Solutions Manual For Operations Management An Integrated Approach 8th Edition By Dan Reid, Nada Sand

Reid & Sanders Operations Management, 8th Edition

Solutions Manual

Solutions Manual Chapter 2: Operations Strategy and Competitiveness 1. Two workers have the job of placing plastic labels on packages before the packages are shipped out. The first worker can place 1,000 labels in 30 minutes.

The second worker can place 850 labels in 20 minutes. Which worker is more productive?

Answer: Productivity of worker 1 = 1000 labels/30 minutes = 33.3 labels per minute Productivity of worker 2 = 850 labels/20 minutes = 42.5 labels per minute Worker 2 is more productive.

Section: Productivity Subsection: Interpreting Productivity Measures LO: 2.5 Define productivity and identify productivity measures.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

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Reid & Sanders Operations Management, 8th Edition

Solutions Manual

2. Last week a painter painted four houses in three days. This week she painted six houses in four days. In which week was the painter more productive?

Answer: Productivity in worker 1 = 4 houses / 3 days = 1.33 houses per day Productivity in worker 2 = 6 houses / 4 days = 1.5 houses per day The painter was more productive in week 2.

Section: Productivity Subsection: Interpreting Productivity Measures LO: 2.5 Define productivity and identify productivity measures.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

Reid & Sanders Operations Management, 8th Edition

Solutions Manual

3. One type of bread-making machine can make six loaves of bread in five hours. A new model of the machine can make four loaves in two hours. Which model is more productive?

Answer: Productivity of old model machine = 6 loaves / 5 hours = 1.2 loaves per hour Productivity of new model machine = 4 loaves / 2 hours = 2.0 loaves per hour The new model is more productive.

Section: Productivity Subsection: Interpreting Productivity Measures LO: 2.5 Define productivity and identify productivity measures.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

Reid & Sanders Operations Management, 8th Edition

Solutions Manual

4. A company that makes kitchen chairs wants to compare productivity at two of its facilities. At facility #1, five workers produced 225 chairs. At facility #2, six workers produced 263 chairs during the same time period. Which facility was more productive?

Answer: Productivity at facility #1 = 225 chairs / 5 workers = 45 chairs per worker Productivity at facility #2 = 263 chairs / 6 workers = 43.8 chairs per worker Facility #1 was more productive.

Section: Productivity Subsection: Interpreting Productivity Measures LO: 2.5 Define productivity and identify productivity measures.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

Reid & Sanders Operations Management, 8th Edition

Solutions Manual

5. A painter is considering using a new high-tech paint roller. Yesterday he was able to paint three walls in 45 minutes using his current roller. Today he painted two walls of the same size in 20 minutes. Is the painter more productive using the new paint roller?

Answer: Productivity using old method = 3 walls / 45 minutes = 0.07 walls per minute Productivity using new method = 2 walls / 20 minutes = 0.10 walls per minute The painter is more productive using the new paint roller.

Section: Productivity Subsection: Interpreting Productivity Measures LO: 2.5 Define productivity and identify productivity measures.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

Reid & Sanders Operations Management, 8th Edition

Solutions Manual

6. Aztec Furnishings makes hand-crafted furniture for sale in its retail stores. The furniture maker has recently installed a new assembly process, including a new sander and polisher. With this new system, production has increased to 95 pieces of furniture per day from the previous 55 pieces of furniture per day. The number of defective items produced has dropped from 8 pieces per day to 2 per day. The production facility operates strictly eight hours per day. Evaluate the change in productivity for Aztec using the new assembly process.

Answer: Using only the non-defective production, productivity has increased from (55 – 8) = 47 pieces per day to (95 – 2) = 93 pieces per day.

Change in productivity = (93 – 47)/47 × 100% = 97.9% There is a 97.9% increase in productivity using the new assembly process.

Section: Productivity Subsection: Interpreting Productivity Measures LO: 2.5 Define productivity and identify productivity measures.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

Reid & Sanders Operations Management, 8th Edition

Solutions Manual

7. Howard Plastics produces plastic containers for use in the food packaging industry. Last year its average monthly production included 20,000 containers produced using one shift five days a week with an eight-hour-a-day operation. Of the items produced 15 percent were deemed defective. Recently, Howard Plastics has implemented new production methods and a new quality improvement program. Its monthly production has increased to 25,000 containers with 9 percent defective. a) Compute productivity ratios for the old and new production system. b) Compare the changes in productivity between the two production systems.

Answer: a) Using only the nondetective production, productivity increased from (20,000 × 0.85) = 17,000 units/month to (25,000 × 0.91) = 22,750 units/month. b) Change in productivity = (22,750 – 17,000)/17,000 × 100% = 33.8% There is a 33.8% increase in productivity using the new production method.

Section: Productivity Subsection: Interpreting Productivity Measures LO: 2.5 Define productivity and identify productivity measures.

Bloom’s: Application AACSB: Analytic Time on task: 4 minutes

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Reid & Sanders Operations Management, 8th Edition

Solutions Manual

8. Med-Tech labs is a facility that provides medical tests and evaluations for patients, ranging from analyzing blood samples to performing magnetic resonance imaging (MRI). Average cost to patients is $85 per patient. Labor costs average $18 per patient, materials costs are $22 per patient, and overhead costs are averaged at $20 per patient. a) What is the multifactor productivity ratio for Med-Tech? What does your finding mean? b) If the average lab worker spends four hours for each patient, what is the labor productivity ratio?

Answer: a) Multifactor productivity = $85/($18 + $22 + $20) = 1.42 This means that the lab is charging approximately 42% over the expenses of labor, materials, and overhead. b) Labor productivity = $85/4 hours = $21.25 per hour Section: Productivity Subsection: Measuring Productivity LO: 2.5 Define productivity and identify productivity measures.

Bloom’s: Application AACSB: Analytic Time on task: 4 minutes

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Reid & Sanders Operations Management, 8th Edition

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9. Handy-Maid Cleaning Service operates five crews with three workers per crew.

Different crews clean a different number of homes per week and spend a differing amount of hours. All the homes cleaned are about the same size. The manager of Handy-Maid is trying to evaluate the productivity of each of the crews. The following data have been collected over the past week.

Work Crew Hours Homes Cleaned Anna, Sue, and Tim 35 10 Jim, Jose, and Andy 45 15 Dan, Wendy, and Carry 56 18 Rosie, Chandra, and Seth 30 10 Sherry, Vicky, and Roger 42 18 Assuming the quality of cleaning was consistent between crews, which crew was most productive?

Answer: Productivity of Anna, Sue, and Tim = 10 homes/35 hours = 0.29 homes/hour Productivity of Jim, Jose, and Andy = 15 homes/45 hours = 0.33 homes/hour Productivity of Dan, Wendy, and Carry = 18 homes/56 hours = 0.32 homes/hour Productivity of Rosie, Chandra, and Seth = 10 homes/30 hours = 0.33 homes/hour Productivity of Sherry, Vicky, and Roger = 18 homes/42 hours = 0.43 homes/hour The crew of Sherry, Vicky, and Roger was the most productive.

Alternative answer: Productivity of Anna, Sue, and Tim = 35 hours/10 homes = 3.5 hours per home Productivity of Jim, Jose, and Andy = 45 hours/15 homes = 3 hours per home Productivity of Dan, Wendy, and Carry = 56 hours/18 homes = 3.1 hours per home Productivity of Rosie, Chandra, and Seth = 30 hours/10 hours = 3 hours per home Productivity of Sherry, Vicky, and Roger = 42 hours/18 hours = 2.3 hours per home The crew of Sherry, Vicky, and Roger was the most productive.

Section: Productivity Subsection: Interpreting Productivity Measures LO: 2.5 Define productivity and identify productivity measures.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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Solutions Manual

10. Two nurses must each visit and attend to patients in the main hospital. The first nurse, on average, can visit 50 patients in an 8-hour day. The second nurse tends to work double shifts, and, on average, cares for 96 patients during the 15 hours she is working. Which nurse is more productive?

Answer: Productivity of nurse 1 = 50 patients / 8 hours = 6.25 patients per hour Productivity of nurse 2 = 96 patients / 15 hours = 6.40 patients per hour Nurse 2 is more productive.

Section: Productivity Subsection: Interpreting Productivity Measures LO: 2.5 Define productivity and identify productivity measures.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

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Reid & Sanders Operations Management, 8th Edition

Solutions Manual

11. A gallery owner typically sells 10 works of art per month. If she invests in a popup gallery for 14 days in a local mall, she expects to sell 8 works of art. Would she be more productive if she invested in a pop-up gallery presence?

Answer: Productivity of traditional gallery = 10 works of art / 30 days = 0.33 works of art per day or approximately one work of art every 3 days.

Productivity of pop-up gallery = 8 works of art / 14 days = 0.57 works of art per day or approximately one work of art every 2 days.

She would be more productive with the pop-up gallery Section: Productivity Subsection: Interpreting Productivity Measures LO: 2.5 Define productivity and identify productivity measures.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

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Reid & Sanders Operations Management, 8th Edition

Solutions Manual

12. Tech-Ed Source is a company that administers standardized certification tests in the information technology field. Those who pass obtain industry certifications that are important for advancement in their field. Average cost to test-takers is $50 per test. Labor costs for the testing facility averages $10 per test, materials costs are $5 per test-taker, and overhead costs are averaged at $15 per test-taker. a) What is the multifactor productivity ratio for Tech-Ed Source? What does your finding mean? b) If the average facility proctor spends two hours on each test-taker, what is the labor productivity ratio?

Answer: a) Multifactor productivity = $50/($10 + $5 + $15) = 1.67 This means that the company is charging approximately 67% over the expenses of labor, materials, and overhead. b) Labor productivity = $50/2 hours = $25 per hour Section: Productivity Subsection: Interpreting Productivity Measures LO: 2.5 Define productivity and identify productivity measures.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

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Solutions Manual Chapter 3: Product Design and Process Selection 1. See-Clear Optics is considering producing a new line of eyewear. After considering the costs of raw materials and the cost of some new equipment, the company estimates fixed costs to be $38,000 with a variable cost of $52 per unit produced. a) If the selling price of each new product is set at $90, how many units need to be produced and sold to break even? Use both the graphical and algebraic approaches. b) If the selling price of the product is set at $85 per unit, See-Clear expects to sell 1,800 units. What would be the total contribution to profit from this product at this price? c) See-Clear estimates that if it offers the product at the original target price of $90 per unit, the company will sell about 1,600 units. Will the pricing strategy of $90 per unit or $85 per unit yield a higher contribution to profit?

Answer: a) Total Cost = $38,000 + $52Q Total revenue = $90Q Break – even Quantity: Q = Fixed Cost / (Selling Price – Variable Cost) Q = $38,000 / ($90 $52) Q = 1,000, so break – even in exceeded at 1,001 units.

Total Dollars

Break Even 150000 100000 50000 0 400

800

1200

Quantity Sold Total Cost

Column1

b) Contribution to Profit = Total Revenue – Total Cost SP(Q) – [FC + VC(Q)] = $85,(1,800) – [$38,000 + $52(1,800)] $21,400 c) Contribution to Profit = Total Revenue – Total Cost SP(Q) – [FC + VC(Q)] = $90,(1,600) – [$38,000 + $52(1,600)] $22,800 Selling at $90 per unit is the better alternative, as it yields $1,400 additional revenue.

Section: The Product Design Process Subsection: Break-Even Analysis: A Tool for Product Screening LO: 3.3 Use break-even analysis as a tool in deciding between alternative products.

Bloom’s: Application AACSB: Analytic Time on task: 4 minutes

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2. Med-First is a medical facility that offers outpatient medical services. The facility is considering offering an additional service, mammography screening tests, on-site. The facility estimates the annual fixed cost of the equipment and skills necessary for the service to be $120,000. Variable costs for each patient processed are estimated at $35 per patient. If the clinic plans to charge $55 for each screening test, how many patients must it process a year in order to break even?

Answer: Break – even Quantity = Fixed Cost / (Selling Price – Variable Cost) = $120,000 / ($55 – $35) = 6.000 patients Section: The Product Design Process Subsection: Break-Even Analysis: A Tool for Product Screening LO: 3.3 Use break-even analysis as a tool in deciding between alternative products.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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3. Tasty Ice Cream is a year-round take-out ice cream restaurant that is considering offering an additional product, hot chocolate. Considering the additional machine, it would need plus cups and ingredients, it estimates fixed costs to be $250 per year and the variable cost to be $0.30. If it charges $1.50 for each hot chocolate, how many hot chocolates does it need to sell in order to break even?

Answer: Break – even Quantity = Fixed Cost / (Selling Price – Variable Cost) = $250 / ($1.50 – $0.30) = 208.33 or 209 hot chocolates Section: The Product Design Process Subsection: Break-Even Analysis: A Tool for Product Screening LO: 3.3 Use break-even analysis as a tool in deciding between alternative products.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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4. Slick Pads is a company that manufactures laptop notebook computers. The company is considering adding its own line of computer printers as well. It has considered the implications from the marketing and financial perspectives and estimates fixed costs to be $500,000. Variable costs are estimated at $200 per unit produced and sold. a) If the company plans to offer the new printers at a price of $350, how many printers does it have to sell to break even? b) Describe the types of operations considerations that the company needs to consider before making the final decision.

Answer: a. Break – even Quantity = Fixed Cost / (Selling Price – Variable Cost) = $500,000/($350  $200) = 3,333.3, so 3,334 printers exceeds the breakeven b. The company needs to consider how the consumers purchasing behavior for laptops influences printer purchasing demand. Other factors to consider relate to the technology of the printers being wireless to support the need for a printer for a laptop.

A final consideration is the cost of ink production and sales for the printers.

Section: The Product Design Process Subsection: Break-Even Analysis: A Tool for Product Screening LO: 3.3 Use break-even analysis as a tool in deciding between alternative products.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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5. Perfect Furniture is a manufacturer of kitchen tables and chairs. The company is currently deciding between two new methods for making kitchen tables. The first process is estimated to have a fixed cost of $87,000 and a variable cost of $70 per unit. The second process is estimated to have a fixed cost of $110,000 and a variable cost of $60 per unit. a) Graphically plot the total costs for both methods. Identify which ranges of product volume are best for each method. b) If the company produces 500 tables a year, which method provides a lower total cost?

Answer: a) 350000

Total Dollars

300000 250000 200000 150000 100000 50000 0 1000

2000

3000

Quantity Process 1 Total Cost

Series 3

Process 1 is lower in total cost when the quantities are fewer than approximately 2,300 tables. Process 2 becomes lower in total cost with quantities higher than approximately 2,300 tables.

Solution: $87,000 + $70(Q) = $110,000 + $60(Q) ($70(Q) – $60(Q)) = ($110,000 – $87,000) $10(Q) = $23,000 (Q) = 2,300 units indifference point b) From indifference point: For quantities lower than 2300 units, Process 1 will have a lower total cost.

Mathematically: Process 1 has a total cost of $87,000 + 500 × $70 = $122,000  lower total cost Process 2 has a total cost of $110,000 + 500 × $60 = $140,000 Section: The Product Design Process Subsection: Break-Even Analysis: A Tool for Product Screening LO: 3.3 Use break-even analysis as a tool in deciding between alternative products.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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6. Harrison Hotels is considering adding a spa to its current facility to improve its list of amenities. Operating the spa would require a fixed cost of $25,000 a year. Variable cost is estimated at $35 per customer. The hotel wants to break even if 12,000 customers use the spa facility. What should be the price of the spa services?

Answer: Break-even Quantity = Fixed Cost / (Selling Price – Variable Cost) rearranges to: Selling Price = (Fixed Cost / Quantity) + Variable Cost = ($25,000/12,000) + $35 = $37.083 Section: The Product Design Process Subsection: Break-Even Analysis: A Tool for Product Screening LO: 3.3 Use break-even analysis as a tool in deciding between alternative products.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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7. Kaizer Plastics produces a variety of plastic items for packaging and distribution. One item, container #145, has had a low contribution to profits. Last year, 22,000 units of container #145 were produced and sold. The selling price of the container was $18 per unit, with a variable cost of $16 per unit and a fixed cost of $60,000 per year. a) What is the break-even quantity for this product? Use both graphic and algebraic methods to get your answer. b) The company is currently considering ways to improve profitability by either stimulating sales volumes or reducing variable costs. Management believes that sales can be increased by 30 percent of their current level or that variable cost can be reduced to 95 percent of their current level. Assuming all other costs are equal, which alternative would lead to a higher profit contribution?

Answer: a) Break – even Quantity = Fixed Cost / (Selling Price – Variable Cost) = $60,000 / ($18  $16) = 30,000 units of product

Break Even Total Dollars

800000 600000 400000 200000 0 20000

30000

40000

Quantity Sold Total Cost

Series 3

b) Increase Sales: 22,000 (1.30) = 28,600 units Contribution to Profit = Total Revenue – Total Cost SP(Q) – [FC + VC(Q)] = 28,600($18) – [$60,000 + $16(28,600)] = $2,800 Still losing money, even with more sales.

Reduce Variable Costs: 0.95 ($16) = $15.20 Contribution to Profit = Total Revenue – Total Cost SP(Q) – [FC + VC(Q)] = 22,000($18) – [$60,000 + $15,20(22,000)] = $1,600 Reducing the variable costs contributes more to profits.

Section: The Product Design Process Subsection: Break-Even Analysis: A Tool for Product Screening LO: 3.3 Use break-even analysis as a tool in deciding between alternative products.

Bloom’s: Application AACSB: Analytic Time on task: 6 minutes

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8. George Fine, owner of Fine Manufacturing, is considering the introduction of a new product line. George has considered factors such as costs of raw materials, new equipment, and requirements of a new production process. He estimates that the variable costs of each unit produced would be $8 and fixed costs would be $70,000. a) If the selling price is set at $20 each, how many units have to be produced and sold for Fine Manufacturing to break even? Use both graphical and algebraic approaches. b) If the selling price of the product is set at $18 per unit, Fine Manufacturing expects to sell 15,000 units. What would be the total contribution to profit from this product at this price? c) Fine Manufacturing estimates that if it offers the product at the original target price of $20 per unit, the company will sell about 12,000 units. Which pricing strategy—$18 per unit or $20 per unit—will yield a higher contribution to profit? d) Identify additional factors that George Fine should consider in deciding whether to produce and sell the new product.

Answer: a) Break – even Quantity = Fixed Cost / (Selling Price – Variable Cost) = $70,000/($20  $8) = 5,833.3 5,834 units have to be produced and sold to break even.

Total Dollars

Break Even 200000 Total Cost

100000 0 2000

Total Revenue 4000

6000

8000

Quantity

b) Contribution to Profit = Total Revenue – Total Cost SP(Q) – [FC + VC(Q)] = $18(15,000) – [$70,000 + $8(15,000)] = $80,000 c) Contribution to Profit = Total Revenue – Total Cost SP(Q) – [FC + VC(Q)] = $20(12,000) – [$70,000 + $8(12,000)] = $74,000 Setting the price at $18 is more profitable by $6,000. d) Additional factors to consider include the accuracy of the demand forecasts and types of marketing techniques that might effectively promote sales.

Section: The Product Design Process Subsection: Break-Even Analysis: A Tool for Product Screening LO: 3.3 Use break-even analysis as a tool in deciding between alternative products.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes ©2023 John Wiley & Sons, Inc. All rights reserved.

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9. Handy-Maid Cleaning Service is considering offering an additional line of services to include professional office cleaning. Annual fixed costs for this additional service are estimated to be $15,000. Variable costs are estimated at $45 per unit of service. If the price of the new service is set at $75 per unit of service, how many units of service are needed for Handy-Maid to break even?

Answer: Break – even Quantity = Fixed Cost / (Selling Price – Variable Cost) = $15,000/($75  $45) = 500 units of service Section: The Product Design Process Subsection: Break-Even Analysis: A Tool for Product Screening LO: 3.3 Use break-even analysis as a tool in deciding between alternative products.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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10. Easy-Tech Software Corporation is evaluating the production of a new software product to compete with the popular word processing software currently available. Annual fixed costs of producing the item are estimated at $150,000, and the variable cost is $10 per unit. The current selling price of the item is $35 per unit, and the annual sales volume is estimated at 50,000 units. a) Easy-Tech is considering adding new equipment that would improve software quality.

The negative aspect of this new equipment would be an increase in both fixed and variable costs. Annual fixed costs would increase by $50,000 and variable costs by $3.

However, marketing expects the better-quality product to increase demand to 70,000 units. Should Easy-Tech purchase this new equipment and keep the price of their product the same? Explain your reasoning. b) Another option being considered by Easy-Tech is the increase in the selling price to $40 per unit to offset the additional equipment costs. However, this increase would result in a decrease in demand to 40,000 units. Should Easy-Tech increase its selling price if it purchases the new equipment? Explain your reasoning.

Answer: a) For current system: Contribution to Profit = Total Revenue – Total Cost SP(Q) – [FC + VC(Q)] = $35(50,000) – [$150,000 + $10(50,000)] = $1,100,000 Purchase of new equipment: Contribution to Profit = Total Revenue – Total Cost SP(Q) – [FC + VC(Q)] = $35(70,000) – [$200,000 + $13(70,000)] = $1,340,000 Easy-Tech will have $240,000 more profit with the new equipment. b) Contribution to Profit = Total Revenue – Total Cost SP(Q) – [FC + VC(Q)] = $40(40,000) – [$200,000 + $13(40,000)] = $880,000 Easy-Tech should not consider the price increase as profits will be less than with the existing price.

Section: The Product Design Process Subsection: Break-Even Analysis: A Tool for Product Screening LO: 3.3 Use break-even analysis as a tool in deciding between alternative products.

Bloom’s: Application AACSB: Analytic Time on task: 7 minutes

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11. Zodiac Furniture is considering the production of a new line of metal office chairs. The chairs can be produced in-house using either process A or process B. The chairs can also be purchased from an outside supplier. Specify the levels of demand for each processing alternative given the costs in the table.

Process A Process B Outside Supplier

Fixed Cost $25,000 $35,000 $ 0

Variable Cost $35 $20 $45

Answer: Indifference Quantity between Process A and Process B Total Cost A = Total Cost B = [FC + VC(Q)]A = [FC + VC(Q)]B The equation rearranges to: Q = (FCB  FCA)/(VCA  VCB) = ($35,000  $25,000) / ($35  $20) = 666.7 or 667 chairs Process A is better than Process B when demand is below 667 chairs.

Indifference Quantity between Process A and Outsourcing Total Cost A = Total Cost O = [FC + VC(Q)]A = [FC + VC(Q)]O Q = $25,000/ ($45 – $35) = 2,500 chairs Process A is better than outsourcing when demand is over 2,500 chairs.

Indifference Quantity between Process B and Outsourcing Total Cost B = Total Cost O = [FC + VC(Q)]B = [FC + VC(Q)]O Q = $35,000/ ($45 – $20) = 1,400 chairs.

Process B is better than outsourcing when demand is over 1,400 chairs.

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In summation: Demand Ranges 0 to 1,400 1,400 and more

Cheapest Production Outsource Process B

Process A is never the best alternative.

Section: The Product Design Process Subsection: Break-Even Analysis: A Tool for Product Screening LO: 3.3 Use break-even analysis as a tool in deciding between alternative products.

Bloom’s: Application AACSB: Analytic Time on task: 7 minutes

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12. NewApp Software Production is deciding whether to take on a new, large client, for whom additional hires will need to be made and another workspace will need to be rented. Estimates of the rents and facilities costs per year are $550,000, and the new developers the company must hire cost $75,000 per year, including benefits. One new developer must be hired for each app produced per year. The company estimates that five developers will need to be hired based on initial estimates of the scope of work, which entails five completed apps currently priced at $150,000 each. a) What is the minimum total value of the contract that NewApp must secure in order to break even? Use the algebraic approach. b) How should NewApp price each completed app to ensure a total contribution of $1 million?

Answer: a) Break – even Quantity = Fixed Cost / (Selling Price – Variable Cost) = $550,000/($150,000  $75,000) = 7.33 or 8 Apps b) Contribution to Profit = Total Revenue – Total Cost = SP(Q) – [FC + VC(Q)] Contribution to Profit + [FC + VC(Q)] = SP(Q) {Contribution to Profit + [FC + VC(Q)]}/Q = SP = [$1,000,000+ ($550,000 + $75,000×5)] / 5 = $385,000 NewApp must raise the price of each new app to $385,000 to ensure a contribution of $1 million.

Section: The Product Design Process Subsection: Break-Even Analysis: A Tool for Product Screening LO: 3.3 Use break-even analysis as a tool in deciding between alternative products.

Bloom’s: Application AACSB: Analytic Time on task: 7 minutes

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13. NewApp Software Production, from Problem 12, has decided to produce a new type of app. The app can be made with the current equipment in place. However, the company is considering investing in new software that would allow the developers to produce the app more quickly. The fixed cost would be raised to $650,000 per year, but the variable cost of labor per app would be reduced to $65,000 per app. The company still plans to sell the completed apps at $150,000 each. Should NewApp produce the app with the new or current software described in Problem 12? Specify the volume of demand for which you would choose each process.

Answer: It is known from Problem 12, money is lost from this new client if demand is under 7 apps. Given they will proceed: Breakeven Between Old and New Equipment: FCO + VCO(Q) = FCN + VCN(Q) $550,000 + $75,000(Q) = $650,000 + $65,000(Q) $10,000(Q) = $100,000 Q = 10 Use the existing software package if demand is at most 10 apps. Purchase the new software package if demand is over 10 apps.

Section: The Product Design Process Subsection: Break-Even Analysis: A Tool for Product Screening LO: 3.3 Use break-even analysis as a tool in deciding between alternative products.

Bloom’s: Application AACSB: Analytic Time on task: 7 minutes

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14. Jacob’s Baby Food Company must go through the following steps to make mashed carrots: (1) unload carrots from truck; (2) inspect carrots; (3) weigh carrots; (4) move to storage; (5) wait until needed; (6) move to washer; (7) boil in water; (8) mash carrots; (9) inspect. Draw a process flow diagram for these steps.

Answer: Unload Truck  Inspect Carrots  Isolate Damage for Claim  Weigh Good Carrots  Move to Storage  Wait until Carrots are Needed  Move to Washer  Boil in Water  Mash  Inspect for Packaging  Package  Rework or Dispose Section: Designing Processes Subsection: [untitled] Introduction LO: 3.5 Understand how to use a process flowchart.

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15. Draw a process flow diagram of your last trip to a fast-food drive-through. Identify bottlenecks. Did any activities occur in parallel?

Answer: Enter fast food line  Wait to Order  Order  Wait to Pay  Pay  Wait to receive Food  Receive Food  Depart Bottlenecks occur during the waiting periods.

Activities such as pre-preparation of the most popular foods during some of the waiting times or utilizing more than one ordering station may be performed in parallel.

Section: Designing Processes Subsection: [untitled] Introduction LO: 3.5 Understand how to use a process flowchart.

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16. Oakwood Outpatient Clinic is analyzing its operation in an effort to improve performance. The clinic estimates that a patient spends on average 3.5 hours at the facility. The amount of time the patient is in contact with staff (e.g., physicians, nurses, office staff, lab technicians) is estimated at 40 minutes. On average the facility sees 42 patients per day. Their standard has been 40 patients per day. Determine process velocity and efficiency for the clinic.

Answer: Process Velocity = Throughput Time / Value – Added Time = 3.5 / 0.6667 = 5.25 3.5 hours / (convert 40 mins to hours) 3.5 / 0.6667 = 5.25 actual output 42 Efficiency = = = 105% st andard out put 40 Section: Process Performance Metrics Subsection: [untitled] Introduction LO: 3.6 Understand how to use process performance metrics.

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17. Oakwood Outpatient Clinic rents a magnetic resonance imaging (MRI) machine for 80 hours a month for use on its patients. Last month the machine was used 65 hours out of the month. What was machine utilization?

Answer: Utility = Used/Available = (65/80) × 100% = 81.25% or 81% Section: Process Performance Metrics Subsection: [untitled] Introduction LO: 3.6 Understand how to use performance metrics.

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18. NewApp Software estimates that it takes a full year for each developer to produce one app. However, after a recent process audit, it was discovered that each developer, on average, spends only 25 hours a week directly working on the app. Assume developers work 9 hours a day, 5 days a week, 48 weeks a year. What is the process velocity?

Answer: Process Velocity = Throughput Time/Value – Added Time = 9 × 5 × 48/(25 × 48) = 2,160/1,200 = 1.8 Section: Process Performance Metrics Subsection: [untitled] Introduction LO: 3.6 Understand how to use process performance metrics.

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Solutions Manual Chapter 4: Supply Chain Management 1. Gabriela Manufacturing must decide whether to insource or outsource a new toxic-free miracle carpet cleaner that works with its Miracle Carpet Cleaning Machine. If it decides to insource the product, the process would incur $250,000 of annual fixed costs and $1.25 per unit of variable costs. If it is outsourced, a supplier has offered to make it for an annual fixed cost of $90,000 and a variable cost of $2.05 per unit in variable costs. a) Given these two alternatives, determine the indifference point (where total costs are equal). b) If the expected demand for the new miracle cleaner is 400,000 units, what would you recommend that Gabriela Manufacturing do?

Answer: Indifference Point: Total Cost of Insourcing = Total Cost of Outsourcing Total Cost = FC + VC(Q) a) $250,000 + $1.25(Q) = $90,000 + $2.05(Q) Q = 200,000 units b) Since the demand is expected to be over the indifference point, insourcing is cheaper.

The total cost for insourcing would be $750,000 and the total cost for outsourcing would be $910,000. The actual difference may be computed to be $160,000. $250,000 + $1.25(Q) $250,000 + $1.25(400,000) = $750,000 $90,000 + $2.05(Q) $90,000 + $2.05(400,000) = $910,000 $910,000 – $750,000 = $160,000 Section: Sourcing Decisions Subsection: Insourcing versus Outsourcing Decisions LO: 4.4 Illustrate how sourcing decisions are made.

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2. Gabriela Manufacturing was able to find a new supplier that would provide the item for $1.50 per unit with an annual fixed cost of $300,000. Should Gabriela Manufacturing insource or outsource the item? Given the expected demand is 400,000 units.

Answer: Indifference Point: Total Cost of Insourcing = Total Cost of Outsourcing Total Cost = FC + VC(Q) The variable cost from a new supplier is $1.50.

The new outsourced fixed cost is $300,000. $300,000 + $1.50(Q) > $250,000 + $1.25(Q) $300,000 + $1.50(400,000) > $250,000 + $1.25(400,000) $900,000 > $750,000 Since the fixed cost and variable cost are both more expensive than insourcing, it would be a better choice to insource the work. However, consideration would be the cost, relevancy, and reliability of the raw materials, equipment, and labor as that can tend to shift. These considerations may be a reason to outsource.

Section: Sourcing Decisions Subsection: Insourcing versus Outsourcing Decisions LO: 4.4 Illustrate how sourcing decisions are made.

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3. Downhill Boards (DB), a producer of snow boards, is evaluating a new process for applying the finish to its snow boards. Durable Finish Company (DFC) has offered to apply the finish for $160,000 in fixed costs and a unit variable cost of $0.60. Downhill Boards currently incurs a fixed annual cost of $135,000 and has a variable cost of $0.95 per unit. Annual demand for the snow boards is 170,000. a) Calculate the annual cost of the current process used at Downhill Boards. b) Calculate the annual cost if Durable Finish Company applies the finish. c) Find the indifference point for these two alternatives. d) How much of a change in demand is needed to justify outsourcing the process?

Answer: a) Total Cost = FC + VC(Q) = $135,000 + $0.95(170,000) = $296,500 b) Total Cost from Durable = $160,000 + $0.60(170,000) = $262,000 c) Indifference point: Total Cost of Insourcing = Total Cost of Outsourcing $135,000 + $0.95(Q) = $160,000 + $0.60(Q) Q = 71,429 snow boards (rounded) ($135,000 – $135,000) + ($0.95(Q) – $0.60(Q)) = ($160,000 – $135,000) + ($0.60(Q) – $0.60(Q)) 0.35(Q) = $25,000 Q = $25,000 / $0.35 = 71,428.57 OR 71,429 d) Since demand is 170,000, which is greater than the indifference point of 71,429 snow boards, then outsourcing is the cheaper process. Demand would need to fall by 58 percent to justify bringing the snowboard production process in-house.

Section: Sourcing Decisions Subsection: Insourcing versus Outsourcing Decisions LO: 4.4 Illustrate how sourcing decisions are made.

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4. Fast Finish Inc. (FFI) has made a technological breakthrough in snow board finish application. FFI will apply the finish for $0.23 per unit in variable costs plus a fixed annual cost of $230,000. Use the cost and demand information given in Problem 3 for Downhill Boards to evaluate this proposal. a) What will it cost Downhill Boards to outsource the finishing process? b) At what demand level does it make sense economically to outsource the finishing process? c) What additional factors should be considered when making this outsourcing decision?

Answer: a) Using the current demand of 170,000 units: Total Cost from FFI = $230,000 + $0.23(Q) = $269,100 Therefore, it will save Downhill Boards $296,500 – $269,100 = $27,400 if they outsource the finishing process.

The demand from the previous question was 170,000 $230,000 + $0.23 (170,000) $230,000 + $39,100 = $269,100 b) Indifference point: Total Cost of Insourcing = Total Cost of Outsourcing $135,000 + $0.95(Q) = $230,000 + $0.23(Q) Q = 131,944.4 Therefore, it is better to outsource when demand is 131,945 or more. c) Additional factors that need to be considered when making this outsourcing decision include the economic stability of FFI, the technical ability of FFI to produce a quality product, the ability of FFI to “partner,” the ability of FFI to deliver on time, and the impact of outsourcing on remaining employees.

Section: Sourcing Decisions Subsection: Insourcing versus Outsourcing Decisions LO: 4.4 Illustrate how sourcing decisions are made.

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5. Henri of Henri’s French Cuisine (HFC), a chain of 12 restaurants, is trying to decide if it makes sense to outsource the purchasing function. Currently, Henri employs two buyers at an annual fixed cost of $100,000. Henri estimates that the variable cost of each purchase order placed is $18. Value-Buy (VB), a group of purchasing specialists, will perform the purchasing function for a fixed annual fee of $120,000 plus $5.50 for each purchase order placed. Last year, HFC placed 1,575 purchase orders. a) What was the cost last year to HFC when doing the purchasing in-house? b) What would the cost have been last year had HFC used Value-Buy? c) What is the indifference point for the two alternatives? d) If HFC estimates it will place 1,700 purchase orders next year, should it use VB? e) What additional factors should be considered by HFC? Answer: a) Total Cost Last Year = FC + VC(Q) = $100,000 + $18(1,575) = $128,350 b) Total Cost from VB = $120,000 + $5.50(1,575) = $128,662.50 c) Indifference point: Total Cost of Insourcing = Total Cost of Outsourcing $100,000 + $18(Q) = $120,000 + $5.50(Q) Q = 1,600 orders d) Yes, VB is a cheaper alternative whenever demand exceeds 1,600 orders, so a volume of 1,700 orders would be cheaper with VB. e) Additional factors that should be considered include the economic stability of VB, the ability of VB to manage quality, the ability of VB to “partner,” the ability of VB to deliver on time, the lead time and quantity order requirements of VB, and the impact of outsourcing on remaining employees.

Section: Sourcing Decisions Subsection: Insourcing versus Outsourcing Decisions LO: 4.4 Illustrate how sourcing decisions are made.

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6. Cal’s Carpentry is considering outsourcing its accounts receivable function. Currently, Cal employs two full-time clerks and one part-time clerk to manage accounts receivable.

Each full-time clerk has an annual salary of $36,000 plus fringe benefits costing an additional 30 percent of their salary. The part-time clerk makes $18,000 per year and has no fringe benefits. Total salary plus fringe cost is $111,600. Cal estimates that each account receivable incurs a $10 variable cost. The Small Business Accounts Receivables Group (SBARG) specializes in handling accounts receivable for small- to medium-size companies. Doris Roberts from SBARG has offered to do the accounts receivable for Cal’s Carpentry at a fixed cost of $75,000 per year plus $30 per account receivable. Next year, Cal expects to have 2,000 accounts receivable. a) Calculate the cost for Cal’s Carpentry to continue doing accounts receivable in-house. b) Calculate the cost for Cal’s Carpentry to use SBARG to handle the accounts receivable. c) If the fixed annual cost offered by SBARG is nonnegotiable but it is willing to negotiate the variable cost, what variable cost from SBARG would make Cal indifferent to the two options? d) What other alternatives might Cal consider in terms of his current staffing for accounts receivable? e) What additional criteria should Cal consider before outsourcing the accounts receivable?

Answer: a) Total Cost = FC + VC(Q) = $111,600 + $10(2,000) = $131,600 b) Total cost for SBARG = $75,000 + $30(2,000) = $135,000 c) $111,600 + $10(2,000) = $75,000 + VC(2,000) = $28.30 d) Cal might investigate if he could do more with part time employees or if the variable costs could be reduced from the existing $10.

5 part-time ($18,000) = $90,000 $111,600 – $90,000 = $21,600 saved e) Before outsourcing the accounts receivable, Cal should consider the following:  How stable is the future of SBARG?  How good is the estimated 2,000 accounts receivable?  How reliable is SBARG compared to his own staff?

Section: Sourcing Decisions Subsection: Insourcing versus Outsourcing Decisions LO: 4.4 Illustrate how sourcing decisions are made.

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7. A U.S.-based device manufacturer is considering outsourcing production of the specialty batteries that power its home medical devices. The fixed costs related to the battery production are $275,000 per year and $0.50 in variable costs. A nearby battery manufacturer has offered to make the batteries for an annual fixed cost of $150,000 and a variable cost of $0.80 per unit in variable costs. a) Given these two alternatives, determine the indifference point (where total costs are equal). b) If the expected demand for the home medical device is 300,000 units, what would you recommend that the device manufacturer do?

Answer: Indifference Point: Total Cost of Insourcing = Total Cost of Outsourcing Total Cost = FC + VC(Q) a) $275,000 + $0.50(Q) = $150,000 + $0.80(Q) Q = 416,667 units (rounded) ($275,000 – $150,000) + ($0.50(Q) – $0.50(Q)) = ($150,000 – $150,000) + ($0.80(Q) – $0.50(Q)) 0.30(Q) = $125,000 Q = $125,000 / $0.30 = 416,666.66666667 OR 416,667 b) Since the demand is expected to be under the indifference point, outsourcing is cheaper. If sales equal demand, the total cost for insourcing would be $425,000 and the total cost for outsourcing would be $390,000. The actual difference may be computed to be $35,000.

Section: Sourcing Decisions Subsection: Insourcing versus Outsourcing Decisions LO: 4.4 Illustrate how sourcing decisions are made.

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8. The same device manufacturer was able to find a Chinese supplier that would provide the batteries for $0.40 per unit with an annual fixed cost of $315,000. Should they insource or outsource the item?

Answer: Indifference Point: Total Cost of Insourcing = Total Cost of Outsourcing Total Cost = FC + VC(Q) The variable cost from a new supplier is $0.40.

The new fixed cost is $315,000. $275,000 + $0.50(Q) = $315,000 + $0.40(Q) Q = 400,000 units is the indifference point Since the demand of 300,000 is lower than the indifference point, insourcing is now a cheaper alternative. The total cost for outsourcing now becomes $435,000. The actual difference may be computed to be $10,000 in favor of producing in house.

Section: Sourcing Decision Subsection: Insourcing versus Outsourcing Decisions LO: 4.4 Illustrate how sourcing decisions are made.

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9. Hawthorne and Sons (H&S) is a lumber mill in Virginia that produces rough-cut boards and then runs the boards through a sequence of machines to create green (uncured) lumber of different dimensions. Finally, the boards are kiln-dried to become building products. H&S also owns a piece of equipment that applies a coating to some of the boards to produce prefinished flooring. The fixed costs of this machine are $145,000 a year and the variable costs are $4.50 per bundle of planks. Another lumber mill in the same Virginia county is willing to apply this finish to H&S’s boards for $75,000 in fixed costs and $4.50 per bundle of planks. Unfortunately, choosing this option would also require transportation of the boards in each direction—to the second lumber mill for coating, and then back to H&S for packaging and customer delivery. H&S estimated this transport cost at $0.50 per bundle. If H&S chose to outsource, it could also sell its coating machine at salvage for a one-time sale price of $150,000. Demand for the bundles of planks was expected to be 130,000. a) Calculate the annual cost of the current process used at H&S. b) Calculate the annual cost if the second lumber mill applies the coating. c) Find the indifference point for these two alternatives. d) How much of a change in demand is needed to justify outsourcing the process?

Answer: a) Total Cost at H&S = FC + VC(Q) = $145,000 + $4.50(130,000) = $730,000 b) Total Cost from second mill = $75,000 + $5.00(130,000) = $725,000 While the salvage value of the machine has an immediate value of $150,000, it is a year one benefit only, and is not part of a direct comparison of the ongoing costs. c) Indifference point: Total Cost of Insourcing = Total Cost of Outsourcing $145,000 + $4.50(Q) = $75,000 + $5.00(Q) Q = 140,000 bundles of flooring planks ($145,000 – $75,000) + ($4.50(Q) – $4.50(Q)) = ($75,000 – $75,000) + ($5(Q) – $4.50(Q)) $70,000 = 0.50(Q) Q= 140,000 d) Since demand is 130,000, which is less than the indifference point of 140,000 bundles of flooring planks, then outsourcing is the cheaper process. Demand would need to rise by 7.7 percent in order to justify continuing to insource the coating process.

Section: Sourcing Decision Subsection: Insourcing versus Outsourcing Decisions LO: 4.4 Illustrate how sourcing decisions are made.

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10. A commercial bakery produces 150,000 cakes per month. Of these, 10,000 are partially customized in some way, for example, with the words “Happy Birthday” written across the top, with room for a name or personalized message to be added later. Currently, the fixed costs attributable to this detailed icing application total $35,000 per year, with a variable cost of $0.30 per cake. Should the bakery outsource this function to another supplier, at a fixed cost of $45,000 per year and a variable cost of $0.25 per cake? a) What will it cost the bakery to outsource the detailed icing process? b) At what demand level does it make sense economically to outsource the icing process? c) What additional factors should be considered when making this outsourcing decision?

Answer: a) Using the current demand of 120,000 cakes = 10,000 cakes × 12 months: Total Cost from outsourcing = $45,000 + $0.25(Q) = $45,000 + $30,000 = $75,000 Add: in house costs = 120,000 cakes (.30 per cake) + $35,000 = $71,000 Therefore, it will cost the bakery $75,000 – $71,000 = $4,000 more to outsource the finishing process. b) Indifference point: Total Cost of Insourcing = Total Cost of Outsourcing $35,000 + $0.30(Q) = $45,000 + $0.25(Q) Q = 200,000 ($35,000 – $35,000) + ($0.30(Q) – $0.25(Q)) = ($45,000 – $35,000) + ($0.25(Q) – $0.25(Q)) $0.05(Q) = 10,000 Q = 200,000 Therefore, it is better to outsource when demand is 200,000 cakes per year or more. c) Additional factors that need to be considered when making this outsourcing decision include the economic stability and quality control of the potential icing service provider, lead times, and the impact of the decision on the current staff.

Section: Sourcing Decision

Subsection: Insourcing versus Outsourcing Decisions LO: 4.4 Illustrate how sourcing decisions are made.

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11. A global insurance company, Global, sells auto insurance online. The company uses its own underwriters and proprietary software algorithms to make determinations on pricing and coverage. Currently, the company also services the policies in case of a claim.

Global’s Chief Operating Officer, Jane, is trying to decide if it makes sense to outsource the claims function, which is currently provided at a fixed cost of $1.65 million and a variable cost of $500 per claim. Jane estimates that if she were to outsource her claims servicing function to another provider, she would incur a fixed annual fee of $1 million plus $550 per claim. Last year, 1.2 million claims were serviced. a) What was the cost last year to Jane when doing the claims management in-house? b) What would the cost have been last year had she outsourced this function? c) What is the indifference point for the two alternatives? d) If Jane estimates her claims will rise to 1.3 million next year, should she outsource the function? e) What additional factors should be considered by Jane?

Answer: a) Total Cost Last Year = FC + VC(Q) = $1.65M + $500(1.2M) = $601.65M b) Total Cost outsourcing = $1M + $550(1.2M) = $661M c) Indifference point: Total Cost of Insourcing = Total Cost of Outsourcing $1.65M + $500(Q) = $1M + $550(Q) Q = 13,000 claims d) No, insourcing is a cheaper alternative whenever demand exceeds 13,000 claims. e) Additional factors that should be considered include the strategic direction of the insurance firm, and whether claims management represents an important part of their value proposition. Also important are the economic stability of the partner, the ability of the partner to manage quality, the ability of the potential partner to be collaborative, the ability of the partner to deliver on time, the lead time and quantity order requirements of the partner, and the impact of outsourcing on remaining employees.

Section: Sourcing Decision Subsection: Insourcing versus Outsourcing Decisions LO: 4.4 Illustrate how sourcing decisions are made.

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12. ZootSuit, a marketing firm, produces an average of 15,000 pieces of marketing collateral per year (e.g., flyers, web pages, presentations, logos, etc.) with a variable cost of $300 per piece. Company leadership wants to increase the output. ZootSuit is considering a subscription in new cloud-based software costing an incremental increase of $30,000 per year. This investment would incrementally decrease the variable cost per piece of marketing collateral by an estimated $50. Current firm overhead related to the production of marketing collateral is $400,000. a) What does it cost ZootSuit to produce at the current rate? What about with the software? b) Should ZootSuit invest in the new software?

Answer: a) Using the current demand of 15,000 pieces of marketing collateral: Total Cost As-is = $400,000 + $300(Q) = $4.9M Total Cost with Software = $430,000 + $250(Q) = $4.18M b) Yes, because the total cost is $0.72M lower and will permit employees to increase output if demand rises. $4.9M – $4.18M = $0.72M Section: Sourcing Decision Subsection: Insourcing versus Outsourcing Decisions LO: 4.4 Illustrate how sourcing decisions are made.

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Solutions Manual Chapter 5: Total Quality Management 1. A smartphone has three components that all must function for the player to work. The average reliability of each component is 0.85. What is the reliability of the smartphone?

Answer: Reliability of smartphone = (0.85)3 = 0.61413 or 61.413% Section: The Philosophy of TQM Subsection: Reliability LO: 5.4 Identify features of the TQM philosophy.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes 2. A jet engine has 10 components in series. The average reliability of each component is 0.998. What is the reliability of the engine?

Answer: Reliability of engine = (0.998)10 = 0.980 or 98.0% Section: The Philosophy of TQM Subsection: Reliability LO: 5.4 Identify features of the TQM philosophy.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes 3. An office copier has four main components in a series with the following reliabilities: 0.85, 0.92, 0.94, and 0.88. Determine the reliability of the copier.

Answer: Reliability of copier = (0.85) (0.92) (0.94) (0.88) = 0.6469 or 64.69% Section: The Philosophy of TQM Subsection: Reliability LO: 5.4 Identify features of the TQM philosophy.

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4. An engine system consists of three main components in a series, all having the same reliability. Determine the level of reliability required for each of the components if the engine is to have a reliability of 0.998.

Answer: (Reliability)3 = 0.998 Therefore, reliability of each component = 0.9993 or 99.93% Section: The Philosophy of TQM Subsection: Reliability LO: 5.4 Identify features of the TQM philosophy.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes 5. A bank loan processing system has three components with individual reliabilities as shown: R1 =

0.82

R2 = 0.991

R3 = 0.98

What is the reliability of the bank loan processing system?

Answer: Reliability of bank loan processing system = (0.82) (0.991) (0.98) = 0.7963 or 79.63% Section: The Philosophy of TQM Subsection: Reliability LO: 5.4 Identify features of the TQM philosophy.

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6. What would be the reliability of the bank system above if each of the three components had a backup with a reliability of 0.80? How would the total reliability be different?

Answer: The backup is drawn in parallel to the original component.

The reliability of the first component = 0.82 + [0.80(1 – 0.82)] = 0.964 or 96.4% The reliability of the second component = 0.91 + [0.80(1 – 0.91)] = 0.982 or 98.2% The reliability of the third component = 0.98 + [0.80(1 – 0.98)] = 0.996 or 99.6% Therefore, the total reliability of the system = (0.964) (0.982) (0.996) = 0.943 or 94.3% The difference in total reliability = 0.943 – 0.7313 = 0.212 or 21.2% Section: The Philosophy of TQM Subsection: Reliability LO: 5.4 Identify features of the TQM philosophy.

Bloom’s: Application AACSB: Analytic Time on task: 4 minutes 7. An LCD projector in an office has a main light bulb with a reliability of 0.94 and a backup bulb, the reliability of which is 0.60. The system looks as follows:

What is the reliability of the system?

Answer: Reliability of LCD projector = 0.60 + (0.94(1 – 0.60)] = 0.976 or 97.6% Section: The Philosophy of TQM Subsection: Reliability LO: 5.4 Identify features of the TQM philosophy.

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8. A university Web server has five main components, each with the same reliability. All five components must work for the server to function as intended. If the university wants to have a 95 percent reliability, what must be the reliability of each of the components?

Answer: (Reliability)5 = 0.95 Therefore, reliability of each component = 0.9898 or 98.98% Section: The Philosophy of TQM Subsection: Reliability LO: 5.4 Identify features of the TQM philosophy.

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9. BioTech Research Center is working to develop a new vaccine for the West Nile Virus.

The project is so important that the firm has created three teams of experts to work on the project from different perspectives. Team 1 has an 86 percent chance of success, team 2 an 87 percent chance of success, and team 3 a 65 percent chance. What is the probability that BioTech will develop the vaccine?

Answer: Probability of developing vaccine = 1 – [(1 – 0.86) (1 – 0.87) (1 – 0.65)] = 0.9936 or 99.36% Section: The Philosophy of TQM Subsection: Reliability LO: 5.4 Identify features of the TQM philosophy.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 10. The following system of components has been proposed for a new product. Determine the reliability of the system.

Answer: The reliability of the first component = 0.90 or 90% The reliability of the second component = 0.85 + [0.85(1 – 0.85)] = 0.9775 or 97.75% The reliability of the third component = 0.90 + [0.90(1 – 0.90)] = 0.99 or 99% The reliability of the fourth component = 0.95 or 95% Therefore, the total reliability of the system = (0.90) (0.9775) (0.99) (0.95) = 0.8274 or 82.74% Section: The Philosophy of TQM Subsection: Reliability LO: 5.4 Identify features of the TQM philosophy.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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11. A new app called "MedAware" is in Beta testing mode. It launches successfully 95% of the time, but only 70% of users are able to operate it the first time they open it. What is the probability of successful first-time operation?

Answer: 95% chance of a successful launch × 70% chance the user will operate correctly on first use = 66.5% chance of a successful first use. This can also be expressed as 0.95 × 0.7 = 0.665 or 66.5%.

Section: The Philosophy of TQM Subsection: Reliability LO: 5.4 Identify features of the TQM philosophy.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes 12. To correctly manufacture a 4-ton heat pump, an HVAC equipment manufacturer must create and then assemble components. The reliability of the processes are as follows:

What is the total reliability of the system?

Answer: 99% reliability of manufacturing components correctly × 97% reliability of assembling components correctly = 96.03% chance of producing reliable equipment, or: 0.99 × 0.97 = 0.9603 or 96.03%.

Section: The Philosophy of TQM Subsection: Reliability LO: 5.4 Identify features of the TQM philosophy.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

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13. That same HVAC manufacturer also offers installation services. A junior installer is deployed first, and is able to complete the install 80% of the time without calling for assistance from a backup senior installer. Senior installers are able to complete the install 98% of the time. What is the end-to-end reliability of the total system, from manufacture to install?

Answer: First, calculate the reliability for the installation process as follows: The junior installer has a reliability rate of 80%, but has a backup senior installer he or she can call with a reliability rate of 98%. The reliability rate of the installers, taken together, is (80% likelihood of the junior installer completing the install) added to (98% likelihood of senior installer completing the install if called to do so × 20% likelihood of requiring the backup.) This can be expressed as: 0.80 + [0.98 × (10.80)] = 0.996 or 99.6%.

Next, multiply the reliability of the manufacturing process (from the previous problem) by the reliability of the installation process to compute total end-to-end reliability as: 96.03% × 99.6% = 95.646% or 0.9603 × 0.996 = 0.95646 or 95.646%.

Answers can be intuitively cross-checked by noting that the reliability of the manufacturing process (96.03%) is decreased only slightly by the installation process (99.6%), so the final answer makes sense at 95.646%.

Section: The Philosophy of TQM Subsection: Reliability LO: 5.4 Identify features of the TQM philosophy.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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14. A piece of medical equipment goes through three factory processes that create a precise point. The processes each meet variance requirements at a rate of 0.98, 0.99, and 0.96.

What percentage of the finished product must be discarded?

Answer: First, calculate the total reliability of the system in producing the precise point as required for the medical equipment, at: 0.95 × 0.99 × 0.96 = 0.93139 or 93.139%.

Therefore 93.139% of the time, the combined process can reliably produce the medical equipment within specifications. Subtracting this from 100%, we see: 100% ˗ 93.139% = 6.861%.

This means that 6.861% of the time, the medical equipment does not meet specifications and must be scrapped.

Section: The Philosophy of TQM Subsection: Reliability LO: 5.4 Identify features of the TQM philosophy.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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Solutions Manual Chapter 6: Statistical Quality Control 1. A quality control manager at a manufacturing facility has taken four samples with four observations each of the diameter of a part. a) Compute the mean of each sample. b) Compute an estimate of the mean and standard deviation of the sampling distribution. c) Develop control limits for 3 standard deviations of the product diameter.

1 5.8 5.9 6.0 6.1

2 6.2 6.0 5.9 5.9

3 6.1 5.9 6.0 5.8

4 6.0 5.9 5.9 6.1

Answer: a) Mean of sample 1 = (5.8 + 5.9 + 6.0 + 6.1)/4 = 5.95 Mean of sample 2 = (6.2 + 6.0 + 5.9 + 5.9)/4 = 6.0 Mean of sample 3 = (6.1 + 5.9 + 6.0 + 5.8)/4 = 5.95 Mean of sample 4 = (6.0 + 5.9 + 5.8 + 6.1)/4 = 5.97 b) The mean of the sampling distribution is the average of the sample means . x = Mean = ( 5.95 + 6.0 + 5.95 + 5.975 ) / 4 = 5.97 The standard deviation of the sampling distribution is computed as

. The σ n population standard deviation σ can be estimated from the 4 samples using the equation: , where n = 16, and ; therefore the estimate is n x = 5.97 2 å i=1( xi - x )

n -1 0.1138.

The standard deviation of the sampling distribution of the sample means is equal to 0.0569 which is estimated using , where σ = 0.1138, and n = 4 (i.e., the number of σ n observations in each sample). c) Center Line

( CL) = x = 5.97

σ = 5.97 + 3 ( 0.0569 ) = 6.14 n σ LCL = x - 3 = 5.97 – 3 ( 0.0569 ) = 5.80 n

UCL = x + 3

Parts a & b Section: Descriptive Statistics ©023 John Wiley & Sons, Inc. All rights reserved.

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Subsection: 1) The Mean and 2) The Range and Standard Deviation LO: 6.3 Explain the use of descriptive statistics in measuring quality characteristics.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes Part c Section: Control Charts for Variables Subsection: Mean (x-Bar) Charts LO: 6.5 Identify the differences between x-bar and R-charts.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 2. A quality control inspector at the Beautiful Shampoo Company has taken three samples with four observations each of the volume of shampoo bottles filled. The data collected by the inspector and the computed means are shown here: Observation 1 2 3 4 Mean

Samples of Shampoo Bottle Volume in Ounces 1 2 18.7 19.2 20.4 20.1 18.6 18.6 20.4 20.5 19.53 19.60

3 18.9 18.3 20.6 20.0 19.45

If the standard deviation of the shampoo bottle-filling operation is 0.2 ounces, use the information in the table to develop control limits of 3 standard deviations for the operation.

Answer: CL = x = (19.53 + 19.60 + 19.45 ) / 3 = 19.53 UCL = x + 3 LCL = x - 3

σ n

= 19.53 + 3 ( 0.1 ) = 19.83

σ = 19.53 – 3 ( 0.1 ) = 19.23 n

Section: Control Charts for Variables Subsection: Mean (x-Bar) Charts LO: 6.5 Identify the differences between x-bar and R-charts.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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3. A quality control inspector has taken four samples with five observations each at the Beautiful Shampoo Company, measuring the volume of shampoo per bottle. If the average range for the four samples is 0.4 ounces and the average mean of the observations is 19.8 ounces, develop 3-sigma control limits for the bottling operation.

Answer: x = 19.8 ounces, R = 0.4 ounces, A2 = 0.58 for n = 5 CL = x = 19.8 UCL = x + A2 R = 19.8 + 0.58 ( 0.4 ) = 20.03

LCL = x - A2 R = 19.8 – 0.58 ( 0.4 ) = 19.57

Section: Control Charts for Variables Subsection: Mean (x-Bar) Charts LO: 6.5 Identify the differences between x-bar and R-charts.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 4. A production manager at Ultra Clean Dishwashing Company is monitoring the quality of the company’s production process. There has been concern relative to the quality of the operation in accurately filling the 16 ounces of dishwashing liquid. The product is designed for a fill level of 16.00 ± 0.30. The company collected the following sample data on the production process: Sample 1 2 3 4 5 6 7 8 9 10

1 16.41 15.95 15.90 16.10 15.80 15.40 16.33 15.50 16.15 15.80

Observations 2 16.10 16.10 16.00 16.21 16.25 15.49 16.21 15.90 16.21 16.43

3 15.85 16.10 16.00 15.5 16.34 15.70 15.99 16.10 16.05 16.50

4 15.76 15.81 .90 15.95 16.41 15.92 16.00 16.03 16.00 15.97

a) Are the process mean and range in statistical control? b) Do you think this process is capable of meeting the design standard?

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Answer: a) Sample 1 2 3 4 5 6 7 8 9 10

1 16.41 15.95 15.90 16.10 15.80 15.40 16.33 15.50 16.15 15.80

2 16.10 16.10 16.00 16.21 16.25 15.49 16.21 15.90 16.21 16.43

Observations 3 4 15.85 15.76 16.10 15.81 16.00 15.90 15.50 15.95 16.34 16.41 15.70 15.92 15.99 16.00 16.10 16.03 16.05 16.00 16.50 15.97 Mean:

Mean 16.03 15.99 15.95 15.94 16.20 15.63 16.13 15.88 16.10 16.18 16.00

Range 0.65 0.29 0.10 0.71 0.61 0.52 0.34 0.60 0.21 0.70 0.47

x = 16.00 R = 0.47 A2 = 0.73

Control limits for x-bar chart:

CL = 16.00 UCL = 16.00 + 0.73 ( 0.47 ) = 16.34 LCL = 16.00 – 0.73 ( 0.47 ) = 15.66

R = 0.47 D3 = 0 D4 = 2.28

Control limits for R-chart:

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CL = 0.47 UCL = D4 R = 2.28(0.47) = 1.07 LCL = D3 R = 0(0.47) = 0

R-chart 1.20 1.00

Range

0.80 0.60 0.40 0.20 0.00

1

2

3

Range

4 CL-R

5

6 UCL-R

7

8

9

10

LCL-R

The sample mean for sample 6 is below the LCL. Therefore, the process mean is not in statistical control. b) From the x-bar chart, we can conclude that the process is not capable of meeting the design standards. Design standards dictate that fill levels range between 16.3 ounces and 15.7 ounces. Alternatively, the process standard deviation can be estimated from the data to be 0.246 ounces. The process capability index of 0.41 would also show the process is not capable of meeting the specification.

From the R-chart, we can conclude that the process is capable of meeting the design standards.

Section: Control Charts for Variables Subsection: Mean (x-Bar) Charts LO: 6.5 Identify the differences between x-bar and R-charts.

Bloom’s: Application AACSB: Analytic Time on task: 15 minutes

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5. Ten samples with five observations each have been taken from the Beautiful Shampoo Company plant to test for volume dispersion in the shampoo bottle-filling process. The average sample range was found to be 0.3 ounces. Develop control limits for the sample range.

Answer: R = 0.3 ounces, D4 = 2.11, D3 = 0, n = 5 CL = R = 0.3 ounces

UCL = D4 R = 2.11 ( 0.3 ) = 0.633

LCL = D3 R = 0 ( 0.3 ) = 0

Section: Control Charts for Variables Subsection: Range (R) Charts LO: 6.5 Identify the differences between x-bar and R-charts.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 6. The Awake Coffee Company produces gourmet instant coffee. The company wants to be sure that the average fill of coffee containers is 12.0 ounces. To make sure the process is in control, a worker periodically selects at random a box of six containers of coffee and measures their weight. When the process is in control, the range of the weight of coffee samples averages 0.5 ounces. a) Develop an R-chart and an -chart for this process. x b) The measurements of weight from the last five samples taken of the six containers follow: x Sample 1 12.0 0.6 2 11.2 0.5 3 12.7 0.5 4 11.3 0.4 5 11.7 0.8 Is the process in control? Explain your answer.

Answer: x = 11.78 R = 0.56 Control limits for x-bar chart: CL = 12.0, using the desired mean as the CL UCL = 12.0 + 0.48 ( 0.56 ) = 12.27 LCL = 12.0 – 0.48 ( 0.56 ) = 11.73

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X-bar Chart 13

Ounces

12.5 12 11.5 11 10.5 10

1

2

3

4

5

Sample X-bar

CL

UCL

LCL

Process mean is not in control. Sample mean 3 is above the UCL, and sample means 2, 4 and 5 are below the LCL. R = 0.56 D3 = 0 D4 = 2.00 Control limits for R chart: CL = 0.5 UCL = 2.00 ( 0.56 ) = 1.12 LCL = 0 ( 0.56 ) = 0

Because all five samples lie within the control limits, the process range is in control.

However, since the x-bar chart is not in control, the process is not in control.

R-chart 1.2 1 0.8 0.6 0.4 0.2 0

1

2

3 R

CL

UCL

4

5

LCL

Section: Control Charts for Variables Subsection: Mean (x-Bar) Charts LO: 6.5 Identify the differences between x-bar and R-charts.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes

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7. A production manager at a Contour Manufacturing plant has inspected the number of defective plastic molds in 5 random samples of 20 observations each. Following are the number of defective molds found in each sample: Sample 1 2 3 4 5 Total

Number of Defects 1 2 2 1 0 6

Number of Observations in Sample 20 20 20 20 20 100

Construct a 3-sigma control chart (z = 3) with this information.

Answer: CL = p = 6 / 100 = 0.06 UCL = 0.06 + 3 LCL = 0.06 - 3

0.06 (1 - 0.06 ) 20

= 0.22

0.06 (1 - 0.06 )

= -0.10 = 0 20 (rounded to zero since the LCL value is negative)

Section: Control Charts for Attributes Subsection: p-Charts LO: 6.6 Identify the differences between p- and c-charts.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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8. A tire manufacturer has been concerned about the number of defective tires found recently. To evaluate the true magnitude of the problem, a production manager selected 10 random samples of 20 units each for inspection. The number of defective tires found in each sample are as follows: Sample 1 2 3 4 5 6 7 8 9 10

Number Defective 0 2 3 2 5 2 1 1 2 3

a) Develop a p-chart with a z = 3. b) Suppose that the next four samples selected had 5, 2, 4, and 1 defects. What conclusion can you make?

Answer: a) CL = p = 21 / 200 = 0.105 UCL = 0.105 + 3 Ö éë0.105 ´ (1 - 0.105 ) / 20 ùû = 0.311 1 - 0.105 ù é LCL = 0.105 - 3 ê0.105 ´ = -0.100 = 0 20 úû ë (rounded to zero since the LCL value is negative)

P-chart 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00

1

2

3

4

5

Fraction Defective

6

7 CL

8

9

10

UCL

LCL

Since the points lie within the control limits, the process is in control.

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b)

Solutions Manual

CL = p = 33 / 280 = 0.12

0.12 (1 - 0.12 )

UCL = 0.12 + 3 LCL = 0.12 - 3

20

= 0.34

0.12 (1 - 0.12 )

= -0.098 = 0 20 (rounded to zero since the LCL value is negative)

P-chart 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00

1

2

3

4

5

Fraction Defective

6

7 CL

8

9

10

UCL

LCL

Since the points lie within the control limits, the process is in control.

Section: Control Charts for Attributes Subsection: p-Charts LO: 6.6 Identify the differences between p- and c-charts.

Bloom’s: Application AACSB: Analytic Time on task: 8 minutes

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9. U-learn University uses a c-chart to monitor student complaints per week. Complaints have been recorded over the past ten weeks. Develop 3-sigma control limits using the following data: Sample 1 2 3 4 5 6 7 8 9 10

Number of Complaints 0 3 1 1 0 0 3 1 1 2

Answer: CL = c = 12 / 10 = 1.2 UCL = 1.2 + 3 1.2 = 4.49 LCL = 1.2 - 3 1.2 = -2.09 = 0 (rounded to zero since the LCL value is negative) Section: Control Charts for Attributes Subsection: c-Charts LO: 6.6 Identify the differences between p- and c-charts.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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10. University Hospital has been concerned with the number of errors found in its billing statements to patients. An audit of 100 bills per week over the past 12 weeks revealed the following number of errors: Week 1 2 3 4 5 6 7 8 9 10 11 12

Number of Errors 6 8 5 5 1 4 8 5 7 8 11 4

a) Develop control charts with z = 3. b) Is the process in control?

Answer: a) CL = c = 72 / 12 = 6 UCL = 6 + 3 6 = 13.35 LCL = 6– = -1.35 = 0 (rounded to zero since the LCL value is negative)

Control chart 20 15 10 5 0

1

2

3

4

5

6

7

8

9

10

11

12

Number Defective CL

UCL

LCL

b) As none of the samples have more errors than the UCL (13.35), the process is in control.

Section: Control Charts for Attributes Subsection: c-Charts LO: 6.6 Identify the differences between p- and c-charts.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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11. Three ice-cream packing machines at the Creamy Treat Company are being evaluated for their capability. The following data are recorded: Packing Machine A B C

Standard Deviation 0.2 0.3 0.05

If specifications are set between 15.8 and 16.2 ounces, determine which of the machines can produce within specifications.

Answer:

C P ( Machine A ) =

USL - LSL 16.2 - 15.8 = = 0.33 6σ 6 ( 0.2 )

CP ( Machine B ) =

USL - LSL 16.2 - 15.8 = = 0.22 6σ 6 ( 0.3 )

USL - LSL 16.2 - 15.8 = = 1.33 6σ 6 ( 0.05 ) Machine C is the only capable machine since its CP value is greater than 1.

CP ( Machine C ) =

Section: Process Capability Subsection: Measuring Process Capability LO: 6.7 Explain the meaning of process capability and the process capability index.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 12. Compute the Cpk measure of process capability for the following machine and interpret the findings. What value would you have obtained with the Cp measure?

Machine data: USL = 120 LSL = 60 Processσ = 5 Processμ = 85 Answer:

æ USL - μ μ - LSL ö C pk = min ç , = min (2.33, 1.67) = 1.67 3σ ÷ø è 3σ This process is capable since the Cpk value is greater than 1.

Looking at the Cp index by itself (1.43) would lead us to believe that the machine is not capable.

Section: Process Capability Subsection: Measuring Process Capability LO: 6.7 Explain the meaning of process capability and the process capability index.

Bloom’s: Application

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AACSB: Analytic Time on task: 3 minutes

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13. Develop an OC curve for a sampling plan in which a sample of n 5 items is drawn from lots of N = 1,000 items. The accept/reject criteria are set up in such a way that we accept a lot if no more than one defect (c = 1) is found.

Answer: Construct an OC chart for n = 5 and c = 1 using the cumulative binomial table in Appendix C.

Alternative solution is to use the Excel BINOM.DIST(c,n,p,TRUE) function.

For example, BINOM.DIST(1,5,0.05,TRUE) = 0.9774 BINOM.DIST(1,5,0.5,TRUE) = 0.1875 BINOM.DIST(1,5,1.0,TRUE) = 0.0 Section: Acceptance Sampling Subsection: Operating Characteristic (OC) Curves LO: 6.9 Explain the process of acceptance sampling and describe the use of operating characteristics (OC) curves.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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14. Quality Style manufactures self-assembling furniture. To reduce the cost of returned orders, the manager of its quality control department inspects the final packages each day using randomly selected samples. The defects include wrong parts, missing connection parts, parts with apparent painting problems, and parts with rough surfaces. The average defect rate is four per day. a) Which type of control chart should be used? Construct a control chart with 3-sigma control limits. b) Today the manager discovered 11 defects. What does this mean?

Answer: a) A c-chart should be used.

CL = c = 4 defects / day

( 4 ) = 10 LCL = 4 - 3 ( 4 ) = -2 = 0 UCL = 4 + 3

(rounded to zero since the LCL value is negative) b) The process appears to be out of control since 11 defects exceeds the upper control limit (UCL) value of 10.

Section: Control Charts for Attributes Subsection: c-Charts LO: 6.6 Identify the differences between p- and c-charts.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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15. Develop an OC curve for a sampling plan in which a sample of n = 10 items is drawn from lots of N = 1,000. The accept/reject criteria are set up in such a way that we accept a lot if no more than one defect (c = 1) is found.

Answer:

Using the cumulative binomial table in Appendix C or the Excel BINOM.DIST(c,n,p,TRUE) function, the table below is obtained.

Percentage of Items Defective (p) Proportion of Defective Items in the Lot Probability of Accepting the Lot 0.05 0.9139 0.10 0.7361 0.15 0.5443 0.20 0.3758 0.25 0.2440 0.30 0.1493 0.35 0.0860 0.40 0.0464 0.45 0.0233 0.50 0.0107 Section: Acceptance Sampling Subsection: Operating Characteristic (OC) Curves LO: 6.9 Explain the process of acceptance sampling and describe the use of operating characteristic (OC) curves.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes

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16. The Fresh Pie Company purchases apples from a local farm to be used in preparing the filling for its apple pies. Sometimes the apples are fresh and ripe. Other times they can be spoiled or not ripe enough. The company has decided that it needs an acceptance sampling plan for the purchased apples. Fresh Pie has decided that the acceptable quality level is 1 defective apple per 100, and the lot tolerance proportion defective is 5 percent.

Producer’s risk should be no more than 5 percent and consumer’s risk 10 percent. a) Develop a plan that satisfies the above requirements. b) Determine the AOQL for your plan, assuming that the lot size is 1,000 apples.

Answer: AQL = 1 per 100 a) Using trial and error and the BINOM.DIST(c,n,p,TRUE) function, we find two plans that satisfy the above requirements.

When n = 132 and c = 3, Alpha: 1-BINOM.DIST(3,132,0.01,TRUE) = 0.044 Beta: BINOM.DIST(3,132,0.05,TRUE) = 0.099 When n = 133 and c = 3, Alpha: 1-BINOM.DIST(3,133,0.01,TRUE) = 0.045 Beta: BINOM.DIST(3,133,0.05,TRUE) = 0.096 b) Using any of the two plans, AOQL is approximately 1.26%, which occurs at about 2.00% defects.

Section: Acceptance Sampling Subsection: Operating Characteristic (OC) Curves LO: 6.9 Explain the process of acceptance sampling and describe the use of operating characteristic (OC) curves.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 17. A computer manufacturer purchases microchips from a world-class supplier. The buyer has a lot of tolerance proportion defective of 10 parts in 5,000, with a consumer’s risk of 15 percent. If the computer manufacturer decides to sample 2,000 of the microchips received in each shipment, what acceptance number, c, would it want?

Answer: Given that the lot tolerance proportion defective is 10/5000 = 0.002 and n = 2,000.

When c = 1: BINOM.DIST(1,2000,0.002,TRUE) = 0.0914.

When c = 2: BINOM.DIST(2,2000,0.002,TRUE) = 0.2378.

Hence we want c = 1.

Section: Acceptance Sampling Subsection: Operating Characteristic (OC) Curves LO: 6.9 Explain the process of acceptance sampling and describe the use of operating characteristic (OC) curves.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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18. Joshua Simms has recently been placed in charge of purchasing at the Med-Tech Labs, a medical testing laboratory. His job is to purchase testing equipment and supplies. MedTech currently has a contract with a reputable supplier in the industry. Joshua’s job is to design an appropriate acceptance sampling plan for Med-Tech. The contract with the supplier states that the acceptable quality level is 2 percent defective. Also, the lot tolerance proportion defective is 5 percent, the producer’s risk is 5 percent, and the consumer’s risk is 10 percent. a) Develop an acceptance sampling plan for Joshua that meets the stated criteria. b) Draw the OC curve for the plan you developed. c) What is the AOQL of your plan, assuming a lot size of 1000?

Answer: a) Using trial and error, we get that n = 306 and c = 10 because alpha: =1-BINOM.DIST(10,306, 0.02, TRUE) = 0.046 (where 0.02 corresponds to AQL=2%) beta: =BINOM.DIST(10,306,0.05,TRUE) = 0.0987 (where 0.05 corresponds to LTPD) b)

c) AOQL is 0.0148 when p = 0.025 to 0.027 Section: Acceptance Sampling Subsection: Operating Characteristic (OC) Curves LO: 6.9 Explain the process of acceptance sampling and describe the use of operating characteristic (OC) curves.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes

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19. Breeze Toothpaste Company makes tubes of toothpaste. The product is produced and then pumped into tubes and capped. The production manager is concerned whether the filling process for the tubes of toothpaste is in statistical control. The process should be centered on 6 ounces per tube. Six samples of five tubes were taken and each tube was weighed.

The weights are: Sample 1 2 3 4 5 6

Ounces of Toothpaste per Tube 1 2 3 5.78 6.34 6.24 5.89 5.87 6.12 6.22 5.78 5.76 6.02 5.56 6.21 5.77 5.76 5.87 6.00 5.89 6.02

4 5.23 6.21 6.02 6.23 5.78 5.98

5 6.12 5.99 6.10 6.00 6.03 5.78

a) Develop a control chart for the mean and range for the available toothpaste data. b) Plot the observations on the control chart and comment on your findings.

Answer: a) Ranges in samples 1 through 6 are 1.11, 0.34, 0.46, 0.67, 0.27, and 0.24, respectively.

Control limits for X-bar chart: CL = 5.95 UCL = 5.95 + 0.58 ( 0.515 ) = 6.25 LCL = 5.95 – 0.58 ( 0.515 ) = 5.65

Control limits for R-chart: CL = 0.515 UCL = 2.11 ( 0.515 ) = 1.087 LCL = 0 ( 0.515 ) = 0

b) X-bar Chart 6.30 6.20 6.10 6.00 5.90 5.80 Ounces 5.70 5.60 5.50 5.40 5.30

6.2495 5.9523 5.6552

1

2

3

4

5

6

Sample Number

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R-chart 1.2 1 0.8 0.6 0.4 0.2 0

1

2

3 R

CL

UCL

4

5

LCL

The mean is in control, but the range is not. (Sample 1 is above the UCL) Section: Control Charts for Variables Subsection: Mean (x-Bar) Charts LO: 6.5 Identify the differences between x-bar and R-charts.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes 20. Breeze Toothpaste Company has been having a problem with some of the tubes of toothpaste leaking. The tubes are packed in containers with 100 tubes each. Ten containers of toothpaste have been sampled. The following number of toothpaste tubes were found to have leaks: Sample 1 2 3 4 5

Number of Leaky Tubes 6 10 11 13 8

Sample 6 7 8 9 10 Total

Number of Leaky Tubes 16 6 8 4 9 91

Develop a p-chart with 3-sigma control limits and evaluate whether the process is in statistical control.

Answer: The ten samples have a total of 91 leaky tubes.

Therefore, the proportion of leaky tubes in the population = 91/1000 = 0.091 CL = p = 0.091 UCL = 0.091 + 3 Ö éë0.091 ´ (1 - 0.091 ) / 100 ùû = 0.177 LCL = 0.091 – 3 Ö éë0.091 ´ (1 - 0.091 ) / 100 ùû = 0.005

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P-chart 0.20 0.18 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0.00

1

2

3

4

5

Fraction Defective

6

7 CL

8

9

10

UCL

LCL

Since none of the p-bars exceeds 0.177, the process is in control.

Section: Control Charts for Attributes Subsection: p-Charts LO: 6.6 Identify the differences between p- and c-charts.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 21. The Crunchy Potato Chip Company packages potato chips in a process designed for 10.0 ounces of chips with an upper specification limit of 10.5 ounces and a lower specification limit of 9.5 ounces. The packaging process results in bags with an average net weight of 9.8 ounces and a standard deviation of 0.12 ounces. The company wants to determine if the process is capable of meeting design specifications.

Answer:

(10.5 - 9.5 ) = 1.39 Cp = 6 ( 0.12 )

, which means the process is in control

However, upon calculating the Cpk; æ 10.5 - 9.8 9.8 - 9.5 ö C pk = min ç , = min(1.94, 0.833) = 0.833 ç 3 ( 0.12 ) 3 ( 0.12 ) ÷÷ è ø The process is not capable.

Section: Process Capability Subsection: Measuring Process Capability LO: 6.7 Explain the meaning of process capability and the process capability index.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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22. The Crunchy Potato Chip Company sells chips in boxes with a net weight of 24 ounces per box (680 grams). Each box contains ten individual 2.4-ounce packets of chips.

Product design specifications call for the packet-filling process average to be set at 69.0 grams so that the average net weight per box will be 690 grams. Specification width is set for the box to weigh 680±10 grams. The standard deviation of the packet-filling process is 8.0 grams per box. The target process capability ratio is 1:33. The production manager has just learned that the packet-filling process average weight has dropped down to 68.0 grams. Is the packaging process capable? Is an adjustment needed?

Answer: The computation Cp = (690-670)/(6×8) = 0.4167. The Cpk would yield the same values as 680 grams is in the center of the process. To be in control, the process variability must be improved.

Section: Process Capability Subsection: Measuring Process Capability LO: 6.7 Explain the meaning of process capability and the process capability index.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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Solutions Manual Chapter 7: Just-in-Time and Lean Systems 1. Jason Carter works for a producer of soaps that come in packages of 6 each. His job is to fill the packages with soap, and he is expected to process 30 packages an hour. The facility where Jason works uses a kanban production system in which each container holds 5 packages of soap. It takes 20 minutes to receive the packages he needs from the previous workstation. How many kanbans are needed for the filling process?

Answer: D = 30 packages per hour T = 20 minutes = 1/3 hour C = 5 packages per container N=

DT ( 30 )(1/3 ) = = 2 kanbans C 5

Section: Just-in-Time Manufacturing Subsection: Kanban Production LO: 7.3 Explain the key elements of JIT manufacturing.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

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2. A manufacturer of thermostats uses a kanban system to control the flow of materials. The packaging center processes 15 thermostats an hour and receives completed thermostats every 20 minutes. Containers hold 5 thermostats each. a) How many kanbans are needed for the packaging center? b) If management decides to keep ten thermostats as safety stock, how many kanbans will be needed?

Answer: a) D = 15 thermostats per hour T = 20 minutes = 1/3 hour C = 5 thermostats per container N = DT/C = (15 × 1/3 ) / 5 = 1 kanban b)

S = 10 N = ( DT + S ) / C = ( (15 × 1/3)+10 ) / 5 = 3 kanbans

Section: Just-in-Time Manufacturing Subsection: Kanban Production LO: 7.3 Explain the key elements of JIT manufacturing.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

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3. A production cell at Canderberry Candle facility operates 5 hours per day and uses a pull method to supply wicks to the assembly line. The wicks are used at a rate of 300 per day.

Each container holds 20 wicks and usually waits 20 minutes in the production cell.

Management wants a safety stock of 10 percent. How many containers should be used at the Canderberry Candle facility for purposes of pull production?

Answer: D = ( 300 wicks/day )(1 day/5 hours )(1 hour/60 minutes ) = 1 wick/minute T = 20 minutes C = 20 per container S = 0.1 (1 × 20 ) = 2 wicks N=

DT + S (1.0 )( 20 ) + 2.0 = = 1.1, rounded to 2 containers C 20

Section: Just-in-Time Manufacturing Subsection: Kanban Production LO: 7.3 Explain the key elements of JIT manufacturing.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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4. Carlos Gonzales is production manager at an assembly plant that manufactures cordless telephones. The company is planning to install a pull system. The process is being planned to have a usage rate of 45 pieces per hour. Each container is designed to hold 8 pieces. It takes an average of 45 minutes to complete a cycle. a) How many containers will be needed? Note: Make sure to round up b) How will the number of needed containers change as the system improves?

Answer: a) D = ( 45 units/hour )(1 hour/60 minutes ) = 0.75 units/minute T = 45 minutes C = 8 per container N = DT /C = ( 0.75 × 45 ) / 8 = 4.22, rounded to 5 containers b) As cycle time decreases, the number of containers needed also decreases.

Section: Just-in-Time Manufacturing Subsection: Kanban Production LO: 7.3 Explain the key elements of JIT manufacturing.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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5. A dye cell at the Acme Clothing Factory uses 500 pounds of dye each day. The dye is moved in vats at a rate of approximately one per hour. Each vat holds 10 pounds of dye.

Management has set safety stock at 10 percent. The facility operates 8 hours per day. How many vats should be used?

Answer: D = ( 500 pounds/day )( l day/8 hours ) = 62.5 pounds/hour T = 1 hour C = 10 per vat S = 0.1 ( 62.5 × 1 ) = 6.25 pounds of dye N=

DT + S ( 62.5 )(1 ) + 6.25 = = 6.875 , rounded to 7 vats C 10

Section: Just-in-Time Manufacturing Subsection: Kanban Production LO: 7.3 Explain the key elements of JIT manufacturing.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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6. Anna works on an assembly line where it takes her 60 minutes to produce 15 units of a product needed to fill a container. It takes her an additional 5 minutes to transport the container to Josh, who works at the next station. The company uses a safety stock of 15 percent. The current assembly line uses 7 kanbans between Anna’s and Josh’s stations.

Compute the demand for the product.

Answer: T = 60 minutes, plus 5 minutes to transport = 65 minutes C = 15 per container N = 7 containers S = 15% = 0.15 ( DT ) = 9.75D 7 = éë( D )( 65 ) + 9.75 ( D ) ùû / 15 D = 1.40 units per minute or 81.28 units per hour Proof: [65(1.4) + (0.15(65)(1.40))]/15 = 6.98, rounded to 7 containers Section: Just-in-Time Manufacturing Subsection: Kanban Production LO: 7.3 Explain the key elements of JIT manufacturing.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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7. Robert produces 300 units of a product per hour and 30 units are needed to fill a container. It takes 15 minutes to receive the materials needed from the previous workstation. The company currently uses a safely stock of 10 percent. Determine the number of kanbans needed between Robert’s station and the previous process.

Answer: D = ( 300 units/hour )(1 hour/60 minutes ) = 5 units/minute T = 15 minutes C = 30 units/container S = 0.1 ( 5 × 15 ) = 7.5 units N=

DT + S ( 5.0 )(15 ) + 7.5 = = 2.75, rounded to 3 kanbans C 30

Section: Just-in-Time Manufacturing Subsection: Kanban Production LO: 7.3 Explain the key elements of JIT manufacturing.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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8. Consider the information from Problem 7. Determine how the number of kanbans and the inventory level will be affected if the time required for Robert to receive the material increases to 20 minutes.

Answer: D = ( 300 units/hour )(1 hour/60 minutes ) = 5 units/minute T = 20 minutes C = 30 units/container S = 0.1 ( 5 × 20 ) = 10 units N = ( DT + S ) / C = éë( 5 × 20 ) + 10 ùû / 30 = 3.67, rounded to 4 kanbans If the time increases by 33%, the number of kanbans also increases in the same proportion. As cycle time is increased by 33%, the safety stock is also increased by 33%.

Section: Just-in-Time Manufacturing Subsection: Kanban Production LO: 7.3 Explain the key elements of JIT manufacturing.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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9. Consider the information from Problem 7. Determine how the number of kanbans and the inventory level will be affected if: a) The container size is decreased to 15 units. b) The container size is increased to 40 units. c) The safety stock is increased to 20 percent.

Answer: a) N=

(5.0 )(15 ) + 7.5 = 5.5, rounded to 6 kanbans

N=

(5.0 )(15 ) + 7.5 = 2.06, rounded to 3 kanbans

15 Number of kanbans increases. Inventory level stays the same. b)

40 Number of kanbans stays the same. Inventory level increases. c)

D = ( 300 units/hour )(1 hour/60 minutes ) = 5 units/minute S = 0.2 ( 5 × 15 ) = 15 units N=

(5.0 )(15 ) + 15 = 3 kanbans

30 Number of kanbans stays the same. Inventory level increases.

Section: Just-in-Time Manufacturing Subsection: Kanban Production LO: 7.3 Explain the key elements of JIT manufacturing.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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10. Kelly Glebe works for a facility that manufactures a popular over-the-counter medication that are shipped in cases of 20 bottles. She fills an average of 45 cases an hour. The kanbans her facility uses hold 4 finished bottles. It takes 5 minutes to receive the bottles she needs from the previous workstation. How many kanbans are needed for the filling process?

Answer: D = 45 packages per hour

T = 5 minutes = 1/12 hour or 0.825 C = 20 packages per container N = DT /C = ( 45 ´ 0.825 ) / 20 = 0.19, rounded to 1 kanban

Section: Just-in-Time Manufacturing Subsection: Kanban Production LO: 7.3 Explain the key elements of JIT manufacturing.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

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11. A pizza box manufacturer uses a kanban system to control the flow of materials. The boxes are stamped (cut to size) and inked at a rate of 1,500 pizza boxes an hour. The next machine on the assembly line then receives the partially completed boxes every 15 minutes and automatically folds them into boxes. Containers hold 250 flat, inked pizza boxes each.

A) How many kanbans are needed for the folding machine? b) If management decides to keep 500 partial pizza boxes as safety stock, how many kanbans will be needed?

Answer: a) D = 1,500 boxes per hour T = 15 minutes = 1/4 hour or 0.25 C = 250 boxes per container N = DT/C = (1,500 × 1/4) / 250 = 1.5, rounded to 2 kanbans b) S = 500 N = (DT + S) / C = ((1,500 × 0.25) + 500) / 250 = 3.5, rounded to 4 kanbans Section: Just-in-Time Manufacturing Subsection: Kanban Production LO: 7.3 Explain the key elements of JIT manufacturing.

Bloom’s: Application AACSB: Analytic Time on task: 4 minutes

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12. A production cell at Stylish Bags facility operates 8 hours per day and uses a pull method to supply purse handles to the assembly line. The purse handles are used at a rate of 200 per day. Each container holds 10 handles and on average remains 30 minutes in the cell before being passed on to the assembly-line team that is responsible for attaching the handles to the finished purses. Safety stock is usually 25 percent. How many containers should the manufacturing cell deploy at the Stylish Bags facility for handles, in order to meet demand?

Answer: D = ( 200 handles/day )(1 day/8 hours )(1 hour/60 minutes ) = 0.42 handles/minute T = 30 minutes C = 10 per container S = 0.25 ( 0.42 × 30 ) = 3.15 N = ( DT + S ) / C = éë( 0.42 × 30 ) + 3.15 ùû / 10 = 1.58, rounded to 2 containers Section: Just-in-Time Manufacturing Subsection: Kanban Production LO: 7.3 Explain the key elements of JIT manufacturing.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

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13. Radha Gupta is production manager at an assembly plant that manufactures personal activity trackers. The company is planning to install a pull system. Radha plans to have a usage rate of 60 pieces per hour. Each container is designed to hold 15 pieces. It takes an average of 30 minutes to complete a cycle. a) How many containers will be needed? b) What will happen to the number of containers required if the system can only process 45 pieces per hour under the new system?

Answer: a) D = ( 60 units/hour )(1 hour/60 minutes ) = 1 units/minute T = 30 minutes C = 15 per container N = DT /C = (1 × 30 ) / 15 = 2 containers b) If the cycle time increases, the number of containers needed also increases.

Section: Just-in-Time Manufacturing Subsection: Kanban Production LO: 7.3 Explain the key elements of JIT manufacturing.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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Solutions Manual Chapter 8: Forecasting 1. Sales for a product for the past three months have been 200, 350, and 287. Use a threemonth moving average to calculate a forecast for the fourth month. If the actual demand for month 4 turns out to be 300, calculate the forecast for month 5.

Answer: F4 = ( A1 + A2 + A3 ) / 3 = ( 200 + 350 + 287 ) / 3 = 279.0

F5 = ( A2 + A3 + A4 ) / 3 = ( 350 + 287 + 300 ) / 3 = 312.33

Section: Time Series Methods Subsection: Simple Moving Average LO: 8.4 Describe time series models.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes 2. Lauren’s Beauty Boutique has experienced the following weekly sales: Week 1 2 3 4 5

Sales 420 382 425 465 395

Forecast sales for week 6 using the naïve method, a simple average, and a three-period moving average.

Answer: Naïve Method:

F6 = A5 = 395 5-Period moving Average: F6 = ( A1 + A2 + A3 + A4 + A5 ) / 5 = (420 + 382 + 425 + 465 + 395 / 5 = 417.4 3-Period moving average:

F6 = ( A3 + A4 + A5 ) / 3 = ( 425 + 465 + 395 ) / 3 = 428.3

Section: Time Series Methods Subsection: Naïve and Simple Moving Average LO: 8.4 Describe time series models.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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3. Hospitality Hotels forecasts monthly labor needs. a) Given the following monthly labor figures, make a forecast for June using a threeperiod moving average and a five-period moving average.

Month January February March April May

Actual Values 32 41 38 39 43

b) What would be the forecast for June using the naïve method? c) If the actual labor figure for June turns out to be 41, what would be the forecast for July using each of these models? d) Compare the accuracy of these models using the mean absolute deviation (MAD). e) Compare the accuracy of these models using the mean squared error (MSE).

Answer: a) 3-Period Moving Average: FJune = ( AMarch + A April + AMay ) / 3 = ( 38 + 39 + 43 ) / 3 = 40 5-Period Moving Average: FJune = ( AJanuary + AFebruary + AMarch + A April + AMay ) / 5 = ( 32 + 41 + 38 + 39 + 43 ) / 5 = 38.6 b) Naïve:

FJune = AMay = 43

c) 3-Period Moving Average: FJuly = ( A April + AMay + AJune ) / 3 = ( 39 + 43 + 41 ) / 3 = 41 5-Period Moving Average: FJuly = ( AFebruary + AMarch + A April + A May + AJune ) / 5 = ( 41 + 38 + 39 + 43 + 41 ) / 5 = 40.4 Naïve:

FJuly = AJune = 41

d) Month

Actual

January February March April May June

32 41 38 39 43 41

3-period moving average

Absolut e Error

37 39.33 40

MAD ( 3-period moving average ) =

5-period moving average

2 3.67 1

38.6

å actual - forecast

© 2023 John Wiley & Sons, Inc. All rights reserved.

n

Absolute Error

2.4

Naïve

Absolut e Error

32 41 38 39 43

9 3 1 4 2

= ( 2 + 3.67 + 1 ) / 3 = 2.22

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MAD ( 5-period moving average ) = MAD ( Naïve ) =

å actual - forecast n

= 2.4 / 1 = 2.4

å actual - forecast

= ( 9 + 3 + 1 + 4 + 2 ) / 5 = 3.8 n The 3-period moving average provides the best historical fit using the MAD criterion and would be better to use.

e) Month Actual 3-period Error moving average January 32 February 41 March 38 April 39 37 4 May 43 39.33 3.67 June 41 40 1

Squared 5-period Error Squared Naïve Error moving Error average

16 13.47 1

MSE ( 3-period moving average ) =

38.6

2.4

5.76

32 41 38 39 43

Error

Squared Error

-9 3 -1 -4 2

81 9 1 16 4

å(actual - forecast)2 = (16 + 13.47 + 1 ) / 3 = 10.15 n

å(actual - forecast)2 MSE ( 5-period moving average ) = = 5.76 / 1 = 5.76 n å(actual - forecast)2 MSE ( Naïve ) = = ( 81 + 9 + 1 + 16 + 4 ) / 5 = 22.20 n The 5-period moving average provides the best historical fit using the MSE criterion, but it only measures one error term.

Parts a & b Section: Time Series Methods Subsection: Simple Moving Average LO: 8.4 Describe time series models.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes Parts c & d Section: Measuring Forecast Accuracy Subsection: Forecast Accuracy Methods LO: 8.6 Computer forecast accuracy.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes

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4. The following data are monthly sales of jeans at a local department store. The buyer would like to forecast sales of jeans for the next month, July.

Month January February March April May June

Sales 37 35 42 38 48 45

a) Forecast sales of jeans for March through June using the naïve method, a two-period moving average, and exponential smoothing with an α = 0.2. (Hint: Use naïve to start the exponential smoothing process.) b) Compare the forecasts using MAD and decide which is best. c) Using your method of choice, make a forecast for the month of July.

Answer: a) Month

Sales

January Februar y March April May June

37 35

Naïve Forecast

42 38 48 45

Absolut e Error

2-period moving average

Absolute Error

Exponential smoothing

Absolute Error

37 35 42 38 48

7 4 10 3

36 38.5 40 43

6 0.5 8 2

36.6 37.68 37.74 39.80

5.4 0.32 10.26 5.2

Exponential smoothing: Naïve: FFeb = AJan = 37

FαA t +1 =

t

+ α(1 -F ) t FMAR = 0.2 ´ 35 + (1 - 0.2 ) ´ 37 = 36.60

FApril = 0.2 ´ 42 + (1 - 0.2 ) ´ 36.60 = 37.68 FMay = 0.2 ´ 38 + (1 - 0.2 ) ´ 37.68 = 37.74 FJune = 0.2 ´ 48 + (1 - 0.2 ) ´ 37.74 = 39.80 b) MAD ( Naïve ) =

å actual - forecast n

MAD ( 2-period moving average ) = MAD ( exponential smoothing ) =

© 2023 John Wiley & Sons, Inc. All rights reserved.

= ( 7 + 4 + 10 + 3 ) / 4 = 6 å actual - forecast

n å actual - forecast n

= ( 6 + 0.5 + 8 + 2 ) / 4 = 4.125

= ( 5.4 + 0.32 + 10.26 + 5.2 ) / 4 = 5.30

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The 2-period moving average and the Naïve approach provide the best historical fit using the MAD criterion and would be better to use. c) 2-period moving average:

Exponential smoothing:

FJuly = ( 48 + 45 ) / 2 = 46.5

FJuly = 0.2 ´ 45 + (1 - 0.2 ) ´ 39.80 = 40.84

Part a Section: Time Series Methods Subsection: Naïve and Simple Moving Average LO: 8.4 Describe time series models.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes Parts c & d Section: Measuring Forecast Accuracy Subsection: Forecast Accuracy Methods LO: 8.6 Computer forecast accuracy.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes 5. The manager of a small health clinic would like to use exponential smoothing to forecast demand for laboratory services in the facility. However, she is not sure whether to use a high or low value of α. To make her decision, she would like to compare the forecast accuracy of a high and low α on historical data. She has decided to use α = 0.7 for the high value and α = 0.1 for the low value. Given the following historical data, which do you think would be better to use?

Week

Demand (lab requirements) 330 350 320 370 368 343

1 2 3 4 5 6 Answer: Week 1 2 3 4 5

Demand (lab requirements) 330 350 320 370 368

Exponential Smoothing (α = 0.1)

Absolute Error

Exponentia l Smoothing (α = 0.7)

Absolute Error

330 332 330.8 334.72

20 12 39.2 33.28

330 344 327.2 357.16

20 24 42.8 10.84

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343

338.05 4.952 å actual - forecast MAD ( exponential smoothing α = 0.1 ) = n = ( 20 + 12 + 39.2 + 33.28 + 4.952 ) /5 = 21.89 MAD ( exponential smoothing α = 0.7 ) =

364.748

21.748

å actual - forecast

= ( 20 + 24 + 42.8 + 10.84 + 21.748 ) /5 = 23.88

n

Using α = 0.1 provides a better historical fit based on the MAD criterion.

Section: Time Series Methods Subsection: Exponential Smoothing LO: 8.4 Describe time series models.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes Section: Measuring Forecast Accuracy Subsection: Forecast Accuracy Methods LO: 8.6 Computer forecast accuracy.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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6. The manager of the health clinic in Problem 5 would also like to use exponential smoothing to forecast demand for emergency services in the facility. As in Problem 5, she is not sure whether to use a high or low value of α. To make her decision, she would like to compare the forecast accuracy of a high and low α on historical data. This time, she has decided to use α = 0.6 for the high value and α = 0.2 for the low value. a) Given the following historical data, which value of α do you think would be better to use?

Week 1 2 3 4 5 6

Demand (in patients serviced) 425 299 355 450 490 375

b) Is your answer the same as in Problem 5? Why or why not?

Answer: a) Week 1 2 3 4 5 6

Demand (in Exponential patients serviced) Smoothing (α = 0.2) 425 299 425 355 399.8 450 390.84 490 402.672 375 420.1376

Absolute Error

Exponential Smoothing (α = 0.6)

Absolute Error

126 44.8 59.16 87.328 45.1376

425 349.4 352.76 411.104 458.4416

126 5.6 97.24 78.896 83.4416

MAD ( exponential smoothing α = 0.2 ) =

å actual - forecast n

= (126 + 44.8 + 59.16 + 87.328 + 45.1376 ) /5 = 72.49 MAD ( exponential smoothing α = 0.6 ) =

å actual - forecast n

= (126 + 5.6 + 97.24 + 78.896 + 83.4416 ) /5 = 78.24 Using α = 0.2 provides a better historical fit based on the MAD criterion. d) The answer is the same since an α value of 0.2 provides a better historical fit. Both data sets do not exhibit a trend and the variation in the data appears to be random.

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Section: Time Series Methods Subsection: Exponential Smoothing LO: 8.4 Describe time series models.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes Section: Measuring Forecast Accuracy Subsection: Forecast Accuracy Methods LO: 8.6 Computer forecast accuracy.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 7. The following historical data have been collected representing sales of a product.

Compare forecasts using a three-period moving average, exponential smoothing with α = 0.2, and linear regression. Using MAD and MSE, which forecasting model is best? Are your results the same using the two error measures?

Week 1 2 3 4 5 6

Demand 20 31 36 38 42 40

Answer: Week

Demand

1 2 3 4 5 6

20 31 36 38 42 40

3-period moving average

29 35 38.67

Absolute Error

9 7 1.33

Squared Error

Exponentia Absolute Squared l smoothing Error Error

81 49 1.77

20 22.2 24.96 27.57 30.454

11 13.8 13.04 14.43 9.55

121 190.44 170.04 208.28 91.12

MAD ( 3-period moving average ) = ( 9 + 7 + 1.33 ) / 3 = 5.77 MSE ( 3-period moving average ) = ( 81 + 49 + 1.77 ) / 3 = 43.92 MAD ( exponential smoothing ) = (11 + 13.8 +13.04 + 14.43 + 9.55 ) / 5 = 12.36 MSE ( exponential smoothing ) = (121 + 190.44 + 170.04 + 208.28 + 91.12 ) / 5 = 156.18

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Regression Model:

Total

Time (X) 1 2 3 4 5 6 21

Demand (Y) 20 31 36 38 42 40 207

X2

XY

1 4 9 16 25 36 91

20 62 108 152 210 240 792

Regression Line 24.857 28.714 32.571 36.428 40.285 44.142

Absolute Error 4.857 2.286 3.429 1.572 1.715 4.142

Squared Error 23.591 5.224 11.755 2.469 2.938 17.163

X = 21 / 6 = 3.5 Y = 207 / 6 = 34.5 å XY - nXY 792 - ( 6 )( 3.5 )( 34.5 ) b= = = 3.857 å X 2 - nX 2 91 - 6(3.5)2

a = Y - bX = 34.5 - 3.857 ( 3.5 ) = 21

MAD = ( 4.857 + 2.286 + 3.429 + 1.572 + 1.715 + 4.142 ) / 6 = 3.00

MSE = ( 23.591 + 5.224 + 11.755 + 2.469 + 2.938 + 17.163 ) = 10.52

The linear regression model provides the best historical fit using the MAD and the MSE criteria.

Section: Time Series Methods Subsection: Exponential Smoothing LO: 8.4 Describe time series models.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes Section: Causal Models Subsection: Linear Regression LO: 8.5 Describe causal modeling using linear regression.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes Section: Measuring Forecast Accuracy Subsection: Forecast Accuracy Methods LO: 8.6 Computer forecast accuracy.

Bloom’s: Application AACSB: Analytic

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Time on task: 5 minutes

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8. A manufacturer of printed circuit boards uses exponential smoothing with trend to forecast monthly demand of its product. At the end of December, the company wishes to forecast sales for January. The estimate of trend through November has been 150 additional boards sold per month. Average sales have been around 1,100 units per month.

The demand for December was 1,250 units. The company uses α = 0.25 and β = 0.10.

Make a forecast including trend for the month of January.

Answer: ADec = 1,250 units/month

SNov = 1,100 units/month

TNov = 150 units/month α = 0.25 β = 0.10

Step 1: Smoothing the level of the series

SαA Dec =

Dec

+α(1 –S )( Nov T + Nov ) = 0.25 (1,250 ) + (1 - 0.25 )(1,100 +150 ) = 1,250 units

Step 2: Smoothing the trend Tβ = ( Dec S – Nov ) +β(1T– ) Nov = 0.10 (1,250 –1,100 ) + 0.90 (150 ) = 150 units Dec S Step 3: Forecast including trend FIT = SDec + TDec = 1,250 + 150 = 1, 400 units Section: Time Series Methods Subsection: Trend-Adjusted Exponential Smoothing LO: 8.4 Describe time series models.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 9. Demand at Nature Trails Ski Resort has a seasonal pattern. Demand is highest during the winter, as this is the peak ski season. However, there is some ski demand in the spring and even fall months. The summer months can also be busy as visitors often come for summer vacation to go hiking on the mountain trails. The owner of Nature Trails would like to make a forecast for each season of the next year. Total annual demand has been estimated at 4,000 visitors. Given the last two years of historical data, what is the forecast for each season of the next year?

Season Fall Winter Spring Summer

Visitors Year 1 Year 2 200 230 1,400 1,600 520 580 720 831

Answer:

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Step 1: Average demand for each season: Year 1: 2,840/4 = 710 Year 2: 3,241/4 = 810.25 Step 2: Seasonal index for each season: Season Fall Winter Spring Summer

Year 1 200/710 = 0.282 1,400/710 = 1.972 520/710 = 0.732 720/710 = 1.014

Year 2 230/810.25 = 0.284 1,600/810.25 = 1.975 580/810.25 = 0.716 831/810.25 = 1.026

Step 3: Average seasonal index for each season: Fall Winter Spring Summer

0.283 1.973 0.724 1.020

Step 4: Average demand per season = 4,000/4 = 1,000 Step 5: Multiply next year’s average seasonal demand by each seasonal index.

Season Fall Winter Spring Summer

Forecast 283 1,973 724 1,020

Section: Time Series Methods Subsection: Trend-Adjusted Exponential Smoothing LO: 8.4 Describe time series models.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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10. Rosa’s Italian restaurant wants to develop forecasts of daily demand for the next week.

The restaurant is closed on Mondays and experiences a seasonal pattern for the other six days of the week. Mario, the manager, has collected information on the number of customers served each day for the past two weeks. If Mario expects total demand for next week to be around 360, what is the forecast for each day of next week?

Day Tuesday Wednesday Thursday Friday Saturday Sunday

Number of Customers Week 1 Week 2 50 49 37 29 42 32 85 100 96 95 60 73

Answer: Step 1: Average demand for each week: Week 1: 370/6 = 61.667 Week 2: 378/6 = 63 Step 2: Seasonal index for each week: Day Tuesday Wednesday Thursday Friday Saturday Sunday

Week 1 0.8108 0.6000 0.6811 1.3784 1.5568 0.9730

Week 2 0.7778 0.4603 0.5079 1.5873 1.5079 1.1587

Step 3: Average daily index for each season: Day Tuesday Wednesday Thursday Friday Saturday Sunday

Average Daily Index 0.7943 0.5302 0.5945 1.4828 1.5323 1.0659

Step 4: Average demand per day = 360/6 = 60

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Step 5: Multiply average demand per day by each average daily index Day Tuesday Wednesday Thursday Friday Saturday Sunday

Forecast 47.6577 31.8095 35.6705 88.9704 91.9408 63.9511

Section: Time Series Methods Subsection: Forecasting Seasonality LO: 8.4 Describe time series models.

Bloom’s: Application AACSB: Analytic Time on task: 8 minutes 11. The president of a company was interested in determining whether there is a correlation between sales made by different sales teams and hours spent on employee training. These figures are shown.

Sales (in thousands) 25 40 36 50 11

Training Hours 10 12 12 15 6

a) Compute the correlation coefficient for the data. What is your interpretation of this value? b) Using the data, what would you expect sales to be if training was increased to 18 hours?

Answer: å XY = 1,978; å X 2 = 649; å Y 2 = 6,142; å X = 55; å Y = 162; n = 5 ∑ X ¿2 ¿ ∑ Y ¿2 ¿ n (∑Y 2)−¿ n(∑ X 2)−¿ √¿ n (∑ XY ) – (∑ X ) ( ∑Y ) Correlation coefficient r = ¿

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55¿2 ¿ 162¿2 ¿ 5 ( 6,142 )−¿ 5 × 649−¿ √¿ 5 ( 1,978 )−( 55 ) ( 162 ) ¿ ¿ The correlation coefficient is 0.9887. This high correlation indicated that there is a strong positive linear association between sales and training hours.

X = 55 / 5 = 11

Y = 162 / 5 = 32.4 å XY - nXY 1,978 - ( 5 )(11 )( 32.4 ) b= = = 4.455 å X 2 - nX 2 649 - 5(11)2

a = Y - bX = 32.4 - 4.45 (11 ) = –16.55

Y = a + bX = -16.55 + 4.455 (18 ) = 63.59 Sales ( in thousands ) = 63.59 Section: Causal Models Subsection: Correlation Coefficient and Linear Regression LO: 8.5 Describe causal modeling using linear regression.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 12. The number of students enrolled at Spring Valley Elementary has been steadily increasing over the past five years. The school board would like to forecast enrollment for years 6 and 7 in order to better plan capacity. Use a linear trend line to forecast enrollment for years 6 and 7.

Year 1 2 3 4 5

Enrollment 225 260 275 292 315

Answer: X = 15 / 5 = 3 Y = 1367 / 5 = 273.4 å XY - nXY 4313 - ( 5 )( 3 )( 273.4 ) b= = = 21.2 X 2 - nX 2 55 - 5(3)2

a = Y - bX = 273.4 - 21.2 ( 3 ) = 209.8

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Year 6 forecast: 209.8 + 21.2 (6 ) = 337 Year 7 forecast: 209.8 + 21.2 (7 ) = 358.2 or 358 Section: Causal Models Subsection: Linear Regression LO: 8.5 Describe causal modeling using linear regression.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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13. Happy Lodge Ski Resorts tries to forecast monthly attendance. The management has noticed a direct relationship between the average monthly temperature and attendance. a) Given five months of average monthly temperatures and corresponding monthly attendance, compute a linear regression equation of the relationship between the two.

If next month’s average temperature is forecast to be 45 degrees, use your linear regression equation to develop a forecast.

Month 1 2 3 4 5

Average Temperature 24 41 32 30 38

Resort Attendance (in thousands) 43 31 39 38 35

b) Compute a correlation coefficient for the data and determine the strength of the linear relationship between average temperature and attendance. How good a predictor is temperature for attendance?

Answer: a) ∑XY = 6,021; ∑X2 = 5,625; ∑Y2 = 7,000; ∑X = 165; ∑Y = 186 X = 165 / 5 = 33 Y = 186 / 5 = 37.2 å XY - nXY 6021 - ( 5 )( 33 )( 37.2 ) b= = = –0.65 å X 2 - nX 2 5625 - 5(33)2

a = Y - bX = 37.2 - ( -0.65 )( 33 ) = 58.65

Resort attendance forecast when the average temperature is 45 degrees: = 58.65 + (–0.65) (45) = 29.4 thousand attendees b) ∑XY = 6,021; ∑X2 = 5,625; ∑Y2 = 7,000; ∑X = 165; ∑Y = 186; n = 5 Correlation coefficient r =

n(å XY ) - (å X ) ( å Y ) n(å X 2 ) - (å X )2 * n( å Y 2 ) - ( å Y )2

= – 0.97

The correlation coefficient is –0.97. Since this value is very close to negative 1, it indicates that the average temperature is a strong predictor of resort attendance. Note that since the sign is negative, it indicates that an inverse or negative relationship exists between the two variables.

Section: Causal Models Subsection: Linear Regression and Correlation Coefficient LO: 8.5 Describe causal modeling using linear regression.

Bloom’s: Application AACSB: Analytic

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Time on task: 8 minutes 14. Small Wonder, an amusement park, experiences seasonal attendance. It has collected two years of quarterly attendance data and made a forecast of annual attendance for the coming year. Compute the seasonal indexes for the four quarters and generate quarterly forecasts for the coming year, assuming annual attendance for the coming year to be 1,450.

Quarter Fall Winter Spring Summer

Park Attendance (in thousands) Year 1 Year 2 345 375 150 230 475 505 335 375

Answer: Step 1: Average demand for each quarter for year 1 = ( 345+150+475+335 ) /4 = 326.25

Average demand for each quarter for year 2 = ( 375+230+505+375 ) /4 = 371.25

Step 2: Compute a seasonal index for every season of every year: Quarter Fall Winter Spring Summer

Year 1 345/326.25 = 1.06 150/326.25 = 0.46 475/326.25 = 1.46 335/326.25 = 1.03

Year 2 375/326.25 = 1.01 230/326.25 = 0.62 505/326.25 = 1.36 335/326.25 = 1.01

Step 3: Calculate the average seasonal index for each season: Quarter Fall Winter Spring Summer

Average Seasonal Index (1.06+1.01)/2 = 1.03 (0.46+0.62)/2 = 0.54 (1.46+1.36)/2 = 1.41 (1.03+1.01)/2 = 1.02

Step 4: Calculate the average demand per season for next year = 1450/4 = 362.5

Step 5: Multiply next year’s average seasonal demand by each seasonal index: Quarter Fall Winter Spring Summer

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Forecast (362.5)(1.03) = 374.75 (362.5)(0.54) = 195.62 (362.5)(1.41) = 510.44 (362.5)(1.02) = 369.19

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Section: Time Series Methods Subsection: Forecasting Seasonality LO: 8.4 Describe time series models.

Bloom’s: Application AACSB: Analytic Time on task: 8 minutes 15. Burger Lover Restaurant forecasts weekly sales of cheeseburgers. Based on historical observations over the past five weeks, make a forecast for the next period using the following methods: simple average, three-period moving average, and exponential smoothing with α = 0.3, given a forecast of 328 cheeseburgers for the first week.

Week

Cheeseburger Sales 354 345 367 322 356

1 2 3 4 5

If actual sales for week 6 turn out to be 368, compare the three forecasts using MAD. Which method performed best?

Answer: Period 6 Forecasts: Simple average: F6 = 348.8 3-Period moving average: F6 = 348.33 Exponential smoothing: Using the first period forecast of 328, F6 = 342.62

Period

Forecast ⍺ = .30

Value 1 2 3 4 5 6

354 345 367 322 356

Absolute Difference

328.00 333.10 343.27 336.89 342.62

17.00 33.90 21.27 19.11

Looking at the error only in period 6, using absolute values MAD (simple average) = 368 – 348.8 = 19.20 MAD (3-period moving average) = 368 – 348.33 = 19.67 MAD (Exponential smoothing) = 368 – 342.62 = 25.38 The simple average provides the best historical fit using the MAD criterion.

Section: Time Series Methods Subsection: Simple Average and Exponential Smoothing © 2023 John Wiley & Sons, Inc. All rights reserved.

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LO: 8.4 Describe time series models.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes Section: Measuring Forecast Accuracy Subsection: Forecast Accuracy Methods LO: 8.6 Computer forecast accuracy.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 16. A company uses exponential smoothing with trend to forecast monthly sales of its product, which show a trend pattern. At the end of week 5, the company wants to forecast sales for week 6. The trend through week 4 has been 25 additional cases sold per week.

Average sales have been 95 cases per week. The demand for week 5 was 100 cases. The company uses α = 0.20 and β = 0.15. Make a forecast including trend for week 6.

Answer: Given: T4 = 25, A5 = 100, S4 = 95 α = 0.20 β = 0.15 Step 1: Smoothing the level of the series:

S5 = αA5 + (1 – α )( S4 + T4 ) = 0.20 (100 ) + 0.80 ( 95 + 25 ) = 116 Step 2: Smoothing the trend:

T5 = β ( S5 – S4 ) + (1 - β ) T4 = 0.15 (116 – 95 ) + 0.85 (25 ) = 24.4 Step 3: Forecast including trend:

FIT6 = S5 + T5 = 116 + 24.4 = 140.4 Section: Time Series Methods Subsection: Trend-Adjusted Exponential Smoothing LO: 8.4 Describe time series models.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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17. The number of patients coming to the Healthy Start maternity clinic has been increasing steadily over the past eight months. Given the following data, use a linear trend line to forecast attendance for months 9 and 10.

Month

Clinic Attendance (in thousands) 3.4 3.9 4.5 5.0 5.8 5.9 6.5 6.7

1 2 3 4 5 6 7 8 Answer:

X = 36 / 8 = 4.5 Y = 41.7 / 8 = 5.21

b=

å XY - nXY = å X 2 - nX 2

208.2 - ( 8 )( 4.5 )(5.2125 )

= 0.489

204 - 8(4.5)2

a = Y - bX = 5.2125 - 0.489 ( 4.5 ) = 3.011 F9 = 3.011 + 0.489 ( 9 ) = 7.412 attendees ( in thousands )

F10 = 3.011 + 0.489 (10 ) = 7.901 attendees ( in thousands ) Section: Causal Models Subsection: Linear Regression LO: 8.5 Describe causal modeling using linear regression.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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18. Given the following data, use exponential smoothing with α = 0.2 and α = 0.5 to generate forecasts for periods 2 through 6. Use MAD and MSE to decide which of the two models produced a better forecast.

Period 1 2 3 4 5 6

Actual 14 16 16 11 12 17

Forecast 17

Answer: Period

Actual

1 2 3 4 5 6

14.00 16.00 16.00 11.00 12.00 17.00

Forecast (α=0.2) 17.00 16.4 16.32 16.26 15.20 14.56

Absolute Error 3.00 0.40 0.32 5.26 3.20 2.44

Squared Error 9.00 0.16 0.10 27.67 10.24 5.95

Forecast (α=0.5) 17.00 15.5 16.2 16.16 13.63 13.60

Absolute Error 3.00 0.5 0.2 5.16 1.63 3.40

Squared Error 9.00 0.25 0.04 26.63 2.65 11.56

MAD ( α = 0.2 ) = ( 3 + 0.40 + 0.32 + 5.26 + 3.20 + 2.44 ) / 6 = 2.44 MAD (⍺=0.5) = (3 + 0.50 + 0.20 + 5.16 + 1.63 + 3.40)/6 = 2.315 Exponential smoothing using α = 0.5 yields lower MAD.

MSE ( α = 0.2 ) = ( 9 + 0.16 + 0.10 + 27.67 + 10.24 + 5.95 ) / 6 = 8.85 MSE ( α = 0.5 ) = 9 + 0.25 + 0.04 + 26.63 +2.65 + 11.56) = 8.35 Exponential smoothing using α = 0.5 yields lower MSE. Section: Time Series Methods Subsection: Exponential Smoothing LO: 8.4 Describe time series models.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes Section: Measuring Forecast Accuracy Subsection: Forecast Accuracy Methods LO: 8.6 Computer forecast accuracy.

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Bloom’s: Application AACSB: Analytic Time on task: 8 minutes

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19. Pumpkin Pies Galore is trying to forecast sales of pies for the month of December.

Demand for pies in September, October, and November has been 230, 304, and 415, respectively. Edith, the company’s owner, uses a three-period weighted moving average to forecast sales. Based on her experience, she chooses to weight September as 0.1, October as 0.3, and November as 0.6. a) What would Edith’s forecast for December be? b) What would her forecast be using the naïve method? c) If actual sales for December turned out to be 420 pies, which method was better (use MAD)?

Answer: a) Forecast using a weighted moving average = 230(0.1) + 304(0.3) + 415(0.6) = 363.2 b) Forecast using the naïve approach = 415 c)

MAD ( weighted moving average ) = 420 – 363.2 = 56.8 MAD ( naïve method ) = 420 – 415 = 5 The naïve approach is better.

Section: Time Series Methods Subsection: Naïve and Weighted Moving Average LO: 8.4 Describe time series models.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes Section: Measuring Forecast Accuracy Subsection: Forecast Accuracy Methods LO: 8.6 Computer forecast accuracy.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 20. A company has used three different methods to forecast sales for the past five months.

Use MAD and MSE to evaluate the performance of the three methods. a) Which forecasting method performed best? Do MAD and MSE give the same results?

Period 1 2 3 4 5

Actual 12 7 13 8 10

Method A 10 11 12 13 14

Method B 7 11 10 6 13

Method C 8 12 7 13 8

b) Which of these is actually the naïve method?

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Answer: a) Period 1 2 3 4 5

Actual Method A Absolute Error 12 10 2 7 11 4 13 12 1 8 13 5 10 14 4 MADA = ( 2 + 4 + 1 + 5 + 4 ) / 5 = 3.2

Squared Error 4 16 1 25 16

Actual Method B Absolute Error 12 7 5 7 11 4 13 10 3 8 6 2 10 13 3 MADB = ( 5 + 4 + 3 + 2 + 3 ) / 5 = 3.4

Squared Error 25 16 9 4 9

MSEA = ( 4 + 16 + 1 + 25 + 16 ) / 5 = 12.4

Period 1 2 3 4 5

MSEB = ( 25 + 16 + 9 + 4 + 9 ) / 5 = 12.6

Period 1 2 3 4 5

Actual

Method C

Absolute Error 12 8 4 7 12 5 13 7 6 8 13 5 10 8 2 MADC = ( 4 + 5 + 6 + 5 + 2) / 5 = 4.4

Squared Error 16 25 36 25 4

MSEC = (16 + 25 + 36 + 25 + 4 ) / 5 = 21.1

Both Method A and Method B provide the best historical fit using the MAD and MSE criteria. b) Forecast method C is the naïve method.

Section: Measuring Forecast Accuracy Subsection: Forecast Accuracy Methods LO: 8.6 Computer forecast accuracy.

Bloom’s: Application AACSB: Analytic

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Time on task: 5 minutes

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21. Two different forecasting models were used to forecast sales of a popular soda on a college campus. Actual demand and the two sets of forecasts are shown. Use MAD to explain which method provided a better forecast.

Period 1 2 3 4 5 6

Actual Demand 90 87 92 95 98 98

Forecast 1 78 85 84 92 100 102

Forecast 2 87 88 90 97 102 101

Answer: Period

Actual Demand

1 2 3 4 5 6

90 87 92 95 98 98

Forecast 1 78 85 84 92 100 102

Absolute Error

Forecast 2 87 88 90 97 102 101

12 2 8 3 2 4

Absolute Error 3 1 2 2 4 3

MAD ( Forecast 1 ) = (12 + 2 + 8 + 3 + 2 + 4 ) / 6 = 5.17 MAD ( Forecast 2 ) = ( 3 + 1 + 2 + 2 + 4 + 3 ) / 6 = 2.50 Forecast 2 provides a better historical fit using the MAD criterion.

Section: Measuring Forecast Accuracy Subsection: Forecast Accuracy Methods LO: 8.6 Computer forecast accuracy.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 22. A producer of picture frames uses a tracking signal with limits of ±4 to decide whether a forecast should be reviewed. Given historical information for the past four weeks, compute the tracking signal and decide whether the forecast should be reviewed. The MAD for this item was computed as 2.

Weeks

Actual Sales

Forecast

1 2 3 4

10 12 13 10

11 13 14 14

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Deviation

Cumulative Deviation 6

Tracking Signal 3

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Answer: The tracking signal is the sum of the forecast errors (the cumulative deviation) divided by the MAD. In this problem MAD has been computed as 2. You need to compute the cumulative deviation for each period and divide by the MAD. Weeks

Actual Sales

Forecast

Deviation

1 2 3 4

10 12 13 10

11 13 14 14

1 1 1 4

Cumulative Deviation 6 7 8 9 13

Tracking Signal 3.00 3.50 4.00 4.50 6.50

The forecast should be reviewed in week 3 because the tracking signal has exceeded +4.

Section: Measuring Forecast Accuracy Subsection: Tracking Signal LO: 8.6 Computer forecast accuracy.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 23. Mop and Broom Manufacturing has tracked the number of units sold of their most popular mop over the past 24 months. This is shown.

Month 1 2 3 4 5 6 7 8

Sales 239 248 256 260 271 280 295 305

Month 9 10 11 12 13 14 15 16

Sales 310 335 348 353 355 368 379 358

Month 17 18 19 20 21 22 23 24

Sales 369 378 367 383 394 393 405 412

a) Develop a linear trend line for the data. b) Compute a correlation coefficient for the data and evaluate the strength of the linear relationship. c) Using the linear trend line equation, develop a forecast for the next period, month 25.

Answer: a)

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X = 300 / 24 = 12.5; Y = 8,061 / 24 = 335.875

b=

å XY - nXY 109,300 - ( 24 )(12.5 )( 335.875 ) = = 7.424 å X 2 - nX 2 4,900 - 24(12.5)2

a = Y - bX = 335.875 - 7.424 (12.5 ) = 243.07 Sales = 243.07 + 7.424 (month) b) Correlation coefficient r =

n(å XY ) - ( å X ) ( å Y ) n(å X 2 ) - (å X )2 * n(å Y 2 ) - ( å Y )2

= 0.97

It indicates a strong, positive linear relationship. c) Forecast for month 25 = 243.07 + (7.424)25 = 428.68 or 429 units Section: Causal Models Subsection: Linear Regression and Correlation Coefficient LO: 8.5 Describe causal modeling using linear regression.

Bloom’s: Application AACSB: Analytic Time on task: 8 minutes 24. Use the sales data below to generate forecasts for months 7–24 using a six-period and a three-period moving average. Use MAD to compare the forecasts. Which forecast is more stable? Which is more responsive and why?

Month 1 2 3 4 5 6 7 8

Sales 120 122 135 138 129 136 139 143

Month 9 10 11 12 13 14 15 16

Sales 145 150 144 154 160 155 162 160

Month 17 18 19 20 21 22 23 24

Sales 158 149 170 165 171 158 173 170

Answer: Month (X)

Sales (Y)

6-period moving average

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Absolute error

3-period moving average

Absolute Error

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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

120 122 135 138 129 136 139 143 145 150 144 154 160 155 162 160 158 149 170 165 171 158 173 170

130.00 133.17 136.67 138.33 140.33 142.83 145.83 149.33 151.33 154.17 155.83 158.17 157.33 159.00 160.67 162.17 161.83 164.33

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9.00 9.83 8.33 11.67 3.67 11.17 14.17 5.67 10.67 5.83 2.17 9.17 12.67 6.00 10.33 4.17 11.17 5.67

125.67 131.67 134.00 134.33 134.67 139.33 142.33 146.00 146.33 149.33 152.67 156.33 159.00 159.00 160.00 155.67 159.00 161.33 168.67 164.67 167.33

12.33 2.67 2.00 4.67 8.33 5.67 7.67 2.00 7.67 10.67 2.33 5.67 1.00 1.00 11.00 14.33 6.00 9.67 10.67 8.33 2.67

MAD ( 6 - period moving average ) = 151.33 / 18 = 8.41 MAD ( 3 - period moving average ) = 136.33 / 21 = 6.49 The very nature of averaging makes the 3-period moving average is more responsive. The 6-period moving average has more weight on older (and lower) sales data. The lower MAD confirms the 3-period moving average was better in forecasting the increasing sales in this data set.

Section: Measuring Forecast Accuracy Subsection: Forecast Accuracy Methods LO: 8.6 Computer forecast accuracy.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes 25. The following data were collected on the study of the relationship between a company’s retail sales and advertising dollars: Retails Sales ($) 29,789 35,434 38,732 43,585 46,821 49,283 © 2023 John Wiley & Sons, Inc. All rights reserved.

Advertising ($) 16,893 18,398 20,376 22,982 25,732 27,281 SM 8-30

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52,271 55,289 57,298 58,293

32,182 35,298 36,281 38,178

a) Obtain a linear regression line for the data. b) Compute a correlation coefficient and determine the strength of the linear relationship. c) Using the linear regression equation, develop a forecast of total retail sales for advertising dollars of $40,000.

Answer: a)

X = 466,795 / 10 = 46,679.5; Y = 273,601 / 10 = 27,360.1 b=

å XY - nXY 13,433,416,842 - (10 )( 46,679.5 )( 27,360.1 ) = = 0.784 å X 2 - nX 2 22,634,058,671 - 10(46,679.5)2

a = Y - bX = 27,360.1 - 0.784 ( 46,679.5 ) = -9,232.57

Value of sales = –9,232.57 + 0.784 (advertising dollars) b) Correlation coefficient r =

n( å XY ) - ( å X ) ( å Y ) n(å X 2 ) - (å X )2 * n(å Y 2 ) - ( å Y )2

= 0.98

It indicates a strong, positive linear relationship. c) Forecast of retail sales = –9,232.57 + 0.784 (40,000) = $22,123.

Total retail sales is $62,123. $22,123 is the difference between sales and advertising dollars. Therefore, the retail sales would be $40,000 + $22,123 = $62,123.

The projected total sales would be $62,123

$ $ $ $ $ $ $ $ $ $ $

Sales 29,789 35,434 38,732 43,585 46,821 49,283 52,271 55,289 57,298 58,293 62,123

Advertising $ 16,893 $ 18,398 $ 20,376 $ 22,982 $ 25,732 $ 27,281 $ 32,182 $ 35,298 $ 36,281 $ 38,178 $ 40,000

Difference $ 12,896 $ 17,036 $ 18,356 $ 20,603 $ 21,089 $ 22,002 $ 20,089 $ 19,991 $ 21,017 $ 20,115 $ 22,123

Section: Causal Models

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Subsection: Linear Regression and Correlation Coefficient LO: 8.5 Describe causal modeling using linear regression.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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Solutions Manual Chapter 9: Capacity Planning and Facility Location 1. Joe’s Tasty Burger has determined that its production facility has a design capacity of 375 hamburgers per day. The effective capacity, however, is 275 hamburgers per day. Lately Joe has noticed that output has been 250 hamburgers per day. Compute both design and effective capacity utilization measures. What can you conclude?

Answer: Actual Output 250 (100% ) = (100% ) = 91% Effective Capacity 275 Actual Output 250 Utilization Design = (100%) = (100%) = 67% Design Capacity 375 Utilization Effective =

The utilization rates show that the facility’s current output is below its design capacity and considerably lower than its effective capacity. If the effective capacity is realistically set, it’s expected the facility will operate over that level for a short time.

Section: Capacity Planning Subsection: Measuring Effectiveness of Capacity Use LO: 9.1 Define capacity planning.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes 2. A manufacturer of printed circuit boards has a design capacity of 1,000 boards per day.

The effective capacity, however, is 700 boards per day. Recently, the production facility has been producing 950 boards per day. Compute the design and effective capacity utilization measures. What do they tell you?

Answer: Actual Output 950 (100% ) = (100% ) = 135.7% Effective Capacity 700 Actual Output 950 Utilization Design = (100% ) = (100% ) = 95% Design Capacity 1000 The utilization rates show that the facility’s current output is slightly below its design capacity and considerably higher than its effective capacity. The facility can probably operate at the effective level for only a short time.

Utilization Effective =

Section: Capacity Planning Subsection: Measuring Effectiveness of Capacity Use LO: 9.1 Define capacity planning.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

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3. Beth’s Bakery can comfortably produce 50 brownies in one day. If Beth takes some unusual measures, such as hiring her two aunts to help in the kitchen and work overtime, she can produce up to 100 brownies in one day. a) What are the design and effective capacities for Beth’s Bakery? b) If Beth is currently producing 55 brownies, compute the capacity utilization for both measures. What can you conclude?

Answer: a) Effective capacity = 50 brownies Design capacity = 100 brownies Actual Output 55 Utilization Effective = (100% ) = (100% ) = 110.00% Effective Capacity 50 b) Actual Output 55 (100% ) = (100% ) = 55% Design Capacity 100 The utilization rates show that the facility’s current output is far below its design capacity and higher than its effective capacity. The design capacity occurs only with extra help.

Utilization Design =

Section: Capacity Planning Subsection: Measuring Effectiveness of Capacity Use LO: 9.1 Define capacity planning.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes 4. The town barber shop can accommodate 35 customers per day. The manager has determined that if two additional barbers are hired, the shop can accommodate 80 customers per day. What are the design and effective capacities for the barber shop?

Answer: Effective capacity = 35 customers/day Design capacity = 80 customers/day Section: Capacity Planning Subsection: Measuring Effectiveness of Capacity Use LO: 9.1 Define capacity planning.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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5. The design and effective capacities for a local paper manufacturer are 1,300 and 800 pounds of paper per day, respectively. At present, the manufacturer is producing 700 pounds per day. Compute capacity utilization for both measures. What can you conclude?

Answer: Actual Output 700 (100% ) = (100% ) = 87.5% Effective Capacity 800 Actual Output 700 Utilization Design = (100% ) = (100% ) = 53.8% Design Capacity 1,300 The computed utilization rates show that the facility’s current output is below its design and effective capacities. This illustrates that the manufacturer is not using capacity to its fullest extent and there is room for improvement.

Utilization Effective =

Section: Capacity Planning Subsection: Measuring Effectiveness of Capacity Use LO: 9.1 Define capacity planning.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes 6. The design and effective capacities for a local emergency facility are 300 and 260 patients per day, respectively. Currently, the emergency room processes 250 patients per day.

What can you conclude from these figures?

Answer: Actual Output 250 (100% ) = (100% ) = 96.2% Effective Capacity 260 Actual Output 250 Utilization Design = (100% ) = (100% ) = 83.3% Design Capacity 300 The utilization rates show that the facility’s current output is below its design and effective capacities.

Utilization Effective =

Section: Capacity Planning Subsection: Measuring Effectiveness of Capacity Use LO: 9.1 Define capacity planning.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

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7. The Steiner-Wallace Corporation has determined that it needs to expand in order to accommodate growing demand for its laptop computers and tables. The decision has come down to either expanding now with a large facility, incurring additional costs and taking the risk that the demand will not materialize, or expanding small, knowing that in three years management will need to reconsider the question.

Management has estimated the following chances for demand:  

The likelihood of demand being high is 0.65.

The likelihood of demand being low is 0.35.

Profits for each alternative have been estimated as follows:   

Large expansion has an estimated profitability of either $125,000 or $50,000, depending on whether demand turns out to be high or low.

Small expansion has a profitability of $30,000, assuming that demand is low.

Small expansion with an occurrence of high demand would require considering whether to expand further. If the company expands at that point, the profitability is expected to be $70,000. If it does not expand further, the profitability is expected to be $45,000.

a) Draw a decision tree showing the decisions, chance events, and their probabilities, as well as the profitability of outcomes. b) Solve the decision tree and decide what Steiner-Wallace should do.

Answer: a)

b)

EVsmall expansion

= 30,000 ( 0.35 ) + 70,000 ( 0.65 ) = $56,000

EVlarge expansion

= 50,000 ( 0.35 ) + 50,000 ( 0.65 ) = $98,750

Steiner-Wallace should opt for the large expansion.

Section: Decision Trees Subsection: [untitled] Introduction LO: 9.3 Explain the usefulness of decision trees in decision making.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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8. The owners of Sweet-Tooth Bakery have determined that they need to expand their facility in order to meet their increased demand for baked goods. The decision is whether to expand now with a large facility or expand small with the possibility of having to expand again in five years.

The owners have estimated the following chances for demand:  

The likelihood of demand being high is 0.70.

The likelihood of demand being low is 0.30.

Profits for each alternative have been estimated as follows:   

Large expansion has an estimated profitability of either $80,000 or $50,000, depending on whether demand turns out to be high or low.

Small expansion has a profitability of $40,000, assuming demand is low.

Small expansion with an occurrence of high demand would require considering whether to expand further. If the bakery expands at this point, the profitability is to be $50,000.

a) Draw a decision tree showing the decisions, chance events, and their probabilities, as well as the profitability of outcomes. b) Solve the decision tree and decide what the bakery should do.

Answer: a)

b)

EVsmall expansion

= 50,000 ( 0.70 ) + 40,000 ( 0.30 ) = $47,000

EVlarge expansion

= 80, 000 ( 0.70 ) + 50, 000 ( 0.30 ) = $71, 000

The bakery should opt for the large expansion.

Section: Decision Trees Subsection: [untitled] Introduction LO: 9.3 Explain the usefulness of decision trees in decision making.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes © 2023 John Wiley & Sons, Inc. All rights reserved.

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9. Demand has grown at Dairy May Farms, and it is considering expanding. One option is to expand by purchasing a very large farm that will be able to meet expected future demand.

Another option is to expand the current facility by a small amount now and take a waitand-see attitude, with the possibility of a larger expansion in two years.

Management has estimated the following chances for demand:  

The likelihood of demand being high is 0.75.

The likelihood of demand being low is 0.25.

Profits for each alternative have been estimated as follows:   

Large expansion has an estimated profitability of either $50,000 or $25,000, depending on whether demand turns out to be high or low.

Small expansion has a profitability of $15,000, assuming that demand is low.

Small expansion with an occurrence of high demand would require considering whether to expand further. If the company expands at that point, the profitability is expected to be $40,000. If it does not expand further, the profitability is expected to be $12,000.

a) Draw a decision tree diagram for Dairy May Farms. b) Solve the decision tree you developed. What should Dairy May Farms do?

Answer: a)

b)

EVsmall expansion

= 40, 000 ( 0.75 ) + 15,000 ( 0.25 ) = $33,750

EVlarge expansion

= 50,000 ( 0.75 ) + 25,000 ( 0.25 ) = $43,750

Dairy May Farms should opt for the large expansion.

Section: Decision Trees Subsection: [untitled] Introduction LO: 9.3 Explain the usefulness of decision trees in decision making.

Bloom’s: Application AACSB: Analytic

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Time on task: 5 minutes 10. Spectrum Hair Salon is considering expanding its business, as it is experiencing a large growth. The question is whether it should expand with a bigger facility than needed, hoping that demand will catch up, or with a small facility, knowing that it will need to reconsider expanding in three years.

The management at Spectrum has estimated the following chances for demand:  

The likelihood of demand being high is 0.70.

The likelihood of demand being low is 0.30.

Estimated profits for each alternative are as follows:   

Large expansion has an estimated profitability of either $100,000 or $70,000, depending on whether demand turns out to be high or low.

Small expansion has a profitability of $50,000, assuming that demand is low.

Small expansion with an occurrence of high demand would require considering whether to expand further. If the business expands at this point, the profitability is expected to be $90,000. If it does not expand further, the profitability is expected to be $60,000.

Draw a decision tree and solve the problem. What should Spectrum do?

Answer:

90,000 ( 0.70 )

+ 50,000 ( 0.30 ) = $78, 000

EVsmall expansion

=

EVlarge expansion

= 100,000 ( 0.70 ) + 70, 000 ( 0.30 ) = $91,000

Spectrum should opt for the large expansion.

Section: Decision Trees Subsection: [untitled] Introduction LO: 9.3 Explain the usefulness of decision trees in decision making.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes © 2023 John Wiley & Sons, Inc. All rights reserved.

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11. Jody of Jody’s Custom Tailoring is considering expanding her growing business. The question is whether to expand with a bigger facility than she needs or with a small facility, knowing that she will have to reconsider expanding in three years.

Jody has estimated the following chances for demand:  

The likelihood of demand being high is 0.55.

The likelihood of demand being low is 0.45.

She has also estimated profits for each alternative:   

Large expansion has an estimated profitability of either $250,000 or $150,000, depending on whether demand turns out to be high or low.

Small expansion has a profitability of $100,000, assuming that demand is low.

Small expansion with an occurrence of high demand would require considering whether to expand further. If the business expands at that point, the profitability is expected to be $120,000. If it does not expand further, the profitability is expected to be $70,000.

Draw a decision tree and solve it. What should Jody’s Custom Tailoring do?

Answer:

EVsmall expansion

= 120,000 ( 0.55 ) + 100,000 ( 0.45 ) = $111,000

EVlarge expansion

= 250, 000 ( 0.55 ) + 150,000 ( 0.45 ) = $205,000

Jody’s Custom Tailoring should opt for the large expansion.

Section: Decision Trees Subsection: [untitled] Introduction LO: 9.3 Explain the usefulness of decision trees in decision making.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

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12. The owners of Speedy Logistics, a company that provides overnight delivery of documents, are considering where to locate their new facility in the Midwest. They have narrowed their search down to two locations and have decided to use factor rating to make their decision. They have listed the factors they consider important and assigned a factor score to each location based on a five-point scale. The information is shown here.

Using the procedure for factor rating, decide on the better location.

Factor Proximity to airport Proximity to road access Proximity to labor source Size of facility

Factor Weight 40 30 10 20

Factor Score at Each Location Location 1 Location 2 5 3 4 1 3 5 2 4

Answer:

Factor Proximity to airport Proximity to road access Proximity to labor source Size of facility

Factor Weight 40

Factor Score at Each Location Location 1 Location 2 5 3

Weighted Score at Each Location Location 1 Location 2 200 120

30

4

1

120

30

10

3

5

30

50

20

2

4

40

80

Total

390

280

Location 1 is the better location because it provides a higher factor rating score than Location 2.

Section: Making Location Decisions Subsection: Factor Rating LO: 9.5 Describe the decision-support tools used in location analysis.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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13. Sue and Joe are a young married couple who are considering purchasing a new home.

Their search has been reduced to two homes that they both like, at different locations.

They have decided to use factor rating to help them make their decision. They have listed the factors they consider important and assigned a factor score to each location based on a five-point scale. The information is shown here. Using the procedure for factor rating, complete the table and help Sue and Joe make their decision.

Factor Proximity to work Proximity to family Size of home Transportation system Neighborhood

Factor Weight 25 20 15 10 30

Factor Score at Each Location Location 1 Location 2 3 2 4 1 4 5 4 2 3 2

Answer: Factor Proximity to work Proximity to family Size of home Transportation system Neighborhood

Factor Weight

Factor Score at Each Location Location Location 2 1

Weighted Score at Each Location Location 1 Location 2

25

3

2

75

50

20

4

1

80

20

15

4

5

60

75

10

4

2

40

20

30

3

2 Total

90 345

60 225

Location 1 is the preferred one.

Section: Making Location Decisions Subsection: Factor Rating LO: 9.5 Describe the decision-support tools used in location analysis.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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14. The Bakers Dozen Restaurant is considering opening a new location. It has considered many factors and identified the ones that are most important. Two locations are being evaluated based on these factors, using factor rating. Each location has been evaluated relative to the factors on a five-point scale. These numbers are shown here. Use factor rating to help the restaurant decide on the better location.

Factor Proximity to customers Proximity to competition Proximity to labor supply Transportation system Quality of life

Factor Weight 30 10 30 20 10

Factor Score at Each Location Location 1 Location 2 5 2 4 2 2 5 5 3 3 5

Answer: Factor Proximity to customers Proximity to competition Proximity to labor supply Transportation system Quality of life

Factor Weight 30

Factor Score at Each Location Location 1 Location 2 5 2

Weighted Score at Each Location Location 1 Location 2 150 60

10

4

2

40

20

30

2

5

60

150

20

5

3

100

60

10

3

5 Total

30 380

50 340

Location 1 is the preferred one.

Section: Making Location Decisions Subsection: Factor Rating LO: 9.5 Describe the decision-support tools used in location analysis.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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15. Joe’s Sports Supplies Corporation is considering where to locate its warehouse in order to service its four stores in four towns: A, B, C, and D. Two possible sites for the warehouse are being considered, one in Jasper and the other in Longboat. The following table shows the distances between the two locations being considered and the four store locations.

Also shown are the loads between the warehouse and the four stores. Use the load– distance model to determine whether the warehouse should be located in Jasper or in Longboat.

Town A B C D

Distance to Jasper 35 15 8 16

Distance to Longboat 18 15 27 20

Load between City and Warehouse 12 8 11 6

Answer: Load-distance score Jasper = å lij dij = (12 )( 35 ) + ( 8 )(15 ) + (11 )( 8 ) + (6 )(16 ) = 724

Load-distance score Longboat = å lij dij = (12 )(18 ) + ( 8 )(15 ) + (11 )(27 ) + ( 6 )( 20 ) = 753 Warehouse should be located in Jasper, unless there are other qualitative reasons to select Longboat—since they are so close.

Section: Making Location Decisions Subsection: The Load-Distance Model LO: 9.5 Describe the decision-support tools used in location analysis.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

16. Given here are the coordinates for each of the four towns to be serviced by the warehouse in Problem 15. Use the information from Problem 15 and the center of gravity method to determine coordinates for the warehouse.

Town A B C D

Coordinates (X, Y) (4,18) (12,2) (10,8) (8,15)

Answer:

Total

Load (li) 15 10 12 8 45

© 2023 John Wiley & Sons, Inc. All rights reserved.

lixi 60 120 120 64 364

liyi 270 20 96 120 506

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Xc. g =

Yc. g =

å li xi 364 = = 8.1 å li 45

å li yi 506 = = 11.2 å li 45

Section: Making Location Decisions Subsection: The Center of Gravity Approach LO: 9.5 Describe the decision-support tools used in location analysis.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 17. Shoeless Joe is a specialty retailer that is deciding where to locate a new facility. The annual fixed and variable costs for each possible site have been estimated as follows: Location A B C D

Fixed Costs $65,000 $39,000 $35,000 $60,000

Variable Costs $2/unit $10/unit $10/unit $3/unit

If demand is expected to be 2,500 units, which location is best?

Answer: Total Cost A = 65,000 + 2 ( 2,500 ) = $70,000

Total Cost B = 39,000 + 10 ( 2,500 ) = $64,000

Total Cost C = 35,000 + 10 (2,500 ) = $60,000 Total Cost D = 60,000 + 3 ( 2,500 ) = $67,500

Shoeless Joe should locate at location C because it provides lowest total cost for the expected demand of 2,500 units.

Section: Making Location Decisions Subsection: Break-Even Analysis LO: 9.5 Describe the decision-support tools used in location analysis.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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18. The Quick Copy center for document copying is deciding where to locate a new facility.

The annual fixed and variable costs for each site it is considering have been estimated as follows: Location A B C D

Fixed Costs $85,000 $49,000 $35,000 $65,000

Variable Costs $2/unit $7/unit $10/unit $6/unit

If demand is expected to be 3000 units, which location is best?

Answer: Total Cost A = 85,000 + 2 ( 3,000 ) = $91,000

Total Cost B = 49,000 + 7 (3,000 ) = $70,000

Total Cost C = 35,000 + 10 ( 3,000 ) = $65,000 Total Cost D = 65,000 + 6 (3, 000 ) = $83,000

Location C is best because it provides the lowest total annual costs.

Section: Making Location Decisions Subsection: Break-Even Analysis LO: 9.5 Describe the decision-support tools used in location analysis.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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19. Jenny Chu can comfortably produce 3 specialty industrial signs in one month. Assume a 20-day work month. If Jenny brings on an apprentice from the design department of the local engineering college, she can produce 4 signs in one month. a) What are the design and effective capacities for Jenny Chu’s sign shop? b) If Jenny begins to produce 3.25 signs per month without the additional help, compute the capacity utilization for both measures. What can you conclude?

Answer: a) Effective capacity = 3 / 20 = 0.15 signs per day Design capacity = 4 / 20 = 0.2 signs per day b) Actual Output 0.1625 Utilization Effective = (100% ) = (100% ) = 108.3% Effective Capacity 0.15 Actual Output 0.1625 Utilization Design = (100% ) = (100% ) = 81.25% Design Capacity 0.2 The utilization rates show that the facility’s current output is below its design capacity and higher than its effective capacity. The design capacity occurs with extra help.

Section: Capacity Planning Subsection: Measuring Effectiveness of Capacity Use LO: 9.1 Define capacity planning.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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20. Raj Venkaraman is a specialty retailer and is deciding where to locate his new auto manufacturing plant. The annual fixed and variable costs for each possible site have been estimated as follows: Location A B C D

Fixed Costs $850,000 $750,000 $800,000 $925,000

Variable Costs $4,000/car $4,500/car $7,200/car $6,600/car

If demand is expected to be 175,000 units, which location is best?

Answer: Total Cost A = 850,000 + 4,000 (175,000 ) = $700,850,000

Total Cost B = 750,000 + 4,500 (175,000 ) = $788,250,000

Total Cost C = 800,000 + 7,200 (175,000 ) = $1,260,800, 000 Total Cost D = 925,000 + 6,600 (175,000 ) = $1,155,925, 000

Venkaraman should locate at location A because it provides lowest total cost for the expected demand of 175,000 units. This analysis assumes other factors are equal.

Section: Making Location Decisions Subsection: Break-Even Analysis LO: 9.5 Describe the decision-support tools used in location analysis.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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Solutions Manual

Solutions Manual Chapter 10: Facility Layout 1. Fresh Foods Grocery is considering redoing its facility layout. The from–to matrix showing daily customer trips between departments is shown in Table 10-10, and their current layout is shown in Figure 10-12. Fresh Foods is considering exchanging the locations of the dry groceries department (A) and the health and beauty aids department (F). Compute the ld score for Fresh Foods current and proposed layouts. Which is better?

Table 10-10: From-To Matrix for Fresh Foods Trips between Departments Department A B C D E A. Dry groceries 30 24 22 32 B. Bread 35 50 30 C. Frozen foods 28 20 D. Meats 40 E. Vegetables F. Health and beauty aids

F 20 28 20 15 25 -

Figure 10-12: Current layout for Fresh Foods

Answer: Exchanging the locations of A and F: Proposed layout: F

B

C

D

E

A

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ld score computations for current and proposed layouts: No. of Trips (l)

Distance (d)

AB AC AD AE AF BC BD BE BF CD

30 24 22 32 20 35 50 30 28 28

1 2 1 2 3 1 2 1 2 3

CE CF DE DF EF Total

20 20 40 15 25

2 1 1 2 1

Departments

Current Layout Proposed Layout Load LoadDistance Distance (d) Distance (ld) (ld) 30 2 60 48 1 24 22 2 44 64 1 32 60 3 60 35 1 35 100 2 100 30 1 30 56 1 28 84 3 84 40 20 40 30 25 684

2 2 1 1 2

40 40 40 15 50 682

The new layout is the preferred one due to a slightly lower ld score.

Section: Designing Process Layouts Subsection: Using Trial and Error LO: 10.3 Describe the steps involved in designing a process layout.

Bloom’s: Application AACSB: Analytic Time on task: 8 minutes 2. Use trial and error to find a better layout for Fresh Foods Grocery in Problem 1. Compute the ld score and compare it to the ld scores computed for Fresh Foods’ current and proposed layouts. Which is best?

Answer: The following departments have the highest number of trips between them and should be located close to each other.

B and D, which have 50 trips between them E and D, which have 40 trips between them

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The new layout switches the final layout from #1 with D and E: F

B

C

E

D

A

ld score computations for proposed layout: Proposed Layout Department s AB AC AD AE AF BC BD BE BF CD CE CF DE DF EF Total

No. of Trips (l)

Distance (d)

Load-Distance (ld)

30 24 22 32

2 1 1 2

60 24 22 64

20 35 50 30 28 28 20 20 40 15 25

3 1 1 2 1 2 3 2 1 2 1

60 35 50 60 28 56 60 40 40 30 25 654

The proposed layout is the preferred one due to a lower ld score.

Section: Designing Process Layouts Subsection: Using Trial and Error LO: 10.3 Describe the steps involved in designing a process layout.

Bloom’s: Application AACSB: Analytic Time on task: 8 minutes

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3. Mason Machine Tools is reevaluating its facility layout. The current layout is shown in Figure 10-13 and the from–to matrix is in Table 10-11. Mason has to leave department C in its current location because relocation costs are too high. It is considering exchanging departments B and D. Evaluate the proposal by computing the ld score for both layouts.

Figure 10-13: Current layout for Mason Machine Tools

Table 10-11: From-To Matrix for Mason Machine Tools Trips between Departments Department A B C D E A 5 20 5 B 30 10 C 20 15 D E F

F 8 10 5 17 -

Answer: Exchanging the locations of B and D: Proposed layout: D

A

B

C

E

F

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ld score computations for current and proposed layouts: Current Layout Proposed Layout Departments No. of Trips Distance (d) Load Distance Distance (d) Load-Distance (l) (ld) (ld) AB 5 1 5 1 5 AC 20 2 40 2 40 AD 5 1 5 1 5 AF 8 2 16 2 16 BD 30 2 60 2 60 BE 10 2 20 2 20 BF 10 3 30 1 10 CD 20 3 60 1 20 CE 15 1 15 1 15 CF 5 2 10 2 10 EF 17 1 17 1 17 Total 278 218 The proposed layout is better based on the ld score.

Section: Designing Process Layouts Subsection: Using Trial and Error LO: 10.3 Describe the steps involved in designing a process layout.

Bloom’s: Application AACSB: Analytic Time on task: 8 minutes 4. Use trial and error to find a better layout for Mason Machine Tools in Problem 3.

Compute the ld score and compare it to Mason’s current and proposed layouts in Problem 3.

Answer: The following departments have the highest number of trips between them and should be located close to each other.

B and D, which have 30 trips between them A and C, or C and D, which have 20 trips between them Proposed layout: A

B

F

C

D

E

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ld score computations for proposed layout: Department s AB AC AD AF BD BE BF CD CE CF EF Total

No. of Trips (l) 5 20 5 8 30 10 10 20 15 5 17

Proposed Layout Distance Load(d) Distance (ld) 1 5 1 20 2 10 2 16 1 30 2 20 1 10 1 20 2 30 3 15 1 17 193

The proposed layout is better based on the ld score.

Section: Designing Process Layouts Subsection: Using Trial and Error LO: 10.3 Describe the steps involved in designing a process layout.

Bloom’s: Application AACSB: Analytic Time on task: 8 minutes 5. Gator Office Systems is comparing two layouts for the design of its office building. It has interviewed managers in order to develop the from–to matrix shown in Table 10-12. The two layouts considered are shown in Figure 10-14. Which layout do you think is better for Gator Office Systems, using the load–distance model?

Table 10-12: From-To Matrix for Gator Office Supplies Trips between Departments Department A B C D E A 30 34 25 B 20 10 C 50 D E F

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F 55 10 5 30 -

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Figure 10-14: Current and proposed layouts for Gator Office Supplies Current Layout

F

E

A

B

C

D

Proposed Layout

C

D

A

B

E

F

Answer: ld score computations for current and proposed layouts: Department s AB AD AE AF BD BE BF CE CF EF Total

No. of Trips (l) 30 34 25 55 20 10 10 50 5 30

Current Layout Distance Load (d) Distance (l) 3 90 1 34 1 25 2 110 2 40 2 20 1 10 1 50 2 10 1 30 419

Proposed Layout Distance Load(d) Distance (ld) 3 90 1 34 2 50 1 55 2 40 1 10 2 20 2 100 3 15 1 30 444

The current layout is better based on the ld score.

Section: Designing Process Layouts Subsection: Using Trial and Error LO: 10.3 Describe the steps involved in designing a process layout.

Bloom’s: Application AACSB: Analytic Time on task: 8 minutes

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6. Use trial and error to develop a better layout for Gator Office Supplies. Which departments do you think need to be in close proximity to one another?

Answer: The following departments have the highest number of trips between them and should be located close to each other.

C and E, which have 50 trips between them A and F, which have 55 trips between them Proposed layout reverses departments A and B: F

E

B

A

C

D

ld score computations for proposed layout: Department s AB AD AE AF BD BE BF CE CF EF Total

No. of Trips (l) 30 34 25 55 20 10 10 50 5 30

Proposed Layout Distance (d) LoadDistance (ld) 3 90 2 68 2 50 1 55 1 20 1 10 2 20 1 50 2 10 1 30 403

The proposed layout is better based on the ld score.

Section: Designing Process Layouts Subsection: Using Trial and Error LO: 10.3 Describe the steps involved in designing a process layout.

Bloom’s: Application AACSB: Analytic Time on task: 8 minutes

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7. T-Shirts Unlimited is a retailer that sells every kind of T-shirt imaginable. The different types of T-shirts are stored in departments that all take up the same amount of space.

Given the available warehouse space (Figure 10-15) and a from–to matrix showing the number of trips to and from each department (Table 10-13), help T-Shirts Unlimited decide where to store each type of T-shirt.

Table 10-13: From-To Matrix for T-Shirts Unlimited Department Category Trips to and from Dock 1 Sports T-Shirts 50 2 Men’s T-Shirts 42 3 Women’s T-Shirts 75 4 Children’s T-Shirts 50 5 Fashion T-Shirts 65 6 Undershirts 48

Figure 10-15: Warehouse storage areas for T-Shirts Unlimited Storage Dock

Storage

Storage

Aisle Storage

Storage

Storage

Answer: To assign departments to specific storage areas, we progressively assign departments with the highest number of trips closest to the dock. Using this logic, the following block plan is developed: 3 Dock

1

2

Aisle 5

4

6

Section: Special Cases of Process Layout Subsection: Storage Areas of Equal Size LO: 10.4 Describe the two special cases of process layout.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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8. David’s Sport Supplies is a store that sells sports equipment and gear for teenagers and young adults. David’s is in the process of assigning the location of storage areas in its warehouse (Figure 10-16) to minimize the number of trips made to retrieve needed items.

Given here in Table 10-14 are the departments that need to be located, the number of trips made per week for each department, and the area needed by each department.

Table 10-14: Department Information for David’s Sport Supplies Department Trips to and from Dock Area Needed 1. Baseball equipment 140 1 2. Football gear 80 1 3. Hockey equipment 135 2 4. Basketball equipment 180 4 5. Sports clothes 250 2 Figure 10-16: Warehouse storage areas for David’s Sport Supplies Storage

Storage

Storage

Dock

Storage

Storage

Storage

Storage

Aisle Storage

Storage

Storage

Answer: Step 1: Compute the ratios of number of trips to the area needed.

Department 1. Baseball equipment 2. Football gear 3. Hockey equipment 4. Basketball equipment 5. Sports clothes

Trips to and from Dock 140 80 135 180 250

Area Needed 1 1 2 4 2

Ratio of Trips to Area Needed 140 80 67.5 45 125

Step 2: Use the ratios from step 1 to assign departments to storage areas.

1

2

Dock

3

4

4

4

4

Aisle 5

5

3

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Section: Special Cases of Process Layout Subsection: Storage Areas of Unequal Size LO: 10.4 Describe the two special cases of process layout.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 9. MMS Associates is a telecommunications service provider. The company is currently redesigning its main office to accommodate six newly hired salespeople. Some of the salespeople are expected to work in teams, so office assignments are very important.

Table 10-15 presents the from–to matrix showing the expected frequency of contacts between members of the new sales staff. The block plan in Figure 10-17 shows the assigned office locations for the six sales members. Assume equal-sized offices and rectilinear distances. How would you evaluate the developed layout? What is the ld score for MMS Associates?

A B C D E F

Table 10-15: Number of Contacts between Sales Staff Sales Person A B C D E 6 12 18 1 4 19 3 5 0 7 -

F 1 0 0 19 0 -

Figure 10-17: Assigned office locations for sales staff at MMS Associates

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Answer: It is reasonable to assume the layout can be improved. A starting point would have D directly next to F, B, and A. Departments AB AC AD AE AF BC BD BE BF CD CE CF DE DF EF Total

No. of Trips (l) 6 12 18 1 1 4 19 3 0 5 0 0 7 19 0

Current Layout Distance (d) Load-Distance (ld) 1 6 2 24 1 18 2 2 3 3 1 4 2 38 1 3 2 0 3 15 2 0 1 0 1 7 2 38 1 0 158

Section: Designing Process Layouts Subsection: Using Trial and Error LO: 10.3 Describe the steps involved in designing a process layout.

Bloom’s: Application AACSB: Analytic Time on task: 8 minutes 10. Michael Marc, the President of MMS Associates, is considering an alternative plan for the sales staff situation described in Problem 9. His alternative plan is shown in Figure 10-18.

What is the ld score for this plan? How does it compare to the original plan considered in Problem 9?

Figure 10-18: Alternative office locations for sales staff at MMS Associates

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Answer: ld score computations for proposed layout: Departments

No. of Trips (l)

AB AC AD AE AF BC BD BE BF CD CE CF DE DF EF Total

6 12 18 1 1 4 19 3 0 5 0 0 7 19 0

Proposed Layout Distance (d) Load-Distance (ld) 2 12 2 24 1 18 1 1 3 3 2 8 1 19 1 3 1 0 3 15 1 0 1 0 2 14 2 38 2 0 155

This plan is slightly better because it yields a lower total ld score. However, this layout is far from the best.

Section: Designing Process Layouts Subsection: Using Trial and Error LO: 10.3 Describe the steps involved in designing a process layout.

Bloom’s: Application AACSB: Analytic Time on task: 8 minutes 11. Use trial and error to find a better layout for MMS Associates. Which salespeople will be your priority to keep together?

Answer: The following departments have the highest number of trips between them and should be located close to each other.

B and D, which have 19 trips between them D and F, which have 19 trips between them A and D, which have 18 trips between them A and C, which have 12 trips between them Proposed layout: B

D

A

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E

F

Solutions Manual

C

ld score computations for proposed layout: Departments AB AC AD AE AF BC BD BE BF CD CE CF DE DF EF Total

No. of Trips (l) 6 12 18 1 1 4 19 3 0 5 0 0 7 19 0

Proposed Layout Distance (d) Load-Distance (ld) 2 12 1 12 1 18 3 3 2 2 3 12 1 19 1 3 2 0 2 10 2 0 1 0 2 14 1 19 1 0 124

Section: Designing Process Layouts Subsection: Using Trial and Error LO: 10.3 Describe the steps involved in designing a process layout.

Bloom’s: Application AACSB: Analytic Time on task: 8 minutes 12. A manufacturing company is designing an assembly line to produce its main product.

The line should be able to produce 60 units per hour. The following data in Table 10-16 give the necessary information.

Task A B C D E F G

Table 10-16: Task Information for Problem 12 Immediate Predecessor Task Time (sec) None 35 A 50 A 21 B 38 C 25 D, E 58 F 15

a) Which task is the bottleneck? b) Draw a precedence diagram for the above information. c) Compute the cycle time with a desired output of 60 units per hour.

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Answer: a) Task F is the bottleneck.

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b)

c) C = (3,600 seconds/hour)/(60 units/hour) = 60 seconds/unit Section: Designing Product Layouts Subsection: Steps 1, 2, & 3 LO: 10.5 Describe the steps involved in designing a product layout.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 13. An assembly line must be designed to produce 50 packages per hour. The following data in Table 10-17 give the necessary information.

A B C D E F

Table 10-17: Task Information for Problem 13 Task Immediate Predecessor Task Time (sec) None 25 A 60 B 35 B 45 B 10 C, D, E 50

a) Draw a precedence diagram. b) Compute the cycle time (in seconds) to achieve the desired output rate. c) What is the theoretical minimum number of stations? d) Which work element should be assigned to which workstation? e) What are the resulting efficiency and balance delay percentages?

Answer: a)

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b) C = (3,600 seconds/hour)/(50 units/hour) = 72 seconds/unit c) TM =

å t 225 = = 3.125 or 4 stations C 72

d) Workstation 1 2 3 4 5 e)

Eligible Task A B E C D F

Task Selected A B E C D F

Task Time 25 60 10 35 45 50

Idle Time 47 12 2 37 27 22

åt 225 (100% ) = (100% ) = 62.5% NC (5 )(72 ) Balance Delay (%) = 37.5% Efficiency =

Section: Designing Product Layouts Subsection: Steps 1, 2, 3, 4, & 5 LO: 10.5 Describe the steps involved in designing a product layout.

Bloom’s: Application AACSB: Analytic Time on task: 12 minutes 14. An assembly line must be designed to produce 40 containers per hour. The following data in Table 10-18 give the necessary information.

A B C D E F G

Table 10-17: Task Information for Problem 14 Task Immediate Predecessor Task Time (sec) None 60 A 12 B 35 A 55 D 10 E 50 F, C 5

a) Draw a precedence diagram. b) Compute the cycle time (in seconds) to achieve the desired output rate. c) What is the theoretical minimum number of stations? d) Which work element should be assigned to which workstation? e) What are the resulting efficiency and balance delay percentages?

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Answer: a)

b) C = (3,600 seconds/hour)/(40 units/hour) = 90 seconds/unit c) TM =

å t 227 = = 2.52 or 3 stations C 90

d) Workstation 1 2 3

e)

Eligible Task A B D E F G C

Task Selected A B D E F G C

Task Time 60 12 55 10 50 5 35

Idle Time 30 18 35 25 40 35 0

åt 227 (100% ) = (100% ) = 84.1% NC ( 3 )( 90 ) Balance Delay (%) = 15.9% Multiple solutions exist for assigning tasks to the three minimum workstations without affecting efficiency.

Efficiency =

Section: Designing Product Layouts Subsection: Steps 1, 2, 3, 4, & 5 LO: 10.5 Describe the steps involved in designing a product layout.

Bloom’s: Application AACSB: Analytic Time on task: 12 minutes

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15. The ABC Corporation is designing its new assembly line. The line will produce 85 units per hour. The tasks, their times, and their immediate predecessors are shown in Table 1019.

Table 10-19: Task Information for ABC Corporation Task Immediate Predecessor Task Time (sec) A None 20 B A 10 C A 17 D B 40 E B,C 12 F C 10 G F 5 H G 15 a) Which task is the bottleneck? b) Compute the cycle time with a desired output of 85 units per hour. c) Use a cycle time of 42 seconds/unit to assign tasks to workstations. d) Compute the theoretical minimum number of stations. Did you end up using more stations than the theoretical minimum? e) Compute the efficiency and balance delay of the line.

Answer: a) Task D is the bottleneck. b) C = (3,600 seconds/hour)/(85 units/hour) = 42.35 seconds/unit c) Workstation Eligible Task Task Selected Task Time 1 A A 20 B B 10 2 3 4

d) TM =

e)

C F D E G H

C F D E G H

17 10 40 12 5 15

Idle Time 22 12 25 15 2 30 25 10

å t 129 = = 3.07 or 4 stations C 42

. åt 129 Efficiency = (100% ) = (100% ) = 76.79% NC ( 4 )( 42) Balance Delay (%) = 23.21% Multiple solutions exist for assigning tasks to the three minimum workstations without affecting efficiency.

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Section: Designing Product Layouts Subsection: Steps 1, 2, 3, 4, & 5 LO: 10.5 Describe the steps involved in designing a product layout.

Bloom’s: Application AACSB: Analytic Time on task: 12 minutes

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16. Kiko Teddy Bear is a manufacturer of stuffed teddy bears. Kiko would like to be able to produce 40 teddy bears per hour on its assembly line. Use the information provided in Table 10-20 to answer the questions that follow.

Work Element A B C D E F G H

Table 10-20: Task Information for Kiko Teddy Bear Task Description Immediate Task Time (sec) Predecessor Cut teddy bear pattern None 90 Sew teddy bear cloth A 75 Stuff teddy bear B 50 Glue on eyes C 20 Glue on nose C 15 Sew on mouth C 35 Attach manufacturer’s label B 15 Inspect and pack teddy bear D, E, F, G 40

a) Draw a precedence diagram. b) What is the cycle time? c) What is the theoretical minimum number of stations? d) Assign tasks to specific workstations using the cycle time you computed in part (b). e) What are the efficiency and balance delay of the line? f) Which task is the bottleneck? g) Compute the maximum output.

Answer: a)

b) C = (3,600 seconds/hour)/(40 units/hour) = 90 seconds/unit c) TM =

å t 222 = = 3.78 or 4 stations C 72

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d) Workstation 1 2 3 4

e)

Eligible Task A D G C F D E H

Task Selected A D G C F D E H

Task Time 90 75 15 50 35 20 15 40

Idle Time 0 15 0 40 5 70 55 15

åt 340 (100% ) = (100% ) = 94.44% NC ( 4 )( 90 ) Balance Delay = 5.56% Efficiency =

f) Task A is the bottleneck. g) Maximum Output = (Available Time)/(Cycle Time) = (3,600 seconds/hour)/(90 units/seconds) = 40 units/hour Section: Designing Product Layouts Subsection: Steps 1, 2, 3, 4, & 5 LO: 10.5 Describe the steps involved in designing a product layout.

Bloom’s: Application AACSB: Analytic Time on task: 12 minutes

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17. Use the longest task time rule to balance the assembly line described in Table 10-21; the line can produce 50 units per hour.

Table 10-21: Assembly-Line Task Information Work Element Immediate Predecessor Task Time (sec) A None 15 B A 20 C A 15 D A 20 E C, D 30 F D 22 G B 14 H G 20 I E, F, H 20 J I 18 K J 16 a) What is the cycle time? b) What is the theoretical minimum number of stations? c) Which work elements are assigned to which workstations? d) What are the resulting efficiency and balance delay percentages?

Answer: a) C = (3,600 seconds/hour)/(50 units/hour) = 72 seconds/unit b) å t 210 TM = = = 2.92 or 3 stations C 72 c) Workstation Eligible Task Task Selected Task Time A A 15 B B 20 1 D D 20 C C 15 E E 30 2 F F 22 G G 14 H H 20 3 I I 20 J J 18 4 K K 16 d)

Idle Time 57 37 17 2 42 20 6 52 32 14 56

åt 210 (100% ) = (100% ) = 72.9% NC ( 4 )(72 ) Balance Delay = 27.1% Efficiency =

Section: Designing Product Layouts Subsection: Steps 1, 2, 3, 4, & 5 © 2023 John Wiley & Sons, Inc. All rights reserved.

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LO: 10.5 Describe the steps involved in designing a product layout.

Bloom’s: Application AACSB: Analytic Time on task: 12 minutes 18. A dress-making operation is being designed as an assembly line. Table 10-22 shows the tasks that need to be performed, their task times, and preceding tasks. If the goal is to produce 30 dresses per hour, answer the questions that follow the table.

Table 10-22: Dress-Making Task Information Work Element Immediate Task Time (sec) Predecessor A. Cut dress body None 30 B. Cut sleeves None 40 C. Cut collar None 20 D. Sew dress body A 100 E. Sew sleeves to dress B, D 25 F. Sew collar to dress C, D 50 G. Hem dress D, E, F 50 H. Package dress G 90 a) Compute the cycle time. b) Which task is the bottleneck? c) What is the maximum output for this line? d) Compute the theoretical minimum number of stations. e) Assign work elements to stations, using the longest task time rule. f) Compute the efficiency and balance delay of your assignment.

Answer: a) C = (3,600 seconds/hour)/(30 units/hour) = 120 seconds/dress b) Task D is the bottleneck. c) Maximum Output = (Available Time)/(Minimum Cycle Time) = (3,600 seconds/hour)/(100 seconds/unit) = 36 dresses per hour d)

TM =

å t 405 = = 3.38 or 4 stations C 120

e) Workstation 1 2 3 4 5

Eligible Task A B C D E F G H

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Task Selected A B C D E F G H

Task Time 30 40 20 100 25 50 20 90

Idle Time 90 50 30 20 95 45 70 30

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f)

Efficiency =

Solutions Manual

åt 405 (100% ) = (100% ) = 67.5% NC (5 )(120 )

Balance Delay = 32.5% Section: Designing Product Layouts Subsection: Steps 1, 2, 3, 4, & 5 LO: 10.5 Describe the steps involved in designing a product layout.

Bloom’s: Application AACSB: Analytic Time on task: 12 minutes 19. Table 10-23 shows the tasks required to assemble an aluminum storm door and the length of time needed to complete each task.

Table 10-23: Task Information for Problem 19 Task Immediate Predecessor Task Time (sec) A None 32 B A 43 C A 12 D A 23 E B, C, D 15 F E 25 G None 20 H F, G 5 a) Calculate the cycle time needed to produce 320 doors in an eight-hour workday. b) What is the minimum number of workstations that can be used on the line and still achieve the desired production rate? Balance the line and calculate its efficiency. c) What is the maximum output possible with these data? The minimum?

Answer: a) C = 320 doors/8 hours = 40 doors/hour = (3,600 sec/hour/40 doors/hour) = 90 sec/door b) TM = 175/90 = 1.94 or 2 workstations Workstation 1 2

Eligible Task A B C D E F G H

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Task Selected A B C D E F G H

Task Time

Idle Time

32 43 12 23 15 25 20 5

58 15 3 67 52 27 7 2

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Efficiency =

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åt 175 (100% ) = (100% ) = 97.2% NC ( 2)( 90 )

c) Maximum output would use the bottleneck task B: (3,600 seconds/hour)/(43 seconds/unit) = 83.72 units per hour Minimum output uses the total time in one workstation: (3,600/175) = 20.57 units per hour Section: Designing Product Layouts Subsection: Steps 1, 2, 3, 4, & 5 LO: 10.5 Describe the steps involved in designing a product layout.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes 20. Use the data from Problem 19 to rebalance the line with a cycle time of 75 seconds. How does the number of workstations change? What happens to the output and the line’s efficiency?

Answer: Workstation 1 2 3

Eligible Task A B C D E F G H

Task Selected A B C D E F G H

Task Time

Idle Time

32 43 12 23 15 25 20 5

43 0 63 40 25 50 30 25

TM = 175/75 = 2.33 or 3 workstations

åt 175 (100% ) = (100% ) = 77.78% NC ( 3 )(75 ) Output = (3,600 seconds/hour)/(75 seconds/unit) = 48 units/hour Efficiency =

Compared to the workstation design and cycle time of question 19, the shorter cycle time increases the number of workstations to 3 and decreases the efficiency to 77.78%. The output, however, is increased to 48 doors per hour.

Multiple solutions exist for assigning tasks to the three minimum workstations without affecting efficiency.

Section: Designing Product Layouts Subsection: Steps 1, 2, 3, 4, & 5 LO: 10.5 Describe the steps involved in designing a product layout.

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Bloom’s: Application AACSB: Analytic Time on task: 8 minutes

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Solutions Manual Chapter 11: Work System Design 1. Given the following information, determine the sample size needed if the standard time estimate is to be within 5 percent of the true mean 97 percent of the time.

Work Element 1 2 3 4

Standard Deviation (minutes) 0.25 0.20 0.10 0.15

Mean Observed Time (minutes) 1.50 1.20 0.80 1.50

Answer: With z = 2.17 and a = 0.05, the number of observations needed for each work element is: 2

é 2.17 0.25 ù Work element 1 : n ³ ê ´ ú = 52.3 ë 0.05 1.5 û 2

é 2.17 0.2 ù Work element 2 : n ³ ê ´ ú = 52.3 ë 0.05 1.2 û 2

é 2.17 0.1 ù Work element 3 : n ³ ê ´ ú = 29.4 ë 0.05 0.8 û 2

é 2.17 0.15 ù Work element 4 : n ³ ê ´ ú = 18.8 ë 0.05 1.50 û Sample size needed is 53.

Section: Work Measurement Subsection: Developing Standards LO: 11.3 Explain work measurement.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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2. Using the information in Problem 1, determine the sample size needed if the standard time estimate is to be within 5 percent of the true mean 99 percent of the time.

Answer: With z = 2.58 and a = 0.05, the number of observations needed for each work element is: 2

é 2.58 0.25 ù Work element 1 : n ³ ê ´ ú = 74 ë 0.05 1.5 û 2

é 2.58 0.2 ù Work element 2 : n ³ ê ´ ú = 74 ë 0.05 1.2 û 2

é 2.58 0.1 ù Work element 3 : n ³ ê ´ ú = 41.6 ë 0.05 0.8 û 2

é 2.58 0.15 ù Work element 4 : n ³ ê ´ ú = 26.6 ë 0.05 1.50 û

Sample size needed is 74.

Section: Work Measurement Subsection: Developing Standards LO: 11.3 Explain work measurement.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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3. Using the following information, determine the sample size needed if the standard time estimate is to be within 5 percent of the true mean 95 percent of the time.

Work Element 1 2 3 4 5 6

Standard Deviation (minutes) 0.60 0.20 1.10 0.85 0.40 0.50

Mean Observed Time (minutes) 2.40 1.50 3.85 2.55 1.60 2.50

Answer:

With z = 1.96 and a = 0.05, the number of observations needed for each work element is: 2

é 1.96 0.60 ù Work element 1 : n ³ ê ´ ú = 97 ë 0.05 2.40 û 2

é 1.96 0.20 ù Work element 2 : n ³ ê ´ ú = 28 ë 0.05 1.50 û 2

é 1.96 1.10 ù Work element 3 : n ³ ê ´ ú = 126 ë 0.05 3.85 û 2

é 1.96 0.85 ù Work element 4 : n ³ ê ´ ú = 171 ë 0.05 2.55 û 2

é 1.96 0.40 ù Work element 5 : n ³ ê ´ ú = 97 ë 0.05 1.60 û 2

é 1.96 0.50 ù Work element 6 : n ³ ê ´ ú = 62 ë 0.05 2.50 û

Sample size needed is 171.

Section: Work Measurement Subsection: Developing Standards LO: 11.3 Explain work measurement.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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4. Using the information in Problem 3, calculate the sample size needed if the standard time estimate is to be within 5 percent of the true mean 99 percent of the time. Calculate the percentage increase in sample size for the higher precision.

Answer: With z = 2.58 and a = 0.05, the number of observations needed for each work element is: 2

é 2.58 0.60 ù Work element 1 : n ³ ê ´ ú = 167 ë 0.05 2.40 û 2

é 2.58 0.20 ù Work element 2 : n ³ ê ´ ú = 48 ë 0.05 1.50 û 2

é 2.58 1.10 ù Work element 3 : n ³ ê ´ ú = 218 ë 0.05 3.85 û 2

é 2.58 0.85 ù Work element 4 : n ³ ê ´ ú = 296 ë 0.05 2.55 û 2

é 2.58 0.40 ù Work element 5 : n ³ ê ´ ú = 167 ë 0.05 1.60 û 2

é 2.58 0.50 ù Work element 6 : n ³ ê ´ ú = 107 ë 0.05 2.50 û

Sample size needed is 296.

Percentage increase in sample size for the higher precision is 73.1%.

Section: Work Measurement Subsection: Developing Standards LO: 11.3 Explain work measurement.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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Use the following information from the Arkade Company for Problems 5–10.

Work Element 1 2 3 4

Mean Observed Time (minutes) 1.60 1.30 0.70 1.90

Performance Rating Factor 0.90 0.95 1.15 1.15

5. Calculate the normal time for each of the work elements.

Answer: Work Element 1 2 3 4

Normal Time (NT) (1.60)(0.90)(1) = 1.44 (1.30)(0.95)(1) = 1.235 (0.70)(1.15)(1) = 0.805 (1.90)(1.15)(1) = 2.185

Section: Work Measurement Subsection: Developing Standards LO: 11.3 Explain work measurement.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes 6. The Arkade Company has decided to use a 15 percent allowance factor based on job time.

Calculate the standard time for each work element and for the total job.

Answer: ST = (NT)(AF), where AF = 1 + PFD = 1 + 0.15 = 1.15 Work Element 1 2 3 4

Standard Time (ST) (1.44)(1.15) = 1.66 (1.235)(1.15) = 1.42 (0.805)(1.15) = 0.93 (2.185)(1.15) = 2.51

STjob = (1.66 + 1.42 + 0.93 + 2.51 ) = 6.52

Section: Work Measurement Subsection: Developing Standards LO: 11.3 Explain work measurement.

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Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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7. Based on the standard time calculated in Problem 6, how many units should an employee operating at 100 percent of standard complete during an eight-hour workday?

Answer: From Problem 6, the sum of the standard times is 6.52 minutes.

Number of units /day = (1 unit/6.52 minutes) × (60 minutes/hour) × (8 hours/day) = 73.62 units/day (60 minutes/6.52 minutes a unit) x 8 hours a day = 9.20245 units per hour x 8 hours = 73.61963 = 73.62 units per 8-hour day Section: Work Measurement Subsection: Developing Standards LO: 11.3 Explain work measurement.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes 8. The Arkade Company is considering switching to a 12 percent allowance based on time worked. Calculate the new standard time for each work element and for the total job.

Answer: ST = (NT)(AF), where

AF = Work Element 1 2 3 4

1 1 = = 1.136 1 - PFD 1 - 0.12

Standard Time (ST) (1.44)(1.136) = 1.64 (1.235)(1.136) = 1.4 (0.805)(1.136) = 0.91 (2.185)(1.136) = 2.48

STjob = (1.64 +1.4 + 0.91 + 2.48 ) = 6.43 minutes Section: Work Measurement Subsection: Developing Standards LO: 11.3 Explain work measurement.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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9. Based on the standard time calculated in Problem 8, how many units should an employee operating at 100 percent of standard complete during an eight-hour workday?

Answer: From Problem 8, the sum of the standard times is 6.43 minutes.

Number of units / day = (1 unit / 6.43 minutes ) ´ ( 60 minutes / hour ) ´ ( 8 hours / day ) = 74.65 units / day (60 minutes/6.43 minutes a unit) × 8 hours a day = 74.65 9.33126 units per hour × 8 hours = 74.65008 = 74.65 units per 8-hour day Section: Work Measurement Subsection: Developing Standards LO: 11.3 Explain work measurement.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes 10. Compare the two standards calculated in Problems 6 and 8. What other factors should be considered in selecting the method for determining the allowance factor?

Answer: The standard time for the total job is greater when the 12 percent allowance factor is based on time worked. Some of the factors that should be considered in selecting the method for determining the allowance factor are the jobs similarities and whether they have the same allowance factors. The allowance factor by time worked is better for jobs with the same PFD allowances. The allowance factor should be determined through actual observation of the various tasks performed and should account realistically for normal variability between worker skills sets and speed, as well as normal variability in performance from day to day.

Section: Work Measurement Subsection: Developing Standards LO: 11.3 Explain work measurement.

Bloom’s: Application AACSB: Reflective Thinking Time on task: 2 minutes

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11. Jake’s Jumbo Jacks has collected the following information to develop a standard time for building jumbo jacks.

Observations Cycle 1 Cycle 2 Cycle 3 Cycle 4 Cycle 5 Cycle 6 Cycle 7 Cycle 8 Cycle 9 Cycle 10 Rating factor Frequency

1 2.18 2.22 2.20 2.18 2.21 2.22 2.17 2.21 2.18 2.17 0.9 1

Element (in minutes) 2 3 4 1.25 1.70 2.74 1.23 1.75 2.66 1.29 1.72 2.60 1.30 1.80 2.56 1.26 1.84 2.58 1.22 1.79 2.58 1.26 1.78 2.60 1.26 1.75 2.58 1.30 1.80 2.60 1.28 1.78 2.58 0.8 1.10 1.05 1 1 1

5 1.57 1.55 1.57 1.57 1.59 1.61 1.57 1.61 1.55 1.57 0.95 1

a) Calculate the mean observed time for each element. b) Calculate the normal time for each element. c) Using an allowance factor of 20 percent of job time, calculate the standard time for each element and for the entire job. d) How many units should be completed each hour if the worker performs at 100 percent of the standard? e) How many units should be completed each hour if the worker performs at 90 percent of the standard?

Answer: a) Element 1 2 3 4 5

Mean Observed Time 2.194 1.265 1.771 2.608 1.576

Element 1 2 3 4 5

Normal Time (NT) (2.194)(0.9)(1) = 1.975 (1.265)(0.8)(1) = 1.012 (1.771)(1.10)(1) = 1.948 (2.608)(1.05)(1) = 2.738 (1.576)(0.95)(1) = 1.497

b)

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c) ST = (NT)(AF), where AF = 1 + PFD = 1 + 0.20 = 1.20 Work Element 1 2 3 4 5

Standard Time (ST) (1.975)(1.20) = 2.370 (1.012)(1.20) = 1.214 (1.948)(1.20) = 2.338 (2.738)(1.20) = 3.286 (1.497)(1.20) = 1.796

STjob = ( 2.370 + 1.214 + 2.338 + 3.286 + 1.796 ) = 11.004 minutes d) Number of units/hour = (60 minutes/hour)(1 unit/11.004 minutes) = 5.45 units/hour e) Number of units/hour = (5.45 units/hour)(0.90) = 4.9 units/hour Section: Work Measurement Subsection: Developing Standards LO: 11.3 Explain work measurement.

Bloom’s: Application AACSB: Analytic Time on task: 8 minutes

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12. Frank’s Fabricators has collected the following information to develop a standard time for producing their high-volume Navigator III, a universal remote control. All of the times are in minutes.

Observations Cycle 1 Cycle 2 Cycle 3 Cycle 4 Cycle 5 Cycle 6 Rating factor Frequency

1 1.10 1.08 1.15 1.16 1.07 1.10 0.95 1.00

2 3.00 3.20 3.10 0.90 0.50

Element 3 4 0.92 1.23 0.88 1.30 0.85 1.26 0.88 1.33 0.90 1.28 0.94 1.30 1.05 1.00 1.00 1.00

5 1.46 1.64 1.55 1.52 1.62 1.60 0.85 1.00

6 1.80 1.78 1.76 1.80 1.82 1.82 1.10 1.00

a) Calculate the mean observed time for each element. b) Calculate the normal time for each element. c) Using an allowance factor of 15 percent of job time, calculate the standard time for each element. d) Calculate the standard time for completing one Navigator III. e) If an employee is able to produce at a rate equal to the standard (100 percent efficiency), how many units should she produce each hour? f) If an employee is working at 90 percent efficiency, how many units should she complete in one hour? g) If a process improvement has changed the mean observed time for element 6 to 1.50 minutes, what is the new standard time for the Navigator III? h) If the company builds 20,000 Navigator IIIs each month, how much less time does it require using the new process?

Answer: a) Element 1 2 3 4 5 6

Mean Observed Time 1.110 3.10 0.895 1.283 1.565 1.797

Element 1 2 3 4 5 6

Normal Time (NT) (1.110)(0.95)(1) = 1.055 (3.10)(0.90)(0.5) = 1.395 (0.895)(1.05)(1) = 0.940 (1.283)(1.00)(1) = 1.283 (1.565)(0.85)(1) = 1.330 (1.797)(1.10)(1) = 1.976

b)

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c) ST = (NT)(AF), where AF = 1 + PFD = 1 + 0.15 = 1.15 Work Element 1 2 3 4 5 6 d)

Standard Time (ST) (1.055)(1.15) = 1.213 (1.395)(1.15) = 1.604 (0.940)(1.15) = 1.081 (1.283)(1.15) = 1.476 (1.330)(1.15) = 1.530 (1.976)(1.15) = 2.273

STjob = (1.213 + 1.604 + 1.081 + 1.476 + 1.530 + 2.273 ) = 9.177 minutes

e) Number of units/hour = (60 minutes/hour)(1 unit/9.177 minutes) = 6.54 units/hour f) Number of units/hour = (6.54 units/hour)(0.90) = 5.88 units/hour g) The standard time for element 6 becomes 1.8975 minutes. The new standard time for the Navigator III is 8.80 minutes/unit. h) Since 0.377 minutes/unit is saved when using the new process, the total time saved for 20,000 Navigator IIIs each month is equal to 0.377(20,000) = 7,540 minutes or 125 hours.

Section: Work Measurement Subsection: Developing Standards LO: 11.3 Explain work measurement.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes

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13. The following information is provided to you for each of five elements performed in building the Aviator model, a basic universal remote control.

Work Element 1 2 3 4 5

Mean Observed Time (minutes) 0.96 1.45 3.33 1.24 1.18

Performance Rating Factor 0.96 1.10 1.00 0.90 1.05

Frequency 1.00 1.00 0.33 1.00 1.00

a) Calculate the normal time for each element. b) If the company uses a 15 percent allowance factor based on time worked, calculate the standard time for each element. c) Calculate the standard hourly output. d) Calculate the expected hourly output at 90 percent of standard.

Section: Work Measurement Subsection: Developing Standards LO: 11.3 Explain work measurement.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes Answer: a) Element 1 2 3 4 5

Normal Time (NT) (0.96)(0.96)(1) = 0.9216 (3.10)(0.90)(0.5) = 1.595 (0.895)(1.05)(1) = 1.110 (1.283)(1.00)(1) = 1.116 (1.565)(0.85)(1) = 1.239

b) ST = (NT)(AF), where

Work Element 1 2 3 4 5

1 1 AF = = = 1.176 1 - PFD 1 - 0.15

.

Standard Time (ST) (0.9216)(1.176) = 1.084 (1.595)(1.176) = 1.876 (1.110)(1.176) = 1.305 (1.116)(1.176) = 1.313 (1.239)(1.176) = 1.458

STjob = (1.084 + 1.876 + 1.305 + 1.313 + 1.458 ) = 7.036 minutes c) Standard hourly output = (60 minutes/hour)(1 unit/7.036 minutes) = 8.528 units/hour © 2023 John Wiley & Sons, Inc. All rights reserved.

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d) Expected hourly output = (8.528 units/hour)(0.90) = 7.675 units/hour Section: Work Measurement Subsection: Developing Standards LO: 11.3 Explain work measurement.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 14. You have 25 observations of university policeman Sgt. Jack B. Nimble during his normal workday. The results are shown here. Assume that the estimated proportion is to be within 5 percent of the true proportion 95 percent of the time.

Activity Observed Doing paperwork On the phone Eating doughnuts Cleaning weapon Idle Not in sight

Number of Times Observed 9 3 3 4 2 4

a) Based on your preliminary observations, how many total observations do you need to estimate the proportion of time Sgt. Nimble spends doing paperwork? b) How many total observations do you need to estimate the proportion of time Sgt. Nimble spends on the phone? c) How many total observations do you need to estimate the proportion of time Sgt. Nimble seems to be unavailable?

Answer: a) Based on the preliminary observations, the estimate of the proportion of time Sgt. Nimble spends doing paperwork is 0.36 (9 observed times/25 observations taken).

2 æ 1.96 ö n=ç ÷ ( 0.36 )(1 – 0.36 ) = 355 observations è 0.05 ø b) Based on the preliminary observations, the estimate of the proportion of time Sgt. Nimble spends on the phone is 0.12 (3 observed times/25 observations taken).

2

æ 1.96 ö n=ç ÷ ( 0.12)(1 – 0.12 ) = 163 observations è 0.05 ø

c) Based on the preliminary observations, the estimate of the proportion of time Sgt. Nimble seems to be unavailable is 0.16 (4 observed times/25 observations taken).

2

æ 1.96 ö n=ç ÷ ( 0.16 )(1 – 0.16 ) = 207 observations è 0.05 ø

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Section: Work Measurement Subsection: Developing Standards LO: 11.3 Explain work measurement.

Bloom’s: Application AACSB: Analytic Time on task: 4 minutes

15. You are given the following information: Observations Cycle 1 Cycle 2 Cycle 3 Cycle 4 Cycle 5 Cycle 6 Cycle 7 Cycle 8 Cycle 9 Cycle 10 Cycle 11 Cycle 12 Cycle 13 Cycle 14 Cycle 15 Rating factor Frequency

1 0.58 0.61 0.59 0.54 0.60 0.57 0.53 0.59 0.63 0.58 0.56 0.55 0.58 0.60 0.62 0.95 1

Element (in minutes) 2 3 4 1.50 0.79 0.30 0.75 0.35 0.73 0.33 0.72 0.35 1.40 0.72 0.30 0.71 0.32 0.80 0.30 0.78 0.28 1.54 0.77 0.35 0.79 0.33 0.72 0.32 0.79 0.34 1.62 0.77 0.29 0.80 0.33 0.74 0.30 0.90 1.00 1.10 0.25 1 1

5

2.00

2.20

2.10 0.90 0.20

a) Develop the mean observed time for each element. b) Calculate the normal time for each element. c) Using an allowance factor of 15 percent of job time, calculate the standard time for each element and for the entire job. d) How many units should be completed each hour if the worker performs at 100 percent of the standard? e) How many units should be completed each hour if the worker performs at 110 percent of the standard?

Answer: a) Element 1 2 3 4 5

Mean Observed Time 0.582 1.515 0.759 0.319 2.100

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b) Element 1 2 3 4 5

Normal Time (NT) (0.582)(0.95)(1) = 0.553 (1.515)(0.90)(0.25) = 0.341 (0.759)(1.0)(1) = 0.759 (0.319)(1.10)(1) = 0.351 (2.100)(0.90)(0.20) = 0.378

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ST = ( NT )( AF ) , where AF = 1 + PFD = 1 + 0.15 = 1.15 Work Element 1 2 3 4 5

Standard Time (ST) (0.553)(1.15) = 0.636 (0.341)(1.15) = 0.392 (0.759)(1.15) = 0.873 (0.351)(1.15) = 0.404 (0.378)(1.15) = 0.435

STjob = ( 0.636 + 0.392 + 0.873 + 0.404 + 0.435 ) = 2.740 minutes d) Number of units/hour = (60 minutes/hour)(1 unit/2.740 minutes) = 21.898 units per hour e) Number of units/hour = (21.898 units/hour)(1.10) = 24.088 units/hour Section: Work Measurement Subsection: Developing Standards LO: 11.3 Explain work measurement.

Bloom’s: Application AACSB: Analytic Time on task: 8 minutes 16. As a class project you have been asked to project the proportion of time a professor spends on various activities. You have decided to use the work-sampling method. Your initial observations are shown.

Activity Observed

Number of Times Observed Grading 4 Administrative paperwork 6 Preparing for class 5 Teaching class 5 Meeting with student(s) 8 On the phone 2 Working on research 6 Unavailable 4 Total 40 You are instructed that your estimates are to be within 5 percent of the true value with 97 percent confidence (z = 2.17).

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a) Based on your initial observations, how many total observations are needed to estimate the proportion of time the professor spends on each activity?

Activity Observed Grading Administrative paperwork Preparing for class Teaching class Meeting with student(s) On the phone Working on research Unavailable Total

Number of Times Observed 30 50 30 30 66 17 45 34 302

b) Determine what proportion of time the professor spends teaching class. c) Determine what proportion of time the professor spends working on research. d) If the professor works approximately 54 hours per week, determine the amount of time that would normally be spent on each activity.

Answer: a) Based on the preliminary observations, the estimate of the proportion of time the professor spends grading is 0.10 (4 observed times/40 observations taken).

2 æ 2.17 ö n=ç ÷ ( 0.10 )(1 – 0.10 ) = 170 observations è 0.05 ø Based on the preliminary observations, the estimate of the proportion of time the professor spends doing administrative paperwork is 0.15 (6 observed times/40 observations taken).

2 æ 2.17 ö n=ç ÷ ( 0.15 )(1 – 0.15 ) = 241 observations è 0.05 ø Based on the preliminary observations, the estimate of the proportion of time the professor spends preparing for class is 0.125 (5 observed times/40 observations taken).

2 æ 2.17 ö n=ç ÷ ( 0.125 )(1 – 0.125 ) = 207 observations è 0.05 ø Based on the preliminary observations, the estimate of the proportion of time the professor spends teaching class is 0.125 (5 observed times/40 observations taken).

2 æ 2.17 ö n=ç ÷ ( 0.125 )(1 – 0.125 ) = 207 observations è 0.05 ø Based on the preliminary observations, the estimate of the proportion of time the professor spends meeting with student(s) is 0.20 (8 observed times/40 observations taken).

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2

æ 2.17 ö n=ç ÷ ( 0.20 )(1 – 0.20 ) = 302 observations è 0.05 ø

Based on the preliminary observations, the estimate of the proportion of time the professor spends on the phone is 0.05 (2 observed times/40 observations taken).

2 æ 2.17 ö n=ç ÷ ( 0.05 )(1 – 0.05 ) = 90 observations è 0.05 ø Based on the preliminary observations, the estimate of the proportion of time the professor spends working on research is 0.15 (6 observed times/40 observations taken).

2 æ 2.17 ö n=ç ÷ ( 0.15 )(1 – 0.15 ) = 241 observations è 0.05 ø Based on the preliminary observations, the estimate of the proportion of time the professor is unavailable is 0.10 (4 observed times/40 observations taken).

2 æ 2.17 ö n=ç ÷ ( 0.10 )(1 – 0.10 ) = 170 observations è 0.05 ø b) The proportion of time the professor spends teaching class is 30/302 = 9.9%. c) The proportion of time the professor spends working on research is 45/302 = 14.9%. d) Activity Observed Grading Administrative paperwork Preparing for class Teaching class Meeting with student(s) On the phone Working on research Unavailable

Time spent (hours) (30/302)(54) = 5.36 (50/302)(54) = 8.94 (30/302)(54) = 5.36 (30/302)(54) = 5.36 (66/302)(54) = 11.80 (17/302)(54) = 3.04 (45/302)(54) = 8.05 (34/302)(54) = 6.08

Section: Work Measurement Subsection: Developing Standards LO: 11.3 Explain work measurement.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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17. Your 20 observations of Dr. Knowitall reveal the following information. Assume that the estimate is to be within 5 percent of the true proportion 95 percent of the time.

Activity Observed With patient Reviewing test results On phone Idle Away on emergency Not available Total

Number of Times Observed 8 5 3 2 4 3 25

a) Calculate the sample size needed to estimate the proportion of time Dr. Knowitall spends away on emergencies. b) Calculate the sample size needed to estimate the proportion of time Dr. Knowitall spends reviewing test results. c) Calculate the minimum number of observations that must be made to complete the work-sampling analysis.

Answer: a) Based on the preliminary observations, the estimate of the proportion of time Dr. Knowitall spends away on emergencies is 0.16 (4 observed times/25 observations taken).

2 æ 1.96 ö n=ç ÷ ( 0.16 )(1 – 0.16 ) = 207 observations è 0.05 ø b) Based on the preliminary observations, the estimate of the proportion of time Dr. Knowitall spends reviewing test results is 0.2 (5 observed times/25 observations taken).

2 æ 1.96 ö n=ç ÷ ( 0.2)(1 – 0.2 ) = 246 observations è 0.05 ø c) The largest observed frequency is 8 (with patient). Based on the preliminary observations, the estimate of the proportion of time Dr. Knowitall spends with patients is 0.32 (8 observed times/25 observations taken). Therefore, the minimum number of observations that must be made to complete the work-sampling analysis is: 2 æ 1.96 ö n=ç ÷ ( 0.32)(1 – 0.32) = 335 observations è 0.05 ø Section: Work Measurement Subsection: Developing Standards LO: 11.3 Explain work measurement.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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18. You need to develop a labor time estimate for a customer order of 20 network installations. It is estimated that the first installation will require 60 hours of labor, and a learning curve of 90 percent is expected. a) How many labor-hours are required for the fifteenth installation? b) How many labor-hours are required for the twentieth installation? c) How many labor-hours are required to complete all 20 installations? d) If the average labor cost is $32, what is the total labor cost for installing the networks?

Answer: a) T = 60 hours, and the coefficient from Table 11-9 is 0.663.

Number of labor-hours = 60(0.663) = 39.78 hours b) T = 60 hours, and the coefficient from Table 11-9 is 0.634.

Number of labor-hours = 60(0.634) = 38.04 hours c) T = 60 hours and the coefficient from Table 11-9 is 14.608.

Total number of labor-hours = 60(14.608) = 876.48 hours d) Total labor cost = (876.48 hours)($32/hour) = $28,047.36 Section: Work Measurement Subsection: Learning Curve LO: 11.3 Explain work measurement.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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19. Students in an operations management class have been assigned eight similar computer homework problems. Alexis needed 30 minutes to complete the first problem. Assuming a 75 percent learning curve, how much total time will Alexis need to complete the assignment?

Answer: T = 30 minutes, n = 8, total learning curve coefficient = 4.802 Total time = 30(4.802) = 144.06 minutes, or 2.401 hours.

Section: Work Measurement Subsection: Learning Curve LO: 11.3 Explain work measurement.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

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20. Your company has received an order for 20 units of a product. The labor cost to produce the item is $9.50 per hour. The setup cost for the item is $60 and material costs are $25 per unit. The item is sold for $92. The learning rate is 80 percent. Overhead is assessed at a rate of 55 percent of unit labor cost. a) Determine the average unit cost for the 20 units if the first unit takes four hours. b) Determine the minimum number of units that need to be made before the selling price meets or exceeds the average unit cost.

Answer: a) n =20, total learning curve coefficient = 10.485 Total cost = setup + material costs + labor cost + overhead cost = 60 + 20(25) + 9.5(4) (10.485) + 0.55(9.5)(4)(10.485) = $1,177.57 Average cost per unit = $1,177.57/20 = $58.88 b) If Total Revenue ≥ Total Cost, the selling price per unit exceeds the average unit cost.

92n ≥ 60 + 25n + 1.55(9.5)(4)(total learning curve coefficient for n) More simply, 67n – 60 = 58.9 (total learning curve coefficient); While several approaches may be taken from here, the simplest may be by trial and error. The production becomes profitable at n = 4.

Section: Work Measurement Subsection: Learning Curve LO: 11.3 Explain work measurement.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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Solutions Manual Chapter 12: Inventory Management 1. Elyssa’s Elegant Eveningwear (EEE) needs to ship finished goods from its manufacturing facility to its distribution warehouse. Annual demand for EEE is 2,400 gowns. EEE can ship the gowns via regular parcel service (3 days transit time), premium parcel service (1 day transit time), or via public carrier (7 days transit time). Calculate the average annual transportation inventory for each alternative.

Answer:

tD ( 3 ) ( 2,400 ) = =19.7 units 365 365 1 2,400 ) AT I premium service = ( ) ( =6.6 units 365 7 2,400 ) AT I public carrier = ( ) ( = 46.0 units 365 AT I regular service =

Section: How Manufacturers Use Inventory Subsection: Transportation or Pipeline Inventory LO: 12.1 Discuss basic inventory principles.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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2. Yasuko’s Art Emporium (YAE) ships art from its studio located in the Far East to its distribution center located on the West Coast of the United States. YAE can send the art either via transoceanic ship freight service (20 days transit) or by air freight (3 days transit time). YAE ships 20,000 pieces of art annually. a) Calculate the average annual transportation inventory when sending the art via transoceanic ship freight service. b) Calculate the average annual transportation inventory when sending the art via air freight. c) What additional information is needed to compare the two alternatives?

Answer:

20 ) ( 20,000 ) =1096 units 365 3 20,000 ) =164.4 units b) ATI = ( ) ( 365 c) Cost related information is needed to compare the two alternatives. For example, transportation costs, insurance costs, costs related to possible theft or damage while in transit and this effect on customer satisfaction, and the speed and reliability of the mode of service delivery are among the important considerations in comparing the two alternatives. a)

ATI = (

Section: How Manufacturers Use Inventory Subsection: Transportation or Pipeline Inventory LO: 12.1 Discuss basic inventory principles.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 3. Joe, the owner of Genuine Reproductions (GR), a company that manufactures reproduction furniture, is interested in measuring inventory effectiveness. Last year the cost of goods sold at GR was $3,000,000. The average inventory in dollars was $250,000. a) Calculate the inventory turnover for GR. b) Calculate the weeks of supply. Assume 52 weeks per year. c) Calculate the days of supply. Assume that GR operates 5 days per week.

Answer: a) Inventory turnover = COGS/Average Inventory = $3,000,000/$250,000 = 12 inventory turns b) Weeks of supply = Average Inventory/Average Weekly Sales = $250,000/($3,000,000/52) = $250,000/$57,692.31 = 4.33 weeks c) Days of supply = Average Inventory/Average Daily Sales = $250,000/($3,000,000/(5×52) = 21.67 days Section: Inventory Management Objectives Subsection: Minimum Inventory Investments LO: 12.2 Describe inventory management objectives.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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4. Genuine Reproductions (GR) plans on increasing next year’s sales by 30 percent but expects to increase average inventory in dollars of $300,000. a) Calculate the expected inventory turnover for next year. b) Calculate the expected weeks of supply.

Answer: a) Inventory turnover = COGS/Average Inventory = ($3,000,000 × 1.30)/$300,000 = $3,900,000/$300,000 = 13 times b) Weeks of supply = Average Inventory/Average Weekly Sales = $300,000/($3,900,000/52) = $300,000/$75,000 = 4 weeks of supply Section: Inventory Management Objectives Subsection: Minimum Inventory Investments LO: 12.2 Describe inventory management objectives.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 5. What is the inventory turnover for Genuine Reproductions from Problems 3 and 4 if sales actually increase 20 percent but the average inventory rises to $325,000?

Answer: Inventory turnover = COGS/Average Inventory = ($3,000,000 × 1.20)/$325,000 = $3,600,000/$325,000 = 11.08 times Section: Inventory Management Objectives Subsection: Minimum Inventory Investments LO: 12.2 Describe inventory management objectives.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

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6. Frederick’s Farm Factory (FFF) currently maintains an average inventory valued at $2,900,000. The company estimates its capital cost at 8 percent, its storage cost at 4 percent, and its risk cost at 3 percent. a) Calculate the annual holding cost rate for FFF. b) Calculate the total annual holding costs for FFF. Answer: a) Annual holding cost rate = (8 + 4 + 3) percent = 15 percent b) Annual holding costs = $2,900,000 × 0.15 = $435,000 Section: Relevant Inventory Costs Subsection: [untitled] Introduction LO: 12.3 Explain the relevant inventory costs.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes 7. The Federal Reserve Board has just increased the interest rate. FFF in Problem 6 now has to pay 12 percent for its capital. Calculate the impact on total annual holding costs for FFF. Answer: Annual holding cost rate = (12 + 4 + 3) percent = 19.0 percent Annual holding costs = $2,900,000 × 0.19 = $551,000 Annual holding cost will increase by $116,000 or 26.67 percent [($551,000-435,000)/ $435,000] Section: Relevant Inventory Costs Subsection: [untitled] Introduction LO: 12.3 Explain the relevant inventory costs.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

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8. A technology problem has rendered some of the inventory at FFF (Problem 6) obsolete. FFF estimates that the risk cost of its inventory is now 10 percent. a) Calculate the new annual holding cost rate. b) Calculate the new total annual holding costs for FFF. Answer: a) Annual holding cost rate = (8 + 4 + 10) percent = 22 percent b) Annual holding costs = $2,900,000 × 0.22 = $638,000 Section: Relevant Inventory Costs Subsection: [untitled] Introduction LO: 12.3 Explain the relevant inventory costs.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes 9. Custom Computers Inc. assembles custom home computer systems. The heat sinks needed are bought for $12 each and are ordered in quantities of 1,300 units. Annual demand is 5,200 heat sinks, the annual inventory holding cost rate is $3 per unit, and the cost to place an order is estimated to be $50. Calculate the following: a) Average inventory level b) The number of orders placed per year c) The total annual inventory holding cost d) The total annual ordering cost e) The total annual cost Answer: a) Average inventory level = Q/2 = 1,300/2 = 650 units b) Number of orders placed per year (N) = D/Q = 5,200/1,300 = 4 orders c) Annual inventory holding cost = (Q/2)H = (1,300/2) × $3 = $1,950 d) Total annual ordering cost = (D/Q)S or N × S = (5,200/1,300) × $50 or 4 × $50 = $200 e) Total annual cost = (D/Q)S + (Q/2)H = $200 + $1,950 = $2,150 Not counting the sinks which is (5,200 × $12) = $62,400.

Section: Determining Order Quantities Subsection: Calculating Inventory Policy Costs LO: 12.6 Calculate order quantities.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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10. Custom Computers Inc. from Problem 9 is considering a new ordering policy. The new order quantity would be 850 heat sinks. Recalculate Problem 9, parts (a) through (e), and compare results.

Answer: a) Average inventory level = Q/2 = 850/2 = 425 units b) Number of orders placed per year (N) = D/Q = 5,200/850 = 6.12 orders c) Annual inventory holding cost = (Q/2)H = (850/2) × $3 = $1,275 d) Total annual ordering cost = (D/Q)S = (5,200/850) × $50 = $305.88, or N × S = 6.12 × $50 = $306 e) Total annual cost = (D/Q)S + (Q/2)H = $305.88 + $1,275 = $1,580.88, or $306 + 1,275 = $1,581 An order quantity of 850 units instead of the original or current quantity of 1,300 units will reduce the annual total cost (outside of the purchase price) by $2,150 ˗ $1,581 = $569.

Section: Determining Order Quantities Subsection: Calculating Inventory Policy Costs LO: 12.6 Calculate order quantities.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 11. Bill Maze, recently hired by Custom Computers Inc., has suggested using the economic order quantity for the heat sinks. Using the information in Problem 9, calculate the following: a) Economic order quantity b) Average inventory level c) The number of orders placed per year d) The total annual ordering cost e) The total annual holding cost f) The total annual cost g) Compare these results with the costs calculated in Problems 9 and 10.

Answer:

√

√

2 ( 5200 ) ( 50 ) 2 DS = =416 units H 3 b) Average inventory level = Q/2 = 416/2 = 208 units c) Number of orders placed per year N = D/Q = 5,200/416 = 12.5 = 13 d) Total annual ordering cost = (D/Q)S = (5,200/416 = 12.5 = 13) × $50 = $650 or 12.49 × $50 = $624.50 e) Total annual holding cost = (Q/2)H = (416/2 = 208) × $3) = $624 or 208.17 × $3 = $624.51 f) Total annual cost = (D/Q)S + (Q/2)H = $624.50 + $624.50 = $1,249 (Not counting the cost of the heat sinks, which is $62,400.) g) The original total cost, when 1,300 units were ordered is $2,150. The total cost when 850 units are ordered each time is $1,581. Under the economic order quantity model, the total cost is $1,249. The economic order quantity provides the lowest total annual costs, excluding the purchase price which would be the same under each model. a)

Q=

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Section: Determining Order Quantities Subsection: Calculating Inventory Policy Costs LO: 12.6 Calculate order quantities.

Bloom’s: Application AACSB: Analytic Time on task: 7 minutes 12. A local nursery, Greens, uses 1,780 bags of plant food annually. Greens works 50 weeks per year. It costs $12 to place an order for plant food. The annual holding cost rate is $6 per bag.

Lead time is one week. a) Calculate the economic order quantity. b) Calculate the total annual costs. c) Determine the reorder point.

Answer:

√

√

2 ( 1780 ) ( 12 ) 2 DS = =84 units H 6 b) Total annual cost = (D/Q)S + (Q/2)H = (1,780/84) × $12 + (84/2) × $6 = $254.29 + $252.00 = $506.29 c) Reorder point R= dL = (1,780 bags /50 weeks) × (1 week) = 36 units a)

Q=

Section: Determining Safety Stock Levels Subsection: [untitled] Introduction LO: 12.7 Calculate the appropriate safety stock level.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 13. Rapid Grower, the supplier of plant food for Greens in Problem 12, has offered the following quantity discounts. If the nursery places orders of 60 bags or less, the cost per bag is $25. For orders of 60 to 100 bags, the cost per bag is $23. For orders over 100 bags, the cost is $22 per bag. Greens estimates its holding cost to be 25 percent of the unit price. Determine the most cost-effective ordering policy for Greens.

Answer:

√

2 ( 1,780 ) ( 12 ) =83 bags ( infeasible quantity ) 25 ( 0.25 ) This quantity is an infeasible quantity as the price of $25 per bag holds only if fewer than 60 bags are ordered.

EOQ at price of $25/bag =

√

2 ( 1,780 ) ( 12 ) =86 bags ( feasible quantity ) 23 ( 0.25 ) This quantity is feasible as the price of $23 per bag holds if between 60 and 100 bags are ordered.

EOQ at price of $23/bag =

√

2 ( 1,780 ) ( 12 ) =88 bags ( infeasible quantity ) 22 ( 0.25 ) This quantity is an infeasible quantity as the price of $22 per bag holds only when more than 100 bags are ordered.

EOQ at price of $22/bag =

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As the only feasible quantity is 86 bags, the total cost, at that quantity is = (D/Q)S + (Q/2)H = (1,780/86)($12) + (86/2)($23 × 0.25) = (20.70 × $12) + (43 × $5.75) = $247.25 + $248.40 = $495.65 Add to this the purchase price of 1,780 × $23 = $40,940 the total cost is $40,940 + 495.65 = $41,435.65 Total cost at Q = 100: (100/2)(23 × 0.25) + (1,780/100)(12) + (23)(1780) = $41,441.10 Step 6: After EOQ cost is found, the total annual cost for all of the prices that are lower with the purchase quantity being the minimum required to get that price.

If Q=101, the price is $22. At that quantity, TC = (1780/101)($12) + (101/2)($22 × 0.25) = (17.62 × $12) + (50.5 × $5.50) = $211.49 + $277.75 = $489.24 Add to this the purchase price of 1,780 × $22 = $39,160 the total cost is $39,160 + $489.24 = $39,649.24 The order quantity of 101 bags reduces the total cost by $1,786.41 than the EOQ total cost, so this should be the order quantity.

Section: Determining Order Quantities Subsection: Quantity Discount Model LO: 12.6 Calculate order quantities.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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14. In an effort to reduce its inventory, Rapid Grower is offering Greens, a local nursery (Problems 12 and 13), two additional price breaks to consider. If the nursery orders a threemonth supply, the cost per bag is $16. If Greens orders a six-month supply, the cost per bag is $14.50. Should Greens change its order quantity calculated in Problem 13?

Answer:

√ √

( 2 ) ( 1780 ) ( 12 ) =108.6 bags ( infeasible quantity ) ( 14.50 ) ( 0.25) 2 1780 ) ( 12 ) EOQ at price $16.00/bag = ( ) ( =103.3 bags ( infeasible quantity ) ( 16.00 ) ( 0.25 ) EOQ at price $14.50/bag =

Price

Holding Cost (Price × .25)

3-month supply (1,780/4 = 445)

$16.0 0

$4.00

103.34 Infeasible $28,893.38

6-month supply (1,780/2 = 890)

$14.5 0

$3.63

108.56 Infeasible $26,203.52

EOQ

Total Cost

Because both quantities are not feasible the EOQ quantities no longer apply. The total cost for each, if the required quantities were ordered would be: Ordering Q = 445 ((1780/445) × 12) + ((445/2) × 4) + (16 × 1780) = $29,418.00 Ordering Q = 890 ((1780/890) × 12) + ((890/2) × 3.63) + (14.5 × 1780) = $27,447.13 When the required quantities, rather than EOQ quantities are ordered, in both cases the total cost is greater than the total cost at the respective EOQ level. However, the total cost is lower than the EOQ cost at the original price break quantities from Problem 13, with the lowest cost being at the 6-month ordering level.

Therefore, Rapid Grower should change its ordering policy to the 6-month supply option, with the lowest overall cost of $27,447 Section: Determining Order Quantities Subsection: Quantity Discount Model LO: 12.6 Calculate the appropriate safety stock level.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 15. In a further attempt to liquidate its inventory, Rapid Grower has offered Greens, the local nursery, an option to buy the entire year’s supply at one time. The cost per bag would be $12.

Should Greens take advantage of this offer?

Answer:

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EOQ at price $12/bag =

√

Solutions Manual

( 2 ) ( 1780 ) ( 12 ) = 119.33 bags ( infeasible quantity) ( 12 ) ( 0.25 )

Total cost when Q is 1,780 bags: [(1,780/2)(12 × 0.25)] + (1,780/1,780)($12) + ($12)(1,780) = $24,042 Greens should accept this latest offer since total cost would be lower than all other options.

Section: Determining Order Quantities Subsection: Quantity Discount Model LO: 12.6 Calculate the appropriate safety stock level.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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16. Sam’s Auto Shop services and repairs a particular brand of foreign automobile. Sam uses oil filters throughout the year. The shop operates 52 weeks per year, and weekly demand is 150 filters Sam estimates that it costs $20 to place an order and his annual holding cost rate is $3 per oil filter. Currently, Sam orders in quantities of 650 filters. Calculate the total annual costs associated with Sam’s current ordering policy.

Answer: Total annual costs = D = 7,800 (150 × 52 weeks) (D/Q)S + (Q/2)H = [(7,800/650) × $20] + (650/2) × $3 = $240 + $975 = $1,215 Section: Determining Order Quantities Subsection: Calculating Inventory Policy Costs LO: 12.6 Calculate the appropriate safety stock level.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 17. Using the information in Problem 16, calculate the following: a) The economic order quantity b) The total annual costs using the EOQ ordering policy c) The penalty costs Sam is incurring by using his current policy Answer: a)

√

√

2 ( 7,800 ) ( 20 ) 2 DS = = 323 units H 3 b) Total annual costs = (D/Q)S + (Q/2)H = (7,800/323) × $20 + (323/2) × $3 = $482.97 + 484.50 = $967.47 c) Penalty costs incurred by Sam = $1,215 – $967.47 = $247.53 Q=

Section: Determining Order Quantities Subsection: Calculating Inventory Policy Costs LO: 12.6 Calculate the appropriate safety stock level.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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18. The local Office of Tourism sells souvenir calendars. Sue, the head of the office, needs to order these calendars in advance of the main tourist season. Based on past seasons, Sue has determined the probability of selling different quantities of the calendars for a particular tourist season.

Demand for Calendars 75,000 80,000 85,000 90,000 95,000

Probability of Demand 0.15 0.25 0.30 0.20 0.10

The Office of Tourism sells the calendars for $12.95 each. The calendars cost Sue $5 each.

The salvage value is estimated to be $0.50 per unsold calendar. Determine how many calendars Sue should order to maximize expected profits.

Answer: Based on the information provided, a payoff table is developed to determine the expected profit with each possible order quantity.

Demand 75,000

80,000

85,000

90,000

95,000

75,000

596,25 0

596,250

596,250

596,250

596,250

80,000

573,75 0

636,000

636,000

636,000

636,000

85,000

551,25 0

613,500

675,750

675,750

675,750

90,000

528,75 0

591,000

653,250

715,500

715,500

95,000

506,25 0

568,500

630,750

693,000

755,250

Amount Ordered

Amount Ordered = Demand, payoff = [Demand × (Selling$ - Unit Cost)] Amount Ordered > Demand, payoff = [Demand × (Selling$ – Unit Cost)] – [(Ordered Demand) × (Unit Cost – Salvage $) Amount Ordered < Demand, payoff = [Ordered × (Selling$ – Unit Cost)] Selected Calculations: Order = Demand: 75,000 × ($12.95 – $5.00) = $596,250

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Order > Demand: 80,000 ordered, 75,000 demand = (75,000 × ($12.95 – $5.00) – [(80,00075,000) × ($5.00 – $0.50)) = $573,750 Order < Demand: 75,000 ordered, 80,000 demand = 75,000 × ($12.95 – $5.00) = $596,250 Expected Profit = Sum of (Payoff × Probability) Probability 0.15

0.25

0.30

0.20

0.10

Demand Ordered

75,000

80,000

85,000

90,000

95,000

Expected Profit

75,000

596,250

596,250

596,250

596,250

596,250

$ 596,250

80,000

573,750

636,000

636,000

636,000

636,000

$ 626,663

85,000

551,250

613,500

675,750

675,750

675,750

$ 641,513

90,000

528,750

591,000

653,250

715,500

715,500

$ 637,688

95,000

506,250

568,500

630,750

693,000

755,250

$ 621,413

The expected profit is equal to the sum of the respective Payoff times its probability. The first two Expected Profit calculations are: = (596,250 × .15) + (596,250 × .25) + (596,250 × .30) + (596,250 × .20) + (596,250 × .10) = (573,750 × .15) + (636,000 × .25) + (636,000 × .30) + (636,000 × .20) + (636,000 × .10) Once the expected profit for each of the possible order quantities is determined, the order quantity with the highest expected profit is selected. Therefore, an order amount of 85,000 calendars maximizes expected profits.

Section: Determining Order Quantities Subsection: Single-Period Inventory Model LO: 12.6 Calculate the appropriate safety stock level.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes 19. The Office of Tourism (Problem 18) has decided to heavily promote local events this year and anticipates more tourists this season. Sue has changed the probability of selling different quantities of calendars as shown. Given the new probabilities, determine how many calendars Sue should order to maximize expected profits.

Demand for Calendars 75,000 80,000 85,000 90,000 95,000

Probability of Demand 0.05 0.20 0.25 0.30 0.20

Answer:

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Based on the information provided, a payoff table is developed to determine the expected profit with each possible order quantity.

75,000

80,000

Demand 85,000

90,000

95,000

Amount Ordered 75,000

$

596,250

$

596,250

$ 596,250

$ 596,250

$ 596,250

80,000

$

573,750

$

636,000

$ 636,000

$ 636,000

$ 636,000

85,000

$

551,250

$

613,500

$ 675,750

$ 675,750

$ 675,750

90,000

$

528,750

$

591,000

$ 653,250

$ 715,500

$ 715,500

95,000

$

506,250

$

568,500

$ 630,750

$ 693,000

$ 755,250

When Amount Ordered = Demand then payoff = [Demand × (Selling $ - Unit Cost)] When Amount Ordered > Demand then payoff = [Demand × (Selling $ - Unit Cost)] [(Ordered - Demand) × (Unit Cost - Salvage $) When Amount Ordered < Demand then payoff = [Ordered × (Selling $ - Unit Cost)] Selected Calculations: Order = Demand: 75,000 × ($12.95 - $5.00) = $596,250 Order > Demand: 80,000 ordered, 75,000 demand = (75,000 × ($12.95 - $5.00) – [(80,000-75,000) × ($5.00 - $0.50)) = $573,750 Order < Demand: 75,000 ordered, 80,000 demand = 75,000 × ($12.95 - $5.00) = $596,250 Expected Profit = Sum of (Payoff × Probability) Probability 0.05 0.20 0.25 Demand Ordered

75,000

80,000

85,000

0.30

90,000

0.20 Expected Profit

95,000

75,000

$

596,250

$

596,250

$ 596,250 $

596,250

$ 596,250

$

596,250

80,000

$

573,750

$

636,000

$ 636,000 $

636,000

$ 636,000

$

632,888

85,000

$

551,250

$

613,500

$ 675,750 $

675,750

$ 675,750

$

657,075

90,000

$

528,750

$

591,000

$ 653,250 $

715,500

$ 715,500

$

665,700

95,000

$

506,250

$

568,500

$ 630,750 $

693,000

$ 755,250

$

655,650

The expected profit is equal to the sum of the respective Payoff times its probability.

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The first two Expected Profit calculations are: = (596,250 × .05) + (596,250 × .20) + (596,250 × .25) + (596,250 × .30) + (596,250 × .20) = (573,750 × .05) + (636,000 × .20) + (636,000 × .25) + (636,000 × .30) + (636,000 × .20) Once we generate the expected profit for each of the possible order quantities, we select the order quantity with the highest expected profit. Therefore, an order amount of 90,000 calendars maximizes expected profits.

Section: Determining Order Quantities Subsection: Single-Period Inventory Model LO: 12.6 Calculate the appropriate safety stock level.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes 20. Given the following list of items, a) Calculate the annual usage cost of each item. b) Classify the items as A, B, or C. Item

Annual Demand

101 102 103 104 105 201 202 203 204 205

500 1,500 5,000 250 1,500 10,000 1,000 1,500 500 100

Ordering Cost ($) 10 10 25 15 35 25 10 20 40 10

Holding Cost (%) 20 30 30 25 35 15 20 25 25 15

Unit Price ($) 0.50 0.20 1.00 4.50 1.20 0.75 1.35 0.20 0.80 2.50

Answer: a) Item

Annual Demand

Unit Price ($)

101 102 103 104 105 201 202 203

500 1,500 5,000 250 1,500 10,000 1,000 1,500

0.50 0.20 1.00 4.50 1.20 0.75 1.35 0.20

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Annual Usage (Demand * Unit Price) $ 250 $ 300 $5,000 $1,125 $1,800 $7,500 $1,350 $ 300

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204 205

500 100

0.80 2.50

$ 400 $ 250

Item

Annual Usage ($)

% of Total Dollars

201 103 105 202 104 204 203 102 101 205 Total

7,500 5,000 1,800 1,350 1,125 400 300 300 250 250 18,275

41.0% 27.4% 9.8% 7.4% 6.2% 2.2% 1.6% 1.6% 1.4% 1.4% 100%

b) Cumulative % of Total Dollars 41.0% 68.4% 78.2% 85.6% 91.8% 94.0% 95.6% 97.3% 98.6% 100.0%

Item Classification A A B B B C C C C C

Section: ABC Inventory Classification Subsection: [untitled] Introduction LO: 12.4 Explain the ABC inventory classification model.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes

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21. Using the information provided in Problem 20, a) Calculate the economic order quantity for each item. (Round to the nearest whole number.) b) Calculate the company’s maximum inventory investment throughout the year. c) Calculate the company’s average inventory level.

Answer: a) EOQ Formula: Q=

√

2 DS H

Example formula: Item #101 Q=

Item

Annual Demand 500 1,500 5,000 250 1,500 10,000 1,000 1,500 500 100

101 102 103 104 105 201 202 203 204 205

Ordering Cost ($) 10 10 25 15 35 25 10 20 40 10

√

2 ( 500 ) 10 = 0.5×0.2

Holding Cost (%) 20 30 30 25 35 15 20 25 25 15

√

10,000 =316.23 0.1

Unit Price ($) 0.50 0.20 1.00 4.50 1.20 0.75 1.35 0.20 0.80 2.50

EOQ 316.23 707.11 912.87 81.65 500 2,108.19 272.17 1,095.45 447.21 73.03

EOQ (rounded) 317 708 913 82 500 2,109 273 1,096 448 74

EOQ values should always be rounded up regardless of the decimal value. b) Item

EOQ (rounded)

Unit Price ($)

101 102 103 104 105 201 202 203 204 205

317 708 913 82 500 2,109 273 1096 448 74

0.50 0.20 1.00 4.50 1.20 0.75 1.35 0.20 0.80 2.50

Product Cost (EOQ*Unit Price) 158.50 141.60 913.00 369.00 600.00 1,581.75 368.55 219.20 358.40 185.00

Using the EOQ number times the unit cost, the company’s maximum inventory investment throughout the year occurs for Item 201 with an inventory level of 2,109 units and value of $1,581.75.

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c) Item

EOQ (rounded)

Average Inventory (EOQ/2)

101 102 103 104 105 201 202 203 204 205

317 708 913 82 500 2,109 273 1,096 448 74

158.50 354 456.50 41 250 1054.50 136.50 548 224 37

Average Inventory Dollar Value (Avg Inv * Unit Price) $ 79.25 70.80 456.50 184.50 300.00 790.88 184.28 109.60 179.20 92.50 $ 2,447.50

Average inventory in dollars = $2,447.50 Section: ABC Inventory Classification Subsection: [untitled] Introduction LO: 12.4 Explain the ABC inventory classification model.

Bloom’s: Application AACSB: Analytic Time on task: 15 minutes 22. Tax Preparers Inc. works 280 days per year. The company uses adding machine tape at a rate of six rolls per day. Usage is believed to be normally distributed with a standard deviation of three rolls during lead time. The cost of ordering the tape is $15, and holding costs are $0.25 per roll per year. Lead time is two days. a) Calculate the economic order quantity. Round number up to the nearest whole number. b) What reorder point will provide an order-cycle service level of 97 percent? Round number up to the nearest whole number. c) How much safety stock must the company hold to have a 97 percent order-cycle service level? Round number up to the nearest whole number. d) What reorder point is needed to provide an order-cycle service level of 99 percent?

Round number up to the nearest whole number. e) How much safety stock must the company hold to have a 99 percent order-cycle service level? Round number up to the nearest whole number.

Answer:

√

√

2 6 × 280 ) ( 15 ) 2 DS = ( )( =449 H 0.25 b) R = dL + zσ = (2)(6) + (1.88)(3) = 17.64 or 18 units c) SS = zσ = (1.88)(3) = 5.64 units d) R = dL + zσ = 12 + (2.33)(3) = 18.99 or 19 units e) SS = zσ = (2.33)(3) = 6.99 units a)

Q=

Section: The Periodic Review System Subsection: [untitled] Introduction LO: 12.8 Describe the periodic review approach.

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Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 23. Healthy Plants Ltd. (HP) produces its premium plant food in 60-pound bags. Demand for the product is 150,000 pounds per week. HP operates 50 weeks per year and can produce 250,000 pounds per week. The setup cost is $200 and the annual holding cost rate is $0.55 per bag. Currently, HP produces its premium plant food in batches of 1,000,000 pounds. a) Calculate the maximum inventory level for HP. b) Calculate the total annual costs of this operating policy.

Answer: a)

I Max =Q

( ) 1−

d p

=1,000,000

(

1−

150,000 250,000

)

=400,000 pounds

b) Total annual costs = (D/Q)S + (IMax/2)H = (7,500,000/1,000,000)(200) + (400,000/2)(0.55/60) = $1,500 + 1,833.33 = $3,333.33 Section: Determining Order Quantities Subsection: Calculating EPQ LO: 12.6 Calculate order quantities.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 24. Using the data provided in Problem 23, determine what will happen if HP uses the economic production quantity model to establish the quantity produced each cycle. a) Calculate the economic production quantity (EPQ). b) Calculate the maximum inventory level using the EPQ. c) Calculate the total annual cost of using the EPQ. d) Calculate the penalty cost HP is incurring with its current policy.

Answer: a) b)

Q=

√ ( )( () ) ( 2 ) ( 7,500,000 ) ( 200 ) 150,000 ( 0.55 / 60 ) 1− 250,000

I Max =Q

1−

d p

=904 ,537

1−

=904,537 pounds or 15,076 bags

150,000 250,000

=361,813 pounds or 6,031 bags

c) Total annual costs = (D/Q)S + (IMax/2)H = (7,500,000/904,537)(200) + (361,813/2)(0.55/60) = $1,658.26 + 1,658.36 = $3,316.62 d) Penalty cost = $3,333.33 – $3,316.62 = $16.71

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Section: Determining Order Quantities Subsection: Calculating EPQ LO: 12.6 Calculate order quantities.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 25. Greener Pastures Inc. (GPI) produces a high-quality organic lawn food and weed eliminator called Super Green (SG). Super Green is sold in 50-pound bags. Monthly demand for Super Green is 75,000 pounds. Greener Pastures has capacity to produce 24,000 50-pound bags per year. The setup cost to produce Super Green is $300. Annual holding cost is estimated to be $3 per 50-pound bag. Currently, GP is producing in batches of 2,500 bags.

A) Calculate the total annual costs of the current operating policy at GPI. b) Calculate the economic production quantity (EPQ). c) Calculate the total annual costs of using the EPQ. d) Calculate the penalty cost incurred with the present policy.

Answer: a)

b) c)

(

)

18,000 =625 24,000 Total annual costs = (18000/2500)(300) + (625/2)(3) = $3,097.50 I Max =2,500

Q=

1−

√ (( ))

( 2 ) ( 18,000 ) ( 300 ) =$3,794.73, or $ 3,795 3

I Max =3,795

18,000 24,000 18,000 1− 24,000

1−

=948.75, or 949

Total annual costs = (18,000/3,795)(300) + (949/2)(3) = $2,846.42 d) Penalty cost = $3,097.50 – $2,846.42 = $251.08 Section: Determining Order Quantities Subsection: Calculating EPQ LO: 12.6 Calculate order quantities.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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26. Lissette Jones, the materials manager for an upscale retailer, wants to measure her customer service level. She has collected the following representative data.

Order Number 1 2 3 4 5 6 7 8 9 10 Totals

Number of Line Items 4 8 2 6 10 8 8 4 5 5 60

Dollar Value of Order 1,000 1,440 1,600 920 1,800 1,200 2,700 1,560 1,780 1,000 15,000

Assuming that orders 1–6 and 8–10 shipped on schedule: a) Calculate the customer service level using the percentage of orders that shipped on schedule. b) Calculate the customer service level using the percentage of line orders that shipped on schedule. c) Calculate the customer service level using the percentage of dollar volume that shipped on schedule. d) Which of these measures would you recommend to Lissette?

Answer: a) A total of 6 + 3 = 9 orders were shipped on schedule.

Customer service level = (9/10)100 = 90% a) Summing the line items for the orders that were shipped on schedule: = 4 + 8 + 2 + 6 + 10 + 8 + 4 + 5 + 5 = 52 Customer service level = (52/60)100 = 86.7% b) Summing the line items for the orders that were shipped on schedule: = $15,000 – 2,700 = $12,300 (since order number 7 is the only order not shipped on schedule) Customer service level = (12300/15000)100 = 82% d) Given these orders vary in value, I would recommend the dollar volume measure as this measure recognizes both line item and dollar value differences..

Section: Inventory Management Objectives Subsection: Customer Service LO: 12.2 Describe inventory management objectives.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

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27. Your new company has decided to use a periodic review system. You have learned that average weekly demand is 48 units per week with a standard deviation of 8 units. You believe that your cycle-service level should be 94 percent. Lead time is two weeks. Initially, you believe that you should do a review every Friday. a) Determine the required safety stock and the target inventory level. b) How would this procedure change if the cycle-service level needed to be 98 percent? c) What is the impact of changing the review period from every Friday to every other Friday, assuming that the cycle-service level is 94 percent?

Answer: a) The z for 94% is found to be approximately 1.55 (the area above it is the risk of stock outs).

SS= z ( σ √ RP + L =1.55 ( 8 √ 1+ 2 =21.48, or 22 units TI =d ( RP + L ) + SS = 48 ( 1+ 2 ) +22=166 units

)

)

b) The z for 98% is found to be approximately 2.05.

SS = 2.05 ( 8 √ 1+ 2 =28.40, or 29 units TI = 48 ( 1+ 2 ) + 29=173 units

)

c)

SS= 2.05 ( 8 √ 2 + 2 = 24.80, or 25 units TI = 48 ( 2+ 2 ) + 25=217 units

)

Section: The Periodic Review System Subsection: [untitled] Introduction LO: 12.8 Describe the periodic review approach.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 28. Michael’s Office Supply (MOS) sells office furniture, equipment, and supplies. This week the company has received 50 customer orders. Each order has an average of five line items.

The average dollar amount of each order is $1,200. MOS was able to ship 47 of the 50 orders on schedule. a) Using the percentage of orders shipped on time, calculate the customer service level. b) If Michael’s calculates customer service level by using the percentage of line items shipped on schedule, how many line items must be shipped to achieve the same customer service level calculated in part (a)? c) If Michael’s calculates customer service level by the percentage of dollar volume shipped, how many dollars of product must be shipped to achieve the same customer service level calculated in part (a)? d) What factors determine the customer service level measure that MOS should use?

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Answer: c) Customer service level = (47/50)100 = 94% d) Number of line items = 47 × 5 = 235 e) Number of dollars of products shipped = 47 × $1200 = $56,400 f) Measurement with percentage of orders should have similar orders in size and value.

Percentage of lines should have orders with similar number of lines and lines similar in value. Percentage by dollar amount requires the most effort, but would best reflect customer service when orders are dissimilar.

Section: Inventory Management Objectives Subsection: Customer Service LO: 12.2 Describe inventory management objectives.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 29. My Kitchen Delights (MKD), a regional producer of gourmet jams and jellies, uses approximately 24,000 glass jars each month during its production. Because of space limitations, MKD orders 5,000 jars at a time. Monthly holding cost is $0.08 per jar, and the ordering cost is $60 per order. The company operates 20 days per month. a) What penalty cost is the company incurring by its present replenishment policy? b) MKD would prefer to order eight times each month but needs to justify any change in order size. How much would ordering cost need to be reduced to justify a lot size of 3,000 jars? c) If MKD can reduce its ordering cost to $30, what is the optimal replenishment order quantity?

Answer: a) Total cost at Q = 5,000 = ((24,000/5,000) × 60) + ((5,000/2) × 0.08) = $488 per month 2 24,000 ) ( 60 ) EOQ = ( ) ( = $6,000 per month 0.08 Total cost = ((24,000/6,000) × 60) + ((6,000/2) × 0.08) = $480 per month

√

Penalty cost = $488 – $480 = $8 per month b) With Q = 3000, holding costs are reduced to $120 ((3,000/2) × 0.08). To keep the total cost at $480, the order costs would be $360 ($480 – $120). Solving 8(S) = $360, order costs would have to be reduced to $45 to stay at the same expense as the EOQ. c)

√

EOQ =

2 ( 24,000 ) ( 30 ) =4,242.6∨4,243 jars 0.08

Section: Determining Order Quantities Subsection: Calculating Inventory Policy Costs LO: 12.6 Calculate order quantities.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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30. My Kitchen Delights (MKD) is considering two new suppliers for the jars used in the production process. The quality at both suppliers is equal. Assume that the annual holding cost is 30 percent of the unit price. Monthly demand averages 20,000 jars. Ordering cost with these two suppliers is $30 per order. The price lists for the suppliers are as follows: Supplier A Quantity Unit Price 1-2499 $3.00 2,500-3,499 2.90 3,500-4,999 2.80 5,000 or more 2.70 Supplier B Quantity Unit Price 1-1,999 $3.50 2,000-2,999 3.15 3,000-3,999 2.85 4,000-4,999 2.75 5,000 or more 2.60 a) Determine the optimal order quantity when using Supplier A. b) Determine the optimal order quantity when using Supplier B. c) Given MKD’s lack of space, which supplier do you recommend be used? Justify your answer.

Answer: a) EOQ at the best price = 4,216.4 or 4,216 for Supplier A (infeasible quantity) EOQ at the second best price = 4,140.4 or 4,140 for Supplier A (feasible quantity) b) EOQ at the best price = 4,296.7 for Supplier B (infeasible quantity) EOQ at the second best price = 4,177.9 for Supplier B (feasible quantity) c) Supplier A at Best Price: Total Cost = 2,500(0.81) + 48(30) + 2.70(240,000) = $651,465 Supplier A at EOQ: Total Cost = 57.96(30) + 2070.50(0.84) + 2.80(240,000) = $675,477.93 Supplier B at Best Price: Total Cost = 2,500(0.78) + 48(30) + 2.60(240,000) = $627,390 Supplier B at EOQ: Total Cost = 57.44(30) + 2089(0.825) + 2.60(240,000)= $663, 446.72 Both suppliers have lowest costs at 5,000 jars.

Their cheapest option is to order 5,000 jars at a time from Supplier B. The annual savings of $36,056.72 has to be weighed against the problems created with the shipments of 822 extra jars.

Section: Determining Order Quantities Subsection: Quantity Discount Model LO: 12.6 Calculate order quantities.

Bloom’s: Application AACSB: Analytic Time on task: 8 minutes © 2023 John Wiley & Sons, Inc. All rights reserved.

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Solutions Manual Chapter 13: Aggregate Planning Use the following data to solve Problems 1 to 7.

Problem Data Cost data Regular-time labor cost per hour Overtime time labor cost per hour Subcontracting cost per unit (labor only) Holding cost per unit per period Back-order cost per unit per period Hiring cost per employee Firing cost per employee Capacity data Beginning workforce Beginning inventory Labor standard per unit Regular time available per period Overtime available per period Subcontracting maximum per period in units of 500 Subcontracting minimum per period Demand data Period 1 Period 2 Period 3 Period 4 Period 5 Period 6

$12.00 $16.00 $75.00 $8.00 $15.00 $500.00 $350.00 210 employees 400 units 6 hours 160 hours 30 hours 1,000 units 500 units Units 5,500 4,600 7,600 5,800 6,560 4,000

1. The BackPack Company produces a line of backpacks. The manager, Tameka Nicholas, is interested in using a level aggregate plan. Inventories and back orders will be used to handle demand fluctuations. She has asked you to develop such a plan. a) Calculate the aggregate production rate. b) Calculate the appropriate workforce given the aggregate production rate. If Tameka will need to hire workers, include this. c) Show what would happen if this plan were implemented. d) Calculate the costs of this plan. e) Evaluate the plan in terms of cost, customer service, operations, and human resources.

Answer: a) Total aggregate production =

å demand = 34,060 units 6

i -1

i

Production rate per period = ( 34,060 – 400 ) / 6 = 5,610 units

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b) Workforce = (5,610 units × 6 hours per unit)/160 hours per employee = 210.375, or 211 employees. This means she has barely enough employees to implement her plan.

She may wish to consider hiring more. c) Period Demand net of beginning inventory Cumulative demand Production per period Cumulative production Cum. Dem. Minus Cum. Prod Ending inventory Back orders

1

2

3

4

5

6

Total

5,100

4,600

7,600

5,800

6,560

4,000

33,660

5,100

9,700

17,300

23,100

29,660

33,660

5,610

5,610

5,610

5,610

5,610

5,610

5,610

11,220

16,830

22,440

28,050

33,660

-510

-1,520

470

660

1610

0

510

1,520

0

0

0

0

2,030

0

0

470

660

1610

0

2,740

33,660

d) Back order cost: 2,740 units × $15 per unit = $41,100 Holding cost: 2,030 units × $8 per unit = $16,240 Regular-time labor cost: $12 per hour ×160 hours per period × 6 periods × 211 employees = $2,430,720 Total cost: $2,488,060 e) The costs appear very high, considering the customer is served very poorly. However, the level aggregate plan is good for operations and human resources.

Section: Developing the Aggregate Plan Subsection: Aggregate Plans for Companies with Tangible Products LO: 13.4 Develop aggregate plans.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes 2. Tameka has decided that the BackPack Company must have very good customer service.

She has asked you to develop a level aggregate plan using inventories but not back orders.

All demand must be met each period. You must: a) Calculate the aggregate production rate. b) Calculate the appropriate workforce given the aggregate production rate. c) Show what would happen if this plan were implemented. d) Calculate the costs of this plan. e) Evaluate the plan in terms of cost, customer service, operations, and human resources.

Answer: a) Production rate per period = 5,932 units (refer to chart for part c); answer is the highest value for “Cumulative demand / Number of periods) as highlighted. © 2023 John Wiley & Sons, Inc. All rights reserved.

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b) Workforce =

Solutions Manual

(5,932 units ´ 6 hours per unit ) hours per employee = 222.45 or223 employees 160

c) Period Demand less beginning inventory Cumulative demand Cumulative demand/periods Production per period Cumulative production Cum. Dem. Minus Cum. Prod Ending inventory

1

2

3

4

5

6

Total

5,100

4,600

7,600

5,800

6,560

4,000

33,660

5,100

9,700

17,300

23,100

29,660

33,660

5,100

4,850

5,767

5,775

5,932

5,610

5,932

5,932

5,932

5,932

5,932

5,932

5,932

11,864

17,796

23,728

29,660

35,592

-832

-2,164

-496

-628

0

-1,932

832

2,164

496

628

0

1,932

35,592

6,052

d) Regular-time labor cost: $12 per hour × 160 hours per period × 6 periods × 223 employees = $2,568,960 Holding cost: 6,052 units × $8 per unit = $48,416 Hiring cost: 13 employees × $500 per employee = $6,500 Total cost: $2,623,876 e) The cost of this plan is $135,816 higher than a level aggregate plan using inventories and back orders. Under this plan, about 32% of the inventory holding cost is incurred in Period 6. However, in terms of customer service, this plan has a high degree of customer service since no orders are placed on back order. Also, under this plan the inventory and hiring costs are higher than a level aggregate plan using inventories and backorders.

Section: Developing the Aggregate Plan Subsection: Aggregate Plans for Companies with Tangible Products LO: 13.4 Develop aggregate plans.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes

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3. Although the BackPack Company has always used a level aggregate plan, Tameka is interested in evaluating chase aggregate plans also. She has asked you to calculate how many hires and fires would be necessary to adjust capacity to meet demand exactly each period. If necessary, incur some undertime. Calculate the number of workers needed each period.

Answer: Period Demand net of beginning inventory Production per period Employees needed in period

1

2

3

4

5

6

Total

5,100

4,600

7,600

5,800

6,560

4,000

5,100

4,600

7,600

5,800

6,560

4,000

33,660

192

173

285

218

246

150

1,264

Example of calculation: 5100 units × 6 hours per unit/160 hours per employee = 191.25, rounded up to 192 employees Section: Developing the Aggregate Plan Subsection: Aggregate Plans for Companies with Tangible Products LO: 13.4 Develop aggregate plans.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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4. Now that you have calculated the number of workers needed each period in Problem 3, Tameka wants to see how the plan would actually work. You need to: a) Show what would happen if this plan were implemented. b) Calculate the costs associated with this plan. c) Evaluate the plan in terms of cost, customer service, operations, and human resources.

Answer: a) Period Demand net of beginning inventory Production per period Employees needed in period Number to hire Number to fire

1

2

3

4

5

6

Total

5,100

4,600

7,600

5,800

6,560

4,000

5,100 192 0 18

4,600 173 0 19

7600 285 112

5,800 218 0 67

6,560 246 28

4,000 150 0 96

33,660 1,264 140 200

b) Regular-time labor cost: $12 per hour × 160 hours per period × 1264 employees = $2,426,880 Hiring cost: 140 employees × $500 per employee = $70,000 Firing cost: 200 employees × $350 per employee = $70,000 Total cost: $2,566,880 c) This plan yields no inventory holding cost since it produces exactly what is needed.

The total costs are lower than the level aggregate plan in Problem 2. Moreover, customer service is high since no backorders are needed. However, from a planning perspective this plan would be difficult to implement due to the high variability of resources needed for each period. Moreover, employee morale would tend to be fairly low with such a plan due to lack of job security.

Section: Developing the Aggregate Plan Subsection: Aggregate Plans for Companies with Tangible Products LO: 13.4 Develop aggregate plans.

Bloom’s: Application AACSB: Analytic Time on task: 7 minutes

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5. Tameka Nicholas is concerned about BackPack’s corporate image and has decided against using hires and fires. Instead, she has asked you to consider a chase aggregate plan using the current workforce supplemented by either overtime or undertime to change capacity to match demand exactly. She has asked for the following information from you: a) How many production hours would be required each period to produce the exact quantity needed? b) How many regular-time production hours are available each period? c) How many overtime production hours would be needed each period? Explain any problems that may occur. d) Show what would happen if this plan were implemented. e) Calculate the costs associated with this plan. f) Evaluate the plan in terms of cost, customer service, operations, and human resources.

Answer: a) Period Demand per period, net of beginning inventory Production hours required (Demand x 6 hours per unit)

1

2

3

4

5

6

5,100

4,600

7,600

5,800

6,560

4,000

30,600

27,600

45,600

34,800

39,360

24,000

b) Amount of regular-time production available each period with current workforce level = (160 hours per employee per period)(210 employees) = 33,600 hours per period c) Period Demand net of beginning inventory Production hours required Regular-time hours available Overtime hours needed

1

2

3

4

5

6

5,100

4,600

7,600

5,800

6,560

4,000

30,600

27,600

45,600

34,800

39,360

24,000

33,600

33,600

33,600

33,600

33,600

33,600

0

0

12,000

1,200

5,760

0

Total

18,960

The maximum overtime hours available, with the current workforce is 210 × 30 hours = 6,300 overtime hours. Unfortunately, with this plan, the number of overtime hours required in period 3 is double what is allowed.

d)

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Period Demand net of beginning inventory Production hours required Regular-time hours of capacity Overtime hours needed Undertime hours Overtime hours per employee (Overtime hours needed / 210 employees)

Solutions Manual

1

2

3

4

5

6

Total

5,100

4,600

7,600

5,800

6,560

4,000

30,600

27,600

45,600

34,800

39,360

24,000

33,600

33,600

33,600

33,600

33,600

33,600

0

0

12,000

1,200

5,760

0

18,960

3,000

6,000

0 57.1

0 5.7

0 27.4

9,600

18,600

e) Regular labor cost: $12 per hour ×160 hours per period × 6 periods × 210 employees = $2,419,200 Overtime labor cost: 18,960 hours × $16 per hour = $303,360 Total cost: $2,722,560 f) The customer demand is being met, but costs are $155,680 over the costs of the hiring and firing plan in Problem 3. During some periods there is a great deal of idle time.

Once again, this chase plan is disruptive to operations and to human resources.

Certainly, if this plan is implemented, arrangements must be made to increase available overtime.

Section: Developing the Aggregate Plan Subsection: Aggregate Plans for Companies with Tangible Products LO: 13.4 Develop aggregate plans.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes 6. Tameka Nicholas believes there must be a better aggregate plan. She has suggested a hybrid plan, using a permanent workforce of 195 employees and subcontracting as needed. Once again, Tameka has requested that you provide the following information: a) Calculate the regular-time production possible each period given a workforce of 195. b) Show what would happen if this plan were implemented. c) Calculate the costs associated with this plan. d) Evaluate the plan in terms of cost, customer service, operations, and human resources.

Answer: a) Regular-time production = (195 employees × 160 hours per employee per period)/(6 hours per unit) = 5,200 units/period b) Assuming no reduction in workforce or in work hours per period.

Note: The backorders could be eliminated if inventory was built with more subcontracting units earlier in the plan.

Production Plan

1

© 2023 John Wiley & Sons, Inc. All rights reserved.

2

3

4

5

6

Total

SM 13-7

Reid & Sanders Operations Management, 8th Edition

Calculations Demand Beginning Inventory Net Demand after Available Inventory Regular-time Production Sub-contracting Units Needed Sub-Contracting Production (Max=1,000) in batches of 500 Ending Inventory Backorder Units

Solutions Manual

5,500 400 5,100

4,600 100 4,500

7,600 700 6,900

5,800 0 5,800

6,560 0 6,560

4,000 0 4,000

34,060

5,200 0

5,200 0

5,200 1,700

5,200 600

5,200 1,360

5,200 0

31,200 3,660

0

0

1,000

500

1,000

0

2,500

100 0

700 0

0 700

0 100

0 360

1,200 0

2,000 1,160

32,860

c) Regular-time production cost: 31,200 units × 6 hours per unit × $12 per hour = $2,246,400 Subcontracting cost: 2,500 units × $75 per unit = $187,500 Back-order cost: 1,160 units × $15 per unit = $17,400 Inventory holding cost: $2,000 × $8 = 16,000 Total cost: $2,467,300 d) From an operations perspective, this plan provides stable use of resources throughout the production period. However, periods 3, 4, and 5 will experience backordering costs. Therefore, this plan will not provide a high level of customer satisfaction. The total cost of this plan is lower than the level plan in Problem 2, but it is higher than the level plan that allowed backorders.

Section: Developing the Aggregate Plan Subsection: Aggregate Plans for Companies with Tangible Products LO: 13.4 Develop aggregate plans.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

SM 13-8

Reid & Sanders Operations Management, 8th Edition

Solutions Manual

7. Tameka wants you to consider a hybrid aggregate plan, using up to the maximum overtime per employee for any period where demand cannot be satisfied with the current regular-time production and the available inventory. Back orders can occur. a) Show what would happen if this plan were implemented. b) Calculate the costs associated with this plan. c) Evaluate the plan in terms of cost, customer service, operations, and human resources.

Answer: a) Given the production rate, 1/6 unit per hour (from 6 hours per unit), the production from the maximum of 30 hours of overtime for each of the 210 employees is computed to be 1,050 units (30 OT hours × 210 employees / 6 hours per unit).

Regular-time production = (210 employees × 160 hours per employee per period)/(6 hours per unit) = 5, 600 units per period Production Plan Calculations Demand Beginning Inventory Net Demand after Available Inventory Regular-time Production Overtime Production Needed Ending Inventory

1 5,500 400

2 4,60 0 500

5,100 5,600 0 500

4,100 5,600 0 1,500

3

4

5

6

TOTAL

7,600 1,500

5,800 6,560 4,000 0 0 0

34,060

6,100 5,600 500 0

5,800 6,560 4,000 5,600 5,600 5,600 200 960 0 0 0 1,600

33,600 1,660 3,600

b) Regular-time labor cost = 33,600 units × $12 per hour × 6 hours per unit = $2,419,200 Overtime labor cost: 1,660 units × $16 per hour × 6 hours per unit = $159,360 Inventory holding cost = 3,600 × $8 = $28,800 Total cost = $2,607,360 c) This hybrid aggregate plan does not require back orders. Overtime costs are still a significant amount at $153,600 but this expense is far less than using only overtime to meet demand. This plan fits the existing available overtime hours and should therefore not be too disruptive to operations.

Section: Developing the Aggregate Plan Subsection: Aggregate Plans for Companies with Tangible Products LO: 13.4 Develop aggregate plans.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

SM 13-9

Reid & Sanders Operations Management, 8th Edition

Solutions Manual

Use the information shown here for Problems 8 to 13. The Draper Tax Company provides tax services to local businesses. Draper chooses to meet all demand as it occurs because customers are unwilling to accept back orders. The company has provided the following cost, capacity, and demand information.

Draper Tax Company Problem Data Cost data Regular-time labor cost per hour $25.00 Overtime labor cost per hour $37.50 Temporary worker cost per hour $40.00 Hiring cost per permanent worker $2,000.00 Firing cost per permanent worker $1,200.00 Backorder cost $500.00 Capacity data Beginning workforce 12 employees Labor standard per service 12 hours Regular-time hours per period 40 hours Overtime hours per period 8 hours Demand data Week 1 48 clients Week 2 36 clients Week 3 50 clients Week 4 40 clients Week 5 38 clients Week 6 48 clients 8. Calculate the size of the workforce needed for the company to meet average weekly demand.

Answer: Average weekly demand = 43.33 clients Workforce (assuming no overtime) = (43.33 clients × 12 hours/client)/40 hours per employee = 13 employees Section: Developing the Aggregate Plan Subsection: Aggregate Plans for Companies with Nontangible Products LO: 13.4 Develop aggregate plans.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

SM 13-10

Reid & Sanders Operations Management, 8th Edition

Solutions Manual

9. Develop a level aggregate plan for the Draper Tax Company if back orders are permitted. a) Show what would happen if this plan were implemented. b) Calculate the costs associated with this plan. c) Evaluate the plan in terms of cost, customer service, operations, and human resources.

Answer: a) Assuming no overtime and using 13 employees (answer from previous problem which requires the hiring of 1 employee): Average weekly demand = 43.33 clients Average hourly demand = 43.33 × 12 = 520 hours Available time = 13 × 40 = 520 Period Number of clients Hours required Backorders carried forward Total hours required Regular hours available Back orders (hours)

1 48 576 0

2 36 432 56

3 50 600 0

4 40 480 80

5 38 456 40

6 48 576 0

Total 260 3,120 176

576 520

488 520

600 520

560 520

496 520

576 520

3,120

56

0

80

40

0

56

b) Hiring cost: (1 employee)($2,000 per employee) = $2,000 Regular-time labor cost: (3,120 hours)($25 per hour) = $78,000 Total cost: $80,000 Note that this total cost does not incorporate any consequences of the unmet hours. c) This plan provides a uniform availability of weekly hours. However, in weeks 1, 3, 4, and 6, there are unmet client hours. If the clients are unwilling to wait for their services, this could result in potential revenue loss to the firm. Also, there are excess service hours in weeks 2 and 5, which cannot be inventoried.

Section: Developing the Aggregate Plan Subsection: Aggregate Plans for Companies with Nontangible Products LO: 13.4 Develop aggregate plans.

Bloom’s: Application AACSB: Analytic Time on task: 7 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

SM 13-11

Reid & Sanders Operations Management, 8th Edition

Solutions Manual

10. Develop a level aggregate plan for the Draper Company if no back orders are permitted. a) Show what would happen if this plan were implemented. b) Calculate the costs associated with this plan. c) Evaluate the plan in terms of cost, customer service, operations, and human resources.

Answer: a) Number of employees needed to meet peak demand (assuming no overtime): (50 clients × 12 hours per client)/ 40 hours per employee = 15 employees Regular hours available: 15 employees × 40 hours = 600 hours Period Number of clients Hours required Regular-time hours Undertime hours

1 48 576 600 24

2 36 432 600 168

3 50 600 600 0

4 40 480 600 120

5 38 456 600 144

6 48 576 600 24

Total 3,120 3,600 480

b) Hiring cost: (3 employees × $2,000 per employee) = $6,000 Regular-time labor cost: (3600 hours)($25 per hour) = $90,000 Total cost: $96,000 c) This plan provides a uniform use of resources and meets customer service needs by ensuring there are no back orders. Therefore, this plan will be easy to implement from an operational perspective. However, about 13.3% (480 / 3,600) of the regular-time hours available are not fully utilized. The underutilized hours could perhaps be used to perform other administrative tasks to improve productivity.

Section: Developing the Aggregate Plan Subsection: Aggregate Plans for Companies with Nontangible Products LO: 13.4 Develop aggregate plans.

Bloom’s: Application AACSB: Analytic Time on task: 7 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

SM 13-12

Reid & Sanders Operations Management, 8th Edition

Solutions Manual

11. Develop a chase aggregate plan using hires and fires to adjust the capacity for Draper. All demand must be met each period. a) Show what would happen if this plan were implemented. b) Calculate the costs associated with this plan. c) Evaluate the plan in terms of cost, customer service, operations, and human resources.

Answer: a) Period Number of clients Hours required Number of employees needed (#clients x 12 hours per client)/40 hours Employees needed, rounded up Current number of employees Number of hires Number of fires Overtime hours required

1 48 576

2 36 432

3 50 600

4 40 480

5 38 456

6 48 576

Total 260 3,120

14.4

10.8

15

12

11.4

14.4

15

11

15

12

12

15

80

12 3 0 0

15 0 4 0

11 4 0 0

15 0 3 0

12 0 0 0

12 3 0 0

10 7 0

b) Hiring cost: 10 employees × $2,000 per employee = $20,000 Firing cost: 7 employees × $1,200 per employee = $8,400 Regular-time labor cost: 80 employees × 40 hours × $25 per hour = $80,000 Total cost: $108,400 c) Customer demand is met at a cost of $108,400. The hiring and firing would be very difficult for operations and for the employees. However, customer service objectives are met as there are no back orders.

Section: Developing the Aggregate Plan Subsection: Aggregate Plans for Companies with Nontangible Products LO: 13.4 Develop aggregate plans.

Bloom’s: Application AACSB: Analytic Time on task: 7 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

SM 13-13

Reid & Sanders Operations Management, 8th Edition

Solutions Manual

12. Develop a chase aggregate plan for Draper using a permanent workforce of 12 employees supplemented by overtime. All demand must be met each period. a) Show what would happen if this plan were implemented. b) Calculate the costs associated with this plan. c) Evaluate the plan in terms of cost, customer service, operations, and human resources.

Answer: a) Period Number of clients Hours required Number of employees required Regular-time labor hours Overtime hours needed Undertime hours

1 48 576 12

2 36 432 12

3 50 600 12

4 40 480 12

5 38 456 12

6 48 576 12

Total 576 3,120 72

480

480

480

480

480

480

2,880

96

0

120

0

0

96

312

0

48

0

0

24

0

72

b) Regular-time labor cost: (12 employees × 40 hrs per period × 6 periods) × $25 per hour = $72,000 Overtime cost: 312 hours × $37.50 per hour = $11,700 Total cost: $83,700 c) This plan uses a uniform level of resources. Therefore, it will be easy to implement as it uses the same number of employees each period. Under this plan, supplementing the permanent workforce of 12 employees by over-time yields considerably lower total costs than a level or a chase aggregate plan. The maximum number of overtime hours available each week is 12 employees × 8 overtime hours per employee or 96 overtime hours each week. In the third week a total of 120 overtime hours are needed, which is 24 hours over capacity. To make this plan work, each employee would need to two hours more that week. The undertime hours under this plan is only 2.5% of the total service hours available indicating high productivity.

Section: Developing the Aggregate Plan Subsection: Aggregate Plans for Companies with Nontangible Products LO: 13.4 Develop aggregate plans.

Bloom’s: Application AACSB: Analytic Time on task: 7 minutes

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SM 13-14

Reid & Sanders Operations Management, 8th Edition

Solutions Manual

13. Concerned about the welfare of its workers, Draper has decided to try a strategy without any overtime. Instead of overtime, Draper has decided to supplement the permanent workforce of 12 employees with temporary workers. Any temporary worker must work the entire week. There is no hiring or firing cost associated with temporary workers.

Develop this aggregate plan for Draper. a) Show what would happen if this plan were implemented. b) Calculate the costs associated with this plan. c) Evaluate the plan in terms of cost, customer service, operations, and human resources.

Answer: a) Period Number of clients Hours required Hours available Unmet hours Temporary employees needed (Unmet hours/40) Temporary workers, rounded up

1 48 576 480 96 2.4 3

2 36 432 480 0

3 50 600 480 120 3

4 40 480 480 0

5 38 456 480 0

3

6 48 576 480 96 2.4

Total 260 3,120 2,880 31

3

9

b) 9 temporary employees = 9 x 40 hours = 360 hours × $40 per hour = $14,400 12 permanent employees = 2,880 hours × $25 = $72,000 Total cost = $86,400 c) Customer demand is met at a lower cost than hiring and firing permanent employees.

This plan is good for the permanent employees with minimal operational issues if the temporary employees are readily available and require minimal training time.

Section: Developing the Aggregate Plan Subsection: Aggregate Plans for Companies with Nontangible Products LO: 13.4 Develop aggregate plans.

Bloom’s: Application AACSB: Analytic Time on task: 7 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

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Reid & Sanders Operations Management, 8th Edition

Solutions Manual

14. W. C. Sanders, owner of Fort Engines, a producer of heavy-duty snow blower engines, needs to develop an aggregate plan for the coming year. The company currently uses 20 individuals working 160 regular-time hours each month. Each worker is capable of producing 10 heavy-duty snow blowers per month. Employees are paid $12 per hour.

Overtime is limited to a maximum of 40 hours per month per employee. Holding costs are $5 per unit per period. Back-order cost is $10 per unit per period. The beginning inventory is 40 units. Monthly demand projections are: Month January February March April May June July August September October November December

Demand 250 230 190 170 200 220 220 220 260 260 240 220

a) Develop a hybrid aggregate plan using the initial workforce supplemented by overtime. If demand in any period exceeds regular-time production plus overtime production plus any beginning inventory, the company will use back orders. Calculate the cost of this plan. b) Another alternative is to try a level plan that uses inventory and back orders to absorb fluctuation. Calculate the cost of this plan. c) A third alternative being considered is to use a hybrid plan but also to close down the facility for the entire month of July. Overtime, inventory, and back orders can be used.

Calculate the cost of this plan. d) Compare the three plans in terms of cost, customer service, operations, and human resources.

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Reid & Sanders Operations Management, 8th Edition

Answer: a) Period

Solutions Manual

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Total

Demand

250

230 190 170 200 220 220 220 260 260 240 220 2,680

Beginning Inventory

40

0

Net Production Needed

210

230 190 160 160 180 200 220 260 260 240 220 2,530

Units Produced with Regular hours

200

200 200 200 200 200 200 200 200 200 200 200 2,400

Inventory

0

0

10

40

40

20

0

0

Shortfall

10

30

0

0

0

0

0

20

O/T Hours Needed (Shortfall × 16 hrs)

160

480

0

0

0

0

0

320 960 960 640 320 3,840

O/T hours used (Max = 20 × 40 hrs)

160

480

0

0

0

0

0

320 800 800 640 320 3,520

Final Shortfall Hours

0

0

0

0

0

0

0

0

160 160

0

0

320

Number of Units on Backorder

0

0

0

0

0

0

0

0

10

0

0

20

0

10

40

40

20

0

0

60

0

0

0

150

0

0

0

110

60

40

20

240

10

Each employee can produce 10 units per month with 160 hours available per month. Each unit therefore needs 160/10 = 16 hours. Overtime can produce another 2.5 units per employee.

Regular labor cost: 2,400 units × 16 hours × $12 per hour = $460,800 Overtime labor cost: 3,520 hours × $18 per hour = $63,360 Inventory holdings cost: 110 units × $5 = $550 Backorder cost: 20 units × $10 = $200 Total cost: $524,910 Assuming the overtime rate is time and a half.

© 2023 John Wiley & Sons, Inc. All rights reserved.

SM 13-17

Reid & Sanders Operations Management, 8th Edition

Solutions Manual

b) Period

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Total

Demand

250 230 190 170 200 220 220 220 260 260 240 220 2,680

Beginning Inventory

40

Net Production Needed

210 230 190 160 160 180 200 220 260 260 240 220 2,530

Units Produced with Regular hours

200 200 200 200 200 200 200 200 200 200 200 200 2,400

Units Produced with Overtime

50

50

50

50

50

50

50

50

50

50

50

50

600

Ending Inventory

40

20

60

90

90

70

50

30

0

0

10

30

490

Shortfall = Backorder

0

0

0

0

0

0

0

0

10

10

0

0

0

10

40

40

20

0

0

0

0

150

20

A level production plan would need to produce 260 units per month in order to meet peak demand. However with regular production and the maximum available overtime a total of 250 units is the maximum production capacity, assuming that all available overtime is used.

Regular units produced (20 employees × 160 hours)/16 hours per unit = 200 Maximum overtime production (40 hours × 20 employees)/16 = 50 Regular labor cost = 2,400 units × 16 hours × $12 = $460,800 Overtime labor cost = 600 units × 16 hours × $18 = $172,800 Inventory holdings cost = 490 units × $5 = $2,450 Backorder cost = 20 units × $10 = $200 Total cost: $636,250 Assuming the overtime rate is time and a half.

© 2023 John Wiley & Sons, Inc. All rights reserved.

SM 13-18

Reid & Sanders Operations Management, 8th Edition

Solutions Manual

Assuming no overtime is used: Period

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Total

Demand

250 230 190 170 200 220 220 220 260 260 240 220 2,680

Beginning Inventory

40

Net Production Needed

210 230 190 160 160 180 200 220 260 260 240 220 2,530

Units Produced with Regular hours

200 200 200 200 200 200 200 200 200 200 200 200 2,400

Units Produced through Overtime

0

0

0

0

0

0

0

0

0

0

0

0 0

Ending Inventory

0

0

10

40

40

20

0

0

0

0

0

0

110

Shortfall = Backorder

10

30

0

0

0

0

0

20

60

60

40

20

240

0

0

10

40

40

20

0

0

0

0

0

150

-

Regular labor cost = 2,400 units × 16 hours × $12 = $460,800 Inventory holdings cost = 110 units × $5 = $550 Backorder cost = 240 units × $10 = $2,400 Total cost: $463,750 Of the two level-plan options, using only regular production time, and placing any shortfalls on backorder has the lowest cost. However, customer service objectives are not as well met, and significantly more units are placed on back order.

© 2023 John Wiley & Sons, Inc. All rights reserved.

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Reid & Sanders Operations Management, 8th Edition

Solutions Manual

c) Assuming August production is limited to demand.

Period

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Total

Demand

250 230 190

170

200 220 220 220 260 260 240 220 2,680

Beginning Inventory

40

40

90

160 200 220

0

Net Production Needed

210 200 150

80

40

20

0

220 260 260 240 220 1,900

Units Produced 200 200 200 with Regular hours

200

200 200

0

200 200 200 200 200 2,200

Units Produced with O/T

40

40

40

40

40

40

0

20

50

50

40

20

420

Ending Inventory

30

40

90

160

200 220

0

0

0

0

0

0

740

Shortfall = Backorder

0

0

0

0

0

0

0

10

10

0

0

20

30

0

0

0

0

0

780

If the facility is closing for the month of July, and in order to ensure that July demand is not backordered, Inventory must be built up to be 220 units by the end of June.

Maximum overtime production is 50 units per month. A total of (250 + 230 + 190 + 170 + 200 + 220 + 220) or 1,480, less beginning inventory of 40 units = 1,440 units must be produced from January to June. Regular production can produce (200 × 6) = 1,200 units, leaving 240 units to be produced through overtime. For each of the six months that equals 240/6 = 40 units to be produced through overtime, for the months of January through June only.

Regular labor cost = 2,200 units × 16 hours × $12 = $422,400 Overtime labor cost = 420 units × 16 hours × $18 = $120,960 Inventory holdings cost = 740 units × $5 = $3,700 Backorder cost = 20 units × $10 = 200 Total Cost = $547,260 Assuming the overtime rate is time and a half.

© 2023 John Wiley & Sons, Inc. All rights reserved.

SM 13-20

Reid & Sanders Operations Management, 8th Edition

Solutions Manual

Assuming August production is set to pre-shutdown level.

Period Demand Beginning Inventory Net Production Needed Units Produced with Regular hours Units Produced with O/T Ending Inventory Shortfall = Backorder

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Total 250 230 190 170 200 220 220 220 260 260 240 220 2,680 40

30

40

90

160 200 220

0

210 200 150

80

40

20

0

0

0

800

20

0

220 240 260 240 220 1,880

200 200 200 200 200 200

0

200 200 200 200 200 2,200

40

40

40

40

40

0

40

40

40

40

40

440

30

40

90

160 200 220

0

20

0

0

0

20

780

0

0

0

0

0

0

0

20

0

0

20

40

0

0

Regular labor cost = 2,200 units × 16 hours × $12 = $422,400 Overtime labor cost = 440 units × 16 hours × $18 = $126,720 Inventory holdings cost = 780 units × $5 = $3,900 Backorder cost = 20 units × $10 = 200 Total cost = $553,220 Assuming the overtime rate is time and a half. d) Cost comparison for the various plans:

Total Cost Total Backorders

Part a) Plan $524,910 20

Part b) (Level OT) $636,250 20

Part b) Part c), Post July (no OT limited to overtime) Demand $463,750 $547,260 240 20

Part c) Level OT Production $553,220 20

While 4 of the 5 plans do not have significant backorder numbers, the plan with the lowest cost also has the highest number of backorders, making this plan one of the least preferred, as customer service objectives will not be met and there is a high risk of lost revenue should customers decide to go elsewhere. Of the remaining four plans, each of which has the same level of backorders, the lowest cost is the first plan which using regular production and all available overtime. The next best plan is where the plant is shut down for the month of July and OT post shut down is limited to demand. Shutting down for the month may have additional costs and may impact employee morale; however, if planned and communicated well in advance, these problems should be mitigated.

The company will have to balance customer service, cost and employee impact when deciding which plan to adopt.

© 2023 John Wiley & Sons, Inc. All rights reserved.

SM 13-21

Reid & Sanders Operations Management, 8th Edition

Solutions Manual

Section: Developing the Aggregate Plan Subsection: Aggregate Plans for Companies with Nontangible Products LO: 13.4 Develop aggregate plans.

Bloom’s: Application AACSB: Analytic Time on task: 12 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

SM 13-22

Reid & Sanders Operations Management, 8th Edition

Solutions Manual

Solutions Manual Chapter 14: Resource Planning Use the information given here to solve Problems 1 to 5.

Item Q R S T X Y V Z

Usage per Parent 3 2 1 3 2 1 2

Load Time (weeks) 2 4 3 2 2 3 3 4

1. Will’s Welded Widgets (WWW) makes its Q Model from components R, S, and T. Component R is made from three units of component X and two units of component Y. Component T is made from one unit of component V and two units of component Z. Draw the product structure tree for the Q Model.

Answer:

Q LT=2

R Usage = 3 LT = 4 X Usage = 3 LT = 2

S Usage = 2 LT = 3 Y Usage = 2 LT = 3

T Usage = 1 LT = 2 V Usage = 1 LT = 3

Z Usage = 2 LT = 4

Section: Material Planning Systems Subsection: Bills of Materials LO: 14.3 Describe material planning systems.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 2. Using the given information, calculate the replenishment lead time for the Q Model assuming that you have no beginning inventories.

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Reid & Sanders Operations Management, 8th Edition

Solutions Manual

Answer: The cumulative lead time is 9 weeks (Y R Q). (3 + 4 + 2 = 9).

Section: Material Planning Systems Subsection: Bills of Materials LO: 14.3 Describe material planning systems.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes 3. Using the given information, calculate the gross requirements for each of the components if the company plans to build 100 of its Q Models. Assume that there are no beginning inventories.

Answer: Product Gross Requirements

Q 100

R 300

S 200

T 100

X 900

Y 600

V 10 0

Z 200

Section: Material Planning Systems Subsection: Bills of Materials LO: 14.3 Describe material planning systems.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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SM 14-2

Reid & Sanders Operations Management, 8th Edition

Solutions Manual

4. Using the given information, calculate the gross requirements for each of the components when the company plans to build 80 of its Q Models if you have these inventories: 80 units of component T and 200 units of component R. Answer: Product Gross Requirements

Q 80

R 40

S 160

T 0

X 120

Y 80

V 0

Z 0

Calculation for ‘R’: (Quantity of R used by Q) – Inventory = (80 × 3) – 200 = 40 Calculation for ‘T’: (Quantity of T used by Q) – Inventory = (80 × 1) – 80 = 0 Section: Material Planning Systems Subsection: Bills of Materials LO: 14.3 Describe material planning systems.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 5. Using the given information and the beginning inventories from Problem 4, calculate the minimum replenishment time for the 80 Q Models.

Answer: As the longest Item Load Time given is 4 weeks, all Items can be replenished in 4 weeks or less.

Section: Material Planning Systems Subsection: Bills of Materials LO: 14.3 Describe material planning systems.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

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SM 14-3

Reid & Sanders Operations Management, 8th Edition

Solutions Manual

Use the following information for Problems 6 to 10.

Component A B C D E F G H I J K L M N O P Q

Immediate Parent none A A A A B B D D E E E F F H K K

Usage per Parent 2 1 3 2 4 2 3 2 1 5 2 3 6 2 1 2

Lead Time (weeks) 1 2 6 3 2 2 4 2 4 8 1 4 3 3 4 2 3

Beginning Inventory 0 250 500 750 750 3,000 1,000 5,000 5,000 1,000 5,000 2,500 250 2,560 0 500 1,000

6. Flora’s Fabulous Fountains’ (FFF) top product is its Model A. Using the information given, draw the product structure tree for the Model A. Answer:

Note: numbers inside parentheses represent usage per parent while numbers outside the boxes indicate lead time in weeks.

Section: How MRP Works Subsection: [untitled] Introduction LO: 14.4 Demonstrate how MRP works.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

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7. Using the information given, calculate the replenishment time when no beginning inventory exists.

Answer: From Problem 6: The cumulative lead time, the longest planned length of time to complete the end product is 11 weeks (i.e., path J, E, A); equal to 8 + 2 + 1 = 11.

Section: How MRP Works Subsection: [untitled] Introduction LO: 14.4 Demonstrate how MRP works.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes 8. Flora is preparing for her busy season and is building 2,500 Model A fountains. Calculate the gross requirements for each component assuming that there is no beginning inventory.

Answer: Component A B C D E F G H I J K L M N O P Q

Gross Requirement Calculation 2,500 × 2 2,500 × 1 2,500 × 3 2,500 × 2 (2,500 × 2) × 4 (2,500 × 2) × 2 (2,500 × 3) × 3 (2,500 × 3) × 2 (2,500 × 2) × 1 (2,500 × 2) × 5 (2,500 × 2) × 2 (2,500 × 2 × 4) × 3 (2,500 × 2 × 4) × 6 (2,500 × 3 × 3) × 2 (2,500 × 2 × 5) × 1 (2,500 × 2 × 5) × 2

Gross Requirement 2,500 5,000 2,500 7,500 5,000 20,000 10,000 22,500 15,000 5,000 25,000 10,000 60,000 120,000 45,000 25,000 50,000

Section: How MRP Works Subsection: [untitled] Introduction LO: 14.4 Demonstrate how MRP works.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

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Solutions Manual

9. Using the information given and assuming that 2,500 Model A fountains are scheduled for completion, calculate the gross requirements of each component. Use the beginning inventories given.

Answer: Component A B C D E F G H I J K L M N O P Q

Gross Requirement Available Inventory Before Inventory 2,500 0 5,000 250 2,500 500 7,500 750 5,000 750 20,000 3,000 10,000 1,000 22,500 5,000 15,000 5,000 5,000 1,000 25,000 5,000 10,000 2,500 60,000 250 120,000 2,560 45,000 0 25,000 500 50,000 1,000

Gross Requirement After Inventory 2,500 4,750 2,000 6,750 4,250 17,000 9,000 17,500 10,000 4,000 20,000 7,500 59,750 117,440 45,000 24,500 49,000

Section: How MRP Works Subsection: [untitled] Introduction LO: 14.4 Demonstrate how MRP works.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 10. Calculate the minimum replenishment time for the Model A fountains given the beginning inventories.

Answer: Minimum replenishment time is 11 weeks dictated by path J, E, A; equal to 8+2+1 = 11.

Section: How MRP Works Subsection: [untitled] Introduction LO: 14.4 Demonstrate how MRP works.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

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11. Fill in the partially completed inventory record shown here.

Item: AB500 Lot Size Rule: L4L Lead Time: 2 weeks

Parent: None Children: AB501, AB511, AB521 1

2

3 100

4 200

5 100

6 200

7 100

8 200

9 100

10 200

4 200 200 0 200 9 100 100 0

5 100 100 0 100 10 200 200 0

Gross Requirements: Scheduled Receipts: Projected Available: Planned Orders: Gross Requirements: Scheduled Receipts: Projected Available: Planned Orders: Answer: Item: AB500 Lot Size Rule: L4L Lead Time: 2 weeks Gross Requirements: Scheduled Receipts: Projected Available: Planned Orders: Gross Requirements: Scheduled Receipts: Projected Available: Planned Orders:

Parent: None Children: AB501, AB511, AB521 1

2

0 100 6 200 200 0 200

0 200 7 100 100 0 100

3 100 100 0 100 8 200 200 0 200

Section: How MRP Works Subsection: [untitled] Introduction LO: 14.4 Demonstrate how MRP works.

Bloom’s: Application AACSB: Analytic Time on task: 7 minutes 12. Using the planned orders generated in Problem 11, complete inventory records for components AB501, AB511, and AB521. The lot size rule, lead time, and usage information are shown here.

Component AB501 AB511 AB521

Lot Size Rule L4L FOQ = 300 POQ = 2

Time (weeks) 2 2 2

Lead Usage Factor 2 1 2

Beginning Inventory 1,000 650 650

Answer:

© 2023 John Wiley & Sons, Inc. All rights reserved.

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Reid & Sanders Operations Management, 8th Edition

Product: AB501 Week 1 Gross Requirement 200 s Scheduled 0 Receipts Projected Available 1,000 Planned Orders

Solutions Manual

2

3

4

5

6

7

8

9

10

400

200

400

200

400

200

400

0

0

0

0

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400

200

400

0

0

400

200

0

0

0

0

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0

200

200

400

200

400

1

2

3

4

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8

9

10

100

200

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100

200

100

200

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0

0

0

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0

300

0

0

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250

50

250

50

50

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0

0

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0

300

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3

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200

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0

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0

800

Product: AB511 Week Gross Requirement s Scheduled Receipts Projected Available 650 Planned Orders

Product: AB521 Week 1 Gross Requirements 200 Scheduled 0 Receipts Projected Available 450 650 Planned 550 Orders

0

Section: How MRP Works Subsection: [untitled] Introduction LO: 14.4 Demonstrate how MRP works.

Bloom’s: Application AACSB: Analytic Time on task: 12 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

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13. Using the inventory records completed in Problem 12, calculate the average inventory level of AB501, AB511, and AB521. Assume that any remaining inventory after the gross requirements are satisfied is used in the next production cycle.

Answer: From the inventory records in Problem 12: Average Inventory for AB501 = 1,400/10 = 140 units Average Inventory for AB511 = 1,900/10 = 190 units Average Inventory for AB521 = 1,700/10 = 170 units Section: How MRP Works Subsection: [untitled] Introduction LO: 14.4 Demonstrate how MRP works.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 14. Use the planned orders generated in Problem 11. Calculate the average inventory records if the company decides to switch the lot size rule for AB511 and AB521 to lot-for-lot.

Assume the same starting inventory of 650 units for AB511 and AB521. Compare the number of replenishment orders using the new lot size rules.

Answer: Average Inventory for AB501 = 1,400/10 = 140 units Average Inventory for AB511 = 1,200/10 = 120 units Average Inventory for AB521 = 500/10 = 50 units Using the new L4L rule for AB511, average inventory drops but the number of orders doubles from 2 to 4.

Using the new L4L rule for AB521, average inventory drops significantly but the number of orders doubles from 3 to 6.

Note: From what was explained L4L with 650 available, the projected available each period would be 450 for the first period, 50 for the second period and then 0 for the remaining periods as L4L inventory would bring in exactly what was needed. I added a table as well to outline the information.

Therefore, the average inventory would be the following: AB511 would be 2,650/10 = 265 AB251 would be 4,440/10 = 440 Using the new L4L rule for AB511, average inventory drops but the number of orders doubles from 2 to 4.

Using the new L4L rule for AB521, average inventory drops significantly but the number of orders doubles from 3 to 6.

© 2023 John Wiley & Sons, Inc. All rights reserved.

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AB511 Period Gross Requirement Projected available 650 L4L inventory

1 200 450

2 400 50 350

3 200 0 200

4 400 0 400

5 200 0 200

6 400 0 400

7 200 0 200

8 400 0 400

9 0 0 0

1 0 0 0 0

3 400 0 400

4 800 0 800

5 400 0 400

6 800 0 800

7 400 0 400

8 800 0 800

9 0 0 0

1 0 0 0 0

AB521 Period Gross Req Projected available 650 L4L inventory

1 400 250

2 800 0 550

Section: How MRP Works Subsection: [untitled] Introduction LO: 14.4 Demonstrate how MRP works.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

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15. Using the information given, fill in the partially completed inventory record shown here.

Item: AB500 Lot Size Rule: FOQ = 150 Lead Time: 2 weeks

Parent: None Children: AB501, AB511, AB521 1

2

3 100

4 200

5 100

6 200

7 100

8 200

9 100

10 200

Gross Requirements: Scheduled Receipts: Projected Available: Planned Orders: Gross Requirements: Scheduled Receipts: Projected Available: Planned Orders: Answer: Item: AB500 Lot Size Rule: FOQ = 150 Lead Time: 2 weeks Gross Requirements: Scheduled Receipts: Projected Available: Planned Orders: Gross Requirements: Scheduled Receipts: Projected Available: Planned Orders:

Parent: None Children: AB501, AB511, AB521 1

2 0

3 100 150 50

4 200 150 0

5 100 150 50

0 15 0 6 20 0 15 0 0 15 0

150

150

150

150

7

8

9

10

100

200

100

200

150

150

150

150

50

0

50

0

150

150

Section: How MRP Works Subsection: [untitled] Introduction LO: 14.4 Demonstrate how MRP works.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

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16. Using the planned orders generated in Problem 15, complete the inventory record for components AB501, AB511, and AB521. Use the lot size rule, lead time, and usage information given in Problem 12. Indicate any problems that occur.

Answer: Item: AB501: L4L Gross Requirements Scheduled Receipts Projected Available: 1,000 Planned Orders

1

2

3

4

5

6

7

8

9

10

300

300

300

300

300

300

300

300

0

0

0

0

0

200

300

300

300

300

0

0

700

400

100

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0

0

0

0

0

0

200

300

300

300

300

0

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0

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Item: AB511 FOQ: 300 Gross Requirements Scheduled Receipts Projected Available: 650 Planned Orders

1

2

3

4

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150 0

150 0

150 0

150 0

150 300

150 0

150 300

150 0

0 0

0 0

500

350

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50

200

50

200

50

50

50

0

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0

300

0

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0

Item: AB521 POQ: 2 Gross Requirements Scheduled Receipts Projected Available: 650 Planned Orders

1

2

3

4

5

6

7

8

9

10

300 0

300 0

300 550

300 0

300 600

300 0

300 600

300 0

0 0

0 0

350

50

300

0

300

0

300

0

0

0

550

0

600

0

600

0

0

0

0

0

Section: How MRP Works Subsection: [untitled] Introduction LO: 14.4 Demonstrate how MRP works.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

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17. Fill in the partially completed inventory record shown here.

Item: AB500 Lot Size Rule: POQ = 3 Lead Time: 2 weeks

Parent: None Children: AB501, AB511, AB521 1

2

3 100

4 200

5 100

6 200

7 100

8 200

9 100

10 200

Gross Requirements: Scheduled Receipts: Projected Available: Planned Orders: Gross Requirements: Scheduled Receipts: Projected Available: Planned Orders: Answer: Item: AB500 Lot Size Rule: POQ = 3 Lead Time: 2 weeks Gross Requirements: Scheduled Receipts: Projected Available: Planned Orders: Gross Requirements: Scheduled Receipts: Projected Available: Planned Orders:

Parent: None Children: AB501, AB511, AB521 1

2

0 400 6 200 500 300 0

0 0 7 100 0 200 300

3 100 400 300 0 8 200 0 0 0

4 200 0 100 500 9 100 300 200

5 100 0 0 0 10 200 0 0

Section: How MRP Works Subsection: [untitled] Introduction LO: 14.4 Demonstrate how MRP works.

Bloom’s: Application AACSB: Analytic Time on task: 7 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

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18. Using the planned orders generated in Problem 17, complete inventory records for components AB501, AB511, and AB521. Use the lot size rule, lead time, and usage information given in Problem 12, shown again below. Note any problems that may occur.

Component

Lot Size Rule L4L FOQ = 300 POQ = 2

AB501 AB511 AB521

Time (weeks) 2 2 2

Lead Usage Factor 2 1 2

Beginning Inventory 1,000 650 650

Answer: In periods 1 and 2, given the amount of AB521 on hand, the lot size rule, and lead time, there will not be enough of part AB521 to meet demand.

Item: AB501 Gross Requirements

1

2

3

5

6

7

8

9

10

0

0

600

0

0

0

0

4 1,00 0 800

800

0

0

Scheduled Receipts Projected Available: 1000 Planned Orders

0

0

0

0

600

0

0

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200

200

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0

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800

0

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600

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0

0

Item: AB511 Gross Requirements Scheduled Receipts Projected Available: 650 Planned Orders

1 400 0

2 0 0

3 0 0

4 500 300

5 0 0

6 0 0

7 300 300

8 0 0

9 0 0

10 0 0

250

250

250

50

50

50

50

50

50

50

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300

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0

Item: AB521 Gross Requirements Scheduled Receipts Projected Available: 650 Planned Orders

1 800 0

2 0 0

3 0 150

4 5 1,000 0 1,000 0

6 0 0

7 600 600

8 0 0

9 0 0

10 0 0

-150

-150

0

0

0

0

0

0

0

0

150

1,000 0

0

600

0

0

0

0

0

Section: How MRP Works Subsection: [untitled] Introduction LO: 14.4 Demonstrate how MRP works.

Bloom’s: Application AACSB: Analytic Time on task: 10 minutes

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19. The Yankee Machine Shop has the following orders scheduled in Work Center 111 for week 12. Calculate the capacity needed.

Orders

Quantity

LL110 LL118 LL131 LL140

10 25 100 50

Setup Time (hours) 2.0 4.0 6.0 4.0

Run Time per Piece (hours) 1.2 0.4 0.6 0.2

Answer: Orders LL110 LL118 LL131 LL140

Setup + (Run Time × Quantity) 2.0 + 1.2(10) = 14 4.0 + 0.4(25) = 14 6.0 + 0.6(100) = 66 4.0 + 0.2(50) = 14

Total time: 108 hours Capacity needed: 108 hours Section: Capacity Requirement Planning (CRP) Subsection: [untitled] Introduction LO: 14.5 Demonstrate capacity requirements planning (CRP).

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 20. The Yankee Machine Shop currently has three machines working in Work Center 111, eight hours per day, five days per week, a utilization rate of 90 percent, and an efficiency rate of 90 percent. a) Calculate the available capacity. b) Is the available capacity enough to complete the orders given in Problem 19 that are already scheduled in Work Center 111? If not, how much additional capacity is needed?

Answer: a) Available weekly capacity = (3 machines)(8 hours/day)(5 days/week)(0.90 utilization) (0.90 efficiency) = 97.2 hours b) Capacity needed (from Problem 19) = 108 hours Additional capacity of (108 – 97.2) = 10.8 hours are needed.

Section: Capacity Requirement Planning (CRP) Subsection: [untitled] Introduction LO: 14.5 Demonstrate capacity requirements planning (CRP).

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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21. The Yankee Machine Shop has decided to schedule its workforce to work ten hours per day, five days per week. Does this new policy provide enough capacity to complete the orders shown in Problem 19?

Answer: From the capacity requirement of 108 hours in Problem 19 (and detail from Problem 20): Available weekly capacity = (3 machines)(10 hours/day) (5 days/week)(0.90 utilization) (0.90 efficiency) = 121.5 hours This new policy provides adequate capacity.

Section: Capacity Requirement Planning (CRP) Subsection: [untitled] Introduction LO: 14.5 Demonstrate capacity requirements planning (CRP).

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 22. Unfortunately, after extending the workday from 8 hours to 10 hours, the Yankee Machine Shop has noted that efficiency has decreased to 80 percent. Given this new piece of information, is there enough capacity to complete the orders given in Problem 19?

Answer: Available weekly capacity = (3 machines)(10 hours/day)(5 days/week)(0.90 utilization) (0.80 efficiency) = 108 hours There is just enough capacity to complete the orders.

Section: Capacity Requirement Planning (CRP) Subsection: [untitled] Introduction LO: 14.5 Demonstrate capacity requirements planning (CRP).

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 23. In week 13, the Yankee Machine Shop, has the following orders scheduled for Work Center 111. Calculate the capacity needed.

Orders

Quantity

Setup Time (hours) 4.0 6.0 3.0 8.0

MM078 MM118 MM213 MM240

100 250 100 500

Answer: Orders MM078 MM118 MM213 MM240

Setup + (Run Time × Quantity) 4.0 + 0.3(100) = 34 6.0 + 0.1(250) = 31 3.0 + 0.3(100) = 33 8.0 + 0.1(500) = 58

Run Time per Piece (hours) 0.3 0.1 0.3 0.1

Total time: 156 hours

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Capacity needed: 156 hours Section: Capacity Requirement Planning (CRP) Subsection: [untitled] Introduction LO: 14.5 Demonstrate capacity requirements planning (CRP).

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 24. In an effort to increase capacity in Work Center 111 for week 13, Yankee Machine Shop has authorized overtime. The work center will be staffed 12 hours per day for 6 days.

Because of the additional stress on the three machines, it is expected that the utilization rate will drop to 85 percent. The efficiency rate is expected to fall to 80 percent. a) Calculate the capacity available in Work Center 111 for week 13. b) Will this plan provide sufficient capacity to complete the orders given in Problem 23?

If not, what do you recommend be done?

Answer: a) Available weekly capacity = (3 machines)(12 hours/day)(6 days/week)(0.85 utilization)(0.80 efficiency) = 146.88 hours b) There is insufficient capacity with the extra shifts to complete the orders. Additional capacity needs to be procured through acquiring an additional machine (which would add (1 × 12 × 6 × 0.85 × 0.80) 48.96 hours. This would bring total available capacity to 146.88 + 48.96 = 195.84 which would exceed the capacity requirement for the order. Alternatively, a shift could be added, which would also add 48.96 hours, assuming the three existing machines can handle an additional shift.

Section: Capacity Requirement Planning (CRP) Subsection: [untitled] Introduction LO: 14.5 Demonstrate capacity requirements planning (CRP).

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 25. The Gamma Ray Company produces two products, the Gamma Blaster (GB) and the Gamma Disaster (GD). Each product is made from three components: A, B, and C. The Gamma Blaster is made from the following components: A (2), B (3), and C (4). The Gamma Disaster is made from A (3), B (2), and C (1). All other relevant information is provided. Complete the appropriate inventory records.

Answer: Item

On Hand

GB GD A B C

0 0 250 25 0

Scheduled Receipts 0 0 200, period 4

© 2023 John Wiley & Sons, Inc. All rights reserved.

Lot Size MPS Rule L4L 150, period 8 L4L 100, period 6 FOQ = 200 FOQ = 300 L4L

Lead Time 1 2 4 2 3

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Answer: The 150 Gamma Blasters need a planned order for period 7. The 100 Gamma Disasters need a planned order in period 4.

Component A: Period Gross Requirements Scheduled Receipts Projected Available: 250 Planned Orders Component B: Period Gross Requirements Scheduled Receipts Projected Available: 25 Planned Orders Component C: Period Gross Requirements Scheduled Receipts Projected Available: 0 Planned Orders

1

2

3

250

250

250 200

1

2

3

25

25 300

25

1

2

3

0 100

0

0

4 300 200 150

5

6

150

150

4 200 300 125

5

6

125 600

125

4 100 100 0 600

5

6

0

0

7 300 200 50

8

7 450 600 275

8

7 600 600 0

8

50

275

0

Section: Capacity Requirement Planning (CRP) Subsection: [untitled] Introduction LO: 14.5 Demonstrate capacity requirements planning (CRP).

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 26. Using the information in Problem 25, determine the minimum lead time to satisfy a new order for the Gamma Blaster. Determine the minimum lead time required to satisfy a new order for the Gamma Disaster.

Answer: Since Component A has the longest lead time (4 weeks), the minimum lead time for Gamma Blasters is 5 weeks and the minimum lead time for Gamma Disasters is 6 weeks.

Section: Capacity Requirement Planning (CRP) Subsection: [untitled] Introduction LO: 14.5 Demonstrate capacity requirements planning (CRP).

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

© 2023 John Wiley & Sons, Inc. All rights reserved.

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Solutions Manual Chapter 15: Scheduling 1. Jack, the owner and manager of Jack’s Box Company, wants to monitor usage at Work Center 3, which is a bottleneck in the system. He has collected data on the planned and actual input and output.

Input Information (in hours) 4 5 6 40 50 50 45 45 45

60 55

Output Information (in hours) 4 5 6 70 70 70 60 60 60

70 60

Planned input Actual input Deviation Cumulative deviation

Planned output Actual output Deviation Cumulative deviation Backlog 75 hours

7

8 60 60

7

8 70 60

a) Complete the input/output record. b) Describe your concerns based on the results of the input/output analysis.

Answer: a) Planned input Actual input Deviation Cumulative deviation

Input Information (in hours) 4 5 6 40 50 50 45 45 45 5 -5 -5 5 0 -5

Planned output Actual output Deviation Cumulative deviation Backlog 75 hours

Output Information (in hours) 4 5 6 70 70 70 60 60 60 -10 -10 -10 -10 -20 -30 60 45 30

7 60 55 -5 -10

8 60 60 0 -10

7 70 60 -10 -40 25

8 70 60 -10 -50 25

b) If actual output is consistently below planned output, Work Center 3 may not have enough capacity to meet the planned input for Work Center 4. However, the backlog is consistently decreasing in Work Center 3, as their actual output exceeds the actual inputs in periods 4 through 7.

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Section: Basic Scheduling Concepts Subsection: Monitoring Workflow LO: 15.1 Explain basic scheduling concepts.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 2. Since Jack believes that Work Center 3 is his bottleneck, he has asked you to do the following: a) Calculate the percentage of planned output needed to complete the planned inputs [(planned input + backlog/planned output] at Work Center 3. b) Calculate the percentage of planned input that actually happened [(actual input + backlog/planned input + backlog)]. c) Calculate the percentage of available output that was actually accomplished (actual output/planned output) at Work Center 3. d) Given your results, do you believe that Jack is correct in assuming that Work Center 3 is his bottleneck? Justify your answer.

Answer: a) Planned input Backlog 75 hours Planned output Percentage of planned output

4

5

6

7

8

40 60 70 142.9%

50 45 70 135.7%

50 30 70 114.28%

60 25 70 121.4%

60 25 70 121.4%

Total 260 350

The percentage is declining, so if this continues the backlog will be in control for a few more periods. One has to be concerned if Work Center 2 has enough capacity to keep Work Center 3 running efficiently. Additionally, based on the actual output to planned output, one must be concerned if Work Center 3 can efficiently supply Work Center 4. b) Actual input Backlog 75 hours Planned input Percentage of planned input

4

5

6

7

8

45 60 40 105.0%

45 45 50 94.7%

45 30 50 93.8%

55 25 60 94.1%

60 25 60 100%

4 60 70 85.7%

5 60 70 85.7%

6 60 70 85.7%

7 60 70 85.7%

8 60 70 85.7%

Total 250 260

c) Actual output Planned output Percentage of available output

Total 300 350

d) The work center is producing almost 86% of its total planned output. Since both the planned and the actual inputs are below or equal to the actual output, and are below the planned outputs, the backlogs may be a temporary problem.

© 2023 John Wiley & Sons, Inc. All rights reserved.

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Section: Basic Scheduling Concepts Subsection: Monitoring Workflow LO: 15.1 Explain basic scheduling concepts.

Bloom’s: Application AACSB: Analytic Time on task: 8 minutes 3. Based on your concerns, Jack has gathered the following information for Work Center 2, the work center that directly feeds Work Center 3. Complete the input/output analysis of Work Center 2.

Input Information (in hours) 4 5 6 40 50 50 25 35 35

Planned input Actual input Deviation Cumulative deviation

Planned output Actual output Deviation Cumulative deviation Backlog 75 hours

Output Information (in hours) 4 5 6 40 50 50 45 45 45

7

8

60 40

60 40

7

8

60 55

60 60

a) Complete the input/output record. b) Describe your concerns based on the results of the input/output analysis.

Answer: a) Planned input Actual input Deviation Cumulative deviation

Input Information (in hours) 4 5 6 40 50 50 25 35 35 -15 -15 -15 -15 -30 -45

Planned output Actual output Deviation Cumulative deviation Backlog 75 hours

Output Information (in hours) 4 5 6 40 50 50 45 45 45 5 -5 -5 5 0 -5 55 45 35

7 60 40 -20 -65

8 60 40 -20 -85

7 60 55 -5 -10 20

8 60 60 0 -10 0

b) The work center nearly has the capacity to meet planned input if backlogs did not exist. Actual output is consistently less than the planned output especially in periods

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5, 6, and 7, thus indicating that the work center is not producing with the efficiency expected. Work Center 1 has to be studied regarding its output.

Section: Basic Scheduling Concepts Subsection: Monitoring Workflow LO: 15.1 Explain basic scheduling concepts.

Bloom’s: Application AACSB: Analytic Time on task: 8 minutes 4. Jack has asked you to calculate the following for Work Center 2. a) Calculate the percentage of planned output needed to complete the planned inputs [(planned input + backlog/planned output] at Work Center 2. b) Calculate the percentage of planned input that actually happened [(actual input + backlog)/planned input + backlog)] at Work Center 2. c) Calculate the percentage of available output that was actually accomplished (actual output/planned output) at Work Center 2. d) What insights can you offer Jack about Work Centers 2 and 3?

Answer: a) Planned input Backlog 75 hours Planned output Percentage of planned output

4

5

6

7

8

40 55 40 237.5%

50 45 50 190%

50 35 50 170%

60 20 60 133.3%

60 0 60 100%

4

5

6

7

8

25 55 40 84.2%

35 45 50 84.2%

35 35 50 82.3%

4

5

6

7

8

45 45 40 50 112.5% 90%

45 50 90%

55 60 91.7%

60 60 100%

Total 260 260

b) Actual input Backlog 75 hours Planned input Percentage of planned input

40 20 60 75%

40 0 60 66.7%

Total 175 260

c) Actual output Planned output Percentage of available output

Total 250 260

d) Work Center 2 is producing at 96.2% of its total planned output and appears to have a higher productivity level than Work Center 3. However, the ratio of actual input to planned input is lower for Work Center 2.

Section: Basic Scheduling Concepts Subsection: Monitoring Workflow LO: 15.1 Explain basic scheduling concepts.

Bloom’s: Application

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AACSB: Analytic Time on task: 10 minutes

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5. Henri’s Custom Gowns has six jobs waiting to be processed. Each gown is at the beading work center. The job information is shown here.

Job A B C D E F

Job Time (in days)

Days until Due

8 11 6 8 10 5

35 40 25 22 30 15

Job Time at Other Work Centers 8 4 6 12 10 6

Operations Remaining at Other Work Centers 3 5 2 3 5 2

a) Determine the sequence Henri should follow if he uses the SPT (shortest processing time) priority rule. b) Based on the sequence developed in part (a), calculate the following performance measures: makespan, mean job flow time, average number of jobs in the system, mean job lateness, mean job tardiness, and maximum tardiness.

Answer: a) Job sequence: F, C, D, A, E, B b) Makespan = 48 days Mean job flow time = (5 + 11 + 19 + 27 + 37 + 48)/6 = 24.5 days Average number of jobs in the system = 147/48 = 3.06 jobs Job F C D A E B Total

Completion Date 5 11 19 27 37 48

Due Date 15 25 22 35 30 40

Lateness -10 -14 -3 -8 7 8 -20

Tardiness 0 0 0 0 7 8 15

Mean job lateness = 20/6 = 3.33 days Mean job tardiness = 15/6 = 2.50 days Maximum tardiness = 8 days Section: Developing a Schedule of Operations Subsection: Scheduling Performance Measures LO: 15.2 Develop a schedule of operations.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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6. Henri has decided to try a different priority rule. a) Using the data in Problem 5, determine the sequence Henri should follow if he uses the EDD (earliest due date) priority rule. b) Based on the sequence developed in part (a), calculate the following performance measures: makespan, mean job flow time, average number of jobs in the system, mean job lateness, mean job tardiness, and maximum tardiness.

Answer: a) Job sequence: F, D, C, E, A, B b) Job # (EDD Order)

Job Time (days)

F D C E A B

5 8 6 10 8 11

Job Flow Time 5 13 19 29 37 48 151

5+8 13 + 6 19 + 10 29 + 8 37 + 11 TOTAL

Makespan = 48 days Mean job flow time = 151/6 = 25.17 days Average number of jobs in the system = 151/48 = 3.15 jobs Job A B C D E F Total

Completion Date 37 48 19 13 29 5

Due Date 35 40 25 22 30 15

Lateness 2 8 -6 -9 -1 -10 -16

Tardiness 2 8 0 0 0 0 10

Mean job lateness = -16/6 = -2.67 days Mean job tardiness = 10/6 = 1.67 days Maximum tardiness = 8 Section: Developing a Schedule of Operations Subsection: Scheduling Performance Measures LO: 15.2 Develop a schedule of operations.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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7. Henri has heard of the LPT (longest processing time) priority rule and wonders how that would change the sequence of the jobs listed in Problem 5. a) Determine the sequence Henri should follow if he uses the LPT priority rule. b) Based on the sequence developed in part (a), calculate the following performance measures: makespan, mean job flow time, average number of jobs in the system, mean job lateness, mean job tardiness, and maximum tardiness.

Answer: a) Job sequence: B, E, D, A, C, F b) Job # (LPT Order) B E D A C F

Job Time (days) 11 10 8 8 6 5

Job Flow Time 11 21 29 37 43 48 189

11 + 10 = 21 + 8 = 29 + 8 = 37 + 6 = 43 + 5 = TOTAL

Makespan = 48 days Mean job flow time = (11 + 21 + 29 + 37 + 43 + 48)/6 = 31.5 days Average number of jobs in the system = 189/48 =3.94 jobs Job B E D A C F Total

Completion Date 11 21 29 37 43 48

Due Date 40 30 35 22 25 15

Lateness -29 -9 7 2 18 33 22

Tardiness 0 0 7 2 18 33 60

Mean job lateness = -22/6 = 3.67 days Mean job tardiness = 60/6 = 10 days Maximum tardiness = 33 days Section: Developing a Schedule of Operations Subsection: Using Different Priority Rules LO: 15.2 Develop a schedule of operations.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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8. In an effort to be fair to his customers, Henri has decided to use the FCFS (first come, first served) priority rule. Using the job data from Problem 5, assume that the jobs arrive in order—that is, A first, then B, then C. a) Using the data in Problem 5, determine the sequence Henri should follow if he uses the FCFS (first come, first served) priority rule. b) Based on the sequence developed in part (a), calculate the following performance measures: makespan, mean job flow time, average number of jobs in the system, mean job lateness, mean job tardiness, and maximum tardiness.

Answer: a) Job sequence: A, B, C, D, E, F b) Makespan = 48 days Mean job flow time = (8 + 19 + 25 + 33 + 43 + 48)/6 = 29.33 days Average number of jobs in the system – 176/48 = 3.67 Job A B C D E F Total

Job Time 8 11 6 8 10 5

Job Flow Time 8 + 11 = 19 + 6 = 25 + 8 = 33 + 10 = 43 + 5 =

Due Date

Lateness

Tardiness

35 40 25 22 30 15

-27 -21 0 11 13 33 9

0 0 0 11 13 33 57

8 19 25 33 43 48 176

Mean job lateness = 9/6 = 1.50 days Mean job tardiness = 57/6 = 9.50 days Maximum tardiness = 33 days Section: Developing a Schedule of Operations Subsection: Using Different Priority Rules LO: 15.2 Develop a schedule of operations.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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9. Henri recently learned about global priority rules. He is interested in the slack over remaining operations (S/RO) rule. a) Using the data in Problem 5, determine the sequence Henri should follow if he uses the S/RO priority rule. b) Based on the sequence developed in part (a), calculate the following performance measures: makespan, mean job flow time, average number of jobs in the system, mean job lateness, mean job tardiness, and maximum tardiness.

Answer: a) Job

A B C D E F

Job Time

Remaining Job Time at Other w/c

Due Date

Slack Times

8 11 6 8 10 5

8 4 6 12 10 6

35 40 25 22 30 15

35 ˗ (8 + 8) = 40 - (11 + 4) = 25 - (6 + 6) = 22 - (8 + 12) = 30 - (10 + 10) = 15 - (5 + 6) =

Operations Remaining at Other w/c 3 5 2 3 5 2

19 25 13 2 10 4

S/RO

19/(3 + 1) 25/(5 + 1) 13/(2 + 1) 2/(3 + 1) 10/(5 + 1) 4/(2 + 1)

4.75 4.17 4.33 0.50 1.67 1.33

Job sequence: S/RO = Jobs are arranged in order of slack time, lowest to highest Job sequence = D, F, E, B, C, A b) Job D F E B C A Total

Job Time 8 5 10 11 6 8

Completion Date 8+5= 13 + 10 = 23 + 11 = 34 + 6 = 40 + 8 =

8 13 23 34 40 48 166

Due Date 22 15 30 40 25 35

Lateness

Tardiness

-14 -2 -7 -6 15 13 -1

0 0 0 0 15 13 28

Makespan = 48 days Mean job flow time = 166/ 6 = 27.67 days Average number of jobs in the system = 166/48 = 3.46 jobs Mean job lateness = -1/6 = -0.17 days Mean job tardiness = 28/6 = 4.67 days Maximum job tardiness = 15 days Section: Developing a Schedule of Operations Subsection: Using Different Priority Rules LO: 15.2 Develop a schedule of operations.

Bloom’s: Application AACSB: Analytic Time on task: 7 minutes

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10. Joe’s Twenty-four Seven Laundromat has the following jobs waiting to be processed. The first step of the process includes washing and drying the clothes; the second step is pressing the clothing. Joe wants to minimize the amount of time it takes to do all the jobs.

The five jobs waiting to be processed are shown here.

Job A B C D E

Wash and Dry (hours) 6 3 2 7 4

Press (hours) 4 5 3 5 3

a) Using FCFS, assume the jobs arrive in the order shown (A, then B, then C, etc.).

Show the beginning and ending time for each job. b) Calculate the makespan, the mean job flow time, and the average number of jobs in the system.

Answer: a) A, B, C, D, E Wash 0

6 A

0

9 B

A 6

11

18

C B 10

D C 15

22

26

E D 18

23

E 26

Press b) Makespan = 26 hours Mean job flow time = (10 + 15 + 18 + 23 + 26)/5 = 18.4 hours Average number of jobs in the system = 92/26 = 3.54 jobs Section: Developing a Schedule of Operations Subsection: Sequencing Jobs Through Two Work Centers LO: 15.2 Develop a schedule of operations.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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11. Joe thinks that it is probably more efficient to use SPT (shortest processing time) as his priority rule. a) Develop a sequence using SPT based on processing time for the wash and dry operation. b) Calculate the makespan, the mean job flow time, and the average number of jobs in the system.

Answer: a) C, B, E, A, D Wash 0 C 0

2

5 B C

2

9

15

E B 5

A E 10

22

27

D A 13 15

D 19

22

27

Press b) Makespan = 27 hours Mean job flow time = (5 + 10 + 13 + 19 + 27) / 5 = 14.8 hours Average number of jobs in the system = 74/27 = 2.74 jobs Section: Developing a Schedule of Operations Subsection: Sequencing Jobs Through Two Work Centers LO: 15.2 Develop a schedule of operations.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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12. Joe has asked you to develop a sequence that minimizes makespan for the sequence of jobs given in Problem 10. Compare the makespan, mean job flow time, and average number of jobs in the system to your results in Problems 10 and 11.

Answer: Using Johnson’s rule: C, B, D, A, E Wash 0

2 C

5 B

2

18

D

C 0

12 A

B 5

E

D 10

12

22

A 17

18

E 22

2 5

Press C, B, D, A, E Makespan = 25 hours Mean job flow time = (5 + 10 + 17 + 22 + 25) / 5 = 15.8 hours Average number of jobs in the system = 79 / 25 = 3.16 jobs Comparison: (Problem #10) A, B, C, D, E Makespan = 26 hours Mean job flow time = (10 + 15 + 18 + 23 + 26) / 5 = 18.4 hours Average number of jobs in the system = 92 / 26 = 3.54 jobs (Problem #11) C, B, E, A, D Makespan = 27 hours Mean job flow time = (5 + 10 + 13 + 19 + 27) / 5 = 14.8 hours Average number of jobs in the system = 74/27 = 2.74 jobs Makespan is smallest in Problem 12.

Section: Developing a Schedule of Operations Subsection: Scheduling Performance Measures LO: 15.2 Develop a schedule of operations.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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13. Raquel’s Landscaping Company has contracted for several landscaping jobs. Each job requires preparing the areas (identifying the locations and types of plants, preparing the soil, etc.) and then planting the trees, bushes, and shrubs. The expected time for each of the jobs is shown next.

Job R S T U V W

Preparing the Area (days) 3 1 4 8 6 4

Planting (days) 2 3 5 5 4 3

a) Using FCFS (first come, first served), assume the jobs arrive in the order shown (R, then S, then T, etc.). Show the beginning and ending time for each job. b) Calculate the makespan, the mean job flow time, and the average number of jobs in the system.

Answer: a) R, S, T, U, V, W

b) Makespan = 29 days Mean job flow time = (5 + 8 + 13 + 21 + 26 + 29)/6 = 17 days Average number of jobs in the system = 102/29 = 3.52 jobs Section: Developing a Schedule of Operations Subsection: Sequencing Jobs Through Two Work Centers LO: 15.2 Develop a schedule of operations.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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14. Raquel is concerned with efficiency. She believes it is probably more efficient to use SPT (shortest processing time) as her priority rule. (a) Develop a sequence using SPT based on processing time for preparing the area. (b) Calculate the makespan, the mean job flow time, and the average number of jobs in the system.

Answer: a) S, R, T, W, V, U

b) Makespan = 31 days Mean job flow time = (4 + 6 + 13 + 16 + 22 + 31)/6 = 15.33 days Average number of jobs in the system = 92/31 = 2.97 jobs Section: Developing a Schedule of Operations Subsection: Sequencing Jobs Through Two Work Centers LO: 15.2 Develop a schedule of operations.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 15. Raquel has asked you to develop a sequence that minimizes makespan for the sequence of jobs given in Problem 13. Compare the makespan, mean job flow time, and average number of jobs in the system to your results in Problems 13 and 14.

Answer: S, T, U, V, W, R

Make span = 28 days Mean job flow time = (4 + 10 + 18 + 23 + 26 + 28)/6 = 18.17 days Average number of jobs in the system = 109/28 = 3.89 jobs Johnson’s Rule has produced the minimum makespan. However, this has the highest mean job flowtime and the highest average number of jobs in the system.

Section: Developing a Schedule of Operations Subsection: Sequencing Jobs Through Two Work Centers LO: 15.2 Develop a schedule of operations.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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16. Barb’s Beach Bar operates seven days per week. Barb uses only full-time employees and wants each employee to have two consecutive days off each week. She believes that she needs a minimum of three employees on Monday, Tuesday, Wednesday, Thursday, and Sunday. On Friday and Saturday, she believes that she needs six employees. Develop a workforce schedule for Barb.

Answer: Employe e 1 2 3 4 5 6

Monda y OFF X OFF X X OFF

Tuesda y OFF X OFF X X OFF

Wednesda y X OFF X OFF OFF X

Thursda y X OFF X OFF OFF X

Frida y X X X X X X

Saturda y X X X X X X

Sunda y X X X X X X

Various answers may result given the ties that occur at the start of the exercise.

Section: Scheduling Issues for Service Organizations Subsection: Sequencing Employees LO: 15.4 Describe scheduling issues for service organizations.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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17. Next week is a three-day weekend. Barb believes that she will need a minimum of six employees on Friday, Saturday, and Sunday. The other days will still need a minimum of three workers. Remember that Barb wants each employee to have two consecutive days off each week. Develop a workforce schedule for the holiday weekend.

Answer: Employe e 1 2 3 4 5 6

Monda y OFF X OFF X OFF X

Tuesda y OFF X OFF X OFF X

Wednesda y X OFF X OFF X OFF

Thursda y X OFF X OFF X OFF

Frida y X X X X X X

Saturda y X X X X X X

Sunda y X X X X X X

Various answers may result given the ties that occur at the start of the exercise.

Section: Scheduling Issues for Service Organizations Subsection: Sequencing Employees LO: 15.4 Describe scheduling issues for service organizations.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 18. Marvin’s Beach Cleaners is responsible for keeping the beach clean. Marvin estimates

that he needs a minimum of two people on Monday; three people on Tuesday and Sunday; four people on Wednesday and Thursday; and five people on Friday and Saturday. Contractually, Marvin is required to give each employee two consecutive days off during the week. Develop a workforce schedule for Marvin to use.

Answer: Employe e 1 2 3 4 5 6

Monda y OFF OFF X X OFF X

Tuesda y OFF OFF X X OFF X

Wednesda y X X OFF X X OFF

Thursda y X X OFF X X OFF

Frida y X X X X X X

Saturda y X X X OFF X X

Sunda y X X X OFF X X

Various answers may result from ties that occur after the first four employees are assigned days off.

Section: Scheduling Issues for Service Organizations Subsection: Sequencing Employees LO: 15.4 Describe scheduling issues for service organizations.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes © 2023 John Wiley & Sons, Inc. All rights reserved.

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19. Marvin is preparing for the upcoming three-day weekend. He estimates that he will need three additional workers on Sunday of that week. Using the requirements given in Problem 18 for the other days of the week, develop a workforce schedule for Marvin to use.

Answer: Employe e 1 2 3 4 5 6

Monda y OFF OFF X OFF X X

Tuesda y OFF OFF X OFF X X

Wednesda y X X OFF X OFF X

Thursda y X X OFF X OFF X

Frida y X X X X X OFF

Saturda y X X X X X OFF

Sunda y X X X X X X

Various answers may result from ties that occur after the first four employees are assigned days off.

Section: Scheduling Issues for Service Organizations Subsection: Sequencing Employees LO: 15.4 Describe scheduling issues for service organizations.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 20. Cathy’s Coney Islands operates seven days per week. Demand is relatively constant during the week and tails off on the weekend. She estimates that she needs six employees Monday through Friday, and three employees on Saturday and Sunday. She is committed to giving each employee two consecutive days off during the week. Develop a workforce schedule for Cathy.

Answer: Employe e 1 2 3 4 5 6 7 8

Monda y X X X X OFF X X OFF

Tuesda y X X X X OFF X X X

Wednesda y X X X X X OFF X X

Thursda y X X X X X OFF X X

Frida y X X X X X X OFF X

Saturda y OFF OFF OFF OFF X X OFF X

Sunda y OFF OFF OFF OFF X X X OFF

Various answers may result from ties that occur after the first four employees are assigned days off. Unfortunately, if Cathy remains committed to offering all employees five consecutive days of work and two consecutive days off, then she will have one extra employee (7) on four of the five weekdays. She may need to look for another approach.

Section: Scheduling Issues for Service Organizations

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Subsection: Sequencing Employees LO: 15.4 Describe scheduling issues for service organizations.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 21. Cathy’s Coney Islands has just purchased new equipment. Cathy believes the improved efficiency will reduce the number of people needed Monday through Friday down to five.

She does not believe she can ever have fewer than three persons working, so the weekend requirements remain the same. Develop a workforce schedule for Cathy.

Answer: Employe e 1 2 3 4 5 6 7

Monda y X X X OFF OFF X X

Tuesda y X X X X OFF X X

Wednesda y X X X X X OFF X

Thursda y X X X X X OFF X

Frida y X X X X X X OFF

Saturda y OFF OFF OFF X X X OFF

Sunda y OFF OFF OFF OFF X X X

Various answers may result from ties that occur after the first three employees are assigned days off. Unfortunately, if Cathy still remains committed to offering all employees five consecutive days of work and two consecutive days off, then she will still have one extra employee (6) on four of the five weekdays. She may need to look for another approach.

Section: Scheduling Issues for Service Organizations Subsection: Sequencing Employees LO: 15.4 Describe scheduling issues for service organizations.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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22. Bill’s Bar & Grill is open seven nights a week. Business is busiest on Thursday, when there is a concert in the park across the street. Bill wants a workforce schedule that allows each employee two consecutive days off each week. He believes that he needs four employees every day except on Thursday, when he believes he needs six employees.

Develop a workforce schedule for Bill.

Answer: Employe e 1 2 3 4 5 6 7

Monda y OFF X X X OFF X X

Tuesda y OFF X OFF X OFF X X

Wednesda y X X OFF X X OFF X

Thursda y X X X X X OFF X

Frida y X OFF X X X X OFF

Saturda y X OFF X OFF X X OFF

Sunda y X X X OFF X X X

Various answers may result given the ties that occur at the start of the exercise.

Section: Scheduling Issues for Service Organizations Subsection: Sequencing Employees LO: 15.4 Describe scheduling issues for service organizations.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 23. During the winter, no concerts are given in the park. Bill believes that he needs a minimum of four employees every day of the week. Develop a workforce schedule for Bill that allows each employee two consecutive days off each week. Compare the number of workers he needs during the winter to the number of workers needed during the concert season.

Answer: Employe e 1 2 3 4 5 6

Monda y OFF X X X OFF X

Tuesda y OFF X X X OFF X

Wednesda y X OFF X X X OFF

Thursda y X OFF X X X OFF

Frida y X X OFF X X X

Saturda y X X OFF OFF X X

Sunda y X X X OFF X X

Various answers may result given the ties that occur at the start of the exercise. This winter schedule has one fewer employee, with 2 days out of 7 having 5 employees at work.

Section: Scheduling Issues for Service Organizations Subsection: Sequencing Employees

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LO: 15.4 Describe scheduling issues for service organizations.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 24. Student Premier Painters (SPP) has six house painting jobs in a particular neighborhood.

The houses vary in size, condition, and painting requirements, but each house must be prepped (cleaned, sanded, and primed) first, before it is painted. The relevant information is given.

Days Houses Arnold (A) Blake (B) Coffman (C) Downs (D) Eckels (E) Farber (F)

Prep 4 3 2 3 6 5

Paint 3 6 5 2 4 7

a) Sequence the painting jobs to minimize makespan. b) Calculate the mean job flow time associated with this sequence. c) Calculate the average number of jobs in the system with this sequence.

Answer: a) C, B, F, E, A, D 2 Prepare Paint

C

5

10

B

16

F C

B 7

20

E

A

23 D

F 13

E 20

A 24

27

D 29

b) Mean job flow time = (7 + 13 + 20 + 24 + 27 + 29) / 6 = 20.0 days c) Average number of jobs in the system = 120 / 29 = 4.14 jobs Section: Developing a Schedule of Operations Subsection: Sequencing Jobs Through Two Work Centers LO: 15.2 Develop a schedule of operations.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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25. Using the job information in Problem 24, develop a sequence based on the shortest processing time priority rule. Prioritize based on prep time. In the case of ties, select the house that has the shortest paint time. a) Calculate the makespan. b) Calculate the mean job flow time associated with this sequence. c) Calculate the average number of jobs in the system with this sequence.

Answer: a) C, D, B, A, F, E 2 Prepare Paint

C

5 D

8

12

B C

A D

7

17 F

B 9

23

29

E A

15

F 18

E 25

29

Makespan = 29 days b) Mean job flow time = (7 + 9 + 15 + 18 + 25 + 29) / 6 = 17.17 days c) Average number of jobs in the system = 103 / 29 = 3.55 jobs Section: Developing a Schedule of Operations Subsection: Sequencing Jobs Through Two Work Centers LO: 15.2 Develop a schedule of operations.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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26. Using the information in Problem 24, develop a sequence based on the longest processing time priority rule. Prioritize based on prep time. In the case of ties, select the house that has the longest paint time. a) Calculate the makespan. b) Calculate the mean job flow time associated with this sequence. c) Calculate the average number of jobs in the system with this sequence.

Answer: a) E, F, A, B, D, C Prepare

Paint a) Makespan = 34 days b) Mean job flow time = (10 + 18 + 21 + 27 + 29 + 34) / 6 = 23.17 days c) Average number of jobs in the system = 139 / 34 = 4.09 jobs Section: Developing a Schedule of Operations Subsection: Sequencing Jobs Through Two Work Centers LO: 15.2 Develop a schedule of operations.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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Solutions Manual Chapter 16: Project Management Use the following project information for Problems 1 and 2.

Activity A B C D E F G

Activity Time (weeks) 3 4 2 5 4 3 2

Immediate Predecessor(s) none A B B C D E, F

1. Construct a network diagram using AON notation.

Answer:

Section: Project Management Concepts Subsection: Step 2: Diagram the Network LO: 16.2 Describe project management concepts.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes 2. Using the network diagram constructed in Problem 1, a) Calculate the completion time for the project. b) Determine which activities are included on the critical path.

Answer: a) There are only two possible paths for this project: A, B, C, E, G and A, B, D, F, G Duration for the first path: 3 + 4 + 2 + 4 + 2 = 15 weeks Duration for the second path: 3 + 4 + 5 + 3 + 2 = 17 weeks b) Critical path: The critical path is therefore the second path, with the longest duration: A, B, D, F, G Section: Project Management Concepts Subsection: Step 3: Estimate the Project’s Completion Time LO: 16.2 Describe project management concepts.

Bloom’s: Application AACSB: Analytic Time on task: 2 minutes

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3. Jack’s Floating Banana Party Company is planning to add a new party vessel for the upcoming season. Jack has identified several activities that must be finished before the start of the season. Using the following information, Activity A B C D E F

Activity Time (weeks) 6 5 3 3 6 9

Immediate Predecessor(s) none none A B C, D D

a) Draw the network diagram for this project. b) Identify the critical path. c) Calculate the expected project length.

Answer: a)

b) The paths are either: A, C, E or B, D, F and B, D, E The respective durations are: 6 + 3 + 6 = 15 weeks; 5 + 3 + 9 = 17 weeks or 5 + 3 + 6 = 14 weeks.

The critical path is therefore B, D, F. c) Expected project length = critical path = 5 + 3 + 9 = 17 weeks Section: Project Management Concepts Subsection: Step 3: Estimate the Project’s Completion Time LO: 16.2 Describe project management concepts.

Bloom’s: Application AACSB: Analytic Time on task: 4 minutes

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Use the following project information for Problems 4 to 8.

Activity A B C D E F G H I J

Optimistic Time Estimate (weeks) 2 2.5 5 3 4 2 4 6 4 2

Most Likely Time Estimates (weeks) 3 4 7 5 7 5 5 6 8 3

Pessimistic Time Estimates (weeks) 5 5 8 7 14 8 6 8 11 3

Immediate Predecessor(s) none A A B, C B, C D E F G H, I

4. Using the information given, construct a network diagram using AON notation.

Answer:

Section: Project Management Concepts Subsection: Step 2: Diagram the Network LO: 16.2 Describe project management concepts.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 5. Using the information given, calculate the expected time for each of the project activities.

Answer: Expected time = (Optimistic time + {4 × Most likely time} + Pessimistic time) / 6 Activity Expecte d Time

A 3.17

B 3.92

C 6.83

D 5.0

E 7.67

F 5.0

G 5.0

H 6.33

I 7.83

J 2.83

Selected Calculations: A: (2 + {4 × 3} + 5)/6 = 3.1667 B: (2.5 + {4 × 4} + 5)/6 = 3.9167 C: (5 + {4 × 7} + 8)/6 = 6.8333

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D: (3 + {4 × 5} + 7)/6 = 5 Section: Project Management Concepts Subsection: Step 3(b): Probabilistic Time Estimates LO: 16.2 Describe project management concepts.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 6. Using the information given, calculate the variance for each of the project activities.

Answer: Variance = (p-o)/6)2 Activity Variance

A 0.25

B 0.17

C 0.25

D 0.44

E 2.78

F 1.0

G 0.11

H 0.11

I 1.36

J 0.03

Selected Calculations: A: ((5 - 2)/6)2 = 0.25 B: ((5 - 2.5)/6)2 = 0.17 C: ((8 - 5)/6)2 = 0.25 D: ((7 - 3)/6)2 = 0.44 Section: Project Management Concepts Subsection: Step 3(b): Probabilistic Time Estimates LO: 16.2 Describe project management concepts.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 7. Using your results from Problems 4 and 5, a) Calculate the completion time for this project. b) Identify the activities included on the critical path of this project.

Answer: a) The project completion time is equal to the critical path using expected time. There are four possible paths for this project, with completion times of: A-B-D-F-H-J = 3.2 + 3.9 + 5.0 + 5.0 + 6.3 + 2.8 = 26.20 weeks A-B-E-G-I-J = 3.2 + 3.9 + 7.7 + 5.0 + 7.8 + 2.8 = 30.40 weeks A-C-D-F-H-J = 3.2 + 6.8 + 5.0 + 5.0 + 6.3 + 2.8 = 29.10 weeks A-C-E-G-I-J = 3.2 + 6.8 + 7.7 + 5.0 + 7.8 + 2.8 = 33.33 weeks The project completion time is 33.33 weeks. b) Critical path: A, C, E, G, I, J Section: Project Management Concepts Subsection: Step 3(b): Probabilistic Time Estimates LO: 16.2 Describe project management concepts.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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8. Using your results from Problem 6, a) Calculate the probability that the project will be completed in 32 weeks. b) Calculate the probability that the project will be completed in 35 weeks.

Answer: a) The variance of the critical path equal to the sum of the variances for each activity in the critical path = .25 + .25 + 2.78 + .11 + 1.36 + .03 = 4.78.

The standard deviation is then = √4.78 = 2.19 32 - 33.33 P(project £ 32) = P(critical path £ 32) = P( z £ ) = P( z £ - 0.61) = 0.2709 4.78 b)

æ 35 - 33.33 ö P ( project £ 35 ) = P ( critical path £ 35 ) = P ç z £ ÷ = P ( z £ 0.76 ) = 0.7564 4.78 ø è

Section: Estimating the Probability of Completion Dates Subsection: [untitled] Introduction LO: 16.3 Estimate the probability of a project being completed by a specific due date.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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Use the information provided in Table 16-14 and the network diagram in Figure 16-20 to solve Problems 9 to 12.

Activity

Normal Time (weeks)

Normal Cost ($)

A B C D E F G H

3 4 6 6 6 5 7 3

500 600 800 1,500 1,750 200 1500 500

Table 16-14 Crash Time Crash (weeks) Cost ($)

2 3 3 4 4 5 5 3

Maximum weeks Reduced

Crash Cost per Week ($)

1 1 3 2 2 0 2 0

400 100 166.67 325 250 0 150 0

900 700 1,300 2,150 2,250 1,200 1,800 500

Figure 16-20. AON network diagram

9. Using the information given, a) Calculate the completion time of the project. b) Identify the activities on the critical path.

Answer: a) There are four possible paths for this project, with respective completion times: A-B-E-G-H = 3 + 4 + 6 + 7 + 3 = 23 weeks A-C-E-G-H = 3 + 6 + 6 + 7 + 3 = 25 weeks A-C-F-G-H = 3 + 6 + 5 + 7 + 3 = 24 weeks A-D-F-G-H = 3 + 6 + 5 + 7 + 3 = 24 weeks Project completion time = longest completion time = 25 weeks b) Critical path: A, C, E, G, H Section: Project Management Concepts Subsection: Step 3: Estimate the Project’s Completion Time LO: 16.2 Describe project management concepts.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes

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10. Using the information given and the project completion time calculated in Problem 9(a), reduce the completion time of the project by 3 weeks in the most economical way.

Answer: Reduce G by 2 weeks (2 × $150 × $300) and C by 1 week ($166.67) at a total cost of $466.67.

Section: Reducing Project Completion Times Subsection: Crashing Projects LO: 16.4 Demonstrate how to reduce a project’s completion time.

Bloom’s: Application AACSB: Analytic Time on task: 4 minutes 11. Using the information given and the project completion time calculated in Problem 9(a), reduce the completion time of the project by 5 weeks in the most economical way.

Answer: Reduce G by 2 weeks (2 × $150 = $300), C by 1 week ($166.67). At this point, the critical path changes because the critical paths become A, C, E, G, H and A, D, F, G, H, both with a maximum project time of 22 weeks. Since A is common to all paths, reduce A by 1 week ($400). The critical paths are now at 21 weeks, so C and D can each be reduced by 1 week at a cost of $166.67 and $325 respectively to bring A, C, E, G, H to A, D, F, G, H, which balances all paths at 20 weeks. Total cost is $1,358.33 Section: Reducing Project Completion Times Subsection: Crashing Projects LO: 16.4 Demonstrate how to reduce a project’s completion time.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 12. Using the information given and the project completion time calculated in Problem 9(a), calculate the minimum time for completing the project possible.

Answer: Minimum project completion time = 19 weeks.

Reduce G by 2 weeks, C by 1 week. At this point, the critical path changes because the critical paths become A, C, E, G, H and A, D, F, G, H, both with a maximum project time of 22 weeks. Since A is common to all paths, reduce A by 1 week. The critical paths are now at 21 weeks, so C and D can each be reduced by 1 week to bring A, C, E, G, H to A, D, F, G, H, which balances all paths at 20 weeks. There is the opportunity to remove 1 week from tasks B, C, and D bringing the project completion time to 19 weeks.

Section: Reducing Project Completion Times Subsection: Crashing Projects LO: 16.4 Demonstrate how to reduce a project’s completion time.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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Use Figure 16-21 and the following project data for the next two problems.

Figure 16-21. Network diagram

Activity A B C D E F G H

Optimistic Time (weeks) 8 4 4 6 4 6 4 3

Most Likely Time (weeks) 10 10 5 8 7 7 8 3

Pessimistic Time (weeks) 12 16 6 10 12 9 12 3

13. Using the information given, a) Calculate the expected time for each of the project activities. b) Calculate the variance for each of the project activities. c) Evaluate the connected paths through the diagram to determine the expected project completion time.

Answer: a) Activity O

A 10

B 10

C 5

D 8

E 7.33

F 7.17

G 8

H 3

G 1.78

H 0

b) Activity Variance c)

A 0.44

B 4

C 0.11

D 0.44

E 1.78

F 0.25

A, B, D, F, H = 38.17 A, B, D, G, H = 39 A, C, E, F, H = 32.5 A, C, E, G, H = 33.33 Critical Path = A, B, D, G, H Expected project completion time = 39 weeks

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Section: Project Management Concepts Subsection: Step 3(b): Probabilistic Time Estimates LO: 16.2 Describe project management concepts.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 14. Using the information given and expected project completion time from Problem 13(c), a) Calculate the probability of completing the project in 36 weeks. b) Calculate the probability of completing the project in 40 weeks.

Answer: a) Sum of the critical path variance = 0.44 + 4.0 + 0.44 + 1.78 + 0 = 6.66 project variance 36 - 39 P(project £ 36) = P(critical path £ 36) = P(z £ ) = P(z £ -1.16) = 0.1230 6.66 b)

æ 40 - 39 ö P ( project £ 40 ) = P ( critical path £ 40 ) = P ç z £ ÷ = P ( z £ 0.39 ) = 0.6517 6.66 ø è

Section: Estimating the Probability of Completion Dates Subsection: [untitled] Introduction LO: 16.3 Estimate the probability of a project being completed by a specific due date.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes 15. The accounting department at Northeast University is offering a combined five-year B.S./M.S. in accounting. The senior accounting professor has identified the project activities and any precedence relationships, as shown in Table 16-15. Three time estimates for each activity are included. a) Develop a network diagram for this project using AON notation. b) Calculate the expected time for each of the project activities. c) Identify the critical path for the project. d) Calculate the expected project completion time.

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Activit y

Description of Activity

A

Design general curriculum requirements Develop program brochure Identify prospective students Develop advertising campaign Design specific curriculum content Send brochure and student application Evaluate applications

Solutions Manual

Table 16-15 Immediate Predecessor(s )

Optimistic Time Estimates (weeks)

Most Likely Time Estimates (weeks)

Pessimistic Time Estimates (weeks)

None

4

8

16

A

2

3

6

None

3

6

9

B, C

4

7

10

A

8

16

20

D

2

3

4

F

2

4

6

G

1

2

3

H

1

1

7

H

1

2

3

K

Accept students, notify students Schedule rooms for classes Designate professors to teach courses Select texts for courses

J

3

5

7

L

Order and receive texts

K

6

8

17

B C D E F G H I J

Answer: a)

b) Activity Expecte d Time

A 8.67

B 3.33

C 6

D 7

E 15.33

F 3

G 4

H 2

I 2

J 2

K 5

L 9.17

Example for expected time calculations: Activity A = ((O + (4M) + P))/6 4 + (4 × 8) + 16 = 52/6 = 8.67

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c) Critical path: A, B, D, F, G, H, J, K, L d) Expected project completion time = (8.67 + 3.33 + 7 + 3 + 4 + 2 + 2 + 5 + 9.17) = 44.17 weeks Section: Project Management Concepts Subsection: Step 3(b): Probabilistic Time Estimates LO: 16.2 Describe project management concepts.

Bloom’s: Application AACSB: Analytic Time on task: 8 minutes 16. The dean of the business school wants to start offering this program starting 32 weeks from now. Using the information provided in Problem 15, a) Calculate the probability of the program starting on time. b) If the dean needs a 95 percent probability of being done on time, how long can the expected project duration be?

Answer: a) Variance for the critical path: Activity

Optimisti c Time

Pessimisti c Time

Variance ((P-O)/6)2

A B D F G H J K L

4 2 4 2 2 1 1 3 6

16 6 10 4 6 3 3 7 17

4.00 0.44 1.00 0.11 0.44 0.11 0.11 0.44 3.36 10.01

P(project £ 32) = P(critical path £ 32) = P(z £ b) z=

32 - 44.17 ) = P(z £ -3.84) = 0 10.01

Xμσ

(Assuming the variance does not change) 32 - EPD 1.645= 10.01 Expected project duration (EPD) = 26.79 weeks The dean needs to reconsider the starting date. The optimistic times listed in Table 1615 sum to 37 weeks.

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Section: Estimating the Probability of Completion Dates Subsection: [untitled] Introduction LO: 16.3 Estimate the probability of a project being completed by a specific due date.

Bloom’s: Application AACSB: Analytic Time on task: 5 minutes Use the following information for Problems 17 to 19.

Activity

Immediate Predecessor(s)

A B C D E F G H I J K L

none none A, B A B C, D, E F F F H, I J K

Normal Time (weeks) 10 4 6 7 14 3 4 2 5 2 5 10

Normal Cost ($) 16,000 6,000 12,000 7,000 28,000 4,500 7,200 5,000 15,000 7,000 10,000 24,000

Crash Time (weeks) 8 3 3 5 12 3 3 2 4 2 4 10

Crash Cost ($) 20,000 9,000 24,000 10,000 34,000 4,500 9,000 5,000 18,000 7,000 11,000 24,000

17. Draw the network diagram. Determine the normal time it will take to complete this project. Determine the critical path for the project. Calculate the cost of completing the project in normal time.

Answer:

Normal time = 43 weeks Critical path = B, E, F, I, J, K, L Cost of completing the project in normal time = $141,700 Section: Project Management Concepts Subsection: Step 3: Estimate the Project’s Completion Time LO: 16.2 Describe project management concepts.

Bloom’s: Application AACSB: Analytic

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Time on task: 5 minutes 18. Determine the absolute minimum time it will take to complete the project if the project is crashed as far as possible. Calculate the cost associated with this approach to the project.

Answer: The shortest critical path becomes 38 weeks. From the critical path in Problem 17 (B, E, F, I, J, K, L), reducing K by one week, I by one week, B by one week, and E by two weeks reduces the completion time to 38 weeks. However, A, D, F, I, J, K, L would now be the critical path, so A or D must also be reduced by two weeks. A is more expensive to crash, so it makes sense to reduce D. However, after reducing D by 1 day, the path ADFIJKL would be 39 days as will the path ACFIJKL. Reducing A by 1 day will reduce both of these paths to 38 days.

Total cost = $1,000 + $3,000 + $3,000 + $6,000 + $1,500 + $2,000 = $16,500 Section: Reducing Project Completion Times Subsection: Crashing Projects LO: 16.4 Demonstrate how to reduce a project’s completion time.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes 19. If the company wants to complete the project in 40 weeks, which activities should be crashed? Calculate the additional cost incurred to complete the project in 40 weeks.

Answer: If the company wants to complete the project in 40 weeks, K, I, and B should be reduced by one week.

Additional cost incurred = $1,000 + $3,000 + $3,000 = $7,000 Section: Reducing Project Completion Times Subsection: Crashing Projects LO: 16.4 Demonstrate how to reduce a project’s completion time.

Bloom’s: Application AACSB: Analytic Time on task: 3 minutes

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Solutions Manual Supplement A: Spreadsheet Modeling: An Introduction Problem 1: See File Prb_A_01.xls Answers to Questions A. See “BaseCase” worksheet in file Prb_A_01.xls. Base Case profit is ˗$30,000, i.e., a $30,000 loss.

B. Using Goal Seek (“set cell” = B16, “by changing” B9, “to value” 0), the breakeven quantity is 35,000 units.

C. See “BaseCase” worksheet.

D. If sales increase by 35%, to 20,000 × 1.35 = 27,000 units, there will still be a loss of $16,000.

Alternately, if variable costs can be decreased to 90% of the current level, to $18 × 0.90 = $16.20 (with sales remaining at 20,000), then there will be a $6,000 profit. Therefore, given a choice between these two activities, one would choose the alternative which would reduce variable cost.

E. If sales increase to 27,000 units and fixed costs increase by $5,000, there will be a $21,000 loss. Alternately, if variable costs decrease to $16.20 but fixed costs increase by $10,000, there will be a $4,000 loss. Therefore, the variable-cost reduction alternative is better, even though it still results in a loss. However, note that both of these alternatives are better than the base case results.

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual

Section: The Spreadsheet Modeling Process Subsection: all subsections LO: 2 Describe the spreadsheet modeling process.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 15 minutes

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Problem 2: See File Prb_A_02.xls Answers to Questions A. See “BaseCase” worksheet in the file Prb_A_02.xls. At this point, any value for the Sales Price can be entered.

B. Using Goal Seek (see “SolvedCase” worksheet), the breakeven Sales Price is $37.08.

Section: The Spreadsheet Modeling Process Subsection: all subsections LO: 2 Describe the spreadsheet modeling process.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 15 minutes

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Problem 3: See File Prb_A_03.xls Answers to Questions A. See “BaseCase” worksheet in the file Prb_A_03.xls.

B. For an annual demand of 300,000, the total cost for outsourcing is $795,000, and for insourcing is $750,000. Therefore, there is a $45,000 cost advantage for insourcing at this volume.

C. Using Goal Seek, setting the difference cell (B12) to zero, the indifference point is found to be 240,000 units. Above this volume, insourcing costs less than outsourcing, and vice versa.

D. See “BaseCase” worksheet.

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Page 4 of 14

Reid/Sanders, Operations Management, 8th Edition Solutions Manual Section: The Spreadsheet Modeling Process Subsection: all subsections LO: 2 Describe the spreadsheet modeling process.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 15 minutes

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual

Problem 4: See File Prb_A_04.xls Answers to Questions A. See “BaseCase” worksheet in the file Prb_A_04.xls.

B. For 1,450 purchase orders, insourcing is $500 less expensive. Insourcing costs a total of $106,750 while outsourcing costs $107,250.

C. Using Goal Seek, setting the difference cell (B12) to zero, the indifference point is found to be 1,500 purchase orders. At this volume, the cost of each alternative is $107,500. Above this volume, insourcing costs less than outsourcing, and vice versa.

D. See “BaseCase” worksheet.

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual

Section: The Spreadsheet Modeling Process Subsection: all subsections LO: 2 Describe the spreadsheet modeling process.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 15 minutes

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual

Problem 5: See File Prb_A_05.xls Answers to Questions A. See “BaseCase” worksheet in the file Prb_A_05.xls. Location 3 is the optimal location, with a total weighted score of 340. Location 2 has a score of 315, and Location 1 has a score of 310.

B. See “BaseCase” worksheet.

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual

Section: The Spreadsheet Modeling Process and Multiple-Criteria Decision Making Subsection: all subsections LO: 2 Describe the spreadsheet modeling process.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 10 minutes

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual

Problem 6: See File Prb_A_06.xls Answers to Questions A. See “Model” worksheet in the file Prb_A_06.xls. The forecast for week 7 is 360.3, or 360.

B. Week 4 has the largest error, 36.7 units. Week 6 has the smallest error, -9.7 units. Positive errors imply that the actual is larger than the forecast (i.e., the forecast was too low). Negative errors mean the forecasted value was higher than the actual value.

Section: The Spreadsheet Modeling Process Subsection: all subsections LO: 2 Describe the spreadsheet modeling process.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 15 minutes

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Problem 7: See File Prb_A_07.xls Answers to Questions A. See “Model” worksheet in the file Prb_A_07.xls.

B. Since the error for each week is the actual value minus the forecasted value, a positive error implies that the forecasted value is less than the actual, and a negative error means that the forecasted value is more than the actual.

C. See Column E in the model. Here, the mean absolute deviation (MAD) is 24.4 cheeseburgers.

This is the average forecast error.

D. See Column F in the model. The RMSE is 28.1 cheeseburgers.

E. See Column G in the model. The MAPE is 7.1%. Thus, the forecast error is, on average, 7.1% of the actual value.

Section: The Spreadsheet Modeling Process and Multiple-Criteria Decision Making Subsection: all subsections LO: 2 Describe the spreadsheet modeling process.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 15 minutes

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual

Problem 8: See File Prb_A_08.xls Answers to Questions A. See "BaseCase" worksheet in file Prb_A_08.xls.

B. For an order quantity of 1,300 heat sinks, the average inventory level will be 650, they will place 4 orders per year, the total annual holding cost will be $1,950, the annual ordering cost will be $300, and the total annual cost will be $2,250.

C. For an order quantity of 650 heat sinks, the total annual cost is $1,575, and for an order quantity of 1,950 heat sinks, the total annual cost is $3,125.

D. See the “BaseCase” worksheet for the data table and XY chart. From the data table, the Economic Order Quantity (EOQ) is 500 or 525 (to a precision of 25 heat sinks; the more precise value of the EOQ is 510). At this order quantity, the annual cost is $1,530.

E. Ordering at the EOQ costs $1,530. Ordering 1,300 heat sinks at a time costs $2,250, so ordering at the EOQ results in an annual savings of $720.

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual

Section: The Spreadsheet Modeling Process and Multiple-Criteria Decision Making Subsection: all subsections LO: 2 Describe the spreadsheet modeling process.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 15 minutes

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual

© 2023 John Wiley & Sons, Inc. or the author. All rights reserved. Instructors who are authorized users of this course are permitted to display these videos to students in a classroom setting.

Except as permitted herein or by law, no part of these materials should be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise. Advice on how to obtain permission to reuse this material is available at http://www.wiley.com/go/permissions.

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual

Solutions Manual Supplement B: Introduction to Optimization Problem 1: See File Prob_B_01.xls Decision Variables R1 = # radio ads in 0-10 range R2 = # radio ads in 11-20 range R3 = # radio ads in 21-30 range T1 = # TV ads in 0-5 range T2 = # TV ads in 6-10 range T3 = # TV ads in 11-15 range Model Maximize 900*R1 + 700*R2 + 10,000*T1 + 7500*T2 + 5000*T3 Subject to R1 < = 10 (constraints to limit each type of radio ad) R2 < = 10 R3 < = 10 T1 < = 5 (constraints to limit each type of TV ad) T2 < = 5 T3 < = 5 1000*R1 + 900*R2 + 800*R3 + 12,000*T1 + 10,000*T2 + 8000*T3 < = 145,000 (budget constraint) R1, R2, R3, T1, T2, T3 > = 0 Note: You might include the two constraints: R1 + R2 + R3 < = 30 T1 + T2 + T3 < = 15 Including these is not wrong, but with the first six constraints (which are needed), these two constraints are redundant.

Summary of Answers a) See “Formulation” worksheet in file Prob_B_01.xls. b) This worksheet shows the model and it contains the Solver settings. c) The optimal solution is to purchase 20 radio ads and 12 TV ads. Doing so will result in the maximum number of new customers reached, 113,500. The rationale for the solution is that ads are effectively chosen based on the number of new customers reached per advertising dollar. For example, the first 10 radio ads each reach 900 new customers, each at a cost of $1,000; there, 0.90 new customers are reached per dollar. The highest-yielding ads are put into the solution until the budget is used. d) See “Model_d” worksheet in file Prob_B_01.xls. As show, the solution is unrealistic for two reasons. The main reason is that the TV ads 11-15 all get selected (5 are selected), but only

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual

2.6 TV ads in the 6-10 range get selected. The other reason is that we cannot purchase a fractional TV ad. Solver is correct; it is solving the model we are giving it. With the changed parameters, indeed the third TV variable is more attractive than the second. The problem is that we would need to add logic to the model to ensure that no ads in the 11-15 range get selected before all ads in the 6-10 range get selected.

Section: Optimization; Algebraic Formulation; Spreadsheet Model Development; Solver Basics; and Interpreting the Solution Subsection: All subsections

©2023 John Wiley & Sons, Inc. All rights reserved.

Page 2 of 18

Reid/Sanders, Operations Management, 8th Edition Solutions Manual LO: 1. Recognize decision-making situations that may benefit from an optimization modeling approach; 2.

Formulate algebraic models for linear programming problems; 3. Develop spreadsheet models for linear programming problems; 4. Use Excel’s Solver add-in to solve linear programming problems.; 5. Interpret the results of models and perform basic sensitivity analysis.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 15 minutes

©2023 John Wiley & Sons, Inc. All rights reserved.

Page 3 of 18

Reid/Sanders, Operations Management, 8th Edition Solutions Manual

Problem 2: See File Prob_B_02.xls Decision Variables X1 = tons of paper stock 1 to use X2 = tons of paper stock 2 to use X3 = tons of paper stock 3 to use X4 = tons of paper stock 4 to use Model Minimize 150*X1 + 110*X2 + 90*X3 + 50*X4 Subject to X1 + X2 + X3 + X4 = 500 (size of order) (8*X1 + 6*X2 + 5*X3 + 3*X4)/500 >= 7 (strength requirement) (9*X1 + 7*X2 + 5*X3 + 4*X4)/500 >= 5 (color requirement) (8*X1 + 5*X2 + 6*X3 + 5*X4)/500 >= 6 (texture requirement) X1, X2, X3, X4 >= 0 Note: The strength, color and texture constraints can be written in several different ways. The following ways (strength constraint use for example): (8*X1 + 6*X2 + 5*X3 + 3*X4) >= 7*500 (8*X1 + 6*X2 + 5*X3 + 3*X4)/(X1 + X2 + X3 +X4) >= 7 (8*X1 + 6*X2 + 5*X3 + 3*X4) >= 7*(X1 + X2 + X3 +X4) The second of these methods causes difficulties if implemented this way in Excel/Solver. The reason is that we have a nonlinear expression of variables (one combination of variables divided by another combination). The remedy is to multiply both sides by the denominator, resulting in the third expression.

Summary of Answers a) See “Formulation” worksheet in file Prob_B_02.xls. b) This worksheet shows the model and it contains the Solver settings. c) The optimal solution is to use 400 tons of stock 1 and 100 tons of stock 4, resulting in a total cost of $65,000. The rationale is based on stock 1 more than meeting the strength, color, and texture requirements, and stock 4 being the least expensive stock. Essentially, as much as possible of stock 4 is used to keep the cost down, but most of the blend needs to be Stock 1 in order to meet the requirements.

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual

Section: Optimization; Algebraic Formulation; Spreadsheet Model Development; Solver Basics; and Interpreting the Solution Subsection: All subsections

©2023 John Wiley & Sons, Inc. All rights reserved.

Page 5 of 18

Reid/Sanders, Operations Management, 8th Edition Solutions Manual LO: 1. Recognize decision-making situations that may benefit from an optimization modeling approach; 2.

Formulate algebraic models for linear programming problems; 3. Develop spreadsheet models for linear programming problems; 4. Use Excel’s Solver add-in to solve linear programming problems.; 5. Interpret the results of models and perform basic sensitivity analysis.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 15 minutes

©2023 John Wiley & Sons, Inc. All rights reserved.

Page 6 of 18

Reid/Sanders, Operations Management, 8th Edition Solutions Manual

Problem 3: See File Prob_B_03.xls Decision Variable Xij = truckloads to send from plant I (i=1, 2, 3 for Atlanta, Tulsa, Springfield, respectively) to distribution center j (j = A, B,C, D) Model Minimize 800*X1A + 1300*X1B + 400*X1C + 700*X1D + 1100*X2A + 1400*X2B + 600*X2C + 1000*X2D + 600*X3A + 1200*X3B + 800*X3C + 900*X3D subject to X1A+X1B+X1C+X1D <= 13 (capacity Atlanta) X2A+X2B+X2C+X2D <= 18 (capacity Tulsa) X3A+X3B+X3C+X3D <= 12 (capacity Springfield) X1A + X2A + X3A >= 10 (demand distribution center A) X1B + X2B + X3B >= 10 (demand distribution center B) X1C + X2C + X3C >= 10 (demand distribution center C) X1D + X2D + X3D >= 10 (demand distribution center D) Xij >= 0 Summary of Answers a) See “Formulation” worksheet in the file Prob_03.xls. b) This worksheet shows the model and it contains the Solver settings. c) The optimal solution is to send 3 trucks from Atlanta to distribution center C, 10 trucks from Atlanta to D, 8 trucks from Tulsa to B, 7 trucks from Tulsa to C, 10 trucks from Springfield to A, and 2 trucks from Springfield to B. All demands are met exactly. Atlanta and Springfield operate at capacity, and Tulsa operates 3 truckloads below capacity. The total transportation cost is $32,000.

©2023 John Wiley & Sons, Inc. All rights reserved.

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual

Section: Optimization; Algebraic Formulation; Spreadsheet Model Development; Solver Basics; and Interpreting the Solution Subsection: All subsections

©2023 John Wiley & Sons, Inc. All rights reserved.

Page 8 of 18

Reid/Sanders, Operations Management, 8th Edition Solutions Manual LO: 1. Recognize decision-making situations that may benefit from an optimization modeling approach; 2.

Formulate algebraic models for linear programming problems; 3. Develop spreadsheet models for linear programming problems; 4. Use Excel’s Solver add-in to solve linear programming problems.; 5. Interpret the results of models and perform basic sensitivity analysis.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 15 minutes

©2023 John Wiley & Sons, Inc. All rights reserved.

Page 9 of 18

Reid/Sanders, Operations Management, 8th Edition Solutions Manual

Problem 4: See File Prob_B_04.xls Decision Variables Xij = tons newsprint to ship from mill i (i=A,B,C) to market j (j=1,2,3,4,5) Model Minimize 210*(XA1+XA2+XA3+XA4+XA5) + 225*(XB1+XB2+XB3+XB4+XB5) + 220*(XC1+XC2+XC3+XC4+XC5) + 20*XA1 + 25*XA2 + 30*XA3 + 15*XA4 + 35*XA5 + 30*XB1 + 20*XB2 + 32*XB3 + 28*XB4 + 19*XB5 + 25*XC1 + 18*XC2 + 28*XC3 + 23*XC4 + 31*XC5 subject to XA1+XA2+XA3+XA4+XA5 <= 1200 (capacity mill A) XB1+XB2+XB3+XB4+XB5 <= 1500 (capacity mill B) XC1+XC2+XC3+XC4+XC5 <= 900 (capacity mill C) XA1 + XB1 + XC1 >= 600 (demand market 1) XA2 + XB2 + XC2 >= 100 (demand market 2) XA3 + XB3 + XC3 >= 500 (demand market 3) XA4 + XB4 + XC4 >= 800 (demand market 4) XA5 + XB5 + XC5 >= 500 (demand market 5) Xij >= 0 Summary of Answers a) See "Formulation" worksheet in file Prob_B_04.xls. b) This worksheet shows the model and it contains the Solver settings. c) The optimal solution is to produce 1200 tons at Mill A, 500 at Mill B, and 800 at Mill C. Thus, Mill A will operate at capacity and Mill C will be 100 tons under capacity, while Mill B will be 1000 tons below capacity. Mill A's production will be split between Market 1 (400 tons) and Market 4 (800 tons). Mill B's production will go completely to Market 5 (500 tons). Mill C's production will be split between Markets 1, 2, and 3 (200, 100, and 500 tons respectively).

©2023 John Wiley & Sons, Inc. All rights reserved.

Page 10 of 18

Reid/Sanders, Operations Management, 8th Edition Solutions Manual

©2023 John Wiley & Sons, Inc. All rights reserved.

Page 11 of 18

Reid/Sanders, Operations Management, 8th Edition Solutions Manual Section: Optimization; Algebraic Formulation; Spreadsheet Model Development; Solver Basics; and Interpreting the Solution Subsection: All subsections LO: 1. Recognize decision-making situations that may benefit from an optimization modeling approach; 2.

Formulate algebraic models for linear programming problems; 3. Develop spreadsheet models for linear programming problems; 4. Use Excel’s Solver add-in to solve linear programming problems.; 5. Interpret the results of models and perform basic sensitivity analysis.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 15 minutes

©2023 John Wiley & Sons, Inc. All rights reserved.

Page 12 of 18

Reid/Sanders, Operations Management, 8th Edition Solutions Manual

Problem 5: See File Prob_B_05.xls Decision Variables Ri = regular production (in feet) in month i, i=1,...,7 Vi = overtime production (in feet) in month i, i=1,...,7 li = ending inventory (in feet) in month i, i=1,...,7 Model Minimize 1.25*(R1 + R2 + R3 + R4 + R5 + R6 + R7) + 1.50*(V1 + V2 + V3 + V4 + V5 + V6 + V7) + 0.15*(I1 + I2 + I3 + I4 + I5 + I6 + I7) subject to Ri <= 75,000 for i=1,...,7 (regular production capacity) Vi <= 15,000 for i=1,...,7 (overtime production capacity) I_i-1 + Ri + Vi - Di = Ii for i=1,...,7 (production/inventory/demand balance) Ii >= 0 for i=1,...,7 (ending inventory must be >= 0; forces demand to be met) I7 = 10,000 (to force September ending inventory to be 10,000) I0 = 5,000 (to correctly set ending February inventory to be 5,000) Ri >= 0 for i=1,...,7 Vi >= 0 for i=1,...,7 Note: Di here is used to represent the forecasted demand for month i. For example, D1=40,000, D2=60,000, etc.

Summary of Answers a) See "Formulation" worksheet in the file Prob_B_05.xls. b) This worksheet shows the model and it contains the Solver settings c) The optimal solution is to produce just to meet demand in March and April. Starting in May, we produce more than needed, in order to meet the higher demands in the later months. Regular production is 35,000 in March, 60,000 in April, and at 75,000 (capacity) from May through September. Overtime production is at 10,000 in June, and at 15,000 (capacity) from JulySeptember. The production/inventory plan results in an overall total cost of $675,250 and ends September with 10,000 feet of pipe on hand.

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual

©2023 John Wiley & Sons, Inc. All rights reserved.

Page 14 of 18

Reid/Sanders, Operations Management, 8th Edition Solutions Manual Section: Optimization; Algebraic Formulation; Spreadsheet Model Development; Solver Basics; and Interpreting the Solution Subsection: All subsections LO: 1. Recognize decision-making situations that may benefit from an optimization modeling approach; 2.

Formulate algebraic models for linear programming problems; 3. Develop spreadsheet models for linear programming problems; 4. Use Excel’s Solver add-in to solve linear programming problems.; 5. Interpret the results of models and perform basic sensitivity analysis.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 15 minutes

©2023 John Wiley & Sons, Inc. All rights reserved.

Page 15 of 18

Reid/Sanders, Operations Management, 8th Edition Solutions Manual

Problem 6: See File Prob_B_06.xls Decision Variables Si = Number employees assigned to shift type i, i=1,...,6 Note: Shift type 1 works 12midnight-8am, shift type 2 works 4am-noon, and so on through shift type 6 which works 8pm-4am Model Minimize S1 + S2 + S3 + S4 + S5 + S6 subject to S1 + S6 >= 3 (midnight-4am requirements) S1 + S2 >= 6 (4am-8am requirements) S2 + S3 >= 13 (8am-noon requirements) S3 + S4 >= 15 (noon-4pm requirements) S4 + S5 >= 12 (4pm-8pm requirements) S5 + S6 >= 9 (8pm-midnight requirements) Si >= 0 Summary of Answers a) See "Formulation" worksheet in the file Prob_B_06.xls. b) This worksheet shows the model and it contains the Solver settings. c) The optimal solution requires 30 employees. Five employees should work midnight-8am, 1 from 4am-noon, 12 from 8am-4pm, 3 from noon-8pm, and 9 from 4pm-midnight. The result is that every four-hour period has exactly the number of employees needed, except for the 12midnight-4am time period, where five employees will be working (3 are needed). The reason for the overage is that we are constrained to 8-hour shifts. If 4-hour shifts were possible, we could meet each requirement exactly. Note: There may be alternate optimal solutions to this problem; that is, solutions having 30 employees, but with different numbers of employees working the various shift types.

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual

Section: Optimization; Algebraic Formulation; Spreadsheet Model Development; Solver Basics; and Interpreting the Solution Subsection: All subsections

©2023 John Wiley & Sons, Inc. All rights reserved.

Page 17 of 18

Reid/Sanders, Operations Management, 8th Edition Solutions Manual LO: 1. Recognize decision-making situations that may benefit from an optimization modeling approach; 2.

Formulate algebraic models for linear programming problems; 3. Develop spreadsheet models for linear programming problems; 4. Use Excel’s Solver add-in to solve linear programming problems.; 5. Interpret the results of models and perform basic sensitivity analysis.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 15 minutes

© 2023 John Wiley & Sons, Inc. or the author. All rights reserved. Instructors who are authorized users of this course are permitted to display these videos to students in a classroom setting.

Except as permitted herein or by law, no part of these materials should be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise. Advice on how to obtain permission to reuse this material is available at http://www.wiley.com/go/permissions.

©2023 John Wiley & Sons, Inc. All rights reserved.

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual

Solutions Manual Supplement Chapter C: Waiting Line Models Answers to Problems 1.

a. Average system utilization. λ 8 ρ= = =0.80 μ 10 b. Average number of customers in the system.

L=

λ 8 = =4.0 μ− λ 10−8

c. Average number of customers waiting in line.

L Q =¿ ρL = (0.80)(4.0) = 3.2 Section: Waiting Line Performance Measures Subsection: Single-Server Waiting Line Model and Multiserver Waiting Line Model LO: 2 Use waiting line models to estimate system performance.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 6 minutes

2.

a. The average time a customer spends in the system.

W=

1 1 = =0.50 hour μ−λ 10−8

b. The average time a customer spends waiting in line.

W Q =¿ ρW = (0.80)(0.50) = 0.40 hour

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual

c. The probability of having more than three customers in the system.

P 0 = ( 1− ρ ) ρ0 = ( 1−0.80 ) 1=0.20 P 1 = ( 1− ρ ) ρ1 = ( 1−0.80 ) 0.80 =0.16

P 2 = ( 1− ρ ) ρ2 =( 1−0.80 ) 0.80 2 =0.13 P 3 = ( 1− ρ ) ρ3 = ( 1−0.80 ) 0.80 3 =0.10

The probability of more than three customers in the system is 0.41 or 41% (1 – (0.20 + 0.16 + 0.13 + 0.10)) d. The probability of having more than four customers in the system.

P 4 = ( 1− ρ ) ρ4 = ( 1−0.80 ) 0.80 4 =0.08 The probability of more than four customers in the system is 0.33 or 33% (0.41 – 0.08) Section: Waiting Line Performance Measures Subsection: Single-Server Waiting Line Model and Multiserver Waiting Line Model LO: 2 Use waiting line models to estimate system performance.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 8 minutes

3.

a. Average system utilization. λ 10 ρ= = =0.63 μ 16 b. Average number of customers in the system.

L=

λ 10 = =1.67 μ− λ 16−10

c. Average number of customers waiting in line.

L Q =¿ ρL = (0.63)(1.67) = 1.05 Section: Waiting Line Performance Measures Subsection: Single-Server Waiting Line Model and Multiserver Waiting Line Model

©2023 John Wiley & Sons, Inc. All rights reserved.

Page 2 of 9

Reid/Sanders, Operations Management, 8th Edition Solutions Manual LO: 2 Use waiting line models to estimate system performance.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 6 minutes

4.

a. The average time a customer spends in the system.

W=

1 1 = =0.16 hour μ−λ 16−10

b. The average time a customer spends waiting in line.

W Q =¿ ρW = (0.63)(0.16) = 0.10 hour

c. The probability of having more than three customers in the system.

P 0 = ( 1− ρ ) ρ0 = ( 1−0.63 ) 1=0.38 P 1 = ( 1− ρ ) ρ1 = ( 1−0.63 ) 0.631 =0.23 P 2 = ( 1− ρ ) ρ2 =( 1−0.63 ) 0.632 =0.15 3 3 P 3 = ( 1− ρ ) ρ = ( 1−0.63 ) 0.63 =0.09 The probability of more than three customers in the system is 0.15 or 15% (1 – (0.38 + 0.23 + 0.15 + 0.09)) d. The probability of having more than four customers in the system.

P 4 = ( 1− ρ ) ρ4 = ( 1−0.63 ) 0.634 =0.057

The probability of more than four customers in the system is approx. 0.10 or 10% (0.15 – 0.05) Section: Waiting Line Performance Measures Subsection: Single-Server Waiting Line Model and Multiserver Waiting Line Model LO: 2 Use waiting line models to estimate system performance.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 6 minutes

5.

a. Average system utilization. λ 12 ρ= = =0.75 μ 16

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Page 3 of 9

Reid/Sanders, Operations Management, 8th Edition Solutions Manual

b. Average number of customers in the system.

L=

λ 12 = =3 μ− λ 16−12

c. Average number of customers waiting in line.

L Q =¿ ρL = (0.75)(3) = 2.25 Section: Waiting Line Performance Measures Subsection: Single-Server Waiting Line Model and Multiserver Waiting Line Model LO: 2 Use waiting line models to estimate system performance.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 6 minutes

6.

a. The average time a customer spends in the system.

W=

1 1 = = 0.25hours μ−λ 16−12

b. The average time a customer spends waiting in line.

W Q = ρW (0.75)(0.25)=0.1875 hours c. The probability of having more than four customers in the system.

P 0 = ( 1− ρ ) ρ0 = ( 1−0.75 ) 1=0.25 P 1 = ( 1− ρ ) ρ1 = ( 1−0.75 ) 0.751 =0.19 P 2 = ( 1− ρ ) ρ2 =( 1−0.75 ) 0.752 =0.14 P 3 = ( 1− ρ ) ρ3 = ( 1−0.75 ) 0.753 =0.11 P 4 = ( 1− ρ ) ρ4 = ( 1−0.75 ) 0.754 =0.08

The probability of more than three customers in the system is 0.24 or 24% (1 – (0.25 + 0.19 + 0.14 + 0.11 + 0.08)) Section: Waiting Line Performance Measures Subsection: Single-Server Waiting Line Model and Multiserver Waiting Line Model LO: 2 Use waiting line models to estimate system performance.

Bloom’s: Application AACSB: Analytic

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Page 4 of 9

Reid/Sanders, Operations Management, 8th Edition Solutions Manual Difficulty level: Medium Time on task: 6 minutes

©2023 John Wiley & Sons, Inc. All rights reserved.

Page 5 of 9

Reid/Sanders, Operations Management, 8th Edition Solutions Manual

7.

a. Average system utilization. ρ=

λ 24 = =0.75 sμ (2)(16) b. Average number of customers in the system.

P 0=

[

2−1

∑

n= 0

0.1429 LQ =

24 16

+

24 16 1!

1

+

24 16 2!

2

1 1−0.75

2

2 ! 1−0.75

W Q=

]

( ) ( ) ( )( ) (( ) ) 24 16 0!

0

−1

=0.1429

×0.75 2

=1.9286

L Q 1.9286 = =0.0803 λ 24

1 1 W =W Q + =0.0803+ = 0.1428 μ 16 L= λ × W =24× 0.1428=3.428 c. Average number of customers waiting in line:

L Q =1.9286

d. Average time a customer spends in the system:

W =0.1428

e. Average time a customer spends waiting in line:

W Q =0.0803

f. Probability of having more than 4 customers:

Pn=

{

n

( λ ⁄ µ ) for n≤ s

n! n µ) ( λ ⁄ n−s for n> s s! s

The probability of more than 4 customers in the system is 0.2712 or 27% = (1 – (P0 + P1 +P2 + P3 + P4)) = 1 – (0.1429 + 0.2143 + 0.1607 + 0.1205 + 0.0904) = 0.2712

Section: Waiting Line Performance Measures Subsection: Single-Server Waiting Line Model and Multiserver Waiting Line Model

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual LO: 2 Use waiting line models to estimate system performance.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 6 minutes

8. We want to find the arrival rate at which LQ = 4. There are two ways to do this. One way is to use the formula for LQ, shown below. λ / μ¿s ρ ¿ 1− ρ¿2 s !¿ P0 ¿ L Q =¿ The idea is to set LQ = 4, and solve for . However, ρ = /µ, and so solving for  becomes fairly difficult algebraically.

An easier way is to simply use the flexible spreadsheet model, with Goal Seek. Doing so with the Goal Seek settings shown below, results in a  value of 26.86. See file Prb_C_08.xlsx.

©2023 John Wiley & Sons, Inc. All rights reserved.

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual

Rho (=B5/B12) P0 =(@INDEX(H9:H109,B7)+(((F5^B7)/F6)*((1)/(1-B15))))^(-1) Lq =(B16*(F5^B7)*B15)/(@INDEX(G9:G109,B7+1)*(1-B15)^2) Section: Waiting Line Performance Measures Subsection: Single-Server Waiting Line Model and Multiserver Waiting Line Model LO: 2 Use waiting line models to estimate system performance.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 6 minutes

© 2023 John Wiley & Sons, Inc. or the author. All rights reserved. Instructors who are authorized users of this course are permitted to display these videos to students in a classroom setting. Except as permitted herein or by law, no part of these materials should be reproduced, stored in a retrieval system, or

©2023 John Wiley & Sons, Inc. All rights reserved.

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual

transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise.

Advice on how to obtain permission to reuse this material is available at http://www.wiley.com/go/permissions.

©2023 John Wiley & Sons, Inc. All rights reserved.

Page 9 of 9

Reid/Sanders, Operations Management, 8th Edition Solutions Manual

Solutions Manual Supplement D: Master Scheduling and Rough-Cut Capacity Planning Answers to Problems 1.

a. and b.

Week Forecast Proj Avail 15 MPS = 72

1 20 67 72

2 35 32 0

3 50 54 72

4 50 4 0

5 45 31 72

6 40 63 72

c. Maximum number of units in inventory is 67 in week 1. d. Number of MPS orders needed is 4.

Section: Developing an MPS and Using the MPS Subsection: Rough-Cut Capacity Planning and Evaluating and Accepting the MPS LO: 2 Develop a master production schedule and project the capacity needed using rough-cut capacity planning and 3 Calculate available-to-promise quantities.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 6 minutes

2.

a. and b.

Week Forecast Proj Avail 15 MPS = 120

1 20 115 120

2 35 80 0

3 50 30 0

4 50 100 120

5 45 55 0

6 40 15 0

c. This plan provides higher average inventory per period and lower set up costs than the MPS developed in Problem 1.

Section: Developing an MPS and Using the MPS Subsection: Rough-Cut Capacity Planning and Evaluating and Accepting the MPS LO: 2 Develop a master production schedule and project the capacity needed using rough-cut capacity planning and 3 Calculate available-to-promise quantities.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 6 minutes

©2023 John Wiley & Sons, Inc. All rights reserved.

Page

Reid/Sanders, Operations Management, 8th Edition Solutions Manual

3

a. and b.

Week Forecast Proj Avail 15 MPS = 48

1 20 43 48

2 35 8 0

3 50 6 48

4 50 4 48

5 45 7 48

6 40 15 48

c. This plan provides lowest average inventory per period, and the highest set up costs, compared to plans in Problem 1 and Problem 2.

Section: Developing an MPS and Using the MPS Subsection: Rough-Cut Capacity Planning and Evaluating and Accepting the MPS LO: 2 Develop a master production schedule and project the capacity needed using rough-cut capacity planning and 3 Calculate available-to-promise quantities.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 6 minutes

4.

Standard Corkscrew Forecast Proj. Avail 2000 MPS

1 3,000 3,000 4,000

2 3,500 3,500 4,000

3 5,000 2,500 4,000

4 4,000 500 2,000

Deluxe Corkscrew Forecast Proj. Avail 2000 MPS

1 2,000 1,000 1,000

2 1,500 500 1,000

3 1,000 500 1,000

4 3,000 500 3,000

Note: The MPS for the first and fourth periods may be changed with no change in the average total inventory. One might question if the holding costs would be the same if those period MPS numbers were changed.

Section: Developing an MPS and Using the MPS Subsection: Rough-Cut Capacity Planning and Evaluating and Accepting the MPS LO: 2 Develop a master production schedule and project the capacity needed using rough-cut capacity planning and 3 Calculate available-to-promise quantities.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 6 minutes

©2023 John Wiley & Sons, Inc. All rights reserved.

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Reid/Sanders, Operations Management, 8th Edition Solutions Manual

5.

Capacity - Standard Direct Labor Machine Time

1 800 400

2 800 400

3 800 400

4 400 200

Total 2,800 1,400

Capacity - Deluxe Direct Labor Machine Time

1 500 300

2 500 300

3 500 300

4 1,500 900

Total 3,000 1,800

Section: Developing an MPS and Using the MPS Subsection: Rough-Cut Capacity Planning and Evaluating and Accepting the MPS LO: 2 Develop a master production schedule and project the capacity needed using rough-cut capacity planning and 3 Calculate available-to-promise quantities.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 6 minutes

6.

a. Machine time required using existing machine: Period Machine Time (hours)

1 700

2 700

3 700

4 1100

Total 3,200

3 720

4 1360

Total 3,520

Machine time required using new machine: Period Machine Time (hours)

1 720

2 720

b. Total machine time requires 3,520 hours with new machine compared to 3,200 hours with old existing machine. Therefore, new machine will prove to be not beneficial to WAI. Section: Developing an MPS and Using the MPS Subsection: Rough-Cut Capacity Planning and Evaluating and Accepting the MPS LO: 2 Develop a master production schedule and project the capacity needed using rough-cut capacity planning and 3 Calculate available-to-promise quantities.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 6 minutes

©2023 John Wiley & Sons, Inc. All rights reserved.

Page

Reid/Sanders, Operations Management, 8th Edition Solutions Manual

7.

Hours Needed - Dept. 101 Direct Labor (40%) Machine Time (60%) Hours Needed - Dept. 102 Direct Labor (60%) Machine Time (40%)

1 520 420 1 780 280

2 520 420 2 780 280

3 520 420 3 780 280

4 760 660 4 1,140 440

Total 2320 1,920 Total 3,480 1,280

Section: Developing an MPS and Using the MPS Subsection: Rough-Cut Capacity Planning and Evaluating and Accepting the MPS LO: 2 Develop a master production schedule and project the capacity needed using rough-cut capacity planning and 3 Calculate available-to-promise quantities.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 6 minutes

8.

Period Forecast Customer Orders Projected Available 50 Available-to-promise MPS

1 30 35 15 15

2 30 15 85 47 100

3 30 20 55

4 40 18 15

5 40 12 75 88 100

6 40 0 35

7 45 15 90 85 100

8 45 0 45

Section: Developing an MPS and Using the MPS Subsection: Rough-Cut Capacity Planning and Evaluating and Accepting the MPS LO: 2 Develop a master production schedule and project the capacity needed using rough-cut capacity planning and 3 Calculate available-to-promise quantities.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 6 minutes

©2023 John Wiley & Sons, Inc. All rights reserved.

Page

Reid/Sanders, Operations Management, 8th Edition Solutions Manual

9.

Additional Orders Period Forecast Customer Orders Projected Available 50 Available-to-promise MPS

25 1 30 60 -10 -10

2 30 15 60 37 100

10 3 30 30 30

4 40 18 -10

40 5 40 52 38 28 100

20 6 40 20 -2

7 45 15 53 85 100

8 45 0 8

The company cannot accept additional order in Week 1, because ATP quantity is negative.

If the company does not accept the order in week 1, the MPS record below illustrates that ATP quantity does not turn negative, and all other additional orders are possible.

Additional Orders Period Forecast Customer Orders Projected Available 50 Available-to-promise MPS

0 1 30 35 15 15

0 2 30 15 85 37 100

10 3 30 30 55

0 4 40 18 15

40 5 40 52 75 28 100

20 6 40 20 35

0 7 45 15 90 85 100

0 8 45 0 45

Section: Developing an MPS and Using the MPS Subsection: Rough-Cut Capacity Planning and Evaluating and Accepting the MPS LO: 2 Develop a master production schedule and project the capacity needed using rough-cut capacity planning and 3 Calculate available-to-promise quantities.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 6 minutes

©2023 John Wiley & Sons, Inc. All rights reserved.

Page

Reid/Sanders, Operations Management, 8th Edition Solutions Manual

10.

Period Forecast Customer Orders Projected Available 50 Available-to-promise MPS

1 30 35 15 15

2 30 40 75 12 100

3 30 30 45

4 40 18 5

5 40 77 28 3 100

6 40 20 -12

7 45 15 43 85 100

8 45 0 -2

Since there are no negative numbers in the available-to-promise row, WAI will be able to meet delivery requirements.

Section: Developing an MPS and Using the MPS Subsection: Rough-Cut Capacity Planning and Evaluating and Accepting the MPS LO: 2 Develop a master production schedule and project the capacity needed using rough-cut capacity planning and 3 Calculate available-to-promise quantities.

Bloom’s: Application AACSB: Analytic Difficulty level: Medium Time on task: 6 minutes

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