

Project Information Volunteers 35 15 40 18 42 25 45 30

Forecasting 27 35 38 42

ANSWER:

Team A works with the forecast, Team B works with volunteers, Team C works with budgeting, and Team D works with information. The total time is 131.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.19.02 - 19.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic © Cengage. Testing Powered by Cognero.

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CH 19 - Solution Procedures for Transportation and Assignment Problems TOPICS: KEYWORDS:

19.2 Assignment Problem: A Special-Purpose Solution Procedure Bloom's: Evaluate

36. A manufacturer of electrical consumer products, with its headquarters in Burlington, Iowa, produces electric irons at Manufacturing Plants 1, 2, and 3. The irons are shipped to Warehouses A, B, C, and D. The shipping cost per iron, the monthly warehouse requirements, and the monthly plant production levels are as follows:

Plant 1 Plant 2 Plant 3 Monthly warehouse requirements (units)

A $0.20 0.15 0.15

Warehouse B C $0.25 $0.15 0.30 0.20 0.20 0.20

D $0.20 0.15 0.25

12,000

8000

5000

15,000

Monthly Plant Production (units) 10,000 20,000 10,000

How many electric irons should be shipped per month from each plant to each warehouse to minimize monthly shipping costs? a. b. c.

Use the minimum cost method to find an initial feasible solution.

Can the initial solution be improved?

Compute the optimal total shipping cost per month.

ANSWER:

a. The minimum cost method (and breaking ties by choosing the cell corresponding to the arc over which the most units can be shipped) found the solution shown below.

Origin Plant 1 Plant 2 Plant 3 Demand

Warehouse A 0.20

12,000

Destination Warehouse B Warehouse C 0.25 0.15 10,000

0.15 0.15

12,000

0.30

8000

0.20

8000

3000 2000

0.20

Warehouse D 0.20

5000

0.20

15,000

0.15 0.25

Supply 10,000 20,000 10,000

5000

b. The solution cannot be improved. It is optimal. c. Total monthly shipping cost is $6650.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Evaluate 37. Al Bergman, staff traffic analyst at the corporate headquarters of Computer Products Corporation (CPC), is developing a monthly shipping plan for the El Paso and Atlanta manufacturing plants to follow next year. These plants manufacture specialized computer workstations that are shipped to five regional warehouses. Al has developed the following estimated requirements and costs: Plant

Chicago

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Dallas

Warehouse Denver New York

San Jose

Monthly Plant Production (units) Page 14

CH 19 - Solution Procedures for Transportation and Assignment Problems Atlanta El Paso Monthly warehouse requirements (units)

$35 50

$40 30

$60 35

$45 95

$90 40

75

100

25

150

150

200 300

a. Use the minimum cost method to find an initial feasible solution. b. Use the transportation simplex method to find an optimal solution. c. Compute the optimal total shipping cost.

ANSWER: a. The initial feasible solution found using the minimum cost method is shown below. Total cost is $20,500.

Origin Atlanta

Chicago 35 75 50

El Paso Demand

75

Dallas 40

Destination Denver New York 60 45 125

30

100

35

25

100

25

25

95

150

San Jose 90

150

40

Supply 200 300

150

b. The optimal solution found using MODI and stepping-stone methods is shown below.

Origin Atlanta

Chicago 35 50

El Paso

25

Demand

50 75

Dallas 40

Destination Denver New York 60 45 150

30

100 100

35

25 25

95 150

San Jose 90

150

40

Supply 200 300

150

c. Total monthly shipping cost is $19,625.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Evaluate 38. Consider the transportation problem below.

Origin A B C Demand a. b. c.

1 $0.50 0.80 0.90 300

Destination 2 $0.90 1.00 0.70 800

3 $0.50 0.40 0.80 400

Supply 100 500 900

Use the minimum cost method to find an initial feasible solution.

Can the initial solution be improved?

Compute the optimal total shipping cost.

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CH 19 - Solution Procedures for Transportation and Assignment Problems ANSWER:

a.

The minimum cost method provided the solution shown below.

Origin

b. c.

0.80

1.00

1

A

100

B

100

C

100

Demand

0.50

Destination 2 0.90

0.90

300

3

400

0.70

800 800

Supply 0.50 0.40 0.80

100 500 900

400

The solution cannot be improved. It is optimal.

Total shipping cost is $940.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Evaluate 39. Five customers needing their tax returns prepared must be assigned to five tax accountants. The estimated profits for all possible assignments are shown below. Only one accountant can be assigned to a customer, and all customers' tax returns must be prepared. What should the customer–accountant assignments be so that estimated total profit is maximized? What is the resulting total profit?

Customer A B C D E ANSWER:

1 $500 625 825 590 450

2 $525 575 650 650 750 Customer A B C D E

Accountant 3 $550 700 450 525 660

4 $600 550 750 690 390

Accountant 5 3 1 4 2 Total profit

5 $700 800 775 750 550 Profit $ 700 700 825 690 750 $3665

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.19.02 - 19.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic © Cengage. Testing Powered by Cognero.

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CH 19 - Solution Procedures for Transportation and Assignment Problems TOPICS: KEYWORDS:

19.2 Assignment Problem: A Special-Purpose Solution Procedure Bloom's: Evaluate

40. Four jobs must be assigned to four work centers. Only one job can be assigned to each work center, and all jobs must be processed. The cost of processing each job at each work center is shown below. Determine which jobs should be assigned to which work center to minimize total processing cost. Compute the total processing cost.

Job A B C E

1 $50 25 65 55

Work Center 2 3 $45 $50 40 35 60 55 65 75

ANSWER:

Job A B C D

4 $65 20 65 85 Work Center 2 4 3 1 Total cost

Cost $ 45 20 55 55 $175

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.19.02 - 19.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 19.2 Assignment Problem: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Evaluate 41. Four employees must be assigned to four projects. Only one employee can be assigned to each project, and all projects must be completed. The cost of each employee completing each project is shown below. Determine which employee should be assigned to which project to minimize total project completion cost. Be sure to compute the total project completion cost.

Project Employee Al Ben Cal Dan ANSWER:

POINTS: DIFFICULTY:

1 $300 400 350 400

2 $325 525 400 350 Employee Al Ben Cal Dan

3 $500 575 600 450 Project 4 1 2 3 Total cost

4 $350 600 500 450 Cost $ 350 400 400 450 $1600

1 Challenging

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CH 19 - Solution Procedures for Transportation and Assignment Problems LEARNING OBJECTIVES: IMS.ASWC.19.19.02 - 19.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 19.2 Assignment Problem: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Evaluate 42. A large screen printer is faced with six jobs due on Tuesday. The plan is to do the jobs on Monday so they will be ready on time. The shop has six worker-machine pairs that can work on any of the six jobs. Because of differing experience levels and machine capabilities, processing times differ. The processing times presented in the table below are in minutes. What is the optimal assignment of jobs to worker-machine pairs that minimizes total processing time?

Job A B C D E F

1 250 350 410 380 395 250

Worker-Machine Pair 3 4 175 425 375 410 325 275 350 375 280 390 410 385

2 375 310 450 245 250 285

5 225 275 315 210 410 300

6 350 225 275 350 375 295

ANSWER: Job A B C D E F

Worker-Machine Pair Time 3 175 6 225 4 275 5 210 2 250 1 250 Total time = 1385

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Evaluate 43. A company ships products from four factories to four warehouses. The factory capacities, warehouse requirements, and per-unit shipping costs are shown below.

Warehouse Factory A Factory B Factory C Factory D © Cengage. Testing Powered by Cognero.

1

2

3

4

Monthly Factory Capacity (units)

$11 12 19 7

$13 10 16 6

$9 7 15 4

$6 9 21 9

5000 10,000 10,000 5000 Page 18

CH 19 - Solution Procedures for Transportation and Assignment Problems Monthly warehouse requirement (units)

3000

8000

12,000

5000

How many products should the company ship from each factory to each warehouse to minimize monthly shipping costs?

What will the monthly shipping cost be if the shipping plan is followed? (Use the minimum cost method to find an initial feasible solution and the transportation simplex method to find an optimal solution.)

ANSWER:

Factory A to Warehouse 4 = 5000 units Factory B to Warehouse 3 = 10,000 units Factory C to Warehouse 2 = 8000 units Factory D to Warehouse 1 = 3000 units Factory D to Warehouse 3 = 2000 units Total cost = $257,000 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Evaluate 44. The Des Moines plant of Tri-B Corp. has three fabrication departments with each producing a single unique product with equipment that is dedicated solely to its product. The three products are moved to four assembly departments where they are assembled.

Although any of the three products can be processed in any of the assembly departments, the materials handling and assembly costs are different because of the varying distances between departments and because of different equipment.

Each fabrication and assembly department has a different monthly capacity, and it is desirable that each department operate at capacity. The variable costs and capacity for each department are shown below.

Fabrication Dept. A Fabrication Dept. B Fabrication Dept. C Monthly assembly dept. capacity (units)

1 $1.20 0.70 0.50

Assembly Department 2 3 $0.70 $0.50 0.50 0.50 0.70 0.80

4 $0.60 0.60 1.20

3000

10,000

12,000

15,000

Monthly Fabrication Dept. Capacity (units) 9000 17,000 14,000

How many units of each product should be moved from each fabrication department to each assembly department to minimize total monthly costs? (Use the minimum cost method to find an initial feasible solution and the transportation simplex method to find an optimal solution.) Compute the optimal total monthly cost.

ANSWER:

Dept. A to Dept. 4 = 9000 units Dept. B to Dept. 3 = 14,000 units Dept. B to Dept. 4 = 3000 units Dept. C to Dept. 1 = 3000 units Dept. C to Dept. 2 = 10,000 units Dept. C to Dept. 3 = 1000 units Total shipping cost = $23,500.00

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CH 19 - Solution Procedures for Transportation and Assignment Problems POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.19.01 - 19.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 19.1 Transportation Simplex Method: A Special-Purpose Solution Procedure KEYWORDS: Bloom's: Evaluate

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CH 20 - Minimal Spanning Tree True / False 1. The arcs in a minimal spanning tree problem can be measured in terms of criteria other than distance. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.20.01 - 20.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 20.1 A Minimal Spanning Tree Algorithm KEYWORDS: Bloom's: Remember 2. Cases in which a greedy algorithm provides the optimal solution are rare. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.20.01 - 20.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 20.1 A Minimal Spanning Tree Algorithm KEYWORDS: Bloom's: Understand 3. The minimum spanning tree allows a person to visit every node without backtracking. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.20.01 - 20.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 20.1 A Minimal Spanning Tree Algorithm KEYWORDS: Bloom's: Remember 4. In the minimal spanning tree algorithm, you must consider the unconnected nodes that can be reached from any of the connected nodes, rather than arbitrarily considering only one of the connected nodes. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.20.01 - 20.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 20.1 A Minimal Spanning Tree Algorithm © Cengage. Testing Powered by Cognero.

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CH 20 - Minimal Spanning Tree KEYWORDS:

Bloom's: Understand

5. Cases in which a greedy algorithm provides the optimal solution are rare, but greedy algorithms are excellent heuristics. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.20.01 - 20.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 20.1 A Minimal Spanning Tree Algorithm KEYWORDS: Bloom's: Understand 6. The minimal spanning tree algorithm will lead to an optimal solution regardless of which node is chosen at the start of the algorithm. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.20.01 - 20.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 20.1 A Minimal Spanning Tree Algorithm KEYWORDS: Bloom's: Understand 7. A minimal spanning tree problem involves using the arcs of the network to reach all nodes of the network such that the total length of all the arcs used is maximized. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.20.01 - 20.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 20.1 A Minimal Spanning Tree Algorithm KEYWORDS: Bloom's: Understand 8. For a network consisting of N nodes, a spanning tree will consist of N – 1 arcs. a. True b. False ANSWER: POINTS: DIFFICULTY:

True 1 Easy

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CH 20 - Minimal Spanning Tree LEARNING OBJECTIVES: IMS.ASWC.19.20.01 - 20.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 20.1 A Minimal Spanning Tree Algorithm KEYWORDS: Bloom's: Understand Multiple Choice 9. Consider a minimal spanning tree problem in which pipe must be laid to connect sprinklers on a golf course. When represented with a network, a. the pipes are the arcs and the sprinklers are the nodes. b. the pipes are the nodes and the sprinklers are the arcs. c. the pipes and the sprinklers are the tree. d. each sprinkler must be connected to every other sprinkler.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.20.01 - 20.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 20.1 A Minimal Spanning Tree Algorithm KEYWORDS: Bloom's: Understand 10. The minimal spanning tree algorithm is considered to be a(n) a. greedy algorithm. b. arc algorithm. c. non-optimal algorithm. d. non-feasible algorithm.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.20.01 - 20.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 20.1 A Minimal Spanning Tree Algorithm KEYWORDS: Bloom's: Understand 11. The minimal spanning tree algorithm has connected nodes 8 and 9. Node 8 could be connected to nodes 11 (distance 6) and 12 (distance 5), and node 9 could be connected to node 12 (distance 3) and node 13 (distance 2). Which of the following will you do next? a. connect 8 to 11 b. connect 8 to 12 c. connect 9 to 12 d. connect 9 to 13 ANSWER: d POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.20.01 - 20.1 © Cengage. Testing Powered by Cognero.

Page 3

CH 20 - Minimal Spanning Tree NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 20.1 A Minimal Spanning Tree Algorithm KEYWORDS: Bloom's: Apply 12. For a network consisting of N nodes, a minimal spanning tree will consist of a. N − 2 arcs. b. N − 1 arcs. c. N arcs. d. N + 1 arcs.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.20.01 - 20.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 20.1 A Minimal Spanning Tree Algorithm KEYWORDS: Bloom's: Apply 13. The minimal spanning tree algorithm is considered a greedy algorithm that, when taking the best action at each stage, will a. sometimes fail to produce a feasible solution. b. always produce a feasible, but not necessarily optimal, solution. c. always produce an optimal solution. d. always produce an optimal, but not necessarily feasible, solution.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.20.01 - 20.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 20.1 A Minimal Spanning Tree Algorithm KEYWORDS: Bloom's: Understand Subjective Short Answer 14. For the following eight cities with the given distances, find the minimal spanning tree path.

From City 1 2 3 4 5 6 7 ANSWER: POINTS: DIFFICULTY:

To City 1 – 5 2 6 – – –

2 3 5 2 – – – – 4 3 6 – – 12 – – 1-3-4-2-5-6-7 1 Challenging

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4 6 4 3 – 9 8 –

5 – 6 – 9 – 0 14

6 – – 12 8 0 – 4

7 – – – – 14 4 –

Page 4

CH 20 - Minimal Spanning Tree LEARNING OBJECTIVES: IMS.ASWC.19.20.01 - 20.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 20.1 A Minimal Spanning Tree Algorithm KEYWORDS: Bloom's: Analyze 15. Find the minimal spanning tree for this network.

ANSWER:

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.20.01 - 20.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 20.1 A Minimal Spanning Tree Algorithm KEYWORDS: Bloom's: Analyze 16. Find the minimal spanning tree for this network.

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CH 20 - Minimal Spanning Tree ANSWER:

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.20.01 - 20.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 20.1 A Minimal Spanning Tree Algorithm KEYWORDS: Bloom's: Analyze 17. The numbers on this network represent times to distribute a message. Use the minimal spanning tree algorithm to determine how messages should be passed in order to minimize total time.

ANSWER:

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CH 20 - Minimal Spanning Tree POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.20.01 - 20.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 20.1 A Minimal Spanning Tree Algorithm KEYWORDS: Bloom's: Evaluate 18. Ethernet cable costs $25 per foot to install. The table below gives the approximate direct routing distance in feet between various pairs of machines.

WS1 WS2 LP1 LP2 CC

WS1 0

WS2 46 0

LP1 16 81 0

LP2 23 44 45 0

CC 100 67 57 38 0

FM 29 98 36 73 52

a. Use the minimal spanning tree algorithm to determine the least cost method of connecting all machines. b. How much cheaper would it be if the company decided not to hook up the second printer to this system?

ANSWER: a.

WS1-LP1, WS1-LP2, WS1-FM, LP2-WS2, LP2-CC; 150 feet; Total cost = $3750 b.

WS1-LP1, WS1-WS2, WS1-FM, FM-CC; 143 feet; Total cost = $3575; $175 cheaper POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.20.01 - 20.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 20.1 A Minimal Spanning Tree Algorithm KEYWORDS: Bloom's: Evaluate 19. Wondercamp is planning a new resort for the urban professional who wishes to "get back to nature." For a hefty fee, it plans to transport clients up the Crocodile River by canoe and then have the clients camp at one of eight designated campsites.

It must build a 1000-yard trail from the river to the first campsite as there are no feasible alternatives. It then wishes to build a sequence of trails (of minimum total length) so that every campsite can be reached from any other campsite. A study of the area's terrain has yielded the following possibilities for trails between campsites. Which trails should be built?

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CH 20 - Minimal Spanning Tree

ANSWER:

There are two alternative optimal solutions: 1. (River-1),(1-4),(4-6),(4-5),(6-7),(4-3),(3-8),(3-2) 2. (River-1),(1-4),(4-6),(4-5),(6-7),(4-3),(3-8),(4-2) Total yards = 2050 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.20.01 - 20.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 20.1 A Minimal Spanning Tree Algorithm KEYWORDS: Bloom's: Evaluate 20. Griffith's Cherry Preserve is a combination wild animal habitat and amusement park. Besides its phenomenally successful wild animal safari tour, there are eight different theme areas in the amusement park.

One problem encountered by management is to develop a method by which people can efficiently travel between each area of the park. Management has learned that a people mover can be constructed at a cost of $50 per foot.

If the following network represents the distances (in feet) between each area of the park for which a people mover is possible, determine the minimum cost for such a system.

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CH 20 - Minimal Spanning Tree

ANSWER:

Optimal tree = (1-3),(1-4),(3-2),(3-6),(6-7),(6-8),(8-5),(8-9) Total distance = 9110 feet Total cost = $455,500 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.20.01 - 20.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 20.1 A Minimal Spanning Tree Algorithm KEYWORDS: Bloom's: Evaluate

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CH 21 - Dynamic Programming True / False 1. Dynamic programming requires that its subproblems be independent of one another. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.21.01 - 21.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.1 A Shortest-Route Problem KEYWORDS: Bloom's: Remember 2. Dynamic programming, when used for the shortest-route problem, requires complete enumeration of paths from the beginning to ending node. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.21.01 - 21.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.1 A Shortest-Route Problem KEYWORDS: Bloom's: Remember 3. The solution of stage k of a dynamic programming problem is dependent upon the solution of stage k−1. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.21.02 - 21.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.2 Dynamic Programming Notation KEYWORDS: Bloom's: Understand 4. The output of stage k is the input for stage k−1. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.21.02 - 21.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.2 Dynamic Programming Notation © Cengage. Testing Powered by Cognero.

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CH 21 - Dynamic Programming KEYWORDS:

Bloom's: Understand

5. State variables are a function of a state variable and a decision. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.21.02 - 21.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.2 A Dynamic Programming Notation KEYWORDS: Bloom's: Understand 6. The return function for a shortest-route problem refers to two directional arcs between nodes. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.21.01 - 21.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.1 A Shortest-Route Problem KEYWORDS: Bloom's: Remember 7. In solving a shortest-route problem using dynamic programming, the stages represent how many arcs you are from the terminal node. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.21.01 - 21.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.1 A Shortest-Route Problem KEYWORDS: Bloom's: Understand 8. As opposed to a specific technique such as linear programming, dynamic programming is considered a general approach. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.21.01 - 21.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking © Cengage. Testing Powered by Cognero.

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CH 21 - Dynamic Programming TOPICS: KEYWORDS:

21.1 A Shortest-Route Problem Bloom's: Remember

9. Dynamic programming must only involve a finite number of decision alternatives and a finite number of stages. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.21.01 - 21.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.1 A Shortest-Route Problem KEYWORDS: Bloom's: Understand 10. Dynamic programming is a general approach used when it is possible to break a large problem into interrelated smaller problems, with stage decisions proceeding recursively, solving one of the smaller problems at each stage. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.21.01 - 21.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.1 A Shortest-Route Problem KEYWORDS: Bloom's: Understand 11. In a production and inventory control problem, the states can correspond to the amount of inventory on hand at the beginning of each period. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.21.04 - 21.4 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.4 A Production and Inventory Control Problem KEYWORDS: Bloom's: Understand 12. In a knapsack problem, if one adds another item, one must completely resolve the problem in order to find a new optimal solution. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Easy © Cengage. Testing Powered by Cognero.

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CH 21 - Dynamic Programming LEARNING OBJECTIVES: IMS.ASWC.19.21.03 - 21.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.3 The Knapsack Problem KEYWORDS: Bloom's: Understand 13. The subscripts used in dynamic programming notation refer to states. a. True b. False ANSWER: False POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.21.02 - 21.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.2 Dynamic Programming Notation KEYWORDS: Bloom's: Remember 14. In order to use dynamic programming, one must be able to view the problem as a multistage decision problem. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.21.02 - 21.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.2 Dynamic Programming Notation KEYWORDS: Bloom's: Understand 15. Finding the optimal solution to each stage of a dynamic programming problem will always lead to an optimal solution to the total problem. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.21.02 - 21.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.2 Dynamic Programming Notation KEYWORDS: Bloom's: Understand 16. The stage transformation function identifies which state one reaches at the next stage for a given decision. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Moderate © Cengage. Testing Powered by Cognero.

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CH 21 - Dynamic Programming LEARNING OBJECTIVES: IMS.ASWC.19.21.02 - 21.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.2 Dynamic Programming Notation KEYWORDS: Bloom's: Understand 17. An input state variable and an output state variable together define the condition of the process at the beginning and end of a stage. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.21.02 - 21.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.2 Dynamic Programming Notation KEYWORDS: Bloom's: Remember 18. A knapsack problem is so named because the objective is to find the number of N items, each of which has a different weight and value, that can be placed in a knapsack with limited weight capacity so as to maximize the total value of the items placed in the knapsack. a. True b. False ANSWER: True POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.21.03 - 21.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.3 The Knapsack Problem KEYWORDS: Bloom's: Remember Multiple Choice 19. Stages of a dynamic programming solution procedure a. represent parts of a large mathematical model. b. often represent a sequence of decisions made over time. c. are usually not independent of each other. d. All of these are correct.

ANSWER: d POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.21.01 - 21.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.1 A Shortest-Route Problem KEYWORDS: Bloom's: Understand © Cengage. Testing Powered by Cognero.

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CH 21 - Dynamic Programming 20. The stage transformation function a. transforms the input into the output. b. transforms a stage into a state. c. is a different function for each stage. d. None of these are correct.

ANSWER: a POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.21.02 - 21.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.2 Dynamic Programming Notation KEYWORDS: Bloom's: Understand 21. State variables in a shortest-route problem represent a. decisions. b. locations in the network. c. the minimum distance between nodes. d. None of these are correct.

ANSWER: b POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.21.01 - 21.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.1 A Shortest-Route Problem KEYWORDS: Bloom's: Remember 22. The function that “transforms” the input of the stage into the output of the stage is referred to as a stage transformation function and a. is always linear. b. calculates the return of the transformation. c. determines the output of the stage. d. All of these are correct.

ANSWER: c POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.21.02 - 21.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.2 Dynamic Programming Notation KEYWORDS: Bloom's: Remember 23. A return function is a value such as profit or loss associated with making decision dn at a. stage n for a specific value of output variable xn. b. stage n for a specific value of input variable xn. c. stage n for a specific value of stage m. © Cengage. Testing Powered by Cognero.

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CH 21 - Dynamic Programming d. input n for a specific value of output variable xn.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.21.01 - 21.1 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.1 A Shortest-Route Problem KEYWORDS: Bloom's: Understand 24. If x3 = t4(x4,d4) = x4 − 2d4 and r4(x4,d4) = 16d4, the state variable is a. t. b. x. c. r. d. d.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.21.02 - 21.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 21.2 Dynamic Programming Notation KEYWORDS: Bloom's: Apply 25. If x3 = t4(x4,d4) = x4 − 2d4 and r4(x4,d4) = 16d4, the stage transformation function is a. t. b. x. c. r. d. d.

ANSWER: a POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.21.02 - 21.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 21.2 Dynamic Programming Notation KEYWORDS: Bloom's: Apply 26. If x3 = t4(x4,d4) = x4 − 2d4 and r4(x4,d4) = 16d4, the subscripts refer to a. state. b. stage. c. transformation. d. return.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate © Cengage. Testing Powered by Cognero.

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CH 21 - Dynamic Programming LEARNING OBJECTIVES: IMS.ASWC.19.21.02 - 21.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 21.2 Dynamic Programming Notation KEYWORDS: Bloom's: Apply 27. The knapsack problem is to determine how many units of each item to place in the knapsack to a. minimize total value. b. maximize total value. c. minimize the number of items in the knapsack. d. maximize the number of items in the knapsack.

ANSWER: b POINTS: 1 DIFFICULTY: Moderate LEARNING OBJECTIVES: IMS.ASWC.19.21.03 - 21.3 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.3 The Knapsack Problem KEYWORDS: Bloom's: Remember 28. Solutions in dynamic programming a. are not optimal. b. are unique. c. represent each stage. d. All of these are correct.

ANSWER: c POINTS: 1 DIFFICULTY: Easy LEARNING OBJECTIVES: IMS.ASWC.19.21.02 - 21.2 NATIONAL STANDARDS: United States - BUSPROG: Reflective Thinking TOPICS: 21.2 Dynamic Programming Notation KEYWORDS: Bloom's: Understand Subjective Short Answer 29. Find the shortest path through the following network using dynamic programming.

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CH 21 - Dynamic Programming ANSWER:

STAGE 1 Input Node Decision Arc Shortest Distance 8 8−10 4 9 9−10 3 STAGE 2 Input Node Decision Arc Shortest Distance 5 5−9 6 6 6−8 10 7 7−8 11 STAGE 3 Input Node Decision Arc Shortest Distance 2 2−5 16 3 3−5 13 4 4−6 17 STAGE 4 Input Node Decision Arc Shortest Distance 1 1−4 19 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.21.01 - 21.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 21.1 A Shortest-Route Problem KEYWORDS: Bloom's: Analyze 30. Audio Disks will be opening outlets in the greater Phoenix area. The estimated sales at each store are dependent not only on the store location, but also on the number of sales personnel, as presented in the table below ($1000s/year). Each store requires at least two salespeople, and a pool of nine salespeople is available.

Store 1 Store 2 Store 3

2 60 105 120

Staff Size 3 85 120 145

4 90 130 160

5 100 150 175

a. b. c.

What would the states be in the dynamic programming formulation?

Draw the network that represents the dynamic programming formulation.

Given the above network, solve the sales personnel allocation problem by finding the longest path.

ANSWER: The states represent the number of salespeople available at each a. stage (store).

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CH 21 - Dynamic Programming

b.

c.

Store 1 2 3 Total sales

Alternate optimal solutions: Salespeople Sales Salespeople 3 85 3 3 120 2 3 145 4

Sales 85 105 160

350

350

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.21.02 - 21.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 21.2 Dynamic Programming Notation KEYWORDS: Bloom's: Evaluate 31. Consider the following integer linear program: Max s.t.

5x1 + 7x2 + 9x3 2x1 + 3x2 + 4x3 ≤ 8 x1 ≤ 3 x2 ≤ 2 x1, x2, x3 ≥ 0, integer

a. Set up the network that represents the dynamic programming formulation. b. Solve the problem using dynamic programming.

ANSWER:

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CH 21 - Dynamic Programming

a.

b.

x1 = 1, x2 = 2, x3 = 0

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.21.02 - 21.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 21.2 Dynamic Programming Notation KEYWORDS: Bloom's: Create 32. A driver wants to make a trip from city 1 to city 7. The road mileage between cities is given below. Find the shortest route.

From City 1 2 3 4 5 6 7

1 − − − − − − −

2 2 − − − − − −

3 2 − − − − − −

To City 4 4 − − − − − −

5 − 4 12 10 − − −

6 − 10 6 − − − −

7 − − − − 14 2 −

ANSWER: 1-3-6-7; 13 miles POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.21.01 - 21.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 21.1 A Shortest-Route Problem KEYWORDS: Bloom's: Evaluate 33. The owner of a small construction firm is excavating at three sites. He wishes to assign his five additional trucks in such a way as to minimize his total costs. Each site can use zero to three additional trucks; no site can use more than three trucks efficiently. The following site total costs are known.

Number

Cost of Excavating

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CH 21 - Dynamic Programming of Trucks 0 1 2 3 a. b.

Site 1 $10,000 10,000 9,200 8,500

Site 2 $15,000 14,000 13,250 12,750

Site 3 $20,000 18,000 17,500 17,250

Use dynamic programming to find the assignment of the additional trucks that minimizes total cost.

If the owner had only four trucks to assign, what would be the optimal assignment and total cost?

ANSWER:

a.

The minimal cost truck assignments are: 0 trucks to site 1 3 trucks to site 2 2 trucks to site 3 at a cost of $40,250.

b.

There are two optimal assignments: Site 1 0 trucks 0 trucks Site 2 3 trucks or 2 trucks Site 3 1 truck 2 trucks at a cost of $40,750.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.21.02 - 21.2 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 21.2 Dynamic Programming Notation KEYWORDS: Bloom's: Evaluate 34. A number of types of items are to be shipped as cargo. The total available weight in the truck is 10 tons. Determine the number of units of items to be shipped to maximize profit.

Item A B C

Weight (tons) 3 4 2

Profit ($1000s) 5 7 3

ANSWER:

Ship 0 tons of item A. Ship 2 tons of item B. Ship 1 ton of item C. Total profit is 17.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.21.03 - 21.3 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 21.3 The Knapsack Problem KEYWORDS: Bloom's: Evaluate 35. A cargo company has a set of delivery patterns for its goods from its location at city 1 to a series of cities 2, 3, 4, 5, and 6. The delivery times between cities are given in hours below. Find the shortest route. © Cengage. Testing Powered by Cognero.

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CH 21 - Dynamic Programming To City: From City: 1 2 3 4 5 6 1 − 6 10 7 − − 2 6 − 4 − 5 − 3 10 4 − 4 2 4 4 7 − 5 − − 8 5 − 5 3 − − − 6 − − 4 8 − − 7 − − − − 7 5 ANSWER: 1-2-5-7; 18 hours POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.21.01 - 21.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 21.1 A Shortest-Route Problem KEYWORDS: Bloom's: Evaluate

7 − − − − 7 5 −

36. Unidyne Corporation is currently planning the production of red dye number 56 for the next four months. Production and handling costs, as well as production and storage capacity, vary from month to month. These data are given in the table below.

Production and holding costs are in $1000s per batch, and production levels and storage capacities are in batches.

Holding costs are based on inventory on hand at the end of the month. The number of orders for batches the sales department has received over the four-month period is also given.

Month February March April May

Production Cost 11 15 16 9

Maximum Production 3 4 3 2

Holding Cost 3 2 2 1

Storage Capacity 4 3 5 2

Orders Received 2 4 2 3

Unidyne does not wish to have any inventory of the dye at the end of May. Its current inventory is two batches.

Determine a production schedule for the next four months.

ANSWER:

Produce 3 batches in February.

Produce 1 batch in March.

Produce 3 batches in April.

Produce 2 batches in May. Total cost is $125,000.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.21.04 - 21.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 21.4 A Production and Inventory Control Problem KEYWORDS: Bloom's: Evaluate 37. Marvelous Marvin is planning his annual "Almost Everything Must Go" inventory clearance sale. Marvin has decided to allocate 18 shelf feet to the cooking section. He is considering offering up to five items for sale in this category. © Cengage. Testing Powered by Cognero.

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CH 21 - Dynamic Programming Item A B C D E

Shelf Feet Required 1 2 3 5 7

Expected Profit $ 10 25 40 70 100

If Marvin wants at least one item A and one item B on sale, what stock should he have on sale, and what is the total expected profit?

ANSWER: Three solutions: (1) 2-A, 2-E, 1-B (2) 1-A, 1-B, 3-D (3) 1-A, 1-B, 1-C, 1-D, 1-E Total profit is $245.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.21.04 - 21.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 21.4 A Production and Inventory Control Problem KEYWORDS: Bloom's: Evaluate 38. Franklin Plate Company manufactures commemorative plates for holidays. The company is currently planning its production schedule for this year's Thanksgiving plate.

Discussions with the sales manager have indicated that sales of the plate will run from August through November. The production manager has determined the manufacturing cost per plate for each of these months as well as the sales demand (in 1000s) and manufacturing capacity (in 1000s). The data are as follows: Month August September October November

Demand 2 4 6 3

Manufacturing Cost per Plate 2 3 3 4

Maximum Production 6 6 4 3

The maximum inventory level possible at the end of any month is 5000 plates. The storage cost per plate for each plate in inventory at the end of the month is $0.50.

Franklin has no inventory on hand at the beginning of August and wishes to have no inventory left after November. Use dynamic programming to determine an optimal production schedule.

ANSWER: There are four alternative optimal solutions giving a total of $45,000: Month Solution 1 Solution 2 Solution 3 Solution 4 August 6 6 6 6 September 2 3 4 5 October 4 4 4 4 November 3 2 1 0 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.21.04 - 21.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 21.4 A Production and Inventory Control Problem © Cengage. Testing Powered by Cognero.

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CH 21 - Dynamic Programming KEYWORDS:

Bloom's: Evaluate

39. Walt's Custom Boats manufactures luxury yachts. Walt is phasing out production of his 42-foot Sportfisher that he plans to replace with a 44-foot Sportfisher. Walt currently has orders for the next four months for the 42-foot model after which he will cease production. The following data (with costs in $1000s) are for the next four months:

Month May June July August

Production Cost per Boat 25 27 30 29

Maximum Production Level 4 3 2 3

Holding Cost per Boat in Inventory at End of Month 1 1 2 1

Maximum Storage Capacity 6 5 3 4

Number of Boats to Be Delivered in Month 1 3 3 4

If Walt's current inventory of 42-foot Sportfishers is one boat, how should Walt schedule production of the 42-foot Sportfisher over the next four months to minimize the total production and inventory holding costs?

ANSWER:

Produce 4 boats in May. Produce 3 boats in June.

Produce 0 boats in July.

Produce 3 boats in August.

Total cost is $278,000.

POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.21.04 - 21.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 21.4 A Production and Inventory Control Problem KEYWORDS: Bloom's: Evaluate 40. Mission Bay Development Corporation is developing an exclusive 25-acre parcel of property. It has received bids from five builders to purchase construction lots. Because of the different nature of the builders, they desire different sized lots. The data are as follows: Builder 1 2 3 4 5

Lot Size Requested 3 acre 1 acre 6 acre 20 acre 5 acre

Offered Price per Lot $ 50,000 15,000 105,000 350,000 91,000

Maximum Number of Lots Desired 6 5 8 1 2

How should Mission Bay sell the land to maximize total sales revenue?

ANSWER: Builder 1: 1 lot; Builder 3: 2 lots; Builder 5: 2 lots; Total revenue = $442,000 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.21.04 - 21.4 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 21.4 A Production and Inventory Control Problem KEYWORDS: Bloom's: Evaluate © Cengage. Testing Powered by Cognero.

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CH 21 - Dynamic Programming 41. In the following shortest-route problem, travel is only permitted from left to right. Use dynamic programming to determine the shortest path from node 1 to: a. node 15. b. node 14.

ANSWER:

a. 1-2-7-12-15: Distance = 36 b. 1-4-9-13-14: Distance = 37 POINTS: 1 DIFFICULTY: Challenging LEARNING OBJECTIVES: IMS.ASWC.19.21.01 - 21.1 NATIONAL STANDARDS: United States - BUSPROG: Analytic TOPICS: 21.1 A Shortest-Route Problem KEYWORDS: Bloom's: Evaluate

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