Operations Management Creating Value Along the Supply Chain, 7th Edition Solution Manual

Type:

Solution Manual

Operations Management Resource: Creating Value Along the Supply Chain Edition:

7th Edition

Author(s):

Roberta S. Russell Bernard W. Taylor III

1 - Introduction to Operations and Supply Chain Management Answers to Questions 1-1.

The operations function involves organizing work, selecting processes, arranging layouts, locating facilities, designing jobs, measuring performance, controlling quality, scheduling work, managing inventory, and planning production. Operations interacts with marketing in product development, forecasting, production planning, and customer service. Operations and finance interact in capital budgeting, cost analysis, production and inventory planning, and expansion and technology plans.

Operations and human resources work together recruiting, training and evaluating workers, designing jobs and working with unions. IT and operations work together daily on e-commerce, enterprise resource planning and supply chain management systems.

1-2.

a.

b. c.

d.

Operations at a bank involves transferring funds, processing funds, providing checks, cashing checks, preparing monthly statements, reconciling statements, approving loans, loaning money, keeping track of loan payments, approving credit cards, and more.

Operations at a retail store involves purchasing goods, stocking goods, selling goods, keeping track of inventory, scheduling workers, laying out the store, locating the store, forecasting demand, and more.

Operations at a hospital involves preparing the rooms, scheduling doctors, nurses and other workers, processing paperwork, ordering supplies, caring for patients, maintaining the facility, laying out the facility, ensuring quality and more.

Operations at a cable TV company involves taking orders, installing equipment, maintaining equipment, keeping the shows on the air, scheduling work, processing statements and payments, and more.

1-3.

Inventions during the industrial revolution brought workers together under one roof in a factory setting where division of labor and interchangeable parts encouraged the formation of separate worker and management jobs. Ideas from the scientific management era made work more efficient. Human relations theorists emphasized the importance of the human element in operations management. The management science era saw many advances in quantitative techniques and their application. The quality revolution focused management on meeting customer expectations and emphasized quality over quantity. The Internet brought numerous opportunities to do work faster and better. It also opened doors to new markets worldwide. Today’s successful companies compete worldwide for both market access and production resources.

1-4.

Competitiveness is the degree to which a nation can produce goods and services that meet the test of international markets. A country’s competitiveness is measured by its GNP, import/export ratio, and increases in productivity. Industry competitiveness can be measured by the number of major players in the industry, average market share, and average profit margin. Measures of a firm’s competitiveness include market share, earnings per share, revenue growth, and profit margins. The Internet has opened new avenues of trade so that more firms compete for a larger, global market. The ease with which consumers can compare products and prices online has also increased competitiveness.

1-5.

Student answers will vary. The information can be accessed directly from the Internet or through the hyperlinks provided in Chapter 1 of the text’s homepage located at www.wiley.com/college/russell.

1-6.

Student answers will vary.

1-7.

Students can begin this assignment by accessing Fortune’s homepage and referring to the Fortune 500 or Global 500 by industry. The leaders in each industry are listed and there is usually some discussion of industry concerns. Individual data on companies can be found at Hoover’s website (www.hoovers.com).

1-8.

Student answers will vary.

1-9.

Student answers will vary.

1-10. Student answers will vary. The information can be accessed directly from the Internet or through the hyperlinks provided in Chapter 1 of the text’s homepage located at www.wiley.com/college/russell.

1-11. The WTO is an international organization that works to establish and enforce rules of trade between nations.

WTO agreements are ratified by the governing bodies of the nations involved. WTO’s dispute settlement process interprets agreements and rules on violations, thereby avoiding political or military conflict. The group promotes free trade and more recently, has helped developing nations enter the trade arena on more equitable grounds. Currently, there are 147 member nations. Membership is achieved by meeting certain environmental, human rights, and trade criteria, agreeing to abide by the rules of the organization, and being approved by two-thirds of the existing membership. See www.wto.org 1-12. Student answers will vary. Access www.executiveplanet.com 1-13. Student answers will vary. Access www.transparency.org 1-14. Student answers will vary. Access http://www.justice.gov/criminal/fraud/fcpa/ for basic information.

1-15. Students will find a variety of answers for this question. In general, it is easy to find mission or vision statements, but more difficult to find evidence of the mission or vision being applied.

1-16. Strategy formulation consists of four basic steps: (1) defining a primary task—what is the purpose of the firm? What the firm is in the business of doing? (2) assessing core competencies—what does a firm do better than anyone else? (3) determining order winners and order qualifiers—what wins orders in the marketplace?

What qualifies a product or service to be considered for purchase? (4) positioning the firm—what one or two important things should the firm choose to concentrate on? How should the firm compete in the marketplace?

Student answers will vary. Most start-ups try too much too soon. It’s difficult to stick with what you do best.

1-17. Core competencies are the essential capabilities that create a firm’s sustainable competitive advantage. They have usually been built up over time and cannot be easily imitated. For example, First National Bank, one of our local banks, is known as a risk taker. Its core competence is its ability to size up the potential of investment opportunities. Through its familiarity with local businesses and its experience in loan making, the bank has developed the ability to predict which loans are worth taking extra risks.

Bonomo’s, a successful retail store, is known for having just the right item in stock for special occasions.

The store stocks a variety of stylish women’s clothing, but not too much of each style. They specialize in knowing individual customers and even keep track of evening wear purchases so that no one else at a particular party will be wearing the same dress.

Toyota emphasizes superior quality at a price below its competitors with its Lexus line of automobiles. To establish a special reputation for quality over the lifetime of the car, the company set up separate sales and service facilities. When it is time for servicing, Lexus owners can have their vehicle picked up and delivered to their home or place of business. The car returns the same day, washed and vacuumed, often with a gift certificate inside for a night on the town complements of the dealer.

1-18. While the answers to this question vary considerably, most students feel competent in the technical areas of their major, but uncomfortable with their communication skills (both oral and written) and their ability to

make decisions. This opens the way for more project-oriented assignments from the instructor. The question also helps students prepare for the inevitable interview question—what are your strengths and weaknesses?

1-19. Order qualifiers are characteristics of a product or service that qualify it to be considered for purchase by a customer. An order winner is the characteristic of a product or service that wins orders in the marketplace— the final factor in the purchasing decision.

1-20. a.

b.

c.

d. e. f.

Most companies approach quality in a defensive or reactive mode; quality is confined to minimizing defect rates or conforming to design specifications. To compete on quality, companies must view quality as an opportunity to please the customer, not just as a way to avoid problems or to reduce rework costs.

The manufacturer of Rolex watches competes on quality.

Companies that compete on cost relentlessly pursue the elimination of all waste. The entire cost structure is examined for reduction potential, not just direct labor costs. High volume production and automation may or may not provide the most cost-effective alternative. Wal-Mart competes on cost.

Flexibility includes the ability to produce a wide variety of products, to introduce new products and to modify existing products quickly, and, in general, to respond to customer needs. National Bicycle Industrial Company competes on flexibility.

Competing on speed requires a new type of organization characterized by fast moves, fast adaptations, and tight linkages. Citicorp competes on speed.

Competing on dependability requires a stable environment, adequate resources, high standards for performance, and tight control. Maytag competes on dependability.

Competing on service requires closeness to the customer, availability of resources, attention to detail, and flexible operations. Ritz-Carlton competes on service.

1-21. Operations can play two important roles in corporate strategy: (1) it can provide support for the strategy of a firm (help with order qualifiers), and (2) it can serve as a firm’s distinctive competence (win orders).

1-22. Strategic decisions in operations and supply chain management involve products and services, processes and technology, capacity and facilities, human resources, quality, sourcing, and operating systems.

1-23. Policy deployment tries to focus everyone in an organization on common goals and priorities by translating corporate strategy into measurable objectives down through the various functions and levels of the organization. As a result, everyone in the organization should understand the strategic plan, be able to derive several goals from the plan, and determine how each goal ties into their own daily activities.

1-24. The balanced scorecard examines a firm’s performance in four critical areas – its finances, customers, processes and capacity for learning and growing. Although operational excellence is important in all four areas, the tools in operations are most closely associated with process.

1-25. Student answers will vary.

1-26. Student answers will vary. The balanced scorecard worksheet in Table 2.1 is helpful. Finances might refer to future income, customers to potential employers who are interested in both grades and experience, processes to how students will raise their grades and gain experience, and learning and growing to developing skills in several areas.

Solutions to Problems 1-1.

The Blacksburg store is the most productive.

Store Sales volume Labor hours Productivity

1-2.

Blacksburg $12,000 60 $200

b.

Based on productivity, the Blacksburg store should be closed. Other factors to consider include total revenue, potential for growth, and options for reducing costs.

Annandale $40,000 250 $6.75 $1,800 $11

Blacksburg $12,000 60 $6.50 $2,000 $5

Danville $25,000 200 $5.50 $800 $13

Last yr 4 12 20 60

This yr 6 15 25 62.5

Productivity could be measured by total account dollars per hour worked, new account dollars per hour worked or existing account dollars per hour worked. Bates is the most productive based on total output.

Albert and Duong have the most new accounts, and thus the greater potential returns in the future. However, Duong cannot work many more hours a week and Bates is only working half time. Bates has the potential to sell more if he works more hours.

Albert $100,000 $40,000 40 $3,500.00 $2,500.00 $1,000.00

Bates $40,000 $40,000 20 $4,000.00 $2,000.00 $2,000.00

Cressey $80,000 $150,000 60 $3,833.33 $1,333.33 $2,500.00

Duong $200,000 $100,000 80 $3,750.00 $2,500.00 $1,250.00

The U.S. is the most productive.

U.S. Germany Japan 1-6.

Charlottesville $60,000 500 $6.00 $1,200 $14

By number, Jim was more productive last year. By weight, Jim was more productive this year.

Agents New accounts Existing accounts Labor hours Total $/hr $ New accts/hr $ Existing accts/hr 1-5.

Danville $25,000 200 $125

Charlottesville is the most productive.

Hours fishing Bass caught Average weight Bass/hr 1-4.

Charlottesville $60,000 500 $120

a.

Sales volume Labor hours Labor cost/hr Rent Productivity 1-3.

Annandale $40,000 250 $160

Labor Hours 89.5 83.6 72.7

Units of Output 136 100 102

Productivity 1.52 1.20 1.40

Omar should probably close the plant in Guadalajara because its multifactor productivity is the lowest, its labor productivity is the second lowest, and its output is the least of the four plants.

Units (in 000’s) Finished goods Work-in-process

1-7.

Cinncinnati 10,000 1,000

Guadalajara 5,000 3,000

Bejiing 8,000 6,000

Costs (in 000’s) Labor costs Material costs Energy costs Transportation costs Overhead costs

$3,500 $3,500 $1,000 $250 $1,200

$4,200 $3,000 $1,500 $2,500 $3,000

$2,500 $2,000 $1,200 $2,000 $2,500

$800 $2,500 $800 $5,000 $500

Labor productivity Total productivity

3.14 1.16

3.38 1.00

3.20 0.78

17.50 1.46

Hall is the most productive in terms of rushing yards and touchdowns per carry. However, Dayne has highest number of rushing yards and touchdowns. Using “carries” as the input variable skews the results.

Productivity is not always the best measure of performance.

Candidates Rushing yards # Carries # Touchdowns Yards/carry Touchdowns/carry

1-8.

Frankfurt 12,000 2,200

Hall 2,110 105 15

Walker 3,623 875 20

Dayne 6,925 1,186 70

20.10 0.14

4.14 0.02

5.84 0.06

1 1,225 4 3 102.08

2 1,435 3 5 95.67

3 2,500 5 6 83.33

Productivity decreases from week to week.

Installation Square Yards # workers # hours Square yds/hr

1-9.

Center Pieces processed Workers/hr Hourly wage rate Overhead/hr Multifactor productivity a. b. c.

1 1,000 10 $5.50 $10 15.38

2 2,000 5 $10 $25 26.67

3 3,000 2 $12 $50 40.54

3c.

5,000 2 $12 $80 48.08

Work center # 3 is the most productive.

With a 10% raise in center 1, productivity goes down to 14.18 pieces per dollar spent.

With new equipment in center 3, productivity goes up to 48 pieces.

Install the new equipment.

1-10. Material productivity is stable over the 4 weeks. Labor productivity increases in week 2 and decreases in weeks 3 and 4.

Week Units of output # workers Hours per week Labor cost per hour Material (lbs.)

Material cost per lb.

1 2,000 4 40 $10 286 $4

2 4,000 4 48 $10 570 $4

3 5,000 5 56 $10 720 $4

4 7,000 6 70 $10 1,000 $4

Labor productivity Material productivity Multifactor productivity

1.25 1.75 0.73

2.08 1.75 0.95

1.79 1.74 0.88

1.67 1.75 0.85

1-11. John is the most productive.

# ads sold # hours spent Output/hr

Jake

Josh

Jennifer

John

100 40 2.50

50 15 3.33

200 85 2.35

35 10 3.50

1-12. Choose Cold Case.

Purchase cost Daily energy consumption (kwh) Cost per kwh

Alaskan Seal

Brr Frost

Cold Case

Deep Freeze

$3,270

$4,000

$4,452

$5,450

3.61

3.88

6.68

29.07

$0.10

$0.10

$0.10

$0.10

Daily energy cost

$0.36

$0.39

$0.67

$2.91

Daily purchase cost

$2.99

$3.65

$4.07

$4.98

Total cost

$3.35

$4.04

$4.73

$7.88

Volume (cu ft)

25

35

49

79

Productivity (cu ft/$)

7.47

8.66

10.35

10.02

Cost/cu ft

$0.13

$0.12

$0.10

$0.10

CASE SOLUTION 1.1: Visualize This 1.

It is difficult to follow the four steps of strategy formulation for this case. Students will be able to easily identify VT’s core competency but will struggle with its primary task, and without a product, it’s impossible to determine an order winner and order qualifiers. “Developing the next generation of visualization tools” is probably not a marketable task. Students will come up with a variety of ideas from their Internet search.

2.

Student answers will vary depending on how the primary task is derived in question 1.

3.

That’s the crux of the problem for this case. Isaac needs to find a way to keep his business going to obtain the capital to pursue his dream. Great for class discussion.

4.

(1) and (3) are more in keeping with VT’s earlier projects but require more hardware and do not promise future business. (2) is the most time-consuming, least challenging, but most sustainable. (4) and (5) are the most lucrative but do not advance VT’s knowledge of the field.

5.

The selection of projects should reinforce the strategy determined by the student. This case is based on an actual situation. The company chose projects (1) and (3). The museum job consumed so much time and resources that the company had to turn down the bank training job. Without a “product” and no immediate repeat business, the company folded and the owner went back to academe. A student took on project (5), became quite successful and now has operations in three states.

CASE SOLUTION 1.2: Whither an MBA at Strutledge?

1.

The board of Regents should look at the proposal carefully and identify first what they are trying to achieve with this new program. If the program fits within their mission, and if they have the resources to pursue it, they need to assess the likelihood of their success or failure. It doesn’t appear that the board has sufficient information or insight to make the decision. A lot of questions remain. The focus of the program (i.e., interdisciplinary, problem solving, etc.) doesn’t seem like much of a focus at all. The desire to “try anything” to get more students is troublesome. A new program that Strutledge can’t support would damage their reputation. Strutledge needs to gather more information before a decision can be made.

2.

Strutledge should go through the process of identifying its primary task. This would include the type of students it wishes to serve and their future role in society (i.e., community, state, regional, national, global).

A clear assessment of Strutledge’s core competence is also needed. What special resources does the college have? What is it best known for? How does it compare to other institutions of similar size and mission?

After those issues have been settled, the college needs to find out what its customers (i.e., students) look for when deciding where to go to school. What are some basic requirements that Strutledge should meet (i.e., order qualifiers)? What factor prompts the final determination of which school to attend (i.e., order winner)? If, as is hinted in the case, the ability to find employment upon graduation is important to prospective students, then the college should gather information from potential employers about their needs. It may very well be that an MBA program is needed in the area, but this needs to be determined from data. Only after the determination has been made, that the area needs another MBA program, should Strutledge explore the possibility of providing it. If the college concludes that it has the skills and resources necessary to pursue the task, then it should try to position itself properly in the market and find a special niche for its particular MBA program.

CASE SOLUTION 1.3 – Weighing Options at the Weight Club

A Balanced Scorecard for the Weight Club Key Performance Indicator

Goal

Revenue Growth

Attract new customers

% increase in customers

25%

Quality

Meet or exceed customer needs

% customers satisfied

100%

Retention

Build sustainable customer base

% membership renewals

75%

# exercise classes/week

12

Fitness

Incr. participation in exercise classes Increase use of personal trainers

# client hours/week

100

Processes

Finances

Objectives Generate revenue for firstclass facility

Customers

Dimension

Enhance client experience Client services

Learning & Growing

Facilitate use of services Equipment maintenance

Maintain equipment in top working condition

Program development

Develop professional staff

Facility development

Provide first-class facilities and equipment

Organizational development

Develop management and administrative skills

% increase in revenue

30%

% participation in customer orientation # massage appointments/week Time required for check-in Hours of child care/week % fully operational % on regular maintenance schedule % new classes # innovative suggestions % equipment new or updated Months until facility expanded/ renovated

75% 200 1 min 90% 95% 60% 25 30 100% 6

# persons on Board of Directors

6

# full-time managers

3

Supplement 1 Operational Decision-Making Tools: Decision Analysis S1-1. a. Minimin: South Korea 15.2 China 17.6 Taiwan 14.9 Poland 13.8 Mexico 12.5  minimum Select Mexico b. Minimax: South Korea 21.7 China 19.0  minimum Taiwan 19.2 Poland 22.5 Mexico 25.0 Select China c. Hurwicz ( = 0.40 ) : South Korea: 15.2 ( 0.40 ) + 21.7 ( 0.60 ) = 19.10 China: 17.6 ( 0.40 ) + 19.0 ( 0.60 ) = 18.44 Taiwan: 14.9 ( 0.40 ) + 19.2 ( 0.60 ) = 17.48  minimum Poland: 13.8 ( 0.40 ) + 22.5 ( 0.60 ) = 19.02 Mexico: 12.5 ( 0.40 ) + 25.0 ( 0.60 ) = 20.0 Select Taiwan d. Equal likelihood: South Korea: 21.7 ( 0.33) + 19.1( 0.33) + 15.2 ( 0.33 ) = 18.48 China: 19.0 ( 0.33) + 18.5 ( 0.33) + 17.6 ( 0.33) = 18.18 Taiwan: 19.2 ( 0.33) + 17.1( 0.33) + 14.9 ( 0.33) = 16.90  minimum Poland: 22.5 ( 0.33) + 16.8 ( 0.33) + 13.8 ( 0.33 ) = 17.52 Mexico: 25.0 ( 0.33) + 21.2 ( 0.33) + 12.5 ( 0.33) = 19.37 Select Taiwan S1-2.

EV (South Korea ) = 21.7 (.30 ) + 19.1(.40 ) + 15.2 (.30 ) = 18.71

EV ( China ) = 19.0 (.30 ) + 18.5 (.40 ) + 17.6 (.30 ) = 18.38 EV ( Taiwan ) = 19.2 (.30 ) + 17.1(.40 ) + 14.9 (.30 ) = 17.07  minimum EV ( Poland ) = 22.5 (.30 ) + 16.8 (.40 ) + 13.8 (.30 ) = 17.61

EV ( Mexico ) = 25.0 (.30 ) + 21.2 (.40 ) + 12.5 (.30 ) = 19.73 Select Taiwan

Expected value of perfect information = 19 (.30 ) + 16.8 (.40 ) + 12.5 (.30 ) = 16.17 EVPI = 16.17 − 17.07 = $ − 0.9 million

The EVPI is the maximum amount the cost of the facility could be reduced (.9 million) if perfect information can be obtained.

S1-3. a. Maximax criteria: Office building 4.5  maximum Parking lot 2.4 Warehouse 1.7 Shopping mall 3.6 Condominiums 3.2 Select office building b. Maximin criteria: Office building 0.5 Parking lot 1.5  maximum Warehouse 1.0 Shopping mall 0.7 Condominiums 0.6 Select parking lot c. Equal likelihood: Office building: 0.5 ( 0.33) + 1.7 ( 0.33) + 4.5 ( 0.33) = 2.21  maximum Parking lot: 1.5 ( 0.33) + 1.9 ( 0.33) + 2.4 ( 0.33 ) = 1.91 Warehouse: 1.7 ( 0.33) + 1.4 ( 0.33) + 1.0 ( 0.33 ) = 1.35 Shopping mall: 0.7 ( 0.33) + 2.4 ( 0.33) + 3.6 ( 0.33) = 2.21  maximum Condominiums: 3.2 ( 0.33) + 1.5 ( 0.33) + 0.6 ( 0.33) = 1.75 Select office building or shopping mall d. Hurwicz criteria ( = 0.3) : Office building: 4.5 ( 0.3) + 0.5 ( 0.7 ) = 1.70 Parking lot: 2.4 ( 0.3) + 1.5 ( 0.7 ) = 1.77  maximum Warehouse: 1.7 ( 0.3) + 1.0 ( 0.7 ) = 1.21 Shopping mall: 3.6 ( 0.3) + 0.7 ( 0.7 ) = 1.57 Condominiums: 3.2 ( 0.3) + 0.6 ( 0.7 ) = 1.38 Select parking lot S1-4. a) EV ( Office building ) = .5 (.50 ) + 1.7 (.40 ) + 4.5 (.10 ) = 1.38

EV ( Parking lot ) = 1.5 (.50 ) + 1.9 (.40 ) + 2.4 (.10 ) = 1.75 EV ( Warehouse ) = 1.7 (.50 ) + 1.4 (.40 ) + 1.0 (.10 ) = 1.51

EV ( Shopping mall ) = 0.7 (.50 ) + 2.4 (.40 ) + 3.6 (.10 ) = 1.67 EV ( Condominiums ) = 3.2 (.50 ) + 1.5 (.40 ) + .06 (.10 ) = 2.26  maximum Select Condominium project b) EVPI = Expected value of perfect information–expected value without perfect information

= 3.01–2.26 = $0.75 million S1-5. a. Maximax: Risk fund, maximax payoff = $167,000 b. Maximin: Savings bond maximin payoff = $30,000 c. Equal likelihood: Bond fund, maximum payoff = $35,000 d. Best decision, given probabilities: Bond fund, maximum payoff = $35,000 S1-6. a. Maximax: Pass for a gain of 20 yd b. Maximin: Option for a loss of 1 yd c. Equal likelihood: Option for a gain of 7 yds. d. Plays ranked best to worst: Play Pass Option Toss sweep Off tackle Screen Draw

EV 6.4 5.3 4.8 3.2 2.3 1.6

With a 60% chance of a blitz they should run the option, with an expected value of 11.5 yd.

S1-7. a.

Product Widget

Expected Value 160, 000 ( 0.2 ) + 90, 000 ( 0.5 ) − 50, 000 ( 0.3) = $62, 000

Hummer

70, 000 ( 0.2 ) + 40, 000 ( 0.5 ) + 20, 000 ( 0.3 ) = $40, 000

Nimnot

45, 000 ( 0.2 ) + 35, 000 ( 0.5) + 30, 000 ( 0.3) = $35,500

The best option is to introduce the widget. b. EV given perfect information: 160, 000 ( 0.2 ) + 90, 000 ( 0.5 ) + 30, 000 ( 0.3) = $86, 000.

EV without perfect information: Widget at $62,000.

Value of perfect information: $86, 000 − $62, 000 = $24, 000 The company would consider this a maximum; since perfect information is rare, it would probably pay less than $24,000. c. Maximax: Introduce widget, maximax payoff = $160,000 Maximin: Introduce nimnot, maximin payoff = $30,000.

Minimax regret: Introduce widget, Minimax regret = $80,000 Equal likelihood: Introduce widget, maximum payoff = $66,000 S1-8. a. Maximax: Major physical revision, maximum payoff = $972,000 b. Maximin: Paperback, maximum payoff = $68,000 c. Equal likelihood: Major content revision, maximum payoff = $419,430 d. Hurwicz: Major content revision, maximum payoff = $273,900

S1-9.

Publication Decision Paperback Similar revision Major content revision Major physical revision

Expected Value $216,290 386,340 468,780 405,970

Best decision = major content revision Overall “best” decision appears to be a “major content revision” EVPI = (.23)(68,000) + (.46)(515,000) + (.31)(972,000) − 468,780 = $85,080 This is the maximum amount Wiley would pay an “expert” for additional information about the future competitive market.

S1-10. a. Maximax: Singapore, maximum payoff = $71 million b. Maximin: Kaohsiung, maximum payoff = -$15 million c. Equal likelihood: Kaohsiung, maximum payoff = $28.05 million d. Hurwicz: Singapore, maximum payoff = $37.8 million e. Minimax regret: Singapore, minimum regret = $9 million S1-11. Expected value Port Hong Kong Singapore Shanghai Busan Kaohsiung

Expected Value $22.99 34.52 24.54 28.30 33.66

a. Best decision = Singapore b. Singapore appears to be the best “overall” decision.

S1-12. Expected value Lease Decision 1 – year 2 – year 3 – year 4 – year 5 – year

Expected Value $65,980 103,010 133,810 154,300 114,210

S1-13. EVPI = (.17)(1,228,000) + (.34)(516,000) + (.49)(−551,000) − 154,300 = $237,740 This is the maximum amount the restaurant owner would pay an energy “expert” for additional information about future energy prices.

S1-14. a. Maximax: Food court, maximum payoff = $87,000 b. Maximin: Child care center, maximum payoff = $17,000 c. Hurwicz: Lockers and showers, maximum payoff = $32,250 d. Equal likelihood: Lockers and showers, maximum payoff = $34,980 S1-15 Service Facility Child care center Swimming pool

Expected Value $30,560 7,610

Lockers and showers Food court Spa

44,150 15,440 20,580

Best decision = Lockers and showers S1-16. a.

Payoff table:

Stock (lb) 20 21 22 23 24

20 0.10 20.00 18.50 17.00 15.50 14.00

21 0.20 20.00 21.00 19.50 18.00 16.50

Demand 22 0.30 20.00 21.00 22.00 20.50 19.00

23 0.30 20.00 21.00 22.00 23.00 21.50

24 0.10 20.00 21.00 22.00 23.00 24.00

EV ( 20 ) = $20 EV ( 21) = 18.50 ( 0.1) + 21( 0.2 ) + 21( 0.3) + 21( 0.3) + 21( 0.1) = $20.75

EV ( 22 ) = 17 ( 0.1) + 19.50 ( 0.2 ) + 22 ( 0.3) + 22 ( 0.3) + 22 ( 0.1) = $21.00 EV ( 23) = 15.50 ( 0.1) + 18 ( 0.2 ) + 20.50 ( 0.3 ) + 23 ( 0.3 ) + 23 ( 0.1) = $20.50 EV ( 24 ) = 14 ( 0.1) + 16.50 ( 0.2 ) + 19 ( 0.3 ) + 21.50 ( 0.3 ) + 24 ( 0.1) = $19.25

Order 22 lb of apples for a profit of $21.00. b. Maximax: Stock 24 lb for a maximax profit of $24.00.

Maximin: Stock 20 lb for a maximin profit of $20.00.

S1-17. a. Payoff table: Demand Stock (lb) (boxes) 25 26 27 28 29 30

25 0.10 50 49 48 47 46 45

26 0.15 50 52 51 50 49 48

27 0.30 50 52 54 53 52 51

28 0.20 50 52 54 56 55 54

29 0.15 50 52 54 56 58 57

EV ( 25 ) = 50 ( 0.1) + 50 ( 0.15 ) + 50 ( 0.3) + 50 ( 0.2 ) + 50 ( 0.15 ) + 50 ( 0.1) = $50.00

EV ( 26 ) = 49 ( 0.1) + 52 ( 0.15 ) + 52 ( 0.3) + 52 ( 0.2 ) + 52 ( 0.15 ) + 52 ( 0.1) = $51.70 EV ( 27 ) = 48 ( 0.1) + 51( 0.15 ) + 54 ( 0.3) + 54 ( 0.2 ) + 54 ( 0.15 ) + 54 ( 0.1) = $52.95 EV ( 28 ) = 47 ( 0.1) + 50 ( 0.15 ) + 53 ( 0.3) + 56 ( 0.2 ) + 56 ( 0.15 ) + 56 ( 0.1) = $53.30 EV ( 29 ) = 46 ( 0.1) + 49 ( 0.15 ) + 52 ( 0.3) + 55 ( 0.2 ) + 58 ( 0.15 ) + 58 ( 0.1) = $53.05 EV ( 30 ) = 45 ( 0.1) + 48 ( 0.15 ) + 51( 0.3) + 54 ( 0.2 ) + 57 ( 0.15 ) + 60 ( 0.1)

30 0.10 50 52 54 56 58 60

= $52.35 Best decision: Stock 28 boxes, for a profit of $53.30.

b. Expected value under uncertainty: EV = 500 ( 0.10 ) + 52 ( 0.15 ) + 54 ( 0.30 ) + 56 ( 0.20 ) + 58 ( 0.15 ) + 60 ( 0.10 ) = $54.90 EVPI = $54.90 − $53.30 = $1.60

S1-18. a) Stock 25, maximum of minimum payoffs = $50 b) Stock 30, maximum of maximum payoffs = $60 c) 25 : 50 (.4 ) + 50 (.6 ) = 50;

26 : 52 (.4 ) + 49 (.6 ) = 50.2;

27 : 54 (.4 ) + 48 (.6 ) = 50.4;

28 : 56 (.4 ) + 47 (.6 ) = 50.6; 29 : 58 (.4 ) + 46 (.6 ) = 50.8; 30 : 60 (.4 ) + 45 (.6 ) = 51; stock 30 boxes. d) Stock 28 or 29 boxes; minimum regret = $4.

S1-19. EV ( press ) = 40, 000 (.4 ) − 8, 000 (.6 ) = $11, 200;

EV ( lathe ) = 20, 000 (.4 ) + 4, 000 (.6 ) = $10, 400; EV ( grinder ) = 12, 000 (.4 ) + 10, 000 (.6 ) = $10,800; purchase press.

S1-20.

S1-21. a. Maximax = Gordon b. Maximin = Jackson c. Hurwicz ( = 0.25 )

Morris = 4.4 ( 0.25) + ( −3.2 )( 0.75 ) = −$1.3M O’Neil = 6.3 ( 0.25 ) + ( −5.1)( 0.75 ) = −$2.3M

Jackson = 5.8 ( 0.25 ) + ( −2.7 )( 0.75 ) = −$0.58M Gordon = 9.6 ( 0.25 ) + ( −6.3)( 0.75 ) = −$2.33M Select Jackson d. Equal likelihood Morris = 4.4 ( 0.33) + (1.3)( 0.33) + ( −3.2 )( 0.33) = $.83M O’Neil = 6.3 ( 0.33) + (1.8 )( 0.33) + ( −5.1)( 0.33) = +$.99M

Jackson = 5.8 ( 0.33) + ( 0.7 )( 0.33) + ( −2.7 )( 0.33) = +$1.254M

Gordon = 9.6 ( 0.33) + ( −1.6 )( 0.33) + ( −6.3)( 0.33) = $.561M Select Jackson e.

EV ( Morris ) = ( −3.2 )( 0.15 ) + (1.3)( 0.55 ) + ( 4.4 )( 0.30 ) = $1.56M EV(O’Neil) = ( −5.1)( 0.18 ) + (1.8 )( 0.26 ) + ( 6.3)( 0.56 ) = $3.08M

EV ( Jackson ) = ( −2.7 )( 0.21) + ( 0.7 )( 0.32 ) + ( 5.8 )( 0.47 ) = $2.38M EV ( Gordon ) = ( −6.3)( 0.30 ) + ( −1.6 )( 0.25 ) + ( 9.6 )( 0.45 ) = $2.03M Select O’Neil.

S1-22. a. Maximax = Real Estate b. Maximin = Nursing c. Equal Likelihood: select Real Estate Graphic design = $170,000 Nursing = $187,500 Real Estate = $202,500 Medical Technology = $195,000 Culinary technology = $170,000 Computer information technology = $186,250 d. Hurwicz (alpha = 0.25): select Nursing Graphic design = $141,250 Nursing = $161,250 Real Estate = $158,750 Medical Technology = $157,500 Culinary technology = $136,250 Computer information technology = $158,750 S1-23. EV(Graphic design) = $164,250 EV(Nursing) = $183,500 EV(Real Estate) = $174,400 EV(Medical Technology) = $187,500 EV(Culinary technology) = $149,250 EV(Computer information technology) = $174,750 S1-24. a. Maximax = Juan Ramon

b. Maximin = Alan Rodriguez c. Equal likelihood: Garcia = 106.92 Ramon = 119.46 Terry = 103.29 Rodriguez = 96.03 Washburn = 92.73 d. Hurwicz: Garcia = 91.95 Ramon = 95.10 Terry = 94.55 Rodriguez = 95.75 Washburn = 84.35

SELECT

SELECT

S1-25. a. EV(Garcia) = 100.3 EV(Ramon) = 112.4 SELECT EV(Terry)= 98.2 EV(Rodriguez) = 91.6 EV(Washburn) = 85.2 b. Probably Terry; he seems to have the best tradeoff between low cost and wins. However, this is an objective opinion depending on the degree of risk the decision maker is willing to take on. c. EV(Garcia) = 109.71 EV(Ramon) = 109.74 EV(Terry)= 106.81 EV(Rodriguez) = 100.00 EV(Washburn) = 93.48

SELECT

S1-26. a. Maximax = Hong Kong b. Maximin = Pusan c. Equal likelihood: Shanghai = $0.44 billion Singapore = $0.37 billion Pusan = $0.43 billion Kaoshiung = $0.41 billion Hong Kong = $0.47 billion d. Hurwicz (alpha = .55): Shanghai = $0.47 billion Singapore = $0.41 billion Pusan = $0.46 billion Kaoshiung = $0.54 billion Hong Kong = $0.77 billion S1-27. EV(Shanghai) = $0.608 billion EV(Singapore) = $0.606 billion EV(Pusan) = $0.502 billion EV(Kaoshiung) = $0.487 billion EV(Hong Kong) = $0.724 billion

S1-28.

EV ( snow shoveler ) = $30 (.12 ) + 60 (.19 ) + 90 (.24 ) + 120 (.22 ) + 150 (.13) + 180 (.08 ) + 210 (.02 ) = $101.10

The cost of the snow blower ($575) is much more than the annual cost of the snow shoveler, thus on the basis of one year the snow shoveler should not be purchased. However, the snow blower could be used for an extended period of time such that after approximately 6 years the cost of the snow blower would be recouped. Thus, the decision hinges on weather or not the decision maker thinks 6 years is too long to wait to recoup the cost of the snow blower.

S1-29.

Since cost of installation ($900,000) is greater than expected value of not installing ($552,000), do not install an emergency power generator

S1-30.

Select strategy 3; Change oil regularly; EV = $98.80

S1-31.

Select Strategy 4; Change oil and sample; EV = $716.40

S1-32. a.

b.

.98 9.2 x + 1.5 (1 − x ) + (.02 )(1.5) = 3.810 .98  7.7 x + 1.5 + .030 = 3.810

7.546 x + 1.47 + .030 = 3.810 7.546 x = 2.31

x = .306 probability of winning in overtime

S1-33.

S1-34. The following table includes the medical costs for all the final nodes in the decision tree.

Expense 100 500 1,500 3,000 5,000 10,000

E (1) = 954 E ( 2 ) = 976.5 E ( 3) = 810 Select plan 3

Plan 1 481 884 984 1,134 1,334 1,834

Plan 2 160 560 1,290 1,440 1,640 2,140

Plan 3 318 438 738 1,188 1,788 3,288

S1-35.

S1-36.

CASE S1.1: Whither an MBA at Strutledge? -Continued a. Maximax: IT, maximum payoff = $517,000 b. Maximin: Health Administration, maximum payoff = −$75,000 c. Equal likelihood: Nursing, maximum payoff = $114,500 d. Hurwicz: Nursing, maximum payoff = $86,000 e. They do not have sufficient insight into the probability of the future success of the programs to indicate either optimism or pessimism; or for “political” reasons they feel it is imprudent to express a “preference.” f. Best decision = Nursing Graduate Program MBA Computer Science Information Technology Nursing Health Administration

Expected Value −27,470 −45,000 10,790 126,760 124,250

g. Nursing appears to be the best overall decision. h. Depends on student answer.

CASE S1.2 : Transformer Replacement at Mountain State Electric Service The decision tree solution for this problem is shown below. The decision should be to retain the existing transformer; the cost of replacement ($85,000) is greater than the cost of retention ($61,000).

CASE S1.3: Evaluating Projects at Nexcom Systems Project 1 2 3 4 5

EV 404,368 434,976 442,891 344,490 262,252

2 Quality Management Answers to Questions 2-1.

Consumers perceive quality to be how well a product meets its intended use—that is, how well it does what it is supposed to do—whereas from the product’s perspective, quality is how well the product conforms to its design during the production process.

2-2.

1. Performance: operating characteristics of a product 2. Features: extra items added to basic characteristics 3. Reliability: probability that a product will operate properly 4. Conformance: the degree to which a product meets standards 5. Durability: how long the product lasts 6. Serviceability: ease and speed of repair and courtesy of repair person 7. Aesthetics: how a product looks, feels, sounds, smells, or tastes 8. Safety: Assurance that the customer will not suffer harm; especially important for autos.

9. Other: subjective perceptions based on brand name or advertising

2-3.

Quality of design is the degree to which quality characteristics are designed into a product, whereas quality of conformance is how effectively the production process is able to conform to the specifications required by design.

2-4.

The cost of quality assurance is the cost of maintaining an effective quality program and includes prevention and appraisal costs. The cost of nonconformance, or poor quality, is the result of internal and external failures. These two costs react oppositely to each other; as the cost of quality assurance increases, the cost of poor quality decreases.

2-5.

Internal failure costs are incurred when poor quality is discovered before the product is delivered to the customer, whereas external failure costs are incurred after a customer receives a poor-quality product. Internal failure costs include scrap, rework, process failure, and downtime, whereas external failure costs include customer complaints, product returns, warranty claims, product liability, and lost sales.

2-6.

The contractor could be experiencing low productivity yields and have extensive internal failure costs, including scrap, rework, process failure, and downtime costs.

2-7.

a. From the consumer’s (e.g., student or parent) perspective, quality is probably determined by whether the college education provides the job opportunity expected and whether the graduate perceives he or she has acquired an anticipated level of knowledge that will enable the graduate to perform the job effectively. From the producer’s (e.g., university) perspective, quality is how effectively it is able to deliver knowledge (i.e., required courses) and provide the quality of life experience expected by the student. b. The education achieved by the student provides the job opportunities expected and a level of knowledge that enables the graduate effectively to perform the job achieved. c. Quality-assurance costs include the cost of hiring the best faculty, administrators, and support personnel, the cost of designing and redesigning courses and curriculum to meet changing needs, the cost of providing a good physical and mental environment (i.e., housing, food,

entertainment, security, etc.), the cost of modern technical teaching equipment, the cost of information systems, and the cost of assessing alumni satisfaction with their education. Costs of poor quality include students who fail or drop out, reduced funding from the state or private donors, and fewer enrollments. d. Quality circles could be developed within administrative and operational units and academic departments. Circles might include both faculty or administrators and classified employees.

The normal quality circle stages of training, problem identification, analysis, solution, and presentation could be followed.

2-8.

Improving quality will increase product yield—that is, the number of acceptable units—thus increasing productivity.

2-9.

The cost of poor quality could include external failure costs for customer complaints, returned DVD players to be repaired under warranty, lost future sales, and liability costs if someone is hurt because of the problem. Costs of quality improvement might include improved design costs for the DVD player, process costs, and inspection costs for the final product and at various stages of the production process.

2-10. DVD player: Visual attractiveness, size, weight, clarity of sound and picture, and features for rewind and fast-forward, program search, programming, playback, etc.

Pizza: Size, ingredients, taste, smell, service in delivery, temperature.

Running shoes: Size, weight, comfort, visual attractiveness, durability.

2-11. The input is customer inquiries and the final product is responses that result in customer satisfaction. Associated quality costs might include prevention costs, such as designing a telephone system to ensure prompt connections without waiting and a properly designed computer system to provide accurate customer account information, and training costs to make certain service operators are courteous and knowledgeable. Appraisal costs might include the cost of monitoring service calls to ascertain response rates and operator courtesy. Poor quality might result in complaints from customers and lost accounts.

A quality management program could incorporate a system to monitor calls to ensure prompt, courteous, and knowledgeable service. An employee-involvement program, wherein operators might identify problems, would be beneficial.

2-12. Prevention costs are directed at preventing poor quality products from reaching the customer, thus avoiding the various internal and external failure costs associated with poor quality.

2-13. It is important to have a means for assessing the impact of quality improvement programs on the organization’s profitability and productivity.

2-14. W.E. Deming: Introduced the Japanese to quality management principles and philosophy, embodied in his 14 points.

Joseph Juran: A major contributor to the Japanese quality movement.

Phillip Crosby: Changed general perceptions of cost of quality and promoted zero defects.

Armand Feigenbaum: Introduced the concept of total quality control, a total company-wide approach to quality management.

Kaoru Ishikawa: Introduced quality circles and cause and effect diagrams.

Genichi Taguchi: Developed the Taguchi method for product and process design.

2-15. The Baldrige Award has had a pervasive impact on American companies, in general promoting

quality improvement. Thousands of companies request award applications each year to use simply to establish quality management programs based on Baldrige Award criteria.

2-16. This should be a student project. The journal can be found in most libraries, and the articles are generally easy to read.

2-17. The student could provide many reasons for failure including lack of total commitment, ineffective planning, goals too easy or too difficult to achieve, improper measurement techniques, ineffective leadership, not enough employee training, etc. See G. Salegra and Farzaneh, “Obstacles to Implementing Quality,” Quality Progress, 33, no. 7 (July 2000): 53–57.

2-18. The dimensions of quality for a service company are located in the text. The student should identify these or similar ones for the company they select.

2-19. The two service companies should be in the community and the quality characteristics the students will tend to focus on will include courtesy and quickness of service.

2-20. Although students in this class might suggest that grades are a quality measurement a more realistic approach to evaluation are student evaluations of the class or surveys of students. Quality characteristics might include course organization, presentation of lectures, class environment, physical appearance of the classroom, schedule (i.e., are the lectures completed on time), the quality of supplementary material, physical appearance and demeanor of the instructor, including friendliness and courtesy, the accuracy and completeness of assignments, etc.

2-21. The answer depends on the company selected by the student. For example, there is a particular hotel that has never gotten a room reservation right for us, and, the instructions for ordering tickets at the web site for the 1996 Olympics in Atlanta were littered with pitfalls. Airlines are a favorite example of a poor quality service for students who travel.

2-22. A similar question to 26. Restaurants, retail stores and grocery stores are examples of local businesses that, in our experience, tend to vary in quality. We have never had a bad ordering experience with L.L. Bean although that’s not true of some other mail order operations we have dealt with. In most cases, if a service has been identified by the student it will be because of courteous, helpful employees, while if a manufacturing product has been identified, it will be because of superior physical traits, such as durability.

2-23. TQM tends to give some focus and structure to strategic planning. TQM provides identifiable goals, and many well-documented initiatives for quality improvement such as quality circles, employee training, empowerment, etc. TQM also provides a means for measuring success which is essential in a strategic planning process.

2-24. Many U.S. suppliers cannot do business with companies overseas unless they have ISO certification. In addition, many U.S. companies also desire or request their suppliers to comply with ISO 9000 standards.

2-25. Common characteristics that the students will discover include strong leadership at the top, total company commitment, employee training, involvement and empowerment, challenging goals for quality achievement, focus on customer satisfaction, and extensive use of statistical quality control techniques, among other things.

2-26. Some companies believe their quality is “good enough.” However, primary reasons for not implementing a TQM program are lack of time and the cost involved; some companies do not have the resources available to undertake a TQM program.

2-27. This will depend on the web site the student accesses. In general, they should adapt the attributes described for services.

2-28. This depends on the airline the student selects. Example defects they might mention are flight delays or cancellations, lost luggage or luggage mishandling, discourteous employees, wrong or misleading flight information, uncomfortable seats, etc.

2-29. This will depend on which websites the student selects.

2-30. Categories of possible quality problems might be related to the ordering process, pizza construction, pizza ingredients, packaging/boxing, time to receive order, order accuracy/correctness, and pricing.

2-31. If someone purchases a residence then the dwelling is more of a product. However, renting a dormitory room or an apartment tends to fall into the service category because it is part of an ongoing process or interaction between the owner and renter. In other words, the owner retains responsibility for the product, i.e., the dwelling. As such, the quality of the living accommodation should be assessed according to the dimensions of quality for a service.

2-32. Categories of quality problems for flight delays could relate to employees (not enough to check in, problems with the check-in process, deplaning problems, insufficient maintenance personnel to accomplish plane turn-around, etc.), mechanical problems, luggage problems, maintenance problems, weather, flight controller problems, over booking, over scheduling, etc.

2-33. This depends on the business the student selects. If for example, they selected a restaurant they eat at frequently they would need to identify the categories of quality problems which might include employees, food quality, restaurant environment, waiting time, price, service, menu, etc.

2-34. It should be obvious to the student that the most important defects are the engine problems and faulty brakes. The priority of the quality problems is almost the reverse of the frequencies; faulty brakes are clearly the most significant category of defects. This points out that when applying Pareto analysis the degree of importance must be the same for all defect categories. If not then the categories should be weighted according to their importance in order to adjust the chart.

2-35. Marketing has direct contact with the customer. Marketing is typically responsible for the consumer research that determines the quality characteristics that customers want and need, and the price they are willing to pay for it. Marketing also informs the consumer about the quality characteristics of a product through advertising and promotion. Sales provides feedback information through its interaction with the customer, which is a determinant of product design.

Research and development will explore new ideas for products and be actively involved in product innovation.

Engineering translates the product quality characteristics determined by marketing and top management into a product design, including technical specifications, material and parts requirements, equipment requirements, workplace and job design, and operator training and skills.

Overdesigning the product is a drain on the company’s resources and can erode profits, whereas underdesigned products will generally not meet the customer’s quality expectations. Genichi Taguchi, the Japanese quality expert, estimates that poor product design is the cause of as much as 80 percent of all defective items. It is much cheaper and easier to make changes at the design stage than at the production stage, so companies need to focus on quality at all stages of the design process.

Purchasing must make sure that the parts and materials required by the product design are of

high quality. Quality of the final product will be only as good as the quality of the materials used to make it. Purchasing must select vendors who share the company’s commitment to quality and who maintain their own quality management program for providing high-quality service, materials, and parts.

Human resources is responsible for hiring employees that have the required abilities and skills, and training them for specific job tasks. Employees not well trained in their tasks will probably contribute to poor quality or service. Personnel also have responsibility for educating employees about quality and ways to achieve quality in their tasks. TQM requires that all employees throughout the organization be responsible for quality. Employees, collectively and individually, must not only perform their tasks according to design specifications but also be responsible for identifying poor quality or problems that may lead to poor quality and taking action to correct these problems. Performance appraisal under TQM focuses more on quality improvement and group and company achievement than on individual job performance.

Management at all levels must implement the product design according to quality specifications, controlling labor, materials, equipment, and processes. Failure to manage effectively can result in employee errors, equipment breakdown, bottlenecks, interrupted service, and the like, which contribute to poor quality.

Distribution makes sure that high-quality products are delivered on-time and undamaged to the customer. Packaging methods and materials, storage facilities and procedures, and shipping modes must ensure that final products are protected and that customers receive them on time.

2-36. This answer depends on the award the student selects.

2-37. This answer depends on the company the student selects.

2-38. This answer depends on the article and company the student selects.

2-39. This question was adapted from: L. Fredendall, J. Patterson, C. Lenhartz and B. Mitchell,” What Should Be Changed?”, Quality Progress 35 (1; January 2002): p. 50–59. This is an excellent article about the use of cause and effect diagrams students can be referred to.

2-40. Black Belt—the leader of a quality improvement project, which is a full-time position Green Belt—a project team member, which is a part-time position Master Black Belt—a teacher and mentor for Black Belts which, is also a full-time position. A Black Belt would have to have led several successful projects before being certified as a Master Black belt.

2-41. Breakthrough Strategy: 1. Define the process including who the customers are and what their problems are.

2. Measure the process and collect data.

3. Analyze the data in order to develop information that provides insight into the process, including causes of defects.

4. Improve the process by making changes and measuring the results.

5. Control the improved process by monitoring it and making sure the desired performance level is sustained.

2-42. This answer depends on the project the student selects.

2-43. In general, the ACSI model is a set of causal equations that link customer expectations, perceived quality, and perceived value to customer satisfaction (ACSI). In turn, satisfaction is linked to consequences as defined by customer complaints and customer loyalty—measured by price tolerance and customer retention. There are two menu items on the ACSI website that describe, in general terms, how the ACSI is determined—“What it measures,” and “Methodology.” The student should refer to these descriptions. As an example, the student could select two fast food restaurant

chains in the fast food industry and compare the company with the highest score with the lowest scoring company and explain the reasons for the difference in scores.

2-44.

Answer depends on the personal health improvement project the student selects.

2-45.

Answer depends on the personal improvement project the student selects.

2-46.

Answer depends on the infirmary process the student selects.

2-47.

Answer depends on the registration process at the student’s university.

2-48.

In general, the Japanese recognized that even though high quality might cost more in the “short run,” in the long run it would help them gain market share, which would increase long term profits. This is something American companies did not recognize. The Japanese economic climate and business and management culture was also more conducive to quality management programs than American companies.

2-49.

The student should go to the ISO website at www.iso.org to determine these steps.

2-50.

Answer depends on the store the student selects.

2-51.

The student should refer to the ISO website at www.iso.org and the Baldrige Award website at www.nist.gov to answer this question.

2-52.

The answer should include references to some form of customer feedback, such as a “voice of the customer (VOC)” process and surveys. Since assessing customer satisfaction is a critical part of the Baldrige Award criteria, the summaries of Baldrige Award winning companies at www.nist.gov are a good source for students to learn how companies and organizations assess customer satisfaction.

2-53.

In general, this will be a subjective answer on the part of the student. A good source for this question is an article by David Goldhill titled “How American Health Care Killed My Father,” that appeared in the September 2009 online edition of The Atlantic magazine (www.theatlantic.com).

Solutions to Problems 2-1.

a. Failure costs as percentage of quality costs: 157.7 = 0.8424, or 84.24% 187.2 161.8 2007 2004 : = 0.8022, or 80.22% 201.7 153.6 2008 2005 : = 0.7288, or 72.88% 212.5 127.2 2009 2006 : = 0.6560, or 65.6% 193.9 97.3 2010 2007 : = 0.5830, or 58.3% 166.9

2006 2003 :

The failure costs decrease as a percentage of total quality costs. This may be attributed to an increase in product monitoring and inspection. Fewer defective products are reaching the consumer, as evidenced by the sharp decline in external failure costs. b.

Prevention costs as % of quality Appraisal costs as % of quality costs costs: 3.2 = 0.0171, or 1.71%; 187.2 10.7 2007 2004 : = 0.0530, or 5.3%; 201.7 28.3 2008 2005 : = 0.1332, or 13.32%; 212.5 42.6 2009 2006 : = 0.2197, or 21.97%; 193.9 50 2010 2007 : = 0.2996, or 29.96%; 166.9

2006 2003 :

26.3 = 0.1404, or 14.04% 187.2 29.2 = 0.1448, or 14.48% 201.7 30.6 = 0.144, or 14.4% 212.5 24.1 = 0.1243, or 12.43% 193.9 19.6 = 0.1174, or 11.74% 166.9

The increase in prevention costs as a percentage of total quality costs indicates that Backwoods American is placing more emphasis on prevention of defects rather than correction of them.

Perhaps they are spending more in the areas of quality planning, product design, process, training, and information. This is contributing to a decline in the need for inspection and testing, equipment testing, and operators to test quality; thus appraisal costs decline, both absolutely and as a percentage of total costs. Prevention also contributes to the decline in external and internal failures, because fewer defective products are produced to begin with. Increases in prevention expenditures will result in a decrease in all other quality costs. c.

2006 2007 2008 2009 2010

Quality Sales Index 6.93 7.50 7.85 6.90 5.79

Quality-Cost Index 44.48 47.64 50.04 44.46 38.32

These index values do not provide much information regarding the effectiveness of the quality assurance program. They are, however, useful in making comparisons from one period to the next and in showing trends in product quality over time. d. Examples of quality-related costs: • Prevention: Market research, that is, producing what consumers want; purchasing only highquality down and other materials, designing an efficient and effective manufacturing process; training employees in making quality products.

• Appraisal: Inspection of raw materials, work-in-process, and finished product; equipment testing (pattern cutter, sewing machines, etc.), inspection.

• Internal failure: Wasted materials and labor, defective products discovered during inspection, use of inefficient processes, equipment downtime, poorly trained employees.

• External failure: Defective products, customer complaints, warranty costs, lost sales, loss of good will.

2-2.

a. Product yield 2003 : 20, 000 ( 0.83) + 20, 000 (1 − 0.83)( 0.20 ) = 16, 600 + 680

= 17, 280 parkas 2004 : 20, 000 ( 0.85 ) + 20, 000 ( 0.15 )( 0.20 ) = 17, 000 + 600

= 17,600 parkas 2005 : 20, 000 ( 0.87 ) + 20, 000 ( 0.13)( 0.20 ) = 17, 400 + 520

= 17,920 parkas 2006 : 20, 000 ( 0.89 ) + 20, 000 ( 0.11)( 0.20 ) = 17, 800 + 440

= 18, 240 parkas 2007 : 20, 000 ( 0.91) + 20, 000 ( 0.09 )( 0.20 ) = 18, 200 + 360

= 18,560 parkas

b.

Manufacturing cost per good parka: 2003 : 2004 :

2005 : 2006 : 2007 :

420, 900 + 12 ( 680 ) 17, 280 423, 400 + 12 ( 600 ) 17, 600

424, 700 + 12 ( 520 ) 17, 920 436,100 + 12 ( 440 ) 18, 240 435, 500 + 12 ( 360 ) 18, 560

=

429, 060 = $24.83 17, 280

=

430, 600 = $24.47 17, 600

=

430, 940 = $24.05 17, 920

=

441, 380 = $24.20 18, 240

=

439, 820 = $23.70 18, 560

Improving the quality assurance program has resulted in fewer defective parkas, lower rework costs, and greater productivity. This has lowered the per-unit manufacturing costs without additional capital investment.

2-3.

a. y = ( I )( %G ) + ( I )(1 − %G )( % R ) = (150 )( 0.83) + (150 )(1 − 0.83)( 0.60 ) = 139.8 file cabinets

b. 145 = (150 )( %G ) + (150 )(1 − %G )( 0.60 ) 145 = 150G + (150 − 150G )( 0.60 )

145 = 150G + 90 − 90G 55 = 60G 55 G= = 0.916 = 91.6% 60

2-4.

27 (150 ) + 8 (15 )

4, 050 + 120 = 139 139 4,170 = = $30 per cabinet if quality is 83%, 139 Note: if don’t round until the end, $29.85 instead of $30 Cost =

27 (150 ) + 8 ( 9 )

4,122 144 144 = $28.63 per cabinet if quality is 90%

Cost =

2-5.

=

Manufacturing cost per good product: 2008 2005 : Yield = 32, 000 ( 0.78) + 32, 000 ( 0.22 )( 0.25 ) = 26,720 Product Cost = ( 278, 000 + 3520 )  26, 720 = $10.54

In this case, total direct manufacturing cost = $278, 000, and total direct rework cost = 32, 000 ( 0.22 )( 0.25 )( 2 ) = $3,520 2009 2006 : Yield = 34, 600 ( 0.83) + 34, 600 ( 0.17 )( 0.25 ) = 30,188.50 Product Cost = ( 291, 000 + 2,941)  30,188.50 = $9.74

2010 2007 : Yield = 35, 500 ( 0.9 ) + 35, 500 ( 0.1)( 0.25 ) =32,837.50 Product Cost = ( 305, 000 + 1, 775 )  32,837.50 = $9.34

Percentage change: 2008 - 2009: -7.60% 2009 - 2010: -4.10%

2-6.

a. Product yield = 300 ( 0.87 )( 0.91)( 0.94 )( 0.93)( 0.93)( 0.96 ) = 185 cabinets

b. For a yield of 300, input would have to be

I ( 0.87 )( 0.91)( 0.94 )( 0.93)( 0.93)( 0.96 ) = 300

I ( 0.6179 ) = 300 I = 486 cabinets

2-7.

a. Alt. 1: 300 ( 0.93)( 0.91)( 0.94 )( 0.93)( 0.93)( 0.96 ) = 198 Alt. 2 : 300 ( 0.87 )( 0.96 )( 0.94 )( 0.97 )( 0.93)( 0.96 ) = 204 Greatest yield Alt. 3 : 300 ( 0.87 )( 0.91)( 0.94 )( 0.93)( 0.97 )( 0.98 ) = 197 Alt. 4 : 300 ( 0.87 )( 0.97 )( 0.94 )( 0.93)( 0.93)( 0.96 ) = 198

b. Alternative 2 will result in the highest yield and will be the most effective.

2-8.

320 (1 − 0.12 )(1 − 0.08)(1 − 0.04 ) = 320 ( 0.88)( 0.92 )( 0.96 ) = 248 errorless orders

2-9.

( 585 + 16 )(100 ) = 5.11 650 (18 ) + 16 ( 3.75) ( 720 + 20 )(100 ) b. QPR = = 5.11 800 (18) + 20 ( 3.75 ) ( 585 + 16 )(100 ) c. QPR = = 5.58 650 (16.50 ) + 16 ( 3.20 ) ( 604 + 11)(100 ) d. QPR = = 5.24 650 (18 ) + 11( 3.75 ) a. QPR =

2-10. a. QPR = b. QPR =

250 (100 )

250 ( 47 ) + 33 (16 )

= 2.04

320 (100 )

320 ( 42 ) + 19.2 (12 )

2-11. a. Product cost = =

= 2.34

( K d ) ( I ) + ( K r )( R ) Y

( $6.15)( 680 ) + (1.75)( 2.72 ) 655.52

4186.76 = 655.52 = $6.39

b. Product cost =

( $6.20 )( 680 ) + (1.75)( 0.68 ) 673.88

4217.19 = 673.88 = $6.26

Cost savings = $0.13/ order

Annual savings = $0.13/ order  680 orders / day  365 days / year = $32, 266

c. It is likely that some customers who receive defective orders will not return, thus, fewer defective orders will retain more customers and also increase the number of orders.

2-12. a. QPR = b. QPR =

655.52

( 680 )( 6.15) + ( 2.72 )(1.75)

(100 ) = 15.66

673.88 (100 ) = 15.98 ( 680 )( 6.20 ) + ( 0.68)(1.75)

2-13.

With defects: v=

cf p − cv

$350, 000 1, 000 − 600 = 875 units or $875,000 in sales

=

Without defects (six sigma): $350, 000 1, 000 − 540.05 = 760.952 or $760,952 in sales

v=

The slope of the line is steeper with a reduced break-even point; the company can make more money without selling additional units.

2-14.

With a 0% defect rate: Sales Variable costs Fixed costs Profit

$151,200 61,200 31,000 $59,000

With zero defects: Sales $151,200 Variable costs 56,305 Fixed costs 31,000 Profit 63,895 Six sigma results in an 8% reduction in variable costs and a corresponding 8% increase in profit.

The return on the six sigma investment would be: 100(4, 200 − 695) 25, 000 = 14%

Return =

2-15.

A possible version of the cause and effect matrix:

2-16: missing?

CASE 2.1: Designing a Quality Management Program for the Internet at D4Q This can be an instructive, hands-on case project. The students should first search the Internet for different web sites at which retail items can be ordered. They should next develop a list of quality characteristics or dimensions to focus on. These might include the visual appearance of the web site, the friendliness of the language used, the accuracy of instructions, the availability of e-mail or telephone support, etc. An attractive web site should probably include photos of the catalog items, for example book jackets, CD covers and video jackets, instead of just item titles. Instructions for ordering should be detailed and accurate with help icons located at every step. Customer support should be easy to access with e-mail or by telephone. Responses to requests for support should be quick. Service measurement is difficult in this type of operation. If an order center is used then the company can count the number of customers who enter the center then abort as the result of poor instructions. Follow up surveys of customers who place orders or request hard copy catalogs is a good way to evaluate service. From the server end, the server responses to customer inquiries can be monitored for accuracy, completeness and timeliness. These are just a few of the possible quality initiatives you might suggest to develop a high quality web site ordering system.

CASE 2.2: Quality Management at State University In general the student should respond to this case by attempting to go through the chapter and to discuss each major topic in terms of a university environment. This will require that they first identify the product and the process, obviously the student and the educational process. However, in a university environment, is the product also the customer? This is an interesting question to begin with.

An initial step should be to develop a customer definition of quality—that is, what are the dimensions of quality in an education that parents, students, and legislators expect? This step can be accomplished by surveying alumni, potential employers, parents, students, and legislators. It would also be beneficial to see what the competition (other colleges and universities) does.

The various support functions in the university should be identified in terms of a production process. For example, the admissions office is analogous to the purchasing department. A key problem here in a QM approach is that the university has little, if any, control over suppliers (high schools). As a result, admissions must institute inspection and process control procedures to ensure high-quality raw material (i.e., students), is admitted.

The product-design function, or curriculum design, is typically decentralized in a university among various colleges and departments. In some cases, the university administration will design a core curriculum for the first two years and college and departments will design the curriculum for the last two.

The production process is the movement of students through the curriculum to graduation. Discussion should focus on how to institute process control in order to avoid final product “defects.” This obviously requires a definition as to what a defective item is—a student who enters but fails to graduate, a student who graduates but does not gain employment, or a graduate who indicates disappointment with his or her education five years from now. An area on which to focus is the degree to which quality control tools such as brainstorming, quality circles, histograms, check sheets, and fishbone diagrams can be used to evaluate the process.

Customer service would seem to be an integral part of a QM approach in the university. This service would focus on support services such as dining, recreation, housing, advising, counseling, extracurricular

activities, entertainment, placement, alumni services, etc.

Depending on the time designated to spend on this case, students might interview various administrators at their own university to determine where QM can be applied in the university and obstacles to a TQM approach.

CASE 2.3: Quality Problems at the Tech Bookstore a. Mr. Watson’s organization of the customer survey categorized the two bookstores and types of customer, i.e., students and non-students. He differentiated between the two stores because they carried different products, and it also was likely that a different population of customers visited the two stores since they were in different locations. Also, the two stores had different managers, staff and employees; in effect they were separate entities. He differentiated between the two customer groups because he knew they likely had different characteristics and different service expectations. Also, it was probable that they shopped for different items. Student would be primarily interested in textbooks, school supplies, computer items and apparel, whereas non-students would have not been as interested in textbooks and school supplies, and they would have had a higher interest in trade books.

A customer survey was probably the best way to start in order to see if there was a quality problem and its extent. He might also analyze different processes in the store, such as employee floor service, checkout, etc. He could have probably gotten the help of an OM class on campus to help him analyze various service processes in the store. He could also benchmark other “successful” college bookstores. b. It is possible to develop 9 different Pareto charts—a chart each for students and non-students at the campus store, and a chart each for students and non-students at the off-campus store; a combined chart for students at both stores and a combined chart for non-students at both stores; a chart for each store combining the two customer groups at each one; and a summary chart combining all the data for both stores and both customer categories. Following is a summary of the survey data. c. Using a fitness for use definition, quality should be prompt, knowledgeable and courteous customer service; a pleasant shopping environment with prompt and courteous customer checkout; and quality products at a competitive (or lower) price. d. The most pronounced problem is the discrepancy between the student and non-student groups in their perception of service. This is likely due to the different expectations of the two groups. Non-students are likely to be older and less patient. The student employees’ attitude and demeanor is probably more familiar to other students, i.e., what they are used to, and they are probably more patient with the student employees than the non-student group. Also it is less likely that students will ask questions than the non-student group so the problem of not being helpful or knowledgeable does not come up as much with students as with non-students. It also may be that students visit the bookstores more often and know where to find items, and they may also be more familiar with bookstore policies. Many of the non-student customers could be visitors.

However, this discrepancy does not hide two other potential quality problems—that the off campus store has poorer quality service across all categories than the on-campus store, and there seems to be a significant problem with employee training at both stores. The relatively low percentage of customers who think employees are knowledgeable and helpful, reinforced by the graduate student evaluators and the complaint incidents to the Board, clearly indicates that the student employees are not adequately prepared to do their job. They are probably in need of more training, however it is likely the bookstore has been hesitant to provide additional training because of the high turnover rate, i.e., the return on training investment might be perceived by management to be low when student employees leave after a semester or two. The off campus store may have poorer quality service, in general, because of the management staff, and the fact that the Executive Director resides at the on campus store.

The much lower percentage of students who think that the cost of purchases at the bookstores is reasonable is probably due to the fact that students purchase textbooks, which are expensive. The bookstore probably does not do a very good job of publicizing the fact that it has a very low mark up on textbooks. This is something that could be highlighted on the store web site, which the students access more than non-students.

e. Since this is a service several of the costs of poor quality that relate to manufactured products such as scrap, product rework, returned products, etc., do not apply. Thus, the two primary costs of poor quality are lost sales and customer complaint costs. The costs the bookstore is incurring to conduct the customer survey, hire the graduate student evaluators, and analyze the results are all costs of poor quality. From the limited information provided it is difficult to address the question of lost sales.

However, it is noted that that the town and university have been growing while sales have remained steady. Given the ideal location of the on campus store in particular, and the fact that the football team is very successful should mean much higher sales of licensed apparel. It could be that the quality problems are having a very negative impact on sales. f. While the bookstores would benefit from a complete QM program, the most immediate need is for a more extensive employee training program. The bookstore needs to establish a plan for improving its quality that includes employee training as a top priority. The plan needs tangible objectives. For example, every customer question should be answered as promptly as possible. Some form of reward system for employees might be beneficial. They likely would benefit from a set of guidelines for employees for addressing customer questions. Having a resource person available that has intimate knowledge of all aspects of store policies, and knows where all items are in the store, that the student employees could contact at any time would be a good personnel investment. In addition, management needs to establish a process for monitoring employee performance on a routine, daily basis.

Benchmarking other college bookstores could provide insight into ways to solve quality problems. A process for measuring customer and employee satisfaction on a regular basis needs to be put into effect, and performance measures could be tied to these surveys. g. Because the bookstores are quasi non-profit, government-type entities, the revenue pressures that a business might feel are not present. Thus, the motivation for improved quality is primarily the store’s reputation. Another factor is the university-invoked policy of hiring students on a part-time basis. This means that to solve the quality problems associated with part time students, bookstore management must look to other solutions besides hiring more-experienced, full-time employees. The fact that students will be serving adults is a situation that cannot be changed. h. The most probable benefit would be an increase in sales and revenue. However, employee satisfaction would also likely increase as a result of a quality management program. Also, the Board of Directors would likely receive fewer complaints.

Campus Store

Off Campus Store

Student

Non-student

Student

Non-student

Total

Yes

No

Yes

No

Yes

No

Yes

No

Yes

No

%

%

%

%

%

%

%

%

%

Were employees courteous and friendly?

86

14

67

33

75

25

% 52

48

73

27

Were employees knowledgeable and helpful?

79

21

52

48

77

23

41

59

65

35

Was the overall service good?

86

14

63

37

71

29

55

45

71

29

Did you have to wait long for service?

11

89

45

55

11

89

46

54

25

75

Did you have to wait long to checkout?

12

88

46

54

16

84

45

55

27

73

Was the item you wanted available?

92

8

84

16

90

10

90

10

89

11

Was the cost of your purchase(s) reasonable?

58

42

90

10

60

40

92

8

72

28

Have you visited the store web site?

50

50

12

88

58

42

5

95

41

59

42

5

95

41

59

CASE SOLUTION 2.4 - Product Yield at Continental Luggage Company Average Weekly Yield

Stage 1 yield : Y1 = ( I )( %G ) + I (1 − %G )( % R )

= 500 ( 0.94 ) + 500 ( 0.06 )( 0.23)

= 470 + 6.9

Y1 = 476.9 Stage 2 yield : Y2 = ( 476.9 )( 0.96 ) + 476.9 ( 0.04 )( 0.91)

= 457.82 + 17.36

Y2 = 475.2 Stage 3 yield : Y3 = 475.2 (.95 ) + ( 475.2 )(.05 )(.67 )

= 451.44 + 15.92 = 467.6 Stage 4 yield : Y4 = 467.6 (.97 ) + ( 467.6 )(.03)(.89 )

= 453.57 + 12.48 = 466.1 Stage 5 yield : Y5 = 466.1(.98 ) + 466.1(.02 )(.72 )

= 456.78 + 6.71

= 463.5 Increasing Good Quality Yield by 1% at Each Stage

Stage 1 yield : Y1 = 500 ( 0.95) + 500 ( 0.05)( 0.23)

Y1 = 480.75 Stage 2 yield : Y2 = 480.75 ( 0.97 ) + 480.75 ( 0.03)( 0.91)

Y2 = 479.45 Stage 3 yield : Y3 = 479.45 ( 0.96 ) + 479.45 ( 0.04 )( 0.67 )

Y3 = 473.12

Stage 4 yield : Y4 = 473.11( 0.98 ) + 473.11( 0.02 )( 0.89 )

Y4 = 472.07 Stage 5 yield : Y5 = 472.07 ( 0.99 ) + 472.07 ( 0.01)( 0.72 )

Y5 = 470.75 Difference in yields = 470.75 − 463.5 = 7.25 units Percentage increase =

7.25 = 1.56% 436.2

3 Statistical Process Control Answers to Questions 3-1.

In attribute control charts (such as p-charts and c-charts), the measures of quality are discrete values reflecting a simple decision criterion such as good or bad. The quality measures used in variable-control charts (such as x -charts and R-charts) are continuous variables reflecting measurements such as weight, time, or volume.

3-2.

An R-chart reflects the process variability, whereas an x -chart indicates the tendency toward a mean value; thus, the two complement each other. That is, it is assumed that process average and variability must be in control for the process to be in control. When they are used together, the control limits are computed as

x + A2 R. 3-3.

A pattern test is used to determine if sample values from a process display a consistent pattern that is the result of a nonrandom cause, even though control charts may show the process to be in control.

3-4.

Width is determined by the size of the z value used; the smaller the value of z, the narrower the control limits.

3-5.

A c-chart is used when it is not possible to determine a proportion defective (for a p-chart), for example, when counting the number of blemishes on a sheet of material. In a p-chart it must be possible to distinguish between individual defective and non-defective items.

3-6.

Tolerances are product-design specifications required by the customer, whereas control limits are the upper and lower bands of a control chart indicating when a process is out of control.

3-7.

Management usually selects 3-sigma limits because if the process is in control they want a high probability that the sample will fall within the control limits. With wider limits management is less likely to erroneously conclude that the process is out of control when points outside the control limits are due to normal, random variations.

3-8.

Process control charts could be used to monitor service time in a restaurant, bank, hospital, store, etc.

3-9.

For example, in a fast food restaurant a control chart could be used to measure service times, defective menu items, out of stock menu items, customer complaints, cleanliness, and order errors, among other things.

3-10. The process capability ratio

( C p ) indicates if the process is capable of meeting design specifications. The

process capability index indicates if the process mean is off-center and has shifted toward the upper or lower design specifications.

3-11.

Cp =

tolerance range .14 = = 1.00 process range .14

The tolerance range and process range are equal so the process is capable but some defects will result.

Solutions to Problems 3-1.

p=

=

453 = 0.151; ( 30 )(100 ) p (1 − p ) = n

( 0.151)( 0.849 ) = 0.0358

UCL = p + z

100

p (1 − p ) n

= 0.151 + 3 LCL = p − z

( 0.151)( 0.849 ) = 0.258 100

p (1 − p ) n

= 0.151 − 3

( 0.151)( 0.849 ) = 0.044 100

The process does not seem to be out of control, although the decreasing number of defects from sample 8 to sample 17 should probably be investigated to see why the steady improvement occurred; likewise, the steadily increasing number of defects from sample 17 to sample 25 should probably be investigated to see why the quality deteriorated.

3-2.

p = 0.153;  =

p (1 − p ) 0.153 ( 0.847 ) = = 0.036 100 100

UCL = p + z = 0.153 + 2 ( 0.036 ) = 0.225 LCL = p − z = 0.153 − 2 ( 0.036 ) = 0.081

In general, the proportion of defectives increases from sample 6 to sample 20, where it is eventually above the upper control limit. This indicates the process is moving out of control.

3-3.

p = 0.053;  =

p (1 − p ) 0.053 ( 0.947 ) = = 0.016 n 200

UCL = p + 3 = 0.053 + 3 ( 0.016 ) = 0.101 LCL = p − 3 = 0.053 − 3 ( 0.016 ) = 0.005

The process does not seem to be out of control.

3.4

UCL = p + z

p (1 − p ) n

0.03 = 0.02 + z

0.02 ( 0.98 ) n

n=

2 0.02 ( 0.98 )

n = ( 28 ) n = 784

0.01 2

3-5.

a.

742 = 24.73 30 UCL = c + z c c=

= 24.73 + 3 24.73 = 39.65 LCL = c − z c = 24.73 − 3 24.73 = 9.81

With three points outside the control limits, the process appears to be out of control.

3-6.

b.

Nonrandom factors that might cause the process to move out of control could include (among other things) problems with the telephone order system, inexperienced operators taking orders, computer system problems, or shipping problems and delays.

a.

86 = 4.3 20 UCL = c + z c c=

= 4.3 + 3 4.3 = 10.52 LCL = c − z c = 4.3 − 3 4.3 = −1.92 or 0.0 ( since the control chart cannot go below zero )

The process was not strictly out of control; however, from sample 10 to sample 20, the sample values were above the average and exhibited increasingly nonrandom behavior.

3-7.

256 = 10.67 24 UCL = c + z c c=

= 10.67 + 2 10.67 = 17.20 LCL = c − z c = 10.67 − 2 10.67 = 4.14

The process does not appear to be out of control, but sample 21 is close to the UCL and the process should be investigated.

3-8.

c=

219 = 7.3 30

Control limits using z = 3.00:

UCL = c + z c = 7.3 + 3 7.3 = 7.3 + 8.11 = 15.41 LCL = c − z c = 7.3 − 3 7.3 = 7.3 − 8.11  0 All the sample observations are within the control limits suggesting that the invoice errors are in control.

3-9.

255 = 12.75 20 UCL = c + z c = 12.75 + 3 12.75 = 23.46

c=

LCL = c − z c = 12.75 − 3 12.75 = 2.04 All the sample observations are within the control limits suggesting that the delivery process is in control.

3-10.

Sample 1 2 3 4 5 6 7 8 9 10

p=

Proportion Defective .028 .044 .072 .034 .050 .082 .036 .038 .052 .056

Sample 11 12 13 14 15 16 17 18 19 20

Proportion Defective .076 .048 .030 .024 .020 .032 .018 .042 .036 .024

421 421 = = .0421 ( 20 )( 500 ) 10, 000

UCL = p + z

p (1 − p ) n

= .0421 + 3.00

.0421(1 − .0421) 500

= .0421 + .027 = .069 p (1 − p ) LCL = p − z n .0421(1 − .0421) 500 = .0421 − .027 = .015

= .0421 − 3

Samples 3 and 11 are above the upper control limit indicating the process may be out of control.

3-11.

Sample 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

x 2.00 2.08 2.92 1.78 2.70 3.50 2.84 3.26 2.50 4.14 2.12 4.38 2.84 2.70 3.56 2.96 3.34 4.16 3.70 2.72 60.20

R 2.3 2.6 2.7 1.9 3.2 5.0 2.2 4.6 1.3 3.5 3.0 4.0 3.3 1.1 5.6 3.1 6.1 2.4 2.5 2.9 63.3

 R 63.3 = = 3.17 k 20  x 60.20 x= = = 3.01 20 20

R=

R-chart

D3 = 0, D4 = 2.11, for n = 5

UCL = D4 R = 2.11( 3.17 ) = 6.69 LCL = D3 R = 0 ( 3.17 ) = 0 There are no R values outside the control limits, which would suggest the process is in control.

x -chart A2 = 0.58

UCL = x + A2 R = 3.01 + 0.58 ( 3.17 ) = 4.85 LCL = x − A2 R = 3.01 − 0.58 ( 3.17 ) = 1.17 There are no x values outside the control limits, which suggests the process is in control.

3-12.

 = 9 in; σ = 0.06 in; n = 10 a.

 σ  UCL =  + z    n  0.06  = 9 + 3   10  = 9.057    LCL =  − z    n  0.06  = 9 − 3   10  = 8.943

b. c.

Yes, it appears to be.

The control limits become narrower, but increasing the sample size will not affect the results in part b.

3-13.

Sample 1 2 3 4 5 6 a.

R 0.67 0.69 0.93 0.52 0.64 0.71

Sample 7 8 9 10 11 12

 R 6.87 = = 0.57 k 12 From Table 3.1 in the text, D3 = 0 and D4 = 2.11 R=

UCL = D4 R = 2.1( 0.57 ) = 1.21

LCL = D3 R = 0 ( 0.57 ) = 0

R 0.45 0.17 0.32 0.99 0.65 0.15

b.

The process variability is within the control limits.

3-14.

Sample 1 2 3 4 5 6 7 8 9 10

a.

R 8.5 5.8 8.1 6.4 7.1 6.0 9.9 2.5 1.6 9.4

Sample 11 12 13 14 15 16 17 18 19 20

 R 12.7 = = 6.24 k 20 From Table 3.1, D3 = 0.0 and D4 = 2.11 R=

UCL = D4 R = 2.11( 6.24 ) = 13.17

LCL = D3 R = 0.0 ( 6.24 ) = 0

b.

The temperature is within the control limits.

R 8.1 4.4 5.8 3.9 6.2 5.4 3.6 10.9 7.5 3.6

3-15.

R=2 a.

From Table 3.1, D3 = 0 and D4 = 2.28

UCL = D4 R = 2.28 ( 2 ) = 4.56

LCL = D3 R = 0 ( 2 ) = 0 Sample 1 2 3 4 5 6 7 8 9 10

R 2.1 4.5 7.0 3.0 5.0 4.3 6.5 2.0 3.2 4.0

The process clearly seems to be out of control. There are three of the sample points above the UCL, and all other sample values are above the center line for R , indicating nonrandom variations.

3-16.

Sample 1 2 3 4 5 6

x 8.89 8.88 8.99 9.19 9.04 8.71

Sample 7 8 9 10 11 12

x 9.05 9.16 8.97 9.06 9.09 9.01

 x 108.04 = = 9.00 12 12 From Table 3.1, A2 = 0.58. x=

UCL = x + A2 R

= 9.00 + 0.58 ( 0.57 ) = 9.33

LCL = x − A2 R

= 9 − 0.58 ( 0.57 ) = 8.67

The process appears to be in control from both the x and R charts, although sample 6 is close to the LCL and perhaps should be investigated.

3-17.

Sample 1 2 3 4 5 6 7 8 9 10

x 43.9 39.7 37.2 40.4 39.0 41.8 39.4 40.7 41.6 39.0

Sample 11 12 13 14 15 16 17 18 19 20

x 39.2 39.7 42.9 37.8 36.6 37.6 39.9 40.7 38.2 39.0

x = 39.7 From Table 3.1, A2 = 0.58.

UCL = x + A2 R

= 39.7 + 0.58 ( 6.24 ) = 43.32

LCL = x − A2 R

= 39.7 − 0.58 ( 6.24 ) = 36.08

The process appears to be in control, with sample 1 seeming to be an aberration, however, the process should still be checked.

3-18.

Sample 1 2 3 4 5 6 7 8 9 10

x 37.0 35.8 35.3 36.1 36.1 34.3 36.3 35.6 37.7 34.5

x = 35.9 From Table 3.1,

A2 = 0.73.

UCL = x + A2 R

= 35.9 + 0.73 ( 2 ) = 37.36

LCL = x − A2 R

= 35.9 − 0.73 ( 2 ) = 34.44

The process may be out of control (sample 6 and 9), although overall the x -chart does not reflect the magnitude of the out of control situation, as does the R-chart in Problem 3-11. The process should be investigated to determine a cause for the out of control samples.

3-19.

Sample 1 2 3 4 5 6 7 8 9 10 11 12

Above/Below B B B A A B A A B A A A

Up/Down — D U U D D U U D U U D

Zone B B C B C A C B C C C C

There are no discernible nonrandom patterns.

3-20.

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

Above/ Below A B A A B B B B A A A A A A A

Up/ Down — D U U D U D D U U D U D D D

Zone C B A A B C C B B A C A A A C

Sample 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Above/ Below B B B B A A B B A A A B B B B

Up/ Down D D U U U U D U U U U D D U D

Zone B A B B C B A B C C A C A B B

The zone pattern test is violated for samples 10 through 14 (2 out of 3 consecutive points in zone A but within limits).

3-21.

Sample 1 2 3 4 5 6 7 8 9 10

Above/ Below A B A A A B A B B A

Up/ Down — D U D U D U D D U

Zone C C C C C C B B A B

Sample 11 12 13 14 15 16 17 18 19 20

Above/ Below A B B B A B B A A B

Up/ Down D D U D U D D U D D

Zone C C C B C C B A C B

Samples 7 through 10 appear to violate a zone pattern test (four out of five points in zone B or beyond) so there may be a nonrandom pattern.

3-22.

Sample 1 2 3 4 5 6 7 8 9 10

Above/ Below A — B A B A B A A B

Up/ Down — D D U D U D U U D

Zone A C A C C B C C B C

Sample 11 12 13 14 15 16 17 18 19 20

Above/ Below B B B B B B B A B B

Up/ Down U U U D D U U U D U

Zone C C A B A B C C B C

There are several instances where zone pattern test rules are violated; samples 1 to 3, and sample 13 through 16. Thus, nonrandom patterns may exist.

3-23.

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

Above/ Below B B B A A A B A A A B B B B B

Up/ Down — U D U U U D U D D D U U D D

Sample 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Above/ Below B B B B B A A A A A A A A A A

Up/ Down D D U U U U U U D U D U U D U

There are eight consecutive points on one side of the center line on two occasions—samples 11-20 and samples 21-30—indicating nonrandom patterns may exist.

3-24.

146 = 7.3 20 UCL = c + z c

c=

= 7.3 + ( 3) 7.3 = 7.3 + ( 3)( 2.70 ) = 15.41 LCL = c − z c = 7.3 − ( 3)( 2.70 ) 0

The process appears to be in control however the process may be moving toward an out-of-control situation.

3-25.

x = 3.25, R = 3.86, A2 ( n = 5) = 0.58 UCL = x + A2 R

= 3.25 + ( 0.58 )( 3.86 ) = 5.48

LCL = x − A2 R

= 3.25 − ( 0.58 )( 3.86 ) = 1.01

UCL = D4 R = 2.11( 3.86 ) = 8.14 LCL = D3 R = 0 ( 3.86 ) = 0

The process appears to be in control.

3-26.

718 = 0.36 2000 p (1 − p ) UCL = p + 3 100 = 0.36 + ( 3)( 0.048 ) p=

= 0.504 LCL = p − ( 3)( 0.048 ) = 0.36 − 0.144 = 0.216

The process appears to be out of control.

3-27.

326 = 16.3 20 UCL = c + z c c=

= 16.3 + ( 3)( 4.03) = 28.412 LCL = c − z c = 16.3 + ( 3)( 4.03) = 4.188 The process is in-control.

3-28.

x = 7.28, R = 4.25, A2 ( n = 5) = 0.58 UCL = x + A2 R

= 7.28 + ( 0.58 )( 4.25 ) = 9.75

LCL = x − A2 R = 7.28 − 2.47 = 4.81 UCL = D4 R = 2.11( 4.25 ) = 8.97 LCL = D3 R = 0 ( 4.25 ) = 0

While the process appears to be in control the mean of 7.28 appears to be significantly lower than the objective of 8 chips per cookie that management has established. Thus, the company should adjust their process to increase the number of chips and construct a new control chart.

3-29.

p=

105 = 0.22 ( 30 )(16 )

p (1 − p ) n = 0.22 + ( 3)( 0.0756 )

UCL = p + 3

= 0.447 LCL = 0.22 − ( 3)( 0.0756 )  0.216 Although the process appears to be in control, the average proportion of “defects’’ among leaving patients, p = 0.22, seems high and the hospital should probably adopt some quality improvement measures.

3-30.

202 = 10.1 20 UCL = c + z c

c=

= 10.1 + ( 3)( 3.18 ) = 19.6 LCL = c − z c = 10.1 − ( 3)( 3.18 ) = 0.57

The process appears to be in control.

3-31.

x = 3.17, R = 3.25, A2 ( n = 5) = 0.58

UCL = x + A2 R

= 3.17 + ( 0.58 )( 3.25 ) = 5.04

LCL = x − A2 R = 3.17 − 1.89 = 1.29 UCL = D4 R = 2.11( 3.25 ) = 6.86 LCL = D3 R = 0 ( 3.25 ) = 0

The process appears out of control for sample 10, however, this could be an aberration since there are no other apparent nonrandom patterns or out-of-control points. Thus, this point should probably be “thrown out’’ and a new control chart developed with the remaining eleven samples.

3-32.

x = 3.17, days R = 3.25 3 = A2 R = ( 0.58 )( 3.25 ) = 1.89 Cp =

2.00 2.00 = = 0.53 2 (1.89 ) 3.78

 3.17 − 2.00 4.00 − 3.17  C pk = minimum  ,  1.89  1.89 = minimum (.62, .44 ) = .44 The process is not capable of meeting the company’s design specifications and defects (i.e., late deliveries) will occur.

3-33.

x = 7.28, R = 4.25 3 = ( 0.58 )( 4.25 ) = 2.47 Cp =

4.00 4.00 = = .81 2 ( 2.47 ) 4.94

 7.28 − 6.00 10 − 7.28  C pk = minimum  , 2.47   2.47 = minimum (.52, 1.10 ) = .52 The process is not capable of meeting the design specifications and it appears that cookies will be produced with too few chips.

3-34.

x = 9.00 R = 0.57 3 = A2 R = ( 0.58 )( 0.575 ) = .33 1.00 = 1.52 .66  0.5 0.5  C pk = minimum  ,  .33 .33  = minimum (1.52, 1.52 ) Cp =

= 1.52 The process is capable of meeting design specifications.

3-35.

Cp =

420 420 = = 1.27 6 ( 55 ) 330

1, 050 − 915 1,335 − 1, 050  C pk = minimum  ,  3 ( 55 )  3 ( 55 )  = minimum (.82, 1.72 ) = .82 While C p = 1.27 indicates the process is capable, C pk = .82 indicates the process mean has shifted toward the lower design specification, and defective (shorter lived) bulbs will be generated.

3-36.

Cp =

.048 .048 = = 1.00 6 (.008 ) .048

1.281 − 1.251 1.299 − 1.281  C pk = minimum  ,  3 (.008 )   3 (.008 ) = minimum (1.25, 0.75 ) = 0.75

C p = 1.00 means the tolerance range and the process range are virtually the same indicating that some defective parts will occur. C pk = 0.75 indicates that the process mean has shifted toward the upper specification indicating the process will result in some parts that are defective (too large).

3-37. The process mean would need to be shifted back toward the nominal design value of 1,125 hours. To achieve six sigma quality the process range would need to be reduced to one-half of the tolerance range. Since the tolerances are 210 hours, the tolerance range is 420 hours. Thus, the process range would need to be 210 hours, which are 3 control limits of 105 hours. Thus, the process mean would need to be 1,125 hours with an upper control limit of 1,230 hours, and a lower control limit of 1,020 hours.

3-38. Machine 1:

.030 = 1.25 .024  .0995 − .082 .112 − .0995  C pk = minimum  ,  .012 .012  = minimum (1.46, 1.04 ) Cp =

= 1.04 Machine 1 is capable of meeting the design specifications.

Machine 2:

.030 = 0.56 .054  .1002 − .0820 .1120 − .1022  C pk = minimum  ,  .027 .027  = minimum (.67, .44 ) Cp =

= .44 Machine 2 is not capable of meeting the design specifications.

Machine 3:

.030 = 1.00 .030  .0951 − .0820 .1120 − .0951  C pk = minimum  ,  .015 .015  Cp =

= minimum (.87, 1.13) = .87

Machine 3 is capable of meeting the design specifications but the process center has shifted too far toward the lower design specification.

3-39.

x

Month 1 2 3 4 5 6 7 8 9 10 11 12 Avg

7.02 8.16 8.18 9.18 10.32 9.54 6.96 10.38 8.12 10.26 9.66 9.10 8.91

R 5.6 3.7 4.9 5.4 4.1 4.1 3.5 9.0 8.9 6.9 9.2 4.3 5.8

x = 8.91,

R = 5.8, A2 = .58, D3 = 0, D4 = 2.11

X-bar chart

R-Chart

UCL = 12.27 LCL = 5.54

UCL = 12.24 LCL = 0

The process is in control according to both control charts.

3-40.

p = 0.48 UCL = p + 3

p 1− p = 0.48 + 3 n

(

)

0.48(1 − 0.48) = 0.60 150

LCL = p - 3

p 1− p = 0.48 - 3 n

(

)

0.48(1 − 0.48) = 0.36 150

The process is not in control and does not meet the target value of 90%. It appears that improvement occurred in week 7, but the improvement was not consistent and was significantly below the target value. The hospital should reevaluate the process improvements it has implemented using quality tools and, after implementing a new program, a new control chart should be developed.

3-41.

Sample1 1 2 3 4 5 6 7 8 9 10

x 145.88 144.86 144.22 145.82 143.10 147.82 143.04 141.44 148.72 142.42

R 6.9 8.6 9.4 5.2 6.7 6.2 5.5 5.9 5.9 8.5

Sample 11 12 13 14 15 16 17 18 19 20

x 144.54 145.48 145.26 148.78 143.58 146.48 145.22 144.80 143.46 145.92

R 6.7 5.5 8.5 4.8 5.1 3.1 2.4 4.6 5.4 7.1

x = 145.06 R = 6.1 A2 = 0.58

x -chart: UCL = 145.06 + 0.58 ( 6.1)

= 148.60 LCL = 145.06 − 0.58 ( 6.1)

R -chart:

UCL = D4 R = ( 2.11)( 6.1) = 12.87 LCL = D3 R = ( 0 )( 6.1) = 0

= 141.52 Sample 8 is slightly below the LCL and sample 14 is slightly above it.

Cp =

149 − 142 7.00 = = .99 2 ( 3.54 ) 7.08

145.06 − 142 149 − 145.06  C pk = minimum  ,  3.54 3.54  = minimum (.86, 1.04 ) C pk = .86 The process is not capable of meeting design specifications. Since C p is very close to 1.00, some defective baseballs will be generated, and, C pk = 0.86 indicates they will typically not weigh enough.

3-42. To achieve six sigma quality the process range would need to be reduced to one-half of the tolerance range.

The tolerance range is 7 gms therefore the process range must be 3.5 gms which are 3 control limits of 1.75 gms. Thus, the process mean would need to be 145.5 gms with an upper control limit of 147.25 and a lower control limit of 143.75.

3-43.

Sample 1 2 3 4 5

x

R 7.1 6.4 12.7 6.8 12.4

7.72 6.58 12.90 8.76 11.06

Sample 6 7 8 9 10

x 8.90 11.52 8.02 9.58 9.12

R 6.0 14.6 4.7 8.5 5.6

x = 9.42, R = 8.48 A2 = 0.58 R -chart:

x -chart: x = 9.42 UCL = 14.31 LCL = 4.52

R = 8.48 UCL = 17.94 LCL = 0

The process is in control according to both control charts.

Cp =

12 − 6 6 = = .61 14.31 − 4.52 9.79

 9.42 − 6 12 − 9.42  C pk = minimum  , 4.92   4.92 = minimum ( 0.70, 0.52 ) = 0.52 The process is not capable of meeting the design specifications and the customer waiting times will be greater than the upper specification and lower than the lower specification. Although the lower times might be considered good, it could be that customer service representatives are not devoting enough time to each customer.

3-44. (a) Sample 1 2 3 4 5 6 7 8 9 10

x 21.4 27.0 19.0 24.4 26.6 20.8 24.8 26.4 29.6 25.4 245.4

R 9 23 8 19 33 15 14 23 11 9 164

 R 164 = = 16.4 k 10  x 245.4 x= = = 24.54 10 10

R=

R -chart

D3 = 0, D4 = 2.11, for n = 5

UCL = D4 R = ( 2.11)(16.4 ) = 34.604

LCL = D3 R = ( 0 )(16.4 ) =0 There are no R values outside the control limits, which suggests the process is in control.

x -chart

A2 = 0.58

UCL = x + A2 R = 24.54 + (.58 )(16.4 ) = 34.05 LCL = x − A2 R = 24.54 − (.58 )(16.4 ) = 15.03 There are no x values outside the control limits, which suggests the process is in control. (b)

upper specification limit − lower specification limit 6 30 − 20 = 19.02 = .52  x − lower specification, upper specification − x  C pk = minumum   3 3  

Cp =

 24.54 − 20 30 − 24.54  = minimum  , 9.51   9.51 = minimum (.47, .57 ) = .47 Since C p = .52, which is less than 1.0, the process range is greater than the tolerance range and the process is not capable of producing within the design specifications all the time. Since Cpk = .47 is less than 1.0, the process has moved closer to the lower design specification and will generate defects.

3-45. (a) Sample 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

x 4.83 5.38 5.38 5.75 5.00 6.40 6.65 4.57 6.18 5.32 4.02 4.57 4.95 5.05 4.47 78.52

R 3.6 3.3 1.3 3.3 2.1 2.9 2.7 3.5 4.6 1.2 4.5 4.2 3.7 2.1 3.6 46.6

 R 46.6 = = 3.107 k 15  x 78.52 x= = = 5.23 15 15 R -chart R=

D3 = 0, D4 = 2.00, for n = 6

UCL = D4 R = ( 2.00 )( 3.107 ) = 6.21 LCL = D3 R = ( 0 )( 3.107 ) = 0 There are no R values outside the control limits, which suggest the process is in control.

x -chart

A2 = .48

UCL = x + A2 R = 5.23 + (.48 )( 3.107 ) = 6.73 LCL = x − A2 R = 5.23 − (.48 )( 3.107 ) = 3.74 There are no x values outside the control limits, which suggest the process is in control. (b)

upper specification limit − lower specification limit 6 6−4 2 = = 2.98 2.98 = .67

Cp =

 x − lower specification, upper specification − x  Cpk = minumum   3 3    5.23 − 4 6 − 5.23  = minimum  , 1.49   1.49 = minimum (.83, .51) = .51 Since C p = .51, which is less than 1.0, the process range is greater than the tolerance range and the process is not capable of producing within the design specifications all the time. Since Cpk = .51 is less than 1.0, the process has moved closer to the lower design and will generate defects.

3-46.

64 = 2.667 24 UCL = c + z c c=

= 2.667 + 3 2.667 = 7.57 LCL = c − z c = 2.667 − 3 2.667 =0 The process is only “out of control” in month 5 when there were zero falls. This month should be investigated to see if the circumstances that resulted in no falls can be repeated.

Any number of falls would seem to be poor quality, so even though this “process” is technically “in control,” the process should be improved with a six sigma goal of zero defects. While 2.667 falls per month is not a lot, it is likely too many.

3-47.

(a) Sample 1 2 3 4 5 6 7 8 9 10 11 12

x 88.33 93.33 82.50 91.00 89.83 87.00 89.67 86.00 84.00 93.67 92.83 85.50

 R 230 = = 19.17 k 12  x 1063.67 x= = = 88.64 12 12

R=

R 18 12 28 22 13 15 15 34 31 15 14 13

R -chart

D3 = 0, D4 = 2.00, for n = 6 UCL = D4 R = (2.00)(19.17) = 38.41 LCL = D3 R = (0)(19.17) = 0 There are no R values outside the control limits, which suggests the process is in control. x -chart

A2 = .48 UCL = x + A2 R = 88.64 + (.48)(19.17) = 97.89 LCL = x − A2 R = 88.64 − (.48)(19.17) = 79.38 There are no x values outside the control limits, which suggest the process is in control. (b)

upper specification limit − lower specification limit 6 98 − 92 6 = = (88.64 − 79.38) 18.71 = .32

Cp =

 88.64 − 92 98 − 88.64  C pk = minimum  , 9.27   9.27 = minimum ( − 0.36, 1.01) = −0.36 Since C p = 0.32 (1.0), the service department is not currently capable of consistently achieving the desired customer satisfaction score. Since C pk = −0.36 is less than 1.0 the service department will continue to generate less than desired customer scores. The service department would need to make process improvements in order to consistently achieve the desired customer satisfaction scores.

3-48.

(a) Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

x

R 143.00 139.00 112.33 170.67 165.33 135.33 149.33 113.67 143.67 138.00 118.33 176.33 167.67 157.00 165.67 144.67 113.00 167.67 197.33 115.67 148.33 160.33 159.67 107.33 129.67 178.33 155.33 153.67 145.00 132.33 4403.7

40 33 31 88 26 55 32 18 22 57 62 56 92 87 61 31 47 19 34 67 27 81 46 38 61 46 21 78 17 51 1424

 R 1424 = = 47.47 k 30  x 4403.7 x= = = 146.79 30 30

R=

R -chart

D3 = 0, D4 = 2.574, for n = 3 UCL = D4 R = (2.574)(47.47) = 122.18 LCL = D3 R = (0)(47.47) = 0 There are no R values outside the control limits, which suggest the process is in control.

x -chart

A2 = 1.023 UCL = x + A2 R = 146.79 + (1.023)(47.47) = 195.35 LCL = x − A2 R = 146.79 − (1.023)(47.47) = 98.23 There is one x value outside the control limits on day 19, which at least suggests a problem on that day should be investigated, although there is no consistent pattern of being out of control. (b)

upper specification limit − lower specification limit 6 135 − 105 30 = = 97.12 97.12 = .31

Cp =

146.79 − 105 135 − 146.79  C pk = minimum  , 48.56   48.56 = minimum (.86, − 0.24) = −0.24 Since C p = 0.31 and C pk = −0.24, the bed turnaround time is not capable of achieving the hospital’s goal of 120 minutes. Process improvements would be necessary to achieve the desired bed turnaround times.

CASE SOLUTION 3.1: Quality Control at Rainwater Brewery This is basically a discussion question; therefore, the student responses might vary.

The owners have stated in the case description that the chances of a batch being spoiled—and, thus, an unhealthy batch—are very unlikely. However, even a very slight risk of a contaminated batch of 1,000 bottles might be too much, given the health consequences of a spoiled batch. Testing only a small sample of even a few bottles would indicate if the batch was, in fact, bad, so a simple testing procedure such as opening and testing 5 to 10 bottles might be prudent.

Most of the quality control efforts should focus on process control procedures at the various stages of the brewing process. Obvious candidates are x and R-charts to monitor temperature, specific gravity, and pH during the fermentation and aging stages. Some type of process control testing of the final bottled product probably is warranted also.

Quality control methods can also be used at the beginning of the brewing process for checking materials such as bottles and caps and ingredients such as yeast, hops, and grain. Bottles and caps that are not completely clean and sterile can result in a spoiled batch, and poor-quality ingredients can obviously contribute to an “off’’ brew.

CASE SOLUTION 3.2: Quality Control at Grass, Unlimited 249 defects = 4.15 60 samples z = 2.00 UCL = c + Z c = 4.15 + 2 4.15 = 8.22

c=

LCL = c − Z c = 4.15 − 2 4.15 = 0.076 The chart exceeds the control limits for samples 12 and 14, however, the chart appears to have been in control prior to sample 12. Although, the reasons for the out-of-control occurrences must be investigated, the chart based on samples 1 through 11 could be used, or, several observations could be collected once the process is brought back in control and used with samples 1 through 12. Thus, this chart could be implemented for continued use.

Other examples of control charts that could be used include p-charts for the number of errors in a sample of orders, or the number of customer complaints for a sample survey of customers. A c-chart could be used for the number of defects found (for cleanliness) during an inspection of facilities.

CASE SOLUTION 3.3: Improving Service Time at Dave’s Burgers x = 1.70, R = 1.40, A2 ( n = 6 ) = 0.48 UCL = x + A2 R

= 1.70 + ( 0.48 )(1.40 ) = 2.37

LCL = x − A2 R

= 1.70 − ( 0.48 )(1.40 ) = 1.03

UCL = D4 R = 2.00 (1.40 ) = 2.80 LCL = D3 R = 0 (1.40 ) = 0

The process is not in control, thus the control chart cannot be used on a continuing basis.

The out-of-control situation should be investigated and upon correction, new data should be gathered to establish a revised control chart. If that control chart is valid (i.e., in control), it may be used to monitor quality on a continuing basis.

Dave may want to chart % of customers orders completed correctly (p-chart) or the number of customer complaints (c-chart).

Answers from students, of course, may vary.

4 - Product Design Answers to Questions 4-1.

A firm’s products and services define its customers, as well as its competitors. New products and services often involve new markets and require new processes. The design process is the most obvious driver of change in an organization. It capitalizes on a firm’s core competencies and determines what new competencies need to be developed. Organizations can gain a competitive edge through designs that (1) bring ideas to the market quickly, (2) do a better job of satisfying customer needs, and (3) are easier to manufacture, use, and repair than existing products.

4-2.

Quality of design—size; location and amount of lighting; size of blackboards; heat; presentation equipment; Quality of conformance—no. of students crowded into room; overhead projector or lights don’t work; too hot or too cold; instructor doesn’t speak loud enough or students are too loud.

4-3.

Student answers will vary. Bad designs fail to take the user’s point of view, are too confusing to operate, are too complex, and go against the natural tendency of the user.

4-4.

Student answers will vary.

4-5.

Student answers will vary.

4-6.

Student answers will vary.

4-7.

Perceptual maps compare customer perceptions of a company’s products with competitor products. Students will construct a variety of maps in response to this question. Typical dimensions include: (a) cost and reputation, (b) ease of use and features, (c) cost and variety.

4-8.

Benchmarking refers to finding the best-in-class part, product or process, measuring one’s performance against it, and making recommendations for improvement based on the results. Student answers will vary.

Most of the benchmarking studies compare processes, rather than products, across industries.

4-9.

Again, the answers to this question will vary considerably. Some possible characteristics are: size of student body, size of faculty and staff, salary levels, research dollars, publications, number of degrees, types of degrees, SAT or GMAT scores, number of National Merit Scholars, graduation rate, employment rate, and average salary of graduates.

4-10. A market analysis is conducted to assess whether or not sufficient demand for the proposed product exists to merit investment in its further development. If the demand potential exists, an economic analysis is conducted which looks at estimates of production and development costs and compares them to estimated sales volume. Finally, technical and strategic analyses are completed which answer such questions as: Does the new product require new technology? Is the risk or capital investment excessive? Does the company have sufficient labor and management skills to support the technology required? Is sufficient capacity available for production? Does the new product provide a competitive advantage for the company? Does it draw on corporate strengths? Is it compatible with the core business of the firm?

4-11. Performance specifications tell how a product is to perform. Design specifications detail the measurements and standards to which a product is to be built so that it meets performance specifications. Manufacturing specifications outline how the processes are supposed to operate in order to produce the product to meet design specifications.

4-12. Reliability is the probability that a given part or product will perform its intended function for a specified

length of time under normal conditions of use. Maintainability refers to the ease and/or cost with which a product is maintained or repaired. Maintainability and reliability are closely related. For example, if a product is cheap to manufacture and priced so low that customers throw it away when it fails (such as calculators, telephones, and watches), maintainability may be a moot issue. Similarly, if a product is so reliable that it rarely breaks down, then ease of repair many not be important. On the other hand, it may be less costly to make a product easy to maintain than to increase its reliability. And for some products, both reliability and maintainability are very important (e.g., office machines, computers).

4-13. Simplification attempts to reduce the number of parts and assemblies in a design and make the remaining parts compatible. Fewer parts and better fitting parts provide fewer chances for error in manufacture and assembly. Standardization makes possible the interchangeability of parts among products, resulting in higher volume production and purchasing, lower investment in inventory, easier purchasing and material handling, fewer quality inspections, and less difficulties in production. Modular design consists of combining standardized building blocks or “modules’’ in a variety of ways to create unique finished products. Thus, even though the parts may be standardized, the finished product is unique.

4-14. Design teams can obtain input from a variety of sources before erroneous decisions are made. There is often a synergistic effect of people working together. Studies have found that the key to a good design is the involvement and interaction of the “create, make, and market’’ functions from the beginning of the design project. That said, working in teams can be difficult. It is usually easier and less conflicting to work solo.

Team members must be convinced that their joint effort will produce better results than individual efforts.

4-15. Concurrent design changes the design process from a sequential one, where decisions are made by separate departments, to simultaneous decision making by design teams. Concurrent design attempts to integrate product design and process planning into a common activity. It helps improve the quality of early design decisions and thereby reduces the length and cost of the design process. In a group project, students are responsible for completing an assigned portion of the project. If the project is well designed with clear expectations, this division of labor will work. If not, considerable rework will be required.

4-16. DFM identifies product design characteristics that are inherently easy to manufacture, focuses on the design of component parts that are easy to fabricate and assemble. Techniques for DFM include: design for assembly (DFA), failure mode and effect analysis (FMEA), fault tree analysis (FTA), and value analysis (VA).

4-17. Failure Mode and Effects Analysis (FMEA) is a systematic approach to analyzing the causes and effects of product failures. It begins with listing the functions of the product and each of its parts. Failure modes are then defined and ranked in order of their seriousness and likelihood of failure. Each failure is addressed one by one, causes are hypothesized, and design changes are made to reduce the chance of failure. The objective of FMEA is to anticipate failures and design them out of the system. Fault tree analysis (FTA) is similar to FMEA except that it emphasizes the interrelationship between failures, and presents the analysis more graphically. Value analysis (also known as value engineering) was developed by General Electric in 1947 to eliminate unnecessary features and functions in product designs. It has re-emerged as an excellent technique for use by multifunctional design teams. An example fault tree analysis for a term paper is given on the next page.

4-18. Students will find references to design for environment (DFE), extended producer responsibility (under the title of product stewardship), green design, remanufacturing, and sustainable development. The best source of environmental policy for other countries is the European Union’s information site at http://europa.eu 4-19. Student responses will vary.

4-20. The ISO 14000 family of international standards offers a wide-ranging portfolio of standardized sampling, testing and analytical methods to deal with specific environmental challenges. It has developed more than 350 international standards for monitoring the quality of air, water and soil. These standards provide business and government with scientifically valid data on the environmental effects of economic activity, and serve in a number of countries as the technical basis for environmental regulations. ISO 14000 also provides a strategic approach to environmental management systems (EMS) that can be implemented in any type of organization in either public or private sector (companies, administrations, public utilities). ISO 14000 was developed in response to discussions of “sustainable development’’ at the United Nations Conference on Environment and Development in Rio de Janeiro in 1992. The International Standards Organization (ISO) put together a committee of scientists to develop the standards which were first published in 1996. A company obtains certification from a 3rd party registrar who follows ISO 14000 document guidelines and performs audits and site visits. Certification allows companies to use Environmental Labeling, reduces insurance, regulatory and operating costs, and enhances market appeal.

Although voluntary, ISO 14000 standards are often referenced in technical regulations and other types of legal documents. In addition, companies in certain industries require their suppliers to be ISO 14000 certified. For example, Ford suppliers had to have at least one of their plants certified by the end of 2001 and the remainder by 2003. Globally, Japan has the largest number of ISO 14000 registered companies, followed by the UK, Germany, Sweden, and the U.S. 4-21. Quality Function Deployment (QFD) is a structured process that translates the voice of the customer to technical requirements at every stage of design and manufacture. QFD forces management to spend more time defining the new product changes and examining the ramifications of those changes. More time spent in the early stages of design means less time is required later to revise the design and make it work.

4-22. A robust design is one able to withstand variations in environmental and operating conditions and still operate as intended. Example—a cell phone that works after repeatedly being dropped.

4-23. CAD has been especially useful in testing designs (CAE) and as a means of integrating design and manufacture (CAD/CAM). Technology has enabled us to design new products more quickly, consider many more alternatives than would be possible manually, test the design on a computer screen without building a prototype, document the design in different forms, and automatically transfer the design to manufacturing.

Design centers or team members can be located in diverse geographic locations and still communicate freely.

Experts can collaborate on designs and make changes from a distance via the Internet.

Solutions to Problems 4-1.

Most students will not find these instructions clear. For example, should the center line be folded lengthwise or widthwise? Which way should the airplane be folded in half? How exactly do you fold back the wings? A diagram of each step would definitely help in communicating the design. It’s fun to see how many different prototypes the students come up with from the same design instructions. After the class agrees on a set procedure, revise the instructions, draw a diagram, and test them on an unsuspecting student.

4-2.

Folding the paper lengthwise in alternate directions could be made clearer if you added “like an accordion.’’

Words such as “fan out’’ and “small fold’’ in “nose’’ of plane are imprecise. It is also unclear whether to fold the paper lengthwise or widthwise (either way will do). As with the first design, a diagram would be helpful. This plane was easier to construct but did not look as much like a plane! In terms of flying ability, this plane is supposed to fly further (like a glider). The first plane flies higher and faster.

4-3.

This system has both parallel and series components. Calculate the reliability of the parallel components first:

4-4.

a.

x5 = 0.95 broadcast success x = 0.9898 subsystem reliability

b.

x5 = 0.98 broadcast success x = 0.9960 subsystem reliability

c.

x5 = 0.99 broadcast success x = 0.9980 subsystem reliability

4-5.

RS = 1 − (1 − R1 )(1 − R2 ) (1 − R3 ) = 1 − ( 0.10 )( 0.20 )( 0.30 ) = 0.9440 4-6.

a.

Vendor 1: .94  .90 + .10 (.86 )  .93 = .8620

Vendor 2: .85  .93 + .07 (.88)  .95 = .8007 Vendor 3: .92  .95 + .05 (.90 )  .90 = .8239 Choose vendor 1. b.

Vendor 1: .94  .86  .90  .93 = .6766 Vendor 2: .85  .88  .93  .95 = .6609 Vendor 3: .92  .90  .95  .90 = .7079 Yes, choose vendor 3.

4-7.

.98  .97  .95  .96  .99 = 0.8583 No, the probability of failure is (1 − .8583) = 0.1417

a.

b. 5 0.96 = 0.9919 Each component would need a reliability of .9919 4-8. a. System A reliability = .93 x .93 = .8649 Component cost = $1500 x 2 = $3000 Failure cost = (1-.8649) x $10,000 = $351 Total cost of system A = $4351 b. System B reliability = .95 x .95 x .95 = .8574 Component cost = $2000 x 3 = $6000 Failure cost = (1-.8574) x $10,000 = $1426 Total cost of system B = $7426 System A is more reliable and less costly than System B.

c. System C reliability = [.95 + (.05)(.90)] x .95 x [.95 + (.05)(.90)] = .9405 Component cost = ($1000 x 2) + ($2000 x 3) = $8000 Failure cost = (1-.9405) x $10,000 = $595 Total cost of system C = $8595

d. System D reliability = [.90 + (.10)(.90)] x [.90 + (.10)(.90)] x [.90 + (.10)(.90)] = .9703 Component cost = $1000 x 6= $6000 Failure cost = (1-.9703) x $10,000 = $297 Total cost of system D = $6297 System D is more reliable and less costly than System C. This is somewhat surprising because we usually think of a more reliable system as being more costly.

4-9.

a.

System Basic Standard Professional

Component Reliability 0.80 0.90 0.99

No. Components 5 5 5

System Reliability 0.3277 0.5905 0.9510

Purchase Cost $1,000 $2,000 $5,000

Failure Cost $50,000 $50,000 $50,000

Failure Probability 0.6723 0.4095 0.0490

Total Cost $34,616.00 $22,475.50 $7,450.50

No. Components 5 5 5

System Reliability 0.8154 0.9510 0.9995

Purchase Cost $2,000 $4,000 $10,000

Failure Cost $50,000 $50,000 $50,000

Failure Probability 0.1846 0.0490 0.0005

Total Cost $11,231.37 $6,450.50 $10,025.00

Choose the professional system. b.

System Basic Standard Professional

Component w/backup 0.960 0.9900 0.9999

Choose the standard system.

4-10.

4-11.

SA = MTBF / (MTBF + MTTR) = 100/(100+24) = .8065

4-12.

Provider Able Copy Business Mate Copy Whiz

MTBF (hours) 40 80 240

MTTR (hours) 1 4 8

System Availability 0.9756 0.9524 0.9677

Choose Able Copy

4-13.

Provider Airway Bellular CyCom

No. of failures per 8 hrs 10 8 3

MTBF (mins) 48 60 160

MTTR (mins) 2 4 10

System Availability 0.9600 0.9375 0.9412

Choose Airway

4-14.

Provider JCN Bell Comtron

No. of problems 50 100 250

MTBF (hours) (8 x 40) / 50 = 6.4 (8 x 40) / 100 = 3.2 (8 x 40) / 250 = 1.3 Choose JCN

Mean Time to Reach Customer 3 2 1

Mean Time to Fix Customer 2 1 0.5

MTTR (hours) 5 3 1.5

System Availability 0.5614 0.5161 0.4604

4-15.

4-16. Choose Yourizon.

ISP Xceptional

No. Failures 12

Time to regain service, MTTR 2

Failure Rate, MTBF

Uptime as %, SA

40

0.952

Yourizon

4

4

120

0.968

Zelltell

3

10

160

0.941

4-17. Answers will vary. Here is a representative solution.

4-18. Answers will vary. This is a sample set up for the problem.

4-19. Students will come up with some interesting examples. Here is a sample worksheet, also available as an excel file on the text website.

CASE SOLUTION 4.1 - Lean and Mean

CASE SOLUTION 4.2 – Greening Product Design 1.

2.

3.

4.

Hal expresses the reaction of many who view green design as “the next big thing” in management circles, perhaps a passing fad. However, he does have a point that changing a design so significantly is not a simple process, and that he needs time to explore the consequences of these changes. Perhaps he also needs time to explore the benefits of green design.

A good way to justify green design is to quantify the effects on both the environment and the bottom line, shortterm and long-term.

Some companies, such as Patagonia, incorporate green design into their mission, i.e., who they are as a company. Green design spurs innovation and becomes a competitive advantage. Other companies, such as WalMart, have found such large cost savings from green design that they also gain an advantage. Customers also pressure companies to act responsibly and their actions can affect sales. These are the most common reasons, other than legal requirements, why companies go green.

It is generally easy to purchase green commodity products, like light bulbs. The challenge is choosing a sustainable product when it is more costly, less convenient or less attractive. An interesting Harvard Business School report, entitled “Sweatshop Labor is Wrong Unless the Jeans are Cute” (http://hbswk.hbs.edu/item/6126.html) says that ethical behavior is influenced by the desirability of the product. Thus, we might buy a green product that costs a little more, but not a lot; or we might opt for a nongreen product if it is “cute.”

5 - Service Design Answers to Questions 5-1.

Services are acts, deeds, performances or relationships that produce time, place, form or psychological utilities for customers.

5-2.

(1) & (2) Services are intangible and perishable. This makes design more difficult. Intangibles are harder to specify and perishable items have to be redesigned frequently. (3) & (4) Services have high customer contact and variable output. Because people are usually involved in services (both on the side of the customer and the service provider), unexpected things can happen. Designs must be flexible, but must also include contingency plans. (5) Consumers do not separate the service from the delivery of the service, so everything about the service environment (layout, attitude of employees, etc.) is part of the service design. (6) Services tend to be decentralized and geographically dispersed. Thus, a design must be “workable” in any locale, or be adapted to the specific environment in which it must operate. Also, service providers have more authority and may provide the service differently. (7) & (8) Services are “consumed” more often than products, and can be easily emulated. This means that improvements to service design are undertaken more frequently than with products, and the results are more immediate. Risks can be taken on a small scale.

5-3.

Bank—types of accounts, credit cards, loans, branches, hours of operation. ATMs, degree of customization, convenience, decor, information on-line, mortgages, automatic deposit and withdrawal, etc. Airline—flights to particular destinations, reservations, check in, baggage handling, accessibility, types of planes, crowding of planes, service personnel, meals, comfort, disruptions, arrives and departs on time, length of flights, number of direct flights, timing of flights, etc. Lawn Service—frequency of service, jobs (cutting lawn, fertilizing, watering, planting, landscaping), equipment, evaluation of lawn needs, demeanor and performance of personnel, ease of payment, etc.

5-4.

Automated banking—more readily available, accessible from home, more services available (stamps, airline tickets, concert tickets), etc. Higher education—break out of semester segments, self-paced learning, access from home or job site, distance learning, expert-a-day (i.e., more than one instructor), etc. Healthcare—more available, access from schools, records more accessible, availability and instruction for selfcare and self-evaluation, etc.

5-5.

SSME (Services Science, Management and Engineering) is “the application of scientific, management, and engineering disciplines to the tasks or services that one organization beneficially performs for and with another.” IBM recognizes the need for a systematic multidisciplinary study of service systems as industrialized economies are increasingly based on services. In the U.S., services account for 80% of employment, 94% of job growth, and 70% of GDP.

5-6.

Examples might include a number of retail stores, banks, movie theaters, service stations, university operations and functions, computer access, telephone calls, etc.

5-7.

The ratio of the arrival rate to the service rate must be less than 1, which means that the service rate must be greater than the arrival rate. If customers arrive faster than they can be served, the system eventually develops an infinite queue.

5-8.

Examples might include a doctor’s office, a faculty advisor, a machine shop, a number of breakdown/repair systems in which the operating units are finite, etc.

5-9.

These elements are queue discipline, calling population, arrival rate, and service rate.

5-10. Waiting lines are an integral part of a multitude of business operations, and their efficient design and management are an important part of these businesses.

5-11. The mean effective service rate is the number of servers multiplied by the service rate, and it must exceed the arrival rate.

5-12. a. Hair salon: multiple-server; first-come, first-served or appointment; calling population can be finite (appointments only) or infinite (off-the-street business). b. Bank: multiple-server; first-come, first-served; infinite calling population. c. Laundromat: multiple-server; first-come, first-served; infinite calling population. d. Doctor’s office; single- (or multiple-) server; appointment (usually); finite calling population. e. Advisor’s office: single-server; first-come, first-served or appointment; finite calling population. f. Airport runway: single-server; first-come, first-served; finite calling population. g. Service station: multiple-server; first-come, first-served; infinite calling population. h. Copy center: single- or multiple-server; first-come, first-served; infinite calling population. i. Team trainer: single-server; first-come, first-served or appointment; finite calling population. j. Mainframe computer: multiple-server; first-come, first-served (or priority level); infinite calling population.

5-13. The addition of a new counter created two queues. The multiple-server model is for a single queue with more than one server.

5-14. The traditional cost relationship for waiting line analysis is to provide a level of service that minimizes the total cost of providing service and the cost of customers waiting, which is similar to the traditional quality cost relationship to provide a level of quality that minimises the total cost of poor quality and the cast of prevention. However, the recent emphasis on quality management suggests that the level of service should be greater than the minimum cost level, approaching a 100 percent service level.

5-15. a. Single-channel, single-phase—a single server; an example is a post office with one postal clerk. b. Multiple-channel, single-phase—two or more servers in parallel; an example is a post office with two or more clerks. c. Single-channel, multiple-phase—a series of single-servers that the customer must pass through; an example is a medical treatment facility (e.g., hospital) wherein a patient must first see a nurse and then see a doctor. d. Multiple-channel, multiple-phase—a series of two or more servers in parallel; an example is a medical facility with two or more series of nurses and doctors.

5-16. Total waiting line cost is the sum of the cost of providing service and the cost of customer waiting, which generally have an inverse relationship to each other. Thus, as the level of service is increased, the cost of service increases, whereas the cost of waiting decreases. The optimal level of service coincides with the minimum point on the total waiting line cost curve.

5-17. a. False. The operating characteristic values may be higher or lower, depending on the magnitude

of the standard deviation compared to the mean of the exponentially distributed service time. b. True. Since there is no variability the operating characteristics would always be lower.

5-18. If a service facility has a Poisson input with parameter  and an exponential service distribution with parameter  (where    ), then each successive facility in a multiplephase system will have a Poisson input with  . This enables each phase to be analyzed independently; the aggregate operating characteristics at each service facility are summed.

5-19. When arrivals are random, in the short run more customers may arrive than the serving system can accommodate.

5-20. An example is when customers are served according to a prearranged schedule or alphabetically or are picked at random.

5-21. When  =  .

5-22. Automated equipment or robots.

Solutions to Problems 5-1.

=

60 = 16.67 customers per hr 3.6

=

60 = 25 customers per hr 2.4

(16.67 ) 2 Lq = = = 1.33 customers waiting  (  −  ) ( 25 )( 25 − 16.67 ) 2

Wq =

(16.67 )  = = 0.08 hr or 4.80 minutes waiting in line  (  −  ) ( 25 )( 25 − 16.67 )

This is not very good service. Although another window (server) might reduce waiting time, the service time in excess of 2 minutes is still too long; it should probably be less than 1 minute.

Thus, the process of filling orders needs to be improved.

5-2.

 = 8/hr  = 12 / hr Wq =

 8 = = 0.167 hr (10 min )  (  −  ) 12 ( 4 )

= 5-3.

 8 = = 0.667  12

 = 5/hr  = 10 / hr

( 5) = 0.5 cars 2 =  (  −  ) 10 ( 5 ) 1 1 W= = = 0.20 hr (12 min )  − 5 2

a.

Lq =

Wq =

 5 = = 0.10 hr ( 6 min )  (  −  ) 10 ( 5)

b. If the arrival rate is increased to 12 per hour, the arrival rate would exceed the service rate; thus, an infinite queue length would result.

5-4.

The arrival rate must be on an hourly basis.

=

60 = 7.5 per hour 8

 = 10 per hour

( 7.5) = 2.25 parts 2 Lq = =  (  −  ) 10 ( 2.5 ) 2

 = 0.75, I = 0.25 5-5.

=

 

 = 10 per hr  = 0.90  and  = 9 per hour, or 1 part every 6.67 minutes.

Therefore, 0.90 = 10

5-6.

 = 5 per hour =

60 = 7.5 per hour 8

( 5) = 1.33 2 = a. Lq =  (  −  ) 7.5 ( 2.5 ) 2

b. Wq =

c.

W=

 5 = = 0.26 hr (16 min )  (  −  ) 7.5 ( 2.5) 1 1 = = .4 hr ( 24 min )  −  7.5 − 5

45 , or 0.75, utilization factor.  cannot exceed 0.75. d. 45 minutes per hour is a 60

=

 5 = = 0.67 at present  7.5

Therefore, a new air traffic controller is not needed.

5-7.

 = 12 per hour =

60 = 20 per hour 3

One window:

Wq =

 12 = = 0.075 hr ( 4.5 min )  (  −  ) 20 (8)

Two windows:

 = 20 per hour (does not change) However, the arrival rate for each window is now split.

 = 6 per hour Wq =

 6 = 0.021 hr (1.29 min ) =  (  −  ) 20 (14 )

4.5 − 1.29 = 3.21 min reduction in waiting time 3.21 $2, 000 = $6420 Cost of window = $20,000 $6420  $20,000; Therefore, a second drive-in window should not be installed according to

cost alone. However, the existing wait time of 4.5 minutes seems excessive, thus in terms of providing quality service, the second window should probably be installed.

5-8.

 = 10 per hour

=

a.

60 = 12 per hour 5

L=

  −

=

10 =5 2

(10 ) = 4.17 2 Lq = =  (  −  ) 12 ( 2 ) 2

5-9.

W=

1 1 = = 0.5 hr ( 30 min )  − 2

Wq =

 10 = = 0.42 hr ( 25 min )  (  −  ) 12 ( 2 )

=

 10 = = 0.83 = 83%  12

 = 120 per day  = 140 per day

(120 ) = 5.14 trucks 2 = a. Lq =  (  −  ) 140 ( 20 ) 2

W=

1 1 = = 0.05 day  −  20

8 hr / day  60 min / hr = 480 min 0.05  480 = 24 min.

W = 24 min

Wq =

 120 = = 0.043 day ( 20.6 min )  (  −  ) 140 ( 20 )

b. 24 − 15 = 9 min

9 minutes  $10, 000 = $90, 000 per year loss presently.

With a new set of scales the arrival rate would be split.

 = 60 per day per scale W = 1/ ( − ) =   (−) =  day 8 hr/day * 60 min/hr * 0.0125 day = 6 minutes The entire $90,000 would be saved.

Since the scales cost $50,000 per year, they should be installed.

5-10.

Arrival rate = 120 trucks per hour Service rate = 140 trucks per hour P(0) = 0.143 P(1) = 0.122 P(2) = P(3) = P(4) = P( < = 4) = 0.537 P( > = 5) = P( 1 - < = 4) = 0.463 This solution assumes that arriving trucks will pass by the station if they see 4 trucks waiting in a line. Since four trucks are waiting, one truck is being served for a total of 5 trucks in the system.

5-11.

 = 10 per hour  = 12 per hour s=2

P0 =

1

2  1 1  10  1  10   ( 2 )(10 )      +      0 n !  12   2!  12   24 − 12  1 =  1  10 0 1  10 1  1  10 2  20     +   +      0!  12  1!  12   2  12   12  = .412 n

s

      L=  P0 + 2  ( s − 1)!( s −  )

2 10 )(12 )( 0.83) ( = ( 0.412 ) + 0.93 2 1!( 24 − 10 )

=

7265 ( 0.36 ) + 0.83 = 1.01 1024

 = 1.01 − .83 = .177 

Lq = L −

W=

Wq =

L



=

Lq



1.01 = .101 hr ( 6.05 min ) 10

=

.175 = .0175 hr (1.05 min ) 10

The student will probably recommend adding the advisor.

5-12.

 = 0.25 per hour per patient  = 6 per hour N = 15 patients

P0 =

1 N



N!

 ( N − n )!    

n =0

n



=

1 15

15!  0.25   (15 − n )!  6  n =0

n

= 0.423

 +  0.25 + 6  Lq = N −   (1 − P0 ) = 15 −   (1 − 0.423)     0.25  = 0.592 patients L = Lq + (1 − P0 ) = 0.592 + (1 − 0.423) = 1.168 patients

Wq =

Lq

( N − L) 

=

0.592 (15 − 1.168)( 0.25)

= 0.171 hr or 10.27 minutes waiting W = Wq +

1



= 0.171 +

1 6

= 0.338 hr or 20.27 minutes in system

The nurse is correct, the waiting time for a patient who calls does average about 10 minutes.

However, the nurse is currently idle a little over 40 percent of the time, thus, the supervisor cannot achieve both objectives, i.e., a reduced waiting time but no more idle time. If the hospital is quality-conscious they would add a second nurse to reduce waiting time regardless of idle time.

Perhaps the nurse could be shared with another ward or group of patients.

5-13.

Without fax operator:

 = 20/hr  = 30 / hr L = 2.0 employees Lq = 1.33 employees W = 0.10 = 6 min Wq = 0.067 = 4 min With fax operator:

 = 20/hr  = 40 / hr L = 1.0 employees Lq = 0.50 employees W = 0.05 = 3 min Wq = 0.025 = 1.5 min The system operating characteristic required to perform the decision analysis for this problem is L, the mean number in the system. Note that by reducing the mean service time by half a minute, from 2 minutes to 1.5 minutes, which is 25 percent reduction, we have reduced the mean number in the system by 50 percent, from 2 to 1 employee in the system on the average.

Alternative 1. No fax operator

Service Cost/Hour $0

2. Use fax operator

$8

Waiting Cost/Hour

( 2 )(10.20 ) = $20.40 (1)(10.20 ) = $10.20

Total Cost/Hour $20.40 $18.20

The decision analysis, using the cost figures provided, can be summarized as follows, where waiting cost includes time waiting in line and time at fax machines: If the number of working hours per day is 8, this analysis can be converted to a daily basis by multiplying all figures by 8, yielding $163.20 per day expected total waiting cost of alternative 1 and $145.60 per day expected total cost of service and waiting for alternative 2. Thus, the firm’s management should assign an operator to the fax machine, with an expected daily savings of $17.60.

5-14.

 = 90 per hour  = 60 per hour s=2

1

P0 =

 1 1  90 n  1  90 2  2 ( 60 )      +      0 n !  60   2!  60   120 − 90 

= .143

s

    2   ( 90 )( 60 )(1.67 )  L=  P0 + = (.143) + 1.50 = 3.43 2  1!( 30 ) ( s − 1)!( s −  )2

 = 3.43 − 1.50 = 1.93 

Lq = L −

W=

L

3.43 = .038 hr ( 2.29 min ) 90

Lq

1.93 = .021 hr (1.29 min ) 90



Wq =

5-15.

=



=

=

60 = 3.75 parties per hour 16

=

60 = 0.75 parties per hour 80

s=6

P0 =

1  5 1  3.75 n  1  3.75 6   6 ( 3.75 )     +     0 n !  0.75   6!  0.75   ( 6 )(.750 ) − 3.75 

L = 7.94

Lq = L −

Wq =

Lq



 3.75 = 7.94 − = 2.938  .750 =

2.938 = .783 hr or 47 minutes waiting 3.75

= 0.005

W=

L



=

7.94 = 2.11 hr or 127 minutes at the restaurant 3.75

A 47 minute waiting time may seem long, but actually restaurant customers sometimes perceive a waiting line and a reasonably long waiting time as an indicator of “quality.” Thus, in this instance lengthy service may be considered as good quality.

5-16.

 = 300  = 120 s=3

P0 =

1 3  2 1  300  1  300   360     +      0 n !  120   3!  120   60  n

= 0.045

s

    3   ( 300 )(120 )( 2.5)  L=  P0 + = ( 0.045) + 2.5 = 6.01 2  2!( 60 ) ( s − 1)!( s −  )2 Lq = L −

Wq =

Lq



 = 6.01 − 2.5 = 3.51  =

3.51 = .0117 hr ( 0.70 min ) 300

No, another wrapper is not needed.

5-17.

 = 6.5 per hour  = 4 per hour s=2

P0 =

1 1

n

2

1  6.5  1  6.5   8     +      0 n !  4   2!  4   1.5  s

= 0.103

    2   ( 6.5)( 4 )(1.625 )  L=  P0 + = ( 0.103) + 1.625 = 4.78 2  1!(1.5 ) ( s − 1)!( s −  )2

Lq = L −

Wq =

Lq



 = 4.78 − 1.625 = 3.155  =

3.155 = 0.485 hr ( 29.14 min ) 6.5

For s = 3,

P0 =

1 3  2 1  6.5  1  6.5     +   ( 2.18)  0 n !  4   3!  4  n

= 0.182

s

    3   ( 6.5)( 4 )(1.625)  L=  P0 + = (.182 ) + 1.625 = 1.96 2 2  s − 1 ! s  −  2!

5.5 ( )( ) ( ) Lq = L −

Wq =

Lq



 = 1.96 − 1.625 = 0.33  =

0.33 = 0.05 hr ( 3.1 min ) 6.5

Hire a third doctor.

5-18.

 =9  =6 For s = 2 :

P0 = 0.143 s

1     s  Pw =    P0 s !     s  −   2 1  9   ( 2 )( 6 )  =     ( 0.143) = 0.643 2!  6   ( 2 )( 6 ) − 9 

Thus, two salespeople are not enough.

For s = 3 :

P0 = 0.211 3 1  9   ( 3)( 6 )  Pw =     ( 0.211) = 0.237 3!  6   ( 3)( 6 ) − 9 

Thus, three salespeople are sufficient to meet the company policy that a customer should have to wait no more than 30 percent of the time.

5-19.

 = 35  = 15 For s = 3;

P0 = 0.064 s

3   35     35 15 ( )( )      15  0.064 + 35 = 4.47 L=  P + = ( ) 0  2! ( 3)(15 ) − 35 2 15 ( s − 1)!( s −  )2  

Lq = L −

Wq =

Lq



 35 = 4.47 − = 2.14  15 =

2.14 = .061 hr = 3.67 min 35

Three servers should be sufficient.

5-20.

 = 4 per day  =?

If total repair time is 4 days,

W=

1 ;  −

1

1 days 4.33



=

3=

Average repair time = 5-21.

1 ; 3 − 12 = 1 ; 3 = 13 ;  = 4.33  −4

8 hour  1.85 hour, or 110.8 minutes.

4.33

Arrival rate:  = 40 units per hour

Processing times: 1. Without additional employees:

1

1

= 1.2 min per unit ; Thus, 1 = 50 units per hour

2. With additional employees:

1

2

= 0.9 min per unit

Thus, 2 = 66.67 units per hour In-process inventory = number in process and waiting to be processed =

L= number in system

=



 −

1. Without additional employees:

L1 =

40 40 = = 4 units in system 50 − 40 10

2. With additional employees:

L2 =

40 40 = = 1.5 units in system 66.67 − 40 26.67

Decision analysis: The cost of in-process inventory is 1. Without additional employees: ( 4 )( $31) = $124.00 / day 2. With additional employees: (15)( $31) = $4650 / day

Difference = $124.00 − $46.50 = $77.50 / day Thus, the optimal decision is add additional employees at a cost of $52.00 per day, yielding a net expected savings of $77.50 − $52.00 = $25.50 / day.

5-22.

a.

 =5  =2 s=3

P0 =

1  1  5 0 1  5 1 1  5 2  1  5 3 ( 3)( 2 )    +   +   +    0!  2  1!  2  2!  2   3!  2  ( 3)( 2 ) − 5 5 ( 5)( 2 )   2

L=

Wq = W=

3

( 3 − 1)! ( 3)( 2 ) − 5

Lq = 6 −

= 0.045

2

( 0.045) +

5 = 6.0 2

5 = 3.5 2

3.5 = 0.70 hr, or 42 min 5

6.0 = 1.20 hr, or 72 min 5

b.  = 5

1



= 25 min

 = 2.4 s=3

P0 = 0.0982

L=

5 ( 5)( 2.4 )    2.4 

( 3 − 1)! ( 3)( 2.4 ) − 5

Lq = 3.18 − Wq =

3

2

( 0.0982 ) +

5 = 3.18 2.4

5 = 1.1 2.4

1.1 = 0.22 hr, or 13.2 min, waiting time 5

The improvement in average waiting time per truck is 42 − 13.2 = 28.8 minutes. The estimated value of this time savings is ( $750 )( 28.8) = $21, 600. Since the cost of achieving the improved service is only $18,000, the firm should implement the improved system, yielding an expected savings of $21, 600 − $18, 000 = $3, 600.

c. Alternative 1: Add a fourth loading/unloading location at the dock, yielding four locations where each location has a mean service rate of  = 2 per hour.

Alternative 2: Add extra employees and equipment at the existing three dock locations to reduce loading/ unloading times from 30 minutes to 23 minutes per truck, yielding  = 2.6 per hour.

Decision analysis: The tendency of the student will be to compare the waiting time for both alternatives. However, this is not required, since the alternatives can be evaluated using the concept of “effective service rate,” which is determined by multiplying the number of servers by the mean service rate. The purpose of this part of the problem is to introduce this concept; thus, the instructor may wish to give the student a hint before assigning this problem.

Computing the effective service rate for each alternative yields Alternative 1: ( no. servers )( mean service rate ) = ( 4 )( 2 ) = 8 trucks per hour Alternative 2: ( no. servers )( mean service rate ) = ( 3)( 2.6 ) = 7.8 trucks per hour Since the cost of each alternative is approximately equal, alternative 1, to add a fourth dock location, is superior because it increases the effective service rate to 8 trucks per hour, whereas adding extra resources to the existing dock increases the effective service rate to only 7.8 trucks per hour.

5-23.

 = 11/hr  = 13.33/ hr

(11) 2 Lq = = = 1.94 drivers 2 (  −  ) ( 2 )(13.33)(13.33 − 11) 2

Wq = 5-24.

Lq



=

1.94 = 0.176 hr, or 10.6 min 11

 = 50 / hr = .83/min  = 2/min

(.83) 2 Lq = = = 0.149 people in line 2 (  −  ) 2 ( 2 )( 3 − .83) 2

Wq =

Lq



=

0.149 = 0.68 min waiting .83

n

5-25.

 P ( n  10 ) = 1 −  Pn = 1 −     P0 n n  n

 4  = 1−     (.0762 ) = 1 − .5818 10  4.33 

P ( n  10 ) = .4182 5-26.

The appropriate model is Poisson arrivals with constant service times:

 = 149 + 30 = 179 trucks/day  24-hr day 1



= 8 min

Thus,

1  

 =   ( 60 )( 24 ) = 180 / day 8 2

 179    180   Lq = = 89.003 trucks  179  2 1 −   180  Wq =

Lq



=

89.003 = 0.4972 days 179

0.4972 ( 24 ) = 11.93 hr Yes, the port authority can assure the coal company that their trucks will not have to wait longer than 12 hours each on the average.

5-27.

7:00 A.M. to 9:00 A.M.:  = 8

 = 2.5

s must equal at least 4 for the mean effective service rate to exceed the arrival rate.

Wq = 0.30 minutes with 4 cashiers, so 4 is sufficient.

9:00 A.M. to Noon:

=4  = 2.5

s must equal at least 2.

Wq = 0.711 minutes with 2 cashiers, so 2 is sufficient.

Noon to 2:00 P.M.:

 = 14  = 2.5

s must equal at least 6 cashiers.

Wq = 0.823 minutes with 6 cashiers, so 6 is sufficient.

2:00 P.M. to 5:00 P.M.:

 =8  = 2.5

s must equal at least 4 cashiers. Wq = 0.298 minutes with 4 cashiers, so 4 is sufficient.

5-28.

=4  = 1.333 a.

s=4

P0 = 0.038 L = 4.53

Lq = 1.53 W = 1.13 hr = 67.97 min

Wq = 0.382 hr = 22.96 min b.

s=5

P0 = 0.0466 L = 3.36

Lq = .355 W = .838 hr = 50.33 min

Wq = .088 hr = 5.32 min Although the customer waiting time is reduced from 22.96 minutes to 5.32 minutes, 2.3 minutes does not seem excessive for a hair stylist; thus, the impact of adding a fifth stylist may not be significant.

5-29.

 = .00139/hr  = .08333/ hr

Servers = 1 ; N = 8 P0 = .8690

1 − P0 = .1310

−  Lq = N −   (1 − P0 ) = .02    L = Lq + (1 − P0 ) = .1476

Wq =

Lq

= 1.52 hr

1

= 13.52 hr

( N − L) 

W = Wq +



A patrol car is out of service an average of 13.52 hours when being repaired. Whether or not this is adequate repair service depends on how busy the police department is.

5-30.

 = 8.57/hr  = 3.0 / hr s=4 s = 12.0

P0  0.046 ( from Table 17.2 )

1 − P0 = .954 s

     L= P + 2 0  ( s − 1)!( s −  )

 

=

8.57  (8.57 )( 3.0 )    3.0 

( 4 − 1)! ( 4 )( 3.0 ) − 8.57 

Wq =

L



=

Lq



2

( 0.07 ) +

 8.57 = 3.98 − = 1.127  3.0

Lq = L −

W=

4

3.98 = .465 hr = 27.89 min 8.57

=

1.127 = 0.132 hr = 7.89 min 8.57

8.57 = 3.98 3.0

5-31.

The appropriate model for this problem is the finite calling population, exponential service times.

1



1 6

= 6 days

 = per day

= 2 days

 = per day

1



1 2

N =6 First, compute the following values needed in the formula calculations: n

0 1 2 3 4 5 6

   

N! ( N − n )!

720 / 720 = 1 720 /120 = 6 720 / 24 = 30 720 / 6 = 120 720 / 2 = 360 720 /1 = 720 720 /1 = 720

P0 =

N!  ( N − n )!  

1 1/3 1/9 1/27 1/81 1/243 1/729

1 N

n

N!



 ( N − n )!    

n =0

= n

n

1

6/3 = 2 30 / 9 = 3.333 120 / 27 = 4.444 360 / 81 = 4.444 720 / 243 = 2.963 720 / 729 = 0.987 Sum = 19.171

1 1 = = 0.05216 Sum 19.171

 1

1

+ + 6 Lq = N − (1 − P0 ) = 6 − 1 2 (1 − 0.05216 ) = 6 − 4 ( 0.94784) = 6 − 3.791 = 2.209  6  2.2 units expected in the queue ( waiting to be repaired ) L = Lq + (1 − P0 ) = 2.2 + 0.94784

 3.15 units expected in the system ( being repaired and waiting to be repaired ) Wq =

Lq

=

2.2

( N − L )  ( 6 − 3.15) 1

6

=

2.2 2.2 = 2.85 0.475 6

 4.63 days expected time in the queue ( waiting to be repaired ) W = Wq +

1

= 4.63 + 2

 = 6.63 days expected time in the system ( being repaired and waiting to be repaired )

Since the breakdown rate is 16 per day and the number of working days per year is 250, the

= 41.67. expected number of breakdowns per year is 250 6 The expected time at the repair shop per breakdown times the number of breakdowns per year is 6.63  41.67 = 267.27 machine-down days per year.

The annual cost of machine downtime is 276.27  50 = $13,813.50.

The net difference in the cost of the new service agreement under consideration is

$15,000 − $3,000 = $12,000.

Therefore, the company should select the new service agreement at a net expected cost savings of

$13,813.50 − $12,000 = $1,813.50.

 = 40/hr  = 12 / hr

5-32.

s=4 s = 48 P0 = 0.021

1 − P0 = .979 s

4  40      ( 40 )(12 )       12  L= P + = 2 0  ( 4 − 1)! ( 4 )(12 ) − 40  2 ( s − 1)!( s −  )  

Lq = L −

W=

Wq =

L

 Lq



=

 40 = 6.62 − = 3.289  12 6.62 = .166 hr = 9.93 min 40

=

3.289 = .082 hr = 4.93 min 40

(.021) +

40 = 6.62 12

a.  =

 40 = = .833 s 48

Idle time 1 − .833 = .167; thus, the postal workers are idle 16.7 percent of the time, which does not seem excessive. b. Wq = 4.93 minutes and Lq = 3.28 customers, neither of which seems excessive although the waiting time borders on too long. c. A customer can expect to walk in and get served without waiting approximately 2 percent (i.e., P0 = .021 ) of the time.

Overall, the system seems moderately satisfactory from a customer services perspective.

However, the post office might want to analyze the system with five stations instead of four because of the somewhat long waiting time.

5-33.

Lq = 16.35 manuscripts L = 27.78 manuscripts Wq = 2.09 weeks W = 3.47 weeks U = 0.952

5-34.

a. By testing several different numbers of servers (teams) for the multiple server model using Excel it is determined that at least 4 teams are required to be within the two week waiting period. The operating characteristics are,

Lq = 1.128 jobs L = 3.98 jobs Wq = 0.28 weeks W = 1.00 weeks b. This can be determined in two ways. First, if the average number of jobs arriving each week is 4, at $1,700 apiece they will generate $6,800 in revenue per week whereas 4 painting teams will cost $2,000 per week for a difference of $4,800 per week. Alternatively, if there are approximately 4 jobs in the system ( L = 3.98 ) over a one week period (W = 1.00 ) then that will result in $6,800 in revenue for two weeks with the 4 teams paid $2,000 for two weeks or approximately $4,800 per week.

5-35.

Regular copier

High-speed copier

W = 20 minutes

W = 4.62 minutes

 = 7 / hr.  = 10 / hr.

Wq = 14 minutes

 = 7 / hr.  = 20 / hr.

Wq = 1.62 minutes

Cost of waiting, regular copier = $3.33 per employee = 7  3.33 = $23.31 per hour Cost of waiting, hi-speed copier = 10  $0.77 = $.71 7 = $5.39 per hour The lease cost is $8 per hour higher for the highspeed copier, which would still make it more economical than the regular copier, i.e., $13.39 versus $23.31.

5-36.

Single server with finite calling population Current repair service:

 = 0.05 / day  = 1/ day

W = 2.6 days Downtime Cost = 2.6 ( 24 )( $5 ) = $312 New repair service:

 = 0.05 / day  =2

W = .77 days Downtime Cost = (.77 )( 24 )(15) = $277.20 Even at $10 per hour more, the new service is better.

5-37.

 = 6.5 cabs/hr.  = 7.05 cabs / hr.

In this finite queue system the hotel guests are treated as the servers and the cabs are the customers. The guests have a service rate of 8.5 minutes (5 minutes to arrive and 3.5 minutes to load), thus  = 7.05 cabs per hour are “served.” The cabs arrive at the cab line at the rate of 6 per hour. The system size is 6 customers or cabs. a. Wq = the average time a cab must wait for a fare

= 17.34 minutes b. Probability ( x  6 ) = .0006 5-38.

One drive-through window:

 = 10 per hour  = 15 per hour

a. No, Wq = 8 minutes b. Yes,  = 24 and Wq = 2.50 minutes c. No, 5-39.

 = 20,  = 24 and Wq = .53 minutes

 = depends on the number of registers  = customers / hr./ register

Excel is used to analyze a system with a number of single server queues. For example, if 2 registers are open then  = 140 customers (per register). If 4 registers are open,  = 20 and W = 8.12 minutes which meets the service goal of W = 12 minutes. Thus 4 registers need to be opened.

5-40.

Multiple server model:

 = 40 / hr.  = 15 / hr. s=3 Wq = 9.57 minutes At $2 per minute, a 9.57 minute wait “costs” $19.14 per customer. The cost of an employee is only $0.20 per minute thus the hotel should hire enough clerks so there is virtually no wait, i.e. 5 servers will result in almost no wait.

5-41.

Finite calling population: .7  = = .04375 calls per room per day 16

 = 2 calls / hour average utilization = 1 − P0 = 1 − .5776 = .4224 Wq = .31 hr. = 19.1 min.

W = .81 hr. = 49.1 min.

The system seems adequate. 19.1 minutes is somewhat long to wait for service, however the staff person is only busy 42.2% of the time as is thus adding another person seems excessive.

5-42.

Finite Queue Model

=

14 = 1.17 customers / month 12

5 weeks = 1.25 months

=

1 = .80 piece / month 1.25

W = 8.92 months

Po = 0.011 1- Po = 0.989 = 98.9% busy Pm = 0.323 1-Pm = .6765

5-43.

Finite calling population

 = 0.0625 / hour  = 0.67 / hour N = 10 (a) Lq = 1.177 athletes waiting

Wq = 2.33 hours waiting W = 3.82 hours in the system Utilization = .7551 The system does not seem effective. An athlete must wait 2.33 hours which seems too long, and Judith does not have much time ( P0 = .24 ) to work on her own studies. (b) By reducing the number of athletes Judith is responsible for to 6 (i.e., N = 6 ) the waiting time is reduced to .89 hour (53.4 minutes) which seems more reasonable, and Judith has 51 percent of her time free (i.e., P0 = .5109 ).

5-44.

(a)  = 45 passengers per hour  = 52 passengers per hour

Lq =

2  ( −  )

(45) 2 (52)(52 − 45) = 5.56 passengers =

Wq =

  ( −  )

45 52(52 − 45) = 7.42 minutes =

(b) A multiple server model with 3 security scans can accommodate this increased passenger traffic level.

 = 125 passengers per hour

 = 52 passengers per hour c=3

Po = .066   ( /  )c   L= P + 2 o   (c − 1)!(c  −  )   (125)(52)(125/52)3  125 = .066 + 2  52  (2)!(156 − 125)  = 5.52 passengers Lq = L −

 

= 5.52 − (125/52) = 3.121 passengers Lq Wq =



3.121 125 = 1.50 minutes

=

5-45.

A multiple server model with 4 cranes will accommodate the level of container traffic at the inland port.

 = 9 containers per hour  = 60/25 = 2.4 containers per hour c = 4 cranes Po = .007 L = 16.73 containers Lq = 12.98 containers W = 1.86 hours Wq = 1.44 hours

5-46. Constant service time model  = 36 passengers/(4.1 + 3.5 minutes) = 4.74 passengers/minute

 = 4.4 passengers/minute

Lq =

2 2 (  −  )

(4.4)2 = 2(4.74)(4.74 − 4.4) = 6.006 groups of 36 passengers waiting = (6.006)(36) = 216.21 passengers

CASE SOLUTION 5.1 – Streamlining the Refinancing Process

1. Current Process:

Most people assume that bringing in more specialists would speed up the process. But in this case, the loan passes through 7 person’s hands and the customer interacts with several different service providers.

A more efficient operation would have only one individual as the point of contact with the customer, and reduce the number of people involved with handling the customer file. Notice that each time the file is send to another person, there is a good probability it will wait. Service blueprints will differ based on assumptions and level of detail.

2. Revised Process:

Closing agent handles all tasks except for preliminary loan approval. Electronic approval and transmission of information helps speed process. Loans meeting certain criteria are automatically approved. Attorney performs tasks and verifies closing date. Having a closing date target helps to coordinate activities. Delays are still possible but with fewer handoffs are reduced considerably.

CASE SOLUTION 5.2: Herding the Patient 1. Current process

Analysis: Why does the patient register twice?

Can patient pre-register or bring in a hospital card that is scanned?

Can undressing/dressing be performed nearer to the exam room?

Can patient enter hospital nearer to the department?

Can appointment times minimize wait?

2. Revised process

Set up two exam rooms back-to-back so that technician can alternate between the two of them while patients are undressing/dressing. Out-patients can enter diagnostic imaging department directly without having to go through the hospital. Pre-registration by phone, online, or at referring Dr’s office reduces check-in time. As estimates of service time become more accurate, appintment scheduling can be tightened, thereby reducing waiting time. The servicescape should include proper signage describing services and procedures. The rooms should combine a clinical cleanliness with mirrors, comfortable chairs or benches and places to put clothing. A screen or curtain for changing is needed.

CASE SOLUTION 5.3: The College of Business Copy Center A multiple-server queuing model must be evaluated for a center with two copiers and three copiers for the normal academic year and the summer.

Normal Academic Year Two Copiers  = 7.5  =5 c=2 W = 0.457 hr = 27.42 min • In an 8-hr day, there are 60 jobs ( 8  7.5 = 60 )

Three Copiers  = 7.5  =5 c=3 W = 0.232 hr = 13.92 min • In an 8-hr day, there are 60 jobs ( 8  7.5 = 60 )

• 60 jobs  0.457 hr = 27.42 hr of total secretarial time in the college spent on copying jobs

• 60 jobs  0.232 hr = 13.92 hr of total secretarial time in the college spent on copying jobs

• $8.50 hr  27.42 hr = $233.07 in secretarial wages spent daily for copying in the college

• $8.50 hr 13.92 hr = $118.32 in secretarial wages spent daily for copying in the college

• 177 days in the normal academic year  $233.07 = $41, 253.39 per year

• 177 days in the normal academic year  $118.32 = $20.942.64 per year

Summer Months Two Copiers  = 3.75  =5 c=2

Three Copiers  = 3.75  =5 c=3

W = 0.233 hr = 13.98 min • Jobs / day = 30 • 30 jobs  0.233 hr = 6.99 hr • $8.50 hr  6.99 hr = $59.42 / day • 70 days  59.42 = $4.159.05

W = 0.204 hr = 12.24 min • Jobs / day = 30 • 30 jobs  0.204 hr = 6.12 hr • $8.50 hr  6.12 hr = $52.02 • 70 days  52.02 = $3,641.40

The current cost of wages for copying is $41, 253.39 + 4,159.05 = $45, 412.44.

The cost with three copiers is $20,942.64 + 3, 641.40 = $24,584.04.

The total annual wage savings by adding a third machine are $45, 412.44 − $24,584.00 = $20,828.40 per year.

Since a copying machine cost $36,000 and has maintenance costs of $8,000 per year, the total cost over the life of the copier will be $84,000 (with no present-value discounting). Over the same six-year period, adding a third copier would save $124,970.40 in copying wage costs, which outweighs the cost of a new copier.

However, Dr. Moore may still not be able to convince Dr. Burris. The savings in wages are not really savings to the college but are a measure of secretarial time that could be reallocated to other tasks within the departments. The college would not save any money; it would simply incur the cost of the copier. The departments could argue that other tasks the secretaries might perform instead of copying would be a more efficient use of $20,828.40 in annual wages, but Dr. Burris would probably be hard to convince with this argument.

CASE SOLUTION 5.4: Northwoods Backpacker There are four system configurations to be considered, as follows.

1. 5-day, 8-hour per day service 2. 7-day, 8-hour per day service 3. 5-day, 16-hour per day service 4. 7-day, 16-hour per day service In each case the first step is to determine the number of servers that are required to make the system feasible, i.e., c  . Remember, the current system has 5 operators (servers), and,  = 60 / 3.6 = 16.67 customers per hour.

5-day, 8-hour service:  = 175,  = 16.67; c   /  or c  175 /16.67 = 10.49. Thus, at least 11 total operators are required for this (the current) system to be feasible. Since the current physical facility can only accommodate a maximum of 10 work stations, this alternative is eliminated.

7-day, 8-hour service:  = 125,  = 16.67; c   /  or c  125 /16.67 = 7.49. Thus, at least 8 operators are required for this system to be feasible.

5-day, 16-hour service:  = 875,  = 16.67; c   /  or c  875 /16.67 = 5.24. Thus, at least 6 operators are required for this system to be feasible.

7-day, 16-hour service:  = 625,  = 16.67; c   /  or c  625 /16.67 = 3.74. Thus, at least 4 operators are required for this system to be feasible.

Therefore, only the first configuration (the current one) is not feasible and is eliminated.

Next the costs of the remaining 3 alternatives are evaluated.

7-day, 8-hour service

 = 125,  = 16.67 Cost for 7 day service = $3, 600 Cost per extra operator = $3,800 Recall that at least 8 operators are required for this configuration to be feasible. Thus, starting at this point we must compute the waiting times for different numbers of operators until the goal of one-half minute waiting time is achieved.

8 operators:

Wq = 5.76 minutes

9 operators:

Wq = 1.20 minutes

10 operators: Wq = 0.42 minutes

( Probability of waiting = 0.18)

Since 5 extra operators are required to reach the waiting time goal the cost of this alternative is

$3, 600 + ( 5)( 3,800 ) = $22, 600.

5-day, 16-hour service

 = 87.5,  = 16.67 Cost for 16 hour service = $11,500 Cost per extra operator = $4, 700 At least 6 operators are required for this configuration to be feasible.

6 operators:

Wq = 3.24 minutes

7 operators:

Wq = 0.78 minutes

8 operators:

Wq = 0.27 minutes

( Probability of waiting = 0.21) Since 3 extra operators are required to reach the waiting time goal the cost of this alternative is $11,500 + ( 3)( 4, 700 ) = $25, 600.

7-day, 16-hour service

 = 62.5,  = 16.67 Cost for 16-hour service = $11,500 Cost for 7-day service = $7, 200 Cost per extra operator = $6,300 At least 4 operators are required for this configuration to be feasible, however, since 5 operator stations already exist, the starting point is 5 operators.

5 operators: Wq = 1.32 minutes

6 operators:

Wq = 0.36 minutes

( Probability of waiting = 0.10 ) Since only I extra operator is required to reach the waiting time goal the cost of this alternative is $11,500 + 7, 200 (1)( 6,300 ) = $25, 000.

The 7-day, 8-hour service configuration has the lowest cost. However, all three alternatives are very close according to cost. All three also meet the goal of a customer getting immediate service at least 80 percent of the time. Thus, other factors may be taken into consideration. For example, both of the 16-hour service alternatives might be more convenient for customers who work during the day.

CHAPTER 6 - Processes and Technology Answers to Questions 6-1.

In creating a process strategy, a firm must define its goals for capital intensity (mix of capital and labor), process flexibility (ease of adjusting to changes in demand, technology, products or services, and resources), vertical integration (extent to which it produces the inputs and controls the output of each stage of production), and customer involvement (the role of the customer in the production process).

Student responses will vary. The following questions will start students thinking: (1) capital intensity— does the project involve use of computers, overheads, special presentation requirements, new books, or software? (2) process flexibility—can the project be approached from different angles or must everyone follow the same procedure? (3) vertical integration—does data need to be collected as well as analyzed? (4) customer involvement—does the professor meet with the project team to go over their progress?

6-2.

Six factors affecting the make-or-buy decision are: (1) Cost—Would it be cheaper to make the item or buy it, to perform the service in-house or subcontract it out? (2) Capacity—Companies that are operating at less than full capacity usually opt to make components rather than buy them. (3) Quality—The capability to consistently provide quality parts is certainly an important consideration, whether the components are made in-house or come from a supplier. (4) Speed—Sometimes components are purchased because a supplier can provide goods in shorter periods of time than the manufacturer. (5) Reliability—Suppliers need to be reliable both in terms of the quality and timing of parts that are supplied. (6) Expertise—Companies that are especially good at making or designing certain items may want to keep control over their production and produce their components in-house.

The sourcing continuum ranges from vertical integration, to joint ventures with suppliers, to strategic alliances, to single contracts. Vertical integration involves making all of the items for a product in-house, from the raw material to finished product. With joint ventures, such as Ford-Mazda, a company owns equity in the supplier’s business. Strategic alliances involve long-term relationships with trusted suppliers. single contracts views each purchasing decision as separate from the next.

6-3.

Projects involve large sums of money and last a considerable length of time. They represent one-of-a-kind production for an individual customer. Cutting edge technology, project teams, and close customer contact make project work exciting. But projects can also be risky with their large investment in resources, huge swings in resource requirements, limited learning curves, and dependence on a small customer base.

Batch production is used to process many different jobs through the production system at the same time in groups (or batches). Products are made to customer order, volume (in terms of customer order size) is low, and demand fluctuates. To allow for a variety of items to be produced, the equipment tends to be general purpose and the workers highly skilled. Batch production is flexible but not very efficient.

Mass production is used to produce large volumes of a standard product for a “mass” market. Product demand is stable and product volume is high. Because of the stability and size of demand, the production system can afford to dedicate equipment to the production of a particular product. Thus, this type of system tends to be capital intensive, with specialized equipment and limited labor skills. Mass production is efficient but not very flexible.

Continuous production is used for very high volume commodity products that are very standardized. The system is highly automated (the worker’s role is to monitor the equipment) and is typically in operation continuously 24 hours a day. The output is also continuous, not discrete—meaning individual units are measured, rather than counted. Refined oil, treated water, paints, chemicals and foodstuffs are produced by continuous production. Continuous production is highly efficient but very expensive and inflexible.

6-4.

Equipment costs, facilities cost, material cost, labor cost. Break-even analysis is especially useful when evaluating different degrees of automation in process selection. More automated processes have higher fixed costs, but lower variable costs. The selection of the “best” process depends on the anticipated volume of demand for the product, and the trade-offs between fixed and variable costs.

6-5.

a. An assembly chart is a schematic diagram that shows the relationship of each component part to its parent assembly, the grouping of parts that make up a subassembly, and the overall sequence of assembly. b. An operations chart adds a series of operations to every item listed in the assembly chart that describe how each item is to be fabricated. c. A process flow chart classifies processes into the broad categories of operations, inspection, transportation, storage, or delay. The time required to perform each process and the distance between processes may also be included. With this data, the chart can be used to improve the efficiency of operations.

6-6.

The output of process planning is a “work package” of several reports (i.e., routing sheets, operation sheets, etc.) that prescribe how and where to manufacture an item. For mass production and continuous production, a process plan may be developed only once when the assembly line is set up or the process plant built. For batch production, a process plan must be developed for every job that enters the shop or part that is produced. For projects, process plans are usually associated with each activity in the project network.

6-7.

Process innovation involves redesigning a critical business process from scratch. The initial step is to establish the goals and specifications for process performance. Data from the existing process is used as a baseline to which benchmarking data on best industry practices, customer requirements data, and strategic directives are compared. If redesign is necessary, a project team is chartered. The team creates a high level process map by focusing on the performance goal (stated in customer terms) and working backwards through the desired output, subprocesses and initial input requirements. After the general concept of redesign is agreed upon, a detailed process map is prepared for each subprocess or block in the high level map. The detailed map includes key performance measures and guidelines for the allocation of resources and work methods. After the detailed map has been validated, a pilot study is conducted and the process modified if necessary. A successful pilot study leads to full scale implementation. The process innovation project is complete when the new process consistently reaches its objective.

6-8.

a. Z belongs below the line because it is constructed of straight, not curved, lines. b. S and S come next, for six and seven (O T T F F are the first letters in one, two, three, four, and five.)

Acting under the same mind set as part a, most people will say E E because these letters are composed of four straight lines and O T T F F had no straight lines, two and three straight lines respectively. c. The dots can be connected if you do not feel compelled to stop at all the dots, or complete a closed figure.

Here is the arrangement:

d. There is three mistakes in this sentence.

6-9.

Factors often overlooked in the financial analysis of technology include: (1) The add-ons to purchase price, such as the cost of special tools and fixtures, installation, and engineering or programming adjustments (i.e., debugging). (2) Operating expenses other than direct labor, such as indirect labor (e.g., for programming, setups, material handling, training, etc.), power and utilities, supplies, tooling, property taxes and insurance, and maintenance. (3) Annual savings from a reduction in the amount of material or machine time used, a reduction in rework costs, or reduced safety costs. (4) Revenue enhancements such as expanded capacity,

improvements in product quality, price reductions due to decreased costs, and more rapid, flexible or dependable delivery. (5) The opportunity cost of not investing in new equipment when upcoming technology will make the equipment obsolete. (6) The synergistic benefits of a well-designed technology plan.

6-10. E-manufacturing (eM) involves the automated production and management of products, processes, and manufacture through information technology. For products, eM includes CAD, CAE, group technology, product life cycle management, collaborative product commerce, product definition and product data management. For processes, eM involves CAD/CAM and CAPP systems, STEP protocol, sourcing and eprocurement. For manufacture, technologies such as CNC machines, flexible manufacturing system, robots, automated material handling systems and process control systems are coordinated and controlled electronically in cells and centers within individual plants, as well as among multiple factories and suppliers around the world. The IT systems that support these endeavors include B2C and B2B e-commerce, bar codes, RFID, EDI and XML. Collaboration occurs through the Internet, company intranets, extranets and satellites. ERP, SCM, and CRM systems provide the framework upon which these interactions are managed.

Advances in DSS, ES and AI support managerial decision making. Collaborative manufacturing involves sharing real-time data with trading partners and customers, and making collaborative decisions about production based on that data.

6-11. Student answers will vary. Some common resources for outsourcing information are Business Week, Fortune, and The Outsourcing Institute. Several additional weblinks are provided on the student companion website.

6-12. A flowchart for the process of drawing a flowchart.

Solutions to Problems 6-1.

Students will come up with a variety of answers for this question. See if they can relate their ideas for improvement directly to the process flow chart. In many cases, the flow charts do not show everything the student knows about the process. A process flowchart worksheet is posted on the text website.

6-2.

An Operations Chart for Making Pancakes Oper.

No. 1 2

3

4 5

6 7 8

9 10 11

12

6-3.

Description Heat electric skillet to 375  Gather tools and pancake mix Read instructions and choose no. of pancakes to make Gather ingredients: milk, eggs, oil Combine pancake mix, milk, oil, and eggs. Stir until large lumps disappear Pour 1/4 cup batter onto hot skillet Cook Turn pancake when bubbles begin to form on surface Cook Remove and Serve Repeat from step 6 until batter is Gone Unplug skillet and allow to cool

Equipment/ Tools Electric skillet

Time 5 mins.

Bowl, mixing Spoon, measuring cup, spatula Pancake mix box

2 mins.

—

2 mins.

2 mins.

Bowl, spoon, Measuring cup

5 mins.

Measuring cup, Skillet Skillet Spatula

1/2 min.

Skillet Spatula, plate

1 1/2 mins.

1 min.

—

Skillet

c f = 5, 000

cv = $5 CD p = $15 CD cf 5, 000 5, 000 v= = = = 500 p − cv 15 − 5 10 Mikey needs 500 CDs or $7,500 in sales to break even.

1 1/2 mins.

1/2 min.

—

15 mins.

6-4.

a.

c f = $9, 000

cv = $2 CD p = $15 CD

v= b.

9, 000 = 692.31 15 − 2

5, 000 + 5 x = 9, 000 + 2 x 3x = 4, 000

x = 1,333 CDs If the sales volume is expected to exceed 1,333 units, choose the classier studio. Otherwise, choose the first studio.

6-5.

c f = $25, 000

cv = $10 doll p = $50 doll

v= 6-6.

a.

25, 000 25, 000 = = 625 dolls 50 − 10 40

c f = $5, 000

cv = $15 doll p = $50/doll

= b.

25,000 25,000 = = 142.85 or 143 dolls 50 − 15 35

25, 000 + 10 x = 5, 000 + 15 x

20, 000 = 5x 4, 000 = x If demand is expected to be less than 4,000 dolls, choose the new process.

6-7.

a.

c f = $15, 400 cv = $0.25 shirt

p = $1.10 shirt

v=

15, 400 15, 400 = = 18,117.65 or 18,118 1.10 − 0.25 0.85

David needs to press 18,118 shirts to break even. b. At 50 shirts a day, it would take David 18,118 50 = 362.36 or 363 days to break even.

At 200 shirts a day, David would break even in 18,118 200 = 90.59 or 91 days. c.

c f = 15, 400

cv = 0.25 p = 0.99

v=

15, 400 15, 400 = = 20,810.81 shirts 0.99 − 0.25 0.74 20,811 = 83.24 days 250

At 250 shirts a day, David would break even in 84 days. He should purchase the press and lower his price.

6-8.

Make c f = $200

Buy c f = $75

cv = $0.30 slice% cv = $1.125 slice 200 + 0.30 x = 75 + 1.125 x 125 = 0.825x 151.51 = x If more than 151 students want pizza for lunch, the cafeteria should make its own. Otherwise, they should buy from Pizza Den.

6-9.

Supplier vs. Process A

$20 x = $8,000 + $10 x $10 x = $8, 000 x = 800 Process A vs. Process B

$8, 000 + $10 x = $20, 000 + $4 x $10 x = $12, 000 + $4 x $6 x = $12, 000 x = 2, 000 Use the supplier when demand is less than 800 keyboards.

Use process B when demand is over 2,000 keyboards.

Use Process A otherwise.

6-10. Subcontracting vs. Small Facility

$60 x = $200, 000 + $40 x $20 x = $200, 000 x = 10, 000 Small Facility vs. Larger Facility

$200, 000 + $40 x = $600, 000 + $20 x

$40 x = $400, 000 + $20 x $20 x = $400, 000

x = 20,000 Subcontract if demand is less than 10,000 items.

Use the larger facility when over 20,000 item are needed.

Otherwise, use the smaller facility.

6-11. Occasional vs. Frequent

0.50 x = 55 + ( x − 70 ) 0.33 0.50 x = 31.90 + 0.33x 0.17 x = 31.90 x = 187.65 min Switch to frequent-user plan when airtime exceeds 187 min a month.

Frequent vs. Executive

55 + ( x − 70 ) 0.33 = 75 + ( x − 100 ) 0.25 31.90 + 0.33x = 50 + 0.25 x 0.08 x = 18.10

x = 226.25 min Switch to executive user plan when airtime exceeds 226 min a month.

6-12. a. Make: 25, 000 + 40 ( 300 ) = 25, 000 + 12, 000 = $37, 000 Buy: 50 ( 300 ) = $15, 000* Buy the part. b.

$50 (100 ) + 45 ( 300 − 100 ) = 5, 000 + 9, 000 = $14, 000* Buy from the New Supplier.

c. Make: 25, 000 + 40 ( 2, 000 ) = $105, 000 Old: 50 ( 2, 000 ) = 100, 000 New: 5, 000 + 45 ( 2000 − 100 ) = $90,500* Same decision: Buy from the New Supplier.

Make: 25, 000 + 40 ( 5, 000 ) = $225, 000* Old: 50 ( 5, 000 ) = 250, 000

New: 50 (100 ) + 45 ( 5, 000 − 100 ) = $225,500 Decision changes: Make. d. Make vs. New Supplier

25, 000 + 40 x = 5, 000 + 45 ( x − 100 ) 25,000 + 40 x = 5,000 + 45 x − 4,500 25, 000 + 40 x = 500 + 45 x 24,500 = 5x 4,900 = x Old vs. New Supplier

50 x = 5, 000 + 45 ( x − 100 ) 50 x = 5, 000 + 45 x − 4500 5 x = 500 x = 100 •

If demand for component  100, choose old supplier.

•

If demand for component  100 but  4,900, choose new supplier.

•

If demand for component  4,900, make it.

6-13. Provider 1 vs. Provider 3

300 = 200 + 10x 100 = 10x 10 = x Provider 2 vs. Provider 3

100 + 30 x = 200 + 10 x 20 x = 100

x=5

Choose provider 1 if more than 10 claims are made per month, provider 2 if fewer than 5 claims are made per month, and provider 3 otherwise.

6-14. Old System vs. System 1

10,000 + 25 x = 40,000 + 10 x

15 x = 30, 000 x = 2, 000 System 1 vs. System 2

40,000 + 10 x = 100,000 + 5 x

5 x = 60, 000 x = 12, 000 Purchase system 1 when demand exceeds 2,000; Purchase system 2 when demand exceeds 12,000.

6-15. Labor vs. Automated

10, 000 + 14 x = 50, 000 + 8 x

6 x = 40, 000 x = 6, 666.66 Labor vs. Fully Automated

10,000 + 14 x = 300,000 + 2 x

12 x = 290, 000 x = 24,166.66 Automated vs. Fully Automated

50,000 + 8 x = 300,000 + 2 x

6 x = 250,000

x = 41, 666.66 Choose a labor-intensive process if demand is 6,666 or less.

Choose an automated process if demand is between 6,666 and 41,666.

Choose a fully automated process if demand is greater than or equal to 41,666.

6-16. B. B. Lean vs. Sea’s End

400 + 6 x = 460 + 4 x 2 x = 60

x = 30 Spoogle’s vs. Sea’s End

500 + 3x = 460 + 4 x 40 = x If the number of items to be ordered is 30 or less, choose B. B. Lean. If the number of items is between 30 and 40, choose Sea’s End. If the number of items is 40 or greater, choose Spoogle’s.

6-17. Arrange the alternatives from the lowest fixed cost to the highest fixed cost.

Monitor A vs. Monitor B

$700, 000 + $250 x = $1, 000, 000 + $125 x

$125 x = $300, 000 x = 2400 units Monitor B vs. Monitor C

$1, 000, 000 + $125x = $1,500, 000 + $100 x $25 x = $500, 000 x = 20, 000 units Recommendation: If the demand for monitors is less than 2400 units, choose monitor A. If demand is greater than 20,000 units, choose monitor C. Otherwise, choose B. 6-18. This problem involves some thought before running the numbers. Examining the variable costs, making the pies will be preferable at high levels of demand, and buying from the local bakery will be preferable at very low levels of demand. The regional bakery falls in between the two. To find the exact quantities where each alternative is preferred, the regional bakery needs to be compared to both the local and in-house bakeries.

The minimum purchase for the regional bakery does not come into play until the regional and local bakeries are compared.

Regional bakery vs. In-house

$3 x = $80 + $1x $2 x = $80

x = 40 pies

We know that the regional bakery will always be preferred to the local bakery above 25 pies, so we simply equate the cost of the local bakery with the regional bakery cost of 25 pies.

Local bakery vs. Regional bakery

$4 x = $3 ( 25) x = 18.75 or 19 pies Recommendation: Make the pies in-house if demand is more than 40 pies. Buy the pies from the local bakery if the demand is less than 19 pies. Otherwise, purchase the pies from the regional bakery. A graph of the problem appears below.

6-19. Almost Free vs. Best Movies

$40 + $5 x = $65 + $4 x x = 25 Best Movies vs. Choice Cinema

$65 + $4 x = $100 + $3x x = 35 Recommendation: Choose Almost Free Flicks if Keisha orders less than 25 movies. Choose Choice Cinema is Keisha orders more than 35 movies. Otherwise, choose Best Movies.

6-20.

6-21.

CASE SOLUTION 6.1: A Manager’s Woes Current Process Flow Chart Date:

9/25/2011

Location:

Zelmart

Analyst:

K. Peschken

Process:

Electronics check-out

Revised Process Flow Chart Date:

9/25/2011

Location:

Zelmart

Analyst:

K. Peschken

Process:

Revised electronics check-out

CASE SOLUTION 6.2 – Wrong Meds, Again!

1

1.

A process map of the medication list with possible failure points follows.

2.

Often, managers will blame individuals for process errors, even though the same errors are made with new or different staff. While individuals can and do make mistakes, a properly designed process can eliminate many opportunities for error and should make it easier to do the job correctly. This view is reflected in a recent article appearing in Pharmacy Times1: “When analyzing medication errors, the trend in the past has been to place possible negligence on the health care provider. Yet, today we realize that many medication

Greta Pelegrin, “Medication Errors in Hospitals: An Analysis,” Pharmacy Times, Oct. 2004, p. 38.

errors result from inadequate systems leading to serious mistakes by providers. Errors can occur during any stage of the medication process. Rather than upholding a punitive approach, however, now the focus is to concentrate on “prevention” and to devise strategies to minimize errors and adverse medication events.

Although providers are still held to a high standard and must be responsible for the decisions they make, placing blame on an individual seldom leads to positive outcomes.”

3.

The case does not explicitly state how data is stored and transmitted, so student assumptions will vary. In general, the hospital is the most sophisticated, followed by the ambulance service, then the nursing home. This case is based on an actual situation with the following scenario.

The nursing home records data manually on medications given each day on a printout provided by an outside pharmacy. The pharmacy can make daily deliveries if needed, but generally fills prescriptions for a month’s worth of meds at a time. The electronic file of each patient’s medication list is updated at irregular intervals. All transmittals of patient records are either faxed or hand delivered to physicians by patient families.

Electronic transmittal is problematic because of the manual recording of data.

The ambulance has audio contact with the local hospital but no Internet access. Data is recorded manually, relayed by phone, and in some cases recorded in a PDA type of device to be uploaded at a later time to a central system.

The emergency room maintains both a paper and electronic copy of patient information. Patient information is entered by the admitting clerk and the nurses at the nurse’s station. On the hospital floor, all data is recorded and transmitted electronically on PCs rolled into each patient room. The information is uploaded to the central data base nightly.

It would be easier if all parties used a common electronic mode of communication, or if compatible systems could be used. For example, bar codes on patient wrist bands and medication are common. Perhaps a computer chip (RFID-type tag) could be used to record information about the patient and the dispensing of medication.

The hospital and ambulance service could read the tag and obtain up-to-date information. While this solution would be ideal, it is expensive and could still result in technology glitches. Other suggestions include: involvement of a pharmacist and patient (or patient’s POA) in verifying medication; clear dating of data entry; standardized formats for data entry and transmission of data; automatic check on data entry and dosages by weight of patient or patient diagnosis; more rapid updating of system data; verification of data readiness before releasing patient from emergency room to hospital room.

4.

Melanie’s reaction is quite common -- take care of the incident and the problem is solved. But the root cause of the problem remains and errors will continue in the future if the process is not changed. Melanie seems unduly defensive and dismissive (perhaps she is worried about a lawsuit). Again, this is an industry wide problem.

Hospitals nationwide are exploring and developing systems for the purpose of reducing medication administration errors. The Institute for Clinical Systems Improvement (www.icsi.org) recommends the following, none of which can take place in a punitive or dismissive environment: (a) become familiar with the actual errors — how, when, and why they were committed; (b) establish a “non-punitive” approach to encourage the reporting of errors or “near-miss errors”; (c) identify areas of concentrated errors; (d) standardize steps in the identification of errors; and (e) select the proper technology to correct these errors.

SUPPLEMENT 7 Operational Decision-Making Tools: Facility Location Models Answers to Questions S7-1. For manufacturing and service companies it is often important to be close to materials and have adequate distribution and supply routes and modes of transport. Facilities are usually smaller and less costly for service-related businesses and require a sufficient customer population in the vicinity. Although access to customers is a critical location factor for service businesses, for a manufacturing facility important factors include labor climate and wage rates, energy availability and cost, and proximity to other company facilities.

S7-2. Service operations tend to locate near malls and interstate interchanges, where there is heavy customer flow, whereas manufacturing facilities locate in Sunbelt regions, often in more rural areas or in smaller towns.

S7-3. Factors include lower wage rates, less organized labor, warmer climate, and aggressive marketing tactics by Sunbelt communities.

S7-4. Lower wage rates, fewer government regulations, especially related to the environment, and new markets are some of the positive features. Negative features include lack of suppliers, poor modes of distribution and transportation, lack of skilled labor, and lack of markets.

S7-5. Customer traffic, access to distribution centers, a labor pool, customer demographics (e.g., income level, youth population), parking or foot traffic, location environment/neighborhood, etc., all are factors.

S7-6. Each of these responses will depend on the community in which the students live.

S7-7. A labor force that has a strong work ethic, is self-reliant and neighborly. They also have quality health services, low crime rates, a solid infrastructure, quality education and open spaces to expand.

S7-8. This answer depends on the facility the student selects. The student should mention a selection of location factors relevant to the type of business it is.

S7-9. This answer depends on the 3 sites selected, although proximity to customer markets, traffic, site size, roads, and cost should be mentioned.

S7-10. This answer depends on the sites, however, obvious location factors include available space, proximity to student housing concentrations, student traffic, parking, terrain, infrastructure, etc.

S7-11. This answer depends on the student selections.

Solutions to Problems S7-1.

Location Factor College proximity Median income Vehicle traffic Mall quality and size Other shopping Total score

Mall 1 12.00 18.75 15.00 9.00 8.00 62.75

Score Mall 2 Mall 3 18.00 27.00 20.00 18.75 22.50 19.75 10.00 8.00 3.00 6.00 73.50 79.50

Mall 4 15.00 17.50 18.50 9.00 7.00 67.00

Mall 3 has the highest location factor score.

S7-2.

Location Factors Political stability Economic growth Port facilities Container support Land and construction cost Transportation/distribution Duties and tariffs Trade regulations Airline service Area roads Total Score

Shanghai 12.50 16.20 9.00 5.00 7.20 4.00 4.90 3.50 1.20 1.20 64.70

Weighted Scores Hong Kong Singapore 20.00 22.50 14.40 13.50 14.25 13.50 8.00 9.00 4.00 2.40 6.40 5.60 6.30 6.30 4.75 4.75 1.60 1.40 1.40 1.60 81.10 80.55

Select Hong Kong.

S7-3.

Location Factors Proximity to housing Student traffic Parking availability Plot size, terrain Infrastructure Off-campus accessibility Proximity to dining Visitor traffic Landscape/aesthetics Total Score

South 16.10 16.50 14.40 9.60 5.00 5.40 3.00 2.80 1.00 73.80

Select the West A campus site.

S7-4.

Weighted Scores West A West B 20.70 14.95 17.60 13.20 9.60 12.80 8.40 10.80 6.00 4.00 4.20 4.20 4.00 3.50 3.20 2.60 0.80 1.20 74.50 67.25

East 18.40 18.70 11.20 9.00 7.80 4.20 4.00 2.20 1.40 73.90

Location Factors Work ethic Quality of life Labor laws/unionization Infrastructure Education Labor skill and education Cost of living Taxes Incentive package Government regulations Environmental regulations Transportation Space for expansion Urban proximity Total Score

Abbeton 14.40 12.00 10.80 6.00 6.40 5.25 4.20 3.25 4.50 1.20 1.95 2.70 1.80 1.20 75.65

Weighted Scores Bayside Cane Creek 16.20 12.60 13.60 15.20 7.20 7.20 5.00 6.00 7.20 6.80 4.55 4.90 4.80 5.10 3.50 2.75 4.75 3.50 1.50 1.95 1.80 2.10 2.40 2.85 1.90 1.80 1.80 1.40 76.20 74.15

Dunnville 13.50 14.40 8.40 7.00 7.60 5.60 4.50 3.00 4.00 1.65 2.40 2.40 1.80 1.60 77.85

Select Dunnville.

S7-5.

S ( C ) = 76 S ( E ) = 75

S ( D ) = 74 S ( B ) = 70 S ( A ) = 69 S7-6. S ( Tysons ) = 85.27

S ( Fairfax ) = 80.94 S ( Alexandria ) = 80.55

S ( Manassas ) = 77.00 S ( Dupont ) = 76.39

S7-7.

Score Location Factors Elderly population Income level Land availability Average age Public transportation Crime rate Total score Select Dowling

Ashcroft 75 65 90 80

Brainerd 80 75 70 70

Caffee 65 90 90 80

Dowling 75 85 80 75

95 95 77.5

55 70 75.5

75 85 74.25

95 90 78.75

S7—8.

Location Factors Soccer Interest Entertainment competition Playing facility Population (age 15 to 40) Media market Income level (age 15 to 40) Tax incentives Airline transportation Cultural diversity General sports interest Local government support Community support Total score

Atlanta 70 33 50 100 100 80 20 100 100 75 30 20 65.0

Weighted Scores Birmingham Charlotte 40 75 45 65 70 65 70 40 70 80 95 60 35 55.8

60 70 90 95 80 75 95 90 85 75 50 76.4

Durham 90 95 85 25 40 60 60 65 75 65 90 75 74.3

Select Charlotte

S7-9

Solution depends on the weights the student assigned to the location factors.

S7-10. Solution depends on the weights the student assigned to the location factors.

S7-11.

S7-12. x =

y=

(14 )(17, 000 ) + ( 20 )(12, 000 ) + (30 )(9, 000 ) 38, 000

( 30 )(17, 000 ) + (8)(12, 000 ) + (14 )(9, 000 ) 38, 000

= 19.68

= 19.26

S7-13. a.

1 x1 = 100 y1 = 300 w1 = 35

x=

=

2 x2 = 210 y2 = 180 w2 = 24

3 x3 = 250 y3 = 400 w3 = 15

4 x4 = 300 y4 = 150 w4 = 19

5 x5 = 400 y5 = 200 w5 = 38

 xiWi  Wi

(100)( 35) + ( 210)( 24 ) + ( 250)(15) + (300 )(19 ) + ( 400 )(38) 35 + 24 + 15 + 19 + 38

x = 253.36 y=

=

 yiWi  Wi

( 300)(35) + (180)( 24 ) + ( 400 )(15) + (150 )(19 ) + ( 200 )(38)

y = 238.70

131

S7-14. Site A:

d1 =

(100 − 350)2 + (300 − 300)2 = 250.00

d2 =

( 210 − 350)2 + (180 − 300)2 = 184.39

d3 =

( 250 − 350)2 + ( 400 − 300)2 = 141.4

d4 =

( 300 − 350 )2 + (150 − 300 )2 = 158.11

d5 =

( 400 − 350)2 + ( 200 − 300)2 = 111.80

Site B:

d1 = 70.71 d2 = 92.20 d3 = 180.28 d4 = 180.28 d5 = 254.95

Site C:

d1 = 150.00 d2 = 126.49 d3 = 100.00 d4 = 158.11 d5 = 180.28

LD ( A) = 35 ( 250 ) + 24 (184.39 ) + 15 (141.4 ) + 19 (158.11) + 38 (111.8) = 22,549.4

LD ( B ) = 35 ( 70.71) + 24 ( 92.9 ) + 15 (180.28 ) + 19 (180.28 ) + 38 ( 254.95 ) = 20, 505.1 LD ( C ) = 35 (150 ) + 24 (126.49 ) + 15 (100 ) + 19 (158.11) + 38 (180.28 ) = 19, 640.5 Site C has the lowest load-distance value and would minimize transportation costs.

Site C is a little more centrally located as the site determined in Problem S7-8; it is closer to site 3, which has the lowest annual truck shipments.

S7-15. a. 1. Four Corners x1 = 30

y1 = 60 w1 = 8.5 3. Russellville x3 = 10

y3 = 70 w3 = 7.3

2. Whitesburg x2 = 50

y2 = 40 w2 = 6.1 4. Whistle Stop x4 = 40

y4 = 30 w4 = 5.9

x=

 xiWi ( 30 )( 8.5) + ( 50 )( 6.1) + (10 )( 7.3) + ( 40 )( 5.9 ) = = 31.3  Wi 8.5 + 6.1 + 7.3 + 5.9

y=

 yiWi ( 60 )(8.5 ) + ( 40 )( 6.1) + ( 70 )( 7.3) + ( 30 )( 5.9 ) = 51.9 =  Wi 27.8

b.

S7-16. x = 1, 665.4 y = 1,562.9

S7-17. x = 21.0 y = 16.2

S7-18. a. x =

(15)(160) + ( 42 )(90 ) + (88)(105) + (125)(35) 160 + 90 + 105 + 35 + 60 + 75

+

(135)( 60 ) + (180 )(75 ) 160 + 90 + 105 + 35 + 60 + 75

= 78.8

y=

(85)(160) + (145)(90 ) + (145)(105) + (140 )(35) 160 + 90 + 105 + 35 + 60 + 75

+

(125)(60 ) + (18 )(75 ) 160 + 90 + 105 + 35 + 60 + 75

= 106.0 b.

The closest town to the grid coordinates is Seagrove. However, Ashboro is probably a better location at coordinates (76,116) since it is at the confluence of several highways that lead to Wilmington, Raleigh, Charlotte and Winston-Salem.

S7-19. From a map of North Carolina, Fayetteville is at (145, 72) and Statesville is at (10,125).

Fayetteville:

d ( Charlotte ) = 130.7 d ( W − S) = 126.2

d ( Greensboro ) = 92.6 d ( Durham ) = 70.9 d ( Raleigh ) = 53.9

d ( Wilmington ) = 64.4 d ( Charlotte ) = 40.3

Statesville:

d ( W − S) = 37.7 d ( Greensboro ) = 80.5 d ( Durham ) = 116.0 d ( Raleigh ) = 125 d ( Wilmington ) = 200.9 LD ( Fayetteville ) = (160 )(130.7 ) + ( 90 )(126.2 ) + (105 )( 92.6 )

+ ( 35 )( 70.9 ) + ( 60 )( 53.9 ) + ( 75 )( 64.4 ) = 52,533.9

LD ( Statesville ) = (160 )( 40.3) + ( 90 )( 37.7 ) + (105 )(80.5 ) + ( 35 )(116 ) + ( 60 )(125 ) + ( 75 )( 200.9 ) = 44,925.4 The best site, according to the load-distance technique is Statesville.

S7-20. x = 265.23 y = 363.84 S7-21. LD ( Site 1) = 123,859.5

LD ( Site 2 ) = 120,995.3 LD ( Site 3) = 115, 007.4 Site 3 is the best site.

S7-22.

x = 1559.8 y = 1766.8

S7-23.

Site C is the best

LD(Site A) = 298,169 LD(Site B) = 273, 731 LD(Site C) = 247,556

CASE S7.1: Selecting a European Distribution Center Site for American International Automotive Industries Students will need to use a map of Europe to determine location coordinates. Bern, Switzerland was used as a reference point (0, 0) to set up coordinates to plot the coordinates of the seven plant sites and five potential distribution center sites.

Plant sites (x, y) Vienna (300, 60) Leipzig (180, 225) Budapest (390, 50) Prague (240, 160) Krakow (400, 170) Munich (150, 60) Frankfurt (40, 160)

Load 160 100 180 210 90 120 50

Distribution center sites (x, y) Dresden (225, 225) Lodz (420, 250) Hamburg (90, 340) Gdansk (370, 360) Frankfurt (40, 160)

Using the load-distance technique:

LD ( Dresden ) = (160 )(181.2 ) + (100 )( 45) + (180 )( 240.5) + ( 210 )( 66.7 ) + ( 90 )(183.4 ) + (120 )(181.2 ) + ( 50 )(196.1) = 138,865.1

LD ( Lodz ) = (160 )( 224.7 ) + (100 )( 241.3) + (180 )( 202.2 ) + ( 210 )( 201.2 ) + ( 90 )(82.5 ) + (120 )( 330.2 ) + ( 50 )( 390.5 ) = 205,315.2 LD ( Hamburg ) = (160 )( 350 ) + (100 )(146.0 ) + (180 )( 417.3 ) + ( 210 )( 234.3) + ( 90 )( 353.6 ) + (120 )( 286.4 ) + ( 50 )(186.8 ) = 270, 436.5 LD ( Gdansk ) = (160 )( 308.1) + (100 )( 233.1) + (180 )( 310.6 ) + ( 210 )( 238.5) + ( 90 )(192.4 ) + (120 )( 372.0 )

+ ( 50 )( 385.9 ) = 259,854.1 LD ( Frankfurt ) = (160 )( 278.6 ) + (100 )(154.4 ) + (180 )( 366.9 ) + ( 210 )( 200 ) + ( 90 )( 360.1) + (120 )(148.7 ) + ( 50 )( 0 ) = 218, 296.1 Dresden is clearly the best site using the load-distance technique. Observing a European map it is also centrally located among AIAI’s customers. However, several other factors should be taken into consideration. For example, AIAI currently ships into the port of Hamburg so a warehouse/distribution center there would eliminate transport from the port of entry to a distribution center at Dresden or anywhere else. According to the load-distance technique, Hamburg is the worst site, but its proximity to the port may make it much more attractive. However, the attractiveness of Hamburg as a port could be offset by Dresden’s closer proximity to the plants at Vienna and Budapest that require continuous replenishment. A negative location factor for Dresden is that it’s in the old East Germany and its infrastructure and other facilities may be less than ideal.

Students might be encouraged to research for other location factors, such as transportation between the sites, between country trade regulations, costs, etc.

8 Human Resources in Operations Management Answers to Questions 8-1.

For companies committed to quality management, their TQM program must be very closely related to their strategic plan. Employee involvement is an integral part of TQM, thus human resources is an important component of strategic planning.

8-2.

It reflects the fact that companies now view employees more of a valuable “resource” than simply a replaceable part in the production process.

8-3.

Break down jobs into their simplest elements and motions, render the elements as efficient as possible by eliminating unnecessary motions, and then divide the tasks up among several workers so that each task would require only minimal skill; in effect, simplify job designs to the greatest degree possible.

8-4.

F. W. Taylor’s contributions were in the development of the principles of scientific management and the use of motion analysis and the stopwatch to design jobs and measure job performance. Frank Gilbreth originated motion study as a means to determine the “one best way” to perform a work task, which evolved into widely used principles of motion study.

8-5.

Horizontal job enlargement refers to expanding the scope of a job to include all tasks necessary to complete a product or process, providing the worker with a sense of having made something. Vertical job enlargement transfers some supervisory responsibilities for a job from management to the worker, allowing for more job self-determination and control.

8-6.

Tasks are individual activities consisting of one or more elements, which in turn encompass several motions or basic physical movements.

8-7.

The creation of a mass product market and high output volume made the breakdown of work and fragmentation of jobs (and the increase in number of workers) required by scientific management costeffective and economical.

8-8.

Industrialization occurred more rapidly in Japan and bypassed the skilled industrial artisan phase that engendered scientific management, plus job flexibility and factory work teams that included management and maintenance were the cultural norm.

8-9.

See figure 8-1. Japanese job design includes such characteristics as horizontal and vertical job enlargement, as opposed to the task specialization and repetition of scientific management. Job design in Japan also encompasses extensive job training, responsibility and control, cross training, job rotation, higher skill levels, and employee involvement, whereas scientific management encompasses minimal job training, low skill levels, tight control and minimal responsibility.

8-10. Jobs were designed so that workers could be easily replaced and trained at low cost; jobs required minimal skills, resulting in a large labor pool; also unskilled, uneducated workers were allowed to gain employment based on their willingness to physically work hard at jobs that were mentally undemanding.

8-11. Advantages include horizontal and vertical job enlargement, individual responsibility for job reliability and quality, job rotation, and communication between workers and workers and management.

8-12. It has shifted more responsibility for quality to the worker and promoted empowerment of the worker to alert management to quality problems and individually act to correct quality problems without fear of reprisal.

8-13. Americans have tended to adopt Japanese management principles such as quality management, JIT systems, and quality circles without adopting their principles of job design, such as job flexibility, cross training, job training and job responsibility. Some of the Japanese job design features and work environments are cultural and cannot be easily adopted. However, some aspects of job design have been successfully adopted by some U.S. companies, including employee involvement programs and quality circles, worker empowerment and job responsibility, more extensive training, and job flexibility.

8-14. The three elements of job design are task analysis, the determination of how to do each task and how all the tasks fit together; worker analysis, the characteristics and responsibilities of the worker who performs the job; and environmental analysis, which involves the physical location of the job and the conditions that must exist to do the job.

8-15. A process flowchart is used to analyze the sequential steps of a job or how a set of jobs fit together to form a production process, whereas a worker-machine chart illustrates the amount of time a worker and a machine are working or idle in a job.

8-16. Depends on student selection of activity.

8-17. For the employee telecommuting allows for a more flexible work schedule, the opportunity to attend to personal matters within a work schedule, reduced travel expenses and time. Disadvantages are a lack of personal contact with co-workers and management, and it is a possible obstacle to advancement because of lack of contact with superiors.

For the manager advantages include a means to motivate employees, and it saves work space and real estate costs. Disadvantages include a feeling of a lack of control because the employee cannot be seen working and some employees may not be self-motivated to work without direct personal supervision.

8-18. Empowerment allows employees to make quality decisions and corrections immediately without delaying to consult with supervisors. It gives employees more control over quality in their job. Empowerment can be abused by the employee and some employees may not be suited to have responsibility and control.

8-19. Depends on article selected.

8-20. The Baldrige Award site includes brief summaries of all award winning companies. Depends on company selected.

8-21. Depends on the workplace the student selects.

8-22. Depends on the job selected by the student.

8-23. Learning curves are useful primarily for measuring work improvement for nonrepetitive, complex jobs that require a relatively long time to complete, as opposed to short, repetitive, routine jobs that show little improvement over time.

8-24. The learning curve reflects the fact that each time the number of units produced doubles, the processing time per unit decreases by a constant percentage.

8-25. Product modifications during the production process can negate the learning curve effect; improvement can result from sources other than worker learning such as modifications of work methods or new equipment; and industry-developed learning rates may not be appropriate for individual companies.

Answers to Problems 8-1.

tn = t1nb t100 = (48)(100)ln(0.88) ln 2 = 20.5 minutes

800 new forms per week will require

(800)(20.5 minutes) = 16,424.1 minutes per week to process 1 employee works (6 hr) (60 mins) (5 days) = 1,800 minutes per week

New employees required =

16.424.1 = 9.1 1,800

8-2.

The time required for the first 10 exams is shown below.

Grading Time 12.0 10.8 10.15 9.72 9.39 9.13 8.92 8.74 8.59 8.45

Cumulative Time 12.0 22.8 32.95 42.67 52.07 61.20 70.13 78.88 87.47 95.93

Since the first 10 exams will require approximately 96 minutes, Professor Cook will have 300 − 96 = 204 minutes left. The remaining 25 exams will require 8.46 minutes apiece, or 211.5 minutes to complete. Thus, unless Dr. Cook hurries she will not complete them on time.

8-3.

tn = t1nb

t60 = 80(60)ln(0.92) ln 2 = 48.89 hr t120 = 80(120)ln(0.92) ln 2 = 44.98 hr 8-4.

tn = t1nb t30 = 1, 600(30)ln(0.87) ln 2 = 807.88 hr

t60 = 1, 600(60)ln(0.87) ln 2 = 702.86 hr 8-5.

Using Excel the learning coefficient is .9024.

8-6.

Using Excel the total cumulative time required to produce all 120 computers is 1232.9 hours. Thus, it will take the 8 employees, working 20 hours per week, 7.71 weeks to complete the order.

8-7.

tn = t1nb t80 = 126(80)ln(0.85) ln 2 = 45.1 seconds

8-8.

Using Excel, the minimum time of 32 minutes is reached at approximately the 53 rd room, i.e., 53 rooms must be cleaned before complete proficiency is attained. The cumulative time required to clean the first 53 rooms is 1,946.30 minutes. Given that a housekeeper works 390 minutes per day (i.e., 6.5 hours  60 minutes per hour), it will take approximately 5 days, or one work week, to become fully proficient.

8-9.

8 books at 17,000 lines per book is 136,000 lines. Thus, after 136,000 lines the scribe is a craftsman.

tn = t1nb

t136,000 = 15(136,000)ln(0.91) ln 2 = 3 minutes per line for an experienced scribe

The apprentice scribe could copy 50 lines at 15 minutes per line which means the work day was 50 15 = 750 minutes long or about 12.5 hours. In a 750 minute work day an experienced scribe could copy 750 3 = 250 lines.

8-10. Using Excel the number of orders required to reach the target value of 3 days is 23.

Case Solution: Maury Mills The following statement about how Anne and Dana reacted when they returned to the business is damning: “They felt that their employees had become complacent and spoiled and weren’t as committed to the business as they had been when they were smaller and were more of a family.” Anne and Dana were accusing their employees of exactly what the two of them had done–they had stepped back from their commitment to running the business and had obviously let others take over. Without them in an on-site leadership role there was nothing to sustain the good employee relations and family atmosphere that had originally existed. It is likely that Maury Mills’s employees were very motivated and committed to the business in the early days because of Anne and Dana’s hands-on, caring managerial style, but as Anne and Dana drifted away from their leadership role, that feeling left. In fact, it appears that most of their original employees may have left the company–“Walking around their facilities they felt they hardly knew anyone that worked for them.”-which should have been a red flag that Anne and Dana did not pick up on.

Some of the problems that Maury Mills encountered like the economic recession and increased competition were beyond their direct control. However, they seemed to be very slow to react to these conditions and did not have a long-term contingency plan for handling changing business conditions. It seems likely that Anne and Dana could have turned to their employees and management team for suggestions about how to turn things around in this situation, but instead they went to a consulting firm. While consultants aren’t bad and are often very helpful, they do not have the same close relationship with the company workforce as the owners. Anne and Dana probably biased the consultants against their employees by telling them of their employees’ “malaise.” Also, it is unlikely the consultant’s would want to openly dispute their client - who they would assume would know more about their employees than they did. Anne and Dana probably should have asked themselves why they thought their employees were not motivated and determined themselves what they expected from their employees to support the business.

By rewarding short-term performance with incentives, Maury Mills received short term results. Product quality seemed to be a cornerstone of Maury Mills’s business and the changes that were made obviously created quality problems. While incentive programs and bonuses can be useful, they can also have detrimental effects on quality if they are not well-thought out and part of an overall quality management program. The incentive system in the warehouse and distribution center likely caused employees to be in too much of a hurry to process orders and get them out the door by the end of the day, and this resulted in the incomplete and messed up orders. The late shipments were likely caused by a lack of coordination with the company’s shipper. Problems may have also resulted from increased sales; the sales force offered discounts to increase sales without coordinating with production and distribution to see if those orders could be filled on schedule. The grocer’s complaint about abrupt phone conversations was a result of the directive to reduce call length. The idea of reducing call time so that operators can handle more calls is not a bad one and may be justified, but it must be coupled with an understanding of its impact on quality measures.

The emphasis on increased productivity in the food processing area could have led to short-cuts and changes that compromised product quality. Incentive plans based on productivity increases need to be coupled with quality controls and oversight. The company also needed to make sure employees understood the company’s commitment to quality. The quality problems were likely exacerbated by the buyers purchasing inferior ingredients and products in order to reduce purchase costs. Empowering buyers and salespeople with broad discretion without a corresponding understanding of quality goals is an invitation to problems.

The conversation of a Maury Mills employee that Anne overheard at Kroger’s and the pending meeting with minority employees are definite signs that employee relations at Maury Mills have eroded. The conversation Anne heard reflects the employee’s lack of pride in the company and the frustration employees were feeling because of the poor quality problems and the perceived state of the company. It seems unlikely that Anne and Dana realized they

had a large minority contingent in their business and they were obviously not sensitive to potential diversity issues.

Since the cutbacks occurred among a racial grouping it is easy to see how the minority employees might perceive that they were unfairly singled out, especially since other work groups (i.e., in food processing) suffered no cutbacks. The cutbacks did not seem to be racially motivated; however there was certainly no consideration about how the minority employees might perceive them. At the least, employees should have been informed about the reason for the cutbacks and why they occurred. There seems to be a lack of communication and diversity management at Maury Mills.

Can Maury Mills’s problems be solved, and if yes, what can Anne and Dana do? Maury Mills could probably have weathered its crisis by employing some belt-tightening measures to ride out the recession, although a change in strategy to cope with increased competition seemed to be a good idea. The human resource changes recommended by the consulting firm only made things worse. However, despite these problems, Maury Mills appeared to be a fundamentally sound company before their problems. Although Anne and Dana did not have a background in business they seemed to have solid business instincts and good leadership skills. Their problems did not result because of their lack of business acumen but because they stepped back from their involvement in the company and failed to apply their abilities. They seemed to exhibit a lack of confidence in their own abilities, but events may have shown them that they are the most qualified to turn their business around. In order to turn the company around Anne and Dana need to be more involved and provide more direct leadership. They should develop a plan of action and goals for turning the company around and communicate those plans and goals to their employees both as a whole and in groups. They should return to their commitment to quality and institute process improvement initiatives.

While buyers should be allowed to shop for the lowest cost items, they should not compromise quality. It’s likely that Anne and Dana will want to rethink the incentive and bonus plans they introduced and develop some alternative compensation plans that combine performance and productivity with quality. Above all the owners need to return to the employee-friendly environment they originally created.

9 Project Management Answers to Questions 9-1.

CPM/PERT is popular because it provides a picture of the steps of a project, it is easy for managers and participants to understand, and it is easy to apply.

9-2.

The goal is to show the precedence relationship of activities in a project.

9-3.

A dummy activity is used in an ADA network to show a precedence relationship without the passage of time.

It is used to complete a precedence relationship so that two activities will not have the same start and end nodes.

9-4.

The critical path is the longest path in the network. It can be computed by summing the activity times along each path and then seeing which path is the longest. It also is the path with no slack available.

9-5.

Slack is the amount of time an activity can be delayed without affecting the overall project duration. It is computed by subtracting the earliest start time for an activity from the latest start time or the earliest finish time from the latest finish time for an activity.

9-6.

The mean activity time is computed as t = (a + 4m + b) 6, where a is the optimistic activity time, m is the most likely time, and b is the pessimistic time. The variance is computed as v 2 = [(b − a) 6]2 .

9-7.

Total project variance is computed by summing the variances of the critical path activities.

9-8.

The purpose of project crashing is to shorten the project duration at the least possible cost.

9-9.

See which activity on the critical path has the minimum crash cost, and reduce this activity duration by the maximum amount or until another path becomes critical. If more than one path is critical, both paths must be reduced by the same amount simultaneously. Repeat this process until the crashing objective is reached.

9-10. The preference should depend on the project, including the perceived degrees of variability in project activities, the ability to determine probabilistic time estimates, and the degree to which probabilistic analysis is required.

9-11. The Gantt chart is a graphical technique with a bar or time line displayed for each activity in the project.

However, while it will indicate the precedence relationships between activities, these relationships are usually not as easy to discern (i.e., visualize) as with a network.

9-12. Indirect costs include the cost for facilities, equipment, and machinery, interest on investment, utilities, labor, personnel costs, etc. Direct costs are financial penalties for not completing a project on time. In general, project crashing costs and indirect costs have an inverse relationship; crashing costs are highest when the project is shortened, whereas indirect costs increase as the project duration increases.

9-13. A heavy reliance by the project manager on the network can mask errors in the precedence relationships or missing activities can be overlooked. Attention to critical path activities can be excessive to the point of neglecting other crucial project activities. Obtaining both deterministic and probabilistic time estimates can be difficult. The time estimates can be overly optimistic or pessimistic.

9-14. For an activity-on-node network, nodes represent project activities and branches show precedence relationships, whereas on an activity-on-arrow network, branches represent activities and nodes are events specifying the end of one activity and the beginning of another.

9-15. The project team consists of a group of individuals selected because of their special skills, expertise, and experience related to the project. Project planning includes the identification of all project activities and their precedence relationships the determination of activity times, the determination of project duration, comparison of the project time with objectives, and the determination of resource requirements to meet objectives. Project control includes making sure all activities have been identified, making sure the activities

are completed in their proper sequence, identifying resources as they are required, and adjusting the schedule to reflect changes.

9-16. The WBS depends on the “project” the student uses. Figure 9.2 should be used as a guideline.

9-17. Work Breakdown Structure for Dinner Project

9-18. Student answer based on project selected.

9-19. Student answer based on country selected.

Solutions to Problems 9-1.

9-2.

Project completion time = 17 weeks

Activity 1 2 3 4 5 6 7 9-3.

a.

Slack (weeks) 12 0 12 4 0 4 0

Project completion time = 23 weeks

Activity 1 2 3 4 5 6 7 8 9 9-3.

Paths:

b.

1→ 3 → 7 4 + 8 + 2 = 14 1→ 3 → 6 →8 4 + 8 + 5 + 6 = 23* 1→ 4 →8 4 + 3 + 6 = 13 2 →5→8 7 + 9 + 6 = 22

2→9 7 + 5 = 12

Slack (weeks) 0 1 0 10 1 0 9 0 11

9-4.

Paths: 2 → 5 → 7 2→4→6→7 1→ 3 → 6 → 7

10 + 4 + 2 = 16 10 + 5 + 3 + 2 = 20* 7 + 6 + 3 + 2 = 18

Activity 1 2 3 4 5 6 7

EF 7 10 13 15 14 18 20

9-5.

Time 7 10 6 5 4 3 2

ES 0 0 7 10 10 15 18

LS 2 0 9 10 14 15 18

Critical path activities have no slack.

Critical path = 2-4-6-7 = 20

LF 9 10 15 15 18 18 20

Slack 2 0 2 0 4 0 0

9-6.

Critical path = 2-6-9-11-12 = 38 months

9-7.

9-8.

a, b, c.

Paths: a-b-d-f a-c-d-f a-c-e-f

3 + 3 + 4 + 2 = 12 3 + 5 + 4 + 2 = 14* 3 + 5 + 3 + 2 = 13

9-9.

9-10.

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

Time 8 12 3 9 3 2 12 7 30 21 20 5 16 17 5 6 3

ES 0 0 0 12 3 3 21 21 21 21 33 51 42 58 58 53 75

EF 8 12 3 21 6 5 33 28 51 42 53 56 58 75 63 59 78

Critical path = 2-4-10-13-14-17 Project completion time = 78 wk.

LS 4 0 9 12 18 19 40 46 23 21 52 53 42 58 70 72 75

LF 12 12 12 21 21 21 52 53 53 42 72 58 58 75 75 78 78

Slack 4 0 9 0 15 16 19 25 2 0 19 2 0 0 12 19 0

9-11.

Critical path = a-b-f-h = 15 weeks 9-12.

Activity

a

m

b

t

ES

EF

LS

LF

Slack

2

1 2 3 4 5 6 7 8 9 10 11

6 2 4 3 7 4 3 5 3 4 2

10 7 8 10 9 12 6 9 20 12 9

15 16 11 15 20 15 9 16 35 16 14

10.16 7.66 7.83 9.66 10.50 11.16 6.00 9.50 19.66 11.33 8.66

0 0 0 10.16 10.16 10.16 19.83 7.83 20.66 7.83 25.83

10.16 7.66 7.83 19.83 20.66 21.33 25.83 17.33 40.33 19.16 34.50

0 18.00 14.33 16.00 10.16 20.50 25.66 22.16 20.66 29.00 31.66

10.16 25.66 22.16 25.66 20.66 31.66 31.66 31.66 40.33 40.33 40.33

0 18.00 14.33 5.83 0 10.33 5.83 14.33 0 21.16 5.83

2.25 5.43 1.35 4.00 4.67 3.35 1.00 3.35 28.41 4.00 4.00

Expected project completion time = 40.33 wk

 = 5.95 Critical path = 1-5-9

9-13.

9-14. a, b, c, d.

Activity

a

m

b

t

ES

EF

LS

LF

Slack

2

1 2 3 4 5 6 7 8 9 10 11 12

4 6 2 1 3 3 2 9 5 7 5 3

8 10 10 4 6 6 8 15 12 20 6 8

12 15 14 13 9 18 12 22 21 25 12 20

8.00 10.16 9.33 5.00 6.00 7.50 7.66 15.16 12.33 18.66 6.83 9.16

0 0 0 8.00 8.00 10.16 9.33 17.66 17.66 13.00 30.00 36.83

8.00 10.16 9.33 13.00 14.00 17.66 17.00 32.83 30.00 31.66 36.83 46.00

3.66 0 8.33 22.33 11.66 10.16 22.33 21.66 17.66 27.33 30.00 36.83

11.66 10.16 17.66 27.33 17.66 17.66 30.00 36.83 30.00 46.00 36.83 46.00

3.66 0 8.33 14.33 3.66 0 13.00 4.00 0 14.33 0 0

1.77 2.25 4.00 4.00 1.00 6.25 2.76 4.67 7.08 9.00 1.35 8.01

e.

Critical path = 2-6-9-11-12 Expected project completion time = 46 mo ;  = 5 mo

f.

9-15. a, b, c, d.

Activity

a

m

b

t

ES

EF

LS

LF

Slack

2

1 2 3 4 5 6 7 8 9 10 11 12

1 1 3 3 2 2 1 1 1 2 1 1

2 3 5 6 4 3 1.5 3 1 4 2 1

6 5 10 14 9 7 2 5 5 9 3 1

2.50 3.00 5.50 6.83 4.50 3.50 1.50 3.00 1.66 4.50 2.00 1.00

0 2.50 2.50 2.50 8.00 8.00 8.00 9.50 12.50 12.50 12.50 17.00

2.50 5.50 8.00 9.33 12.50 11.50 9.50 12.50 14.16 17.00 14.50 18.00

0 7.50 2.50 2.66 10.50 9.00 8.00 9.50 15.33 12.50 15.00 17.00

2.50 10.50 8.00 9.50 15.00 12.50 9.50 12.50 17.00 17.00 17.00 18.00

0 5.00 0 0.16 2.50 1.00 0 0 2.83 0 2.50 0

0.694 0.436 1.35 3.35 1.35 0.689 0.026 0.436 0.436 1.35 0.109 0

e.

Critical path = 1-3-7-8-10-12

f.

Expected project completion time = 18 months ;  = 1.97

9-16.

Z=

x−



=

50 − 46 4 = = 0.80 5 5

From normal table, p = 0.2881

0.5000 − 0.2881 = 0.2119 probability that the project will exceed 50 mo.

9-17.

Activity

a

m

b

t

ES

EF

LS

LF

Slack

2

a b c d e f g h I j k

1 2 1 4 3 10 5 2 1 2 2

2 5 3 10 7 15 9 3 4 5 2

3 8 5 25 12 25 14 7 6 10 2

2.00 5.00 3.00 11.50 7.16 15.83 9.16 3.50 3.83 5.33 2

0 0 0 2.00 2.00 5.00 3.00 13.50 20.83 20.83 26.16

2.00 5.00 3.00 13.50 9.16 20.83 12.16 17.00 24.66 26.16 28.16

7.33 0 8.66 9.33 13.66 5.00 11.66 22.66 22.33 20.83 26.16

9.33 5.00 11.66 20.83 20.83 20.83 20.83 26.16 26.16 26.16 28.16

7.33 0 8.66 7.33 11.66 0 8.66 9.16 1.50 0 0

0.11 1.00 0.436 12.25 2.25 6.25 2.25 0.689 0.689 1.77 0

c.

Critical path = b-f-j-k

d.

Expected project completion time = 28.17 weeks;  = 3.00

e.

Z=

x−



=

35 − 28.16 6.84 = = 2.28 3 3.00

From normal table, p = 0.4887

0.5000 − 0.4887 = 0.0113 probability that the company will be fined.

9-18.

Z=

x−



=

15 − 18 = −1.52 1.97

From normal table, p = 0.4357

0.5000 − 0.4357 = 0.0643 probability that preparations would be in time.

9-19.

Activity

a

m

b

t

ES

EF

LS

LF

Slack

2

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

1 4 20 4 2 8 10 5 1 6 5 5 4 5 5 1

3 6 35 7 3 12 16 9 2 8 8 10 7 7 9 3

5 10 50 12 5 25 21 15 2 14 12 15 10 12 20 7

3.00 6.33 35.00 7.33 3.16 13.50 15.83 9.33 1.83 8.66 8.16 10.00 7.00 7.50 10.16 3.33

0 0 0 3.00 10.33 13.50 13.50 13.50 35.00 10.33 35.00 36.83 36.83 19.00 43.83 54.00

3.00 6.33 35.00 10.33 13.50 27.00 29.33 22.83 36.83 19.00 43.16 46.83 43.83 26.50 54.00 57.33

7.50 14.66 0 10.50 17.83 23.33 21.00 27.50 35.00 41.16 49.16 44.00 36.83 49.83 43.83 54.00

10.50 21.00 35.00 17.83 21.00 36.83 36.83 36.83 36.83 49.83 57.33 54.00 43.83 57.33 54.00 57.33

7.50 14.66 0 7.50 7.50 9.83 7.50 14.00 0 30.83 14.16 7.16 0 30.83 0 0

0.436 1.00 25.00 1.77 0.25 8.01 3.35 2.76 0.029 1.77 1.36 2.77 1.00 1.36 6.25 1.00

Expected completion time = 57.33 days

 = 5.77 Z=

x−



=

67 − 57.33 = 1.68 5.77

P( x  67) = 0.9535

9-20.

Activity a b c d e f g h I j k l m n o

ES 0 5.33 5.33 5.33 10.33 10.33 9.17 9.17 11.67 11.67 21.17 19.17 17.83 26 34.5

EF 5.33 10.33 9.17 11.67 17.83 19.83 21.17 18.33 19.17 26 34 30 25.17 34.5 42.33

LS 0 15 13.67 5.33 27.5 20 17.5 22.33 24 11.67 29.5 31.5 35 26 34.5

LF 5.33 20 17.5 11.67 35 29.5 29.5 31.5 31.5 26 42.33 42.33 42.33 34.5 42.33

Slack 0 9.67 8.33 0 17.17 9.67 8.33 13.17 12.33 0 8.33 12.33 17.17 0 0

Variance 1 1 0.69 1 2.25 3.36 7.11 1.36 3.36 5.44 3.36 2.25 1.78 4.69 0.69

Critical path = a-d-j-n-o Expected project completion time = 42.3 weeks

 = 3.58 Since probability is 0.90, Z = 1.29.

1.29 =

x − 42.3 3.58

x − 42.3 = 4.61

x = 46.92 To be 90 percent certain of delivering the part on time, RusTech should probably specify at least 50.3 or 51 weeks in the contract

9-21.

a b c d e f g h I j k l m n o p q

a b c d e f

Activity Time 15 8.83 24.16 19.5 8.16 13.66 20.16 25 14.66 23 8.66 7.16 5 4.33 7 5.5 20.833

Early Start 0 0 15 39.16 39.16 47.33 61 58.66 58.66 58.66 81.66 83.66 73.33 90.83 90.83 90.33 97.83

1.66 1.16 2.5 2.16 1.16 2.33

Activity std dev g 1.5 h 3.33 I 2 j 2.33 k 1.33 l 1.5

Critical path = a-c-d-h-l-o-q

Early Finish 15 8.83 39.16 58.66 47.33 61 81.16 83.66 73.33 81.66 90.33 90.83 78.33 94.66 97.83 95.83 118.66

m n o p 1

0.66 1 1 0.83 2.5

Late Start 0.0 64.5 15 39.16 51.5 59.66 73.33 58.66 78.16 61.83 84.83 83.66 92.83 93.5 90.83 113.16 97.83

Late Finish 15 73.33 39.16 58.66 59.66 73.33 93.5 83.66 92.83 84.83 93.5 90.83 97.83 97.83 97.83 118.66 118.66

Slack 0 64.5 0 0 12.33 12.33 12.33 0 19.5 3.16 3.16 0 19.5 3.16 0 22.83 0

 = 118.67

 = 5.85 Z=

x−



=

120 − 118.67 = 0.227 5.85

P( x  120) = 0.59 9-22. a.

b.

Present (normal) critical path = 1-4 Normal critical path time = 30 wk Crash critical path (all crash time) = 1-4 Maximum possible project crash time = 20 wk

c.

Normal cost = 3950 Crashed project cost = 4700

Activity a b c d e

Normal Time 20 24 14 10 11

Crash Time 8 20 7 6 5

Normal Cost 1,000 1,200 700 500 550 3950

a b c d e

Crash Cost/pd 40 50 70 80 30

Crash By 10 4 0 0 5

Crashing Cost 400 200 0 0 150 750

Crash Cost 1,480 1,400 1,190 820 730 5620

9-23.

Normal critical path = a-d-h Normal critical path time = 36 wk Project completion time = 36 Normal cost = 14400 Minimum project completion time = 22 Crash cost = 23,250

a b c d e f g h

Normal Time 16 14 8 5 4 6 10 15

a b c d e f g h

Crashing Cost 2,400 800 200 700 750 0 1,000 3,000

Crash Time 8 9 6 4 2 4 7 10

Normal Cost 2,000 1,000 500 600 1,500 800 3,000 5,000

Crash Cost 4,400 1,800 700 1,300 3,000 1,600 4,500 8,000

Crash Cost/pd 300 160 100 700 750 400 500 600

Crash By 8 5 2 1 1 0 2 5

9-24. Project completion time = 23 Normal cost = 1500 Minimum project completion time = 15 Crash cost = 2650 Activity 1 2 3 4 5 6 7

Normal Time 8 10 5 3 6 3 4

Crash Time 5 7 3 1 4 3 3

Normal Cost 100 250 400 200 150 100 300

Activity 1 2 3 4 5 6 7

Crashing Cost/pd 100 50 200 100 75 0 200

Crash By 3 2 2 0 2 0 1

Crashing Cost 300 100 400 0 150 0 200

Crash Cost 400 400 800 400 300 100 500

9-25. Project completion time = 33 Normal cost = 28800 Minimum project completion time = 26 Crash cost = 33,900

a b c d e f g h i j k

Normal Time 9 11 7 10 1 5 6 3 1 2 8

Crash Time 7 9 5 8 1 3 5 3 1 2 6

Critical path = a-d-g-k Crashing cost = $5,100 Total network cost = $33,900

Normal Cost 4,800 9,100 3,000 3,600 0 1,500 1,800 0 0 0 5,000

Crash Cost 6,300 15,500 4,000 5,000 0 2,000 2,000 0 0 0 7,000

Normal Cost 750 3,200 500 700 0 250 200 0 0 0 1,000

Crash By 2 0 0 2 0 0 1 0 0 0 2

Crashing Cost 1,500 0 0 1,400 0 0 200 0 0 0 2,000

9-26.

Project a b c d e f g h I j

Activity Time 23 3 3.167 4.167 2.833 5 1.833 5.833 3.833 4.167 2.167

ES

EF

LS

LF

Slack

0 0 0 3 5.833 10.833 10.833 12.667 16.667 20.833

3 3.167 4.167 5.833 10.833 12.667 16.667 16.5 20.833 23

0 7.667 6.667 3 5.833 15.167 10.833 17 16.667 20.833

3 10.833 10.833 5.833 10.833 17 16.667 20.833 20.833 23

0 7.667 6.667 0 0 4.333 0 4.333 0 0

Probability the project will be completed in 21 days?

Z=

x−



=

21 − 23 = −1.18 1.70

P( x  21) = .119

Standard Deviation 1.70 0.667 0.5 0.833 0.5 1 0.167 0.833 0.5 0.5 0.5

9-27.

Activity a b c d e f g h I j k l m n o p q r s t u

Time 7.17 18.00 10.17 22.17 30.00 8.67 7.00 21.33 20.17 13.33 7.83 12.33 10.17 7.33 18.33 7.17 8.17 7.00 27.83 17.50 8.17

ES 0 0 0 0 7.17 37.17 45.83 18 18 52.83 39.33 38.17 37.17 45.83 66.17 84.5 66.17 66.17 45.83 47.33 64.83

EF 7.17 18 10.17 22.17 37.17 45.83 52.83 39.33 38.17 66.17 47.17 50.5 47.33 53.17 84.5 91.67 74.33 73.17 73.67 64.83 73

The critical path is: a-e-f-g-j-o-p Project duration = 91.667

LS 0 0 27 15 7.17 37.17 45.83 37 33.67 52.83 58.33 53.83 55.83 58.83 66.17 84.5 83.5 84.67 63.83 66 83.5

LF 7.17 33.67 37.17 37.17 37.17 45.83 52.83 58.33 53.83 66.17 66.17 66.17 66 66.17 84.5 91.67 91.67 91.67 91.67 83.5 91.67

Slack 0 15.67 27 15 0 0 0 19 15.67 0 19 15.67 18.67 13 0 0 17.33 18.5 18 18.67 18.67

Variance 0.69 7.11 1.36 12.25 0 1.78 1 7.11 3.36 4 0.69 1.78 1.36 1.78 2.78 0.69 1.36 1 6.25 4.69 1.36

 = 3.3082 From January 20 to April 29 is 101 days.

P( x  101) =

=

x− Z 101 − 91.667 3.3082

= 2.82

P( x  101) = .9976 Activity “n,” send out acceptance letters, has ES = 45.83 (March 6) and LF = 66.17 (March 20), so it appears the club would meet the deadline of March 30 to send out acceptance letters.

Activity “q,” send out schedules, has ES = 66.16 (March 26) and LS = 83.50 (April 14) and LF = 91.67, so it seems likely the club would meet the deadline of April 15 for sending out game schedules.

9-28.

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

Activity time 11.67 36.67 17.67 9.17 9.5 12 16.17 26.67 8 10.67 9.67 11.17 14.67 14.67 17 4.83

Early Start 0 0 11.67 11.67 36.67 29.33 20.83 36.67 46.17 20.83 54.17 31.5 63.33 63.83 63.83 80.83

Project completion time = 85.66 months Project standard deviation = 7.43 P(x<=96) = .947

Early Finish 11.67 36.67 29.33 20.83 46.17 41.33 37 63.33 54.17 31.5 63.83 42.67 78 78.5 80.83 85.67

Late Start 4.83 0 16.5 20.83 36.67 34.17 30 39.5 46.17 42 54.17 52.67 66.17 66.17 63.83 80.83

Late Finish 16.5 36.67 34.17 30 46.17 46.17 46.17 66.17 54.17 52.67 63.83 63.83 80.83 80.83 80.83 85.67

Slack 4.83 0 4.83 9.17 0 4.83 9.17 2.83 0 21.17 0 21.17 2.83 2.33 0 0

Standard Deviation 3 6.67 4.67 3.17 1.5 1.67 3.5 4 1 2 1.33 3.5 1.33 1.67 2.33 .5

9-29.

Project completion = 32.83 σ = 1.86 Activity a b c d e f g h i j k l m

ES 0 1.83 3.83 3.83 5.83 3.83 9.67 1.83 13.67 13.67 17.50 23.83 29.67 Project

EF 1.83 3.83 9.67 5.83 6.83 6.00 13.67 5.67 16.67 17.50 23.83 29.67 32.83 32.833

Critical Path = a-b-c-g-j-k-l-m

Z = (24 – 32.833) / 1.86 = -4.74 P(x<= 6 months or 24 weeks) = 0

Z = (52 – 32.833 / 1.86 = 10.31 P(x<= a year or 52 weeks) = 1.00

LS 0.00 1.83 3.83 10.67 12.67 11.50 9.67 10.67 14.50 13.67 17.50 23.83 29.67

LF 1.83 3.83 9.67 12.67 13.67 13.67 13.67 14.50 17.50 17.50 23.83 29.67 32.83

Slack 0.00 0.00 0.00 7 6.83 7.67 0.00 9 1 0 0.00 0 0 Project Std.dev

Variance 0.027778 0.111111 0.694444

0.44

0.25 1.00 0.69 0.25 3.47 1.86

9-30.

a b c d e f g h i j k l m n

ES 0.00 2.00 5.00 5.00 5.00 5.00 12.00 7.00 36.33 48.67 61.33 63.33 67.17 69.33 Project

EF 2.00 5.00 7.00 19.33 12.00 12.00 36.33 13.33 48.67 61.33 63.33 67.17 69.33 71.17 71.17

LS 0.00 2.00 28.00 22.00 5.00 29.33 12.00 30.00 36.33 48.67 61.33 63.33 67.17 69.33

Critical path = a – b – e – g – i- j – k – l – m – n Z = (52-71.17) / 3.51= - 5.46 P(x<=52) = 0

Z = (76-71.17) / 3.51= 1.376 P(x<=76) = .500 + .4147 = .9147

LF 2.00 5.00 30.00 36.33 12.00 36.33 36.33 36.33 48.67 61.33 63.33 67.17 69.33 71.17

Slack 0.00 0.00 23.00 17.00 0.00 24.33 0.00 23.00 0.00 0.00 0.00 0.00 0.00 0.00 Project Std.dev

Variance 0.11 0.11

1.00 5.44 1.00 4.00 0.11 0.25 0.25 0.03 12.31 3.51

9-31.

Activity a b c d e f g h I j k l m n

Time 8.17 5.83 21.50 31.00 7.00 17.33 5.33 2.17 35.83 6.33 15.17 6.83 6.17 4.00

ES 0 0 0 0 8.17 15.17 21.50 31.00 33.17 69.00 15.17 75.33 82.17 88.33

EF 8.17 5.83 21.50 31.00 15.17 32.50 26.83 33.17 69.00 75.33 30.33 82.17 88.33 92.33

LS 0.67 3.00 6.33 0.00 8.83 15.83 27.83 31.00 33.17 69.00 60.17 75.33 82.17 88.33

The “suggested” network is as follows:

Critical path = d-h-i-j-l-m-n Expected project completion time = 92.33

 = 5.93 P( x  90) =

90 − 92.33 = .39 5.93

P( x  90) = .5000 − .1517 = .348

LF 8.83 8.83 27.83 31.00 15.83 33.17 33.17 33.17 69.00 75.33 75.33 82.17 88.33 92.33

Slack 0.67 3.00 6.33 0 0.67 0.67 6.33 0 0 0 45.00 0 0 0

Variance 1.36 0.69 6.25 13.44 1.00 4.00 1.00 0.25 17.36 1.00 3.36 0.69 1.36 1.00

-32.

Activity a b c d e f g h I j k l m n o p q r s t u v w

Time 9.33 4.83 5.00 14.50 9.50 2.17 5.00 13.67 6.17 22.17 9.67 25.67 12.50 15.83 45.00 6.83 5.00 12.33 4.17 17.17 2.00 0.50 0.50

ES 0 9.33 9.33 14.33 28.83 14.33 16.5 9.33 23 23 9.33 23.00 48.67 61.17 77.00 48.67 55.50 60.50 122.00 55.50 55.50 45.17 126.17

EF 9.33 14.17 14.33 28.83 38.33 16.50 21.50 23.00 29.17 45.17 19.00 48.67 61.17 77.00 122.00 55.50 60.50 72.83 126.17 72.67 57.50 45.67 126.67

LS 0 120.83 96.67 101.67 116.17 118.5 120.67 9.33 119.5 103.5 116 23 48.67 61.17 77 102.17 109.33 114.33 122 109 124.17 125.67 126.17

LF 9.33 125.67 101.67 116.17 125.67 120.67 125.67 23.00 125.67 125.67 125.67 48.67 61.17 77.00 122.00 109.00 114.33 126.67 126.17 126.17 126.17 126.17 126.67

Slack 0 111.50 87.33 87.33 87.33 104.17 104.17 0 96.50 80.50 106.67 0 0 0 0 53.50 53.83 53.83 0 53.50 68.67 80.50 0

Variance 5.44 1.36 0.44 3.36 1.36 0.25 0.44 5.44 1.36 6.25 4.00 18.78 4.69 6.25 25.00 1.36 1.00 1.00 0.69 3.36 0.11 0.00 0.00

Critical path = a-h-l-m-n-o-s-w Expected project duration ( ) = 126.67 days

 = 8.14 days P( x  150 days): Z =

Z=

x−

 150 − 126.67 8.14

Z = 2.87

P( x  150) = .5000 + .4979 = .9979 The “suggested” network is as follows (although the student’s version may vary).

CASE: Bloodless Coup Concert

Activity 1 2 3 4 5 6 7 8 9 10 11 12

a 2 4 3 1 1 2 1 2 2 2 1 1

m 4 5 5 3 2 4 3 3 6 3 2 5

b 7 8 10 8 4 7 5 4 12 6 3 12

t 4.16 5.33 5.50 3.50 2.16 4.16 3.00 3.00 6.33 3.33 2.00 5.5

ES 0 0 4.16 4.16 4.16 7.66 7.66 9.66 6.33 11.83 10.66 6.33

EF 4.16 5.33 9.66 7.66 6.33 11.83 10.66 12.66 12.66 15.16 12.66 11.83

Critical path = 1-4-6-10 Expected project completion time = 15.17 days

 = 1.79 Z=

x−



=

18 − 15.17 = 1.58 1.79

P( x  18) = 0.9429

LS 0 3.50 6.66 4.16 6.66 7.66 10.16 12.16 8.83 11.83 13.16 9.66

LF 4.16 8.83 12.16 7.66 8.83 11.83 13.16 15.16 15.16 15.16 15.16 15.16

Slack 0 3.50 2.50 0 2.50 0 2.50 2.50 2.50 0 2.50 3.33

sd( )

0.83 0.66 1.16 1.16 0.50 0.83 0.66 0.33 1.66 0.66 0.33 1.83

CASE SOLUTION: Moore Housing Contractors

a b c d e f g h i j k l m n o p q r s t u v w x

Activity time 4.16 3.16 3.83 2.16 2 3.83 3.16 4.16 2.83 2.16 5.16 6.5 8.33 3.33 2.33 3.5 4.16 6.33 5.83 4.33 3.33 6.33 5 2.83

Early Start 0 4.16 7.33 7.33 7.33 11.16 9.33 9.33 15 15 12.5 17.83 17.66 24.33 27.66 30 26 33.5 30.16 30.16 30.16 34.5 40.83 40.83

Early Finish 4.16 7.33 11.16 9.5 9.33 15 12.5 13.5 17.83 17.16 17.66 24.33 26 27.66 30 33.5 30.16 39.83 36 34.5 33.5 40.83 45.83 43.66

Late Start 0.0 4.16 7.83 33.83 7.33 11.66 9.33 13.5 21 15.5 12.5 23.83 17.66 30.33 33.66 36 26 39.5 35 30.16 31.17 34.5 40.83 43

Late Finish 4.16 7.33 11.66 36 9.33 15.5 12.5 17.66 23.83 17.66 17.66 30.33 26 33.66 36 39.5 30.16 45.83 40.83 34.5 34.5 40.83 45.83 45.83

Slack 0.0 0.0 0.5 26.5 0.0 0.5 0.0 4.16 6 0.5 0.0 6 0.0 6 6 6 0 6 4.83 0 1 0 0 2.16

Activity std dev 0.5 0.5 0.5 0.5 0.33 0.5 0.5 0.83 0.5 0.5 0.83 0.83 1 0.66 0.66 0.83 0.5 1 1.5 1 0.66 1 1 0.5

Project completion time = 45.83 Project standard deviation = 2.40 Critical path: a-b-e-g-k-m-q-t-v-w.

Notice that the expected completion time is 45.83 days which is very close to the realtor’s due date for completion.

The probability of finishing in 45 days is 0.3647.

36.47% is not a very high probability that the contractor will complete a house within 45 days thus the Moores should probably inflate their bid.

10 Supply Chain Management Strategy and Design Answers to Questions 10-1. McDonalds: The company is supplied by food distributors and restaurant product suppliers (for plates, napkins, etc.). Production is located in retail sites that are usually small and are near large easily accessible customer markets. Their inventory levels are typically small because food cannot be stored in large quantities. Their primary mode of transportation is trucking.

Toyota: Toyota suppliers include raw materials and auto parts. They receive some items on-demand for JIT production and some is stored in warehouses. Production is in large plants with heavy-machinery in close proximity to good transportation sources. Toyota plants are located all over the world as are their distribution systems. Transportation is by all modes of transportation.

10-2. The strategic goals are low cost and customer service. Purchasing from suppliers must be on time or the entire supply chain is delayed, creating late deliveries to customers. Erratic and poor quality supply can also increase costs. If facilities are not located properly it can delay product or service flow through the supply chain, and increase costs for longer deliveries. Production inefficiency and poor quality can cause delays in product or service flow and it also creates the need for more inventory which increases cost. Inefficient transportation can also result in higher inventories to offset delays and raise costs, and, causes delayed delivery to customers.

10-3. Answer depends on the businesses selected.

10-4. Answer depends on the example the student selects. One example is a grocery chain that uses POS terminals to send information sales to distribution center who, in turn, sends mixed loads to the store by truck weekly.

10-5. The answer depends on the company the student selects.

10-6. Answer depends on the e-marketplace selected.

10-7. Answer depends on the ERP provider selected.

10-8. The answer depends on the article the student selects.

10-9. The answer depends on the article the student selects.

10-10. Toyota is one automobile company that has begun a limited BTO program. The student should go tan automobile company website to see what they are doing with BTO options. There are traditionally many more options with automobiles than with a product like a computer, which means more suppliers to work with. Customers must be reconditioned to make selections from a smaller list of options—like colors, seat type and fabric, sound system, trim, tires, etc. Also the component parts for a computer can generally be stored in one location however, component parts for autos tend to be more geographically dispersed.

10-11. Companies like L.L. Bean and Sears had long experience with telephone catalog sales and thus had already developed the basic supply chain to service a catalog-phone operation, which online sales mirrored.

Amazon.com had virtually no supply chain structure for warehousing, transport and distribution when it started—thus it had to begin supply chain development from scratch.

10-12. RFID allows items to be scanned and identified without direct sight lines which enables items to be checked much more rapidly. Since bills-of-lading must not be much more detailed than previously (listing virtually all items in a shipment) this can be accomplished much more quickly and thoroughly with RFID technology and the information can be transported via satellite. This allows for much quicker security approvals, thus allowing ships and planes to load and unload more quickly.

10-13. Figure 10.6 shows some of the advantages that RFID technology would provide for a retailer like Wal-Mart.

The primary obstacle to the use of RFID is (currently) cost.

10-14. It provides a set of performance evaluators and an apparatus for comparing supply chains, which enables generic supply chain evaluation. These performance evaluators apply to supply chains of different sizes and cut across supply chains from different industries, which could allow them to be used to establish standards that can be used as a certification tool.

10-15. Universities have traditional suppliers for MRO goods like supplies, telephone and computer services, furniture, maintenance equipment and products, food and beverages, etc. They also have non-traditional suppliers like high schools that provide students. The producers are the teachers and other academic and support services that educate the student and provide for daily student life and well-being. Distributors include placement services and companies and organizations students are hired by. Inventory exists for MRO goods and services. Students are also a form of inventory as are classrooms, dorm rooms, parking spaces, etc. Students as inventory incur a carrying cost. The longer a student remains in school the greater the cost to the institution or state. Not only is there a direct cost for maintaining a student, but a student who stays too long also takes up a spot for other incoming students, thus creating shortages of facilities like dorm rooms and classrooms, which in-turn require expenditures for capital expansion.

10-16. Answer depends on student responses.

10-17. Answer depends on student responses.

10-18. Answer depends on company student identifies.

10-19. Answer depends on student responses.

10-20. Sustainability to a large extent is about conserving resources, which coincides with the quality management focus of eliminating waste. In the human resources area, quality management focuses on employee satisfaction which also coincides with sustainability goals. To a certain extent, sustainability also supports the quality goal of customer satisfaction since customers these days seem to desire sustainable products.

10-21. Service companies can achieve sustainability by reducing waste and conserving resources, like reducing paper usage, conserving on heat and electricity in workplace facilities, and telecommuting. However, some service companies, like UPS and Wal-Mart, can also reduce greenhouse gas emissions like carbon dioxide by using more energy-efficient vehicles. Manufacturing firms can reduce their carbon footprint, as well as eliminate waste and conserve energy and other sustainable resources.

10-22.

Answer depends on student responses.

10-23.

Answer depends on student responses.

10-24. In a digital supply chain, products move and are distributed electronically so there are no physical products, modes of transport, inventory or facilities; i.e., the things that make up a physical supply chain. However, there are still suppliers of digital content, the movement of products, distributors, stores and end-use customers. The supply chain tends to be embedded in the Internet.

Solutions to Problems 10-1.

Inventory turns =

470 = 14.2 17.5 + 9.3 + 6.4

Days of supply =

33.2 = 25.94 days 1.28

10-2. Average aggregate value of inventory:

Raw material = 82, 500 WIP = 61, 300 Finished goods = 220, 000 363, 800

Inventory turns =

3, 700, 000 = 10.17 363,800

Days of supply =

363,800 = 35.88 ( 3, 700, 000) / ( 365)

10-3. Aggregate value of inventory: Raw materials = 817,500 WIP = 67,800 Finished goods = 64,000 949,300 17,500, 000 Inventory turns = = 18.4 949,300

Weeks of supply =

949,300

(17,500, 000) / ( 50)

= 2.7 weeks

10-4. Average aggregate value of inventory: Raw materials = 281,090 WIP = 3,435,000 Finished goods = 3,087,000 6,803,090 18,500, 000 Inventory turns = = 2.72 6,803, 090

Days of supply =

6,803, 090 = 134.22 18,500, 000 ) / ( 365 ) (

10-5. (a)

Inventory turns Days of supply

1 8.0 45.7

2 8.3 44.1

Year 3 11.1 32.8

4 12.0 30.5

(b) The company’s supply chain performance has marginally improved. However, the number of turns still seems excessive for a computer firm. (c) For a BTO company the average work-in-process and finished goods inventory seems excessive; the company should work to reduce these levels, possibly by working with its suppliers.

10-6.

Inventory turns Weeks of supply

Supplier 1 22.6 2.3

Supplier 2 9.25 5.6

Supplier 1 has the better supply chain performance. Other factors that might influence Delph could be quality and delivery speed.

10-7.

c=

219 = 7.3 30

Control limits using Z = 3.00:

UCL = c + Z c = 7.3 + 3 7.3 = 7.3 + 8.11 = 15.41

LCL = c − Z c = 7.3 − 3 7.3 = 7.3 − 8.11  0 All the sample observations are within the control limits suggesting that the invoice errors are in control.

10-8.

c=

255 = 12.75 20

UCL = c + Z c = 12.75 + 3 12.75 = 23.46 LCL = c − Z c = 12.75 − 3 12.75 = 2.04 All the sample observations are within the control limits suggesting that the delivery process is in control.

10-9.

Sample 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

p=

Proportion Defective .028 .044 .072 .034 .050 .082 .036 .038 .052 .056 .076 .048 .030 .024 .020 .032 .018 .042 .036 .024

421

( 20 )( 500 )

UCL = p + Z

=

421 = .0421 10, 000

p (1 − p ) n

= .0421 + 3.00

.0421(1 − .0421) 500

= .0421 + .027 = .069

LCL = p − Z

p (1 − p )

= .0421 − 3

n

.0421(1 − .0421) 500

= .0421 − .027 = .015

Samples 3 and 11 are above the upper control limit indicating the process may be out of control.

10-10.

x 2.00 2.08 2.92 1.78 2.70 3.50 2.84 3.26 2.50 4.14 2.12 4.38 2.84 2.70 3.56 2.96 3.34 4.16 3.70 2.72 60.20

Sample 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

R=

 R = 63.3 = 3.17

x=

 x = 60.20 = 3.01

k

20

R 2.3 2.6 2.7 1.9 3.2 5.0 2.2 4.6 1.3 3.5 3.0 4.0 3.3 1.1 5.6 3.1 6.1 2.4 2.5 2.9 63.3

20

20

R-chart

D3 = 0, D4 = 2.11, for n = 5

UCL = D4 R = 2.11( 3.17 ) = 6.69 LCL = 0 R = 0 ( 3.17 ) = 0 There are no R values outside the control limits, which would suggest the process is in control.

x -chart

A2 = 0.58

UCL = x + A2 R = 3.01 + 0.58 (3.17 ) = 4.85 LCL = x − A2 R = 3.01 − 0.58 ( 3.17 ) = 1.17 There are no x values outside the control limits, which suggests the process is in control.

10-11. x = 3.17, R = 3.25, A2 ( n = 5) = 0.58

UCL = x + A2 R = 3.17 + ( 0.58)( 3.25) = 5.04 LCL = x − A2 R = 3.17 − 1.89 = 1.29 UCL = D4 R = 2.11( 3.25 ) = 6.86

LCL = D3 R = 0 ( 3.25 ) = 0

The process appears out of control for sample 10, however, this could be an aberration since there are no other apparent nonrandom patterns or out-of-control points. Thus, this point should probably be “thrown out” and a new control chart developed with the remaining eleven samples.

CASE PROBLEM SOLUTION: Somerset Furniture Company’s Global Supply Chain This case is based on a real furniture company the authors are familiar with. The company essentially reinvented itself from a furniture manufacturer to a supply chain management and distribution company. In doing so it outsourced its primary area of expertise, i.e., manufacturing, thereby losing direct control of the thing it knew most about, and adopted the role of supply chain manager, something it had little prior expertise with. The company suffered a number of inadequacies that left it ill prepared to assume its new role. The case as presented addresses many of the problems faced by companies that are confronted by a global supply chain, and provides students with a general platform for discussing supply chain problems and their remedy.

Following is the variable timeline for product lead time: 1. Develop purchase order: 12 to 25 days 2. Process purchase order overseas: 10 to 20 days 3. Manufacturing process: 60 days 4. Transport from plants to ports: 1 to 14 days 5. Arrange for containers and paperwork: 5 to 10 days 6. Wait for containers: 1 to 7 days 7. Load containers: 3 to 6 days 8. Security checks: 7 to 21 days 9. Overseas shipment: 28 days 10. Clear customs and loading trucks: 7 to 14 days 11. Transport to warehouse: 1 to 3 days 12. Receive and warehouse: 1 to 30 days Total lead time: 119 to 238 days Following is a list of discussion points for this case: • The company identified a set of characteristics of their global supply chain that seem to be endemic to many companies, as follows: Lower product cost Greater information requirements Longer delivery times to customers Quality problems Decrease in forecast accuracy Increase in inventory Replacement parts and repair service challenges Complex multi-step, inter-modal logistics Clash of different business cultures Hiring challenges Longer product to cash cycle Collaboration with multiple business partners • Primary among its deficiencies was its lack of expertise in the information technology area and an insufficient computing system. Much of the variability in the company’s supply chain was attributable to information technology problems. After the company made new hires to acquire IT expertise, and upgraded their computing technology, they were able to successfully address many of their problems in logistics and order processing and fulfillment, including the reduction of system variability.

• The company adopted better, more innovative forecasting methods, which improved their scheduling.

• They contracted with an international trade logistics (ITL) company to improve their overseas logistics, which combined with their improved information technology capabilities, allowed them to directly monitor and

control overseas product movement. This reduced scheduling variability.

• In the quality area they worked with their Chinese suppliers to improve quality management including training local plant managers in the use of quality control techniques. More quality auditors were hired to make more frequent visits to the plants, and quality was monitored more closely at various key points in the manufacturing process. In the case of the humidity problem the production of some lines was moved to other plants in China.

• Suppliers were selected based on the following criteria: quality, capacity, delivery and service.

• Many furniture products were redesigned with the objective of creating full container loads. It was discovered that removing even a few inches from the dimensions of a piece of furniture would allow additional pieces to be loaded into a container.

• The company decided to keep its own spare parts inventory in the U.S., ordering spare parts from their Chinese suppliers when the product line is initially produced. This required a more accurate forecast of spare parts demand. However, improved product quality eventually reduced the need for spare parts.

• To improve service the company determined required inventory levels based on customer demand with an objective of on-time delivery while at the same time seeking ways to reduce inventory levels. The reduction in system variability simultaneously worked to reduce inventory requirements and improve customer service.

• The company implemented a warehouse management system (WMS) to improve its local distribution service to customers.

• The company is also exploring moving some production to other countries to limit its vulnerability to anticipated restrictions on furniture imports from China. Also, the SARS epidemic has caused the company to reconsider sourcing its products from only one country.

11 Global Supply Chain Procurement and Distribution Answers to Questions 11-1. McDonalds: Distribution to McDonalds franchises takes place from its regional distribution centers, as well as directly from suppliers.

11-2. A few suppliers or carriers are easier to coordinate. If a company gives one or two suppliers all of their business they can be more demanding on quality and deliveries. Single-sourcing provides the supplier or carriers with economy of scale that enables them to reduce costs.

11-3. The strategic goals are low cost and customer service. Procured items must be delivered on time or the entire supply chain is delayed, creating late deliveries to customers. Erratic and poor quality supply can also increase costs. If facilities are not located properly it can delay product or service flow through the supply chain, and increase costs for longer deliveries. Inefficient transportation can also result in higher inventories to offset delays and raise costs, and, causes delayed delivery to customers.

11-4. Answer depends on the businesses selected.

11-5. The answer depends on example/business the student selects. One example is a grocery chain that ships foodstuffs from several warehouse/distribution centers to various stores.

11-6. Answer depends on the company the student selects.

11-7. Answer depends on the company the student selects.

11-8. Answer depends on the e-marketplace selected.

11-9. Answer depends on the transportation exchange selected.

11-10. Answer depends on the ERP provider selected.

11-11. Answer depends on the international logistics provider selected.

11-12. The answer depends on the article the student selects.

11-13. The answer depends on the article the student selects.

11-14. Companies like L.L. Bean and Sears use sophisticated IT systems and a network of distribution centers to supply customers and stores. One of their biggest problems is “reverse logistics,” i.e. what to do with items returned from the customer.

11-15. RFID allows items to be scanned and identified without direct sight lines which enables items to be checked much more rapidly. Since bills-of-lading must not be much more detailed than previously (listing virtually all items in a shipment) this can be accomplished much more quickly and thoroughly with RFID technology and the information can be transported via satellite. This allows for much quicker security approvals, thus allowing ships and planes to load and unload more quickly. The primary obstacle to the use of RFID is (currently) cost.

11-16. In vendor-managed inventory a supplier monitors and maintains inventory levels for its customer and has the authority to replenish when necessary. Postponement moves some of the final manufacturing steps like assembly into the warehouse or distribution center. Both concepts shift some area of supply chain responsibility away from the manufacturer or retailer to another party—a supplier or a distributor.

11-17. Cross-docking allows a company to move goods coming into a distribution center or warehouse directly to a shipping dock for outgoing orders without storing them first in a warehouse. This increases the speed with which goods flow through the supply chain and it reduces inventory and handling costs.

11-18. Universities have traditional suppliers for MRO goods like supplies, telephone and computer services, furniture, maintenance equipment and products, food and beverages, etc. They also have non-traditional suppliers like high schools that provide students. The producers are the teachers and other academic and

support services that educate the student and provide for daily student life and well-being. Distributors include placement services and companies and organizations students are hired by. Inventory exists for MRO goods and services. Students are also a form of inventory as are classrooms, dorm rooms, parking spaces, etc. Students as inventory incur a carrying cost. The longer a student remains in school the greater the cost to the institution or state. Not only is there a direct cost for maintaining a student, but a student who stays too long also takes up a spot for other incoming students, thus creating shortages of facilities like dorm rooms and classrooms, which in-turn require expenditures for capital expansion.

11-19. Answer depends on student responses.

11-20.

Using more energy efficient transport vehicles, improving transportation routes, reducing the use of packaging materials, and using less energy to heat, cool, and light warehouse and distribution facilities.

11-21. Depends on the article and company the student identifies.

11-22. Quality problems, transportation costs, communication problems, and intellectual property theft.

11-23. Answer depends on student responses.

CASE PROBLEM 11.1 - Somerset Furniture Revisited Following is the variable timeline for product lead time: 1. Develop purchase order: 12 to 25 days 2. Process purchase order overseas: 10 to 20 days 3. Manufacturing process: 60 days 4. Transport from plants to ports: 1 to 14 days* 5. Arrange for containers and paperwork: 5 to 10 days* 6. Wait for containers: 1 to 7 days* 7. Load containers: 3 to 6 days* 8. Security checks: 7 to 21 days* 9. Overseas shipment: 28 days 10. Clear customs and loading trucks: 7 to 14 days* 11. Transport to warehouse: 1 to 3 days* 12. Receive and warehouse: 1 to 30 days Total lead time: 119 to 238 days This case is based on an actual company. The company contracted with an international trade logistics (ITL) company to improve their overseas logistics, which combined with their improved information technology capabilities, allowed them to directly monitor and control overseas product movement. This reduced scheduling variability.

Transportation within China is primarily by truck, although rail is sometimes used. Poor infrastructure means shipments can be delayed or damaged. Transportation to the U.S. is by ship. Delays are primarily at the sending and receiving ports. Transportation within the U.S. is predominantly by truck, although rail is sometimes used.

Shipments are sent directly to dealers, stores, and customers. Few shipments are sent to Somerset directly.

*

*

Areas in which international trade specialists trade logistic companies and Internet exchanges would be helpful.

Areas in which international trade specialists trade logistic companies and Internet exchanges would be helpful.

Areas in which international trade specialists trade logistic companies and Internet exchanges would be helpful. * Areas in which international trade specialists trade logistic companies and Internet exchanges would be helpful. * Areas in which international trade specialists trade logistic companies and Internet exchanges would be helpful. * Areas in which international trade specialists trade logistic companies and Internet exchanges would be helpful. * Areas in which international trade specialists trade logistic companies and Internet exchanges would be helpful. * Areas in which international trade specialists trade logistic companies and Internet exchanges would be helpful. *

SUPPLEMENT 11 Operational Decision-Making Tools: Transportation and Transshipment Models Answers to Problems S11-1.

Bethlehem − Detroit = 130 Bethlehem − St. Louis = 50 Birmingham − Detroit = 30

Birmingham − Norfolk = 220 Gary − St. Louis = 70 Gary − Chicago = 170 Z = $84,100 S11-2.

No effect. The Gary Mill has 70 tons left over as surplus; reducing the capacity at Gary to 260 tons still leaves 20 tons.

S11-3.

Tampa − New York = 160 Tampa − Chicago = 130 Miami − Philadelphia = 140 Fresno − Philadelphia = 70 Fresno − Boston = 180 Z = $6, 220

S11-4.

There is no effect

S11-5.

Phoenix − Louisville = 10 Seattle − Louisville = 5 St. Louis − Charlotte = 15 Detroit − Greensboro = 10 Detroit − Charlotte = 5 Z = $215, 000

S11-6.

Atlanta − New Orleans = 45 Atlanta − Buffalo = 25 St. Louis − Buffalo = 25 St. Louis − Dallas = 90 Charlotte − New Orleans = 35 Charlotte − Pittsburgh = 25 Z = $17, 750

S11-7.

Atlanta − Buffalo = 25 Atlanta − Pittsburgh = 25 St. Louis − Buffalo = 25 St. Louis − Dallas = 90 Z = $10,500

S11-8.

Tampa − GA = 2100 Tampa − SC = 1050

St. Louis − TN = 950 St. Louis − NC = 1700 St. Louis − KY = 2350 Milwaukee − TN = 850 Milwaukee − VA = 1400 Z = $3, 635

S11-9.

No change; the total minimum cost remains $3,635 although are multiple optimal solutions (routes).

S11-10. Richmond − A & M = 280 Richmond − Central = 140 Atlanta − Tech = 560 Atlanta − State = 50 D.C. − State = 140

D.C. − Central = 200 Z = 21, 720 S11-11. Alternative 1: Richmond − A & M = 280 Richmond − Central = 320 Atlanta − Tech = 560 Atlanta − State = 50 D.C. − State = 320 D.C. − Central = 20 Z = $28,920

Alternative 2: Richmond − A & M = 280 Richmond − Central = 60 Atlanta − Tech = 560 Atlanta − State = 50 D.C. − State = 340 Charlotte − State = 20 Charlotte − Central = 28 Z = $25, 600

Select alternative 2; add a warehouse at Charlotte.

S11-12. 1 − B = 60 2 − A = 55 2 − B = 10 2 − C = 40 3 − B = 10 Z = 1660 units

S11-13. 1 − 1 = 27 1− 2 = 5 1 − 4 = 10 2 − 2 = 21 3 − 3 = 18 4 − 4 = 23 5−4 = 4 5 − 5 = 31 6−3 = 4 6−4 = 7 6 − 6 = 16 Z = 467 miles

S11-14. Charlotte − Atlanta = 30 Memphis − St. Louis = 30 Louisville − NY = 30 Z = $159, 000

S11-15. A − 3 = 1 C −1 = 1 D −1 = 1 E −3 =1 F− 2 =1 G − 2 =1 H −1 = 1 Z = 1, 070 Multiple optimal solutions exist.

S11-16. 1 − Bloomington = 1 2 − Columbus = 1 3 − Bloomington = 1 4 − Columbus = 1 5 − Madison = 1 6 − Madison = 1 7 − Columbus = 1 8 − Madison = 1 9 − Bloomington = 1 Z = 28.5 hours

S11-17. North − Addison = 270 South − Beeks = 120 South − Canfield = 190 East − Addison = 130 East − Canfield = 190 West − Daley = 220 Central − Beeks = 280 Z = 20,550 minutes (multiple optimal solutions)

S11-18. North − Addison = 270 South − Beeks = 200 South − Canfield = 110 East − Addison = 80 East − Canfield = 240 West − Daley = 220 Central − Beeks = 150 Central − Daley = 130 Optimal. Total travel time = Z = 21, 050 minutes.

The overall travel time increased by 500 minutes, which divided by all 1,400 students is only an increase of 0.357 minutes per student. This does not seem to be a significantly large increase.

S11-19.

Optimal, Total Profit = $1,528 with multiple optimal solutions at (B, 2), (E, 4) and (B, Dummy)

S11-20. If Atlantic purchased all the baby food demanded at each store from the distributor total profit would be $1,246, which is less than buying it from the other locations as determined in problem S11-19. This profit is computed by multiplying the profit at each store by the demand. In order to determine if some of the demand should be met by the distributor a new source (F) must be added to the transportation tableau from problem S11-19. This source represents the distributor and has an available supply of 150 cases, the total demand from all the stores. The tableau and optimal solution is shown as follows.

Total profit = $1,545 with multiple optimal solutions S11-21. 1 − E = 8 2 − C = 12 3− E = 2 4−D =8 5−A = 9 6−B = 6 Z = 1, 263 hours S11-22. GA1 − NCSW = 2 GA1 − NCW = 2 GA1 − VASW = 10 GA2 − NCSW = 3 GA2 − VAC = 9 SC1 − NCSW = 2 SC1 − NCP = 6 SC1 − VAT = 7 FL1 − NCE = 9 FL1 − NCW = 6 Z = $806, 000 S11-23. Sacremento − St. Paul = 15 Sacremento − Topeka = 1 Bakersfield − Denver = 10 San Antonio − Topeka = 12 Montgomery − Denver = 10 Jacksonville − Akron = 15 Jacksonville − Topeka = 7 Ocala − Louisville = 15 Z = $276, 200

It is cheaper for TransAm Foods to continue to operate it own trucking firm.

S11-24. Increasing the supply at Sacramento, Jacksonville and Ocala to 25 tons would have little effect, reducing

the overall monthly shipping cost to $276,000, which is still higher than the $245,000 the company is currently spending with its own trucks.

Alternatively, increasing the supply at San Antonio and Montgomery to 25 tons per month reduces the monthly shipping cost to $242,500 which is less than the company’s cost with their own trucks.

S11-25. L.A. − Singapore = 150

L.A. − Taipei = 300 Savannah − Hong Kong = 400 Savannah − Taipei = 200 Galveston − Singapore = 350 Shortages − Hong Kong = 200 Z = $723,580 z=$723500

Penalty costs = 200  $800 = $160, 000 S11-26. x1C (New Orleans – Nashville) = 72 x1D (New Orleans – Louisville) = 13 x2D (Jackson – Louisville) = 76 x2I (Jackson – Wilmington) = 34 x3A (Savannah – Atlanta) = 34 x3B (Savannah – Jacksonville) = 39 x4B (Mobile – Jacksonville) = 45 x5A (Birmingham – Atlanta) = 29 x6F (Houston – Roanoke) = 96 x6G (Houston – Norfolk) = 8 x6H (Houston – Memphis) = 8 x7I (Charleston – Wilmington) = 88 x8B (Shreveport – Jacksonville) = 18 x8D (Shreveport – Louisville) = 27 x8E (Shreveport – Richmond) = 41 x8H (Shreveport – Memphis) = 57

Z = $ 15618

S11-27. x1G (Truck 1 – Customer G) = 1 x2B (Truck 2 – Customer B) = 1 x3H (Truck 3 – Customer H) = 1 x4L (Truck 4 – Customer L) = 1 x5K (Truck 5 – Customer K) = 1 x6F (Truck 6 – Customer F) = 1 x7D (Truck 7 – Customer D) = 1 x8A (Truck 8 – Customer A) = 1 Z = $4,260 S11-28. x1G (Truck 1 – Customer G) = 1 x2F (Truck 2 – Customer F) = 1 x3D (Truck 3 – Customer D) = 1 x4L (Truck 4 – Customer L) = 1 x5C (Truck 5 – Customer C) = 1 x6E (Truck 6 – Customer E) = 1 x7J (Truck 7 – Customer J) = 1 x8A (Truck 8 – Customer A) = 1 Z = $4,840 Average capacity = 87.3% S11-29. x17 (Brazil – Mobile) = .3 x19 (Brazil – Savannah) = 5.6 x28 (Columbia – New Orleans) = 4.3 x38 (Indonesia – New Orleans) = 1.2 x3,10 (Indonesia – Jacksonville) = 2.6 x47 (Kenya – Mobile) = 2.8 x4,10 (Kenya – Jacksonville) = 3.9 x59 (Cote d’ Ivoire – Savannah) = 2.5 x68 (Guatemala – New Orleans) = 4.8 x7,11 (Mobile – New York) = 3.1 x8,11 (New Orleans – New York) = 10.3 x9,11 (Savannah – New York) = 8.1 x10,11 (Jacksonville – New York) = 6.5 Z = $2,206,000 S11-30. x14 ( Hamburg − Norfolk ) = 42

x59 ( NY − Chicago ) = 50

x26 ( Marseilles − Savannah ) = 63

x35 ( Liverpool − NY ) = 37

x48 ( Norfolk − St. Louis ) = 42

x15 ( Hamburg − NY ) = 13

x67 ( Savannah − Dallas ) = 60

x68 ( Savannah − St. Louis ) = 3 Z = $77,362

HND = 38

HNS = 17 MSD = 22

S11-31. x16 ( Mexico − Houston ) = 18

x24 ( Puerto Rico − Miami ) = 11 x34 ( Haiti − Miami ) = 23

x47 ( Miami − NY ) = 20 x48 ( Miami − St. Louis ) = 12

x49 ( Miami − LA ) = 2 x69 ( Houston − LA ) = 18 Z = $479 or $479,000

S11-32. (a)

x14 = 72 x25 = 105 x34 = 83

x46 = 75 x47 = 80 x56 = 15 x58 = 90

Z = $10, 043, 000

(b)

Adding a capacity constraint at plants in Indiana and Georgia

x14 = 72 x46 = 40 x25 = 105 x47 = 80 x34 = 48 x56 = 50 x35 = 35 x58 = 90 Z = $10, 043, 000

S11-33.

x1C = 70

xBA = 10

x2B = 80 xCB = 30 x3A = 50 Z = 1, 490 or $14,900

S11-34. x37 ( Italy − Texas ) = 2.1

x15 ( Germany − Mexico ) = 5.2

x26 ( Belgium − Panama ) = 6.3 x59 ( Mexico − Ohio ) = 5.2

x68 ( Panama − Virginia ) = 3.7 x69 ( Panama − Ohio ) = 2.6 Z = $27.12 million

S11-35. xij = potatoes shipped (in bushels) from farm i (where i = 1, 2, 3, 4) to distribution center j (where j = 5, 6, 7) and from distribution center i (where i = 5, 6, 7) to plant j (where j = 8, 9, 10, 11) Minimize Z = 1.09 x16 + 1.26 x17 + .89 x25 + 1.32 x26 + 1.17 x27 + .78x35 + 1.22 x36 + 1.36 x37 + 1.19 x45 + 1.25 x46

+ 1.42 x47 + 4.56 x58 + 3.98x59 + 4.94 x510 + 3.43x68 + 5.74 x69 + 4.65x610 + 5.01x611 + 5.39x78 + 6.35x79 + 5.70 x710 + 4.87x711

Subject to

x16 + x17  1, 600 x25 + x26 + x27  1,100 x35 + x36 + x37  1, 400 x45 + x46 + x47  1900 x25 + x35 + x45  1,800 x16 + x26 + x36 + x 46  2, 200 x17 + x27 + x37 + x 47  1, 600 x58 + x68 + x78 = 1, 200 x59 + x69 + x79 = 900 x510 + x610 + x710 = 1,100 x611 + x711 = 1,500

x25 + x35 + x45 = x58 + x59 + x510 x16 + x26 + x36 + x 46 = x68 + x69 + x610 + x 611 x17 + x27 + x37 + x 47 = x78 + x79 + x710 + x 711 xij  0 Solution:

x16 (Idaho − Kansas) = 1, 600 x27 (Nebraska − Arkansas) = 1,100 x35 (South Dakota − Indiana) = 1, 400 x46 (Michigan − Kansas) = 600 x59 (Indiana − Ohio) = 900 x510 (Indiana − South Carolina) = 500 x68 (Kansas − Maryland) = 1, 200 x610 Kansas − South Carolina) = 600 x611 (Kansas − Alabama) = 400 x711 = 1,100 Z = $25,192

S11-36 xij = containers shipped from European port i (where i = 1, 2, 3) to U.S. Port j (where j = 4, 5, 6, 7); from U.S. Port i (where i = 4, 5, 6, 7) to Inland Port j (where j = 8, 9, 10); from Inland Port i (where i = 8, 9, 10) to distribution center j (where j = 11, 12, 13, 14, 15) minimize Z = 1,725x14 + 1,800 x15 + 2,345x16 + 2,700 x17 + 1,825x24 + 1,750 x25 + 1,945x26

+ 2,320 x27 + 2,060 x34 + 2,175x35 + 2,050 x36 + 2, 475x37 + 825x48 + 545x49 + 320 x410 + 750 x58 + 675x59 + 450 x510 + 325x68 + 605x69 + 690 x610 + 270 x78 + 510 x79 + 1,050 x710 + 450 x811 + 830 x812 + 565x813 + 420 x814 + 960 x815 + 880 x911 + 520 x912 + 450 x913 + 380 x914 + 660 x915 + 1,350 x1011 + 390 x1012 + 1, 200 x1013 + 450 x1014 + 310 x1015 subject to

x14 + x15 + x16 + x17  125 x24 + x25 + x26 + x27  210 x34 + x35 + x36 + x37  160 x48 + x49 + x410  85 x58 + x59 + x510  110 x68 + x69 + x610  100 x78 + x79 + x710  130 x48 + x58 + x68 + x78  170 x49 + x59 + x69 + x79  240 x410 + x510 + x610 + x710  140 x811 + x911 + x1011 = 85 x812 + x912 + x1012 = 60 x813 + x913 + x1013 = 105 x814 + x914 + x1014 = 50 x815 + x915 + x1015 = 120 x14 + x24 + x34 = x48 + x49 + x410 x15 + x25 + x35 = x58 + x59 + x510 x16 + x26 + x36 = x68 + x69 + x610

x17 + x27 + x37 = x78 + x79 + x710 x48 + x58 + x68 + x78 = x811 + x812 + x813 + x814 + x815 x49 + x59 + x69 + x79 = x911 + x912 + x913 + x914 + x915 x410 + x510 + x610 + x710 = x1011 + x1012 + x1013 + x1014 + x1015 xij  0

Solution: x14 (Antwerp – Boston) = 85 x15 (Antwerp – Savannah) = 40 x25 (Barcelona – Savannah) = 70 x26 (Barcelona – Mobile) = 15 x27 (Barcelona – Houston) = 125 x36 (Cherbourg – Mobile) = 85 x410 (Boston – NC) = 85 x59 (Savannah – Texas) = 55 x510 (Savannah – NC) = 55 x68 (Mobile – Ohio) = 100 x78 (Houston – Ohio) = 70 x79 (Houston – Texas) = 55 x811 (Ohio – Phoenix) = 85 x813 (Ohio – KC) = 35 x814 (Ohio – Louisville) = 50 x912 (Texas – Columbus) = 40 x913 (Texas – KC) = 70 x1012 (NC – Phoenix) = 20 x1015 (NC – Memphis) = 120 Z = $1,179,400

S11-37. (a)

The model is the same as problem #S11-32 except the available shipments from the European ports are 5 each:

x14 + x15 + x16 + x17  5 x24 + x25 + x26 + x27  .5 x34 + x35 + x36 + x37  5 and the demand constraints for the U.S. distributor are:

x811 + x911 + x1011 = 1 x812 + x912 + x1012 = 1 x813 + x913 + x1013 = 1 x814 + x914 + x1014 = 1 x815 + x915 + x1015 = 1 Solution:

x24 = 5 x49 = 2 x410 = 3 x911 = 1 x913 = 1 x1012 = 1 x1014 = 1 x1015 = 1 Z = 144 days Phoenix: Barcelona – Boston – Texas (31) Columbus: Barcelona – Boston – North Carolina (28) Kansas City: Barcelona – Boston – Texas (29) Louisville: Barcelona – Boston – Texas (27) Memphis: Barcelona – Boston – Texas (29)

(b)

x14 = 2 x24 = 3 x49 = 2 x410 = 3 x911 = 1 x913 = 1 x1012 = 1 x1014 = 1 x1015 = 1 Z = 154 days

CASE S11.1: Stateline Shipping and Transport Company The total cost of this solution is $2,630 per week. There are multiple optimal solutions. The solution is summarized as follows.

1.

Kingsport

→ 2.

Danville = 16 bbls

1.

Kingsport

→ 3.

Macon = 19 bbls

2.

Danville

→ B.

Los Canos = 80 bbls

3.

Macon

→ C.

Duras = 78 bbls

4.

Selma

→ 3.

Macon = 17 bbls

4.

Selma

→ 5.

Columbus = 36 bbls

5.

Columbus

→ A.

Whitewater = 65 bbls

6.

Allentown

→ 2.

Danville = 38 bbls

CASE S11.2: Global Supply Chain Management at Centrex International European Port to U.S. Port Lisbon − Jacksonville = 2062.5 lb. goat Lisbon − Savannah = 2000 lb. goat Lisbon − Jacksonville = 1350 lb. lamb Marseilles − Savannah = 2500 lb. goat Marseilles − Savannah = 3000 lb. lamb Caracas − New Orleans = 4200 lb. goat Caracas − Jacksonville = 1237.5 lb. goat Caracas − New Orleans = 2375 lb. lamb Caracas − Jacksonville = 475 lb. lamb

U.S. Port to Distribution Center New Orleans − TN = 4200 lb. goat New Orleans–TN = 2375 lb. lamb Jacksonville–OH = 3000 lb. goat Jacksonville–NY = 300 lb. goat Jacksonville − OH = 1825 lb. lamb Savannah − NY = 4500 lb. goat Savannah − OH = 125 lb. lamb Savannah − NY = 2875 lb. lamb

Tanning Factory to Plant Mende − Limoges = 4000 lb. goat Mende − Limoges = 4400 lb. lamb Foggia − Naples = 3000 lb. lamb Saragossa − Madrid = 6500 lb. goat Saragossa − Madrid = 1300 lb. lamb Feira − Sao Paulo = 5100 lb. goat El Tigre − Caracas = 3600 lb. goat El Tigre − Sao Paulo = 2500 lb. lamb El Tigre − Caracas = 3200 lb. lamb

Plant to Port Madrid − Lisbon = 4062.5 lb. goat Madrid − Lisbon = 650 lb. lamb Naples – Marseilles 1500 lb. lamb Limoges − Marseilles = 2500 lb. goat Limoges − Lisbon = 700 lb. lamb Limoges − Marseilles = 1500 lb. lamb

Sao Paulo − Caracas = 3187.5 lb. goat Sao Paulo − Caracas = 1250 lb. lamb Caracas – Caracas = 2250 lb. goat Caracas – Caracas = 1600 lb. lamb Total cost = $606,965.63

12 Forecasting Answers to Questions 12-1.

Production planning, such as scheduling, inventory, process, facility layout and design, work force, and material purchasing, financial planning, including the development of budgets and capital expenditures; and various marketing functions are dependent on forecasting demand.

12-2.

Qualitative forecasting methods are subjective estimates based on judgment, opinion, past experience, and so on, whereas quantitative methods are mathematical, based on formulas.

12-3.

Short-range forecasts typically encompass the immediate future, in other words, several months, and are concerned with daily operations; medium-range forecasts encompass anywhere from several months up to several years and are used for annual budgets and production plans or the development of a project or program; long-range forecasts usually are for periods longer than one or two years and are used for strategic planning, such as new product development or new programs.

12-4.

All the elements of the supply chain including purchasing, inventory, production, scheduling, facility location, transportation and distribution are affected by forecasting. An inaccurate forecast can result in excessive costly inventories or frequent stockouts and late deliveries.

12-5.

Continuous replenishment requires that suppliers replenish a company’s inventory levels as products are demanded. The primary benefit of a continuous replenishment system is minimal inventory. Thus, it requires very accurate forecasts by suppliers to always be able to meet customer demand on very short notice. Without accurate inventories, suppliers must maintain high inventory levels themselves.

12-6.

Quality customer service means having products or services available when customers demand them, and, being able to deliver products and services on time. Without accurate forecasting of customer demand them is difficult to keep the appropriate amount of inventory on hand to meet demand in a timely manner without excessive costs.

12-7.

Qualitative methods are most often used for long-range strategic planning. Often called, “the jury of executive opinion” it uses judgment, expertise and opinion of knowledgeable people in a company. Other methods include consumer research, the Delphi method and consulting firms.

12-8.

The Delphi method uses the informed opinions, expertise and judgments of knowledgeable individuals and experts. A questionnaire is used to develop a consensus forecast of future trends and events. It’s especially useful for predicting technological advances.

12-9.

A trend is a gradual, long-term, up or down movement of demand; a cycle is an undulating up-and-down movement that repeats itself over a lengthy time span; a seasonal pattern is an oscillating movement in demand that occurs periodically and is repetitive.

12-10.

Exponential smoothing is a moving average that weights the most recent past data more strongly than more distant past data.

12-11.

Other ways used to obtain initial forecasts include taking an average of demand in preceding periods or making a subjective estimate. If forecasting has been a continual process, then preceding forecasts might be used.

12-12.

The higher the smoothing constant, the more sensitive the forecast will be to changes in recent demand.

12-13.

Adjusted exponential smoothing is the simple exponential smoothing forecast with a trend adjustment factor added to it.

12-14.

It is a judgmental choice, but in general, a high smoothing constant  reflects trend changes more than a lower 

12-15.

In a linear trend line equation, the independent variable, x, is always time.

12-16.

This question requires an opinion of the student, but in general, the appropriate model is determined primarily by the extent of any trend pattern.

12-17.

A linear trend model will not adjust to a change in trend as the adjusted exponential smoothing model will, thus limiting the trend line method to a shorter time frame.

12-18.

By summing the differences between the actual forecast and demand; a large positive value indicates the forecast is probably consistently low, whereas a large negative value implies the forecast is consistently high.

12-19.

The movement of a tracking signal is compared to control limits, as long as the tracking signal is within the control limits, the forecast is in control and not biased.

12-20.

This question requires a subjective estimate on the part of the student. A particular method might be viewed as being superior because it is easier to use (compared to the other methods), it is easier to interpret, it makes more sense, it can be used alone rather than in comparison, or it seems to fit the data better.

12-21.

Linear regression relates demand to one other independent variable, whereas multiple regression reflects the relationship between a dependent variable and two or more independent variables.

12-22.

y is the dependent variable, x is the independent variable, a is the intercept, and b is the slope of the line.

12-23.

The Delphi method might be an appropriate method to use to forecast technological advances in video equipment, whereas market/consumer research could be used to forecast consumer demand. Various individuals in-house might also be able to assist in developing a forecast.

Solutions to Problems 12-1. a., b.

Month January February March April May June July August September October November December January c.

Sales 9.00 7.00 10.00 8.00 7.00 12.00 10.00 11.00 12.00 10.00 14.00 16.00 —

3-Month Moving Average — — — 8.67 8.33 8.33 9.00 9.67 11.00 11.00 11.00 12.00 13.33

5-Month Moving Average — — — — — 8.20 8.80 9.40 9.60 10.40 11.00 11.40 12.60

3-month MAD = 1.89; 5-month MAD = 2.43. The dealer should use the 3-month forecast for January because the smaller MAD indicates a more accurate forecast.

12-2. a., b.

Month 1 2 3 4 5 6 7 8 9 c.

Sales 5 10 6 8 14 10 9 12 —

3-Month Moving Average — — — 7.00 8.00 9.33 10.67 11.00 10.33

Weighted 3-Month Moving Average — — — 17.20 7.58 11.06 11.08 9.93 10.77

3-month MAD = 2.07; weighted 3-month MAD = 2.49. The 3-month moving average forecast appears to be slightly more accurate.

12-3. a., b., and c.

Quarter 1 2 3 4 5 6 7 8 9 10 11 12 13 d.

Demand 105.00 150.00 93.00 121.00 140.00 170.00 105.00 150.00 150.00 170.00 110.00 130.00 —

3-Quarter Moving Average Forecast — — — 116.00 121.33 118.00 143.67 138.33 141.67 135.00 156.67 143.33 136.67

Error — — — 5.00 18.67 52.00 –38.67 11.67 8.33 35.00 –46.67 –13.33 —

5-Quarter Moving Average — — — — — 121.80 134.80 125.80 137.20 143.00 149.00 137.00 142.00

Error — — — — — 48.20 –29.80 24.20 12.80 27.00 –39.00 –7.00 —

Weighted 3-Quarter Moving Average — — — 113.95 116.69 125.74 151.77 132.40 138.55 142.35 160.00 136.69 130.20

Error — — — 7.15 23.31 44.26 –46.77 17.60 11.45 27.65 –50.00 –6.60 —

Cumulative errors are: 3-quarter moving average, E = 32.0 5-quarter moving average, E = 36.4 Weighted 3-quarter moving average, E = 28.05 The weighted 3-quarter forecast appears to be the most accurate. All the forecasts exhibit a low bias.

There appears to be a slight upward trend in the demand data and a pronounced seasonal pattern with a peak increase during the second quarter each year, followed by a substantial decrease in the third quarter.

12-4.

12-5. a., b.

Semester 1 2 3 4 5 6 7 8 9 c.

Enrollment 270 310 250 290 370 410 400 450 —

3-Semester Moving Average — — — 276.67 283.33 303.33 356.67 393.33 420.00

Exponentially Smoothed Forecast — 270.00 278.00 272.40 275.92 294.74 317.79 334.23 357.38

3-semester MAD = 61.33; exponentially smoothed MAD = 70.42; 3-semester moving average appears to be slightly more accurate.

12-6. a., b.

Month

Demand

Exponentially Smoothed Forecast (  = 0.30 )

October November December January February March April May June July August

800 725 630 500 645 690 730 810 1200 980 —

800.00 800.00 777.50 733.25 663.27 657.79 667.45 686.21 723.35 866.34 900.44

c.

Adjusted Exponentially Smoothed Forecast (  = 0.30,  = 0.20 ) — 800.00 773.00 720.70 639.23 637.46 653.18 678.55 724.64 895.98 930.96

Exponentially smoothed MAPD = 1, 282.86 / 7, 710 = 0.166 = 16.6%; MAPD = 1282.86/6910 = 18.6%

Adjusted forecast MAPD = 1, 264.59 / 7, 710 = 0.1640 = 16.4%. MAPD = 1264.59/6910 = 18.3% Both forecasts appear to be approximately equally accurate.

12-7.

Month

Price

Exponentially Smoothed Forecast (  = 0.40 )

1 2 3 4 5 6 7 8 9 10 11

62.70 63.90 68.00 66.40 67.20 65.80 68.20 69.30 67.20 70.10 —

62.70 62.70 63.18 65.10 65.62 66.25 66.07 66.92 67.87 67.60 68.60

Cumulative Error MAD

Adjusted Exponentially Smoothed Forecast (  = 0.40,  = 0.30 )

Linear Trend Line y = 63.54 + 0.607 x

— 62.70 63.32 65.78 66.25 66.88 66.46 67.45 68.53 67.98 69.17

64.15 64.75 65.36 65.97 66.57 67.18 67.79 68.39 69.01 69.61 70.22

Exponentially Smoothed

Adjusted Exponentially Smoothed

Linear Trend

14.75 1.89

10.73 1.72

— 1.09

The linear trend line forecast appears to be the most accurate.

12-8.

Year

Occupancy Rate

Exponentially Smoothed Forecast (  = 0.20 )

1 2 3 4 5 6 7 8 9 10

.75 .70 .72 .77 .83 .81 .86 .91 .87 —

.75 .75 .74 .74 .74 .76 .77 .79 .81 .82

E MAD

Exponentially Smoothed Forecast .046 .064

Adjusted Forecast .044 .061

Adjusted Exponentially Smoothed Forecast (  = 0.20,  = 0.20 )

Linear Trend Line y = .683 + .024 x

— .75 .74 .73 .74 .76 .77 .80 .82 .83

.71 .73 .76 .78 .80 .83 .85 .88 .90 .92

Linear Trend Trend Forecast — 0.026

The linear trend line forecast appears to be the most accurate.

12-9. a.

3-month moving average forecast for month 21 = 74.67

MAD = 3.12 b.

3-month weighted moving average forecast for month 21 = 75.875

MAD = 2.98 c.

Exponentially smoothed forecast ( = 0.40 ) for month 21 = 74.60

MAD = 2.87 MAD = 2.99 d.

The lowest MAD values are with both the weighted 3-month moving average forecast and the exponentially smoothed forecast.

12-10. Group data into 3-month periods to forecast periods 19, 20 and 21.

Possible models include the following: Exponential smoothing forecasts ( = 0.3) . F19, 20, 21 = 51.67, MAD = 18.93 Linear trend line forecasts

F19, 20, 21 = 69.27, MAD = 1.10

12-11.

Quarter 1 2 3 4 5 6 7 8 9 10 11 12 13

Ice Cream Sales 350 510 750 420 370 480 860 500 450 550 820 570 —

Exponentially Smoothed Forecast (  = 0.50 ) 350.00 350.00 430.00 590.00 505.00 437.50 458.75 659.37 579.69 514.84 532.42 676.21 623.11

E = 26.30 E = 289.336

The forecast seems to be biased low.

12-12. Seasonal factors: Quarter 1: 1,170 / 6, 630 = 0.18 Quarter 2: 1,540 / 6, 630 = 0.23 Quarter 3: 2, 430 / 6, 630 = 0.37 Quarter 4: 1, 490 / 6, 630 = 0.22 Forecast for 2005: y = 1,850 + 180 x = 2,570 Seasonally adjusted forecasts: Quarter 1: 2,570 ( 0.18 ) = 453.53 Quarter 2: 2,570 ( 0.23) = 596.95 Quarter 3: 2,570 ( 0.37 ) = 941.95 Quarter 4: 2,570 ( 0.22 ) = 577.57 The seasonal factor seems to provide a more accurate forecast.

12-13. Seasonal factors: Quarter 1: 395 /1,594 = 0.25 Quarter 2: 490 /1,594 = 0.31 Quarter 3: 308 /1,594 = 0.19 Quarter 4: 401/1,594 = 0.25 Forecast year 4: y = 440.33 + 45.5 x = 622.33

Adjusted Exponentially Smoothed Forecast (  = 0.50,  = 0.50 ) — 350.00 470.00 690.00 512.50 407.50 454.37 757.50 588.91 487.03 527.30 745.55 631.22

Error — 160.00 280.00 –270.00 –142.50 72.50 405.62 –257.50 –138.91 62.97 292.69 –175.55 —

Seasonally adjusted forecasts: Quarter 1: 622.33 ( 0.25) = 155.6 Quarter 2: 622.33 ( 0.31) = 192.9 Quarter 3: 622.33 ( 0.19 ) = 118.2 Quarter 4: 622.33 ( 0.25) = 155.6 12-14.

Day 1 2 3 4 5 6 7 8

Daily Demand 212 182 215 201 158 176 212 188  D = 1,544

S1 (10 A.M. − 3 P.M.) = S2 ( 3 P.M. − 7 P.M.) =

D1

D

=

D2

=

567 = 0.37 1,544



S3 ( 7 P.M. − 11 P.M.) = S4 (11 P.M. − 2 A.M.) =

D

389 = 0.25 1,544

D3

D

=

320 = 0.21 1,544

D4

=

268 = 0.17 1,544

D

Linear trend forecast for day 9: y = 202.54 − 2.12 x = 183.46

Day 9 forecast for 10 A.M. − 3 P.M. : 183.46 ( 0.25) = 45.87

3 P.M. − 7 P.M. : 183.46 ( 0.37 ) = 67.88 7 P.M. − 11 P.M. : 183.46 ( 0.21) = 38.52 11 P.M. − 2 A.M. : 183.46 ( 0.17 ) = 31.18

12-15.

Year

Sales

Exponentially Smoothed Forecast (  = 0.30 )

1 2 3 4 5 6 7 8 9

4,260.00 4,510.00 4,050.00 3,720.00 3,900.00 3,470.00 2,890.00 3,100.00 —

— 4,260.00 4,335.00 4,249.50 4,090.65 4,033.45 3,864.42 3,572.09 —

MAD E

Adjusted Exponentially Smoothed Forecast 431.71 –2.522

Adjusted Exponentially Smoothed Forecast (  = 0.30,  = 0.20 )

Linear Trend Line y = 4, 690 − 211.67 x

— 4,260.00 4,350.00 4,244.40 4,054.80 3,993.34 3,798.52 3,460.91 3,313.19

4,478.33 4,266.67 4,055.00 3,843.33 3,631.67 3,420.00 3,208.33 2,996.67 2,785.00

Linear Trend Line 166.25

The linear trend line forecast appears to be the most accurate.

12-16. a.

Seasonally adjusted forecast

January − March :

D1 = 106.8

April − June :

D2 = 135.6

July − September :

D3 = 109.0

October − December : D4 = 233.6

D1

=

106.8 = 0.18 585

D2

D

=

135.6 = 0.23 585

D3

D

=

109.0 = 0.19 585

D4

=

233.6 = 0.40 585

s1 =



s2 = s3 = s4 =

D

D

Linear trend line forecast for 2011: y = 96.33 + 6.89 x = 137.67 January–March forecast for 2011: SF1 = S1 F6 = 0.18 (137.67 ) = 24.78 April–June forecast for 2005: SF2 = S2 F6 = 0.23 (137.67 ) = 31.66

July–September forecast for 2005: SF3 = S3 F6 = 0.19 (137.67 ) = 26.16 October–December forecast for 2005: SF4 = S4 F6 = 0.40 (137.67 ) = 55.07 b.

Linear trend line forecast for January–March 2005: y = 16.29 + 1.69 x = 26.43

Linear trend line forecast for April–June 2005: y = 21.75 + 1.79 x = 32.49

Linear trend line forecast for July–September 2005: y = 18.35 + 1.15 x = 25.25

Linear trend line forecast for October–December 2005: y = 39.94 + 2.26 x = 53.50

c.

Year/Quarter

Orders

Seasonally Adjusted Forecast

2006 Jan–Mar Apr–June Jul–Sep Oct–Dec 2007 Jan–Mar Apr–June Jul–Sep Oct–Dec 2008 Jan–Mar Apr–June Jul–Sep Oct–Dec 2009 Jan–Mar Apr–June July–Sep Oct–Dec 2010 Jan–Mar Apr–June July–Sep Oct–Dec

18.6 23.5 20.4 41.9 18.1 24.7 19.5 46.3 22.4 28.8 21.0 45.5 23.2 27.6 24.4 47.1 24.5 31.0 23.7 52.8

18.58 23.74 19.61 41.29 19.82 25.32 20.92 44.04 21.06 26.91 22.23 46.80 22.30 28.49 23.54 49.56 23.54 30.08 24.85 52.31

Seasonally adjusted forecast MAD =

Linear Trend Line Forecast

Dt − Ft

0.02 0.24 0.79 0.61 1.72 0.62 1.42 2.26 1.34 1.89 1.23 1.30 0.90 0.89 0.86 2.50 0.96 0.92 1.15 0.49  | Dt − Ft | = 22.07

17.98 23.54 19.50 42.20 19.67 25.33 20.65 44.46 21.36 27.12 21.80 46.72 23.05 28.91 22.95 48.98 24.74 30.70 24.10 51.24

Dt − Ft

0.62 0.04 0.90 0.30 1.57 0.63 1.15 1.84 1.04 1.68 0.80 1.22 0.15 1.31 1.45 1.88 0.24 0.30 0.40 1.56  | Dt − Ft | = 19.08

|D − F | = 22.07 = 1.10

Linear trend forecast for seasons MAD =

1

1

n

20

|D − F | = 19.08 = 0.954 1

n

1

20

Although both forecasts seem to be relatively accurate, the linear trend line forecast for each season is slightly more accurate according to MAD.

S ( Fall ) =

173.3 = .215 806.1

S ( Winter ) =

155.9 = .193 806.1

S ( Spring ) =

203.5 = .252 806.1

S ( Summer ) =

273.4 = .339 806.1

12-17.

y = 195.55 + 2.39 ( 5 ) = 207.5 Forecasts for 2005: Fall: ( 207.5)(.215) = 44.61

( 207.5)(.193) = 40.13 ( 207.5)(.252 ) = 52.38 ( 207.5)(.339 ) = 70.34

Winter: Spring: Summer:

Yes, there does appear to be a seasonal pattern.

12-18.

Year Time 7:00 AM 8:00 9:00 10:00 11:00 Noon 1:00 PM 2:00 3:00 6:00 7:00 8:00 9:00 Total

1 56 31 15 34 45 63 35 24 27 31 25 14 10 410

2 64 41 22 35 52 71 30 28 19 47 35 20 8 472

3 66 37 24 38 55 57 41 32 24 36 41 18 16 485

4 60 44 30 31 49 65 42 30 23 45 43 17 14 493

Linear Trend Line:

410.40 21.03 b= Linear trend line forecast for year 7 = 557.6 a=

Year 7 Forecasts:

5 72 52 19 28 57 75 33 35 25 40 39 23 15 513

6 65 46 26 33 50 70 45 33 27 46 45 27 18 531

Total 383 251 136 199 308 401 226 182 145 245 228 119 81 2904

SF1 (7:00)= SF2 (8:00)= SF3 (9:00)= SF4 (10:00)= SF5 (11:00)= SF6 (noon)= SF7 (1:00)= SF8 (2:00)= SF9 (3:00)= SF10 (6:00)= SF11 (7:00)= SF12 (8:00)= SF13 (9:00) =

73.54 48.19 26.11 38.21 59.14 77.00 43.39 34.95 27.84 47.04 43.78 22.85 15.55

12-19.

Year 1 2 3 4 5 6 9

Pool Attendance 410 472 485 493 513 531

Exponentially Smoothed Forecast 410.00 410.00 428.60 445.52 459.76 475.73 492.31

Error

Trend

62.00 56.40 47.48 53.24 55.27

0.0000 3.7200 6.3600 7.9368 9.5436 10.951

12-20.

Week 1 2 3 4 5 6 7 8 9 10 Total Linear Trend Line: a = 186.93 b = 5.21

Patients per Period Weekend Weekdays 105 73 119 85 122 89 128 83 117 96 136 78 141 91 126 100 143 83 140 101 1277 879

Total 178 204 211 211 213 214 232 226 226 241 2156

Adjusted Smoothed Forecast 410.00 432.32 451.88 467.70 485.28 503.27 MAD =

Error 62.00 52.68 41.12 45.30 45.72 49.364

Linear trend line forecast for Week 11 = 244.27 SF1 (weekend) = 144.68 SF2 (weekdays) = 99.59

 D = 2,156

12-21.

S1 =

1, 277 = .592 2,156

S2 =

879 = .408 2,156

Linear trend line forecast for week 11: y = 186.9 + 5.21x = 244.27 Weekend forecast: ( 244.27 )(.592 ) = 144.68 Weekday forecast: ( 244.27 )(.408) = 99.59

Month

Actual Demand

Forecast Demand

1 2 3 4 5 6 7 8 9 10

160 150 175 200 190 220 205 210 200 220

170 165 157 166 183 186 203 204 207 203

Cumulative error Bias MAD MAPD

86.00 8.60 15.00 0.08

Error

Dt − Ft

–10 –15 18 34 7 34 2 6 –7 17 86

10 15 18 34 7 34 2 6 7 17 150

Running MAD

Cumulative Error

Tracking Signal

10.00 12.50 14.33 19.25 16.80 19.67 17.14 15.75 14.78 15.00

–10 –25 –7 27 34 68 70 76 69 86

–1.00 –2.00 –0.49 1.40 2.02 3.46 4.08 4.83 4.67 5.73

There is really no way to determine if this is an accurate forecast method unless it is compared with some other method.

12-22. See Problem 12-21 solution for tracking signal values.

The forecast appears to be biased low (i.e., actual demand exceeds the forecast).

12-23.  = 239.38; There does not seem to be any high or low bias in the forecast.

12-24.

Year 1 2 3 4 5 6 7 8

Error — 250 –300 –524.4 –154.8 –523.34 –908.52 –360.91

Cumulative Error — 250 –50 –574.4 –729.4 –1,252.54 –2,161.06 –2,521.97

Running MAD — 250.00 275.00 358.13 307.30 350.51 443.51 431.71

Tracking Signal 1.00 –0.18 –1.60 –2.37 –3.57 –4.87 –5.84

The control chart suggests the forecast is not performing accurately and is consistently biased high (i.e., the actual demand is consistently lower than the forecast).

12-26. a.

Month March April May June July August September October November

Demand

Forecast

Error

Dt − Ft

120 110 150 130 160 165 140 155

— 120.0 116.0 129.6 129.7 141.8 151.1 146.7 150.0

— –10.00 34.00 0.40 30.30 23.20 –11.10 8.30 75.10

— 10.00 34.00 0.40 30.30 23.20 11.10 8.30 117.30

Bias MAD MAPD Cumulative error

10.73 16.76 0.1038 75.10

b.

Month

Demand

3-Month Moving Average

March April May June July

120 110 150 130 160

— — — 126.67 130.00

Error

Dt − Ft

— — — 3.33 30.00

— — — 3.33 30.00

Running MAD

Cumulative Error

Tracking Signal

— 10.00 22.00 14.80 18.67 19.58 18.17 16.76

— –10.0 24.0 24.4 54.7 77.9 66.8 75.1

— –1.00 1.09 1.65 2.93 3.98 3.67 4.48

Month

Demand

August September October November

165 140 155

3-Month Moving Average 146.67 151.67 155.00 153.33 8.00 12.67 0.08 276.64 39.99

Bias MAD MAPD MSE Cumulative error

Error

Dt − Ft

18.33 –11.67 0.00 39.99

18.33 11.67 0.00 63.33

The 3-month moving average seems to provide a better forecast. c.

The tracking signal moves beyond the 3 MAD control limit for July and continues increasing indicating the forecast is consistently biased low.

12-26. The 3-month moving average forecast appears to be more accurate.

Month

Demand

Forecast

Dt − Ft

Month

Demand

Forecast

Dt − Ft

January February March April May June July

9 7 10 8 7 12 10

9.00 9.00 8.60 8.88 8.70 8.36 9.09

— 2.00 1.40 0.88 1.70 3.64 0.91

August September October November December January

11 12 10 14 16 —

9.27 9.62 10.09 10.07 10.86 11.88

1.73 2.38 0.09 3.92 5.14 —

Mad

Moving Average Forecast (Prob. 12-1a) 1.89

Exponentially Smoothed (Prob. 12-20) 2.16

12.27. MAD = 1.79 Cumulative error = 12.36

According to these measures, the forecast appears to be fairly accurate.

Year 1 2 3 4 5 6 7 8

Demand 16.8 14.1 15.3 12.7 11.9 12.3 11.5 10.4

Forecast 16.8 16.8 15.7 15.5 14.4 13.4 12.9 12.4

Error 0 −2.7 −0.4 −2.8 −2.5 −1.1 −1.4 −1.6

Running MAD — 2.70 1.55 1.97 2.10 1.90 1.82 1.79

Cumulative Error — −2.7 −3.1 −5.9 −8.4 −9.5 −10.9 −12.5

Tracking Signal — −1.00 −2.00 −3.00 −4.00 −5.00 −6.00 −7.00

The tracking signal goes outside of the control limits beginning in year 4, indicating a forecast that is biased high.

Linear trend model: Year 1 2 3 4 5 6 7 8 9

Demand 16.80 14.10 15.30 12.70 11.90 12.30 11.50 10.80 —

Forecast 15.87 15.10 14.33 13.56 12.79 12.02 11.24 10.47 9.70

MAD = 0.68 The linear trend line forecast appears to be more accurate for MAD. 12-28. y = 49.95 + 0.428 x, where y = occupancy rate and x = wins.

Forecast if the Blue Sox win 85 games: 49.95 + .428 ( 85 ) = 86.3 occupancy rate. yes; r = .8626 12-29. a.

y = .51 + .403x, where y = sales and x = permits.

Forecast if 25 permits are filed: .51 + .403 ( 25 ) = 10.57 b.

12-30. a.

The correlation coefficient is .914 indicating a strong causal relationship. y = −50.21 + 2.056 x, where y = gallons of ice cream and x = temperature. Forecast for temperature of

80 : − 50.21 + 2.056 ( 80 ) = 114.29 gal. b.

The correlation coefficient is 0.833, indicating a strong causal relationship.

12-31. Coefficient of determination = ( 0.833) = 0.694, indicating that 69.4% of the variation of ice cream sales 2

can be attributed to the temperature.

12-32. a.

y = 7039.24 − 0.337 x where y = applications and x = tuition.

If tuition is $10,000 forecast is 7039.24 −.337 (10, 000 ) = 3, 670.12 applications.

If tuition is $7,000, forecast is 7039.24 − .337 ( 7000 ) = 4, 680.86 applications. b. c.

The correlation coefficient is −.808, indicating a fairly strong linear relationship between tuition costs and number of applicants.

Number of class sections, number of dormitory rooms, number of persons per class, plus numerous budgeting decisions.

12-33. y = 15.864 − 0.575 x

r = −0.785

r 2 = 0.616 There seems to be a relatively strong relationship between production time and defects.

Forecast for “normal” production time of 23 minutes:

y = 15.864 − 0.575 ( 23) y = 2.64% defects

12-34. y = 1.317 + 0.151x r = .744, which indicates a fairly strong relationship between hits and orders

r 2 = 0.553, which means 55.3% of the variation in orders can be attributed to the number of web site hits.

At 60,000 hits/month, y = 1.317 + .151( 60 ) =10.4 or 10,400 orders

12-35.

Year 1 2 3 4 5 6 7 8 9 10 11

Application 6,010 5,560 6,100 5,330 4,980 5,870 5,120 4,750 4,615 4,100 —

Linear Trend Line Forecast 6,069.72 5,886.12 5,702.51 5,518.91 5,335.30 5,151.69 4,968.09 4,784.78 4,600.88 4,417.27 4,233.66

y = 6253.33 − 183.606 x

Correlation coefficient = −0.850 a.

The linear regression forecast (from Problem 12-30) has a MAD value of 310 whereas the MAD value for the linear trend line forecast in this problem is 256, indicating that the linear trend line forecast is somewhat better.

b.

The correlation coefficient is −0.850, indicating a strong relationship between applications and time.

12-36. The slope, b = 2.98, indicates the rate of change, that is, the number of gallons sold for each degree increase in temperature.

y = 380.93 + 16.03 x

12-37. a.

y (11) = 380.93 + 16.03 (11)

= 557.26 printers y = −22.07 + .41x

b.

y = −22.07 + .41(1500 )

= 594.10 c.

MAD for the linear trend line forecast in a. equals 85.69 while MAD for the linear regression forecast in b. equals 45.20. In addition, the correlation coefficient for the linear trend is r = 0.426 whereas the correlation coefficient for the linear regression is r = .848. This evidence seems to indicate the forecast model in b is best.

12-38.

Year 1 2 3 4 5 6 7 8 9 10 11

Demand 381 579 312 501 296 415 535 592 607 473

Forecast — 381.00 440.40 401.88 431.62 390.93 398.85 439.21 485.04 521.63 507.04

The exponential smoothing forecast ( MAD = 104.54 ) appears to be less accurate than the linear regression forecast ( MAD = 45.20 ) developed in 12-35a.

12-39. a.

Seasonally adjusted forecast.

Quarter 1: D1 = 306 Quarter 2: D2 = 334 Quarter 3: D3 = 404 Quarter 4: D4 = 348

S1 =

306 = 0.220 1392

S2 =

354 = 0.240 1392

S3 =

404 = 0.290 1392

S4 =

348 = 0.250 1392

Linear trend line forecast for year 6: ; y = 271.2 + 2.4 x ; y ( 6 ) = 271.2 + 2.4 ( 6 ) = 285.6

SF1 = ( 0.220 )( 285.6 ) = 62.78 SF2 = ( 0.240 )( 285.6 ) = 68.53

SF3 = ( 0.290 )( 285.6 ) = 82.82

SF4 = ( 0.250 )( 285.6 ) = 71.40 b.

Quarter 1: y = 52.69 + 0.3973 x

y ( 20 ) = 56.69 + 0.3973 ( 20 )

= 64.64 =60.64 Quarter 2: y = 91.62 − 0.71x

y ( 36 ) = 91.62 − ( 0.71)( 36 )

= 66.06 Quarter 3: y = 73.57 + 0.29 x

y ( 25 ) = 73.57 + 0.29 ( 25 )

= 80.82 Quarter 4: y = 37.47 + 1.06 x

y ( 30 ) = 37.47 + 1.06 ( 30 )

= 69.27 c.

This is an intuitive assessment, which managers must do on occasion. In general, the linear regression forecast provides a more conservative estimate.

12-40. The adjusted exponentially smoothed forecast ( = 0.4,  = 0.4 ) has a first quarter forecast for year 6 of 75.68 percent seat occupancy. It has a E (bias) value of 1.08 and a MAD value of 8.6, which seem low.

Thus, this may be the best overall forecast model compared to the one developed in 12-37a.

12-41.

The following table shows several different forecast models developed using Excel and selected measures of forecast accuracy.

Year 25 Forecast 5.89

MAD 1.58

E (bias) .127

8.22

1.86

0.000

6.64

1.59

−0.330

Exponential smoothing ( = 0.5)

6.13

1.29

.031

Exponential smoothing ( = 0.3,  = 0.4 )

6.24

1.33

−0.020

Exponential smoothing ( = 0.4,  = 0.5)

5.94

1.22

0.003

Forecast Method Moving average ( n = 3) Linear trend line Exponential smoothing ( = 0.3)

Although this selection of forecast models is not exhaustive, it does seem to indicate the exponential smoothing models are the most accurate, especially the adjusted model with (  = 0.4 and  = 0.5 ).

12-42.

y = 13.8399 + .00150 x y (12000 ) = 31.79 r = 0.95

12-43. (a) Forecast of applicants:

y = 13,803.07 + 470.55 x y (11) = 18,979.12 applicants Forecast of % acceptances:

y = 37.72 + .247 x y (11) = 40.44% Estimated offers = 5, 000 / .4044 = 12, 634

% offers = 12,364 /18,979.12 = 65.15%

(b) Forecast of % offers:

y = 83 − 1.68 x y (11) = 64.54% (c) If the forecast of % acceptances is accurate then the number of applicants is not relevant; 12,634 offers will yield 5,000 acceptances.

12-44. (a) y = 381.32 + 68.40 x

y (11) = 1, 270.48 (b) r = .973 There appears to be a very strong linear relationship

12-45. (a) y = 219.27 + 12.28 x

y (16) = 415.67 (b) y = −5349.77 + .147 x

y (39,300) = 438.31 r = .966 y (40, 000) = 541.41 The club should use the linear regression model. The correlation coefficient shows that town population is a good predictor of the growth in the number of club players plus it provides a more favorable forecast for the club.

12-46. (a) y = 116.12 − 1.28 x

y (70) = 116.12 − 1.28(70) = 26 r 2 = .537 (b) y = 116.9 − 1.24 x1 − 0.14 x2

y (70, 40) = 116.9 − 1.24(70) − 0.14(40) = 25 r 2 = .538 Very little difference between the two forecasts. Annual budget appears to replicate endowment.

12-47. y = 1.704 + 0.269 x, where y = sales and x = promotional expenditures.

Correlation coefficient = 0.546 The correlation coefficient indicates a weak linear relationship between sales and promotion, thus a linear regression model should not be used.

12-48. We tested 3 forecasting methods, as follows.

Month

Demand

1 2 3 4 5 6 7 8 9 10 11 12 13 15 16 17 14 18 19 20 21 22 23 24 25

8.20 7.50 8.10 9.30 9.10 9.50 10.40 9.70 10.20 10.60 8.20 9.90 10.30 11.70 9.80 10.80 10.50 11.30 12.60 11.50 10.80 11.70 12.50 12.80 —

Linear Trend Line Forecast

3-Month Moving Average

Adjusted Exponentially Smoothing Forecast (  = 0.30,  = 0.50 )

— — — 7.93 8.30 8.83 9.30 9.67 9.87 10.10 10.17 9.67 9.57 10.23 10.83 10.67 9.47 10.77 10.63 11.57 11.80 11.63 11.33 11.67 12.33

18.20 8.20 7.99 8.02 8.40 8.61 8.88 9.34 9.44 9.67 9.95 9.42 9.57 10.00 10.51 10.30 9.79 10.45 10.70 11.27 11.34 11.18 11.33 11.68 12.29

8.24 8.42 8.59 8.77 8.95 9.13 9.31 9.49 9.67 9.84 10.02 10.20 10.38 10.74 10.92 11.09 10.56 11.27 11.45 11.63 11.81 11.99 12.17 12.34 12.52

Forecasting Alternatives

MAD

E

Linear trend line

0.546

—

3-month moving average

0.825

9.2

Adjusted exponential smoothing

0.817

9.47

All three methods we chose to evaluate appear to be relatively accurate. The student might select another method that will be more accurate.

12-49. a.

y = 745.91 − 2.226 x1 + 0.163x2

b.

r 2 = 0.992

c.

y = 7,186.91

12-50. a.

y = 608.795 + 0.215x1 − 0.985x2 y = 144.67 + 0.371X1 – 0.307X2

b.

r 2 = 0.766

c.

y = 608.795 + 0.215 (1,500 ) − 0.985 ( 300 ) = 635.79 = 144.67 + 0.371(1500)-0.307(300) = 608.50

12-51. a.

y = 219.167 − 0.027 x1 + 233.871x2

b.

r 2 = 0.956

c.

y = 219.67 − 0.027 (10, 000 ) + 233.871( 4 ) = $882.82

12-52. Selected forecast models 5-day moving average forecasts for day 21: 11 − 12 = 82.11, MAD = 12.40 12 − 1 = 128.4, MAD = 28.32

1 − 2 = 93.0, MAD = 15.82

Exponentially smoothed ( = 0.3) forecasts for day 21: 11 − 12 = 82.11, MAD = 12.40 12 − 1 = 129.7, MAD = 26.36 1 − 2 = 99.61, MAD = 14.23

Linear trend line forecasts for day 21: 11 − 12 = 81.86, MAD = 11.25 12 − 1 = 132.42, MAD = 22.14 1 − 2 = 103.5, MAD = 12.44

The “best” forecast model depends on what models are selected for comparison. For the models tested above, they all seem to be relatively close, although the linear trend model consistently had the highest next period forecast and a slightly lower MAD value.

12-53. a.

y = 43.09 + .0007 x1 + 1.397 x2

where y = SOL scores

x1 = average teacher salary

x2 = average tenure b.

r 2 = 0.696 Approximately 70% of the amount of variation in SOL scores can be attributed to teacher salaries and tenure. This is a moderately strong relationship indicating the superintendent is at least partially right.

c.

y = 43.09 + .0007 ( 30, 000 ) + 1.397 ( 9 ) = 76.66 No, the SOL score would only increase to 76.66.

CASE 12.1: Forecasting at State University Forecasting would be appropriate in a number of different areas. The university needs to be able to forecast future applications and enrollments both in the short and long term. A forecast of the college age population that will apply to State is very important for planning purposes. A multiple regression model that related applications to variables such as population, tuition levels, and entrance requirements would probably be most appropriate for this purpose.

Internal forecasts for classroom space, facilities, dormitory space, dining, etc., would enhance the planning process. Times series methods would probably be sufficient for this type of forecasting.

The university might consider using a forecasting model to determine future funding from the state. Several models, such as a multiple regression and perhaps a qualitative technique like the Delphi method might be combined. Forecasts for other sources of funding such as endowments and tuition increases could be forecast using more conventional methods such as regression or time series.

The university’s TQM approach requires a forecast of what customers perceive educational quality to be in the future—that is, a definition of quality according to students, parents, and legislators. In-house forecasting using key administrators, faculty, and students might be appropriate. Surveys and market research techniques of alumni, students, and parents might be useful in determining what quality factors will be important in the future.

CASE 12.2: The University Bookstore Student Computer Purchase Program The following table shows several different forecast models developed using POM for Windows and selected measures of forecast accuracy.

Year 25 Forecast 1,004.66

MAD 96.96

E (bias) 66.00

1,020.07

73.24

0.00

941.53

126.88

108.59

Exponential smoothing ( = 0.5)

1,003.70

104.95

74.72

Exponential smoothing ( = 0.3,  = 0.4 )

983.22

109.58

62.19

Exponential smoothing ( = 0.4,  = 0.5)

1,031.09

105.13

61.31

Forecast Method Moving average ( n = 3) Linear trend line Exponential smoothing ( = 0.3)

Although this selection of different models is not exhaustive, it does seem to indicate that the linear trend line model is the best.

Other forecast models that the bookstore might consider include forecasts of student enrollment and entering freshmen. Also for longer term forecasts the bookstore could investigate which different majors and classes might be moving to more extensive computer usage in the future, thus driving up long run student demand. Additionally forecasts for other products would help the bookstore plan their inventory, warehouse usage and distribution better.

CASE 12.3: Cascades Swim Club Attendance

Week Day

1

2

3

4

5

6

7

8

9

10

11

12

13

Total

M

139

198

341

287

303

242

194

197

275

246

224

258

235

3,139

T

273

310

291

247

223

177

207

273

241

177

239

130

218

3,006

W

172

347

380

356

315

245

215

213

190

161

274

195

271

3,334

Th

275

393

367

322

258

390

304

303

243

308

205

238

259

3,865

F

337

421

359

419

193

284

331

262

277

256

361

224

232

3,956

Sa

402

595

463

516

378

417

407

447

241

391

411

368

317

5,353

Su

487

497

578

478

461

474

399

399

384

400

419

541

369

5,886

Total

2,085

2,761

2,779

2,625

2,131

2,229

2,057

2,094

1,851

1,939

2,133

1,954

1,901

28,539

The seasonal factors for each weekday are as follows:

S1 ( Monday ) =

3,139 = .110 28,539

S2 ( Tuesday ) =

3, 006 = .105 28,539

S3 ( Wednesday ) =

S4 ( Thursday ) = S5 ( Friday ) =

3,334 = .117 28,539

3,865 = .135 28,539

3,956 = .139 28,539

S6 ( Saturday ) =

S7 ( Sunday ) =

5,353 = .188 28,539

5,886 = .206 28,539

The linear trend line equation computed from the 13 weekly totals is, y = 2,598.2308 − 57.5604 x

Using this forecast model to forecast weekly demand for each of the 13 weeks for the next summer and multiplying each weekly forecast by the daily seasonal factors will give the daily forecast for the next summer. For example, the daily forecast for week 1 is computed as,

y = 2,598.2308 − 57.5604 (1) = 2,540.67 Week 1 Forecasts

Monday = (.110 )( 2,540.67 ) = 279.5

Tuesday = (.105 )( 2,540.67 ) = 266.8 Wednesday = (.117 )( 2,540.67 ) = 297.3 Thursday = (.135)( 2,540.67 ) = 343.0

Friday = (.139 )( 2,540.67 ) = 353.2 Saturday = (.188)( 2,540.67 ) = 477.6 Sunday = (.206 )( 2,540.67 ) = 523.4 The remaining 12 weeks of daily forecasts would be developed similarly.

If the board of directors perceived that the pattern of weekly attendance totals would be closely followed next summer—i.e., low demand in the first week followed by high demand in weeks 2, 3 and 4 followed by gradually declining demand for the remaining 9 weeks—then a seasonally adjusted forecast could be used. That is, seasonal factors could be developed for all 13 weeks, and, weekly forecasts could be computed by multiplying these weekly seasonal factors by the projected summer total attendance, rather than using the linear trend like forecast to compute forecasted weekly attendance.

CASE 12.4 – FORECASTING ARIPORT PASSENGER ARRIVALS Seasonal factors: 4-6 am

98,900/677,200 = .146

6-8 am

111,000/677,200 = .164

8-10 am

116,100/677,200 = .171

10- Noon

65,200/677,200 = .096

Noon – 2 pm

80,700/677/200 = .119

2-4 pm

85,300/677,200 = .126

4-6 pm

74,800/677,200 = .110

6-8pm

34,600/677,200 = .051

8-10pm

10,600/677,200 = .016

(A) Linear trend line forecast for year 4 developed by averaging 10 sample days for each year, creating 3 data point: y = 11,413.3 + 5580 x y (4) = 33,733.3 (B) Linear trend line forecast for year 4 developed using all 30 sample data points: y = 14,893 + 503.62 x y (31) = 30,505.2 Seasonally Adjusted Forecast (A)

Seasonally Adjusted Forecast (B)

4-6 am

4,926.50

4455.06

6-8 am

5529.24

5000.12

8-10 am

5783.28

5229.85

10- Noon

3247.80

2937.01

Noon – 2 pm

4019.91

3635.22

2-4 pm

4249.05

3842.43

4-6 pm

3726.01

3369.45

6-8pm

1723.53

1558.60

8-10pm

528.02

477.59

13 Inventory Management Answers to Questions 13-1. In general, independent demand items are final or finished products that are not dependent upon internal production activity; that is, the demand is usually external and beyond the direct control of the organization.

Alternatively, dependent demand is usually a component part or material used to produce a final product. An example of independent demand for a pizza restaurant would be a final product such as a pizza, whereas dependent demand would be any of the ingredients (cheese, tomato sauce, dough, etc.) and perhaps complementary items such as drinks.

13-2. In a fixed-order-quantity system, an order is placed for the same constant amount whenever the inventory on hand decreases to a certain level, whereas in a fixed-time-period system, an order is placed for a variable amount after an established passage of time.

13-3. The customer service level is the ability to meet internal or external demand at a specified level of efficiency.

High quality service is often perceived as always being able to meet demand, which normally requires high inventory levels and can be costly, or efficient management of the inventory system such that demand is met most of the time.

13-4. An ABC system is a method for classifying inventory according to its dollar value. In general, about 5 to 15 percent of all inventory items will account for 70 to 80 percent of the total dollar value of inventory. Each level of inventory requires different levels of inventory control. That is, the higher the value of inventory, the higher the control. Such a system generally requires less record-keeping and focuses managerial attention on the most important inventory items.

13-5. The two basic inventory decisions are how much to order and when to order items for inventory. In a continuous order system, whenever inventory decreases to a specific level (referred to as the reorder point), a new order is placed for a fixed amount. Alternatively, in a periodic inventory system, inventory on hand is counted at specific time intervals and an order is placed for an amount that will bring inventory back to a desired level.

13-6. The categories are ordering cost, carrying cost, and shortage costs. As the order size increases, ordering costs and shortage costs decrease while carrying costs increase.

13-7. The optimal order quantity occurs when the ordering cost equals carrying cost; thus, the order quantity can be determined by equating these two cost functions and solving for the optimal value.

13-8. Demand is known with certainty, shortages are not allowed, lead time for order receipts is constant, and orders are received all at once. These assumptions are limiting to the extent that they eliminate all uncertainty and potential variation in the model.

13-9. In a continuous inventory system, the reorder point is the inventory level at which a new order is placed, and lead time is the time required to receive an order after it has been placed.

13-10. In a noninstantaneous receipt model, the order quantity is received gradually over time and the inventory level is depleted at the same time it is being replenished, whereas in the basic EOQ model orders are received all at once.

13-11. The total purchase price of all items demanded must be included in the model, since the differences in prices for different order sizes must be reflected in the model.

13-12. Price has no real impact on the optimal order size; it is a constant value that would not alter the basic shape of the EOQ total cost curve.

13-13. The noninstantaneous receipt EOQ model would approach the basic EOQ model with instantaneous receipt.

13-14. The service level is the probability that the amount of inventory on hand during the lead time is sufficient to meet expected demand. The safety stock is the amount of inventory to keep on hand necessary to achieve

this probability.

Solutions to Problems 13-1.

D = 1500

Co = $625

Cc = $130 a.

Q=

2 ( 625)(1500 ) 2Co D = = 120.1 Cc 130

b.

TC =

Co D CcQ + Q 2

=

c.

d.

( 625)(1500 ) + (130 )(120.1) = $15, 612.49 120.1

2

D 1500 = = 12.49 orders Q 120.1

364 = 29.14 days 12.49

13-2.

Case a b c d 13-3.

Q 120.1 120.1 132.8 108.6

TC $14,051.25 $17,173.74 $15,534.24 $15,534.24

D = 16,500

Co = $70 Cc = $27 a.

Q=

2 ( 70 )(16,500 ) 2Co D = = 292.5 Cc 27

b.

TC =

Co D CcQ ( 70 )(16,500 ) ( 27 )( 292.5) + = + Q 2 292.5 2

= $7,897.47

13-4.

c.

D 16,500 = = 56.41 orders Q 292.5

d.

320 = 5.67 days 56.41

D = 45, 000

Co = $1,500 Cc = $0.70 Q=

2 (1500 )( 45, 000 ) 2Co D = Cc 0.70

= 13,887.3 yd TC =

=

Co D CcQ + Q 2

( 0.70 )(13,887.3) + (1500 )( 45, 000 ) 2

13,887.3

= $9, 721.11 Number of orders =

D 45, 000 = = 3.24 per year Q 13,887.3

Time between orders = 13-5.

D = 1, 415, 000

Co = $2, 200 Cc = $0.08

365 = 112.6 days 3.24

a.

2Co D Cc

Q=

2 ( 2, 200 )(1, 415, 000 )

=

b.

0.08

TC =

=

= 278,971.3 yd

Co D CcQ + Q 2

( 2, 200 )(1, 415, 000 ) + ( 0.08)( 278,971.3) 278,971.3

2

= $22,317.71

13-6.

c.

D 1, 415, 000 = = 5.07 per year Q 278,971.3

d.

365 = 72.0 days 5.07

D = 6, 000 d = 23.08 / day p = 116 / day Co = $700

Cc = $9 a.

Q=

2 ( 700 )( 6, 000 ) 2Co D =  d  23.08  5 1 − Cc 1 −   116  p  

= 1, 079.41 b.

TC =

=

Co D CcQ  d  + 1 −  Q 2  p

( 700)( 6, 000) + ( 9 )(1, 079.4 ) 1 − 23.08  1079.4

2

 

 116 

= $7, 782.84 D 6, 000 = = 5.96 runs = 5.56 rund Q 1079.41

c.

d.

260 = 46.67 working days 5.96 Q 1079.41 = = 9.31 working days p 116

e.

13-7.

D = (18)( 52 ) = 936

Co = ( $300 )( 0.25) = $75

Cc = $250 2 ( 250 )( 936 ) 2Co D = Cc 75

Q=

= 79 bicycles TC =

=

Co D CcQ + Q 2

( 250 )( 936 ) + ( 75)( 79 ) 79

2

= $5,924.53 13-8.

D = 7, 000

Co = $3,600 Cc = $50 L = 10

Q=

=

2Co D Cc

2 ( 3, 600 )( 7, 000 ) 50

= 1004 gallons

TC =

Co D CcQ + Q 2

 7, 000   1004  = 3, 600   + 50    1004   2  = $50,199.6

 7,000  R = dL =   (10 ) = 225.81 gallons  310  13-9.

D = 5, 000

Co = $80 Cc = $0.50 L=4 a.

Q=

2 ( 80 )( 5, 000 ) 2Co D = = 1, 264.9 boxes Cc 0.50

b.

TC =

Co D CcQ  5,000   1264.9  + = 80   + 0.50   Q 2  1264.9   2 

= $632.46 c.

13-10.

 5, 000  R = dL =   ( 4 ) = 54.79 boxes  365  d = 205 lb / day p = 350 lb / day D = 74,825

Co = $175 Cc = $12

Q=

2 (175)( 74,825) 2Co D = = 2295.18  d  205  12 1 − Cc 1 −   p  350   

TC =

=

Co D CcQ  d + 1 −  Q 2  p

(175)( 74,825) + (12 )( 2295.18) 1 − 205  2295.18

2

 

 350 

= $11, 410.32 13-11.

d = 1,800 p = 3, 000 D = 657, 000

Co = $7,500 Cc = $60

Q=

TC =

2 ( 7,500 )( 657, 000 ) 2Co D = = 20, 263.88  d  1,800  50 1 − Cc 1 −   p  3, 000    Co D CcQ  d + 1 −  Q 2  p

= ( 7,500 )

657, 000 20, 263.88  1,800  + ( 60 ) 1 − 3, 000  20, 263.88 2  

= $486,333.22 13-12. Operates 360 days/year 12 converters 5 tons coal/day/converter

D = ( 5 tons )(12 converters )( 360 days ) = 21, 600 tons / year

Co = $80 Cc = 20%of average $ inventory level

Cc = ( 0.20 )( $12 ) = $2.40 2 ( 80 )( 21, 600 ) 2Co D = = 1, 444, 000 Cc 2.4

Q=

a.

= 1, 200 tons b.

TC =

Cc D Co D ( 2.4 )(1, 200 ) (80 )( 21, 600 ) + = + 2 Q 2 1, 200

= 1, 440 + 1, 440 = $2,880 R=

c.

13-13.

LD 5 ( 21, 600 ) = = 300 tons 360 360

D = 10,000 logs / year T = 250 days / year p = 60 / day

R = 60 ( 250 ) = 15, 000 / year

Co = $1,600 Cc = $15

a.

Q=

2 (1, 600 )(10, 000 ) 2Co D = D   10, 000  Cc 1 −  15 ) 1 − (  R   15, 000 

= 6, 400, 000 = 2,529.8 logs b.

TC =

=

Co D CcQ  D + 1 −  Q 2  R

(1, 600 )(10, 000 ) +  (15)( 2,529.8) 1 − 10, 000  2,592.8

 

2

= 6,324.5 + 6,324.5 = $12, 648

 

 15, 000  

c.

d.

N=

D 10, 000 = = 3.95 = 4 orders per year Q 2,529.8

Tb =

T 250 = = 63.3 days between orders N 3.95

Q = 2,529.8, R = 60 The number of operating days to receive the entire order is

Q 2,529.8 = = 42.2 days r 60 13-14. a.

D = 12, 400

Cc = $3.75 Co = $2,600 Q=

2 ( 2, 600 )(12, 400 ) 2Co D = = 4146.6 Cc 3.75 Co D CcQ + Q 2

TC =

=

( 2, 600 )(12, 400 ) + (3.75)( 4,912.03) 4,912.03

2

= $15,549.92 b.

Co = $1,900

Cc = $4.50 Q=

TC =

=

2 (1,900 )(12, 400 ) 2Co D = = 3235.9 Cc 4.50 Co D CcQ + Q 2

(1,900 )(12, 400 ) + ( 4.50 )(3,833.19 ) 3,833.19

2

= 14,561.59 Select the new location.

13-15. a.

d = 220, 000

Cc = $0.12 / lb Co = 620 p = 305, 000

2 ( 620 )( 220, 000 ) 2Co D = 220  d ( 0.12 ) 1 −  Cc 1 −  p  305  

Q=

= 90,317.52  d maximum level = Q 1 −  = ( 90,317.52 )( 0.2787 ) p 

= 25,170.46 TC =

=

Co D CcQ  d + 1 −  Q 2  p

( 620 )( 220, 000 ) + ( 0.12 )( 90,317.52 ) 0.2787 ( ) 90,317.52

2

= $3020.45 b.

P = 360,000

Q=

TC =

2 ( 620 )( 220, 000 ) = 76, 457.27  220  ( 0.12 ) 1 −   305 

( 620 )( 220, 000 ) + ( 0.12 )( 76, 457.27 ) 0.25 ( ) 76, 457.27

= $3,568.01 No, the total inventory cost increases.

2

D = 1400

13-16.

Co = $7,600 Cc = ?

Q = 120 2Co D Cc

Q=

2 ( 7, 600 )(1400 )

120 =

(120 )2 =

Cc

2 ( 7, 600 )(1400 ) Cc

Cc = $1, 477.78 13-17.

D = 200 / day

Co = $25 Cc = $0.20 / min. = $120 / day a.

Q=

2 ( 25 )( 200 ) 120

= 9.1 orders or 9

The truck should carry approximately 9 orders each time it makes deliveries.

200 = 24 deliveries per day 9

10 = 0.416 hour = every 25 minutes a delivery truck goes out to deliver orders 24 TC =

( 25)( 200 ) + (120 )(8) 8

= $1095.45 b.

Q=6

2

200 = 33.33 or 33 deliveries per day.

6 10 = .30 hr. = every 18 minutes a delivery truck is sent out.

33 TC =

( 25)( 200 ) + (120 )( 6 ) 6

2

= $1193.73 13-18.

Co = $1,700 Cc = $1.25 D = 21, 000 / yr.

P = 30, 000 / yr.

a.

Q=

TC =

2 (1, 700 )( 21, 000 )  21, 000  1.25 1 −   30, 000 

= 13, 798.55

(1, 700 )( 21, 000 ) + 1.25  13, 798.55   1 − 21, 000  13, 798.55

 

2

    30, 000 

= 5,174.46 Number of production runs =

D 21, 000 = = 1.52 Q 11, 062.62

 d Maximum inventory level = Q 1 −  p   21, 000  = 13, 798.55 1 −   30, 000  = 4,139.57 b.

Maximum inventory level = 2,500  21, 000  2,500 = Q 1 −   30, 000 

2,500 = Q (.3) Q = 8,333.33

TC =

(1, 700 )( 21, 000 ) + 1.25  8,333.33  1 − .70 )  ( 

8,333.33

= $5,846.50 13-19.

D = 280, 000 C = $7, 000 Cc = $0.80 / ft.3

Q=

2 ( 7, 000 )( 280, 000 ) 0.80

= 70, 000 ft.3 TC =

( 7, 000 )( 280, 000 ) + 0.80 ( 70, 000 ) 52,915

2

= $56, 000 Number of orders =

280, 000 56, 000

=4 13-20.

D = 40, 000 Co = $800

Cc = $1.90 Without discount:

Q=

TC =

2 ( 800 )( 40, 000 ) 1.90

Co D Q + Cc + PD Q 2

= 5,803.81

2



=

(800 )( 40, 000 ) + (1.90 )(5,803.81) + 3.40 40, 000 ( )( ) 5,803.81

2

= 147, 027.24 with discount of Q = 20, 000

TC = $140, 600 Take discount for Q = 20, 000 13-21.

D = 10, 000 Co = $300

Cc = $1.25 Order Size

P

0–4,999

$8.00

5, 000 +

$6.50

Without discount:

Q=

TC =

=

2 ( 300 )(10, 000 ) 2Co D = = 2,190.9 Cc 1.25 Co D CcQ + + PD Q 2

( 300 )(10, 000 ) + (1.25)( 2,190.9 ) + 10, 000 8 ( ) 2,190.9

= $82, 738.61 With discount:

Q = 5, 000 TC =

Co D CcQ + + PD Q 2

2

=

( 300 )(10, 000 ) + (1.25)(5, 000 ) + 10, 000 6.50 ( ) 5, 000

2

= $68, 725 Select discount; Q = 5, 000.

13-22.

Without discount:

Q = 200 TC = $55, 440 With discount:

Q = 300 P = 52

TC =

=

Co D CcQ + + PD Q 2

(160 )( 900 ) + ( 7.20 )( 300 ) + 52 900 ( )( ) 300

2

= 48,360 Take the discount, Q = 300.

13-23.

D = 1, 700 Co = $120

Cc = ( 0.25)( $38) = $9.50 Q=

2 (120 )(1, 700 ) 2Co D = = 207 Cc ( 9.5)

TC =

(120 )(1, 700 ) + ( 9.50 )( 207 ) + 38 1, 700 ( )( ) 207

= $66,568.76

2

Q = 300 :

(120 )(1, 700 ) + ( 9.31) 300 + 37.24 1, 700 ( )( )

TC =

300

2

= $65,384.80 Q = 500 : TC =

(120 )(1, 700 ) + ( 9.12 ) 500 + 36.48 1, 700 ( )( ) 500

2

= $64, 704

Q = 800 : TC =

(120 )(1, 700 ) + ( 9.025) 800 + 36.10 1, 700 ( )( ) 800

2

= $65, 235 Select Q = 500; TC = $64, 704.

13-24.

Q=

TC =

2 (120 )(1, 700 )

(8)

= 226

(120 )(1, 700 ) + (8)( 226 ) + 38 1, 700 ( )( ) 226

2

= $66, 406.65 Q = 300 : TC =

(120 )(1, 700 ) + (8)( 300 ) + 37.24 1, 700 ( )( ) 300

2

= $65,188 Q = 500 : TC =

(120 )(1, 700 ) + (8)( 500 ) + 36.48 1, 700 ( )( ) 500

= $64, 424

2

Q = 800 :

(120 )(1, 700 ) + (8)(800 ) + 36.10 1, 700 ( )( )

TC =

800

2

= $64,825 Select Q = 500; TC = $64, 424.

13-25.

D = 6,500

Co = $28 Cc = $3 Q=

2 ( 28 )( 6,500 ) 2Co D = = 348.32 = 348 Cc 3

TC =

( 28)( 6,500 ) + ( 3)( 348) + 16 6,500 ( )( ) 348

2

= $105, 045 Q = 1, 000 :

TC =

( 28)( 6,500 ) + ( 3)(1, 000 ) + 14 6,500 ( )( ) 1, 000

2

= $92, 682 Q = 3,000 :

TC =

( 28)( 6,500 ) + ( 3)( 3, 000 ) + 13 6,500 ( )( ) 3, 000

2

= $89,060.67

Q = 6, 000 :

TC =

( 28)( 6,500 ) + ( 3)( 6, 000 ) + 12 6,500 ( )( ) 6, 000

= $87,030.33

2

Select Q = 6, 000; TC = $87,030.33.

13-26.

Q=

( 2 )( 28)( 6,500 ) = 337.26 = 337 boxes

TC =

( 28)( 6,500 ) + ( 3.20 )( 337 ) + 16 6,500 ( )( )

3.20

337

2

= $105, 079.20 Q = 1, 000 :

TC =

( 28)( 6,500 ) + ( 2.80 )(1, 000 ) + 14 6,500 ( )( ) 1, 000

2

= $92,582 Q = 3,000 :

TC =

( 28)( 6,500 ) + ( 2.60 )(3, 000 ) + 13 6,500 ( )( ) 3, 000

2

= $88, 460.67

Q = 6, 000 :

TC =

( 28)( 6,500 ) + ( 240 )( 6, 000 ) + 12 6,500 ( )( ) 6, 000

2

= $85, 230.33 Select Q = 6, 000; TC = $85, 230.33.

13-27.

D = 2,300, 000 /100 = 23, 000 boxes Co = $320

Cc = $1.90 Q=

2 ( 320 )( 23, 000 ) 2Co D = = 2, 783.4  2, 784 boxes Cc 1.90

( 320 )( 23, 000 ) + (1.9 )( 2, 784 ) + 47 23, 000 ( )( )

TC =

2, 784

2

= $1, 086, 228.50 Q = 7, 000 :

( 320 )( 23, 000 ) + (1.9 )( 7, 000 ) + 43 23, 000 ( )( )

TC =

7, 000

2

= $996, 701.43 Q = 12, 000 :

TC =

( 320 )( 23, 000 ) + (1.9 )(12, 000 ) + 41 23, 000 ( )( ) 12, 000

2

= $955, 013.33 Q = 20, 000 :

TC =

( 320 )( 23, 000) + (1.9)( 20, 000) + 38 23, 000 ( )( ) 20, 000

2

= $893,368 Select Q = 20, 000; TC = $893,368.

13-28.

Q=

TC =

2 ( 320 )( 23, 000 ) 2.35

= 2,502.76  2,503 boxes

( 320 )( 23, 000 ) + ( 2.35)( 2,503) + 47 23, 000 ( )( ) 2,503

2

= $1, 086,881.50 Q = 7, 000 :

TC =

( 320 )( 23, 000 ) + ( 2.15)( 7, 000 ) + 43 23, 000 ( )( ) 7, 000

= $997,576.43

Q = 12, 000 :

2

TC =

( 320 )( 23, 000 ) + ( 2.05)(12, 000 ) + 41 23, 000 ( )( ) 12, 000

2

= $955,913.33 Q = 20, 000 :

TC =

( 320 )( 23, 000 ) + (1.90 )( 20, 000 ) + 38 23, 000 ( )( ) 20, 000

2

= $893,368

Select Q = 20, 000 boxes; TC = $893,368.

13-29.

d = 4, 000 L=7

 d = 600 R = dL + Z d L R = 4, 000 ( 7 ) + 1.64 ( 600 ) 7 = 30, 603.42

Safety stock = 2,603.42 yd 13-30.

If safety stock = 2, 000,

Z ( 600 )

( 7 ) = 2, 000, Z = 1.26,

which corresponds to a 90% service level.

13-31.

d = 9,000

 d = 1,900 L=8 Z = 2.05

R = dL + Z d L = ( 9, 000 )( 8 ) + ( 2.05 )(1,900 ) 8

= 83, 016.7 lbs 13-32.

d = 20

d = 4 L = 2 L = 12 Z = 1.28

R = dL + Z d L = ( 20 )( 2 ) + 1.28 ( 4 ) 2 = 40 + 7.30 (20)(12) + 1.28(4)sqrt(12)

= 47.3 gal =257.87 gal Safety stock = 7.3 gal. =17.87 gal.

For service level of 95%

Z = 1.65

Safety stock = Z d L = (1.65 )( 4 ) 2 = 9.33 =(1.65)(4)sqrt(12) = 22.86 gal.

13-33.

R = dL + Z d L = 30 (8) + 0.68 (10 )

( 8 ) = 259.2 gal

For a 95% service level,

Safety stock Z d L = 1.65 (10 )

( 8 ) = 46.67

The reorder point increases to 286.7 gal.

13-34.

d = 3.5

 d = 1.2 L = 25 Z = 1.29

R = dL + Z d L = ( 3.5 )( 25 ) + 1.29 (1.2 ) 25 = 87.5 + 7.74 R = 95.24

Safety stock = 7.74 13-35.

R = dL + Z d L = 3.5 (8) + 1.29 (1.2 )

( 8 ) = 32.38

Decision would be based on inventory holding cost, desire for low inventory, importance of reliable delivery, cost of the monitors from each source, etc.

13-36.

d = 200

tb = 30 L=4

 d = 80 I = 60

Q = d ( tb + L ) + Z d tb + L − I = 200 ( 30 + 4 ) + 2.33 (80 ) 30 + 4 − 60

= 7,826.89oz 13-37.

d =8 tb = 10 L=3  d = 2.5 I =0 Q = d ( tb + L ) + Z d tb + L − I = 8 (10 + 3) + 2.33 ( 2.5 ) 10 + 3 − 0

= 122 pizzas

122 − 5 = 117 pizzas

13-38.

d = 18 tb = 30 L=2 d = 4 I = 25

Q = d ( tb + L ) + Z d tb + L − I = 18 ( 30 + 2 ) + 1.65 ( 4 ) 30 + 2 − 25 = 588.3 bottles 13-39.

Item 25 23 20 22 24 16 5 10 12 2 4 1 27 9 29 26 28 13 30 18 6 7 21 17 19 8 3 11 15 14

Usage 870 30 19 12 24 60 18 67 682 510 300 36 750 344 46 244 45 95 165 270 500 710 910 120 45 80 50 510 820 10 8,342

Unit Cost 105 2,710 3,200 4,750 1,800 610 1,900 440 35 30 45 350 15 28 160 30 110 50 25 15 8 4 3 20 50 26 23 2 1 3

Annual Usage $91,350 81,300 60,800 57,000 43,200 36,600 34,200 29,480 23,870 15,300 13,500 12,600 11,250 9,632 7,360 7,320 4,950 4,750 4,125 4,050 4,000 2,840 2,730 2,400 2,250 2,080 1,150 1,020 820 30 571,957

% Annual Value 15.97% 14.21 10.63 9.97 7.55 6.40 5.98 5.15 4.17 2.68 2.36 2.20 1.97 1.68 1.29 1.28 0.87 0.83 0.72 0.71 0.70 0.50 0.48 0.42 0.39 0.36 0.20 0.18 0.14 0.01 100.00%

% Annual Usage 10.43% 0.36 0.23 0.14 0.29 0.72 0.22 0.80 8.18 6.11 3.60 0.43 8.99 4.12 0.55 2.92 0.54 1.14 1.98 3.24 5.99 8.51 10.91 1.44 0.54 0.96 0.60 6.11 9.83 0.12 100.00%

Class A A A A A A A B B B B B B B B B B C C C C C C C C C C C C C

13-40.

Class A B

Items 4, 5, 6, 13, 20, 22, 23, 30 7, 10, 14,

Quantity 52

$ Value 59,835

% Value 51

77

30,640

26

% Quantity 5.5 8.2

C

13-41.

15, 19, 21 1, 2, 3, 8, 9, 11, 12, 16, 17, 18, 24, 25, 26, 27, 28, 29, 31, 32

811

27,493

23

86.3

940

$117,968

100%

100%

d = 2.6 packages/day  d = 0.8 packages/day

L = 2 days  L = 0.5 days Z = 2.33 R = d L + Z  d2 L +  L2 d 2

= ( 2.6 )( 2 ) + 2.33 (.8) 2 + (.5 ) ( 2.6 ) 2

2

2

= 9.22 packages of paper 13-42.

d = 6/hr.  d = 2.5/hr.

L = 0.5 hr.  L = .133 hr.

Z =? a.

R = d L + Z  d2 L +  L2 d 2

1 = ( 6 )( 0.5) + Z

( 2.5)2 ( 0.5) + (.133)2 ( 6 )2

1 = 3 + Z (1.94 ) −2 = Z (1.94 ) Z = −1.03 Service level = .5000 − .3485 = .1515 = 15.15% b.

R = 3 + ( 2.05)(1.94 ) = 6.977 pizzas

CASE SOLUTION 13.1: The A to Z Office Supply Company

D = $17, 000 / day = $5,185, 000/year ( 305 day year )

Cc = $0.09 / dollar / year = $0.09 Co = $1, 200 / loan + 0.0225Q L = 15 days Optimal loan amount per loan:

2Co D = Cc

Q=

( 2 )(1, 200 )( 5,185, 000 ) 0.09

= $371,842.26 loan amount per loan Memo: The 0.0225Q cost per loan is not included in the calculation of Q since it is paid on the entire dollar amount of the loan regardless of loan size, and thus it is simply an annual cost, i.e., 0.0225  D.

TC = Co

D Q + Cc + 0.0225D Q 2

 5,185, 000   371,842  = (1, 200 )  + ( 0.09 )    2    371,842 

+ ( 0.0225)( 5,185, 000 ) = $150,128.30 total cost of borrowing N=

D 5,185, 000 = = 13.944 loans / year Q 371,842

 14 loans/year for about $371, 000 per loan

r = (15)(17, 000 ) = $255, 000 reorder point When cash balance gets down to $255,000 initiate another loan.

Quantity Discount Analysis: If Q  $500,000;

points = 2%

Since Q is unaffected by points, and Q was $371,842; we know we must set Q = $500, 000 for this alternate option.

TC = Co

D Q + Cc + 0.02 D Q 2

 5,185, 000   500000  = (1, 200 )  + ( 0.09 )     2   500, 000 

+ ( 0.02 )( 5,185, 000 ) = 12, 444 + 22,500 + 103, 700 = $138, 644 Since this option yields a lower TC of $11, 484 (150,128 − 138, 644 ) ; it should be accepted.

CASE SOLUTION 13.2: The Texas Stadium Store The objective of this case problem is to determine the reorder point with variable demand. The first step is to complete the average demand and standard deviation from the data provided in the problem. This is a good opportunity to allow students to use a statistical software package (if they have access to one) to compute these statistics.

d = 42.57 hats per week

 = 10.41 hats per week L = 20 days = 2.86 weeks The first question is, if R = 140, what level of service does this correspond to. Thus, we are seeking Z as follows

140 = dL + Z d L 140 = ( 42.57 )( 2.86 ) + Z (10.41) 2.86

140 = 121.75 + Z (17.6 ) Z = 1.04 This Z value corresponds to a normal probability value of 0.8508, thus, the service level is approximately 85.1 percent. The desired service level is 99 percent ( Z = 2.33) . The reorder point and safety stock for this service level is determined as follows.

R = dL + Z d L R = ( 42.57 )( 2.86 ) + 2.33 (10.41) 2.86

140 = 121.75 + 41.02 R = 162.76 or 163 hats Ms. Jones could determine the order size with EOQ analysis by using the average demand, d, as D in the EOQ formula. However, she would also need the ordering and carrying costs. It is likely that the ordering cost is relatively high as compared to carrying cost since the hats are shipped from Jamaica while it would probably not be very expensive to store hats (given their small size and weight).

CASE SOLUTION 13.3: Pharr Food Company This problem requires the development of a forecast for product demand in year 4 (see chapter 11). A seasonal forecast was developed, as follows.

Year

Jan–March

April–May

June–Aug

Sept

Oct

Nov-Dec

Total

1

607

488

479

256

342

524

2696

2

651

487

660

263

370

537

2968

3

685

539

672

302

411

572

3181

Total

1943

1514

1811

821

1123

1633

8845

S1 =

0.220

S2 =

0.171

S3 =

0.205

S4 =

0.093

S5 =

0.127

S6 =

0.185

Linear trend line forecast: y = 2463.3 + 242.5 x

y ( 4 ) = 3433.33 cases SF1 =

754.21

SF 2 =

587.68

SF 3 =

702.97

SF 4 =

318.68

SF 5 =

435.91

SF 6 =

633.88

Total

3433.33

Q=

2 ( 4700 )( 3433.33) 116

= 527.5

Comparing monthly forecasts (with seasonal pattern) with order size, Q, using order frequency of 2 months:

No. of orders =

D 3433.3 = = 6.5 orders Q 527.5

52 weeks = 8 weeks ( 2 months ) per order 6.5 orders Monthly Forecast 792

528

528

528

528

528

Balance

January

251

541

February

251

289

March

251

566

April

294

272

May

294

506

June

234

272

July

234

565

August

234

331

September

318

541

October

436

105

November

317

316

December

317

−1

Total

3433

Note that the “.5” order was added to the first month. The order size ( Q = 528) seems to be adequate to offset seasonal patterns.

Ingredient orders: Chocolate

D = 108,140 lbs. (3433.3 cases  60 bags/case = 205,980 bags;

205,980 bags 12 bars / bag = 2, 471, 760 bars; Demand = 2, 471, 760 bars  0.70 oz chocolate / bar = 1,730, 252 oz = 108,140 lbs.)

Co = $5,700 Cc = $0.45 / lb.

Q = 52,340

TC (1) = $353,380, Q = 52,340 TC ( 2 ) = $337,159, Q = 52,340 TC ( 3) = 326, 048, Q = 100, 000 TC ( 4 ) = $319, 023, Q = 150, 000*Optimal Nuts

D = 77, 242 lb.

Co = $6,300 Cc = $0.63/ lb.

Q = 39,304 lbs.

TC (1) = $526,834, Q = 39,304 TC ( 2 ) = $507,524, Q = 39,304 TC ( 3) = $488,591, Q = 70, 000*Optimal Filling

D = 61, 794 lb.

Co = $4,500 Cc = $0.55

Q = 31, 799

TC (1) = $110,180, Q = 31, 799 TC ( 2 ) = $101,373, Q = 40, 000*Optimal TC ( 3) = $102, 718, Q = 80, 000 Pharr Foods might experience quality problems with its large orders for ingredients that take advantage of price discounts; ingredients may be in storage for long periods. Also, the demand forecast is treated with certainty; if significant variation occurs it could create shortages and the need for safety stocks. Lead times are considered negligible, which could also create problems along the supply chain if they are significant in reality.

Supplement 13 Operational Decision-Making Tools: Simulation Answers to Questions S13-1. The Monte Carlo technique is a technique for selecting numbers randomly from a probability distribution for use in a trial run of a simulation. Random numbers are organized into ranges to reflect the probabilities of the different possible occurrences of a random variable. When the random numbers are selected in the simulation, they represent the value of a random variable.

S13-2. They are achieved by repeating the simulation many times.

S13-3. In general, simulation provides descriptive, probabilistic information about a system. It can be used to find an optimal set of operating characteristics by searching through all possible sets of operating characteristics until the best set is found.

Solutions to Problems S13-1.

Time Between Calls(hours) 1 2 3 4 5 6

Cumulative Probability 0.05 0.15 0.45 0.75 0.95 1.00

RN 65 71 20 15 48 89

Time Between Calls 4 4 3 2 4 5

Random Numbers 01–05 06–15 16–45 46–75 76–95 96–99, 00

a.

Cumulative Clock 4 8 11 13 17 22

RN 18 83 08 90 05 89 18 08 26 47 94 06 72 40 62

Time Between Calls 3 5 2 5 1 5 3 2 3 4 5 2 4 3 4

Cumulative Clock 25 30 32 37 38 43 46 48 51 55 60 62 66 69 73

b.

=

73 = 3.48 hours between calls 21

EV = 1( 0.05 ) + 2 ( 0.10 ) + 3 ( 0.30 ) = 3.65 + 4 ( 0.30 ) + 5 ( 0.20 ) + 6 ( 0.05 ) The results are different because there were not enough simulations to enable the simulated average to approach the analytical result. c.

21 calls. No, this is not the average number of calls per 3 days. In order to determine this average, this simulation would have to be repeated a number of times in order to get enough observations of calls per 3-day period to compute an average.

S13-2.

Machine Breakdowns per Week 0 1 2 3 4 5

Cumulative Probability 0.10 0.20 0.40 0.65 0.95 1.00

Random Numbers 01–10 11–20 21–40 41–65 66–95 96–99, 00

a.

Week 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

=

b.

RN 20 31 98 24 01 56 48 00 58 27 74 76 79 77 48 81 92 48 64 06

Breakdowns 1 2 5 2 0 3 3 5 3 2 4 4 4 4 3 4 4 3 3 0 59

59 = 2.95 breakdowns per week 20

First three columns are from part a. Select as many RN2’s as there are breakdowns from a different random number stream.

Week 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

RN1 20 31 98 24 01 56 48 00 58 27 74 76 79 77 48 81 92 48 64 06

Breakdowns

RN 2

Repair Time

1 2 5 2 0 3 3 5 3 2 4 4 4 4 3 4 4 3 3 0

58 47, 23 69, 35, 21, 41, 14 59, 28 — 13, 09, 20 73, 77, 29 72, 89, 81, 20, 85 59, 72, 88 11, 89 87, 59, 66, 53 45, 56, 32, 44 08, 82, 55, 27 49, 24, 83, 05 81, 07, 78 92, 36, 53, 04 95, 79, 61, 44 37, 45, 18 65, 37, 30 —

2 2 +1 = 3 2 + 2 +1+ 2 +1 = 8 2 +1 = 3 0 1+1+1 = 3 2 + 2 +1 = 5 2 + 3 + 3 + 1 + 3 = 12 2+2+3= 7 1+ 3 = 4 3+ 2+ 2+ 2 = 9 2+2+2+2 =8 1+ 3 + 2 +1 = 7 2 +1+ 3 +1 = 7 3 +1+ 2 = 6 3 + 2 + 2 +1 = 8 3+ 2+ 2+ 2 = 9 2 + 2 +1 = 5 2 + 2 +1 = 5 0

Total repair time = 111 hours

=

111 = 5.55 hr / wk 20

It could bias the results. Selecting a high random number, such as 98, results in a high number of breakdowns (i.e., 5). If the same random number is used, it will result in a high repair time (i.e., 3 hours). Thus, a relationship will result wherein a high number of breakdowns equals high repair times and vice versa. The effect in this model will not be too bad, since several repair-time random numbers are selected for each breakdown. c.

Total weekly breakdown cost = 111 hours x $50 = $5,550

=

5,550 = $277.50 per week 20

d.

Breakdowns per Week 0 1 2 3 4 5

Week 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Cumulative Probability 0.20 0.50 0.70 0.85 0.95 1.00

RN1 20 31 98 24 01 56 48 00 58 27 74 76 79 77 48 81 92 48 64 06

Random Numbers 01–20 21–50 51–70 71–85 86–95 96–99, 00

Breakdowns 0 1 5 1 0 2 1 5 2 1 3 3 3 3 1 3 4 1 2 0

Total repair time = 77 hours

RN 2 — 58 47, 23, 69, 35, 21 41 — 14, 59 28 13, 09, 20, 73, 77 29, 72 89 81, 20, 85 59, 72, 88 11, 89, 87 59, 66, 53 45 56, 32, 44 08, 82, 55, 27 49 24, 83 —

Repair Time 0 2 2 +1+ 2 + 2 +1 = 8 2 0 1+ 2 = 3 1 1+1+1+ 2 + 2 = 7 1+ 2 = 3 3 3 +1+ 3 = 7 2+2+3= 7 1+ 3 + 3 = 7 2+2+2 = 6 2 2+2+2 = 6 1+ 3 + 2 +1 = 7 2 1+ 3 = 4 0

Total breakdown cost = 77  $50 = $3,850

=

3,850 20

Average weekly breakdown cost = $192.50 Reduction in average weekly repair cost = 277.50 − 192.50 = $85

Since the maintenance program costs $150 and only $85 would be saved, it should not be put into effect.

However, the results should be applied with some hesitancy since they were derived for only one actual simulation. This whole simulation process should be repeated a number of times.

Note: In part d the same random number streams were used as in part a. This was done to replicate as much as possible the conditions of the first simulation, since the results were to be compared. However, if many simulations were conducted, this would not have been necessary.

S13-3

Demand/Week 0 1 2 3 4 5 6

Random Numbers 01–04 05–12 15–40 41–80 81–96 97–98 99, 00

Lead Time 1 2 3

Random Numbers 01–60 61–90 91–99, 00

=

Week 0 1 2 3 4 5 6

RN1 — 39 72 37 87 98 10

RN 2 — 73 75 02 — — 47

5, 020 = $251 20

Demand

Order Placed

Order Received

Balance

Carrying Cost

Order Cost

Stockout Cost

Total

— 2 3 2 4 5 1

— 5 5 5 — — 5

— — — 5 5+5 — —

5 3 0 3 9 4 3

— 120 0 120 360 160 120

— 100 100 100 — — 100

— — — — — — —

— 220 100 220 360 160 220

7 8 9 10 11 12 13 14 15

93 21 97 41 80 67 59 63 87

— 95 69 91 — — — 78 47

4 2 5 3 3 3 3 3 4

— 5 5 5 — — — 5 5

5 — — — 5+5 — 5 — —

4 2 0 0 7 4 6 3 0

160 80 0 0 280 160 240 120 0

— 100 100 100 — — — 100 100

— — 400 400 — — — — 400

160 180 500 500 280 160 240 220 500

Week

RN1 56 22 19 78 03

RN 2 — — 16 — —

Demand

Order Placed

Order Received

Balance

Carrying Cost

Order Cost

Stockout Cost

Total

3 2 2 3 0

— — 5 — —

5+5 — — 5 —

7 5 3 5 5

280 200 120 200 200

— — — — —

— — — — —

16 17 18 19 20

S13-4.

White Sox Play Designation No advance Groundout Double play Long fly Very long fly Walk Infield single Outfield single Long single Double Long double Triple Home run

Cumulative Probability 0.03 0.42 0.48 0.57 0.65 0.71 0.73 0.83 0.86 0.90 0.95 0.97 1.00

Random Numbers (r) 01–03 04–42 43–48 49–57 58–65 66–71 72–73 74–83 84–86 87–90 91–95 96–97 98–99, 00

280 200 120 200 200 5,020

Yankees Play Designation No advance Groundout Double play Long fly Very long fly Walk Infield single Outfield single Long single Double Long double Triple Home run

Cumulative Probability 0.04 0.42 0.46 0.56 0.62 0.69 0.73 0.83 0.87 0.92 0.95 0.96 1.00

Random Numbers (r) 01–04 05–42 43–46 47–56 57–62 63–69 70–73 74–83 84–87 88–92 93–95 96 97–99, 00

Inning 1:

White Sox

Yankees

RN 39 73 72 75 37 02

Play Groundout I. single I. single O. single Groundout Out

87 98 10 47 93 21

Single Home run Groundout Fly Double Groundout

Outs 1

RBI

1 1 1

____ 1 2

1 1 1

____ 2

Inning 2:

White Sox

Yankees

RN 95 97 69 41 91 80 67 59 63

Play Double Triple Walk Groundout L. Double O. Single Walk V. long fly V. long fly

Outs

RBI

78 87 47 56 22

O. single L. single L. fly L. fly Groundout

RN 19 16 78 03

Play Groundout Groundout O. single Out

Outs 1 1

RBI

1

____ 0

04 61 23

Out V. long fly Groundout

1 1 1

1 1

1 1 1

1 1

____ 4

1 1 1

1 ____ 1

Inning 3:

White Sox

Yankees

____ 0

Inning 4:

White Sox

Yankees

RN 15 58 93 78 61

Play Groundout V. long fly L. double O. single V. long fly

Outs 1 1

42 77 65 71 18 12

Groundout O. single Walk I. single Groundout Groundout

1

1

1 1

RBI

1 ____ 1

1 ____ 1

Inning 5:

White Sox

Yankees

RN 17 48 89 18

Play Groundout Out Double Groundout

83 08 90 05 89 18

O. single Groundout Double Groundout Double Groundout

Outs 1 1

RBI

1

____ 0

1 1 1 1 ____ 2

1

Inning 6:

White Sox

Yankees

RN 08 26 47

Play Groundout Groundout Out

94 06 72 62 47

L. double Groundout I. single V. long fly L. fly

Outs 1 1 1

RBI

____ 0

1 1 1 1

____ 1

Inning 7:

White Sox

Yankees

RN 68 60 88 17 36

Play Walk V. long fly Double Groundout Groundout

77 43 28

O. single Double play Groundout

Outs

RBI

1 1 1

2 1

1 ____ 1

____ 0

Inning 8:

White Sox

Yankees

RN 31 06 68 39

Play Groundout Groundout Walk Groundout

71 22 76 81 88 94 76 23 47

I. single Groundout O. single O. single Double L. double O. single Groundout Out

RN 25 79 08 15

Play Groundout O. single Groundout Groundout

Outs 1 1

RBI

1

____ 0

1 1 1 2 1 1 1

____ 5

Inning 9:

White Sox

White Sox Yankees

1 2

4 1

Outs 1 1 1

1 1

Line Score: 0 0 2 1

Cumulative Probability 50 .80 1.00

Random Numbers 01–50 51–80 81–99, 00

0 0

S13-5.

Months to Receive an Order 1 2 3

Demand per Month 1 2 3 4

Cumulative Probability 0.10 0.40 0.80 1.00

Random Numbers 01–10 11–40 41–80 81–99, 00

RBI

0 0

1 0

0 5

0 X

7 12

Reorder Number

RN1 21 41 14 59 28 68 13 09 20 73 77 29 72 89 81 20 85 59 72 88 11

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

Reorder Number 22 23 24 25 26 27 28 29 30

RN1 89 87 59 66 53 45 56 22 49

Months Lead Time

RN 2

1 1 1 2 1 2 1 1 1 2 2 1 2 3 3 1 3 2 2 3 1

70 38 11 70, 52 88 25, 17 18 00 70 04, 11 76, 29 01 11, 42 70, 17, 48 87, 20, 16 56 35, 94, 91 65, 20 97, 46 10, 98, 32 57

Months Lead Time 3 3 2 2 2 1 2 1 1

RN 2 92, 55, 35 57, 99, 20 52, 45 01, 73 58, 24 72 84, 35 41 32

Total demand = 143 Average demand during lead time = 143/30 = 4.76 Thus, should reorder at 5-car level.

Demand per Month 3 2 2 3, 3 = 6 4 2, 2 = 4 2 4 3 1, 2 = 3 3, 2 = 5 1 2, 3 = 5 3, 2, 3 = 8 4, 2, 2 = 8 3 2, 4, 4 = 10 3, 2 = 5 4, 3 = 7 2, 4, 2 = 8 3

Demand per Month 4, 3, 2 = 9 3, 4, 2 = 9 3, 3 = 6 1, 3 = 4 3, 2 = 5 3 4, 2 = 6 3 2 143

S13-6.

Customers/ Day 0 1 2 3

Random Numbers 01–20 21–40 41–90 91–99, 00

Day

Available Cars

1

Duration 1 2 3 4 5

Customers

4

RN1 62

2

2

48

2

3 4

1 2

96 86

3 2

5

2

86

2

6 7 8 9 10

1 0 1 2 3

29 79 22 08 62

1 2 1 0 2

2

Random Numbers 01–10 11–40 41–80 81–90 91–99, 00

RN 2 19 66 27 43 20 92 22 91 46 49 — 66 — 09

Duration/ Car

Car Day Available

2 3 2 3 2 5 2 5 3 3 — 3 — 1 81

3 4 4 5 5 9 6 10 8 9 — 11 — 11 4

Customers Not Served

2

2

14

Probability of not having a car available = customers not served/ total customers = 4/17 = 0.235.

Since almost 24% of customers are not served, expansion would probably be warranted. A simulation model of this system necessary to make a decision would need to perform this simulation for a number of different fleet sizes (in addition to the four cars used in this experiment). The simulation should also include the daily cost of the car to the rental agency and the daily rental price plus some estimate of lost current and potential sales when a customer is turned away. The fleet size selected would be the one that maximized average daily profit.

S13-7.

Time Between Arrivals 5 10 15 20 25 30

Cumulative Probability 0.06 0.16 0.39 0.68 0.86 1.00

Cumulative

RN Ranges 01–06 07–16 17–39 40–68 69–86 87–99,00

RN

Service Doctor (D) Nurse (N) Both (B)

Time 10 15 20 25 30

Patient 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Probability 0.50 0.70 1.00

Doctor Cumulative Probability 0.22 0.53 0.78 0.90 1.00

RN1 — 24 48 27 79 81 64 34 65 22 26 15 52 12 40 27 76 51 38 40

Ranges 01–50 51–70 71–99,00

RN Ranges 01–22 23–53 54–78 79–90 91–99,00

Arrival Clock — 30 50 65 90 115 135 150 170 185 200 210 230 240 260 275 300 320 335 355

Time 5 10 15 20

RN 2 31 01 00 74 77 92 06 53 68 30 43 85 46 26 33 13 55 57 83 71

P ( wait ) =

Nurse Cumulative Probability 0.08 0.32 0.83 1.00

Service D D B B B B D N N D D B D D D D N N B B

15 = 0.75 20

RN 3 98 56 58 76 48 48 94 88 79 53 52 87 47 56 31 06 13 31 79 94

RN Ranges 01–08 09–32 33–83 84–99,00

Time 15 20 25 30 35 40

Both Cumulative Probability 0.07 0.23 0.44 0.72 0.89 1.00

Time of Service (mins.)

Doctor Nurse Wait?

30 20 30 35 30 30 30

30 35 30 30 20 15

15 15 35 15 20 15 10 10 10 35 40

Average waiting time =

35

35 40

15 25 25 35 20 20 15 15 20 35 60 60 60

10 25

RN Ranges 01–07 08–23 24–44 45–72 73–89 90–99,00

Departure clock D N 30 50 80 115 140 170 200

190 205 215 230 265 300 320 335 345

380 420

440 mins = 22 mins 20

It would seem that this system is inadequate given that the probability of waiting is high and the average time is high. Also, observing the actual simulation three customers had to wait an hour and two others had to wait 35 minutes which seems excessive. Of course in order to make a fully informed decision this simulation experiment would need to be extended for more patients and then repeated several hundred times.

S13-8.

Direction

Probability

Cumulative Probability

RN Ranges

80 115 140 170

265

310 330 380 420

( x = +1) ( x = −1) West ( y = +1) North ( y = −1) South East

0.25

0.25

01–25

0.25

0.50

26–50

0.25

0.75

51–75

0.25

1.00

76–99, 00

Monte Carlo Simulation (using 16th row of random numbers from Table S12.3). Considering the city as a grid with an x and y axis with the store at point (0,0) each random number selected indicates a movement of 1 unit (block) in either an x or y direction.

End of Block 1 2

RN 58 47

Trial 1 (x, y) (0, 1) ( −1, 1)

Trial 2 RN (x,y) 68 (0,1) 13 (1,1)

RN 20 85

Trial 3 (x,y) (1,0) (1, − 1)

Trial 4 RN (x, y) 53 (0,1) 45 − ( 1, 1)

RN 24 83

Trial 5 (x, y) (1,0) 1, ( − 1)

3

23

(0,1)

09

(2,1)

59

(1,0)

56

( −1, 2 )

05

( 2, − 1)

4

69

(0,2)

20

(3,1)

72

(1,1)

22

(0,2)

81

( 2, − 2 )

5

35

( −1, 2 )

73

(3,2)

88

(1,0)

49

( −1, 2 )

07

( 3, − 2 )

6

21

(0,2)

77

(3,1)

11

(2,0)

08

(0,2)

78

( 3, − 3)

7

41

( −1, 2 )

29

(2,1)

89

( 2, − 1)

82

(0,1)

92

( 3, − 4 )

8

14

(0,2)

72

(2,2)

87

( 2, − 2 )

55

(0,2)

36

( 2, − 4 )

9

59

(0,3)

89

(2,1)

59

( 2, − 1)

27

( −1, 2 )

53

( 2, − 3)

10

28

( −1, 3)

81

(2,0)

66

(2,0)

49

( −2, 2 )

04

( 3, − 3)

Within 2 blocks?

no

yes

yes

no

no

In 2 of the 5 trials the robber is within 2 blocks of the store. As an example, at the end of 10 blocks in trial 1, the robber is 1 block west and 3 blocks north.

S13-9.

Stock Price Movement (+) Increase Same (0) (−) Decrease

Change in Stock Price 1/8 1/4

Probability 0.45

Cumulative Probability 0.45

RN Ranges 01–45

0.30 0.25

0.75 1.00

46–75 76–99, 00

Probability Increase 0.40 0.17

Cumulative Probability 0.40 0.57

RN Ranges 01–40 41–57

Probability Decrease 0.12 0.15

Cumulative Probability 0.12 0.27

RN Ranges 01–12 13–27

3/8 1/2 5/8 3/4 7/8 1

0.12 0.10 0.08 0.07 0.04 0.02

0.69 0.79 0.87 0.94 0.98 1.00

58–69 70–79 80–87 88–94 95–98 99, 00

0.18 0.21 0.14 0.10 0.05 0.05

0.45 0.66 0.80 0.90 0.95 1.00

28–45 46–66 67–80 81–90 91–95 96–99, 00

Monte Carlo Simulation (using the third column of random numbers from Table S12.3)

Day

RN

Stock Price movement

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

76 47 25 08 76 56 31 94 88 14 77 06 62 92 68 27 76 17 35 96 58 31 30 70 33 88 70 70 07 03

− 0 + + − 0 + − − + − + 0 − 0 + − + + − 0 + + 0 + − 0 0 + +

23

Price +, − Change ( ) −1/ 4

79 15 58

+1/ 2 +1/ 8 −1/ 2

11 31 10 54 24 23

+1/ 8 −3 / 8 −1/ 8 +1/ 4 −1/ 4 +1/ 8

87

−3 / 4

23 28 98 25 53

+1/ 8 −3 / 8 + 7/8 +1/ 8 −1/ 2

07 45

+1/ 8 +1/ 4

69 16

+3 / 8 −1/ 4

37 47

+1/ 8 +1/ 4

RN

Stock Price 61 3/4 61 3/4 62 1/4 62 3/8 61 7/8 61 7/8 62 61 5/8 61 1/2 61 3/4 61 1/2 61 5/8 61 5/8 60 7/8 60 7/8 61 60 5/8 61 1/2 61 5/8 61 1/8 61 1/8 61 1/4 61 1/2 61 1/2 61 7/8 61 5/8 61 5/8 61 5/8 61 3/4 62

In order to expand the model for practical purposes the length of the simulation trial would be increased to one year. Then this simulation would need to be repeated for many trials, i.e., 1,000 trials.

S13-10.

Sales Volume 300 400 500 600 700 800

RN 1 Range 1–12 13–30 31–50 51–73 74–91 91–99, 00

Price $22 23 24 25 26 27

RN2 Range 1–7 8–23 24–47 48–72 73–90 91–99, 00

Variable Cost $8 9 10 11 12

RN3 Range 1–17 18–49 50–78 79–92 93–99, 00

c f = $9, 000

Month

RN1

Sales Volume (V)

RN 2

Price (p)

RN 3

Variable Cost ( Cv )

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

58 69 41 28 09 77 89 85 88 87 53 22 82 49 05 78 53 79 37 65

600 600 500 400 300 700 700 700 700 700 600 400 700 500 300 700 600 700 500 600

47 35 14 68 20 29 81 59 11 59 45 49 55 24 81 92 04 61 45 37

$24 24 23 25 23 24 26 25 23 25 24 25 25 24 26 27 22 25 24 24

23 21 59 13 73 72 20 72 89 66 56 08 27 83 07 36 75 44 18 30

9 9 10 8 10 10 9 10 11 10 10 8 9 11 8 9 10 9 9 9

Z = VP − 9,000 − VCv 0 0 −2, 200 −2,500 −5,100 +800 +2,900 +1,500 −600 +1,500 −600 −2, 200 +2, 200 −2,500 −3, 600 +3, 600 −1,800 +2, 200 −1,500 0

Probability of at least breaking even =

10 = 0.50 20

S13-11.

Attractiveness 1 2 3 4 5

Date 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

RN1 Range 1–27 28–62 63–76 77–85 86–99, 00

RN1 95 93 09 95 56 79 61 14 85 09 60 86 69 96 78 61 04 16 82 03

Intelligence 1 2 3 4 5

Attractiveness 5 5 1 5 2 4 2 1 4 1 2 5 3 5 4 2 1 1 4 1

RN 2 Range 1–10 11–26 27–71 72–88 89–99, 00

RN 2 30 28 54 36 26 14 81 24 49 53 98 74 08 06 22 18 23 20 88 65

Personality 1 2 3 4 5

Intelligence 3 3 3 3 2 2 4 2 3 3 5 4 1 1 2 2 2 2 4 3

RN 3 Range 1–15 16–45 46–78 79–85 86–99, 00

RN 3 59 72 66 98 60 50 84 75 08 45 90 55 10 62 99 45 63 99 85 33

Personality

Average Rating

3 3 3 5 3 3 4 3 1 2 5 3 1 3 5 2 3 5 4 2

3.67 3.67 2.33 4.33 2.33 3.00 3.33 2.00 2.67 2.00 4.00 4.00 1.67 3.00 3.67 2.00 2.00 2.67 4.00 2.00

Average overall rating of Salem dates = 2.92 S13-12. There are several ways to access the accuracy of the results. First the student can determine the expected value for each characteristic and average them to see if this results in a value close to the stimulated result.

E ( Attractiveness ) = 2.50

E ( Intelligence ) = 3.05

E ( Personality ) = 2.77

Average rating =

2.50 + 3.05 + 2.77 = 2.77 3

This is relatively close to the simulated result of 2.92 which tends to verify that result.

Confidence limits can also be developed for the average rating. However, this is best done with Excel.

17 Scheduling Answers to Questions 17-1. Projects are scheduled with network planning techniques such as PERT or CPM, which are powerful enough to handle the numerous interrelated decisions that must be made. Many of the scheduling decisions for mass production are made when the assembly line is set up and balanced. Day-to-day scheduling decisions include increasing or decreasing the output rate and model sequencing. Process scheduling involves determining an appropriate mix of ingredients (a linear programming problem) and the optimum length of a production run (an inventory problem).

17-2. Scheduling a job shop is difficult because of the variety of jobs that are processed, each with distinctive routing and processing requirements. Also, there are usually many jobs in the shop at the same time competing for limited resources.

17-3. A production control department checks the availability of resources, assigns jobs to machines, prioritizes the work for each machine, and monitors the progress of each job as it moves through the system.

17-4. A rescue squad wants to minimize response time. A print shop wants to meet customer due dates. A grocery store might want to minimize the time a customer spends in the system. A manufacturer would be concerned about job lateness as well as efficiency measures such as minimizing inventory or overtime.

17-5. The success of a scheduling system is usually measured in terms of mean flow time (the average amount of time a job is in the system), mean tardiness, maximum tardiness, and percent of jobs completed tardy.

17-6. Loading involves assigning jobs to machines or workers to tasks. Usually there are several machines (or workers) that can perform a task, but not at the same level of efficiency. Load leveling tries to smooth out the initial loading assignments to balance the work-load in a facility. Assignments can be made with the assignment method of linear programming.

17-7. Dispatch lists specify the order in which a list of jobs is to be completed. They are usually computergenerated daily according to a priority rule chosen by management.

17-8. a. SPT is most useful when a shop is highly congested. b. Johnson’s rule gives an optimum sequence for minimizing makespan when jobs are processed through two serial processes. c. DDATE should be used when only small values of tardiness can be tolerated. d. FCFS is fine when operating at low-capacity levels.

17-9. DDATE and SLACK are the most common ways to schedule work. However, some students may prefer to complete short assignments first (SPT); others may tackle large projects first (LPT), work on classes they enjoy or those in which their grade is suffering (customer priority), or simply work on assignments as they are assigned (FCFS).

17-10. Critical ratio uses job due date, remaining processing time, and today’s date to construct a ratio of work remaining to time remaining. Priorities can change as work is completed. SLACK uses the same information but arranges it in a different format, time remaining minus work remaining. The sequence of jobs may differ for the two rules.

17-11. Work packages contain all the instructions for manufacture, such as bills of material, routing sheets, and operations sheets. Hot lists specify what jobs need to be finished immediately. Exception reports point out problems and highlight deficiencies.

17-12. Gantt charts plan, or “map out,” work activities. They can also be used to monitor a job’s progress against the plan. Gantt charts are simple to construct and understand. They are a good means of visual control.

17-13. Input/output control (I/O) monitors the input to and output from each work center. Examining only the output from a work center in comparison to the planned output can lead to some erroneous conclusions about the source of a problem. Reduced output at one point in the production process may have been caused by problems at the current work center, or by problems at previous work centers that feed the current work center. The input rate to a work center can readily be controlled only for the initial operations of a job, called the gateway work centers. Input to later operations, performed at downstream work centers, is difficult to control because it is a function of how well the rest of the shop is operating. The backlog, or queue, in front of a process is a measure of congestion and the effectiveness of a scheduling system.

17-14. Infinite scheduling assumes infinite capacity in the initial loading process. Leveling and sequencing decisions are made after overloads or underloads have been identified. Finite scheduling assumes a fixed maximum capacity and will not load the resource beyond its capacity. Loading and sequencing decisions are made at the same time, so that the first jobs loaded onto a work center are the highest priority.

17-15. The theory of constraints considers priority and capacity simultaneously. In concentrates on bottleneck scheduling and treats process batches and transfer batches differently. Bottleneck resources control the flow of work through a system. They should be scheduled first with no slack. The nonbottleneck resources should be scheduled to support the bottleneck activities. Transfer batches refer to the amount that must accumulate before items are transferred to the next workstation. Ideally, this should be one item. Process batches refer to the amount of a product that is produced at one setting of a machine. They are determined by factors associated with the individual process and should have no influence on how finished items are transferred.

17-16. The drum is the bottleneck, beating to set the pace of production for the rest of the system. The buffer is inventory placed in front of the bottleneck to insure it is always kept busy. The rope is the communication signal that tells the processes upstream from the bottleneck when they should begin production.

17-17. Both TOC and lean production signal upstream production and transfer in small lots. TOC sets the pace of production according to the bottleneck operation. Lean production sets the pace according to customer demand. Lean production is a more comprehensive system that includes other factors such as continuous improvement and cellular layouts. TOC is more applicable to batch production; lean production is more applicable to repetitive systems.

17-18. Employee scheduling must consider the needs of the workplace and the availability and suitability of the worker. Absenteeism, performance, personal schedules, and the right mix of workers all affect the scheduling decision.

17-19. Math programming, heuristics, and expert systems have been applied successfully to employee scheduling.

The heuristic presented in this chapter is Baker-Magazine. Advanced planning and scheduling systems use mathematical programming, genetic algorithms, neural networks, constraint based programming, simulation, network analysis and expert systems.

17-20. Students will find a variety of techniques from such companies as SAP, JDA, and QAD.

Solutions to Problems 17-1.

17-2.

None of the nurses are assigned to the same patients as in problem 17-1. This points out the usefulness of multicriteria decisionmaking. It appears that patients like the nurses spending time with them. Oftentimes the producer and the consumer measure quality and efficiency differently.

Other criteria: seniority, nurse preference, skill set, rotation, previous assignment, location

17-3.

17-4.

17-5. Initial matrix: 18 17 15 19 18 16

17 15 15 16 15 16

14 12 13 18 12 16

19 14 17 18 17 18

19 20 20 18 19 20

18 17 18 20 17 17

Minimization matrix (convert max prob to min prob by assigning zero to max value in table and assigning other elements the value of the difference between their value and the max value) 2 3 5 1 2 4

3 5 5 4 5 4

6 8 7 2 8 4

1 6 3 2 3 2

1 0 0 2 1 0

2 3 2 0 3 3

2 5 5 4 4 4

5 8 7 2 7 4

0 6 3 2 2 2

0 0 0 2 0 0

1 3 2 0 2 3

0 3 3 2 2 2

3 6 5 0 5 2

0 6 3 2 2 2

0 0 0 2 0 0

1 3 2 0 2 3

0 3 3 2 2 2

3 6 5 0 5 2

0 6 3 2 2 2

0 0 0 2 0 0

1 3 2 0 2 3

Row reduction: 1 3 5 1 1 4 Column reduction: 0 2 4 0 0 3

Cover all zeroes: 0 2 4 0 0 3

Modify matrix: 2 2 4 2 0 3

0 1 1 2 0 0

3 4 3 0 5 0

0 4 1 2 0 0

2 0 0 4 0 0

1 1 0 0 0 1

0 1 1 2 0 0

3 4 3 0 5 0

0 4 1 2 0 0

2 0 0 4 0 0

1 1 0 0 0 1

0 1 1 2 0 0

3 4 3 0 5 0

0 4 1 2 0 0

2 0 0 4 0 0

1 1 0 0 0 1

Cover all zeroes: 2 2 4 2 0 3

Make assignments: 2 2 4 2 0 3

There are multiple optimal solutions. Albertson and Finch could change assignments without affecting total performance.

Original Matrix with assignment:

Employee Albertson Bunch Carson Denali Ebersole Finch

Sales

Finance

Logistics

Marketing

Production

Customer Service

18 17 15 19 18 16

17 15 15 16 15 16

14 12 13 18 12 16

19 14 17 18 17 18

19 20 20 18 19 20

18 17 18 20 17 17

Total performance = 18 + 17 + 18 + 18 + 20 + 18 = 109 Avg performance = 109/6 = 18.17

17-6. a. FCFS Sequence FCFS Job A B C D E F Average

Start time 0 2 3 7 10 14

Processing time 2 1 4 3 4 5

Completion Time 2 3 7 10 14 19 9.17

Processing time 1 2 3 4 4 5

Completion Time 1 3 6 10 14 19

Duedate 3 2 12 4 8 10

Tardiness 0 1 0 6 6 9 3.67

Duedate 2 3 4 12 8 10 8.83

Tardiness 0 0 2 0 6 9 2.83

b. SPT Sequence

SPT Job B A D C E F Average c.

Start time 0 1 3 6 10 14

DDATE

DDATE Job B A D E F C Average d.

Job A B D E F C

Start time 0 1 3 6 10 15

Processing time 1 2 3 4 5 4

Completion Time 1 3 6 10 15 19 9.00

SLACK Processing time 2 1 3 4 5 4

Duedate 3 2 4 8 10 12

Slack 1 1 1 4 5 8

Duedate 2 3 4 8 10 12

Tardiness 0 0 2 2 5 7 2.67

SLACK Job A B D E F C Average

Start time 0 2 3 6 10 15

Processing time 2 1 3 4 5 4

Completion Time 2 3 6 10 15 19 9.17

Duedate 3 2 4 8 10 12

Tardiness 0 1 2 2 5 7 2.83

SPT 8.83 2.83 3 9

DDATE 9.00 2.67 4 7

SLACK 9.17 2.83 5 7

Performance Measures Mean Flowtime Mean Tardiness No. of Jobs Tardy Max Tardiness

FCFS 9.17 3.67 4 9

17-7.

FCFS Start Processing Job time time 1 2 5 2 7 10 3 17 4 4 21 21 5 42 14 Average Mean Flowtime*

Completion Time 7 17 21 42 56 28.6 26.6

Duedate 20 33 25 45 32

Tardiness 0 0 0 0 24 4.8

*Mean flowtime is different from avg. completion time for this problem since the start time was 2.

SPT Start Processing Job time time 3 2 4 1 6 5 2 11 10 5 21 14 4 35 21 Average Mean Flowtime

Completion Time 6 11 21 35 56 25.8 23.8

Duedate 25 20 33 32 45

Tardiness 0 0 0 3 11 2.8

DDATE Start Job time 1 2 3 7 5 11 2 25 4 35 Average Mean Flowtime

Processing time 5 4 14 10 21

Completion Time 7 11 25 35 56 26.8 24.8

Duedate 20 25 32 33 45

Tardiness 0 0 0 2 11 2.6

Duedate 20 32 25 33 45

Tardiness 0 0 0 2 11 2.6

SLACK Job 1 5 3 2 4

Processing time 5 14 4 10 21

SLACK Start Job time 1 2 5 7 3 21 2 25 4 35 Average Mean Flowtime

Duedate 20 32 25 33 45

Processing time 5 14 4 10 21

Slack 15 18 21 23 24

Completion Time 7 21 25 35 56 28.8 26.8

Performance Measures Mean Flowtime Mean Tardiness No. of Jobs Tardy Max Tardiness

FCFS 26.6 4.8 1 24

SPT 23.8 2.8 2 11

DDATE 24.8 2.6 2 11

SLACK 26.8 2.6 2 11

Recommendation FCFS completes the most jobs on time; however, the assignment that is late is overdue by 24 days. DDATE and SLACK yield the lowest mean tardiness of assignments. SPT has the lowest mean flowtime. Katie might also want to consider the weight each assignment has in her class, her average grade in each class, her goal in terms of grades, and the penalties for late assignments.

17-8. Today is day 4.

Job D A B E F G C

Processing time 4 3 10 5 8 7 2

Duedate 8 10 12 15 18 20 25

Slack 0 3 -2 6 6 9 19

FCFS Start Job time A 4 B 7 C 17 D 19 E 23 F 28 G 36 Average Mean Flowtime

Processing time 3 10 2 4 5 8 7

Completion Time 7 17 19 23 28 36 43 24.71 20.71

SPT Start Job time C 4 A 6 D 9 E 13 G 18 F 25 B 33 Average Mean Flowtime

Processing time 2 3 4 5 7 8 10

Completion Time 6 9 13 18 25 33 43 21.00 17.00

DDATE Start Job time D 4 A 8 B 11 E 21 F 26 G 34 C 41 Average Mean Flowtime

Processing time 4 3 10 5 8 7 2

Completion Time 8 11 21 26 34 41 43 26.29 22.29

SLACK

Processing

Completion

Start

Duedate 10 12 25 8 15 18 20

Tardiness 0 5 0 15 13 18 23 10.57

Duedate 25 10 8 15 20 18 12

Tardiness 0 0 5 3 5 15 31 8.43

Duedate 8 10 12 15 18 20 23

Tardiness 0 1 9 11 16 21 18 10.86

Job time B 4 D 14 A 18 E 21 F 26 G 34 C 41 Average Mean Flowtime

time 10 4 3 5 8 7 2

Summary of Results Mean Flowtime Mean Tardiness No. of Jobs Tardy Max Tardiness

Time 14 18 21 26 34 41 43 28.14 24.14

FCFS 20.71 10.57 5 23

Duedate 12 8 10 15 18 20 25

SPT 17.00 8.43 5 31

Tardiness 2 10 11 11 16 21 18 12.71

DDATE 22.29 10.86 6 21

SLACK 24.14 12.71 7 21

SPT yields the lowest mean flowtime, mean tardiness, and number of jobs tardy.

17-9. Today’s date is day 5.

Sequencing rule: DDATE Processing Job time A 5 B 8 C 6 D 3 E 10 F 14 G 7 H 3 Start Job time A 5 B 10 C 18 D 24 E 27 F 37 G 51 H 58 Average Mean flowtime

Duedate 10 15 15 20 25 40 45 50 Processing time 5 8 6 3 10 14 7 3

DDATE Performance Measures Mean Flowtime 30.75 Mean Tardiness 8.25 Max Tardiness 13 No. of Jobs Tardy 7

Completion Time 10 18 24 27 37 51 58 61 35.75 30.75

Duedate 10 15 15 20 25 40 45 50

Tardiness 0 3 9 7 12 11 13 11 8.25

Sequencing rule: SPT Processing Job time D 3 H 3 A 5 C 6 G 7 B 8 E 10 F 14 Start Job time D 5 H 8 A 11 C 16 G 22 B 29 E 37 F 47 Average Mean flowtime

Duedate 20 50 10 15 45 15 25 40 Processing time 3 3 5 6 7 8 10 14

SPT Performance Measures Mean Flowtime Mean Tardiness Max Tardiness No. of Jobs Tardy

23.88 9.75 22 5

Sequencing rule: SLACK Processing Job time A 5 B 8 C 6 E 10 D 3 F 14 G 7 H 3

Duedate 10 15 15 25 20 40 45 50

Job A B C E D F G H Average

Start time 5 10 18 24 34 37 51 58

Processing time 5 8 6 10 3 14 7 3

Completion Time 8 11 16 22 29 37 47 61

Duedate 20 50 10 15 45 15 25 40 28.875 23.875

Tardiness 0 0 6 7 0 22 22 21 9.75

Slack 5 7 9 15 17 26 38 47 Completion Time 10 18 24 34 37 51 58 61

Duedate 10 15 15 25 20 40 45 50 36.63

Tardiness 0 3 9 9 17 11 13 11 9.125

Mean flowtime

31.63

SLACK Performance Measures Mean Flowtime 31.63 Mean Tardiness 9.13 Max Tardiness 17 No. of Jobs Tardy 7 Recommendation: SPT is the best rule for mean flowtime and number of jobs tardy.

DDATE is best for mean tardiness and maximum tardiness.

Alice should use whichever rule meets her objectives.

For example, more customers will be happy if she uses SPT; however, jobs B, E, and F will be very late. With DDATE no job is over 2 weeks late.

17-10. a.

SPT Job B D A C

Job B D A C Average

Processing time 10 15 20 30 Start time 0 10 25 45

Processing time 10 15 20 30

Performance Measures Mean Flowtime Mean Tardiness No. of Jobs Tardy Max Tardiness b.

Duedate 15 30 20 50 Completion Time 10 25 45 75 38.75

Duedate 15 30 20 50

Tardiness 0 0 25 25 12.5

38.75 12.5 2 25

SLACK Job A B D C

Job A B D C Average

Processing time 20 10 15 30 Start time 0 20 30 45

Performance Measures

Duedate 20 15 30 50

Processing time 20 10 15 30

Slack 0 5 15 20 Completion Time 20 30 45 75 42.5

Duedate 20 15 30 50

Tardiness 0 15 15 25 13.75

Mean Flowtime Mean Tardiness No. of Jobs Tardy Max Tardiness

42.5 13.75 3 25

Recommendation SPT has the same maximum tardiness as SLACK, but outperforms SLACK on every other measure. Choose SPT. 17-11.

SPT Job B A D C

Job B A D C Average

Processing time 3 5 6 9 Start time 0 3 8 14

Performance Measures Mean Flowtime Mean Tardiness Max Tardiness No. of Jobs Tardy

Duedate 5 8 7 18 Processing time 3 5 6 9

Completion Time 3 8 14 23 12

Duedate 5 8 7 18

Tardiness 0 0 7 5 3

12 3 7 2

DDATE Job B D A C

Job B D A C Average

Processing time 3 6 5 9 Start time 0 3 9 14

Performance Measures Mean Flowtime Mean Tardiness Max Tardiness No. of Jobs Tardy SLACK Processing

Duedate 5 7 8 18 Processing time 3 6 5 9

12.25 3.25 6 3

Completion Time 3 9 14 23 12.25

Duedate 5 7 8 18

Tardiness 0 2 6 5 3.25

Job D B A C

Job D B A C Average

time 6 3 5 9 Start time 0 6 9 14

Performance Measures Mean Flowtime Mean Tardiness Max Tardiness No. of Jobs Tardy

Duedate 7 5 8 18 Processing time 6 3 5 9

Slack 1 2 3 9 Completion Time 6 9 14 23 13

Duedate 7 5 8 18

13 3.75 6 3

Recommendation DDATE outperforms SLACK. Choose DDATE if minimizing maximum tardiness is most important.

Choose SPT otherwise.

17-12. The Johnson’s rule sequence is: 8, 4, 6, 5, 9, 2, 1, 3, 10, 7

17-13. a. FCFS

b. SPT

Tardiness 0 4 6 5 3.75

Recommendation: Use FCFS for this data.

17-14.

Process 1 5 7 3 4 1 3

Process 2 4 3 2 1 2 4

E

F

A

E 1 3

F 4 8

Makespan

24

Job A B C D E F Sequence

Process 1 Process 2

Completion Times A 9 13

B

C

D

B 16 19

C 19 21

D 23 24

17-15. Sequence: E, B, D, C, A

17-16.

House A B C D E

Prep 4 3 6 5 1

Paint 2 6 4 6 4

a. SPT sequence: E, A, B, C, D

b. LPT sequence: D, C, B, A, E

Total 6 9 10 11 5

17-17.

Chapter 1 2 3 4 5 6

Typing 30 90 60 45 75 20

Proofing 20 25 15 30 60 30

17-18.

Job A B C D E

Cutting 4 6 1 2 3

Sewing 2 3 3 4 1

D

B

Sequence C

A

E

A 13 15

E 16 17

OUTPUT

Process 1 Process 2

Completion Times B 9 12

C 1 4

D 3 8

Makespan

17

The order can be shipped in 17 days.

17-19. Work Center 7 Period Planned input Actual input Deviation Planned output Actual output Deviation Backlog

30

1 50 50 0 65 60

2 55 50

3 60 55

4 65 60

5 65 65 0 65 60

−5

−5

−5

−5

65 60

−5

65 60

−5

65 60

−5

−5

20

10

5

5

10

Total 295 280 -15 325 300

−25

It appears that no matter how many units are input, work center 7 can process only 60 units per week. The variation in input has caused the backlog to decrease, but now it is starting to rise again. Unless the input is reduced, the backlog will continue to increase.

17-20. Work Center 6 Period Planned input Actual input Deviation Planned output Actual output Deviation Backlog

1 50 40

10

2 55 50

3 60 55

4 65 60

−10

−5

−5

−5

50 50 0 0

55 50

60 55

65 60

−5

−5

−5

0

0

0

5 65 65 0 65 65 0 0

Total 295 270

−25 295 280

−15

It appears from the planned I/O report that the scheduler wants to maintain a 10 unit backlog. The backlog actually disappears in the first period. Work center 6 is producing everything it can, given the input. The problem is at the feeding work center until week 5.

From this I/O alone, the scheduler should input 10 extra units next week to maintain a 10-unit backlog. If this I/O is considered in conjunction with the one in problem 17-19, however, then this action would aggravate work center 7’s problem. This illustrates the domino effect of scheduling linked systems.

17-21. a.

Nurses needed Johnson Swann Coligny Betts Truong

M 3 O O X X X

T 3 X X O O X

W 4 X X X X O

TH 5 X X X X X

F 4 O X X X X

SA 3 X O O X X

SU 3 X X X O O

c. Revised for consecutive days off: Nurses needed Johnson Swann Coligny Betts Truong

M 3 O O X X X

T 3 X O X X O

W 4 X X X X O

TH 5 X X X X X

F 4 X X O X X

SA 3 X X O O X

SU 3 O X X O X

There are several arrangements that produce two consecutive days off. This is one of them. Note that both M-T and SU-M are consecutive. With everyone working TH, days off would need to include TW, F-SA and SA-SN.

17-22.

Volunteers 1. Haynes 2. Tagliero 3. White 4. Cooke 5. Black 6. Romero

M 4 O O X X X X

T 3 X X O O O X

W 2 O O O X X O

TH 3 X X X O O O

F 6 X X X X X X

SA 4 O O X X X X

SU 2 X X O O O O

Wait Staff 1. A. Russell 2. S. Hiller 3. J. Jones 4. T. Turner 5. E. Trice 6. P. Dubois

M 2 O O O O X X

T 3 O X X X O O

W 4 X O O X X X

TH 4 X X X O O X

F 5 X X X X X O

SA 5 O X X X X X

SU 4 X O O X X X

17-23.

17-24.

CASE SOLUTION 17.1 a. Distance to School (in minutes)

Student Amy Brent Calvin Deidre Eliena Franklin

1 45 10 30 20 30 15

Assignment based on distance

Student Amy Brent Calvin Deidre Eliena Franklin Total

1 0 1 0 0 0 0 1

2 30 30 45 30 45 20

School (P.S. or M.S.) 3 4 5 60 30 45 45 20 15 30 60 20 45 15 45 15 45 15 20 30 20 Total Distance

2 0 0 0 0 0 1 1

6 20 10 30 45 30 60

100

School (P.S. or M.S.) 3 4 5 0 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 0 0 1 1 1

6 1 0 0 0 0 0 1

Total 1 1 1 1 1 1

b. Preference/Need Matchup Student Amy Brent Calvin Deidre Eliena Franklin

School (P.S. or M.S.) 3 4 5 1 2 4 4 1 3 4 1 1 1 3 2 2 3 1 1 3 2

1 2 5 2 2 1 1 2 2 2 2 5 1 3 Scored on a scale of 0 to 5.

One point is given for every preference that matches a need.

6 1 3 2 3 2 1

Assignment based on preferences

Student Amy Brent Calvin Deidre Eliena Franklin Total

c.

1 1 0 0 0 0 0 1

2 0 0 0 0 1 0 1

Score

23

School (P.S. or M.S.) 3 4 5 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 1 1 1

6 0 0 0 1 0 0 1

Total 1 1 1 1 1 1

There are several ways distance can be combined with preferences.

One simple way is to assign a 1 to the closest schools and add that to the preference/need matrix.

Preference /Need Matrix including distance

Student Amy Brent Calvin Deidre Eliena Franklin

1 5 3 1 2 2 2

Assignment

Student Amy Brent Calvin Deidre Eliena Franklin Total

2 2 1 2 2 5 3

3 1 4 4 1 3 1

Score

1 1 0 0 0 0 0 1

2 0 0 0 0 1 0 1

School (P.S. or M.S.) 4 5 2 4 1 3 1 2 4 2 3 2 3 2

6 2 4 2 3 2 1

24

School (P.S. or M.S.) 3 4 5 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 1

6 0 1 0 0 0 0 1

Total 1 1 1 1 1 1

d. To prioritize preferences or needs, assign different weights to each factor.

The "score" for each school for a student would be a sum of the weighted factors.

Alternatively, more points could be assigned to the presence of certain factors.

CASE SOLUTION 17.2 1. Find the bottleneck Machine 1:

A1 C1 D1 D3

4 Machine 2: 5 2 4 15 Machine 2 is the bottleneck.

A3 B2 C2 D4

5 8 4 7 24

Machine 3:

A2 B1 C3 D2

6 4 9 1 20

2. Find the fastest time to the bottleneck A3 10 min B2 4 min C2 5 min D4 7 min Bottleneck sequence: B2, C2, D4, A3 3. Sequence the other machines to support the bottleneck.

Machine 1: D1, C1, D3, A1 Machine 3: B1, D2, C3, A2

Items are transferred one at a time, but processed in batches of 50.

Machine 2 starts 4 minutes after machine 3.

50 X’s can be assembled in 1204 minutes.

Determining the sequence involves trying different arrangements, always keeping the bottleneck busy.