Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

SOLUTIONS

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

SOLUTION MANUAL FOR OPERATION MANAGEMENT 3RD EDITION BY GERARD CACHON, CHRISTIAN TERWIESCH CHAPTER 1 INTRODUCTION TO OPERATIONS MANAGEMENT CONCEPTUAL QUESTIONS 1. Below are a number of slogans used for advertisement. Which dimensions of customer utility do the slogans emphasize?

Answer: Slogan A emphasizes Fit; Slogan B emphasizes Timing; Slogan C emphasizes Price; Slogan D emphasizes Location; and Slogan E emphasizes Performance.

2. Which of the following is not a dimension or subdimension in a customer’s utility function?

Answer: D. Customer Satisfaction 3. The efficient frontier is given by the cheapest company in the industry. True or false?

Answer: False 4. There can be no more than two firms on the efficient frontier. True or false?

Answer: False 5. Two retailers compete on costs and the ambience of their retail stores. They are identical in all other dimensions of customer utility. Retailer A is cheaper than retailer B. Retailer A also has the better ambience. Does this mean that retailer A is on the efficient frontier? Yes or no?

Answer: Yes 6. Which of the following is NOT one of the three system inhibitors?

Answer: C. Fatigue 7. Which of the following is an example of a system inhibitor?

Answer: B. An emergency room doctor has no patients at the moment.

8. Which of the following questions is NOT related to operations management?

Answer: B. How much will the CEO be paid?

9. Which of the following is most related to the operations management of an organic sheep farm?

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Answer: A. The amount of land they rent for grazing sheep.

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

PROBLEMS AND APPLICATIONS 1. What are the subcomponents of inconvenience in a customer utility function?

Answer: C. Location and Time Feedback: Location and timing are the two major subcomponents of inconvenience.

2. Which dimension of the customer utility function is particularly emphasized with the concept of ―custom built‖?

Answer: B. Fit. Feedback: Fit is a subcomponent of the consumption utility that captures how well the product or service matches with the unique characteristics of a given consumer.

3. Which of the following characteristics is a subcomponent of the consumption utility in a customer utility function?

Answer: A. Performance.

Feedback: Performance is a subcomponent of the consumption utility that captures how much an average consumer desires a product or service.

4. Which dimension of the customer utility function is particularly emphasized with the ―to-go‖ section?

Answer: D. Timing.

Feedback: The "to-go" section is designed for customers to purchase food quickly and move on their way to their departure gate. The primary focus is on the speed of service, which addresses the timing element of the customer utility function.

5. Which dimension of the customer utility function is particularly emphasized with the special edition coupe?

Answer: A. Performance.

Feedback: The "special edition" coupe has features that give it a higher level of performance compared to the standard model. As a result, the "special edition" vehicle clearly emphasizes the performance dimension of the customer utility function.

6. Which of these hotels are on the efficient frontier? You may select more than one answer.

Answer: Hotels B, C, and D. Feedback: The only hotel that is Pareto dominated is hotel A - all other are on the efficient frontier. Hotel A is Pareto dominated by hotel B, as B is both cheaper and better.

7. Which of these LTL carriers are on the efficient frontier?

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Answer: Carriers A, B, and C. Feedback: Carrier D is the only one not on the efficient frontier because it is dominated

by Carrier A on both measures.

8. Which of these dry cleaners are NOT on the efficient frontier?

Answer: Dry Cleaner B. Feedback: Dry Cleaner B is the only one not on the efficient frontier because it is dominated by both Dry Cleaner A and C on both measures.

9. Which of the following items would be considered an input in the operations of a soft drink manufacturer?

Answer: C. Empty bottles Feedback: Inputs are the things that a business purchases. Empty bottles would be part of the soft drink product.

10. Which of the following items would be considered a resource in the operations of a soft drink manufacturer?

Answer: B. Bottling machines.

Feedback: Resources are the things in a business that help transform input into output and thereby help provide supply for what customers demand. The bottling machine is a resource used to manufacture the soft drinks. The other items are inputs and are part of the product.

11. Which of the following items would be considered an input in the operations of a doctor’s office?

Answer: C. Needle Feedback: The only item in the list that is a material used in the doctor's office operations is a needle. The rest of the items would be considered resources.

12. Which of the following items would be considered a resource in the operations of a movie theater?

Answer: B. Projector Feedback: The only item in the list that is used to transform inputs to outputs is the projector. The rest of the items in the list would be classified as inputs because they are materials and supplies used in the operations.

13. Which of the following inefficiencies in a grocery store’s operations results from inflexibility?

Answer: D. Employee work schedules set a week in advance Feedback: Inflexibility is the inability of an operation to quickly and cheaply change in response to new information.

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

14. Which of the following inefficiencies in a bank’s operations results from variability?

Answer: C. Customers incorrectly listing information on forms Feedback: Variability corresponds to changes in either demand or supply over time.

15. Which of the following inefficiencies in a bank’s operations results from variability?

Answer: A (When will the demand be fulfilled?) and D. Where will the demand be fulfilled?

Feedback: Convenience refers to the questions of ―when‖ and ―where‖ the demand will be fulfilled.

16. Which of the following inefficiencies in a bank’s operations results from variability?

Answer: B. What are the shipping charges to the customer?

Feedback: The operational efficiency will affect the price that the firm is able to charge for its product and service to maximize its profitability. The shipping charges are part of the price component.

17. Which of the following operational decisions correspond(s) to the consumption utility component of the consumer utility function? Instructions: You may select more than one answer.

Answer: C. What is the product or service to be delivered?

Feedback: The product or service characteristics will affect how much each consumer will like the overall product or service, which is measured by consumption utility.

CASE There is no case for this chapter.

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

CHAPTER 2 INTRODUCTION TO PROCESSES

CONCEPTUAL QUESTIONS 1. From the perspective of process analysis, which of the following could be appropriate flow units for a hardware store?

Answer: C. Number of customers.

Feedback: The number of workers, cash registers, and suppliers are unlikely to change much over the course of a month and do not ―flow‖ through the process of the hardware store.

2. Over the course of a month, which of the following is most likely to describe an appropriate flow unit for a process analysis of a hospital?

Answer: D. The number of patients.

Feedback: Physicians, beds, and square footage are unlikely to change much over the course of a month and do not ―flow‖ through the process of a hospital.

3. At a cruise ship terminal, each day on average 1000 passengers embark on ships. On average, passengers spend 5 days on their cruise before returning to this terminal. If the flow unit is a passenger, then what are the flow rate and flow time of this process?

Answer: The flow rate is 1,000 passengers per day and the flow time is 5 days.

4. It is election day and 1800 voters vote in their precinct’s library during the 10 hours the polls are open. On average, there are 15 voters in the library and they spend on average 5 minutes in the library to complete their voting. What is the inventory of voters, the flow rate, and the flow time for each voter?

Answer: The inventory is 15 voters (this is given), the flow rate is 180 voters per hour or 3 voters per minute and the flow time is 5 minutes.

5. Over the course of a day, fans pour into a NASCAR venue at the rate of 8000 people per hour.

The average rate at which fans leave the venue .

Answer: B. must be exactly 8000 people per hour Feedback: The flow rate into a process must equal the flow rate out of a process (for the process to be stable).

6. A computer server experiences large fluctuations in the amount of data requests it receives throughout the day. Because of this variation, Little’s Law does not apply. True or false?

Answer: False.

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Feedback: Little’s Law applies even if there are fluctuations in inventory, flow rates, and flow times.

PROBLEMS AND APPLICATIONS 1. For the purpose of process analysis, which of the following measures would be considered an appropriate flow unit for analyzing the operation of a coffee shop? Instructions: You may select more than one answer.

Answer: D. Number of customers served each week Feedback: The number of customers is the appropriate flow unit for process analysis. The employees are resources, and the other two measures are unlikely to change from week to week.

2. For the purpose of process analysis, which of the following measures would be considered an appropriate flow unit for analyzing the main operation of a local accounting firm? Instructions: You may select more than one answer.

Answer: B. Number of tax returns completed each week Feedback: The number of tax returns completed each week reflects the main operation of the accounting firm during tax season. The accountants are resources; the customers with past-due invoices reflect the accounts receivable process and not the main operation; and the reams of paper received are a result of the firm’s purchasing policies and not necessarily the main operation.

3. For the purpose of process analysis, which of the following measures would be considered an appropriate flow unit for analyzing the main operation of a gas station? Instructions: You may select more than one answer.

Answer: A (sales dollars) and D (number of customers served per day) are correct Feedback: The gasoline pumps and employees are resources, not flow units.

4. What is the flow rate of callers from 8:00 a.m. to 8:20 a.m.?

Answer: 0.4 callers per minute Feedback: 8 calls divided by 20 minutes = 0.4 calls per minute.

5. What is the flow time of callers from 8:00 a.m. to 8:20 a.m.?

Answer: 4 minutes Feedback: To calculate the flow time of the callers, subtract the caller’s departure time from his or her arrival time. (4+5+2+3+5+8+3+2 = 32) so, 32 total minutes divided by 8 callers = 4 minutes.

6. What is the flow rate of customers from 9:00 a.m. to 10:00 a.m.?

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Answer: 0.1667 customers per minute Feedback: Flow rate = 10 customers divided by 60 minutes = 0.1167 customers per minute.

7. What is the flow time of customers from 9:00 a.m. to 10:00 a.m.?

Answer: 8.6 minutes Feedback: To calculate the flow time of the customers, subtract the customers departure time from his or her arrival time. (6+15+12+5+8+7+8+6+15+4 = 86) 86 total minutes divided by 10 customers = 8.6 minutes.

8. What is the average amount of time that a customer spends in process?

Answer: 4 minutes Feedback: To solve this problem, use Little’s Law. Inventory = Flow rate × Flow time.

The flow rate is 300 customers divided by 120 minutes = 2.5.

10 people in line (average inventory) = 2.5 flow rate x flow time Flow time = 4 minutes 9. How many wafers does the cooling tube hold, on average, when in production (in other words, don’t count the time they are not in production)?

Answer: 90,000 wafers Feedback: Inventory = flow rate x flow time = 100 per second x 60 seconds per minute x 15 minutes = 90,000 wafers 10. How many skiers are riding on the lift at any given time?

Answer: 360 skiers Feedback: 1,800 skiers divided by 60 minutes per hour (flow rate) x 12 minutes (flow time) = 360 skiers 11. Last year, on average, how many visitors were in the park simultaneously?

Answer: 8,539 visitors Feedback: Flow rate = 3,400,000 visitors divided by 365 days = 9,315.07 visitors per day Flow Rate = 22 hours/ 24 hours per day = .9167 day Inventory = 9,315.07 (flow rate) x 0.9167 (flow time) = 8539.12 visitors per day 12. How many patients are taking this drug on average at any given time?

Answer: 900,000 patients Feedback: 6 months (flow time) x 150,000 new patients per month (flow rate)= 900,000 patients 8

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

13. On average, how many chat sessions are active (i.e., started but not completed)?

Answer: 20 chat sessions Feedback: Flow rate = 240 chats divided by 30 employees = 8 chats per employee Flow time = 5 minutes divided by 60 minutes = 0.833 hour Inventory = Flow Rate x Flow Time, 8 x 0.833= 0.6667 x 30 employees = 20 chats 14. How large of an oven is required so that the company is able to produce 4200 units of bread per hour (measured in the number of units that can be baked simultaneously)?

Answer: Each oven should have a capacity of 840 units Feedback: 4,200 units multiplied by (12 minutes/60 minutes) = 840 units 15. How many new skiers are arriving, on average, in LaVilla every day?

Answer: 120 skiers Feedback: 1,200 beds divided by 10 days = 120 new skiers per day.

16. How long did the average passenger have to wait in line?

Answer: 7.5 minutes Feedback: To solve this problem, use Little’s Law. Inventory = Flow rate × Flow time.

30 people in line (average inventory) = 240 customers/ 60 minutes (flow rate) x flow time. Flow time = 7.5 minutes 17. How long is the average associate employed at the consulting company?

Answer: 8 years Feedback: 120 associates = 15 new employees per year x flow time. Flow time = 8 years.

CASE Cougar Mountain Although the analysis of the case is relatively simple, the intuition is not always easy to grasp— many students will intuitively believe that the capacity of the faster lift should be greater than the capacity of the slower lift. The main lesson in this case is to get students to understand why that intuition is not correct.

To begin the case discussion, ask the students their opinion as to who is correct, Mark (unloading capacity should be twice as high on the detachable lift) or Doug (the unloading capacity should be the same on the two lifts). Hopefully there are students who support each opinion.

To resolve the question, begin with the simple process flow diagram:

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Skiers

Skiers

Lift

Ask the question ―Do all of the skiers that get on the lift at the bottom get off the lift at the top?‖ Of course, the answer is ―We would hope so!‖. So ―What does that mean about how the rate of skiers getting on the lift, Ron, is related to the rate of skiers getting off the lift, Roff?‖ And the answer there must be that they are equal! If the rate on where faster than the rate off, the number of people on the lift would grow and grow and grow. We know that can’t happen. Similarly, if the rate off exceeded the rate on, then the number of people on the lift would shrink and shrink and shrink, leaving the lift eventually with nobody. Which also doesn’t happen.

So we can add to our process flow diagram:

Ron

Skiers

Skiers

Lift

Roff

= Now it is time to compare the two lifts. We can draw the process flow for each of them, emphasizing that the rate on for each must equal the rate off:

Rs

Slow Lift

Rs

Rf

Fast Lift

Rf

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Now ask students ―How can we compare the rates across the two types of lifts?‖ The answer is given in the case—we are told that the rate skiers load onto the slow (fixed grip) lift is the same as the rate they load onto the fast (detachable) lift. That means that Rs = Rf. And that means that the rates that they onload skiers at the top must be the same!

Rs

Rs

Slow Lift

= Rf

= Rf

Fast Lift

Thus, Doug is correct—both lifts have the same capacity to unload skiers at the top even though one is faster than the other.

And this brings us to Jessica’s question—so what is the difference between the two lifts? If you ask students this question, the likely first response is that skiers spend less time on the faster lift.

And that is correct. But are there other differences? Actually, there are two additional differences worth mentioning. The first comes from Little’s Law and the 2nd one requires a deeper understanding of this process.

The first obvious difference is the number of skiers on the lift. According to Little’s Law, I = R x T. So if the two lifts have the same R, but the faster lift has a smaller T, then the faster lift must have a smaller I as well:

Tf = seconds on lift

R

Fast Lift

R

If = # of skiers 11

Tf < Ts If = R x Tf < R x Ts = Is

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

So fewer people are on the faster lift and they spend less time on the lift but the faster lift and the slower lift bring skiers to the top at the same rate.

If students can’t get the next difference between the two lifts, then you can prompt them with the following question ―If the faster lift has fewer skiers than the slower lift, then where are the additional skiers?‖ Or put another way: ―If the ski area attracts a certain number of skiers but the faster lift has fewer skiers on it, then where are the other skiers?‖ The answer is that they are on the slopes! That means that adding a faster lift takes skiers off the lift but they don’t disappear. Instead, they are on the only other place they can be, the slopes. Which means, somewhat counter-intuitively, that adding a faster lift makes the slopes more crowded (holding the total number of skiers fixed).

CHAPTER 3 PROCESS ANALYSIS CONCEPTUAL QUESTIONS

1. Which of the following questions would be asked in a process analysis of a college admissions office?

Answer: B. How long does it take the office to process an application?

2. Which of the following items would be considered resources in a restaurant?

Answer: Chefs 3. You are sitting in a restaurant and the waiter brings you the food you ordered a while ago. If you think about you being the flow unit in the process of the restaurant, which step of this process will be downstream relative to your current position in the process?

Answer: Paying the bill 4. What is the relationship between the processing time at a resource and its capacity?

Answer: They are reciprocals of each other.

5. You observe a long line at the airport security. The process currently is: Answer: capacity-constrained.

6. You observe a bank and notice that a customer leaves the bank about every five minutes.

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

These five minutes between customers are:

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Answer: the cycle time.

7. What is the maximum utilization a resource can achieve?

Answer: 1.00 8. Is the capacity of the bottleneck larger than, equal to, or smaller than the capacity of the process?

Answer: Equal to 9. Smartphones are made on a 40-step assembly process. All 40 steps are connected through a conveyor belt and all of the 40 steps must work at the same rate even if some of them have more capacity than others. Is this process a machine-paced process or a worker-paced process?

Answer: Machine-paced 10. You observe a vehicle registration department at your local township. Assume that all employees are ready to work at 9 a.m. You arrive at 9 a.m. sharp and are the first customer. Is your time through the empty process longer or shorter than the flow time averaged across all customers that arrive over the course of the day?

Answer: Shorter than the average flow time PROBLEMS AND APPLICATIONS

1.

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What is the capacity of the barber expressed in customers per hour?

Answer: 4 customers per hour Feedback: 60 minutes/15 minutes = 4 customers per hour

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Assuming the demand for the barber is two customers per hour, what is the flow rate?

Answer: 2 customers per hour

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Assuming the demand for the barber is two customers per hour, what is the utilization?

Answer: 50 percent

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Feedback: 2 demand / 4 capacity = 50%

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Assuming the demand for the barber is two customers per hour, what is the cycle time?

Answer: 30 minutes per customer Feedback: Cycle time = 1/flow rate, 2/ 60 minutes= 30 minutes per customer or since flow rate = 2 customers per hour, then cycle time = ½ hr. = 30 mins. per customer.

2.

•

What is the capacity of the nurse team over the course of a 9-hour workday?

Answer: 72 visits per 9-hour work day Feedback: 12 nurses × 9 hours/1.5 hours per visit = 72 visits

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Assuming the demand for the nurses is 60 patients per day, what is the utilization of the nurse team?

Answer: 83 percent Feedback: Utilization = demand/ capacity = 60/72 = 83%

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Assuming the demand for the nurses is 60 patients per day, what is the cycle time?

Answer: 9 minutes per patient Feedback: (9 hours × 60 minutes) / 60 patients = 9.0 minutes per patient

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

3.

•

Draw a process flow diagram of this process.

Resource 1 6 mins/unit

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Resource 2 3 mins/unit

Resource 3 5 mins/unit

What is the capacity of resource 2?

Answer: 0.333 units per minute Feedback: 1 unit / 3 minutes per unit = 0.333 units per minute.

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What is the bottleneck in this process?

Answer: Resource 1 Feedback: Resource 1 has the longest processing time. Or, calculate flow rates for all resources and the smallest flow rate is the bottleneck resource.

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What is the utilization of resource 2?

Answer: 50% Feedback: Demand = 60 minutes / 6 minutes per unit Output of Resource 1(bottleneck) = 10 units, Capacity of Resource 2 = 60 minutes / 3 minutes per unit = 20 units Utilization R2 = 10 units / 20 units = 50%

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

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How long does it take the process to produce 200 units starting with an empty system, assuming this is a worker-paced process?

Answer: 1,208 minutes Feedback: 200 units × 6 minutes (Resource 1) + 3 minutes (Resource 2) + 5 minutes (Resource 3) = 1,208 minutes, since R1 is the bottleneck and there would be no waiting time for R2 and R3.

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

4.

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Draw a process flow diagram of this process.

Bottling 1 sec/btl

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Labelin g3 secs/btl

Apply a

Lid 3 secs/btl

Packagin g4 secs/btl

What is the capacity (bottles/hour) at the resource ―Apply a lid‖?

Answer: 1,200 bottles per hour

•

What is the bottleneck in the process?

Answer: Packaging

•

Assuming unlimited demand, what would be the flow rate?

Answer: 900 bottles per hour

•

Assuming unlimited demand, what would be the utilization at resource “Apply a lid”?

Answer: 75%

•

Assume the process started empty and that this is a machine-paced process. How long would it take to produce 500 bottles?

Answer: 1,996 + 16 = 2,012 seconds Feedback: There are 4 stations and the cycle time is 4 seconds. Total time to produce 500 units is 4 x 4 + (500-1) x 4 = 2012

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

5.

•

Draw a process flow diagram of this process.

SelfServe 5 min/pt

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Receptionist 5 mins/pt

Dental Assistant 15 mins/pt

Dentist 30 mins/pt

What is the capacity (patients/hour) at the resource ―Dentist‖?

Answer: 20 patients per hour Feedback: 10 workers × (60 minutes per hour/30 minutes per patient) = 20 patients per hour.

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What is the bottleneck in the process?

Answer: The dental assistants Feedback: The dental assistants have the lowest capacity of 12 patients per hour, 3 dental assistants × (60 minutes per hour/ 15 minutes per patient) = 12 patients per hour.

•

Assuming unlimited demand, what would be the flow rate?

Answer: 12 patients per hour Feedback: The Flow Rate = 12 patients per hour, 3 dental assistants × (60 minutes per hour/ 15 minutes per patient) = 12 patients per hour.

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Assuming unlimited demand, what would be the utilization at resource ―Receptionists‖?

Answer: 50%

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Feedback: Receptionists capacity = 2 receptionists × (60 minutes per hour/5 minutes per patient) = 24 patients per hour.

Utilization = 12 patients per hour at the bottleneck/24 patients per hour capacity = 50%

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Assuming the process started empty. How long would it take to serve 20 patients?

Answer: 150 minutes Feedback: Time to produce N units when starting with an empty worker-paced system = Sum of processing times + (N-1) x Cycle time, where Cycle time = 1 / Flow rate, which in this case is 1/(12 patients per hour) = 1/(0.2 patients per minute). Therefore, time to serve 20 patients = (5 min + 5 min + 15 min + 30 min) + (20-1) x (5 min) = 150 minutes.

Note: there are no questions corresponding to parts (g) and (h) in the text.

6.

•

What is the capacity of the baking process step (in batches per hour)?

Answer: 5 batches per hour Feedback: Since there is only one machine at each process step, the capacity is the reciprocal of the activity time. The capacity of the baking process step is 1 / 12 batches per minute * 60 minutes per hour = 5 batches per hour.

•

What is the bottleneck of the manufacturing process?

Answer: Cooling Feedback: Since there is only one machine at each process step, the bottleneck is the step with the most work to do.

•

Assuming unlimited demand, what is the process flow rate (in batches per hour)?

Answer: 3.33 batches per hour

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Feedback: The process flow rate is the capacity of the bottleneck, which is the cooling process. The process flow rate is 1 / 18 batches per minute * 60 minutes per hour = 3.33 batches per hour.

•

Assuming unlimited demand, what is the utilization of the mixing process step?

Answer: 83 percent Feedback: The capacity of the mixing process step is 1 / 15 batches per minute * 60 minutes per hour = 4 batches per hour. The process flow rate is 3.33 batches per hour. The utilization is the flow rate divided by the capacity of the process step = 3.33 batches per hour / 4 batches per hour = 83%.

•

If the manufacturing process is currently full of work-in-process inventory, how long would it take to complete 50 batches of biscuits?

Answer: 15 hours to complete 50 batches.

Feedback: The time to complete X units in a full system is X * Cycle time. The cycle time is 1 / 3.33 hours, so it takes 50 / 3.33 = 15 hours to complete 50 batches.

7.

•

What is the bottleneck of the process?

Answer: Office managers Feedback: The capacity of each resource is the number of employees / activity time.

The capacity for the office managers is 2 / 4 = 0.50 loans per hour, which is less than the capacities of either of the other two resources.

•

Assuming unlimited demand, what is the process flow rate (in loans per week)?

Answer: 20 loans per week Feedback: The flow rate is the capacity of the bottleneck, which is 2 / 4 = 0.50 loans per hour * 8 hours per day * 5 days per week = 20 loans per week.

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

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If the customer demand is 18 loans per week, what is the utilization of the office managers resource?

Answer: 90 percent Feedback: If the demand is 18 loans per week, this is the process flow rate because it is lower than the process capacity. The utilization of the office managers’ resource is Flow rate / Resource capacity = 18 loans per week / 20 loans per week = 90%.

•

Assuming that the office currently has no backlog of loans that it is processing, how long will it take to complete 10 loans?

Answer: 30 hours Feedback: This is a worker-paced process, so the first loan will take 1 + 7 + 4 = 12 hours to complete. The remaining loans have a cycle time of 1 / 0.50 = 2 hours per loan. This means that the remaining loans will take an additional 9 * 2 = 18 hours to complete. The total time to complete 10 loans is 12 + 9 * 2 = 30 hours.

8. Assuming that the process starts out empty, how long will it take (in hours) to complete a batch of 105 units?

Answer: 54 hours Feedback: Step 3 is the bottleneck step of the process, so the machine-paced process will be set at a speed of 1/2 hour per unit at each step. The first unit will take 4 * 1/2 = 2 hours to complete. The remaining 104 units will have a cycle time of 1/2 hour.

The total time required to complete a batch of 105 units is 2 + 104 * 1/2 = 54 hours.

9. Assuming that the process starts out empty and is worker-paced, how long will it take (in minutes) to serve 20 customers?

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Answer: 128 minutes to serve 20 customers Feedback: The bottleneck is the first step of the process, which means that the cycle time for all customers other than the first one is 5 minutes. The first customer takes 5 + 15 + 10 + 3 = 33 minutes to complete service. As a result, it takes a total of 33 + 19 * 5 = 128 minutes to serve 20 customers.

10.

•

If the car wash has a demand of 15 cars per hour, what is the flow rate of the process?

Answer: 12 customers per hour Feedback: The bottleneck is the Wash process, with a capacity of 12 cars per hour.

The flow rate is the minimum of the capacity of the bottleneck and the demand.

•

If the car wash has a demand of 15 cars per hour, what is the utilization of the machine that performs the wax process?

Answer: 60 percent Feedback: Utilization is the flow rate divided by the capacity of the process step. If the demand is 15 cars per hour, the flow rate is 12 cars per hour, because the flow rate is the minimum of the capacity of the bottleneck and the demand. The capacity of the machine that performs the Wax process is 1 / 3 * 60 = 20 cars per hour. The utilization of this machine is 12 / 20 = 60%.

11.

•

Assuming unlimited demand, what is the flow rate of the process in customers per hour?

Answer: 12 per hour Feedback: The bottleneck resource is the license processing station. Each machine can process 4 customers per hour, and there are three machines; thus, the flow rate through this stage (and the process as a whole) is 12 customers per hour. All of the other steps have higher capacities.

•

Assuming unlimited demand, what would the new flow rate be if the center added one server to the bottleneck resource?

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Answer: 15 per hour Feedback: The current bottleneck is the license processing station. Each machine has a capacity of 4 customers per hour, so adding one machine would increase the capacity of this station to 16 customers per hour. The cashiers each have a capacity of 7.5 customers per hour, so the capacity of this resource is only 15 customers per hour. The cashiers will now be the bottleneck, and the process flow rate is 15 customers per hour.

CASE

The chapter’s “Connection side-box” showcases the Tesla plant. The key goal of this mini-case is that students learn how to take a complex process as shown in the video and then abstract this information to something simple as a process flow diagram.

The process flow diagram can take various forms / there does not exist a uniquely right answer.

We suggest a diagram such as the following:

As far as the calculations are concerned: 1. Cycle time: the process operates for 80 hours per week. It produces 500 cars per week. So, the cycle time is 80 hours / 500 cars = 0.16 hours/car = 9.6 minutes/car 2. The flow time of the process is the 3-5 days it takes a car to journey through the process 3. Inventory can exist between steps and at the beginning as raw materials. However, most of the inventory is likely to sit in each of the boxes as work in process inventory. This includes many cars that are half assembled as well pieces of metal that are waiting to be joined together.

4. The total inventory is I = R*T = 100 cars per day * 4 days = 400 cars.

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

CHAPTER 4 PROCESS IMPROVEMENT

CONCEPTUAL QUESTIONS 1. A process is replicated in another country where wages are 50 percent lower. Staffing and processing times are identical. What would be the effect on the costs of direct labor?

Answer: A. Costs of direct labor would be 50 percent lower.

2. You and three of your friends run a car wash for a fund-raiser. Between interior and exterior cleaning, you spend about 40 minutes per vehicle. You are so successful that the next day you invite four more friends to help. How does this impact the labor content?

Answer: B. The labor content stays the same, 3. A nonbottleneck worker currently has an idle time of 20 seconds per unit. Because of the large demand, the company improves the process by adding more capacity to the bottleneck. How does this impact the idle time of the worker?

Answer: A. The idle time would decrease.

4. A group of workers works really hard. In fact, they work so hard that one of them claims to have an average labor utilization of 120 percent. Is that possible?

Answer: B. No 5. The bottleneck resource in a process has the least idle time. True or false?

Answer: A. True 6. Which of the following statements about takt time and cycle time is true?

Answer: A. Takt time only depends on demand, not capacity. Cycle time does depend on capacity.

7. If the takt time is shorter than the cycle time, the process needs to run faster. True or false?

Answer: A. True 8. How does the takt time change as the demand rate increases?

Answer: C. The takt time decreases.

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

9. How does the target manpower change as the demand rate increases?

Answer: A. The target manpower increases.

10. What happens to the target manpower if the labor content is doubled?

Answer: B. The target manpower doubles.

11. Which of the following actions does not relate to off-loading the bottleneck?

Answer: C. Increasing wages for production workers 12. It is possible to increase process capacity by balancing a process. True or false?

Answer: A. True 13. Balancing a process with a fixed sequence of activities will achieve a higher average labor utilization than balancing a process with no fixed sequence of activities. True or false?

Answer: B. False 14. Specialization increases the costs of labor because specialists will command a higher wage rate. True or false?

Answer: B. False 15. Specialization typically lead to a higher average labor utilization. True or false?

Answer: B. False 16. A process has high fixed costs and low variable costs. It is currently capacityconstrained. Will the impact of an efficiency improvement be small or large?

Answer: B. Large Feedback: Low variable costs result in a high unit margin, so every additional customer yields a significant contribution to profitability.

17. A process has low fixed costs and high variable costs. It is currently capacityconstrained. Will the impact of an efficiency improvement be small or large?

Answer: A. Small Feedback: High variable costs result in a low unit margin, which reduces the impact of each additional customer on profitability.

18. When an operation improves its efficiency, its revenue will always stay constant, while its costs will go down. True or false?

Answer: B. False Feedback: Efficiency can result in higher revenue as well because the operation can serve more customers (an increase in throughput rate).

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

PROBLEMS AND APPLICATIONS 1. What is the cost of direct labor for the lawyer expressed in $ per customer?

Answer: $800 per customer Feedback: The direct labor cost is (8 hours × $200 per hour)/2 customers = $800 per customer.

2. (a) What is the cost of direct labor?

Answer: $3.60 per unit Feedback: Resource 1 is the bottleneck center at 6 minutes per unit. The direct labor cost is (6 minutes per resource × 3 resource centers/60 minutes per hour) × $12 per hour = $3.60 (b) What is the labor content?

Answer: 14 minutes per unit Feedback: Labor content = 6 minutes; resource 1 + 3 minutes; resource 2 + 5 minutes; resource 3 = 14 minutes (c) How much idle time does the worker at resource 3 have per unit?

Answer: One. Feedback: Worker at resource 3 has 6 minutes – 5 minutes = 1 minute of idle time per unit since resource 1 is the bottleneck. (d) What is the average labor utilization?

Answer: 77.78 percent Feedback: Labor Utilization = 14 demand/18 capacity = 77.78% (e) Assume the demand rate is 20 units per hour. What is the takt time?

Answer: 3 minutes per unit Feedback: Takt Time = 1/Demand rate = 1/(20 customers/60 minutes) = 3 minutes (f) Assume the demand rate is 20 units per hour. What is the target manpower?

Answer: 4.67 minutes per unit 28

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Feedback: The target manpower = 14 Labor Content/3 minutes Takt Time = 4.67

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

3. (a) What is the labor content?

Answer: 50 minutes per patient (b) What is the cost of direct labor?

Answer: $95.83 per patient Feedback: $1,150 (10 x $100/hr + 5 x $30/hr) per hour / 12 patients per hour = $95.83 per patient 4. (a) What is the labor content?

Answer: 60 minutes per customer Feedback: Labor content = 10 minutes Activity 1 + 10 minutes Activity 2 + 10 minutes Activity 3 + 25 minutes Activity 4 + 5 minutes Activity 5 = 60 minutes (b) What is the average labor utilization?

Answer: 66.67 percent Feedback: 60 minutes labor content / (3 employees x 30 minutes bottleneck) = 66.67% (c) At a wage rate of $20 per hour, what is the cost of direct labor per customer?

Answer: $30 per customer Feedback: $60 per hour / 2 customers per hour = $30 per customer (d) How would this change his cost of direct labor?

Answer: $26.67 per customer Feedback: $80 per hour / 3 customers per hour = $26.67 per customer (e) How would this change his cost of direct labor?

Answer: $25 per customer Feedback: $60 per hour / (60/25) customers per hour = $25 per customer 5. (a) What is the labor content?

Answer: 395 seconds per watch Feedback: Labor content = 68 seconds Station A + 60 seconds Station B + 70 seconds Station C + 58 seconds Station D + 75 seconds Station E + 64 second Station F = 395 seconds 30

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

(b) Assume a demand rate of 50 watches per hour. What is the takt time?

Answer: 72 seconds per watch Feedback: Takt Time = 1/Demand rate = 1/(50 watches/3600 seconds per hour)= 72 seconds (c) Assume a demand rate of 50 watchers per hour. What is the target manpower?

Answer: 5.49 workers Feedback: 395 / 72 = 5.49 (d) How would that impact process capacity?

Answer: 0.0 watches per hour Feedback: There is no impact. (e) How could you increase the capacity of the process by rebalancing it?

Answer: D. Move cosmetic inspection (step 14) to station D 6. (a) What is the cost of direct labor for the bicycle?

Answer: $1.25 per unit Feedback: 5 * 15 / 60 = $1.25 per unit (b) How much profit does the company make per hour?

Answer: $25 per hour Feedback: Revenues are $360 ($6*60) per hour and costs are $335 (FC $200 + $1.25*60 bikes per hour + $1*60 parts) per hour so profits are $25 per hour. (c) What would be the profits per hour if Atlas would be able to source the parts 10 percent cheaper ($0.90 for the parts of one unit)?

Answer: $31 per hour Feedback: Savings = 60 units per hour x $0.10 = $6.00. $6.00 savings + $25 per hour (see part B) = $31.00 or Revenues = $360 per hour and costs are now $329 per hour so profits are now = $31 per hour. (d) What would be the profits per hour if Atlas were able to reduce fixed costs by 10 percent (to $180 per hour)?

Answer: $45 per hour Feedback: $20 fixed cost savings + $25 (see part B) = $45 per hour.

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

(e) What would be the profits per hour if Atlas were able to reduce the processing time at the bottleneck by 5 seconds per unit (assume unlimited demand)?

Answer: $52.27 per hour Feedback: It now takes 55 seconds to make each unit. 3600 seconds per hour divided 55 seconds per unit = 65.4545 units. Revenues are $392.73 - $65.46 parts - $75 labor $200 fixed coast = $52.27.

CASE Xootr The mini-case for this chapter is the Xootr manufacturing process.

The process flow diagrams are as follows:

We compare the two proposals assuming that there are no benefits of specialization.

Proposal

Specialized

One worker does it all

Number of workers

6

6

Wages per hour

6*12

6*12

Processing times

6.5., 6.5, 5.5, 5.5, 4, 4

32

Bottleneck

First resource

First resource

Capacity

1/6.5 units per minute

6/32 units per minute

Cost of direct labor

72 / (60/6.5)

72 / (6/32)

Idle time

0, 0, 1, 1, 2.5, 2,5

0

Labor content

32 minutes per unit

32 minutes per unit

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Labor utilization

=32/(32+7)

1

Variables that might differ: the question is if it would be possible to cut the processing times reflecting specialization gains in the first proposal. Also, the first proposal could be improved if the line could be better balanced

CHAPTER 5 PROCESS ANALYSIS WITH MULTIPLE FLOW UNITS CONCEPTUAL QUESTIONS 1. A process has six resources and three types of flow units. How many rows will there be in the demand matrix?

Answer: C. 6 2. A process has 10 resources and two types of flow units. How many columns will there be in the demand matrix?

Answer: A. 2 3. The total demand rate for a resource is the sum of the individual demand rates that need to be processed by the resource. True or false?

Answer: A. True 4. The resource with the largest total demand rate is the bottleneck. True or false?

Answer: B. False.

Feedback: This really depends on the available capacity.

5. The implied utilization can never be bigger than 100 percent. True or false?

Answer: B. False.

Feedback: When demand exceeds capacity, the implied utilization is bigger than 100%.

6. A resource has an implied utilization of 150 percent. This is the highest implied utilization in the process. This means that the resource is the bottleneck. True or false?

Answer: A. True

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

7. A process has a yield of 50 percent. How many units are needed as inflow to create an outflow of 40 units per day?

Answer: C. 40 / 0.5 = 80 units per day 8. What is the highest possible yield a process can obtain?

Answer: B. 100%

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

9. To find the bottleneck in a process with attrition loss, all we need to do is find the resource with the lowest capacity. True or false?

Answer: B. False.

Feedback: We have to also look at the demand matrix.

10. As the attrition loss in a process increases, what happens to the yield?

Answer: C. The yield decreases.

11. What flow unit do you use when dealing with flow unit–dependent processing times?

Answer: D. Minutes or hours of work 12. The workload matrix has exactly the same number of rows and the same number of columns as the demand matrix. True or false?

Answer: A. True 13. If a flow unit type j does not consume any time at a resource i, the corresponding cell in the workload matrix, WL(i, j), is zero. True or false?

Answer: A. True 14. Rework can increase the costs of an operation but has no impact on the capacity of the process. True or false?

Answer: B. False 15. What is the relationship between a process with rework and a process with scrap?

Answer: D. Scrap is a special case of rework in which flow units have to repeat all resources in the process up to the defect.

Feedback: In a scrap process, units do not visit a resource multiple times whereas that is possible in a rework process.

16. In a process with rework, we find the bottleneck as the resource with the highest implied utilization. True or false?

Answer: A. True

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

PROBLEMS AND APPLICATIONS 1. What is the total demand rate for each of the three resources?

Answer: 50 per week for the administrative support person.

Answer: 10 per week for the senior accountant.

Answer: 40 per week for the junior accountant.

Feedback: Step 1: Compute the demand matrix

D=

(Administrator) (Senior ac.)

(Junior ac.)

(Group 1) 0.2 * 50 0.2 * 50 0

(Group 2) 0.8 * 50 0 0.8 * 50

Step 2: Compute the total demand rates, which are 50 per week for the administrator, 10 for the senior accountant, and 40 for the junior accountant.

2. (a) Which of the three persons is the bottleneck?

Answer: The administrative support person is the bottleneck because he/she is the first step in process. (b) For a mix of 20:80 between new and old cases, what is the flow rate and what is the capacity?

Answer: Flow rate is 40 hours per week * 60 mins / 20 mins per unit = 120 units per week.

3. (a) Where is the bottleneck, according to the current staffing plan?

Answer: The written exam Feedback: We follow the 4-step process outlined in Exhibit 5.2.

Step 1: We find the demand matrix by starting with one good unit (a student that passes).

Given the 40% failure rate, we need 1 / 0.6 = 1.67 students to take the road exam to get one succeed. In the same way, we compute that we need 1.667/ 0.85 = 1.96 students to take the written exam, and 1.96 / 0.9 = 2.18 students to show up. So the demand matrix is: 36

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

D=

(ID) (Written) (Road)

(2.18) (1.96) (1.67)

Step 2: The capacity levels are simply computed as the number of workers divided by the processing times: Capacity(ID) = 4/5 units / minute = 384 units per day Capacity(Written) = 2/3 units / minute = 320 units per day Capacity(Road) = 15/20 units / minute = 360 units per day Step 3: Compute the levels of implied utilization IU(ID) = 2.18 / 0.8 = 2.72 IU(Written) = 1.96 / 0.67 = 2.94 IU(Road) = 1.67 / 0.75 = 2.22 So the bottleneck is at the written exam. (b) What is the capacity of the process (expressed in approved cases per day)?

Answer: 163 good units per day Feedback: Step 4: Find the capacity of the process in terms of good units by dividing 1 good unit of output by the highest implied utilization: Capacity(ID) = 1 / 2.94 = 0.34 good units per minute = 163.2 good units per day

4. We use the 5-step process outlined in Exhibit 5.1. (a)What is the bottleneck?

Answer: Resource 3 Feedback: Step 1: Compute the workload matrix

(Type A)

(Type B)

(Type C)

(Resource 1) (40 * 5

50 * 5

60 * 5)

(Resource 2) (40 * 4

50 * 4

60 * 5)

WL = (Resource 3) (40 * 15 0 (Resource 4) (0

0) 50 * 3 37

60 * 3)

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

(Resource 5) (40 * 6

50 * 6

38

60 * 4)

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Step 2: Add up the workloads for each resource: Resource 1: 750 minutes per day Resource 2: 660 minutes per day Resource 3: 600 minutes per day Resource 4: 330 minutes per day Resource 5: 780 minutes per day

Step 3: Compute the available time per day at each resource Resource 1: 2 * 480 = 960 minutes per day Resource 2: 2 * 480 = 960 minutes per day Resource 3: 1 * 480 = 480 minutes per day Resource 4: 1 * 480 = 480 minutes per day Resource 5: 2 * 480 = 960 minutes per day

Step 4: Compute the implied utilization levels as demand rate in minutes of work divided by the available time in minutes of work: Resource 1: 750 minutes per day / 960 minutes per day = 0.78 Resource 2: 660 minutes per day / 960 minutes per day = 0.69 Resource 3: 600 minutes per day / 480 minutes per day = 1.25 Resource 4: 330 minutes per day / 480 minutes per day = 0.6875 Resource 5: 780 minutes per day / 960 minutes per day = 0.8125

Resource 3 is the bottleneck because it has the highest implied utilization.

(b) What is the flow rate for each flow unit, assuming that demand must be served in the mix described above (i.e., for every four units of A, there are five units of B and six units of C)?

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Answer: The flow rate for A is 32 units per day; B is 40 units per day; and C is 48 units per day.

Feedback: Step 5: The process is capacity constrained. So we have to divide all flows by 1.25: Flow A = 40 / 1.25 = 32 units per day Flow B = 50 / 1.25 = 40 units per day Flow C = 60 / 1.25 = 48 units per day 5. We use the framework outlined in Exhibit 5.4. (a) Where in the process is the bottleneck?

Answer: Step 2 Feedback: Step 1: We compute the workload matrix Good Rework 1 (0.7 * 5 0.3 * [5 + 5]) 2 (0.7 * 6 0.3 * [6 + 6]) WL = 3 (0.7 * 3 0.3 * [3 + 3]) 4 (0.7 * 4 0.3 * 4) Step 2: This gives us a total workload for one good unit for each of the resources as follows: Resource 1: Resource 2: Resource 3: Resource 4:

0.7 * 5 + 0.3 * [5 + 5] = 6.5 minutes per good unit 0.7 * 6 + 0.3 * [6 + 6] = 7.8 minutes per good unit 0.7 * 3 + 0.3 * [3 + 3] = 3.9 minutes per good unit 0.7 * 4 + 0.3 * 4 = 4 minutes per good unit

Step 3: The available time at each resource corresponds to 60 minutes per hour.

Step 4: This gives us the following implied utilizations: Implied utilization (Resource 1) = 6.5 / 60 = 0.11 Implied utilization (Resource 2) = 7.8 / 60 = 0.13 Implied utilization (Resource 3) = 3.9 / 60 = 0.065 Implied utilization (Resource 4) = 4 / 60 = 0.0667 Thus, step 2 is the bottleneck and the highest implied utilization is 0.13.

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

(b) What is the capacity of the process?

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Answer: 7.69 good units per hour Feedback: Step 5: Compute the capacity as 1 good unit per hour / 0.13 = 7.69 good units per hour.

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

CASE Airport Security The process flow diagram looks as follows:

Let M be the total number of passengers arriving and p be the amount of time spent with the CBP officer in the traditional process for a US citizen / permanent resident.

For the 80-10-10 mix, the total workload on the CBP officers would be: Traditional Process: 0.8*M * (0.5* p + 0.5 * 2p)

= 0.8 * M * 1.5 p = 1.2 Mp

NEXUS Process: Mp

= 0.1 * M * 0.075p = 0.0075

+ 0.1 * M * (0.95* 0 + 0.025 p + 0.025 * 2p)

New Kiosk Process: + 0.1 * M * ( 0.5* 0.5*p + 0.5 * p)

= 0.1 * M * 0.75p = 0.075 Mp

TOTAL = 1.2825 Mp

For the 70-10-20 mix, the total workload on the CBP officers would be: Traditional Process: Mp

0.7*M * (0.5* p + 0.5 * 2p)

NEXUS Process:

+ 0.1 * M * (0.95* 0 + 0.025 p + 0.025 * 2p)

43

= 0.7 * M * 1.5p = 1.05 =0.1 * M * 0.075p = 0.0075 Mp

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

New Kiosk Process:

+ 0.2 * M * ( 0.5* 0.5*p + 0.5 * p)

= 0.2 * M * 0.75p = 0.15 Mp

TOTAL = 1.2075 Mp

Reduction in workload for CBP officer: 1.2075 Mp / 1.2825 Mp = 0.94152 => a little less than six percentage reduction in workload

CASE Reverse Logistics Recycle rate = 10% = 0.1

Recovery rate = 80% = 0.8

Lithium (Li) price - $9/kg

Each unit (battery) has 10 kgs. of Li

Year 2035 2036 2037 2038 2039 2040 2041 2042 2043 2044 2045 2046 2047 2048 2049 2050

(incr. 50% / yr.)

Mkt. Size / units 2100000 3150000 4725000 7087500 10631250 15946875 23920313 35880469 53820703 80731055 121096582 181644873 272467310 408700964 613051447 919577170

(10% of mkt.)

Amt. Recycled 210000 315000 472500 708750 1063125 1594688 2392031 3588047 5382070 8073105 12109658 18164487 27246731 40870096 61305145 91957717

(10kg Li /unit) Tot. Amt. of Li 2100000 3150000 4725000 7087500 10631250 15946875 23920313 35880469 53820703 80731055 121096582 181644873 272467310 408700964 613051447 919577170

(80% of Li recvd.)

Amt. Recovered 1680000 2520000 3780000 5670000 8505000 12757500 19136250 28704375 43056563 64584844 96877266 145315898 217973848 326960771 490441157 735661736

So, the revenue in 2050 should be approx. $6.6 billion.

CHAPTER 6 44

Mkt. grows at 50% each year

Li @ $9/kg Revenue / $ $15,120,000 $22,680,000 $34,020,000 $51,030,000 $76,545,000 $114,817,500 $172,226,250 $258,339,375 $387,509,063 $581,263,594 $871,895,391 $1,307,843,086 $1,961,764,629 $2,942,646,943 $4,413,970,415 $6,620,955,623

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

LEARNING CURVES CONCEPTUAL QUESTIONS 1. A bank is underwriting loans for small businesses. Currently, about 5 percent of the underwriting decisions are found to be incorrect when audited by the bank’s quality assurance department. The bank has a goal of reducing this number to 1 percent. What form of an improvement trajectory is most likely to occur?

Answer: B. Exponential decay 2. A bakery produces cookies; however, it makes some defects, leading to occasionally broken or burnt cookies. Presently, the yield of the process is 90 percent (i.e., 9 out of 10 cookies are good). The bakery has a goal of producing 99 percent good cookies. What form of an improvement trajectory is most likely to occur?

Answer: C. Diminishing return growth 3. A regional rail company wants to reduce its delays. Presently, 70 percent of the trains arrive on time. The company’s goal is to improve this to 95 percent. What form of improvement trajectory will most likely occur?

Answer: C. Diminishing return growth 4. A novice rower is practicing for a 2000-meter test on a rowing machine (an ―erg‖). He presently takes 7 minutes and 10 seconds to complete the distance. His goal, in order to be recruited by a good college, is 6 minutes and 30 seconds. What form of improvement trajectory will most likely occur?

Answer: B. Exponential decay 5. The power law describes a learning process with exponential growth in performance. True or false?

Answer: B. False 6. Assume a learning curve following the power law. If you double the initial cost c(1) of a product, the cost of the 21st unit, c(21), will: Answer: C. be doubled.

7. Consider two processes, A and B, with an initial cost of c(1) = 20. Process A has a learning rate of LR = 0.95 and process B has a learning rate of LR = 0.8. After 20 units: Answer: B. process B is cheaper.

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8. A learning curve following the power law has what shape in the log-log graph?

Answer: D. Linearly decreasing 9. When estimating the slope of the line in a log-log graph, we pick two points with coordinates (x1, y1) and (x2, y2). Which of the following statements is true?

Answer: D. The slope of a line can be negative or positive. For a positive rate of learning, however, the slope is always negative.

10. A lower learning rate corresponds to a steeper slope in a log-log graph of the learning curve. True or false?

Answer: A. True 11. A firm has a learning rate almost equal to 1 (just slightly less than 1). What does this imply for the slope of the learning curve in a log-log graph?

Answer: A. The slope is close to 0 12. The learning curve coefficient LCC(10, 0.8) captures: Answer: A. The cost to produce one unit in a process with c(1) = 1, a cumulative output of 10, and a learning rate of 0.8 13. A higher value of the learning rate (LR) leads to a lower value of the learning curve coefficient LCC(x, LR). True or false?

Answer: B. False 14. Instead of using the learning curve coefficient method, it is possible to compute the costs of the Nth unit as c(N) = c(1) × Nb, where b is the slope of the learning curve in the log-log graph. True or false?

Answer: A. True 15. The cumulative learning curve coefficient CLCC(x, LR) is always bigger than the learning curve coefficient LCC(x, LR) for all x > 1. True or false?

Answer: A. True 16. The cumulative learning curve coefficient CLCC(20, 0.8) is defined as follows: Answer: A. The cost to produce 20 units in a process with c(1) = 1 and a learning rate of 0.8 46

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

17. Assume a learning curve following the power law. If you double the initial cost c(1) of a product, the costs of producing an order of 20 units: Answer: C. Are doubled 18. How does the average tenure of an employee relate to the average time an employee spends with the company?

Answer: C. The average tenure is half the average time an employee spends with the company 19. The employee turnover can be computed as: Answer: A. Number of new employees recruited per year / Average number of employees 20. If the employee turnover increases, the average tenure of an employee in the company: Answer: B. Decreases 21. In an environment with high employee turnover, standards are less important than in environments with low turnover. True or false?

Answer: B. False 22. Which of the following is not part of the standard work sheet?

Answer: B. The name of the person in charge of the activity 23. John has been fixing bicycles for 3 years now. He notices that he is getting better with an increase in experience, though he does not necessarily know why. John’s learning is most likely a form of autonomous learning. True or false?

Answer: A. True Feedback: John’s learning most likely is a form of autonomous learning 24. Which of the following activities is not part of the Deming cycle?

Answer: D. Improve 25. A high signal-to-noise ratio makes learning harder. True or false?

Answer: B. False

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

PROBLEMS AND APPLICATIONS 1) Consider the trajectory showing the percentage of customer orders in a restaurant that were handled correctly. What shape would a learning curve have in this setting?

Answer: C. Diminishing return growth.

Feedback: Diminishing returns grow shows a performance trajectory in which the improvement rate decreases over time or experience.

2) Consider the trajectory showing the number of luggage pieces that an airline loses on a flight. What shape would a learning curve have in this setting?

Answer: B. Exponential decay.

Feedback: The rate of performance improvement decreases over time; this is the case for many cost reductions. Because costs cannot become negative, cost reductions tend to exhibit diminishing returns.

3) Consider the trajectory showing the amount of data storage space that comes with the average PC each year. What shape would a learning curve have in this setting?

Answer: A. Exponential growth.

Feedback: Exponential growth shows the improvement trajectory in which the rate of improvement increases over time or experience.

4) Consider a process that makes high-end boards that get mounted on skateboards. The process starts with a unit cost of $20 for the first unit—that is, c(1) = 20—and has a learning rate of LR = 0.95. What will be the unit cost for the 128th unit?

Answer: $13.96675 unit cost Feedback: To reach the 128th unit, we have to double the cumulative output seven times (from 1 to 2, from 2 to 4, from 4 to 8, from 8 to 16, from 16 to 32, from 32 to 64, and from 64 to 128). We can then use the formula: c(after doubling cumulative output 7 times) = c(1) x LR7 = 20 x 0.957 = 13.96675 5) Consider a process restringing tennis rackets. The process starts with a unit cost of $10 for the first unit—that is, c(1) = 10—and a learning rate of LR = 0.9. What will be the unit cost for the 35th unit?

Answer: $5.825 unit cost Feedback: We can directly use the formula: c(35) = c(1) x LR log2 35 = c(1) x LR ln(35)/ln(2) = 10 x 0.9 5.129 = 5.825 49

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

6) An experienced car mechanic is working on changing the exhaust system. In 2010, the mechanic had performed this operation 100 times over her career. She estimates that, on average, it took her 80 minutes to change one exhaust system. By 2014, she had performed that operation 220 times, and it took her about 55 minutes to change one exhaust system. The learning curve in a log-log graph appears to be linear. By how much does the mechanic reduce the processing time of one operation with each doubling of the cumulative output?

Answer: 28% to 29% Feedback: We choose x2 as the cumulative number of operations done by 2014 and get: Ln(processing time) = Ln(55) = 4.007, and Ln(Cumulative operations performed)= Ln(220) = 5.393 Similarly, if we choose x1 as the cumulative number of operations done by 2008, it yields: Ln(processing time) = Ln(80) = 4.382, and Ln(Cumulative operations performed)= Ln(100) = 4.605 So, we can compute the slope as: Slope b = [ln(c(x2)) – ln(c(x1))] / [ln(x2 ) – ln(x1)] = [4.007 – 4.382] / [5.393 – 4.605] = -0.375 / 0.788 = - 0.475 Thus, we obtain a slope of the learning curve in a log-log graph of b = -0.475. We can then use Table 6.2 to find the corresponding learning rate LR, which is between 0.71 and 0.72. So the mechanic reduces the processing time by 28% to 29% every time that she doubles the cumulative output.

7) Consider the preparation of income tax statements. The process starts with an initial cost c(1) = 45 and a learning rate of LR = 0.95, and by now has reached a cumulative output of 100. Using the LCC method, what unit costs do you expect for the 100th unit?

Answer: $32.0045 Feedback: We start by looking up the learning curve coefficient in Table 6.3 for a cumulative output of 100 and a learning rate of LR = 0.95. We get: LCC(100,0.95) = 0.711212 We then compute the costs as: c(after 100 units of cumulative output, LR = 0.95, c(1) = 45) = 45 * LCC(100,0.95) = 32.0045 50

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Thus, we expect a unit cost of $32.0045.

8) Consider again the preparation of income tax statements. The process starts with an initial cost c(1) = 45 and a learning rate of LR = 0.95, and by now has reached a cumulative output of 100. Without using the LCC method, what do you expect production costs to be for the 100th unit?

Answer: $32.0045 Feedback: We first have to look up the slope of the learning curve, b, given the learning rate, LR. In Table 6.2, we see that for LR = 0.95, we get a slope of b = -0.074. We then compute costs directly as c(1) x Nb = 45 x 100-0.074 = 32.0045 Thus, we expect a unit cost of $32.0045. This is within rounding difference of the answer to the tax prep question.

9) Will has just started a small company making special cakes for retirement parties. He just received a large order of five cakes from a local business. Will expects substantial productivity improvements in his work as he moves from the first cake in this order to the fifth cake. His current cost of making a cake is c(1) = 40 and he expects a learning rate of LR = 0.85. What will be his cost for all five cakes combined?

Answer: $161.2434 Feedback: We start by looking up the cumulative learning curve coefficient in Table 6.4 for an output of 5 and a learning rate of LR = 0.85. We get: CLCC(5,0.85) = 4.031086 We then compute the costs of making 5 units as: CC(of making 5 units, LR = 0.85, c(1) = 40) = 40 * CLCC(5,0.85) = 40 * 4.031086 = 161.2434 Thus, we expect a cumulative cost of $161.2434.

10) A company has 2200 employees, on average, and it recruits, on average, 300 employees per year. What is the employee turnover?

Answer: 13.64 percent Feedback: We compute the employee turnover as: Employee turnover = Number of new employees recruited per year / Average number of employees = 300 / 2,200 = 0.1364 51

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

11) A company has 2200 employees, on average, and it recruits, on average, 300 employees per year. What is the average tenure of an employee?

Answer: 3.6667 years Feedback: We first compute the employee turnover as: Employee turnover = Number of new employees recruited per year / Average number of employees = 300 / 2,200 = 0.1364 Next, we compute the average tenure as: Average tenure = 1 / (2 x Employee turnover) = 1 / (2 x 0.1364) = 3.6667 years 12) What are the three elements of a standard work sheet?

Answer: A. They are the processing time for an activity, the work sequence of all steps making up for the activity, and the standard amount of inventory at the resource.

13) Consider the navigation skills of two cab drivers (without a GPS). Driver 1 is driving her routes as demanded by the customer and improves her understanding of the region that way.

Driver 2 is deliberately taking time off from serving customers to explore new shortcuts. Driver 1 is relying on induced learning and Driver 2 is relying on autonomous learning. True or false?

Answer: B. False.

Feedback: Autonomous learning is improvements due to on-the-job learning of employees.

14) What is the difference between the PDCA cycle and the Deming cycle?

Answer: D. There is no difference between the two approaches.

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CASE Ford Plant Employee turnover = 1 / Average time employees spend on average in the company  Average time employees spend on average in the company = 1/3.7 = 0.27 years  Average tenure = 0.5 * 0.27 years = 0.135 * 365 days/year = 49 days Learning curve from 1909 to 1916 We choose x2 as the cumulative output in 1927 and get: Ln(cost) = Ln(3,372) = 8.123, and Ln(Cumulative output)= Ln(15,000,000) =16.5235 Similarly, if we choose x1 as the cumulative output in 1916, which yields: Ln(cost) = Ln(8,084) = 8.9977, and Ln(Cumulative output)= Ln(1,000,000) = 13.815 So, we can compute the slope as: Slope b = [ln(c(x2)) – ln(c(x1))] / [ln(x2 )– ln(x1)] = [8.123 – 8.9977] / [16.5235-13.815] = -0.32288 Thus, we obtain a slope of the learning curve in a log-log graph of b=-0.32288. We can then use table SLOPE to find the corresponding learning rate LR, which is between 0.79 and 0.80. So the process reduces its cost by 19 to 20% every time it doubles its cumulative output. Instead of using table SLOPE, we can also use the equation LR=2Slope which gives us LR=20.32288 =0.799471.

CASE Renewable Energy Q.1

Answer: 711.78

Since the process reduces its cost by 20% every time it doubles its cumulative output. Using the equation LR=2Slope which gives us Slope = ln(LR) / ln 2 =-0.32193.

Learning curve from 2023 to 2030: c(x1) = $100, c(x2) = $58, x1 = 2400 (output in 2023), x2 =? Using Slope b = [ln(c(x2)) – ln(c(x1))] / [ln(x2 )– ln(x1)] -0.32193 = [4.0604 – 4.6052] / [ln(x2) - 7.7832] 54

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

We get ln(x2) = 9.4753, and

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Cumulative output x2 =711.78

CHAPTER 7 PROCESS INTERRUPTIONS CONCEPTUAL QUESTIONS 1. A chemical process involves adding various agents to a vat and then waiting for the agents to react to make a final compound. The vat holds up to 1000 gallons. Chemical agents are added to the vat at a rate of 4.5 gallons per minute. Once the agents have been added to the vat, the reaction takes 85 minutes no matter the actual number of gallons in the batch. After the batch is complete, the vat must be cleaned. Cleaning always takes 15 minutes. What is the setup time of this process?

Answer: C. 100 minutes Feedback: It might seem that pouring the chemical agents into the vat is the ―setup process,‖ but that process actually depends on the amount of agent added to the vat. The reaction time of 85 minutes is independent of the volume in the batch, so it is the setup time. In addition, the 15 minutes of cleaning is also required and is independent of the amount added to the vat. Hence, the total setup time for each production cycle is 85 + 15 = 100.

2. A machine makes three different components used in a gyroscope. Call these components A, B, and C. The following repeating schedule is used to make components: make 100 units of A, make 50 units of B, and then make 200 units of C. How many components are made in the production cycle?

Answer: D. 350.

Feedback: The production cycle is the repeating sequence of A, B, and C. 3. A product requires four parts that are made on the same lathe. Call the parts A, B, C, and D. The product requires 1 unit of A, 2 of B, 4 of C, and 8 of D. Which of the following production cycles is most appropriate?

Answer: D. A:100; B:200; C:400; D:800 Feedback: The ratios of the quantities among the parts should match the ratios of the parts needed for the product. For example, twice as many B parts should be made as A parts.

4. Increasing the batch size on a resource with setups always increases the capacity of the resource. True or false?

Answer: A. True Feedback: A resource’s capacity always increases with its batch size.

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5. Increasing the batch size on a resource with setups always increases the capacity of the process. True or false?

Answer: B. False Feedback: The capacity of the resource with the setup time increases, but the bottleneck may be another process. So increasing the batch size does not necessarily increase the capacity of the process.

6. The batch size is tripled on a resource that has a setup time. What is the likely impact on the resource’s capacity?

Answer: C. Increase by less than 300% Feedback: The resource’s capacity will increase, but by less than a 3 fold increase.

7. If the flow rate of a process increases, then the utilization of a resource with a setup time must also increase. True or False?

Answer: A. True Feedback: Utilization of the resource with a setup time equals the flow rate times the processing time. So an increase in the flow rate means an increase in utilization.

8. Which of the following is likely to increase the utilization of a resource with a setup time?

Answer: A. Increase the flow rate 9. A manager is concerned that there isn’t enough time spent on production and too much time spent on setups. The manager decides to double all production batch sizes. This change has no impact on demand. What impact will this likely have on the average inventory in the process?

Answer: C. Average inventory will increase because larger batches require more time to be completed.

10. Which of the following is most likely to be a concern if batches are very large?

Answer: C. Inventory will be too high.

11. If the batch size is increased, inventory increases, which implies that flow units are likely to spend more time in the process. True or False?

Answer: A. True 12. Suppose a resource has setup times and a setup time must be incurred each time the resource switches production to a different product. Requiring the resource to make a new product (that requires its own setup) is likely to have what effect on the process?

Answer: B. Inventory of all products is likely to increase.

13. Henry Ford proclaimed about the Model T, ―You can have any color you want, as long as it 57

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

is black.‖ Which of the following best reflects his motivation for this position?

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Answer: B. With more than one color, the process would have to switch over time and that could reduce the overall utilization of the process.

14. The primary purpose of the ―Single-Minute Exchange of Die‖ philosophy is to: Answer: A. reduce the setup time for a resource

PROBLEMS AND APPLICATIONS 1. The batch size is cut in half on a resource that has a setup time. What is the likely impact on the resource’s capacity?

Answer: C. Decrease by less than 50 percent Feedback: The total setup time and processing time would not change; therefore, the capacity would drop by less than 50 percent.

2. (a) Suppose the machine rotates between one batch of 1000 units of A and 1000 units of B. In that case, what is the capacity of the machine in component pairs per minute, where a component pair is one unit of A and one unit of B? Answer: 40 Feedback: 250 seconds set-up time for Unit A + (1000 components x 0.5 runtime for Unit A) + 250 seconds set-up time for Unit B + (1000 components x 0.5 runtime for Unit B) =1,500 seconds. 1,500 seconds / 60 seconds = 25 minutes to produce 1,000 component pairs. 1,000 component pairs / 25 minutes = 40 component pairs per minute. (b) Suppose the machine rotates between one batch of 1000 units of A and 1000 units of B. What is the utilization of the machine?

Answer: 50 percent Feedback: Utilization = flow rate x processing time. Flow rate = 30 assembled units per minute, or 30/60=1/2 units per second. Processing time = 2x0.5 seconds = 1 second per unit. Utilization = ½ x 1 = 0.5 (c) Suppose the machine rotates between one batch of 1000 units of A and 1000 units of B. What is the average inventory of B components?

Answer: 375 Feedback: ½ x Batch x [1-(flow rate x processing time)] ½ x 1000 x [1-(0.5 x 0.5)] = 375

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(d) If the production schedule could be adjusted with the goal of minimizing inventory in the process, how many units of A should be produced before switching to component B? Assume the same number of units of B would be produced as well.

Answer: 500 Feedback: Batch size = target capacity x total set-up time/(1-target capacity x processing time) Batch size = 0.5 units/seconds x 500 seconds/ (1- 0.5 units/second x 1 second) = 500. There is 1 unit A per 1 assembled unit. So 500 assembled units equals 500 As. 3. (a) What is the setup time of this process?

Answer: 2.5 minutes Feedback: 3.0 minutes to cook first order – 0.5 minutes to cook additional units = 2.5 minutes (b) If Y&Y operates with batch sizes of 6 units, what is their process capacity (in orders per minute)?

Answer: 1.09 Feedback: First unit 3.0 minutes + (5 orders x 0.5 minutes) = 5.5 minutes to cook 6 orders. 6 orders/5.5 minutes = 1.09 orders per minute (c) If Yum and Yee operates with batch sizes of 10 units, what is the utilization of the wok?

Answer: 0.625 Feedback: Capacity of the wok is 10 / (2.5 + 0.5 x 10 ) = 1.33 orders per minute. The bottleneck is payment. So the flow rate is 1/0.8 minutes =1.25 orders per minute.

Utilization = Flow rate x Processing time = 1.25 x 0.5 = 0.625. (d) Calculate the batch size (in orders) that maximizes the overall flow rate (assume there is ample demand)? Do NOT round the batch size (i.e., assume for this calculation that a noninteger batch size is possible).

Answer: 8.33 Feedback: Batch Size = (Target capacity x total set-up time) / [1 – (target capacity x processing time)] (1.25 x 2.5)/ [1 – (1.25 x 0.5)] = 3.125 / .375 = 8.33 4. (a) Currently, the PCB machine produces 500 boards between setups. Given this operating process, what is the capacity of the PCB machine (in boards per minute)?

Answer: 5.26 Feedback: 20 minute set-up + 500 boards x 0.15 = 95 minutes 60

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

500 units / 95 minutes = 5.26 boards per minute (b) Currently, the PCB machine produces 500 boards between setups. Given this operating process, what is the utilization of the PCB machine?

Answer: 78.95 percent Feedback: Utilization = flow rate x processing time, 5.26 x 0.15 5. How many kgs of ―regular‖ should Sarah produce before switching over to another scent?

Answer: 7,364 Feedback: Total setup time = 1.5 * 4 = 6 hours. Total demand for all four scents per hour = 150 + 120 + 75 + 50 = 395 kgs / hour, hence, the target capacity is 395 kgs / hour. The processing time is 1/450 hours /kg. The desired batch size is 395 x 6 / (1 – 395 x (1/450)) = 19,391 kgs. Regular represents 150/395 of demand, so the regular batch size should be 19,391 x 150 / 395 = 7,364.

6. (a) What is the process capacity in units per hour with a batch size of 100 wafers?

Answer: 100 Feedback: Capacity = 100Batch size / (45 minute set-up + .15 minute processing time x 100 units) = 1.67 units per minute. 1.67 units x 60 minutes = 100 units per hour (b) What is the utilization of depositing if the batch size is 100 wafers?

Answer: 25 percent Feedback: Utilization = flow rate x capacity, 1.67 units per minute x 0.15 = .25 7. (a) What is the process capacity in dolls per hour with a batch size of 500 dolls?

Answer: 200 dolls per hour Feedback: Capacity = 500Batch size / (0 minute set-up + .3 minute processing time x 500 units) = 3.33 units per minute. 3.33 units per minute x 60 minutes = 200 dolls per hour. (b) What is the utilization of molding in dolls per hour with a batch size of 800 dolls?

Answer: 41.67 percent Feedback: The flow rate is equal to the demand rate (4,000 / 40 = 100 units / hour) because this is less than the process capacity of 200 units / hour from part (a).

Utilization = Flow rate * Processing time = 100 units / hour * .25/60 hours / unit = 0.4167. (c) Which batch size would minimize inventory without decreasing the process capacity?

Answer: 300 61

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Feedback: Batch Size = (Target capacity x total set-up time) / [1 – (target capacity x processing time)]. (3.3333 x 15) / [1 – (3.3333 x .25)] = 300. The process capacity = 200 per hour or 3.3333 per minute. (d) Which batch size would minimize inventory without decreasing the current flow rate?

Answer: 67 Feedback: (Target capacity x total set-up time) / [1 – (target capacity x processing time)]. The target capacity = 100 demand rate per hour or 1.67 minute. Painting is the bottleneck. (1.67 x 30) / [1 – (1.67 x .15)] = 67 8. (a) Suppose they operate with a production cycle of 150 kilograms (50 kilograms of fragola, 75 kilograms of chocolato, and 25 kilograms of bacio). What is the capacity of the gelatomaking process (in kilograms per hour)?

Answer: 34 Feedback: Capacity = 150Batch size / (85 minute set-up + 1.2 minute processing time x 150 units) = .566 units per minute. Capacity = .566 units per minute x 60 minutes =34.0 (b) Suppose they operate with a production cycle of 150 kgs (50 kilograms of fragola, 75 kilograms of chocolato, and 25 kilograms of bacio). What is the utilization of the gelatomaking process?

Answer: 60 percent Feedback: The flow rate is equal to the demand rate (30 per hour / 60 minutes = 0.5 units / min) because this is less than the process capacity. Utilization = flow rate x capacity, 0.5 units x 1.2 = .60 (c) Suppose they operate with a production cycle of 150 kilograms (50 kilograms of fragola, 75 kilograms of chocolato, and 25 kilograms of bacio). What is the average inventory of chocolato?

Answer: 26.25 Feedback: The average inventory of chocolato is ½ x Batch size of chocolato x [1-(flow rate of chocolato x processing time of chocolato)] = ½ x 75 x [ 1-(0.25 kgs/min x 1.2 minutes)] = 26.25 (d) Suppose Bruno wants to minimize the amount of each flavor produced at one time while still satisfying the demand for each of the flavors. (He can choose a different quantity for each flavor.) If we define a batch to be the quantity produced in a single run of each flavor, how many kgs should he produce in each batch?

Answer: 106.3

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Feedback: (0.5 Target capacity x 85 minute total set-up time) / [1 – (0.5 target capacity x 1.2 minutes processing time)]. The target capacity = 30 demand rate per hour or 0.5 minute. (e) Given your answer to part (d), how many kgs of fragola should he make with each batch?

Answer: 35.4 Feedback: Fragola = 106.3 batch size x 10 Fragola demand per hour/30 total demand per hour = 35.4 9. (a) What is the maximum capacity (bowls per hour) of this process?

Answer: 9 Feedback: Capacity of cleaning step: 35 / [45 minutes + (40 seconds)/60 seconds / minute x 35] = 0.51 bowls / minute. Capacity of carving step: 15/100 = 0.15 bowls / minute. Capacity of staining step: 35/(80 minutes + 0.5 minutes * 35) = 0.36 bowls / minute. The capacity of this process is 0.15 bowls / minute or 9 bowls / hour. (b) Suppose Wavy wants to operate with the same number of bowls in each batch that are cleaned or stained. For example, if it cleans in batches of 30, then it stains in batches of 30 as well. What batch size (in bowls) minimizes its inventory while allowing the process to produce at the maximum flow rate?

Answer: 13 Feedback: Batch size = (Target capacity x total set-up time) / [1 – (target capacity x processing time)]. The target capacity = 0.15 demand rate per minute. The target capacity is based on the bottleneck location, which is carving. For the stain machine, batch size = [0.15 bowls/min x 80min] / [1 – (0.15 bowls/min x 30 sec/60 sec/min)] = 13.

For the cleaning machine, batch size = [0.15 bowls/min x 45min] / [1 – (0.15 bowls/min x 40 sec/60 sec/min)] = 7.5. Use the larger batch size because the smaller batch size would result in less capacity at the stain machine than desired.

10. (a) If the CNC machine produces 14 housings between setups, what would be its capacity (in housings per hour)?

Answer: 3.5 Feedback: Capacity = 14Batch size / (30-minute set-up + 15 minute processing time x 14 units) = 14/240=.058 per minute or 3.5 housing per hour. (b) Assuming the process is supply-constrained, what is the utilization (as a number between 0 and 1) of the CNC machine if it operates in batches of 14 housings?

Answer: 75 percent 63

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Feedback: Utilization = flow rate x processing time. Flow rate is determined by the assembly step, which has a capacity of 1/20 housings per minute. Hence, 1/20 units per minute x 15 minutes/unit = 0.75. (c) Assuming the process is supply-constrained and operates with batches of 14 housings, what is the average inventory of housings?

Answer: 1.75 Feedback: Average inventory of housings = ½ x 14 x [1-(3x1/4)] = 1.75 (d) Aquatica actually makes five different housings. As already mentioned, the CNC machine must be set up whenever production switches from one housing to the other. In each case, the setup time is 30 minutes and it takes 15 minutes to drill the block for each type of housing.

Demands for the housings are given in Table 7.7.

Answer: 4 Feedback: Batch size = (Target capacity x total set-up time) / [1 – (target capacity x processing time)]. The target capacity = 3 unit demand rate per hour. The target capacity is based on the demand, which is less than the capacity.

Batch size = (3 target demand x 2.5 hours setup time) / [1- (3 target capacity x 0.25 hour processing time)] =30 D700 housings per batch = 30 x (.4 D700 demand per hour/3 total demand per hour) = 4

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CASE Bonaire Salt In this case students are asked to: ▪

Evaluate the capacity of a batch process.

▪

Determine a production schedule (i.e., choose a batch size) to achieve a target flow rate.

▪

Assess how two changes to the setup time in the process influence capacity.

The case provides more information than is typical of a standard ―end of chapter‖ question. As a result, students need to sift through the case to determine what is relevant. Another challenge is that the case doesn’t obviously map into the batch processes considered in the chapter (the milling machine for Xootr or the soup manufacturing process). In particular, in this case, harvesting is the setup process and evaporation is the processing step even though it might seem that the reverse is true because the company doesn’t do any work during evaporation while it does do work during harvesting.

Once a student realizes that harvesting is the setup process and evaporation is the processing step, the first three questions can be answered using the material in the chapter. The fourth question requires the student to think a bit beyond what is in the chapter.

The following table summarizes the calculations for each question: Calculations Total area (m^2) Number of pans Area of each pan (m^2) Salt evaporated (kg) per week per m^2 Salt evaporated (kg) per week per pan Salt evaporated (tons) per week per pan Processing time for evaporation (weeks per ton) Weeks for 1 bulldozer to harvest total area Number of bulldozers Weeks for 1 bulldozer to harvest a pan Weeks to harvest a pan Weeks evaporating Batch size (tons) Capacity per pan (tons per week) Capacity (tons per week) Utilization Target capacity Batch size (tons) Weeks evaporating Price of salt per ton Weekly increase in salt output Weekly increase in salt revenue

Comment

a from case m from case a/m s from case a*s/m r = (a * s / m * 1000) p=1/r t from case n from case t/m h = t / (m * n) e from case b=e*r c = b /(h + b * p) c*m c*p ct bt = ct * h / (1- ct * p) bt / r

Q1, Q2

Q3

Q4

640000 640000 640000 1 1 4 640000 640000 160000 9.8 9.8 9.8 6272000 6272000 1568000 6272 6272 1568 0.000159 0.000159 0.000638 12.5 12.5 12.5 2 3 2 12.5 12.5 3.1 6.25 4.17 1.56 30.0 30.0 30.0 188160 188160 47040 5191 5507 1490 5191 5507 5962 0.83 0.88 0.95 5,700 390,629 62.28 $25 317 7913

Evaporation is the processing step because the amount of salt produced from a pan increases if evaporation is allowed to happen for a longer period of time. Thus, the ―batch size‖ in this process is the amount of salt available to harvest after evaporation. Harvesting is the ―setup‖ in 65

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this process because the time to harvest is independent of the amount of salt in the pan (i.e., independent of the batch size). (In reality, harvesting does depend somewhat on the amount of salt in the pan. But the harvesting time is far less sensitive to the batch size than the evaporation time, so we can approximate harvesting time as if it is independent of the batch size.)

Q1. To determine the capacity of the process we need to determine the capacity of the combination of evaporation and harvesting. Given 640,000 m^2 of area and production rate of 9.8 kg of salt per m^2 per week, the pan produces 6,272 tons of salt per week. If it is allowed to evaporate for 30 weeks, then evaporation produces a batch of 30 weeks x 6,272 tons per week = 188,160 tons.

If 1 bulldozer can harvest the pan in 12.5 weeks, then 2 bulldozers can complete harvesting in 6.25 weeks. So now the standard capacity equation can be used:

𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 =

𝐵𝑎𝑡𝑐ℎ 𝑠𝑖𝑧𝑒 𝑇𝑜𝑡𝑎𝑙 𝑠𝑒𝑡𝑢𝑝 𝑡𝑖𝑚𝑒 + 𝑃𝑟𝑜𝑐𝑒𝑠𝑠𝑖𝑛𝑔 𝑡𝑖𝑚𝑒 𝑥 𝐵𝑎𝑡𝑐ℎ 𝑠𝑖𝑧𝑒

to determine that evaporation/harvesting can produce 5,191 tons per week. The only other process is loading, which can work at the rate of 20,000 tons per week. So evaporation/harvesting is the bottleneck and that determines the flow rate through the process.

Q2. Now we want to determine the batch size that achieves a flow rate of 5,700 tons per week.

Using the equation

𝐵𝑎𝑡𝑐ℎ 𝑠𝑖𝑧𝑒 =

𝑇𝑎𝑟𝑔𝑒𝑡 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑥 𝑇𝑜𝑡𝑎𝑙 𝑠𝑒𝑡𝑢𝑝 𝑡𝑖𝑚𝑒 1 − 𝑇𝑎𝑟𝑔𝑒𝑡 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑥 𝑃𝑟𝑜𝑐𝑒𝑠𝑠𝑖𝑛𝑔 𝑇𝑖𝑚𝑒

we determine that the necessary batch size is 390,629 tons. In terms of week, we divide that figure by the evaporation rate of 6,272 tons per week to yield a required time of 390,629 tons / 6,272 tons per week = 62.3 weeks of evaporation.

Q3. Adding an additional bulldozer effectively reduces the setup time in the process because the time to harvest the pan decreases from 6.25 weeks to 12.5 / 3 = 4.17 weeks. A reduction in the setup time will naturally increase the capacity of the process. Repeating the calculations in Q1 but with the new setup time reveals that adding a bulldozer can increase capacity to 5,507 tons per week. That is a 317 ton increase per week over the case with 2 bulldozers. At $25 per ton, that represents a potential increase in revenue of 317 x $25 = $7,913 per week.

Q4. Bart believes that creating 4 pans will not improve capacity because the total area dedicated to evaporation doesn’t change. While this is logical, it misses a key point. With one pan, during the harvesting there cannot be any evaporation. In contrast, with four pans, while one pan is being harvested, the other three can continue to do evaporation. Thus, while the total area that can potentially evaporate remains the same, the utilization of evaporation is higher when there are multiple pans.

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

With four pans, each one produces 1,490 tons of salt per week (assuming 30 weeks of evaporation and 2 bulldozers). Across 4 pans, that totals to 4 x 1,490 = 5,962 tons per week.

Although the case does not ask for utilization, the utilization of the salt pans can be evaluated with the standard equation:

𝑈𝑡𝑖𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛 = 𝐹𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑥 𝑃𝑟𝑜𝑐𝑒𝑠𝑠𝑖𝑛𝑔 𝑇𝑖𝑚𝑒 As expected, utilization increases if the setup time is reduced (by adding a bulldozer) or if the pan is divided into four pans.

Some students might try to equate ―adding additional pans‖ to ―adding product variety‖. The soup example suggests that adding variety deteriorates performance. So why would adding pans improve performance? The key distinction is that the soup process can only work on one type of soup at a time while multiple pans can evaporate at the same time. Consequently, adding pans increases capacity whereas adding types of soup decreases capacity (or requires substantially larger batches to maintain capacity).

CHAPTER 8 LEAN OPERATIONS AND THE TOYOTA PRODUCTION SYSTEM CONCEPTUAL QUESTIONS 1. An orthopedic surgeon has read an article about the Toyota Production System and now wants to organize her surgery operation based on lean principles. To eliminate all idle time in the operating room, she decides to call in patients 6 hours before the scheduled surgery time. Is this a lean process?

Answer. B. No. Feedback: The surgeon emphasizes the reduction of the waste of the resources (doctor and operating room). However, she does this at the expense of the patient who now wastes six hours of the flow time waiting.

2. Which car company is most often associated with the term lean operations?

Answer. C. Toyota.

Feedback: Toyota was the birthplace of lean and the Toyota Production System is synonymous with lean operations.

3. A worker in charge of assembling a bicycle finds a faster way to mount the pedals. He turns to his supervisor. The supervisor explains that the process had been optimized by the engineering 67

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

department and so the worker should make sure to stick with the existing procedures. Which management scholar’s work does this fit best?

Answer. A. Taylor.

Feedback: Ohno and other scholars of lean would not be pleased by how the intelligence of the worker is wasted by rejecting improvement ideas.

4. Students have to change buildings between classes, which often involves walks of 5 minutes or more. Walking from one building to another is: Answer. B. Feedback: Non-value add time. Such walks are not waste, as they are required given the current lay-out of the campus. However (unless the students take great joy in the walk), they do not add value to the customer; they are thus non-value add time.

5. An executive declares that his company has achieved an OEE of 115 percent. Is this possible?

Answer. B. No. Feedback: The OEE can never exceed 100%. You can simply not use more time productively than you have available.

6. Which of the following does not belong to the seven sources of waste?

Answer. B. Feedback: Over time is not part of the seven sources of waste.

7. In a service operation, we can compute the value-added time by simply looking at the labor content. True or false?

Answer. B. False.

Feedback: The labor content might include processing times that are not value add to the customer.

8. A company is increasing the percentage value-added time in the operation. Its valueadded time and its flow rate remain unchanged. What will happen to its inventory turns?

Answer. A. Inventory turns will go up.

Feedback: If the percentage of value-added time is going up and their value-added time remains constant, then their flow time has to go down. And that means, holding flow rate constant, that the inventory turns have to go up (the inventory level goes down).

9. What are the two pillars of the Toyota Production System?

Answer. C. Built-in-quality and Just-in-time 10. What is the roof in the ―house-shaped‖ representation of the Toyota Production System?

Answer. B. Waste reduction.

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Feedback: The roof is the reduction of waste by better matching supply with demand.

This includes the waste of time of the flow units and the waste of resources.

11. Taichi Ohno built the Toyota Production System with the metaphor of a tiger in mind. Rapid and decisive movements were, in his view, the best way to match supply with demand. True or false?

Answer. B. False.

Feedback: Ohno had the slow but steady movement of the turtle (tortoise) in mind.

12. Which of the following objectives is more important in a pull system?

Answer. A. Producing at the rate of demand 13. What is the relationship between the number of kanban cards in a process and the inventory level?

Answer. B. Feedback: There never can be more inventory in the process than what was authorized via Kanban cards.

14. Which of the following statements about kanban are accurate?

Answer. G. a. Deploying the Kanban system leads to pulling work through the system instead of pushing work; and c. The Kanban system controls the work in process inventory.

15. What happens to the number of Kanban cards in the process if the replenishment time goes up?

Answer. A. The number of Kanban cards increases 16. If customers want to have 1000 units over 10 days, one should aim at producing 100 units per day instead of doing 111 units per day in the beginning and then having a little safety buffer at the end. True or false?

Answer. A. True 17. If the demand rate increases, the takt time: Answer. B. Decreases 18. If the takt time increases, the operation requires: Answer. B. Fewer workers 19. Because of long setup times, a company chooses to run big production batches. Which of the following actions will level the production schedule?

Answer. B. Mixing the models on the final line 69

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

20. A company makes two models, A and B. Which of the following production schedules is more level?

Answer. A. ABABABABAB is more level; it alternates between the two models.

21. The information turnaround time is driven primarily by: Answer. C. The inventory in the process 22. Pulling the andon cord can cause a loss of output. In TPS, an employee should pull the andon cord: Answer. A. Whenever a problem occurs 23. A single inspection system with experts testing the output of the process will help reduce the information turnaround time. True or false?

Answer. B. False Feedback: For a short information turnaround time, many inspection points throughout the process are needed.

24. Kaizen favors automation and investment in technology. True or false?

Answer. B. False Feedback: Kaizen favors continuous improvements driven by the employees.

25. Which of the following terms is not part of the Toyota Production System (TPS)?

Answer. D. Yakimono 26. Genchi genbutsu requires the implementation of a pull system. True or false?

Answer. B. False Feedback: Gathering first-hand information from the situation works in any process environment.

PROBLEMS AND APPLICATIONS 1. You arrive at the airport and wonder to what extent the security checkpoint is a lean operation. Among other things, you observe that it takes about 30 seconds to check an ID, scan a passenger for weapons, and X-ray their luggage. However, the average customer spends about 20 minutes in the process. This observation is an example of which of the following?

Answer. B. Waste of a flow unit’s time 2. As you think about your experience at the last airport security screening, you wonder which of the tasks that you saw the officers do at the security check were waste, which were non-valueadded work, and which were value-added work. The tasks that you observed included officers doing the following: (a) officers checking IDs, (b) officers telling passengers to take off their 70

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

shoes, (c) officers moving bins back to the beginning of the X-ray machines, (d) officers looking at the screen of the X-ray machine, (e) officers putting bags in the X-ray machine for a second time in case they previously contained fluids, and (f) officers waiting for a passenger to arrive.

Identify each of the above activities as waste, value-added work, or non-value-added work.

Answer: (a) value-added work; (b) non-value added work; (c) non-value added work; (d) value-added work; (e) waste; and (f) waste 3. A copy machine is available 24 hours a day. On a typical day, the machine produces 100 jobs.

Each job takes about 3 minutes on the machine, 2 minutes of which is processing time and 1 minute is setup time (logging in, defining the job). About 20 percent of the jobs need to be reworked, in which case the setup time and the processing time have to be repeated. The remainder of the time, the equipment is idle. What is the OEE of the equipment?

Answer. 13.9 percent Feedback: The copy machine is available 60 * 24 = 1,440 minutes a day. The valueadded time is 200 minutes. So, the OEE is: OEE = Value-added time / Time available = 200 / (24 * 60) = 0.139.

4. Determine the exact flow time. What is the value-added percentage of the flow time?

Answer. 0.2168 percent Feedback: The exact flow time is 11:32 hours. Of this, only 1.5 minutes are value-added time. Thus, the percentage of value-added time is: Value-added time of the flow unit / Flow time = 1.5 minutes / 692 minutes = 0.002168 = 0.2168% 5. Which of the following strategies or techniques would reduce inventory in the operation?

Answer. D. Heijunka 6. Which of the following strategies or techniques would use the principle of jidoka to increase quality?

Answer. C. Andon 7. How many kanban containers will be needed to support this system?

Answer. 25 containers Feedback: We first determine the amount of demand during the production lead time.

Demand during lead time = Lead time x Daily demand = 4 days x 2,000 units /day = 8,000 units We translate the 1 day safety stock into a quantity, which is 1 day x 2,000 units / day = 2,000 units.

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

With 10,000 units and a container size of 400 units / container, we need: Number of Kanban containers = (Demand during lead time + Safety stock) / Container size = (8,000 + 2,000) / 400 = 10,000 / 400 = 25 So, we need 25 containers.

8. What is the takt time of the process? What would a level production schedule look like?

Answer. The takt time is 0.22222. A level production schedule would almost alternate between A and B as follows: ABABABABB ABABABABB. Feedback: The total demand across both products is 180 units that we have to process over the course of the 40 hour work week. So, the takt time would be: Takt time = Available time / Demand = 40 hours / 180 units = 0.22222 hours / unit A level production schedule would almost alternate between A and B as follows: ABABABABB ABABABABB… This puts the flow-units in a relationship of 8:10, as is the overall demand mix.

9. What would be the information turnaround time for a defect made at station 2?

Answer. 16 minutes Feedback: It will take 8 cycles of 2 minutes each for the flow unit to be completely tested, yielding an information turnaround time of 16 minutes.

10. Which other Japanese term is most closely related to jidoka?

Answer. A. Andon

CASE Nike Every productivity improvement project has the potential to cut the number of workers that are needed / employed. So, the total number of workers might go down, holding output constant.

This would be a negative for the work-force.

But, there are positives: ▪

Better productivity means a lower cost and thus potential more sales

▪

One factory / region might be able to outperform another region, which might induce Nike to allocate more work to that factory / region

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▪

As workers become more skilled and also empowered to change the process, their wages should increase

▪

The work becomes more sustainable: it is more carefully planned, workers are not treated as machines, and the motions/ergonomics of the work are optimized.

CHAPTER 9 QUALITY AND STATISTICAL PROCESS CONTROL CONCEPTUAL QUESTIONS 1. In the production of pizza, which of the following four variables is not an input variable?

Answer: D. The time that the customer leaves the pizza in the fridge 2. Mary and Tom operate a lemonade stand. They are concerned that the lemonade does not taste fresh at the end of the day. Which of the following variables is an environmental variable in their operation?

Answer: B. The temperature of the air during the day 3. Following the announcement of a major quality problem, the share price of a company falls substantially. This decrease is a result of assignable cause variation. True or false?

Answer: A. True 4. You harvest 50 tomatoes. You notice that the tomatoes vary in size and weight. If some of the tomatoes were exposed to extra sunshine or irrigation, this would be a common cause variation. True or false?

Answer: B. False 5. You measure the exact size of 100 ping pong balls from manufacturer A and then another 100 balls from manufacturer B. The balls of manufacturer B have a much larger variation. Means and specification limits for the two manufacturers are identical. Which of the two manufacturers has a higher process capability index?

Answer: A. Manufacturer A 6. John is a newly minted quality engineer at MakeStuff Inc. His boss tells him to increase the process capability index of their main product. John is somewhat lazy and does not want to tackle the variation in the process, so he decides to simply increase the upper specification limit and reduce the lower specification limit. Does this increase the process capability index?

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Answer: B. Yes, it does, though one might question if the underlying process is really better.

7. If a process has a six-sigma capability, what is the process capability index?

Answer: B. 2 8. For one and the same process, what scenario corresponds to a higher variation?

Answer: B. A three-sigma process 9. What does X-bar stand for?

Answer: A. The average of a sample 10. Which of the following elements is not part of an X-bar control chart?

Answer: C. The specification limits LSL and USL 11. How many standard deviations is the upper control limit, UCL, above the long-run center line, X-bar-bar?

Answer: C. 3 12. What is the difference between a fishbone diagram and an Ishikawa diagram?

Answer: D. There exists no difference.

13. Which of the following statements about Pareto diagrams is correct?

Answer: A. The Pareto diagram shows the possible root causes of a problem alongside with the number of defect occurrences.

14. Which of the following definitions describes a robust process?

Answer: A. A process is robust if it can tolerate variation in input and environmental variables without producing a defect.

15. In a process for which we have to use discrete probabilities to describe whether an activity was carried out correctly or not, which of the following statements determines the defect probability of the entire process?

Answer: D. This really depends on the process, because it is possible that one process is more robust than the other.

16. You look at a process that has three activities; each one can potentially go wrong. How many leaves does the event tree of this process have?

Answer: D. 8 74

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

17. What is the difference between p-charts and attribute-based control charts?

Answer: C. They are one and the same thing.

18. In a p-chart, the upper control limit is how many estimated standard deviations above the average p-bar?

Answer: C. 3 PROBLEMS AND APPLICATIONS 1. Yi is fixing up his road bike to prepare for a triathlon. He looks at his tires and wonders how much he should inflate them. Which of the variables is an outcome variable?

Answer: B. The rolling resistance 2. You start a summer job serving ice cream at the shore. The manager notices that since you started the job, the average portion size given to customers has gone up substantially. This is a result of assignable cause variation. True or false?

Answer: A. True Feedback: Assignable cause variation is the variation that occurs because of a specific change in input or in environmental variables.

3. A company making tires for bikes is concerned about the exact width of its cyclocross tires.

The company has a lower specification limit of 22.8 mm and an upper specification limit of 23.2 mm. The standard deviation is 0.15 mm and the mean is 23 mm. What is the process capability index for the process?

Answer: 0.4444 Feedback: We compute the process capability index as: Cp

USL LSL 6 ˆ

23.2 22.8 6 * 0.15

0.4444

4. Consider again that the company making tires for bikes is concerned about the exact width of its cyclocross tires. The company has a lower specification limit of 22.8 mm and an upper specification limit of 23.2 mm. The standard deviation is 0.15 mm and the mean is 23 mm. What is the probability that a tire will either be too wide or too narrow?

Answer: The probability that a tire will be too narrow is 0.091211. The probability that a tire will be too wide is 0.091211. The probability that a tire will be part detective is 0.182422.

Feedback: The probabilities are as follows: Probability{too narrow} = NORMDIST(22.8, 23, 0.15, 1) = 0.091211 75

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Probability{too wide} = 1 - NORMDIST(23.2, 23, 0.15, 1) = 0.091211 Probability{part defective} = 0.091211 + 0.091211 = 0.182422 5. Consider again that the company making tires for bikes is concerned about the exact width of its cyclocross tires. The company has a lower specification limit of 22.8 mm and an upper specification limit of 23.2 mm. The standard deviation is 0.15 mm and the mean is 23 mm. The company now wants to reduce its defect probability. To what level would it have to reduce the standard deviation in the process to meet this target?

Answer: A value between 0.076923 and 0.08333.

Feedback: We know that the process capability index is: Cp

USL LSL 6 ˆ

Looking at Table SIGMA, the capability index for a defect probability of 0.01 is between 0.8 and 0.86667. If we set Cp = 0.86667, we get: 23.2 22.8 0.4 0.4 ˆ 0.86667 0.86667 0.076923 6 ˆ

6 ˆ

If we set Cp = 0.8, we get: 23.2 22.8 0.4 0.8 0.8 6 ˆ

6 ˆ

ˆ

6 * 0.86667

0.4

0.083333

6 * 0.8

This means that the standard deviation would have to be reduced to a value between 0.076923 and 0.08333.

6. For 120 consecutive days, a process engineer has measured the temperature of champagne bottles as they are made ready for serving. Each day, she took a sample of eight bottles. The average across all 960 bottles (120 days, eight bottles per day) was 46 degrees Fahrenheit.

The standard deviation across all bottles was 0.8 degree. When constructing an X-bar chart, what would be the center line?

Answer: C. 46 Feedback: The center line would be given by the average across all 960 bottles, i.e., it would be 46 degrees.

7. For 120 consecutive days, a process engineer has measured the temperature of champagne bottles as they are made ready for serving. Each day, she took a sample of eight bottles. The average across all 960 bottles (120 days, eight bottles per day) was 46 degrees Fahrenheit.

The standard deviation across all bottles was 0.8 degree. When constructing an X-bar chart, what would be the upper control limit?

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Answer: 46.84853 degrees Fahrenheit

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Feedback: To find the upper control limit, we first have to estimate the standard deviation for the daily X-bar sample. We use the equation: Estimated standard deviation (X-bar)= Standard deviation of all parts / sqrt(n) = 0.8 / sqrt(8) = 0.0282843 We then find the upper control limit as: UCL = X + 3 x ESD(X-bar) = 46 + 3 * 0.0282843= 46.84853 8. Which of the following statements about Pareto diagrams is correct?

Answer: B Feedback: The Pareto diagram shows the defects in order from most commonly occurring to least commonly occurring.

9. A baking operation reduces the amount of variation in the oven temperature. This is an example of robust process design. True or false?

Answer: B. False Feedback: A robust process is a process that can tolerate variation in input variables and environmental variables without leading to a defect.

10. You are thinking about the things that can go wrong on your trip home over the Thanksgiving break. You have booked a flight with US-Scareways. You know that in 30 percent of the cases the company has canceled the flight you were on. Should such a thing occur, there would be no other air travel option home for you. As a backup, your friend Walter has offered you a ride back. However, you know that Walter only has a seat in his car for you with an 80 percent probability. What is the probability of you making it home for the holidays?

Answer: 0.94 Feedback: There are two variables with two outcomes each, so we are dealing with 2 ^ 2 = 4 possible scenarios. In order for you to not make it home, two things have to go wrong: (a) the airline has to cancel the flight (30%) and (b) Walter’s car is full (20%).

So, you will not make it home with a probability of 0.2 * 0.3 = 0.06. This means you will make it home with a probability of 0.94.

11. What would be the center line for a p-chart?

Answer: C. 0.05 Feedback: We first create the center line at 5%.

12. What would be the upper control limit for a p-chart?

Answer: D. 0.142466 Feedback: We compute the estimated standard deviation as follows: 78

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

This estimate is given by the following equation: Estimated standard deviation =

p(1 − p) 0.05 * (1− 0.05) = Sample Size 50

0.03082207

We then compute the upper and lower control limit based on: UCL= p + 3 × Estimated standard deviation = 0.05 + 3 * 0.03082207 = 0.142466

CASE The Production of M&M’s Q1. Assuming a lower specification limit of 47 grams and an upper specification limit of 53 grams, what is the process capability of this process? How many defects would you expect in one million bags?

The process capability is (53-47) / (6* 1.037) = 0.9643.

We can compute the probability that a bag is too light as: Normdist(47, 49.9783, 1.037,1) = 0.00204 The probability that a bag is too heavy is: 1- Normdist(53, 49.9783, 1.037,1) = 0.00179 So the defect probability is 0.00204 + 0.00179 = 0.00383 Q2. And, finally, which M&M product in your view has the highest standard deviation?

We could expect that M&M with peanuts have a higher standard deviation compared to regular chocolate as the weight of the peanut most likely will be exposed to more common cause variation.

CHAPTER 10 INTRODUCTION TO INVENTORY MANAGEMENT CONCEPTUAL QUESTIONS 1. It is costly to hold inventory, but inventory can also be useful in a process because:

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Answer: B. adding inventory to a process can help to reduce situations in which sales are lost due to uncertain demand.

2. A delivery truck from a food wholesaler has just delivered fresh meat and produce to a local restaurant. This meat and produce would be categorized as which type of inventory by the restaurant?

Answer: A. Raw materials inventory 3. Tablets and laptops would be categorized as which type of inventory by an electronics retailer?

Answer: C. Finished goods inventory 4. Which reason for holding inventory enables a firm to reduce the impact of large fixed costs that the firm incurs whenever it places an order to a supplier?

Answer: D. Batching 5. Which reason for holding inventory guards against a reduction in the flow rate when one assembly station is disrupted in a production line?

Answer: C. Buffers 6. If two firms have the same annual inventory turns, they also have the same days-of-supply.

True or false?

Answer: A. True.

Feedback: If two firms have the same annual inventory turns, then they have the same years-of-supply because years-of-supply = 1 / annual inventory turns. If they have the same years-of-supply, then they must have the same days-of-supply as well because Days-of-supply = Years-of-supply x 365.

7. Firms in the same industry will have the same inventory turns. True or false?

Answer: B. False.

Feedback: There can be variation in inventory turns among the same firms within an industry.

8. Which of the following figures from a firm’s financial statements should be used as its flow rate when computing its inventory turns?

Answer: B. Cost of goods sold 9. All else being equal, a larger item is likely to have a higher annual holding cost percentage than a smaller item. True or false?

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Answer: A. True.

Feedback: A larger item requires more storage space, which is costly and this contributes to higher holding costs.

10. Which of the following possible responses by a customer who is faced with a stockout is the most costly to the firm?

Answer: A. Lose the sale and lose the customer 11. Computers lose value as they are stored in inventory. This is an example of which component of a firm’s inventory holding cost?

Answer: D. Obsolescence cost

PROBLEMS AND APPLICATIONS 1. Suppose a retailer turns its inventory of soda 50 times per year. On average, it has 400 bottles of soda on its shelves. What is the retailer’s average daily sales rate? (Assume 365 days per year.)

Answer: 54.8 bottles Feedback: Average daily sales = (400 x 50 turns) / 365 days = 54.8 bottles 2. Suppose a retailer’s annual inventory turns are 7.5. What are its days-of-supply of inventory? (Assume 365 days per year.)

Answer: 48.7 days Feedback: Days-of-supply of inventory = 365 days / 7.5 turns = 48.7 days 3. Apple’s days-of-supply of inventory are 10.5. What are its annual inventory turns? (Assume 365 days per year.)

Answer: 34.8 turns Feedback: Annual inventory turns = 365 days / 10.5 days-of-supply of inventory = 34.8 4. An electronics manufacturer has 25 days-of-supply of inventory for a particular cell phone model. Assuming 365 days per year, what are the annual inventory turns of this cell phone model?

Answer: 14.6 turns Feedback: Turns = 365 days / 25 days-of-supply of inventory = 14.6 5. Suppose that a movie theater snack bar turns over its inventory of candy 3.2 times per month.

If the snack bar has an average of 350 boxes of candy in inventory, what is its average daily sales rate for candy? (Assume that there are 30 days per month.)

Answer: 37.33 boxes per day 81

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Feedback: Boxes per day = (3.2 turns x 350) / 30 =37.33 6. Suppose that a local hardware store turns over its inventory of power tools 7.3 times per year. If the hardware store has an average inventory of 130 power tools, what is its average daily sales rate for power tools? (Assume that there are 365 days per year.)

Answer: 2.6 power tools per day Feedback: Power tools per day = (7.3 turns x 130 inventory) / 365 = 2.6 power tools per day.

7. A grocery chain recently reported annual sales of $89 billion, inventory of $5.8 billion, and annual cost of goods sold of $64 billion. What are the firm’s annual inventory turns?

Answer: 11.0 turns Feedback: Turns = $64 billion annual COGS / $5.8 billion inventory = 11.0 8. A manufacturer of farm equipment has annual turns of four, and its cost of goods sold (COGS) is $44 billion. What is the average inventory it holds?

Answer: $11 billion Feedback: Average inventory = $44 billion COGS / 4 turns = $11 billion 9. A mining company reported annual sales revenue of $75 billion, annual cost of goods sold of $50 billion, and inventory of $15 billion. What are the firm’s annual inventory turns?

Answer: 3.3 turns Feedback: Turns = $50 billion COGS / $15 billion inventory = 3.3 turns 10. A manufacturing company producing medical devices reported $60 million in sales over the last year. At the end of the same year, the company had $20 million worth of inventory of readyto-ship devices. Assuming that units in inventory are valued (based on cost of goods sold) at $1000 per unit and are sold for $2000 per unit, what is the company’s annual inventory turnover?

Answer: 1.5 turns Feedback: Turns = $30 million COGS / $20 million inventory = 1.5 turns 11. A restaurant has annual sales of $420,000, an average inventory of $6000, and an annual cost of goods sold of $264,000. What is the restaurant’s monthly inventory turns?

Answer: 3.67 turns Feedback: Annual turns = $264,000 COGS / $6,000 average inventory = 44 annual turns. Monthly turns = 44 annual turns / 12 months = 3.67 12. A restaurant has annual sales of $420,000, an average inventory of $6000, and an annual cost of goods sold of $264,000. What are the restaurant’s days-of-supply of inventory? (Assume 365 days per year.)

Answer: 8.30 days 82

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Feedback: 264,000 / 365 days = $723.29 COGS per day. $6,000 inventory / $723.29 = 8.3 days.

13. A retailer has annual sales of $500,000 and an average finished-goods inventory of $15,000. If the retailer sells each unit for an average of $25 and purchases the units for $15, what is its annual inventory turnover?

Answer: 20 turns Feedback: COGS = $500,000 x (15/25) = $300,000. Turns = $300,000 COGS / $15,000 average inventory = 20 14. An online shoe retailer’s annual cost of holding inventory is 35 percent. The firm operates with a days-of-supply of 20 days, and assume there are 365 days per year. What is the inventory holding cost (in dollars) for a pair of shoes that the firm purchased for $50?

Answer: $0.96 Feedback: Turns = 365 days / 20 days inventory = 18.25 turns. $50 shoe / 18.25 turns = $ 2.74 average inventory. $2.74 average inventory x 35% = $0.96 holding cost 15. A company’s holding cost is 16 percent per year. Its annual inventory turns are 10. The company buys an item for $40. What is the average cost (in dollars) to hold each unit of this item in inventory?

Answer: $0.64 Feedback: $40 purchase / 10 turns = $4 average inventory. Holding cost = $4 x .16 = $0.64 16. A computer company’s yearly inventory cost is 40 percent (which accounts for the cost of capital for financing the inventory, warehouse space, and the cost of obsolescence). Last year, the company had $400 million in inventory and cost of goods sold of $26 billion. What is the company’s total inventory cost for the year (in millions of dollars)?

Answer: $160 million Feedback: Total inventory cost = $400 million x .4 = $160 million 17. An integrated circuit manufacturer’s annual cost of holding inventory is 48 percent. What inventory holding cost (in dollars) does it incur for an item that costs $300 and has a 1month supply of inventory on average?

Answer: $12 Feedback: Holding costs = ($300 / 12 turns) x .48 = $12 18. A local bookstore turns over its inventory once every 3 months. The bookstore’s annual cost of holding inventory is 36 percent. What is the inventory holding cost (in dollars) for a book that the bookstore purchases for $10 and sells for $18?

Answer: $0.90 Feedback: ($10 COGS / 4 turns) x 0.36 = $0.90 83

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

19. A bicycle manufacturer purchases bicycle seats from an outside supplier for $22 each. The manufacturer’s inventory of seats turns over 1.2 times per month, and the manufacturer has an annual inventory holding cost of 32 percent. What is the inventory holding cost (in dollars) for a bicycle seat?

Answer: $0.49 Feedback: The monthly holding cost is 0.32/12 and the item is held for 1/1.2 months.

Hence the cost to hold a $22 item for 1/1.2 months given the monthly rate of 0.32/12 is ($22/1.2) x (.32/12) = $0.49

CASE Linking Turns to Mountain Margin This case is designed to let students explore the relationship between inventory turns and gross margin. In short, the higher a product’s inventory turns, the less time it spends in inventory, so the less inventory holding costs it accumulates, which means it requires a smaller gross margin to be profitable. Through this analysis the student can better understand the connection between a firm’s inventory and its overall profitability. Furthermore, the case illustrates that firms can differ in their operating strategies – there isn’t a ―one size fits all‖.

Q1. The first question merely asks if there is a pattern between turns and gross margin.

Excluding Ace Hardware and Whole Foods, the pattern is clear – higher inventory turns is associated with lower gross margin.

Q2. To get students to understand why there might be that relationship, ask them what it means to have higher inventory turns. Given that …

Turns = R / I = 1 / T higher turns means lower T, i.e., a product that has a higher turns spends less time in the process.

Next ask the students why spending less time in the process might be related to profitability. It means less time with the capital tied up in inventory, less time spending money on rent to store the inventory and less risk of obsolescence, among other benefits. So in summary …

Higher turns -> Lower inventory related costs The other key cost for an item is the cost to purchase it from a supplier. So the total cost of an item is Purchase cost + Inventory costs Now ask to clarify what gross margin is. It is the markup on the purchase cost that determines the selling price. So the difference between the selling price and the purchase cost increases as the gross margin increases. That difference is called gross profit. And that gross profit needs to be greater than inventory costs for the firm to make a final profit. Hence, a smaller gross margin 84

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

is needed to be profitable if inventory costs are smaller, which occurs, all else being equal, when turns are higher.

Q3. Ace Hardware has low turns and low gross margin. How can they survive if low turns means a product spends a long time in inventory, thereby accumulating inventory costs? The answer must be that the cost of holding inventory per unit of time is not as high for Ace Hardware as it would be for other retailers. For example, if their stores are not in locations with high rents, then the cost of storage might not be high for them. Or, Ace might only stock items that don’t face perishability or obsolescence. Given that they are a hardware store like Home Depot and Lowes, this is unlikely to be the explanation. Similarly, they might have a lower opportunity cost of capital, but again, this is not likely. The best explanation is probably the lower cost of rent means that they don’t need as large a gross margin to remain profitable.

CHAPTER 11 SUPPLY CHAIN MANAGEMENT CONCEPTUAL QUESTIONS 1. Firm A manufactures brake pads, a component of a braking system, and sells them to firm B, which sells braking systems used in vehicles. Firm A is best described as a: Answer: A. Tier 2 supplier.

Feedback: Firm A is a supplier to a car manufacturer’s major component supplier, Firm B. Firm B is a Tier 1 supplier, which makes Firm A a Tier 2 supplier.

2. Which of the following statements best explains why distributors tend to spend less on storage space per square meter per year than retailers?

Answer: C. Distributors can store more items per square meter and require space that is less aesthetically pleasing.

3. Which of the following best explains why distributors help to reduce the space retailers need to store inventory?

Answer: C. Distributors allow retailers to receive smaller quantities of each supplier’s product.

4. What is the relationship between the average inventory and the in-stock probability?

Answer: C. The more inventory, the higher the in-stock probability.

Feedback: As inventory increases, the firm is able to satisfy more demand, thereby increasing the likelihood that all customers will be satisfied.

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5. What is the relationship between a location’s in-stock probability and the lead time that location delivers on shipments to its customers?

Answer: B. There isn’t a definitive relationship—a higher in-stock probability could mean a smaller or a greater lead time.

Feedback: A high in-stock probability can be achieved even with a long lead time if the firm carries enough inventory.

6. Characterize the following decision as either tactical or strategic: Which tier 1 supplier should be used for a particular component system?

Answer: B. Strategic Feedback: Specifying a supplier for a major component is a decision with long-term implications.

7. Characterize the following decision as either tactical or strategic: to operate make-to-stock or make-to-order.

Answer: B. Strategic Feedback: Defining the flexibility of a manufacturing process is a decision with longterm implications.

8. Characterize the following decision as either tactical or strategic: the discounts to offer on inventory available for an end-of-season sale.

Answer: A. Tactical Feedback: End-of-season discounts apply to that one selling period only and are, thus, short-term concerns.

9. Which of the following is the best reason to add a second supplier for a critical component?

Answer: A. The primary supplier is located in a flood prone region while the second supplier’s location is not subject to flooding.

Feedback: A second supplier can help to reduce the risk of a supply chain disruption if the first supplier were to experience flooding.

10. Sorlini Pasta sells pasta throughout Italy. Which of the following is the strongest evidence that Sorlini Pasta is suffering from the bullwhip effect?

Answer: E. The volatility of the orders Sorlini receives from its distributors is greater than the volatility of pasta demand at retailers.

Feedback: The bullwhip effect is related to the increase in demand variability upstream through the supply chain.

11. Retailers order in full truckload quantities from a distributor. Suppose, due to a slowdown in the economy, there is an industrywide decline in demand (i.e., each retailer experiences a reduction in its sales). If retailers continue to order in full truckload quantities, what will happen to the frequency of their orders?

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Answer: A. It will decrease.

Feedback: If the demand rate decreases and the retailers still order full truckloads, each truckload will satisfy demand for a longer period of time. As a result, the orders will become less frequent.

12. Which of the following are likely to be symptoms of the bullwhip effect in a supply chain comprised of consumers, a retailer, a wholesaler, and a factory?

Variability of orders from the retailer to the wholesaler is higher than variability of orders from the wholesaler to the factory.

II. Variability of orders from the wholesaler to the factory is lower than the variability in consumer demand to the retailer.

III. Variability in consumer demand is lower than the variability of orders from the retailer to the wholesaler.

I.

Answer: C. III only Feedback: According to the bullwhip effect, variability in orders increases as firms are further away from the consumer in the supply chain.

13. A company is deciding whether to produce a new gadget at a plant located in a country close to consumers at a higher labor cost and shorter lead time or to outsource it to a country with a low labor cost but a longer lead time. All else being equal, which of the following considerations would provide the most support to produce in the high-cost location?

Answer: C. The in-stock probability requirement for the product is high.

Feedback: Production with a short lead time is best with a product that must carry a substantial amount of inventory due to a high in-stock probability requirement or if it is costly to hold each unit of inventory or if transportation costs from the distant location are high.

14. Over time, consumers have less of a need for a broad product offering. How does this shift in preferences alter the desirability of make-to-stock production relative to make-to-order production?

Answer: A. It increases it; that is, make-to-stock becomes more desirable.

Feedback: As the product variety decreases, make-to-stock becomes more attractive because there is not as much need to make items to meet vastly different customer preferences.

PROBLEMS AND APPLICATIONS 1. For 10 percent of the products in a category, a firm fails to satisfy all demand during the month. What is its in-stock probability?

Answer: 90% Feedback: In-stock probability is 1 – Stockout probability = 1 – 0.10 = 90%.

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2. (a) On average, how many tons does it have on order?

Answer: 28,000 Feedback: The on-order inventory is 10 * 2,800 = 28,000. (b) On average, how many tons does it have on hand?

Answer: 6,800 Feedback: The on-hand inventory is sqrt(10 + 1) * 1,000 * 2.05 = 6,800. (c) If its average inventory was 5000 tons, what would be its average holding cost per week?

Answer: $55,000 Feedback: With an average inventory of 5,000 tons, the average weekly holding cost is $11 * 5,000 = $55,000. (d) If its average inventory was 10,000 tons, what would be its average holding cost per ton of aluminum?

Answer: $39.29 Feedback: With an average inventory of 10,000 tons, the average weekly holding cost is $11 * 10,000 = $110,000. The average demand per week is 2,800 tons, so the average holding cost per ton is 110,000 / 2,800 = $39.29. (e) Suppose its on-hand inventory is 4975 tons, on average. What in-stock probability does it offer to its customers?

Answer: 0.9332 Feedback: The on-hand inventory is sqrt(10 + 1) * 1,000 * z = 4,975. This gives z = 1.50.

Table 11.5 says that a z of 1.50 gives an in-stock probability of 0.9332.

3. (a) What is the per unit holding cost of a laptop with U.S. production?

Answer: $14.18 Feedback: The on-order inventory with U.S. production is 1 * 1,000 = 1,000 units. The on-hand inventory with U.S. production is sqrt(1 + 1) * 800 * 2.25 = 2,546 units. The total average inventory is 1,000 + 2,546 = 3,546 units. The average weekly holding cost is $4 * 3,546 = $14,182. The average holding cost per unit with U.S. production is $14,182 / 1,000 = $14.18. (b) What is the per unit holding cost of a laptop with production in Taiwan?

Answer: $46.90 Feedback: The on-order inventory with production in Taiwan is 8 * 1,000 = 8,000 units.

The on-hand inventory with production in Taiwan is sqrt(8 + 1) * 800 * 2.25 = 88

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

5,400

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units. The total average inventory is 8,000 + 5,400 = 13,400. The average weekly holding cost is $3.50 * 13,400 = $46,900. The average holding cost per unit with production in Taiwan is $46,900 / 1,000 = $46.90. (c) A laptop requires 40 minutes of labor to assemble in the United States or in Taiwan. The total cost of labor in the United States is $40 per hour, while in Taiwan it is $10. What is the change in cost per laptop if it switches production to Taiwan?

Answer: Increases by $16.90 per unit Feedback: If production stays in the United States, the labor cost per unit is $40 * 2/3 hr = $26.67. There is no transportation cost, and the average holding cost per unit is $14.18. The total cost is $26.67 + $14.18 = $40.85 per unit. If production moves to Taiwan, the labor cost per unit is $10 * 2/3 = $6.67. There is a $2 per unit transportation cost and an average holding cost per unit of $46.90. This is a total of $6.67 + $2 + $46.90 = $55.57 per unit. If production moves to Taiwan, the cost increases by $55.57 – $40.85 = $14.72 per unit (mainly due to the additional holding cost).

CASE Timbuk2 This case is a shortened version of ―Where in the World Is Timbuk2? Outsourcing, Offshoring, and Mass Customization,‖ a case written by Gérard Cachon, Kyle Cattani and Serguei Netessine. The longer case is available for use without charge.

With this case students: • Discover the impact of wage differentials across countries on the total cost to produce a product • Use the inventory model described in the chapter to evaluate the amount of inventory that would be held in the supply chain if sourcing is moved to China • Explore the set of issues associated with moving production to another country. For example, do customers care about where their products are manufactured?

The case is organized by three questions. A discussion of the case could be based around those questions.

Q1. To evaluate the total cost of a bag, a student needs to calculate the labor cost per bag and then sum the various cost components: A couple of observations emerge from this table:

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San Francisco China Difference 12.50 1.25 49 63 10.21 1.31 -8.90

(a) Hourly wage ($) (b) Labor content (min) (c = a x b / 60) Direct labor Materials Other manu. Expenses Shipping to SFO Shipping to customers Total

13 1.50 0.00 3.00 27.71

13 0.75 1.00 3.00 19.06

0.00 -0.75 1.00 0.00 -8.65

•

Air shipments from China are not an attractive option because the cost of air shipments is higher than the labor cost savings.

•

Sea shipments do not cost very much ($1) and are much smaller than the labor cost differentials. This helps to explain why production of labor intensive items has tended to move towards countries with low labor costs—the labor cost benefit is much greater than the additional transportation cost

•

It might appear that Timbuk2 can sell a bag to REI for $35 that is made in SF, because the SF cost is ―only‖ $27.71, but that doesn’t leave much of a margin to pay for everything else that Timbuk2 must spend money on to sell bags: e.g., web developers, salaries for managers and designers, and rent of physical space in SF, among other items. Thus, while SF production is viable if each bag is sold for $100, it probably is not viable in a channel that sells them for $35.

Q2. The following table provides calculations for inventory holding costs.

Average weekly demand

MEAN

200

Standard deviation of weekly

STDEV

100

Lead time

L

2

Holding cost per unit

h

$1

Safety factor

z

2.5

demand

Average on-order inventory

𝐼𝑜 = 𝐿 × 𝑀𝐸𝐴𝑁

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Average on-hand inventory

𝐼ℎ = √𝐿 + 1 × 𝑆𝑇𝐷𝐸𝑉

433

× 𝑧 Average total inventory

(Io + Ih)

833

Inventory holding cost per week

h x (Io + Ih)

$833

Inventory holding costs per unit

ℎ × (𝐼𝑜 + 𝐼ℎ)

$4.17

𝑀𝐸𝐴𝑁

Q3. In terms of the typical costs, it is hard to argue against moving production to China. The labor cost savings are substantial and dominate the additional transportation cost ($1) per bag and probably dominate additional holding costs (from Q2). However, there are some costs that have not been included. For example, it would be necessary to select and monitor the supplier in China, which probably involves costly trips. Even with monitoring, will the Chinese supplier produce at a sufficiently high level of quality? And then there is the issue of ―Made in San Francisco.‖ Do customer care enough about where the bag is made to pay more for a bag? This can lead to a fun and lively discussion. Some students will say that they don’t care at all. Some may say that they care but how do we know that they would actually pay more for the bag? (You could try to push them on how much more they would be willing to pay.) In general, the location of where a product is made seems to matter (in terms of customers willing to pay more for the product) when . . .

•

The physical location is tied to actual or perceived quality. This often occurs for food products. For example, would a Napa Valley wine be the same if the winery were in Oregon or Australia?

•

The location is associated with skilled artisans not found in other locations. For example, a scarf made and designed in Italy denotes a sense of ―Italian style,‖ which cannot be replicated by an American. Or, only the Swiss have the skills needed to make a fine watch.

•

Customers have an affinity with the workers producing the product. For example, ―Buy American‖ often appeals to the notion of helping American workers, and ―Fair Trade Coffee‖ appeals to the morality of wanting to treat others with fairness and dignity.

Do any of the above apply to Timbuk2’s bags made in San Francisco? Maybe the third point applies—help American workers rather than Chinese workers. But it is not obvious that customers are willing to pay enough extra per bag to support that preference.

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INVENTORY MANAGEMENT WITH STEADY DEMAND CONCEPTUAL QUESTIONS 1. Which of the following is NOT an assumption of the EOQ model?

Answer: A. It is possible to receive a purchase discount if the order quantity is sufficiently large.

Feedback: The EOQ assumes there are no purchase quantity discounts.

2. The EOQ minimizes the sum of the ordering cost and which of the following costs?

Answer: B. Holding cost Feedback: The EOQ minimizes the sum of ordering and holding costs.

3. If the order quantity doubles, what happens to the frequency of orders (i.e., the number of orders submitted per unit of time)?

Answer: B. Decreases by 50 percent Feedback: The number of orders per unit of time is R / Q, where R is the flow rate and Q is the order quantity. So if Q doubles, then the number of orders per unit of time decreases by 50 percent.

4. If the order quantity doubles but the flow rate remains constant, what happens to the average amount of time a unit spends in inventory?

Answer: E. Increases by more than 50 percent Feedback: Average inventory equals Q / 2, where Q is the order quantity. So average inventory doubles. From Little’s Law, I = R x T. If I doubles and R remains the same, T must double (i.e., increase by 100%), where T is the average time a unit remains in inventory.

5. A firm evaluates its EOQ quantity to equal 180 cases, but it chooses an order quantity of 200 cases. Relative to the order quantity of 180 cases, the order quantity of 200 cases has: Answer: C. Lower ordering cost and higher holding cost Feedback: The ordering cost decreases with the order quantity and the holding cost increases with the order quantity.

6. If the order quantity doubles but the flow rate remains constant, what happens to the sum of ordering and holding costs?

Answer: D. Increases by less than 50 percent Feedback: Assuming the EOQ was initially selected, the ordering cost will decrease by 50 percent, and the holding cost will double. Together, this results in an increase of less than 50 percent.

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7. If a firm wanted to reduce the annual EOQ cost as a percentage of the annual purchase cost by 50 percent, how would the demand rate have to change?

Answer: E. Quadruple 8. Vetox sells industrial chemicals. One of their inputs can be purchased in either jugs or barrels. A jug contains 1 gallon, while a barrel contains 55 gallons. The price per gallon is the same with either container. Vetox is charged a fixed amount per order whether it purchases jugs or barrels. The inventory holding cost per gallon per month is the same with either jugs or barrels. Vetox chooses an order quantity to minimize ordering and holding costs per year.

Would Vetox purchase a greater number of gallons with each order if it purchased with jugs or with barrels?

Answer: D. They might order a greater number of gallons with jugs or with barrels, depending on various factors like the demand rate, ordering cost, and holding cost.

Feedback: It might order a greater number of gallons with jugs or with barrels, depending on various factors like the demand rate, ordering cost, and holding cost. Vetox is likely to be able to order close to the EOQ with jugs. With barrels, Vetox should order close to the EOQ, but might order more or less than the EOQ. Hence, it is not possible to determine whether it orders more with jugs or with barrels.

9. Sarah is a buyer for a department store. A supplier offers her a 5 percent discount if she triples her usual order quantity. Which of the following best explains why Sarah should take the deal?

Answer: C. Sarah knows that even though she may triple her order quantity, this would increase her operating costs by far less than a factor of three.

Feedback: Choice A suggests that the store would lose money on the deal, which is not a good justification for the deal. Choice B is too speculative. Choice D is incorrect—even if operating costs (inventory holding and ordering costs) are small relative to the purchase cost, they should not be ignored. Choice C is correct—large changes in the order quantity generally do not create large changes in operating costs.

PROBLEMS AND APPLICATIONS 1.

(a) What order quantity minimizes Millennium’s annual ordering and holding costs?

Answer: 216 Feedback: Sqrt (2 x 300 x 2340) / (120 x .25) = 216 (b) If Millennium chooses to order 300 cases each time, what is the sum of its annual ordering and holding costs?

Answer: 6,840 94

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Feedback: Orders per year = 2,340 annual cases / 300 = 7.8 orders. Annual order cost = 7.8 x $300 = $2,340. Annual holding cost = .5 (300 x 120 x .25) = $4,500.

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Annual order + holding costs = 2,340 + 4,500 = $6,840. (c) If Millennium chooses to order 100 cases each time, what is the sum of the ordering and holding costs incurred by each case sold?

Answer: 3.64 Feedback: Orders per year = 2,340 annual cases / 100 = 23.4 orders. Annual order cost = 23.4 x $300 = $7,020. Annual holding cost = .5 (100 x 120 x .25) = $1,500.

Annual order + holding costs = 7,020 + 1,500 = $8,520. Annual order + holding costs per case = $8,520 / 2,340 cases = $3.64 (d) If Millennium is restricted to ordering in multiples of 50 cases (e.g., 50, 100, 150, etc.), how many cases should it order to minimize its annual ordering and holding costs?

Answer: 200 Feedback: The order size of 200 is the nearest quantity to the EOQ of 216. To verify the answer, calculate the annual order + holding costs for the order size of 200 and 250. (e) Millennium is offered a 5 percent discount if it purchases at least 1000 cases. If it decides to take advantage of this discount, what is the sum of its annual ordering and holding costs?

Answer: $14,952 Feedback: Orders per year = 2,340 annual cases / 1000 = 2.34 orders. Annual order cost = 2.34 x $300 = $7.02. Annual holding cost = .5 (1000x 114 x .25) = $14,250.

Annual order + holding costs = 702 + 14,250 = $14,952.

2.

(a) If Sarah wants to minimize her annual ordering and inventory holding costs, how much palm oil should she purchase with each order (in kilograms)?

Answer: 5,126 Feedback: Order size = Sqrt (2 x 60 x 260,000) / (4.75 x .25) = 5,126 (b) If Sarah orders 4000 kg with each order, what would be the annual sum of the ordering and holding costs?

Answer: $6,275 Feedback: Orders per year = 260,000 annual kgs / 4000 = 65 orders. Annual order cost = 65 x $60 = $3,900. Annual holding cost = .5 (4000 x 4.75 x .25) = $2,375 Annual order + holding costs = 3,900 + 2,375 = $6,275 (c) If Sarah orders 8000 kg with each order, what would be the sum of the ordering and holding costs per kilogram sold?

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Feedback: Orders per year = 260,000 annual kgs / 8000 = 32.5 orders. Annual order cost = 32.5 x $60 = $1,950. Annual holding cost = .5 (8000 x 4.75 x .25) = $4,750 Annual order + holding costs = 1,950 + 4,750 = $6,700. Annual order + holding costs per kgs = $6,700 / 260,000 kgs = $0.026 (d) Sarah’s supplier is willing to sell her palm oil at a 5 percent discount if she purchases 15,000 kg at a time. If she were to purchase 15,000 kg per order, what would be her annual sum of the ordering and holding costs?

Answer: $9,501 Feedback: Orders per year = 260,000 annual kgs / 15000 = 17.33 orders. Annual order cost = 17.33 x $60 = $1,040. Annual holding cost = .5 (15000 x 4.51 x .25) = $8,461.

Annual order + holding costs = 1,040 + 8,461 = $9,501.

3.

(a) Suppose Joe orders 1000 gallons each time. What is his average inventory (in gal)?

Answer: 500 gallons Feedback: Average inventory = 1,000 x .5 = 500 (b) Suppose Joe orders 1500 gallons each time. How many orders does he place with his supplier each year?

Answer: 8.67 orders Feedback: Orders = (250 per week x 52 weeks) / 1500 = 8.67 (c) How many gallons should Joe order from his supplier with each order to minimize the sum of the ordering and holding costs?

Answer: 1,472 gallons Feedback: Order size = Sqrt (2 x 35 x 250 x 52) / (120 x .35) = 1,472 (d) Suppose Joe orders 2500 gallons each time he places an order with the supplier. What is the sum of the ordering and holding costs per gallon?

Answer: $0.0544 per gallon Feedback: (250 per week x 52 weeks) / 2500 = 5.2 orders. Annual order cost = 5.2 x $35 = $182. Annual holding cost = .5 (2500 x 1.2 x .35) = $525.

Annual order + holding costs = 182 + 525 = $707. (Annual order + holding costs) / annual demand = cost per gallon. $707/ 13,000 = 0.0544 (e) Suppose Joe orders the quantity from part (C) that minimizes the sum of the ordering and holding costs each time he places an order with the supplier. What is the annual cost of the EOQ expressed as a percentage of the annual purchase cost?

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Feedback: (250 per week x 52 weeks) / 1472 = 8.83 orders. Annual order cost = 8.33 x $35 = $309.10. Annual holding cost = .5 (1472 x 1.2 x .35) = $309.12.

Annual order + holding costs = 309.10 + 30912 = $618.22. Annual purchase cost = $1.20 x 13,000 = $15,600. $618.22 Annual order + holding costs / $15,600 = 3.96% (f) If Joe’s supplier only accepts orders that are an integer multiple of 1000 gallons, how much should Joe order to minimize ordering and holding costs per gallon?

Answer: 2,000 gallons Feedback: From part c the EOQ = 1,472. For 1,000 order quantity, Annual order cost = 13 orders x $35 = $455. Annual holding cost = .5 (1000 x 1.2 x .35) = $210.

Annual order + holding costs = 455 + 210 = $665.00.

For 2,000 order quantity, Annual order cost = 6.5 orders x $35 = $227.50. Annual holding cost = .5 (2000 x 1.2 x .35) = $420.

Annual order + holding costs = 227.50 + 420 = $647.50. An order size of 2,000 has the lowest annual order + holding costs. (g) Joe’s supplier offers a 3 percent discount if Joe is willing to purchase 8000 gallons or more. What would Joe’s total annual cost (purchasing, ordering, and holding) be if he were to take advantage of the discount?

Answer: $16,818 Feedback: For 8,000 order quantity, Annual order cost = (13,000/8,000) orders x $35 = $56.88. Annual holding cost = .5 (8000 x 1.164 x .35) = $1,629.60.

Annual order + holding costs + purchasing cost = 56.88 + 1,629.60 + 15,132 = $16,818.48.

4.

(a) Suppose Bruno orders 9000 kg each time. What is his average inventory (in kilograms)?

Answer: 4,500 kgs Feedback: Average inventory = 9,000 / 2 = 4,500. (b) Suppose Bruno orders 7000 kg each time. How many orders does he place with his supplier each year?

Answer: 15.4 orders Feedback: Annual orders = (9,000 per month x 12) / 7,000 per order = 15.43 orders (c) How many kg should Bruno order from his supplier with each order to minimize the sum of the ordering and holding costs?

Answer: 3,464 kgs Feedback: EOQ = Sqrt (2 x 20 order cost x 9,000 monthly demand x 12) / (.03 holding cost per month x 12) = 3464

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(d) If Bruno’s storage vessel can hold only 3000 kg of milk, what would be Bruno’s annual ordering and holding costs?

Answer: $1,260 Feedback: Annual order cost = (9,000 x 12)/3,000) orders x $20 = $720. Annual holding cost = .5 (3000 x .03 x 12) = $540 720 Annual order + 540 holding costs = 1,260 (e) If Bruno’s storage vessel can hold only 6000 kg of milk, what would be Bruno’s annual ordering and holding costs?

Answer: $1,248 Feedback: Annual order cost = (9,000 x 12)/3,464) orders x $20 = $624. Annual holding cost = .5 (3,464 x .03 x 12) = $624 624 Annual order + 624 holding costs = 1,248 (f) Bruno’s supplier’s truck can carry 20,000 kg of milk. The supplier does not want to deliver to more than three customers with each truck. Thus, the supplier requires a minimum order quantity of 6500 kg. If Bruno orders the minimum amount, what would be the sum of his annual ordering and holding costs? Assume he has a storage vessel large enough to hold 6500 kg.

Answer: $1,502 Feedback: Annual order cost = (9,000 x 12)/6,500) orders x $20 = $332.31. Annual holding cost = .5 (6500 x .03 x 12) = $1,170 332.31 Annual order + 1,170 holding costs = 1,502.31 (g) Bruno’s supplier offers a 5 percent discount when a customer orders a full truck, which is 20,000 kg. Assume Bruno can store that quantity and the product will not spoil. If Bruno orders a full truck, what would be the inventory holding and ordering cost incurred per kilogram of milk?

Answer: $0.033 5.

(a) How many kilograms should BZoom order from its supplier with each order to minimize the sum of the ordering and holding costs?

Answer: 2,309 Feedback: Sqrt (2 x 50 order cost x 400 weekly demand x 52) / (1.3 x .3 holding cost) = 2,309 (b) If BZoom orders 4000 kg at a time, what would be the sum of the annual ordering and holding costs?

Answer: $1,040 Feedback: Annual order cost = (400 x 52)/4,000) orders x $50 = $260. Annual holding cost = .5 (4,000 x 1.3 x .3) = $780 260 Annual order + 780 holding costs = 1,040 99

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(c) If BZoom orders 2000 kg at a time, what would be the sum of the ordering and holding costs per kilogram of dye?

Answer: $0.044 Feedback: Annual order cost = (400 x 52)/2,000) orders x $50 = $520. Annual holding cost = .5 (2,000 x 1.3 x .3) = $390 520 Annual order + 390 holding costs = $910 $910 Annual order + holding costs / (400 x 52) annual demand = 0.044 (d) If BZoom orders the quantity from part (A) that minimizes the sum of the ordering and holding costs, what is the annual cost of the EOQ expressed as a percentage of the annual purchase cost?

Answer: 3.33% Feedback: Annual order cost = (400 x 52)/2,309) orders x $50 = $450.41. Annual holding cost = .5 (2,309 x 1.3 x .3) = $450.21 450.41 Annual order + 450.26 holding costs = 900.67, Annual purchase amount 400 x 52 x 1.3) = 27,040.

900.67 Annual order + holding costs / $27,040 annual purchase amount = .0333 (e) BZoom’s purchasing manager negotiated with its supplier to get a 2.5 percent discount on orders of 10,000 kg or greater? What would be the change in BZoom’s annual total cost (purchasing, ordering, and holding) if it took advantage of this deal instead of ordering smaller quantities at the full price?

I. It would decrease by more than $1000.

II. It would decrease by less than $1000.

III. It would increase by less than $1000.

IV. It would increase by more than $1000

Answer: III. It would increase by less than $1,000.

CASE J&J and Walmart With this case, students…

• Use the material from the chapter to evaluate the EOQ quantity and inventory ordering and holding costs for non-EOQ quantities.

• Explore the robustness of the EOQ quantity to errors in demand estimates. This is not discussed in the chapter, but the chapter gives students the tools they need to address this question.

The case is organized by three questions. A discussion of the case could be based around those questions.

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Q1. This question asks students to evaluate the EOQ quantity given the information provided on holding costs and ordering costs. The following table provides the relevant calculations. The EOQ formula returns ―9.23 cases‖ as the optimal order quantity. Because an integer number of cases must be ordered, students can explore whether they should order 9 or 10 cases at a time: Demand Scenario Order quantity Average demand EOQ EOQ rounded down EOQ rounded up Order one tier

R (units per week) 12.3 12.3 12.3 12.3

R (cases Quantity C(Q) / per year) (cases) C(Q) purchase cost 53.3 9.23 $69.26 4.33% 53.3 9.00 $69.28 4.33% 53.3 10.00 $69.48 4.35% 53.3 24.00 $103.33 6.46%

As the table above reveals, 9 cases are better than 10, but the difference between the two possible order quantities is very small.

Q2. With this question students are asked to explore the impact of ordering in an integer multiple of a tier. There are 24 cases per tier, so it should be apparent that they would only order 1 tier at a time (given that the EOQ is much smaller than a tier). Evaluation of the new cost per unit reveals that ordering one tier at a time would raise costs by a noticeable amount – from 4.33% of the product’s cost to 6.46%. The idea proposed in the question is essentially ―cross docking‖—a pallet is labeled at the supplier for destination to a specific retail store so as to minimize the handling of the product in the distribution center. If the benefits of reduced handling are greater than 2.13% of the product’s cost (6.46% – 4.33% = 2.13%), this approach may be beneficial.

Q3. Although the EOQ assumes that demand is steady, demand is rarely steady. The data from the case clearly indicate that demand is seasonal for this product. There are some periods in which demand is about 50% greater than the average (e.g., 18.5 cases per week instead of 12.3), while at other times demand is about 50% less than the average (e.g. 6.2 cases per week instead of 12.3). According to the next table, when demand is 50% higher, the EOQ is 11.31 while if demand is 50%, the EOQ is 6.53.

Demand Scenario Order quantity 50% increase in average demand EOQ Original Q 50% decrease in average demand EOQ Original Q

R (units per week)

R (cases Quantity per year) (cases)

C(Q) / C(Q) purchase cost

18.5 18.5

80.0 80.0

11.31 9.00

$84.83 $87.05

3.54% 3.63%

6.2 6.2

26.7 26.7

6.53 9.00

$48.97 $51.52

6.13% 6.44%

Thus, we can evaluate the ordering and holding costs that would be incurred with the EOQ quantity and with the original 9-case quantity. While the 9-case quantity is not optimal given the higher or lower demand rate, it does not result in a substantial cost penalty. For example, if they use 9 cases instead of the 11.3 EOQ when demand is high, then costs increase from 3.54% to 101

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3.63%, or only 0.09%. Similarly, if they use 9 cases instead of the 6.53 EOQ when demand is low, then costs increase from 6.13% to 6.44%, or only 0.31%.

The conclusion from this exercise is that if an EOQ quantity is selected assuming average demand (e.g., 12.3 cases per week), then that order quantity is effective even if demand turns out to be 50% higher or 50% lower. This is surprising, but remarkably useful. It means that the EOQ model gives a good answer even if our demand estimates are not perfect. Hence, the firm does not need to frequently adjust its order quantities—adjustments are warranted only when demand shifts by a considerable amount.

CHAPTER 13 INVENTORY MANAGEMENT WITH PERISHABLE DEMAND CONCEPTUAL QUESTIONS 1. Which of the following is NOT true about the distribution function for a normal distribution?

Answer: C. It generally has a bell shape when graphed.

Feedback: The density function has a bell shape, but the distribution function is always increasing, so it never has a bell shape.

2. A newsvendor orders the quantity that maximizes expected profit for two products, X and Y. The critical ratio for both products is .8. The demand forecast for both products is 9000 units and both are normally distributed. Product X has more uncertain demand in the sense that it has the larger standard deviation. Of which of the two products does the newsvendor order more?

Answer: A. Product X because it has less certain demand Feedback: Product X has a higher standard deviation of demand, and therefore its optimal order quantity is greater given the same mean and critical ratio.

3. Consider two products, X and Y, that have identical cost, retail price, and demand parameters and the same short selling season (the summer months from May through August).

The newsvendor model is used to manage inventory for both products. Product X is to be discontinued at the end of the season this year and the leftover inventory will be salvaged at 75 percent of the cost. Product Y will be reoffered next summer, so any leftovers this year can be carried over to the next year while incurring a holding cost on each unit left over equal to 20 percent of the product’s cost. The quantity of each product is selected to maximize expected profit. How do those quantities compare?

Answer: B. The quantity of product Y is higher.

Feedback: The salvage value of product X is 75% of its cost. The overage cost, C o, is the difference between product X’s cost and its salvage value, which is then 100% - 75% = 25% of cost. Product Y’s overage cost is 20% of its cost. Both products have the same underage cost, Cu, so product X has the higher overage cost, which means it has the lower critical ratio, Cu / (Co + Cu). Given they have the same demand distribution, 102

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product X’s optimal order quantity must be lower, or product Y’s stocking quantity is higher.

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4. Suppose the newsvendor model describes a firm’s operations decision. Is it possible to have positive stockout probability and positive expected leftover inventory? Choose the best answer.

Answer: D. Yes. A firm does not stock out and have leftover inventory at the same time, but the stockout probability can be positive even though there is positive expected leftover inventory.

Feedback: It is not possible to stockout and have leftover inventory at the same time, but either one is possible, which is why there is a positive probability of a stockout and a positive expectation for leftover inventory.

5. A newsvendor faces normally distributed demand and the critical ratio is .8. If the profitmaximizing quantity is ordered, which of the following statements is true?

Answer: A. Expected sales are less than expected demand.

Feedback: Expected sales is always less than expected demand, i.e., no matter the critical ratio or the demand distribution.

6. A company uses the newsvendor model to manage its inventories and faces normally distributed demand with a coefficient of variation of 0.75. The company decides to order a quantity that exactly equals the mean of its demand forecast. Which of the following is true regarding this company’s performance measures?

Answer: A. There is a 0.50 probability that there is enough inventory to serve all demand Feedback: If the mean of the demand forecast is ordered and demand is normally distributed, then there is a 0.50 probability that all demand is served because there is a 0.50 probability that demand is less than the mean of the forecast.

7. A retailer has two merchandizers, Sue and Bob, who are responsible for setting order quantities for the products they manage. For all of their products, the critical ratio is .7 and the coefficient of variation of their demand forecasts is 0.35. At the end of the season, Sue is proud to report that she has sold the entire inventory she purchased. Bob, on the other hand, sold only about a third of his products. Who is more likely to be choosing quantities that maximize expected profit?

Answer: D. Bob because he is probably ordering more than the mean of the demand forecast Feedback: At the quantity that maximizes expected profit, the probability of stocking out is 1 minus the critical ratio. So in this case the probability of a stockout is less than 0.5.

So Sue is too conservative.

8. Suppose the newsvendor model is used to manage inventory. Which of the following can happen when the order quantity is increased by one unit?

Answer: D. Expected leftover inventory increases by less than one unit Feedback: If the order quantity is increased by one unit, expected sales and expected leftover inventory increase. But they cannot increase by more than one unit – the most 104

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that sales can increase is by one unit and the most that leftover inventory can increase is by one unit.

9. Which of the following changes in the in-stock probability increases the order quantity the most?

Answer: D. An increase in the in-stock probability from 80% to 95% Feedback: The order quantity to achieve an in-stock probability increases in the desired in-stock probability at an increasing rate. Thus, a 15% increase in the in-stock has a larger change in the quantity than a 10% increase in the in-stock. And a 15% increase starting at 80% has a larger increase than a 15% increase starting at 70%.

10. A change in which of the following does not result in a change in the mismatch costs incurred by a newsvendor?

Answer: E. The quality of the product Feedback: Mismatch costs include the loss on inventory that is salvaged and the opportunity cost of demand that is not satisfied due to stockouts. Those are influenced by all of the items except the quality of the product.

11. A change in which of the following results in a change in the maximum profit in a newsvendor setting?

Answer: B. The regular selling price of the product Feedback: The maximum profit depends on the profit earned per unit and the mean of the demand forecast. The profit earned per unit depends on the selling price.

12. For which of the following products is there the highest probability that demand is within 50 percent of the mean of the demand forecast?

Answer: C. Mean = 2000, standard deviation = 300 Feedback: The product with the lowest coefficient of variation will have the highest probability of demand being within 50% of the mean of the demand forecast.

13. Product X’s demand is normally distributed with mean 150 and standard deviation 50.

Product Y’s demand is also normally distributed with a mean of 150 and a standard deviation of 50. The sum of demand for these two products is normally distributed with a mean of 300 and a standard deviation of 50. Which of the following results is most likely?

Answer: A. Demands for these products are negatively correlated.

Feedback: If the products were independent then the standard deviation of total demand would be sqrt(2) x 50. Given that the actual standard deviation of total demand, 50, is less than what it would be if the demands were independent, sqrt(2) x 50, the demands must be negatively correlated.

14. The QBlitz system is best described as: 105

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Answer: B. make-to-order.

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Feedback: QBlitz submits its order after learning demand. It doesn’t assemble the product and customers do not receive unique products.

PROBLEMS AND APPLICATIONS 1.

(a) Dan will consider this book to be a blockbuster for him if it sells more than 400 units. What is the probability that Power and Self-Destruction will be a blockbuster?

Answer: 0.0062 Feedback: z = (400 – 200)/ 80 = 2.5. Using Table 13.4, F(2.5) = 0.9938. The probability that the book is a blockbuster is 1 – F(2.5) = 0.0062. (b) Dan considers a book a ―dog‖ if it sells less than 50 percent of his mean forecast. What is the probability this exposé is a ―dog‖?

Answer: 0.1057 Feedback: The book is a ―dog‖ if it sells less than 100 copies. We compute z = (100 – 200)/ 80 = -1.25. Using Excel, F(-1.25) = 0.1057. Using Table 13.4, F(-1.25) is between F(-1.3) = 0.0968 and F(-1.2) = 0.1151. (c) What is the probability that demand for this book will be within 20 percent of the mean forecast?

Answer: 0.3830 Feedback: If demand for the book is within 20% of the mean forecast, it would be between 160 and 240 copies. This corresponds to z = -0.50 and z = 0.50. Using Table 13.4, F(0.50) – F(-0.50) = 0.6915 – 0.3085 = 0.3830. (d) What order quantity maximizes Dan’s expected profit?

Answer: 240 Feedback: We have Cu = 20 – 12 = 8 and Co = 12 – 8 = 4. This gives the critical ratio of 8/(4 + 8) = 0.6667. Using Table 13.4 and the round-up rule, this corresponds to z = 0.50.

The optimal order quantity is 200 + 0.50 * 80 = 240. (e) If Dan orders the quantity needed to achieve a 95 percent in-stock probability, what is the probability that some customer won’t be able to purchase a copy of the book?

Answer: 0.05 Feedback: The probability that some customer cannot purchase the book is 1 minus the in-stock probability. (f) Suppose Dan orders 300 copies of the book. What is Dan’s expected leftover inventory?

Answer: 107.64

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Feedback: An order quantity of 300 corresponds to z = (300 – 200)/ 80 = 1.25. Using Table 13.4 and the round-up rule, we have I(1.30) = 1.3455. We multiply this by the standard deviation of 80 to get the expected left-over inventory. (g) Suppose Dan orders 300 copies of the book. What are Dan’s expected sales?

Answer: 192.36 Feedback: If the order quantity is 300, the expected inventory is 107.64 (from question 6). The expected sales are 300 – 107.64 = 192.36. (h) Suppose Dan orders 300 copies of the book. What is Dan’s expected profit?

Answer: $1,104 Feedback: Using the answers from questions f and g, the expected profit when the order quantity is 300 is (20 × 192) +( 8 × 108) – (12 × 300) = 1,104. (i) How many books should Dan order if he wants to achieve a 95 percent in-stock probability?

Answer: 336 Feedback: According to Table 13.4 and the round-up rule, a 95% in-stock probability corresponds to z = 1.70. This results in an order quantity of 200 + 1.7 * 80 = 336.

2.

(a) What is the probability that Flextrola’s demand will be within 25 percent of its forecast?

Answer: 0.3231 Feedback: Sales within 25% of the mean forecast would be between 750 and 1,250 units.

This corresponds to z = -0.4167 and z = 0.4167. Using Excel, we have F(0.4167) – F(0.4167) = 0.6615 – 0.3385 = 0.3230. (b) What is the probability that Flextrola’s demand will be more than 40 percent greater than Flextrola’s forecast?

Answer: 0.2525 Feedback: The value that is 40% greater than the mean forecast is 1,400 units. This corresponds to z = (1,400 – 1,000)/ 600 = 0.6667. Using Excel, we have F(0.6667) = 0.7475. The probability that demand is greater than 1,400 units is 1 – 0.7475 = 0.2525. (c) Under this contract, how many units should Flextrola order to maximize its expected profit?

Answer: 1,300 Feedback: We have Cu = 121 – 72 = 49 and Co = 72 – 50 = 22. This gives the critical ratio of 49/(22 + 49) = 0.6901. Using Table 13.4 and the round-up rule, this corresponds to z = 0.50. The optimal order quantity is 1,000 + 0.50 * 600 = 1,300. (d) If Flextrola orders 1200 units, how many units of inventory can Flextrola expect to sell in the secondary electronics market?

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Answer: 378.24 Feedback: An order quantity of 1,200 corresponds to z = (1,200 – 1,000)/600 = 0.3333.

Using Table 13.4 and the round-up rule, we have I(0.40)= 0.6304. We multiply this by the standard deviation of 600 to get the expected left-over inventory. (e) If Flextrola orders 1200 units, what are expected sales?

Answer: 821.76 Feedback: If the order quantity is 1,200, the expected inventory is 378.24 (from question 13). The expected sales are 1,200 – 378.24 = 821.76. (f) If Flextrola orders 1200 units, what is expected profit?

Answer: $31,944.96 Feedback: Using the answers from questions 13 and 14, the expected profit when the order quantity is 1,200 is 121 * 821.76 + 50 * 378.24 – 72 * 1,200 = 31,944.96. (g) Using the more accurate forecast (i.e., the log normal distribution), approximately how many units should Flextrola order to maximize its expected profit?

Answer: 1,125 Feedback: The critical ratio is 0.6901. Using the graph of the distribution function in Figure 13.16, this corresponds to a demand of approximately 1,125 units.

3.

(a) How many kilograms should it place in the warehouse before the growing season?

Answer: 420,000 Feedback: We have Cu = 45 – 8 = 37 and Co = 8 – 3 = 5. This gives the critical ratio of 37/(5 + 37) = 0.8810. Using Table 13.4 and the round-up rule, this corresponds to z = 1.20. The optimal order quantity is 300,000 + 1.20 * 100,000 = 420,000. (b) If it puts 400,000 kg in the warehouse, what is their expected revenue (include both domestic revenue and overseas revenue)?

Answer: 13,450,140 Feedback: An order quantity of 400,000 corresponds to z = (400,000 – 300,000)/100,000 = 1. Using Table 13.4, we have I(1) = 1.0833. We multiply this by the standard deviation of 100,000 to get the expected leftover inventory of 108,330. The expected sales are 400,000 – 108,330 = 291,670. The expected revenue is 45 * 291,670 + 3 * 108,330 = 13,450,140. (c) How many kilograms should it place in the warehouse if it wants to minimize its inventory while ensuring that the stockout probability is no greater than 10 percent?

Answer: 430,000 109

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Feedback: A 10% stockout probability corresponds to a 90% in-stock probability.

According to Table 13.4, this requires z = 1.30. This yields an order quantity of 300,000 + 1.30 * 100,000 = 430,000. (d) What is the maximum profit for this seed?

Answer: 11,100,000 Feedback: The maximum profit is (45 – 8) * 300,000 = 11,100,000.

4.

(a) How many units of each sweater type should Fashionables order to maximize its expected profit?

Answer: 560 Feedback: We have Cu = 70 – 40 = 30 and Co = 40 – 20 = 20. This gives the critical ratio of 30/(20 + 30) = 0.6. Using Table 13.4 and the round-up rule, this corresponds to z = 0.3. The optimal order quantity is 500 + 0.3 * 200 = 560. (b) If Fashionables wishes to ensure a 97.5 percent in-stock probability, what should its order quantity be for each type of sweater?

Answer: 900 Feedback: According to Table 13.4, a 97.5% in-stock probability requires z = 2.0. This yields an order quantity of 500 + 2 * 200 = 900. (c) Say Fashionables orders 725 of each sweater. What is Fashionables’ expected profit?

Answer: 45,945 Feedback: An order quantity of 725 corresponds to z = (725 – 500)/200 = 1.125. Using Table 13.4 and the round-up rule, we have I(1.2) = 1.2561. We multiply this by the standard deviation of 200 to get the expected leftover inventory of 251.22. The expected sales are 725 – 251.22 = 473.78. The expected profit is 70 * 473.78 + 20 * 251.22 – 40 * 725 = 9,189. We have this same expected profit for all five sweaters, so the total expected profit is 9,189 * 5 = 45,945. (d) Say Fashionables orders 725 of each sweater. What is the stockout probability for each sweater?

Answer: 0.1303 Feedback: An order quantity of 725 corresponds to z = (725 – 500)/ 200 = 1.125. Using Excel, the in-stock probability is F(1.125) = 0.8697, which means that the stockout probability is 1 – 0.8697 = 0.1303. Using Table 13.4 and the round-up rule, the in-stock probability is F(1.2) = 0.8849, and the stockout probability is 1 – 0.8849 = 0.1151.

5.

(a) What is the probability this parka turns out to be a ―dog,‖ defined as a product that sells less than half of the forecast?

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Answer: 0.1908 Feedback: 50% of the mean forecast is 1,050 units. This corresponds to z = (1,050 – 2,100)/1,200 = -0.875. Using Excel, F(-0.875) = 0.1908. (b) How many parkas should Teddy Bower buy from TeddySports to maximize expected profit?

Answer: 2,340 Feedback: We have Cu = 22 – 10 = 12 and Co = 10 – 0 = 10. This gives the critical ratio of 12/(10 + 12) = 0.5455. Using Table 13.4 and the round-up rule, this corresponds to z = 0.2. The optimal order quantity is 2,100 + 0.2 * 1,200 = 2,340. (c) If Teddy Bower orders 3000 parkas, what is the in-stock probability?

Answer: 0.7734 Feedback: An order quantity of 3,000 corresponds to z = (3,000 – 2,100)/1,200 = 0.75.

Using Excel, the in-stock probability is F(0.75) = 0.7734. (d) If Teddy Bower orders 3000 parkas, what is the expected leftover inventory?

Answer: 1,104.24 Feedback: An order quantity of 3,000 corresponds to z = (3,000 – 2,100)/1,200 = 0.75.

Using Table 13.4 and the round-up rule, we have I(0.8) = 0.9202. We multiply this by the standard deviation of 1,200 to get 1,104.24. (e) If Teddy Bower orders 3000 parkas, what are expected sales?

Answer: 1,895.76 Feedback: If the order quantity is 3,000, the expected inventory (from question 28) is 1,104.24. The expected sales are 3,000 – 1,104.24 = 1,895.76. (f) If Teddy Bower orders 3000 parkas, what is expected profit?

Answer: 11,706.72 Feedback: Using the answers from questions 28 and 29, the expected profit when the order quantity is 3,000 is 22 * 1,895.76 + 0 * 1,104.24 – 10 * 3,000 = 11,706.72. (g) If Teddy Bower wishes to ensure a 98.5 percent in-stock probability, how many parkas should Teddy Bower order?

Answer: 4,740 Feedback: According to Table 13.4, a 98.5% in-stock probability requires z = 2.2. This yields an order quantity of 2,100 + 2.2 * 1,200 = 4,740.

6.

(a) If Teddy Bower decides to include these boots in its assortment, how many boots should Teddy Bower order from the supplier?

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Answer: 430

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Feedback: We have Cu = 54 – 40 = 14 and Co = 40 – 27 = 13. This gives the critical ratio of 14/(13 + 14) = 0.5185. Using Table 13.4 and the round-up rule, this corresponds to z = 0.1. The optimal order quantity is 400 + 0.1 * 300 = 430. (b) Suppose Teddy Bower orders 380 boots. What is Teddy Bower’s expected profit?

Answer: 2,088.91 Feedback: An order quantity of 380 corresponds to z = (380 – 400)/300 = -0.0667. Using Table 13.4 and the round-up rule, we have I(0) = 0.3989. We multiply this by the standard deviation of 300 to get the expected inventory of 119.67. The expected sales are 380 – 119.67 = 260.33. The expected profit is 54 * 260.33 + 27 * 119.67 – 40 * 380 = 2,088.91. (c) The marketing department insists that its in-stock probability be at least 98 percent. Given this mandate, how many boots does it need to order?

Answer: 1,030 Feedback: According to Table 13.4, a 98% in-stock probability requires z = 2.1. This yields an order quantity of 400 + 2.1 * 300 = 1,030. (d) John Briggs, a buyer in the procurement department, overheard at lunch a discussion of the ―boot problem.‖ He suggested that Teddy Bower ask for a quantity discount from the supplier.

After following up on his suggestion, the supplier responded that Teddy Bower could get a 10 percent discount if it were willing to order at least 800 boots. If the objective is to maximize expected profit, how many boots should Teddy Bower order given this new offer?

Answer: 800 Feedback: In question 32, we determined that the optimal order quantity was 430. This gives z = (430 – 400)/ 30 = 0.1. Using Table 13.4, we have I(0.1) = 0.4509. We multiply this by the standard deviation of 300 to get the expected inventory of 135.27. The expected sales are 430 – 135.27 = 294.73. The expected profit is 54 * 294.73 + 27 * 135.27 – 40 * 430 = 2,367.71. If we order 800 units, the discounted price will be 40 *(1 – 0.10) = 36. The order quantity of 800 corresponds to z = (800 – 400)/300 = 1.3333.

Using Table 13.4 and the round-up rule, we have I(1.4) = 1.4367. We multiply this by the standard deviation of 300 to get the expected inventory of 431.01. The expected sales are 800 – 431.01 = 368.99. The expected profit is 54 * 368.99 + 27 * 431.01 – 36 * 800 = 2,762.73. Therefore, it makes sense to take advantage of the quantity discount and order 800.

7.

(a) How many gallons should Goop purchase to maximize its expected profit?

Answer: 250 Feedback: The salvage value in this case is -5. We have Cu = 25 – 10 = 15 and Co = 10 – 113

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(-5) = 15. This gives the critical ratio of 15/(15 + 15) = 0.5. This means that the order quantity should be the mean of 250.

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(b) Suppose Goop purchases 150 gallons of raw material. What is the probability that it will run out of raw material?

Answer: 0.8413 Feedback: An order quantity of 150 corresponds to z = (150 – 250)/100 = -1. Using Table 13.4, we have F(-1) = 0.1587. The probability that the firm stocks out is 1 – F(-1) = 0.8413. (c) Suppose Goop purchases 300 gallons of raw material. What are the expected sales (in gallons)?

Answer: 230.22 Feedback: An order quantity of 300 corresponds to z = (300 – 250)/100 = 0.5. Using Table 13.4, we have I(0.5) = 0.6978. We multiply this by the standard deviation of 100 to get the expected inventory of 69.78. The expected sales are 300 – 69.78 = 230.22. (d) Suppose Goop purchases 400 gallons of raw material. How much should it expect to spend on disposal costs (in dollars)?

Answer: $764.65 Feedback: An order quantity of 400 corresponds to z = (400 – 250)/100 = 1.5. Using Table 13.4, we have I(1.5) = 1.5293. We multiply this by the standard deviation of 100 to get the expected inventory of 152.93. Each unit costs $5 for disposal, so the total expected disposal cost is 152.93 * 5 = 764.65. (e) Suppose Goop wants to ensure that there is a 92 percent probability that it will be able to satisfy its customers’ entire demand. How many gallons of the raw material should it purchase?

Answer: 400 Feedback: Using Table 13.4 and the round-up rule, we need to have z = 1.5 to ensure a 92% in-stock probability. This corresponds to an order quantity of 250 + 1.5 * 100 = 400.

8.

(a) How much would Land’s Start buy if it chose option 1?

Answer: 288 Feedback: In Option 1, we have Cu = 100 – 65 = 35 and Co = 65 – 53 = 12. This gives the critical ratio of 35/(12 + 35) = 0.7447. Using Table 13.4 and the round-up rule, this corresponds to z = 0.7. The optimal order quantity is 200 + 0.7 * 125 = 287.5 288. (b) How much would Land’s Start buy if it chose option 2?

Answer: 188 Feedback: In Option 1, we have Cu = 100 – 55 = 45 and Co = 55 – 0 = 55. This gives the critical ratio of 45/(55 + 45) = 0.45. Using Table 13.4 and the round-up rule, this 115

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corresponds to z = -0.1. The optimal order quantity is 200 + (-0.1) * 125 = 187.5

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188.

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(c) Which option will Land’s Start choose?

Answer: Option 1 Feedback: We need to determine Land’s Start’s expected profit under each option. In Option 1, I(0.7) = 0.8429, and the expected inventory is 0.8429 * 125 = 105.36.

The expected sales are 288 – 105.36 = 182.64. Land’s Start’s expected profit is 100 * 182.64 + 53 * 105.36 – 65 * 288 = 5,128.08. In Option 2, I(-0.1) = 0.3509, and the expected inventory is 0.3509 * 125 = 43.86. The expected sales are 188 – 43.86 = 144.14. Land’s Start’s expected profit is 100 * 144.14 – 55 * 188 = 4,074. (d) Suppose Land’s Start chooses option 1 and orders 275 units. What is Geoff Gullo’s expected profit?

Answer: $5,907.23 Feedback: An order quantity of 275 corresponds to z = (275 – 200)/125 = 0.6. Using Table 13.4, I(0.6) = 0.7687. The expected inventory is 0.7687 * 125 = 96.09. This represents the expected number of returns that Geoff will receive. His expected profit is (65 – 25) * 275 – 53 * 96.09 = 5,907.23.

9.

(a) How many bagels should the store have at 3 p.m. to maximize the store’s expected profit (from sales between 3 p.m. and closing)? (Hint: Assume day-old bagels are sold for $0.99/6 = $0.165 each; that is, don’t worry about the fact that day-old bagels are sold in bags of six.)

Answer: 75 Feedback: The salvage value per bagel is 0.99/6 = 0.165, but only 2/3 of the bagels are sold. This yields an effective salvage value of 0.165 * 2/3 = 0.11. We have Cu = 0.60 – 0.20 = 0.40 and Co = 0.20 – 0.11 = 0.09. This gives the critical ratio of 0.40/(0.09 + 0.40) = 0.8163. Using Table 13.4 and the round-up rule, this corresponds to z = 1.0. The optimal order quantity is 54 + 1 * 21 = 75. (b) Suppose the store manager has 101 bagels at 3 p.m. How many bagels should the store manager expect to have at the end of the day?

Answer: 48.38 Feedback: An order quantity of 101 corresponds to z = (101– 54)/ 21 = 2.2381. Using Table 13.4 and the round-up rule, I(2.3) = 2.3037. The expected inventory is 2.3037 * 21 = 48.38. (c) Suppose the manager would like to have a .95 in-stock probability on demand that occurs after 3 p.m. How many bagels should the store have at 3 p.m. to ensure that level of service?

Answer: 90 Feedback: Using Table 13.4 and the round-up rule, we see that a 95% in-stock probability requires z = 1.7. The order quantity is 54 + 1.7 * 21 = 89.70 90.

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10.

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(a) Suppose burrito customers buy their snack somewhere else if the Kiosk is out of stock. How many burritos should the Kiosk make for the lunch crowd?

Answer: 18 Feedback: We have Cu = (4 + 0.60) – (2 + 0.05) = 2.55 and Co = (2 + 0.05) – 0.05 = 2.

This assumes that the soda maintains its value if the burrito is not sold; otherwise, the overage cost would be 2.05. This gives the critical ratio of 2.55/(2 + 2.55) = 0.5604.

Using Table 13.9 and the round-up rule, this corresponds to 18. (b) Suppose the Kiosk makes 24 burritos. How many burritos should it expect to discard at the end of the day?

Answer: 6.18 Feedback: From Table 13.9, we have I(24) = 6.18. (c) Suppose the Kiosk makes 24 burritos. How many burritos should it expect to sell?

Answer: 17.82 Feedback: From Table 13.9, we have I(24) = 6.18. The expected sales is 24 – 6.18 = 17.82. (d) Suppose the Kiosk makes 24 burritos. What is the Kiosk’s expected profit, including the profit from the sale of sodas?

Answer: 33.08 Feedback: From questions 49 and 50, the expected inventory is 6.18 and the expected sales are 17.82. The expected profit is 4.60 * 17.82 + 0.05 * 6.18 – 2.05 * 24 = 33.08. (e) Suppose the Kiosk makes 30 burritos. What is the probability that some customer is unable to purchase a burrito?

Answer: 0.0033 Feedback: From Table 13.9, we have F(30) = 0.9967. The probability that a customer cannot purchase a burrito is 1 – F(30) = 0.0033. (f) If the Kiosk wants to be sure it has inventory for its customers with at least a .985 probability, how many burritos should it make?

Answer: 28 Feedback: From Table 13.9 and the round-up rule, the order quantity that will achieve at least a 98.5% in-stock probability is 28 (because F(27) = 0.9827 and F(28) = 0.9897). (g) Suppose that any customer unable to purchase a burrito settles for a lunch of Pop-Tarts and a soda. Pop-Tarts sell for 75¢ and cost the Kiosk 25¢. (As Pop-Tarts and soda are easily stored, the Kiosk never runs out of these essentials.) Assuming that the Kiosk management is interested in maximizing profits, how many burritos should it make?

Answer: 17 119

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Feedback: The kiosk makes a profit of $0.50 on each Pop Tart and $0.55 on each soda.

This represents a profit of $1.05. On each burrito it makes a profit of $2 (and still makes $0.55 on each soda). We still have Co = (2 + 0.05) – 0.05 = 2, but now Cu = 2.55 – 1.05 = 1.50. The critical ratio is 1.50/(2 + 1.50) = 0.4286. From Table 13.9 and the round-up rule, this yields a quantity of 17.

11.

(a) Suppose Larry books the car for two time slots. How likely is he to pay $40 or more in late fees?

Answer: 0.5665 Feedback: If Larry books 2 slots, he will incur at least $40 in late fees if his demand is for four or more slots. We can find this probability by taking 1 – F(3) = 1 – 0.4335 = 0.5665. (b) To minimize his rental costs, how many time slots should Larry reserve?

Answer: 7 Feedback: We have Cu = 20 and Co = 1.50. The critical ratio is 20/(1.5 + 20) = 0.9302.

Using Table 13.10 and the round-up rule, we can determine that the quantity should be 7. (c) Suppose Larry books the car for five time slots. How many time slots can he expect to waste (i.e., they end up being of no use to him)?

Answer: 1.41 Feedback: Using Table 13.10, I(5) = 1.41. (d) Larry hates paying any late penalty fee. Suppose he wants to be 99.9 percent sure that he will not have to pay a late fee. How many slots should he book?

Answer: 11 Feedback: Using Table 13.10, we see that we need to reserve 11 slots to achieve at least a 99.9% in-stock probability (because F(10) = 0.9972 and F(11) = 0.9991).

CASE Le Club Français du Vin This case is a shorter and simpler version of ―Forecasting and Procurement at Le Club Français du Vin‖ by Christian Terwiesch and Antoine Gouze, which is available without cost. It has a broader set of wines and includes more discussion of the forecasting process than the case in the chapter.

With this case students: • Apply the newsvendor model to decide quantities that either maximize expected profit or achieve in-stock probability goals, • Evaluate expected profits given chosen order quantities • Discuss whether an order quantity is appropriate in terms of the service it yields to customers • Discuss how changes in the assortment could help improve their business 120

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performance.

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In terms of applying the newsvendor model, the main issue of the case is to determine the overage and underage costs correctly. This is somewhat challenging because there are some additional costs discussed in the case that are not included in any of the examples in the text.

Nevertheless, students should be able to apply the knowledge from the text to this somewhat different situation.

The case also allows for a discussion of whether the newsvendor’s model recommendations are appropriate (e.g., should some products have a low in-stock probability) and how the assortment could be modified to improve profit.

Q1. To maximize LeClub’s expected profit, how much of each of the wines in Table LECLUB should Zanella order?

To find the optimal order quantity, we need to apply the newsvendor model. Let’s first determine for each wine its cost and salvage value. Then we can determine the overage and underage costs. We finish with the critical ratio and the optimal order quantity.

Le Club purchases each wine for 50% of the price listed in the case table. In addition, each bottle is eventually shipped to a customer for a €1.25 fee. Thus, the total cost for each bottle is the sum of the procurement cost and the shipping cost: Total cost = Procurement cost + Shipping cost The salvage value for overbought bottles includes the discounted sales revenue less the additional expenses associated with the bottle (i.e., the storage cost and the opportunity cost of capital): Salvage value = Discount Price - €1.1 – 15 percent of purchase price Now we are ready to evaluate the underage and overage costs. The underage cost for each wine is relatively straightforward – if a bottle is not ordered but could have been sold, then LeClub loses the bottle’s margin less the shipping cost.

Cu = Sales price – Total cost The overage cost is the consequence of ordering a bottle that is not sold with the main season and is later discounted. It is the difference between the total cost and the salvage value: Co = Total cost – Salvage value To explain, if a bottle is purchased but not sold at full price, then LeClub could have avoided the cost of purchasing the bottle (i.e., total cost), but it also would not have collected the salvage value for the bottle.

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The following table provides the analysis up to the critical ratio: Appellation VDP des Côteaux de L'Ardèche Bordeaux MINERVOIS CÔTES DU VENTOUX CÔTES DE BOURG MADIRAN GIVRY PESSAC LEOGNAN

Retail Purchase Total price Cost price 3.25 1.625 2.875 4.50 2.250 3.500 5.21 2.605 3.855 2.800 4.050 5.60 7.20 3.600 4.850 9.00 4.500 5.750 12.90 6.450 7.700 18.90 9.450 10.700

Salvage value Co 0.76875 2.10625 1.48750 2.01250 1.89575 1.95925 2.12000 1.93000 3.04000 1.81000 4.07500 1.67500 6.31750 1.38250 9.76750 0.93250

Cu 0.375 1.000 1.355 1.550 2.350 3.250 5.200 8.200

Critical Ratio 0.1511 0.3320 0.4088 0.4454 0.5649 0.6599 0.7900 0.8979

The z score for each bottle can be determined from the Standard Normal Distribution Function Table and then order quantity combines the z score with the demand forecast:

Appellation VDP des Côteaux de L'Ardèche Bordeaux MINERVOIS CÔTES DU VENTOUX CÔTES DE BOURG MADIRAN GIVRY PESSAC LEOGNAN

Demand forecast Standard Mean deviation 3500 1280 2900 1080 4000 1430 1200 480 1300 510 12000 3000 900 360 1300 510

Critical Ratio 0.1511 0.3320 0.4088 0.4454 0.5649 0.6599 0.7900 0.8979

z -1.0 -0.4 -0.2 -0.1 0.2 0.5 0.9 1.3

I(z) Q 0.0833 2220 0.2304 2468 0.3069 3714 0.3509 1152 0.5069 1402 0.6978 13500 1.0004 1224 1.3455 1963

Q2. If Zanella orders the quantity that maximizes expected profit, for each wine what is the probability the wine has inventory left over that must be discounted?

The probability a wine will have some bottles to discount equals F(z): Appellation VDP des Côteaux de L'Ardèche Bordeaux MINERVOIS CÔTES DU VENTOUX CÔTES DE BOURG MADIRAN GIVRY PESSAC LEOGNAN

z F(z) -1.0 0.1587 -0.4 0.3446 -0.2 0.4207 -0.1 0.4602 0.2 0.5793 0.5 0.6915 0.9 0.8159 1.3 0.9032

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Q3. From the wines listed in Table LECLUB, which are the most profitable and which are the least profitable? Why? Begin with the evaluation of expected sales and expected inventory: Expected inventory = Standard deviation x I(z) and Expected sales = Q – Expected inventory It follows that Expected profit = Price x Expected Sales + Salvage Value x Expected Inventory – Total Cost x Q Appellation VDP des Côteaux de L'Ardèche Bordeaux MINERVOIS CÔTES DU VENTOUX CÔTES DE BOURG MADIRAN GIVRY PESSAC LEOGNAN

Mean 3500 2900 4000 1200 1300 12000 900 1300

Standard deviation 1280 1080 1430 480 510 3000 360 510

z -1.0 -0.4 -0.2 -0.1 0.2 0.5 0.9 1.3

I(z) Q Exp Inv Exp Sales Profit 0.0833 2220 106.6 2113.4 568 0.2304 2468 248.8 2219.2 1718 0.3069 3714 438.9 3275.1 3578 0.3509 1152 168.4 983.6 1200 0.5069 1402 258.5 1143.5 2219 0.6978 13500 2093.4 11406.6 33565 1.0004 1224 360.1 863.9 3994 1.3455 1963 686.2 1276.8 9830

So, the Madiran is the most profitable, in large part because it has the highest demand forecast. But the size of the demand forecast is not the only driver of profitability. The L’Ardeche has a high demand forecast, 3500 bottles, but is far less profitable than the Givry, which is forecasted to sell only about ¼ as many bottles.

Another way to view profitability is on a per bottle basis. In that case, the Pessac Leognan is the most profitable, due to its high price.

Appellation VDP des Côteaux de L'Ardèche Bordeaux MINERVOIS CÔTES DU VENTOUX CÔTES DE BOURG MADIRAN GIVRY PESSAC LEOGNAN

Q 2220 2468 3714 1152 1402 13500 1224 1963

Profit 568 1718 3578 1200 2219 33565 3994 9830 124

Profit per bottle 0.26 0.70 0.96 1.04 1.58 2.49 3.26 5.01

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

Q4. If Zanella decides that they must have at least a 0.75 in-stock probability with each wine, how much of each wine would they order? Do you recommend that they order those quantities?

If every product has the same in-stock, then the resulting calculations are Appellation VDP des Côteaux de L'Ardèche Bordeaux MINERVOIS CÔTES DU VENTOUX CÔTES DE BOURG MADIRAN GIVRY PESSAC LEOGNAN

z 0.7 0.7 0.7 0.7 0.7 0.7 0.7 0.7

Total Salvage Retail I(z) Q Exp Inv Exp Sales Cost value price Exp Profit 0.8429 4396 1078.9 3317.1 2.875 0.76875 3.25 -1028.5 0.8429 3656 910.3 2745.7 3.500 1.48750 4.50 913.7 0.8429 5001 1205.3 3795.7 3.855 1.89575 5.21 2781.7 0.8429 1536 404.6 1131.4 4.050 2.12000 972.8 5.60 0.8429 1657 429.9 1227.1 4.850 3.04000 7.20 2105.6 0.8429 14100 2528.7 11571.3 5.750 4.07500 9.00 33371.2 0.8429 1152 303.4 848.6 7.700 6.31750 12.90 3993.3 0.8429 1657 429.9 1227.1 10.700 9.76750 18.90 9661.3

The low price wines with the low critical ratios would be stocked ―too much‖ (relative to the optimal quantity) and high price wines would be stocked ―too little‖. Overall, profitability would hurt.

Q5. How could Zanella improve his business?

This question moves the discussion beyond the mechanics of the model to more qualitative issues. For instance, given that shipping costs 1.25 per bottle, no matter if the price of the bottle is 3.25 or 18.90, it is apparent that low priced wines are barely profitable. Consequently, the optimal in-stock is very low for those wines. For example, students could be asked something like ―at the end of the season, how likely is it that some customer could not purchase the L’Ardeche if Le Club orders the profit maximizing quantity?‖ The answer is about 85% (1Critical Ratio). Follow up with something like ―How do you think customers will react?‖. This raises the possibility that customer would be irritated that Le Club doesn’t stock the inexpensive wines, possibly in an effort to force customers into more expensive wines. This could deteriorate long run brand loyalty.

One option for Zanella is to have a minimum in-stock, say 60%. This will ensure better service for the low priced wines at the cost of some lost profit. Alternatively, Zanella could remove the low priced wines from the assortment. Maybe Le Club should stock only wines that are 6 euros or higher.

CHAPTER 14: INVENTORY MANAGEMENT WITH FREQUENT ORDERS CONCEPTUAL QUESTIONS 125

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1. Demand in each period follows the same normal distribution (i.e., there is one demand distribution that represents demand in any single period). Assuming demand is independent across periods, which of the following statements about mean demand over five periods is true?

Answer: C. It equals five times the mean of demand over one period.

Feedback: The mean of demand across multiple periods is the sum of the means in each period. All periods have the same mean demand, so demand over 5 periods is 5 times the demand in a single period.

2. Demand in each period follows the same normal distribution (i.e., there is one demand distribution that represents demand in any single period). Assuming demand is independent across periods, which of the following statements about the standard deviation of demand over five periods is true?

Answer: B. It is greater than the standard deviation of demand over one period, but less than five times the standard deviation of demand over one period.

Feedback: When the same distribution represents demand in each period and demands across periods are independent, the standard deviation of demand across multiple periods is the square root of the number of periods times the standard deviation in a single period. The square root of 5 is greater than 1 but less than 5. Thus, the standard deviation over 5 periods is greater than the standard deviation over a single period, but less than 5 times the single period standard deviation.

3. For products with slow-moving demand—for example, one unit per week—the Poisson distribution is likely to be a better model for demand than the normal distribution because: (choose the best answer) Answer: E. the Poisson distribution does not assign any probability to negative outcomes.

Feedback: Normal distribution assigns positive probabilities to negative demand; however, if the mean is significantly higher than the standard deviation these probabilities become negligibly small. However, if the items are slow moving (mean low compared to standard deviation) there is a significant probability of having negative demand under normal distribution. Poisson distribution does not assign any probability to negative outcomes. Answers in (b), (c), and (d) are incorrect, while the answer in (a) is not relevant.

4. A firm manages its inventory with an order-up-to model. Each period is 1 day, the lead time is 2 days, the order-up-to level is 10, and its inventory position at the start of a day (before it submits an order for that day) is –4. Which of the following statements is definitely true?

Answer: C. There are at least four units backordered.

Feedback: With an inventory position of –4, we know the firm has no on-hand inventory, and the sum of on-order minus backorders is –4. We don’t know how many are on-order or backordered but the firm must have at least 4 units backordered.

5. In the order-up-to model, the standard deviation of the order quantities: 126

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Answer: C. is equal to the standard deviation of demand in one period.

Feedback: In the order-up-to-model items are ordered on a one-for-one basis. Thus the standard deviation of orders equals the standard deviation of demand.

6. Suppose the order-up-to model is used. The probability that you end any given period with no on-hand inventory equals: Answer: E. the stockout probability.

Feedback: Probability of ending a period with no inventory = stockout probability.

7. For a fixed order-up-to level, which definition yields the higher in-stock probability?

Answer: B. The second definition because the firm is more likely to satisfy all demand than to end the day with one unit on hand.

Feedback: Definition II yields higher in-stock because there will be days when there are no items left over but all the demand was satisfied. In such cases, Definition II will be instock, but will be out-of-stock by Definition I. 8. Suppose inventory is managed using the order-up-to model. The inventory position is 20 and demand in the last period was 10. What is the target in-stock probability?

Answer: F. Could be any of the above a–e.

Feedback: The target in-stock is not related to any one order.

9. If the target in-stock probability increases, then the expected time between stockouts: Answer: A. increases.

Feedback: A higher in-stock probability increases the time between stockouts.

10. Demand each period is normally distributed and an order-up-to model is used to decide order quantities. Which of the following influences the chosen order-up-to level (i.e., a change in which of the following would change the chosen order-up-to level)?

Answer: G. A, B, and C. Feedback: The mean in one period influences demand over l+1 periods, which is important for selecting the order up-to level, S. Also important is the target in-stock probability and the standard deviation of demand over l+1 periods.

11. Suppose in the order-up-to model the target in-stock probability is .95 and demand across periods is independent and normally distributed. If the lead time is doubled but the target in- stock probability remains .95, what happens to the order-up-to level?

Answer: C. It increases, but by less than a factor of 2.

Feedback: S = + z × σ , where is the mean of demand over l+1 periods and σ is the standard deviation of demand over l+1 periods. Because the in-stock probability remains the same, z does not change. The lead time doubles, but mean demand over l+1 periods doesn’t double. For example, if the lead time is 4, then demand over l+1 periods 127

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increases

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from 5 to 9 periods. The standard deviation also does not double because it increases by the square root of 2. Thus, the order up-to level increases, but it does so by a factor less than 2.

12. A firm uses the order-up-to model to manage its inventory. It wants to increase its in-stock probability while decreasing its holding costs (i.e., reducing its average inventory). Which of the following actions will help to achieve this goal? (Consider each action independently.)

Answer: A. Only A. Decrease the lead time.

Feedback: Decreasing the lead time will reduce inventory for a given order up-to level and increase the in-stock probability. The others are not correct. If the order up-to level is increased, the in-stock will increase but so will inventory. If the average quantity onorder increases, then it must be that the average lead time increases, which is not good for the in-stock.

13. In the order-up-to model, assume that the mean of demand in a period remains the same and the target in-stock probability is kept at a constant level. If the demand uncertainty (the standard deviation of demand in each period) increases, then: Answer: A. expected on-hand inventory increases.

Feedback: To maintain the same in-stock probability when demand variability increases requires additional on-hand inventory.

14. Suppose the order-up-to model is used to manage inventories. The firm is planning changes that will reduce the lead time to receive replenishments because the firm anticipates that the coefficient of variation of demand will increase even though mean demand per period remains the same. What can be said about the likely change in the firm’s on-order inventory?

Answer: A. It will surely decrease.

Feedback: On-order inventory only depends on the length of the lead time and expected demand. So a reduction in the lead time will reduce on-order inventory.

15. Suppose inventory is managed using the order-up-to model. Which of the following actions will certainly lead to a higher order-up-to level? In all cases, assume the characteristics of the demand process do not change.

Answer: C. A and B Feedback: Both increase the order up-to level.

16. Anna Litic, a new supply chain manager at High Precision Inc. (HP) decides to check some data on the supply chains she manages. She discovers that HP’s in-transit inventory of electronic components from its Tacoma, Washington, factory to its Asian distribution center (DC) has increased from two quarters ago. However, the distribution of demand at the Asian DC has not changed over this period of time. Anna knows that the Asian DC manager is controlling inventory to achieve a fixed in-stock probability target, so she is happy to see that indeed the in- stock inventory at the Asian DC has also not deviated off the target. Anna wonders what has 129

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happened to the Asian DC’s average inventory. What is she likely to discover regarding the Asian DC’s on-hand inventory?

Answer: B. On-hand inventory has increased because the lead time from Tacoma, Washington, to the Asian DC must have increased.

Feedback: If the on-order inventory increases, it means that either the mean demand increased or the lead time increased. Since we know that demand distribution has not changed, it means that the lead time has increased, which leads to higher on-hand inventory.

17. Which of the following is a strategic decision for a grocery retailer?

Answer: B. Whether to open new stores that have twice the square footage of current stores Feedback: Strategic decisions are long-term decisions that influence how business is conducted. Changing the store format is strategic because it cannot be changed quickly and it is important regarding the overall performance of the retailer.

18. Location pooling is most effective at generating which of the following changes to performance objectives?

Answer: D. Decreasing days-of-supply of on-hand inventory Feedback: Location pooling reduces days-of-supply and it does not have a direct effect on gross margin or the target in-stock probability. It has no effect at all on on-order inventory.

19. A firm sells a product whose demand over the next couple of years will be stable (so the order-up-to model is used to manage its inventory). When the demand uncertainty is small (e.g., the weekly demand has a mean of 100 and a standard deviation of 5), the main benefit of reducing the lead time is: Answer: D. to reduce the on-order inventory.

Feedback: On-order inventory is directly proportional to lead time (by Little’s law), and hence is reduced by reducing the lead time. Reducing lead time has a small effect on the expected on-hand inventory when uncertainty is small. Similar arguments apply for gains from pooling.

PROBLEMS AND APPLICATIONS 1. A firm implements the order-up-to model with weekly ordering. In week 11, it observes that demand is much lower than expected demand. At the start of week 12 (before it orders), it decides to change its forecast of demand per week and consequently it lowers its order-up-to level. Its order in week 12 will: Answer: A. be less than the demand they observed in week 11.

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2. Assume a firm implements an order-up-to inventory model with an order-up-to level of 10.

The lead time is two periods and demand is normally distributed. Which of the following is definitely not true?

Answer: C. There are 11 units in on-hand inventory.

Feedback: On-hand inventory cannot exceed the order up-to level.

3. A retailer uses the order-up-to model to manage inventory of an item in a store. The lead time for replenishments is 4 weeks and it can place orders weekly. Weekly demand is Poisson with mean 0.10 unit. Its order-up-to level is five and unfilled demand is backordered. What is the coefficient of variation of its orders?

Answer: 3.16 Feedback: Coefficient of variation = standard deviation / mean; in Poisson distributions, the standard deviation is the square root of the mean.

CV = sqrt(0.10)/0.10 = 3.1622 rounded to 3.16.

4. (a) Suppose it uses an order-up-to level of 301 ounces. What is its expected on-hand inventory?

Answer: 103 Feedback: LT = 4, L+1 = 5, Mean/period = 40, Sigma = 30, Mean over 5 period = 40 × 5 = 200. Sigma over 5 period = 30 × Sqrt(5) = 67.08. Z = (301-200)/67.08 = 1.5056. Use table 14.1 to look up I(z = 1.5) = 1.5293; Expected on-hand inventory = I(z) × sigma =1.5293 × 67.08 = 102.58, rounded to 103. (b) Suppose it uses an order-up-to level of 250 ounces. What is its expected on-order inventory?

Answer: 160 Feedback: LT = 4, Mean/period = 40, Expected On-order inventory = Mean demand per period × lead time = 40 × 4 = 160 (c) Suppose it uses an order-up-to level of 368 ounces. What is its in-stock probability?

Answer: 0.9939 Feedback: LT = 4, Mean/period = 40, Sigma = 30, Mean over 5 periods = 200, Sigma over 5 periods = 30 × Sqrt(5) = 67.08; Z = (368-200)/67.08 = 2.5044. Use Table 14.1 to look up F(z = 2.5) = 0.9939 (d) Suppose it wants a .96 in-stock probability. What should its order-up-to level be?

Answer: 321 Feedback: LT = 4, Mean/period = 40, Sigma = 30, F(z) = 0.96 ==> use Table 14.1 to find the z value that corresponds to F(z) = 0.96. Z = 1.8 has the closest value that is not under 96%. ==> S = 200 + (67.08 × 1.8) = 320.744, rounded to 321.

5.

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(a) Suppose it uses an order-up-to level of 2410. What is its expected on-hand inventory?

Answer: 427 Feedback: LT = 4, Mean/period = 400, Sigma = 152, Mean over (L+1 =) 5 periods = 2000; Sigma over 5 periods = 152 × Sqrt(5) = 339.88 rounded to 340; Z = (24102000)/340 = 1.2059, Use Table 14.1 to look up I(z = 1.2) = 1.2561; On hand inventory = 1.2561 × 340 = 427.074, rounded to 427. (b) Suppose it uses an order-up-to level of 2500. What is its expected on-order inventory?

Answer: 1,600 Feedback: LT = 4, Mean/period = 400; Sigma = 152, Expected on-order inventory = 400 × 4 = 1600 (c) Suppose it uses an order-up-to level of 2000. What is its in-stock probability?

Answer: 0.5 Feedback: LT = 4, Mean/period = 400; Sigma = 152, Mean over 5 periods = 2000; Sigma over 5 periods= 152 × Sqrt(5) = 340; Z = (2000-2000)/340 = 0, Use Table 14.1 to look up in-stock probability with F(z = 0) = 0.5. (d) Suppose it wants a .90 in-stock probability. What should its order-up-to level be?

Answer: 2,442 Feedback: LT = 4, Mean/period = 40; Sigma = 30, F(z) = 0.90 ==> From Table 14.1 find the z value for F(z) = 0.90; Z = 1.30 ==> order-up-to level should be S = 2000 + (340 × 1.30) = 2442 6. (a) Suppose it uses an order-up-to level of 2200. What is its average order quantity?

Answer: 1,068 Feedback: S = 2200, Average order quantity = Average weekly demand (because orders are placed weekly) = 1068 (b) Suppose it uses an order-up-to level of 2480. What is its expected on-hand inventory?

Answer: 344 Feedback: S = 2480 ==> Average demand over L + 1 periods = 1068 × 2 = 2136.

Standard deviation of demand over L + 1 periods = sqrt(2) × 110.2 = 115.9. Z = (2480 – 2136)/155.9 = 2.2067 or 2.2 ==> I(z) = 2.2049 Expected On hand inventory = sigma × I(z) =155.9 × 2.2049 = 344.7 (c) Suppose it uses an order-up-to level of 2600. What is its expected on-order inventory?

Answer: 1,068 132

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Feedback: S = 2600, Expected on–order inventory = average weekly demand × Lead time = 1068 (d) Suppose it uses an order-up-to level of 2697. What is its in-stock probability?

Answer: 0.9998 Feedback: S = 2697 ==> Z = (2697 – 2136)/155.9 = 3.5988 or 3.6 ==> F(z) = .99984 (e) Suppose it uses an order-up-to level of 2400. What is its stockout probability?

Answer: 0.0446 Feedback: S = 2400==> Z = (2400 – 2136)/155.9 = 1.7 ==> F(z) = .9554; 1 – F(z) = 0.0446 (f) Suppose it wants a .945 in-stock probability. What should its order-up-to level be?

Answer: 2,385 Feedback: F(z) = 0.954 ==> Z = NORM.S.INV(0.954) = 1.598; S = 2136 + 1.598 × 55.9 = 2385 7. (a) Suppose it uses an order-up-to level of five. What is its average order quantity?

Answer: 0.8 Feedback: Average order quantity is the same as the prior week's demand, based on frequency of orders. Ordering once a week = 0.80, the mean demand. (b) Suppose it uses an order-up-to level of nine. What is its expected on-hand inventory?

Answer: 6.6 Feedback: Use Appendix 14A to look up the value of mean = 0.8 x (L+1 = ) 3 = 2.4.

I(2.4) with S = 7 is 6.6. (c) Suppose it uses an order-up-to level of six. What is its expected on-order inventory?

Answer: 1.6 Feedback: Expected Order on hand = 0.8 × 2 weeks = 1.6 (does not depend on the order- by-to level) (d) Suppose it uses an order-up-to level of seven. What is its in-stock probability?

Answer: 0.9966 Feedback: With mean = 2.4 and S = 7 using Appendix 14A: F(S) = 0.9967 (e) Suppose it uses an order-up-to level of four. What is its stockout probability?

Answer: 0.0959 133

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Feedback: With mean = 2.4 and S = 4 using Appendix 14A: F(S) = 0.9041; stock out probability = 1 - 0.9041 = 0.0959 (f) Suppose it wants a .97 in-stock probability. What should its order-up-to level be?

Answer: 6 Feedback: F(S) = 0.97, Appendix 14A, S = 6 corresponds to F(S) = 0.9884 while S = 5 corresponds to F(S) = 0.9643. Go with the value that is closest and over, rather than under.

8. (a) Suppose it uses an order-up-to level of 11. What is its expected on-hand inventory?

Answer: 4.1 Feedback: L+1 = 2. Mean = 3.5 × (L+1 ) = 7; Standard deviation for Poisson = sqrt(mean); mean = 7, sqrt(7) = 2.646. The Poisson table for mean 7 is not available, so use the normal distribution with mean 7 and standard deviation 2.646. S = 11, so Z = (11 – 7)/2.646 = 1.512. Use Table 14.1 to look up I(z = 1.5) = 1.5293; Expected onhand inventory = 2.646 × 1.5293 = 4.0465 rounded to 4.1. (b) Suppose it uses an order-up-to level of 12. What is its expected on-order inventory?

Answer: 3.5 Feedback: Expected on-order quantity = 3.5 × 1 = 3.5. (c) Suppose it uses an order-up-to level of 13. What is its in-stock probability?

Answer: 0.9871 Feedback: S = 13, Z = (13 - 7)/2.646 = 2.268 ==> Use Table 14.1 to look up F(Z = 2.3) = 0.9893 (d) Suppose it uses an order-up-to level of 14. What is its stockout probability?

Answer: 0.0047 Feedback: S = 14, Z = 2.645 ==> use Table 14.1 to look up F(Z = 2.6) = 0.9953; 1 – F(Z) = 0.0047. (e) Southern Fresh has decided to make a policy that every product will receive enough shelf space to ensure a .98 in-stock probability. What should its order-up-to level be?

Answer: 13 Feedback: S = 13. F(z) >= 0.98 occurs at Z = 2.1 in Table 14.1. S = 7 + (2.646 × 2.1) = 12.5566, rounded up to 13. (f) What is the probability that on any particular day it doesn’t submit an order?

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Feedback: For them to not submit an order it means that the in-stock probability is 1 or the stockout probability = 0.0. This occurs when Z = 3.9 and a order-up-to level of S = 18 is used. (g) What is the probability that on any particular day it orders four or fewer bags?

Answer: 0.128 Feedback: Probability of ordering 4 or fewer bags = Pr(0) + Pr(1) + Pr(2) + Pr(3) + Pr(4) = 0.004 + 0.008 + 0.017 + 0.036 + 0.063 = 0.128 . (h) What is the probability that on any particular day it orders five or more bags?

Answer: 0.872 Feedback: Probability of ordering 5 or more bags = 1 - Pr(ordering 4 or fewer bags) = 1 – 0.128 = 0.872.

CASE Warkworth Furniture With the Warkworth Furniture case students: • • •

Use the order up to model to evaluate several different supply chain configurations.

Discover the value of lead time pooling (i.e., why adding a distribution center in a supply chain can reduce costs).

Discuss the pros and cons of moving product between stores to address imbalances in supply and demand.

Q1. How much does Mana incur in holding costs each year with their current system of delivering directing from factory to their stores?

The lead time is 10 weeks, so L=10 and L+1 = 11. Average demand is 0.2 per week per store. So demand over L+1 weeks has a mean of (10+1) x 0.2 = 2.2. With such a low demand rate it is appropriate to use a Poisson distribution to model demand.

The following table provides information about the Poisson with mean 2.2.

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Q 0 1 2 3 4 5 6 7 8 9 10

F(Q) 0.1108 0.3546 0.6227 0.8194 0.9275 0.9751 0.9925 0.9980 0.9995 0.9999 1.0000

I(Q) 0.0000 0.1108 0.4654 1.0881 1.9074 2.8349 3.8100 4.8026 5.8006 6.8001 7.8000

To achieve a target in-stock of 0.99, each store needs to operate with a base stock level S = 6: F(5) = 0.9751, which is too low; F(6) = 0.9925, which is sufficient to achieve a 0.99 instock.

With a base stock of S = 6, the average inventory at the store is 3.81 (from the table above).

Therefore across the 30 stores they have 30 x 3.81 = 114.3 desks on average. Holding costs are $150 per year per desk. So their total holding cost for a year is $150 x 114.3 = $17,145 Q2. Say they open a distribution center in southern California. How much would they incur in holding costs each year with that strategy?

With a distribution center in CA, the lead time for each store is 1 week, i.e., L=1. Demand over L+1 weeks would have a mean of (1+1) x 0.2 = 0.4. The following table provides information about the Poisson with mean 0.4.

Q F(Q) I(Q) 0 0.6703 0.0000 1 0.9384 0.6703 2 0.9921 1.6088 3 0.9992 2.6008 4 0.9999 3.6001 5 1.0000 4.6000 6 1.0000 5.6000 7 1.0000 6.6000 8 1.0000 7.6000 9 1.0000 8.6000 10 1.0000 9.6000 Now each store can operate with a base stock level S = 2 to achive a 0.99 instock. Average inventory is then 1.61 per store. Across 30 stores their inventory is 30 x 1.61 = 48.3. At $150 per desk per year, the 48.3 desks result in a total holding cost per year of $150 x 48.3 = $7,245.

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The distribution center would also have to hold inventory. It would experience a 10 week lead time and let’s say it too operates with an in-stock of 0.99. Average weekly demand on the distribution is 6 desks (30 stores x 0.2 desks per store). With a 10 week lead time, average demand over L+1 = 11 weeks is (10+1) x 6 = 66 desks. The sum of Poisson demands is Poisson, so demand at the distribution center over 11 weeks can be modeled with a Poisson demand with mean 66. The standard deviation of the Poisson is the square root of its mean, so the standard deviation is sqrt(66) = 8.12. Because this mean is substantially greater than 20, we can approximate the demand with a normal distribution with mean 66 and standard deviation 8.12.

To achieve a 0.99 in-stock with a standard normal distribution requires a z value of 2.4. Hence the base stock level is S = 66 + 2.4 x 8.12 = 85.5. Average inventory when z =2.4 is sigma x I(2.4) = 8.12 x 2.4027 = 19.5. Inventory in the distribution center costs $60 per year, so the total holding cost for the distribution center per year, with an average inventory of 19.5 units, is $60 x 19.5 = $1,170.

Adding the holding costs for the stores and the distribution center yields a total cost of $7,245 + $1,170 = $8,415. Hence adding the distribution center substantially reduces the cost to hold inventory.

Q3. Say Warkworth opens a distribution center in southern California. How much do they incur in holding costs per desk?

The total holding cost per year with the distribution center is $8,415. Each year they sell 52 weeks per year x 6 desks per week = 312 desks per year. So the holding cost per desk is $8415 / 312 = $26.97 per desk.

Q4. Would you recommend that they consider Karen’s idea of holding all inventory at the stores, but shipping between stores as needed?

It is not immediately obvious how to exactly evaluate Karen’s proposal because we cannot know how much inventory the stores would need to carry if they had the option of receiving inventory from another store and how often they would have to pay the extra $40 per desk that is shipped between stores. But given that with the distribution center the total holding cost per desk is only $26.97 and with that option there is never a need to ship between stores, it seems unlikely that paying an extra $40 per desk is justified.

Q5. Say Warkworth listened to Andy and didn’t hold inventory at the stores. Instead, inventory would be held in a distribution center and shipped to the stores as needed. How much would they save in inventory holding costs with this strategy?

We already evaluated the inventory that would be needed at the distribution center - $19.5 units on average, which would cost $1,170 per year, or $1,170 /312 = $3.75 per desk. We also evaluated that the stores would carry on average 48.3 desks, which would cost $7,245 per year, 137

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or $7,245 / 312 = $23.22 per desk. So if customers are willing to wait, we can save an additional $23.22 per desk.

CHAPTER 15: FORECASTING CONCEPTUAL QUESTIONS 1. When creating a time series–based forecast for the amount of soda to be sold in the cafeteria next week, which data sources can you include in your forecasting process?

Answer: B. Old demand data 2. A company wants to use regression analysis to forecast the demand for the next quarter. In such a regression model, demand would be the independent variable. True or false?

Answer: B. False 3. A pizza chain wants to forecast the demand rate for each store for each hour in the day.

What type of forecasting method is it most likely to deploy?

Answer: A. Automated forecasting 4. What is the definition of a forecast error?

Answer: C. The difference between the forecast and the actual outcome 5. A company has an unbiased forecast for its demand. What does that mean?

Answer: C. The average of all forecast errors is zero 6. The intuition behind the MSE metric to evaluate old forecasts is: Answer: D. To average the squared forecast errors 7. The intuition behind the MAE metric to evaluate old forecasts is: 138

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Answer: E. To average the absolute values of the forecast errors 8. How big is the effect of statistical noise on the naïve forecast?

Answer: A. Large 9. Using the moving average forecast, is it possible to forecast a demand that is bigger than any previously observed demand?

Answer: B. No 10. Using the exponential smoothing forecast, is it possible to forecast a demand that is bigger than any previously observed demand?

Answer: A. Yes 11. Using the double exponential smoothing forecast, is it possible to forecast a demand that is bigger than any previously observed demand?

Answer: A. Yes 12. A startup has a demand that goes up by 50 percent each year. This demand increase is multiplicative. True or false?

Answer: A. True 13. The seasonality index is based on demand fluctuations that are additive. True or false?

Answer: B. False 14. Deseasonalizing old demand data is the process of reintroducing the seasonal effect to forecasted data. True or false?

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Answer: B. False 15. A sales organization creates a new sales forecast by simply taking the average of demand forecasts that each sales manager generated individually. What type of subjective forecasting approach best describes this?

Answer: A. Forecast combination PROBLEMS AND APPLICATIONS 1. Two servers in a restaurant predict how many guests will come for dinner in the next 4 days.

The first server predicts the number of guests to be 23 for day 1, 35 for day 2, 30 for day 3, and 28 for day 4. The second server predicts 26, 27, 28, and 29 for the 4 respective days. The actual attendance turns out to be 30, 22, 31, and 25. Who has the bigger forecast bias? What are the MSE and the MAE for the two servers?

Answer: The calculations in the following table show that Server 1 has an average forecast error of 2, while Server 2 has one of 0.5. So, both of their forecasts are biased.

Server 1 has an MSE of 57 and an MAE of 6. Server 2 has an MSE of 16.5 and an MAE of 4.

2. A police station had to deploy police officers for emergencies multiple times the last four evenings. The numbers of emergencies for Monday, Tuesday, Wednesday, and Thursday were 7, 4, 8, and 11, respectively. What would be the station’s forecast for Friday using a naïve forecasting approach?

Answer: 11 Feedback: The naïve forecast would be 11, as this corresponds to the last observed demand.

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3. A police station had to deploy police officers for emergencies multiple times the last four evenings. The numbers of emergencies for Monday, Tuesday, Wednesday, and Thursday were 7, 4, 8, and 11, respectively. What would be the station’s forecast for the emergencies on Friday using a 2-day moving average approach?

Answer: 9.5 Feedback: With a two day moving average, we would forecast the demand for Friday as the average of the demand of Wednesday and Thursday. That would be: Forecast for Friday = Average(8, 11) = 9.5 4. A police station had to deploy police officers for emergencies multiple times the last four evenings. The numbers of emergencies for Monday, Tuesday, Wednesday, and Thursday were 7, 4, 8, and 11, respectively. What would be the station’s forecast for Friday using an exponential smoothing forecasting approach? Use α = 0.3 and a forecast for Monday of 9.

Answer: 8.45 Feedback: With the exponential smoothing method, we cannot just forecast for Friday.

Instead, we have to start at the beginning and first forecast for Tuesday. We can then, day by day, update our forecast for the next day. This gives us a forecast for Friday of 8.45.

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5. MyApp is a small but growing startup that sees demand for several of its apps increase quickly. For the last 6 months, monthly downloads were 235,000, 290,000, 336,000, 390,000, 435,000, and 498,000. Using a forecast for the first month of 220,000, an initial trend forecast of 40,000, and smoothing parameters of 0.25 for both demand smoothing and trend smoothing, forecast the demand for the next month using double exponential smoothing.

Answer: 519,566.7 Feedback: We forecast the demand for the next month as follows:

So, the forecast for the next period is 519,566.7.

6. Using a smoothing parameter of 0.25 and an initial forecast of 17,000, forecast demand for the four quarters in 2021.

Answer: We forecast the demand for the next year as follows: Quarter 1 2 3 4 Overall

Season average SI 21866.67 1.247 14666.67 0.837 14666.67 0.825 19133.33 1.091 17533.33

Alpha

0.25

Deseasonalized Smoothed demand demand 17720.43 17180.11 17812.27 17338.15 16604.15 17154.65 17411.15 17218.77 16116.77 16943.27 20322.73 17788.14 19997.70 18340.53 17136.24 18039.45 18762.80 18220.29 14465.00 17281.47 15998.16 16960.64 18052.61 17233.63

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Forecast 2021Q1 2021Q2 2021Q3 2021Q4

21492.9 14415.97 14219.39 18806.28

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7. Mary, Susan, and Sarah are running a beach boutique on the boardwalk of Ocean City. Their favorite product is a red lifeguard hoodie. Mary believes it will sell 340 times next season. Susan forecasts sales of 522 and Sarah forecasts 200. What would be the result of a simple forecast combination?

Answer: 354.

Feedback: The forecast combination is simply the average of the three forecasts. In this case, 354.

CASE International Arrivals Q.1 How would you estimate the passenger volume for the coming year?

Two forces seem to be at work here: The forecasts are found as follows: 1.

2.

3.

4.

5.

6.

7.

Compute the average trend from the same month in 2012 to 2013; this gives an average trend factor of 1.041 Detrend the data by dividing the 2013 data by the trend factor Compute the SI index Deseasonalize the detrended data Smooth the data using exponential smoothing Use the smoothed (and detrended and deseasonalized) value for December 2013 as a base Apply the trend factor (1.04^2 as we are making a 2014 prediction) and the SI to get the forecasts

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Q.1

Q.2 What is the role of trends and seasonality?

• •

An overall trend (when comparing the 2013 months with the 2012 months, the 2013 are all a little higher; to be exact, the average growth relative to the same month in the prior year is 4%) Seasonality by month (with the summer months being the most busy; to be exact, the SI’s for July and August are 1.19 and 1.17).

Q.3 What other variables do you expect to influence the number of international travelers?

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Other variables that may influence the number of international travelers could include weather, price of airline tickets, border restrictions etc.

CHAPTER 16: SERVICE SYSTEMS WITH PATIENT CUSTOMERS CONCEPTUAL QUESTIONS 1. A summer camp that offers 1-week programs faces the challenges of long queues as parents try to check in their children each Saturday morning. If they were to add more staff to assist with the check-in process, then which of the following will occur?

Answer: G. All: A. The average time parents wait decreases; B. The maximum time the parents wait decreases. C. The average number of parents waiting to check in their child decreases.

Feedback: Adding capacity reduces the average waiting time, the maximum waiting time, and the average number of parents waiting.

2. Queuing system A has a utilization of 80 percent, and queuing system B has a utilization of 90 percent. Both have a single server. Say the utilization of both systems increases by 5 percent; that is, A increases from 80 percent to 85 percent while B increases from 90 percent to 95 percent.

Which system is likely to experience the bigger change in the average time in queue?

Answer: B. System B. Feedback: Time in queue increases at a faster and faster rate as the utilization is increased. So a 5% increase in utilization has a larger effect from 90% to 95% than from 80% to 85%.

3. The following four graphs display the number of customers in a queuing system (y-axis) over a long period of time (x-axis). Which of the following is most likely a stable system?

Answer: D. Figure D. Feedback: The first three systems display queues that grow large and do not appear to empty. The fourth system displays a queue that grows and shrinks in a repeated pattern, which is indicative of a stable system.

4. Which best reflects pooling capacity to reduce restroom queue lengths?

Answer: C. Convert the separate men’s and women’s rooms into a single unisex restroom (that both men and women can use).

Feedback: Choices A, D, and E are designed to increase capacity, but not to pool capacity. Choice B is designed to shift capacity from one queue to another. Choice C pools capacity because it allows the customers to use all of the servers (toilets).

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5. YourNurse (YN) Inc. uses certified nurses to answer medical queries from customers over the phone. When patients call into YN, they are first asked to provide their zip code, which then allows YN to route their call to the call center nearest to the patient (it operates 10 across the country). Which single suggestion in the following list (and explanation) is most likely to reduce the average time callers wait before speaking with a nurse?

Answer: C. Instead of using callers’ zip codes, route calls to the call center with the fewest callers to help prevent situations in which there are idle nurses at the same time that there are callers on hold.

Feedback: Choice A would increase demand but not capacity, so it increases waiting time. Choice B decreases capacity, so it increases waiting time. Choice D might change the perception of waiting time, but it doesn’t change the actual waiting time.

PROBLEMS AND APPLICATIONS 1. The Shady Farm Milk Company can process milk at a fixed rate of 7500 gal/hr. The company’s clients request 100,000 gallons of milk over the course of 1 day. This demand is spread out uniformly from 8 a.m. to 6 p.m. The company starts producing at 8 a.m. and continues to work until all of the demand has been satisfied. At noon, how many gallons of milk are in the queue to be processed?

Answer: 10,000 Feedback: 100,000/10 = 10,000 per hour. Uniform Distribution from 8 a.m. - 6 p.m.

Total 10 hours, So, per hour's arrival rate is 100,000 / 10 = 10,000. But the process speed is only 2,500 per hour. So hourly queue = 10,000 - 7500 = 2,500 per hour. Total 4 Hours: 4 × 2,500 = 10,000.

2. Using the previous question, change the fixed rate that milk is processed to 6000 gal/hr; all the other factors remain the same. How long does the client requesting milk at 6 p.m. have to wait to have its demand satisfied (in hours)?

Answer: 6.67 Feedback: Time to serve a unit = T × (Demand/Capacity – 1) = 10 × (10000/6000 – 1) =10 × 4/6=40/6.

3. Using question 1, change the fixed rate that milk is processed to 5000 gal/hr; all the other factors remain the same. On average, how long does a client wait (in hours) to receive its product?

Answer: 5.0 Feedback: Average time to serve a unit = ½ × T × (Demand/Capacity – 1) = ½ × 10 × (10000/5000 – 1) = 5 × 1 = 5.

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4. Which of the lines in the graphs best depicts the relationship between utilization and waiting time in a queuing system? (Assume the utilizations depicted are less than 100 percent.)

Answer: D. 5. Huduko Inc. offers a number of computer services. One server handles its own web page. It receives requests for web pages at the rate of 90 per second. The standard deviation of these interarrival times is 0.02 second. What is the coefficient of variation of the interarrival times for this server?

Answer: 1.8 Feedback: Coefficient of variation = Standard deviation / Mean of interarrival time; Mean of interarrival time = 1/90. So 0.02/(1/90) = 90 × 0.02 = 1.8.

6. Max Stamp approves study abroad documents for the university. Students must wait in line with their forms outside Max’s office. One student at a time is allowed in his office and Max takes precisely 15 minutes to evaluate each student’s set of documents. On average, two students per hour go to his office and they spend, on average, 160 minutes trying to get their forms approved (time waiting in queue plus time in Max’s office having him evaluate their documents).

On average, how many students are waiting outside of Max’s office?

Answer: 4.8 Feedback: Time in queue = 160 - 15 = 145 min. Interarrival time = 30 min. The average number of customers waiting in the queue = Time in queue / Interarrival time = 145 / 30 = 4.8.

7. CPU-on-Demand (CPUD) offers real-time high-performance computing services. CPUD owns one supercomputer that can be accessed through the Internet. Its customers send jobs that arrive, on average, every 5 hours. The standard deviation of the interarrival times is 5 hours. Executing each job takes, on average, 3 hours on the supercomputer and the standard deviation of the processing time is 4.5 hours. What is the utilization (as a percent) of CPUD’s supercomputer?

Answer: 60% Feedback: Utilization = Processing time / Interarrival time = 3 / 5 = 0.6 = 60%.

8. Using the information in the previous question, how long does a customer have to wait to have a job completed?

Answer: 10.3 Feedback: Interarrival time = 5. Processing time = 3. Coefficient of variation of the interarrival time = 5/5 = 1. Coefficient of variation of the processing time = 4.5/3 = 1.5. T = 3 + 3 × (0.6 / 0.4) × (1 + 2.25)/2 = 10.3 hours.

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9. The graph plots the number of customers in a queuing system with two servers. The average processing time of each server is 400 seconds, with a standard deviation of 400 seconds.

Customers wait in a single queue. Which of the following average interarrival times, a (in seconds), is consistent with the data observed in the graph?

Answer: E. a = 250 10. Which will happen regarding the average number of passengers waiting in line to check in with the kiosks relative to the average number when the employees did this task?

Answer: C. It will increase.

11. Camile owns Crunch Code, a company that provides quick programming solutions. Clients send Crunch projects via its web page and Crunch bangs out the needed code Page 568as quickly as possible. Camile has five programmers who do all of the coding. On average, a project arrives once every 4.8 hours. Each project is assigned to one programmer and that programmer takes, on average, 19.2 hours to complete each project. Camile wants to know what fraction of the time (as a percent) each of her programmers is busy coding?

Answer: 80% Feedback: Number of servers (programmers); m = 5. Interarrival time: a = 4.8.

Processing time: p = 19.2. Utilization = p / (a × m) = 19.2 / (4.8 × 5) = 0.8 = 80%.

12. In the previous question, a project arrives at Crunch once every 4.8 hours, with a standard deviation of 6.00 hours. Again, each project is assigned to one programmer and that programmer takes, on average, 19.2 hours to complete each project, with a standard deviation of 19.2 hours. How many uncompleted projects does Crunch Code have, on average, at any given time? (Include projects waiting for a programmer as well as those being programmed.)

Answer: 7.0 Feedback: Given interarrival time (a = 4.8 hours), processing time (p = 19.2 hours), number of servers (m = 5 programmers), standard deviation of interarrival time (6 hours, coefficient of variation = 1.25), and standard deviation of processing time (19.2 hours, coefficient of variation = 1), compute time in queue = 14.18. Number of projects waited in queue = Time in queue/a = 14.18/4.8 = 2.95. Number of projects processed = p/a = 19.2/4.8 = 4. Total projects waited and being processed = 2.95 + 4 = 6.95 rounded to 7.

13. Larry Ellison starts a company that manufactures high-end custom leather bags. He hires two employees. Each employee only begins working on a bag when a customer order has been received, and then she makes the bag from beginning to end. The average production time of a bag is 1.8 days, with a standard deviation of 2.7 days. Larry expects to receive one customer order per day on average. The interarrival times of orders have a coefficient of variation of 1.

What is the expected duration, in days, between when an order is received and when production 148

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begins on the bag?

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Answer: 12.6 Feedback: Given interarrival time (a = 1 day), production time (p = 1.8 days), number of servers (m = 2 employees), coefficients of variation of interarrival time (1 day), and standard deviation of production time (2.7 days, coefficient of variation = 1.5), compute time in queue = 12.55. Number of projects waited in queue = Time in queue/a = 12.55/1 = 12.55. Number of projects processed = p/a = 1.8/1 = 1.8. Total projects waited and being processed = 12.55 + 1.8 = 14.35 rounded to 14.

14. Find a Doctor is a small startup that helps people find a physician who best meets their needs (location, insurance accepted, etc.). During a ―slow‖ time for it, it has seven staff members taking calls from customers. On average, one call arrives every 6 minutes (with a standard deviation of 6). Each staff member spends 20 minutes with each customer (on average, with a standard deviation of 30). What is the probability that one of its staff members is busy (as a percent)?

Answer: 48% Feedback: Given processing time (p = 20 min), interarrival time (a = 6 min), and number of servers (m = 7 staff members), the utilization equals 20 / (6 × 7) = 0.476 = 47.6%.

15. Using the information for Find a Doctor in the previous question, how long (in minutes) does one of their customers spend, on average, waiting on hold before he or she can start speaking to a representative?

Answer: 0.96 minute Feedback: Given interarrival time (a = 6 min), service time (p = 20 min), number of servers (m = 7 staff members), standard deviation of interarrival time (6 min, coefficient of variation = 1), and standard deviation of service time (30 min, coefficient of variation = 1.5), compute time in queue = 0.96 min.

16. The organizers of a conference in the Houston Convention Center are evaluating the possibility of setting up a computer area where attendees can check their e-mail on computers provided by the organization. There will be one common queue for all computers, and only one person uses a computer at a time. On average, there are 15 attendee arrivals per hour, and the average time a person spends on the computer is 10 minutes. To ensure that waiting times are not too long, the organizers want to ensure that the utilization of the computers doesn’t exceed 90 percent. At least how many computers do they need to have?

Answer: 3 Feedback: Utilization = p / (a x m), so 0.9 = (1/6) / (1/15 x m), and solve for m. m = 2.778, which rounds up to 3 computers.

17. A small call center normally has five employees answering calls while open. On average, five calls arrive every 4 minutes. Under normal operating conditions, each employee on 150

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average

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handles each call in 3.5 minutes. But today, one employee has the flu, so the center needs to operate with four employees during this time. The manager is nervous that it may be supplyconstrained, so she directs her employees to hurry up their processing of calls today. What is the maximum average processing time (in minutes) for each call that they need to achieve today so that they have a stable queue?

Answer: 3.2 Feedback: Minimum number of servers m = p / a. Given m = 4 and a = 0.8 min, maximum average processing time p = m × a = 4 × 0.8 = 3.2 minutes.

18. Experience the Tour de France (ETF) is a specialty travel agent. It arranges vacations for amateur cyclists who want to experience the Tour de France by riding through one or more stages in the race. It has four people who take calls from clients. Each call lasts 30 minutes, on average, with a standard deviation of 60 minutes. A call arrives, on average, every 20 minutes, with a standard deviation of 20. On average, how many minutes does a caller wait before talking to an agent?

Answer: 3.59 Feedback: Given interarrival time (a = 20 min), service time (p = 30 min), number of servers (m = 4 people), standard deviation of interarrival time (20 min, coefficient of variation = 1), and standard deviation of service time (60 min, coefficient of variation = 2), compute time in queue = 3.598 rounded to 3.60 minutes.

19. CloudRack provides web hosting services on its five servers. When a person requests a page from one of its hosted websites, the server must find the page and send it to the person’s browser.

These requests arrive at the rate of 400 per second. The coefficient of variation of the interarrival times is 2. The processing time for a server is 0.01 second with a coefficient of variation of 1. On average, how much time (in seconds) does a request take to be filled (i.e., include time waiting for a server and the actual processing by the server)?

Answer: 0.0244 Feedback: Given interarrival time (a = 1/400 sec), processing time (p = 0.01 sec), number of servers (m = 5), coefficient of variation of interarrival times = 2, and coefficient of variation of processing time = 1, compute time in queue = 0.0144 sec, then add processing time of 0.01 sec. Total time to complete request = 0.0244 seconds.

20. Huduko operates with a utilization of 30 percent. The interarrival time of jobs is 8 milliseconds (0.008 second) with a coefficient of variation of 1.5. On average, there are 20 jobs waiting in the queue to be served and 60 jobs in process (i.e., being processed by a server rather than waiting to be sent to a server for processing). How many servers does it have in this system?

Answer: 200

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Feedback: Given utilization = 0.3, interarrival time (a = 0.008 sec), and number of jobs in process (Ip = 60), the processing time p = Ip × a = 60 × 0.008 = 0.48. From Utilization = p / (a × m), m = p / (Utilization × a) = 0.48 / (0.008 × 0.3) = 200 servers.

21. A pediatric practice has five physicians. Historically, patients were assigned to one physician, and these patients always were treated by the same physician. For example, Alice’s patients always visited her (for ―well visits‖ and ―sick visits‖) and never interacted with the other physicians. However, the practice has decided to change how it sees patients. Now patients will be seen by whoever is available (i.e., not treating patients). Based on our discussion of queuing theory, which of the three outcomes is/are likely to occur due to this change?

Answer: H. None of them (neither A, B, nor C)

CASE Potty Parity In the Potty Parity case, students: •

•

Connect real world behavior with a service system (e.g., a public restroom) to the specific components in the queuing equation for average waiting time. In other words, students see how a complex looking equation can be used to better understand a process that they all have experience with.

Discuss how to improve a service process either with government regulation or without. For example, does ―pooling‖ help in this situation and would it be an acceptable solution for society?

Everyone can relate to the Potty Parity case. The point of the case is to explore why queues in the women’s restrooms tend to be longer than for men and then to discuss process improvement options. There is no math with this case—only intuition that is motivated and guided by the queuing formula discussed in the chapter.

Q.1 The first question asks why the queues in women’s rooms tend to be longer than for men’s rooms even if they are considered ―equal‖ in terms of physical area. One reason is given in the case—there may be more flushing units in a men’s room. With more ―servers,‖ it is clear that, all else being equal, the queue length for men may be lower. But all else is not equal. Students will offer many theories for the differences. The goal of this discussion is to connect those theories to the waiting time queuing formula:

𝑝 𝑈𝑡𝑖𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛√2(𝑚+1)−1 𝑇 =( )𝑥( 𝑞 𝑚 1 − 𝑈𝑡𝑖𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛

153

) 𝑥 (

𝐶𝑉2 + 𝐶𝑉2

𝑝)

𝑎

2

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

For example, somebody might say that women take longer to ―do their thing‖ than men. That would mean that women have a longer ―p.‖ (A ―bad‖ joke that nevertheless works without offending.) Others might say something like ―women go in groups.‖ Assuming that is so, that would mean that there are batches in their arrival process, which makes the arrival process more variable—a higher CVa. This usually generates the suggestion that women have a more variable processing time—higher CVp. Finally, some may argue that women go to the restroom more frequently. That could mean that their inter-arrival time, a, is shorter. And that would mean a higher utilization, all else equal. Hence, with each of the variables in the Tq formula, it is possible to tell a story that would suggest that the parameter for women is different than the parameter for men in a way that makes Tq greater.

During this discussion somebody often says something that elicits a chuckle or a groan (or both) from others in the classroom. (It is usually a man saying something that the women in the class disagree with.) In the many years of teaching this case, I have not encountered a situation that seemed to ―go too far,‖ in the sense of creating offense or ill will. This may be because there is little debate that women in general suffer through longer queues. The interesting part of the case is to be able to related potential reasons why back to a formula.

Q.2 The second question tests students’ intuition. If the processing time for women were twice as long as for men, would it be sufficient to have twice as many flushing units in the women’s room as the men’s room to equate waiting times? The answer is ―no.‖ Why? Looking at the Tq equation, there are three factors. The first is the ratio of p to m, or the ―capacity factor.‖ That would not change if both p and m are doubled. The last is the ―variability factor,‖ which includes the coefficients of variation. Based on the earlier discussion, it is likely that this factor is higher for women. Furthermore, it is independent of m, so doubling m provides no help to women here.

The middle factor is the ―utilization factor.‖ When p and m both double, utilization stays the same. But while utilization stays the same, the factor also includes m in the exponent. Thus, a doubling of m actually reduces the utilization factor. In other words, all else being equal, doubling p and m actually reduces waiting time. So in this sense women could be better off than men if their processing times were double men’s but they were given twice as many flushing units. In the end, we cannot say for certain if doubling the capacity for women is too little or too much to equate waiting times. If there are big differences in the variability of their arrival and service processes, then it won’t be enough. But if the two coefficients of variation are similar, then doubling capacity for women would be more than enough.

Q.3 The final question asks what else could be done to reduce waiting times. Several ideas may be suggested. An obvious one is to pool capacity, as in unisex restrooms! However, there may be cultural constraints related to this operational improvement. A second idea would be to use flexible capacity. For example, a convention center might have a flexible partition between the two rooms that can be adjusted depending on the event. If the event favors more women, the partition is moved to add capacity to their room. The partition moves in the other direction if the event favors men.

CHAPTER 17: 154

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SERVICE SYSTEMS WITH IMPATIENT CUSTOMERS CONCEPTUAL QUESTIONS 1. If capacity exceeds demand, then all flow units are served. True or false?

Answer: B. False.

Feedback: Some demand may be lost even though capacity exceeds demand.

2. The denial of service probability decreases if more servers are added and nothing else changes. True or false?

Answer: A. True.

Feedback: With more servers, the system is full less often, which means less demand is lost and a lower probability that any one flow unit is lost.

3. In the Erlang loss model, which of the following most likely increases lost demand?

Answer: B. An increase in the average processing time lowers capacity, which increases lost demand.

4. In the Erlang loss model, which best describes the relationship between implied utilization and utilization?

Answer: A. Implied utilization is always greater than utilization.

Feedback: Due to lost demand, utilization is always less than implied utilization.

5. In the Erlang loss model, which of the following is a benefit of reducing the variability of processing times?

Answer: F. None of the above.

Feedback: The variability of processing times has no impact on the Erlang-loss model performance measures.

6. In the Erlang loss model, system A has 5 servers and an implied utilization of 125 percent, while system B has 15 servers and an implied utilization of 125 percent. Which of the following statements is true?

Answer: C. System B has higher utilization because the fraction of lost demand is lower in that system.

Feedback: Given equal implied utilizations, system B has a lower probability of having all servers occupied. Hence, system B loses a smaller fraction of its demand, which means it has a higher flow rate relative to its demand (i.e., a higher utilization).

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7. In the Erlang loss model, pooling is an effective strategy because it always increases implied utilization. True or false?

Answer: B. False.

Feedback: Pooling systems with the same implied utilization results in no change in the implied utilization of the pooled system.

8. Which of the following most directly expresses the motivation behind the expression ―buffer or suffer‖?

Answer: D. If there is variability in the arrival process or in processing times, make sure there is sufficient inventory between stages, otherwise capacity will be reduced.

Feedback: Inventory buffers must be added between stages to prevent (or at least mitigate) blocking and starving, which lead to lower capacity (the ―suffer‖ part).

9. At the Millbrook High School cafeteria, students proceed along a series of stations in a single line: (i) get tray and utensils, (ii) choose food, (iii) select beverage, (iv) pay. The school is concerned that students are taking too long to get their meal. The school has analyzed the capacities of each of the four steps in isolation and found there exists sufficient capacity at each resource in isolation. Which of the following is most likely to be causing the congestion?

Answer: C. This is a sequential process, so blocking and starving is possible.

PROBLEMS AND APPLICATIONS 1.

(a) What is the offered load?

Answer: 2 ( ) Feedback: Offered load(𝑟) = 𝑝 𝑎𝑣𝑒𝑟𝑎g𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠𝑖𝑛g 𝑡𝑖𝑚𝑒

𝑎(𝑎𝑣𝑒𝑟𝑎g𝑒 𝑖𝑛𝑡𝑒𝑟𝑎𝑟𝑟𝑖𝑣𝑎𝑙 𝑡𝑖𝑚𝑒)

=

4(𝑚𝑖𝑛) 60(𝑚𝑖𝑛)×30(𝑐𝑎𝑟𝑠)

=2

(b) What is the implied utilization?

Answer: 0.67 Feedback: Implied utilization = 4(𝑚𝑖𝑛) 2(𝑚𝑖𝑛)×3(𝑝𝑢𝑚𝑝𝑠)

𝑝(𝑎𝑣𝑒𝑟𝑎g𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠𝑖𝑛g 𝑡𝑖𝑚𝑒) 𝑎( 𝑎𝑣𝑒𝑟𝑎g 𝑒 𝑖𝑛𝑡𝑒𝑟𝑎𝑟𝑟𝑖𝑣𝑎𝑙 𝑡𝑖𝑚𝑒)×𝑚(𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑒𝑟𝑣𝑖𝑐𝑒)

= 0.67

(c) What is the capacity of the gas station (cars per hour)?

Answer: 45 ( ) Feedback: Capacity = 𝑚 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑒𝑠𝑜𝑢𝑟𝑐𝑒𝑠

𝑝(𝑎𝑣𝑒𝑟𝑎g𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠𝑖𝑛g 𝑡𝑖𝑚𝑒)

156

=

3(𝑝𝑢𝑚𝑝𝑠)×60(𝑚𝑖𝑛) 4(𝑚𝑖𝑛)

= 45

=

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

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(d) What is the probability that all three pumps are being used by vehicles?

Answer: 0.2105 Feedback: From Table 17A.1 for r = 2 and m(pumps) = 3, Pm0.2105 (e) How many customers are served every hour?

Answer: 18.7 1 Feedback: Flowrate = (1 − 𝑃 ) = 𝑚

𝑎

1

(1 − 0.2105) = 30 × 0.7895 = 23.7

1/30(ℎ𝑜𝑢𝑟)

(f) What is the utilization of the pumps?

Answer: 0.53 Flowrate 23.7 Feedback: 𝑈𝑡𝑖𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛 = = = 0.53 Capacity

45

(g) How many pumps should it have to ensure that it captures at least 98 percent of the demand that drives by the station?

Answer: 6 Feedback: From Table 17A.1 for r = 2 and 1 – Pm = 1 – 0.98 = 0.02, the number of pumps needed lies between 5 and 6. Rounding up, the answer is 6 pumps.

2.

(a) What is the offered load?

Answer: 1.67 p(average processing time) 5(𝑚𝑖𝑛) Feedback: Offered load(𝑟) = = = 1.67 a(average interarrival time)

3(𝑚𝑖𝑛)

(b) What is the implied utilization?

Answer: 0.42 Feedback: Implied utilization = 5(𝑚𝑖𝑛) 3(𝑚𝑖𝑛)×4(𝑜𝑝𝑒𝑟𝑎𝑡𝑜𝑟𝑠)

p(average processing time) a(average interarrival time)×𝑚(𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑒𝑟𝑣𝑒𝑟𝑠)

= 0.42

158

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(c) What is the capacity of this call center (calls per hour)?

Answer: 48 Feedback: Capacity = 𝑚(𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑒𝑠𝑜𝑢𝑟𝑐𝑒𝑠)

=

𝑝(𝑎𝑣𝑒𝑟𝑎g𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠𝑖𝑛g 𝑡𝑖𝑚𝑒)

4(𝑜𝑝𝑒𝑟𝑎𝑡𝑜𝑟𝑠)×60(𝑚𝑖𝑛) 5(𝑚𝑖𝑛)

= 48

(d) What is the probability that all four operators are talking to customers?

Answer: 0.0627 Feedback: From Table 17A.1 for r = 1.67 and m (operators) = 4, Pm = 0.0627 (e) How many customers are served every hour?

Answer: 18.8 1 Feedback: Flowrate = (1 − 𝑃 ) = 𝑚

𝑎

1

(1 − 0.0627) = 20 × 0.9373 = 18.7

1/20(ℎ𝑜𝑢𝑟)

(f) What is the utilization of the operators?

Answer: 0.39 Flowrate 18.7 Feedback: 𝑈𝑡𝑖𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛 = = = 0.39 Capacity

48

(g) How many operators should it have to ensure that it serves at least 97 percent of demand?

Answer: 5 Feedback: From Table 17A.1 for r = 1.67 and 1 – Pm = 1 – 0.97 = 0.03, the number of operators needed lies between 4 and 5. Rounding up the answer is 5 operators.

3.

(a) What is the offered load?

Answer: 6 Feedback: Offered load(𝑟) = p(average processing time) = 3(ℎ𝑜𝑢𝑟𝑠) = 6 a(average interarrival time)

0.5(ℎ𝑜𝑢𝑟𝑠)

(b) What is the implied utilization of LARA’s reporters?

Answer: 1.2 Feedback: Implied utilization = 3(ℎ𝑜𝑢𝑟𝑠) 0.5(ℎ𝑜𝑢𝑟𝑠)×5(𝑟𝑒𝑝𝑜𝑟𝑡𝑒𝑟𝑠)

p(average processing time) a(average interarrival time)×𝑚(𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑒𝑟𝑣𝑒𝑟𝑠)

= 1.2

(c) What is their capacity to complete assignments (in assignments per hour)?

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=

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Answer: 1.67 ( ) Feedback: Capacity = 𝑚 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑒𝑝𝑜𝑟𝑡𝑒𝑟𝑠

𝑝(𝑎𝑣𝑒𝑟𝑎g𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠𝑖𝑛g 𝑡𝑖𝑚𝑒)

= 5(𝑟𝑒𝑝𝑜𝑟𝑡𝑒𝑟𝑠) = 1.67 3(ℎ𝑜𝑢𝑟𝑠)

(d) What is the probability that an incoming text message reporting a viewing cannot be pursued?

Answer: 0.36 Feedback: From Table 17A.1 for r = 6 and m (reporters) = 5, Pm0.3604 (e) What is the flow rate of completed assignments (assignments per hour)?

Answer: 1.28 1 Feedback: Flowrate = (1 − 𝑃 ) = 𝑚

𝑎

1

1

(1 − 0.3604) = ___ × 0.6396 = 1.28

0.5(ℎ𝑜𝑢𝑟𝑠)

0.5

(f) What is the utilization of LARA’s reporters?

Answer: 0.77 Flowrate 1.28 Feedback: Utilization = = = 0.77 Capacity

1.67

(g) How many reporters does LARA need to have on its staff to ensure that it takes advantage of at least 85 percent of the text messages that it receives?

Answer: 8 Feedback: From Table 17A.2 for r = 6 and 1 - Pm = 1 – 0.85 = 0.15, the number of reporters needed lies between 7 and 8. Rounding up the answer is 8 reporters.

4.

(a) What is the offered load?

Answer: 7.5 )×60 Feedback: Offered load(𝑟) = p(average processing time) = 5(ℎ𝑜𝑢𝑟𝑠 = 7.5 a(average interarrival time)

40(𝑚𝑖𝑛𝑢𝑡𝑒𝑠)

(b) What is the implied utilization?

Answer: 0.94 Feedback: Implied utilization = 5(ℎ𝑜𝑢𝑟𝑠)×60 40(𝑚𝑖𝑛)×8(𝑐𝑎𝑟𝑠)

p(average processing time) a(average interarrival time)×𝑚(𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑒𝑟𝑣𝑒𝑟𝑠)

= 0.94

(c) What is the capacity of the process (rentals per hour)?

Answer: 1.6 160

=

Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

(

)

= 8(𝑐𝑎𝑟𝑠) = 1.6 5(ℎ𝑜𝑢𝑟𝑠) 𝑝(𝑎𝑣𝑒𝑟𝑎g𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠𝑖𝑛g 𝑡𝑖𝑚𝑒)

Feedback: Capacity = 𝑚 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑒𝑝𝑜𝑟𝑡𝑒𝑟𝑠

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(d) What is the probability that all eight cars are rented at the same time?

Answer: 0.207 Feedback: From Table 17A.2 for r = 7.5 and m (cars) = 8, Pm = 0.2075 (e) How many customers are served every hour?

Answer: 1.19 1 Feedback: Flowrate = (1 − 𝑃 ) = 𝑚

𝑎

1

(1 − 0.2075) =

40/60(ℎ𝑜𝑢𝑟𝑠)

1

× 0.7925 =

40/60

1.19 (f) What is the utilization of the cars?

Answer: 0.744 Flowrate 1.19 Feedback: Utilization = = = 0.744 Capacity

1.6

(g) How many cars should it have in this lot to ensure that it serves at least 90 percent of demand?

Answer: 10 Feedback: From Table 17A.2 for r = 7.5 and 1 - Pm = 1 – 0.90 = 0.10, the number of cars needed lies between 9 and 10. Rounding up the answer is 10 cars.

5. The Los Angeles Reporters Associated (LARA) is considering a merger with its competitors

Fast Paparazzi of America (FPA). LARA receives two viewings per hour. FPA obtains one viewing per hour. What is the interarrival time (in minutes) for the combined firm, LARA-FPA? Answer: 20 minutes Feedback: 2 viewings per hour (LARA) + 1 viewing per hour; (FPA) = 3 viewings per hour. Interarrival time = 60 (min)/3 = 20 minutes.

6. RideShare is considering combining two lots. One receives a rental request every 45 minutes,

while the other receives a rental request every 90 minutes. What would be the interarrival time (in minutes) of the combined lot? (Assume total demand does not change.)

Answer: 30 minutes Feedback: Lot 1: 4 rental requests per hour; Lot 2: 2 rental requests per hour.

3

Combined lot: 4 + 3

2 3

3

= 2 rental requests per hour. Interarrival time =

60(min)/2 = 30 minutes.

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CASE Bike Sharing Bike sharing is a relatively new and exciting innovation. Through these programs, city residents are given another environmentally friendly option for transportation. However, it is a challenge to balance capacity with demand.

This is a qualitative case (no calculations) but the discussion is informed by the model/math in the chapter.

Students in this case: •

Identify several of the many system inhibitors of good operations that exist in a bike-sharing program.

Discuss options for improving the operations of a bike-sharing program.

Q1. What are some examples of system inhibitors—variability, waste, and inflexibility—present in a bike-sharing program?

There is plenty of variability to discuss. On the demand side, there is seasonal variation – people prefer to bike when it is warmer and there is more daylight. There is also variability during the week and variability due to weather. On the supply side, variability occurs with availability of bicycles, either because they are at the wrong station or because they don’t function (e.g., a flat tire) Waste occurs when a resource’s time is not used to its potential or a flow unit’s time is not effectively used. With bike sharing waste occurs when bikes are idle, or riders can’t find a bike, or riders cannot return a bike to a docking station (because all of the stations are occupied).

Inflexibility refers to resources that cannot adjust to serve the actual demand they receive. For example, some bikes might be idle at one station while they are needed at another (and cannot be moved from one station to the other very quickly). Or too many bikes are demanded on a pleasant summer day, without the flexibility to add more bikes to the system.

In sum, variability, waste and inflexibility are all present in a bike sharing program.

Q2. Is there evidence of blocking and starving in the bike-sharing program?

Definitely! Let’s take a bike to be a resource. A bike can be blocked if it cannot be returned to a station because all of the docks are occupied. A bike is starved whenever there isn’t a rider who wants to use it. If instead you take the rider as a resource, then a rider can be blocked if there are no available docks to return a bike at a station. The rider can also be starved if there are no available bikes at a station.

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Q3. What are some of the options to improve the quality of the service? What are the pros and cons of each option from various perspectives (e.g., the cost of the program, the quality of the service provided, etc.)?

Many suggestions could be offered. In general, a bike sharing program consists of riders, bikes and docking stations.

One problem is an overall lack of bikes or docks in the system. For example, there is much higher demand in the summer than the winter. If the number of bikes in the system matches average demand from throughout the year, then there will be too many in the winter and too few in the summer. If the system purchases enough bikes to satisfy peak summer demand, then the bikes will be considerably idle in the winter, which is costly. If the number of bikes only satisfies winter demand, or even average demand, then their will not be enough bikes for the peak summer months. That is costly too. The same applies to docks.

Beyond the issue of the total number of bikes, there is the issue of too many bikes or docks in one place and too few in another. This can be as big a problem, if not bigger, than the problem of the total number of bikes. As we see in the case, bikes move from residential areas to work areas throughout the day. Consequently, in the late morning there are too few bikes in the residential areas and too many in the work areas. Let’s call this the ―wrong place‖ problem.

One solution to the wrong place problem is to use pricing. For example, charge riders extra to return bikes to certain stations at certain times. This would discourage too many bikes from arriving to a particular zone. However, this pricing scheme is complex (riders need to figure out what the fee will be for their destination) and therefore unlikely to be popular.

A second solution is to build more docks in the locations that tend to have the largest swings in demand throughout the day. While this will help a downtown area absorb more bikes in the morning, it does require that additional real estate is given to the docking stations. And real estate in cities isn’t cheap.

A third solution to the wrong place problem is to have non-riders move bikes. For example, in the NY City system, they have hired workers to take bikes from the downtown work area in the morning (where there are too many bikes) and return them to the midtown residential areas (where the bikes are depleted in the morning). This rebalancing of capacity is costly (you have to pay the worker), but it might be cheaper than buying more bikes and building more docks.

Finally, some students may wonder about the number of stations and the location of docking stations. In general, if more docking stations are added, the average distance between stations decreases. This is desirable from the point of view of customers. However, more docking stations means each serves a smaller amount of demand – if we take total demand as somewhat fixed, doubling the number of stations is likely to cut the average demand per station in half. As these docking stations act like queuing systems, as discussed in the chapter, adding more stations is like the reverse of pooling. Thus, we can expect to have more blocking and starving if the 164

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number of stations increases, and that isn’t desirable for consumers. In short, the average size and the location of the stations is a key decision for the effective operation of a bike sharing programs.

CHAPTER 18: SCHEDULING TO PRIORITIZE DEMAND CONCEPTUAL QUESTIONS 1. Which of the following performance measures depend(s) on how work is sequenced at a resource?

Answer: E. Inventory and Flow Time 2. Suppose there is a set of jobs to be processed at a resource. Shortest processing time always sequences the jobs in a different order than first-come-first-served. True or false?

Answer: B. False.

Feedback: If jobs arrive in ascending order of their processing time, then SPT sequences the jobs in the same order as FCFS. 3. Shortest processing time minimizes the flow time of jobs at a resource because it reduces the processing times of the jobs. True or false?

Answer: B. False.

Feedback: Shortest Processing Time does not change the processing times of the jobs.

4. Depending on how jobs arrive to a resource, it is possible that the average flow time of the jobs with first-come-first-served is lower than that with shortest processing time. True or false?

Answer: B. False.

Feedback: SPT always minimizes the average flow time of the jobs at a resource.

5. The flow time of every job is lower with shortest processing time than with first-come-firstserved. True or false?

Answer: B. False.

Feedback: SPT minimizes the average flow time, but a particular job can have a longer flow time with SPT than with FCFS, especially if the job has a long processing time and arrived early.

6. The difference in the average flow time between shortest processing time and first-come-firstserved is greatest when: 165

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Answer: C. there are many jobs waiting to be processed at the resource.

Feedback: If all jobs have the same processing time, then there is no difference between SPT and FCFS. If utilization is low, then there will be few jobs waiting for processing. If jobs arrive in increasing order of their processing time, then FCFS sequences the jobs in the same order as SPT.

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7. With weighted shortest processing time, jobs are scheduled at a resource in what order?

Answer: D. In decreasing order of the ratio of weight to processing time.

Feedback: With Weighted Shortest Processing Time, the jobs are sequenced in increasing order of their ratio of their weight to their processing time – jobs with high weight and low processing times are processed early.

8. Which of the following statements best explains why an organization would use weighted shortest processing time instead of shortest processing time?

Answer: A. Jobs differ in how important it is for them to be completed quickly. Feedback: WSPT does not minimize the average flow time of jobs (SPT does), it isn’t simple, and it does not minimize the average number of jobs in the system (SPT does).

9. Jobs have different processing times and due dates. If a job is finished earlier than its due date but 1 day sooner than expected, how does this change the job’s lateness?

Answer: A. Lateness decreases by 1 day.

Feedback: Lateness is the difference between completion time and the due date, so it would decrease by one day if the completion time is one day sooner.

10. Jobs have different processing times and due dates. If a job is finished earlier than its due date but 1 day sooner than expected, how does this change the job’s tardiness?

Answer: B. Tardiness does not change.

Feedback: Tardiness (lateness) is the difference between a job’s completion time and its due date if the job is completed past its due date, otherwise tardiness is zero. As the job is finished earlier than its due date in either case, the job’s tardiness does not change (it is zero in each case).

11. The earliest-due-date rule always minimizes the average lateness across jobs. True or false?

Answer: B. False.

Feedback: Sometimes SPT has a smaller average lateness than EDD. 12. The earliest-due-date rule always minimizes the maximum tardiness across jobs. True or false?

Answer: A. True 13. Why does the theory of constraints advocate that a manager’s attention be directed to the bottleneck of a process?

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Answer: B. The bottleneck has considerable influence on the average flow rate of a process.

Feedback: The bottleneck should have a high utilization and an ample buffer of inventory in front of it to ensure that it is not starved of work. The Theory of Constraints is not concerned with fairness across jobs, so FCFS at the bottleneck is not important (and probably not recommended).

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14. The theory of constraints is most applicable in which of the following processes?

Answer: B. A chemical processing plant with numerous pieces of equipment with significant setup times and a large number of products, each requiring different sequences of resources to be completed.

Feedback: The Theory of Constraints applies to complex processes where a bottleneck resource can be identified. A well-balanced assembly line does not have a single bottleneck.

15. If an appointment system evenly spaces out appointments throughout the day, then patients arriving during the second half of the day can expect to wait the same amount of time as patients arriving during the first half of the day. True or false?

Answer: B. False.

Feedback: With evenly spaced out appointments, waiting times tend to increase throughout the day 16. Which of the following is a benefit of overbooking a resource?

Answer: C. It helps to increase the utilization of the resource.

Feedback: Overbooking tends to increase inventory of jobs and therefore the flow time of jobs. It has no impact on actual processing times.

PROBLEMS AND APPLICATIONS 1. An architect has five projects to complete. She estimates that three of the projects will require 1 week to complete, and the other two will require 3 weeks and 4 weeks. During the time these projects are completed (and assuming no other projects are worked on during this time), what is the architect’s flow rate (projects per week)?

Answer: 0.5 Feedback: Five projects will be completed in 3 × 1 + 3 + 4 = 10 weeks.

Five projects in 10 weeks yields an average flow rate of 5/10 = 0.5 projects per week.

2. Given the projects displayed in Table 13.2, if FCFS is used to sequence the jobs, what is the average flow rate of the projects (in projects per hour)?

Answer: 0.36 Feedback: Average flowrate (FCFS) =

8 4+1+2+3+6+5+1+0.5

169

=

8 = 0.36 projects per hour 22.5

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3. Given the projects displayed in Table 18.2, if FCFS is used to sequence the jobs, what is the average flow time of the projects (in hours)?

Answer: 13.4 Feedback: Processing order (FCFS) P1, P2, P3, P4, P5, P6, P7, P8.

Average flowrate (FCFS) =

4 + 5 + 7 + 10 + 16 + 21 + 22 + 22.5 8

170

=

107.5

= 13.4 hours 8

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4. Given the projects displayed in Table 18.21, if FCFS is used to sequence the jobs, what is the average inventory of the projects (in projects)?

Answer: 4.8 Feedback: Inventory (FCFS) = Flow rate × Flow time = 0.36 × 13.4 = 4.8 projects 5. Given the projects displayed in Table 18.2, if SPT is used to sequence the jobs, what is the average flow rate of the projects (in projects per hour)?

Answer: 0.36 Feedback: Average flowrate (FSPT) =

8 4+1+2+3+6+5+1+0.5

=

8 = 0.36 projects per hour 22.5

6. Given the projects displayed in Table 18.21, if SPT is used to sequence the jobs, what is the average flow time of the projects (in hours)?

Answer: 8.4 Feedback: Processing order (SPT): P8, P2, P7, P3, P4, P1, P6, P5.

Average flowrate (SPT) = 0.5+1.5+2.5+4.5+7.5+11.5+6.5+22.5 = 67.0 = 8

8.4 hours

8

7. Given the projects displayed in Table 18.21, if SPT is used to sequence the jobs, what is the average inventory of the projects (in projects)?

Answer: 3.0 Feedback: Inventory (SPT) = Flow rate × Flow time = 0.36 × 8.4 = 3.0 projects 8. Assume the jobs displayed in Table 18.22 need to be processed on a single resource and are sequenced with the WSPT rule. Which job would be processed second?

Answer: C. Job C Feedback: Ratios of weight to processing time are: A – 1.5; B – 0.25; C – 0.625; D – 0.167; E – 0.4.

Sequencing jobs in descending order of those ratios leads to conclusion: job C will be processed second.

9. Assume the jobs displayed in Table 18.22 need to be processed on a single resource and are sequenced with the WSPT rule. What is the average flow rate of the jobs (in jobs per minute)?

Answer: 1/3 Feedback: Average flowrate =

5

= 171

5

1+4+2+3+5

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= 0.3333 jobs per minute

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10. Assume the jobs displayed in Table 18.22 need to be processed on a single resource and are sequenced with the WSPT rule. What is the average flow time of the jobs (in minutes)?

Answer: 7.8 Feedback: Processing order (WSPT): A, C, E, B, D 1+3+8+12+15 39 Average flowrate = = = 7.8 minutes 5

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11. Assume the jobs displayed in Table 18.22 need to be processed on a single resource and are sequenced with the WSPT rule. What is the average inventory of the jobs?

Answer: 2.6 Feedback: Inventory = Flow rate × Flow time = 0.3333 × 7.8 = 2.6 jobs 12. If the jobs displayed in Table 18.23 are processed using the earliest-due-date rule, what would be the lateness of job B? Answer: 2 Feedback: Processing order using earliest-due-date rule: E, A, C, D, B Flow time for job B = 8 weeks. The due date for job B = 6 weeks. The lateness for job B = 8 – 6 = 2 weeks.

13. If the jobs displayed in Table 18.23 are processed using the earliest-due-date rule, what would be the tardiness of job E? Answer: 0 Feedback: Processing order using earliest-due-date rule: E, A, C, D, B Job E is done first and completed before the due date. The tardiness for job E = 0 weeks.

14.

If the jobs displayed in Table 18.23 are processed using the earliest-due-date rule, what is the maximum tardiness?

Answer: 5 Feedback: Processing order using earliest-due-date rule: E, A, C, D, B The tardiness for each of these five jobs equals to: E – 0; A – 0.5; C – 0.5; D – 2.5; B – 2.

The maximum tardiness is 2.5 weeks.

15. If the jobs displayed in Table 18.23 are processed using the earliest-due-date rule, what is the average flow time?

Answer: 2.5 Feedback: Processing order using earliest-due-date rule: E, A, C, D, B 1.5+3.5+4.5+7.5+8 25 Average flow time = = = 5 weeks 5

5

CASE Disney Fastpass The Disney Fastpass case allows students to: • •

Identify how Fastpass changes the performance measures of attractions in the park Understand the relationship between Fastpass and an appointment system

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•

Discuss fairness and other implementation issues related to Fastpass.

This case explores an interesting scheduling innovation by Disney, their Fastpass system. Many students may even be familiar with Fastpass, especially if they visited a Disney park recently. In short, Fastpass allows guests to wait in a virtual queue rather than a physical queue. For example, when a guest requests a Fastpass for an attraction, they are given a time that approximately equals the time they would enter the attraction if they waited in the current queue (which includes guests who are physically waiting as well as guests who have already been assigned a Fastpass time).

Although it might not seem so at first, Fastpass is an appointment system. Consequently, the issues discussed in the chapter regarding appointment systems apply here.

To make Fastpass work well, Disney needs to estimate the number of guests that are waiting and the time to process them. For Disney attractions, processing times are relatively easy to predict.

But the number of guests waiting is not as easy because it can be challenging to count the people entering the queue. And the real challenge is predicting no-shows – customers that are given a Fastpass time but nevertheless do not show up for the attraction. If Disney doesn’t do a good job with these predictions, then either Fastpass guests might have to wait more than desired (given that the program is billed as a way to avoid waiting time) or the capacity of attraction might be wasted (poor utilization).

There are no calculations for this case. Instead, the case requires students to work through a conceptually challenging system.

Q1. Keeping the number of guests in the park constant, by adding Fastpass, does the average number of guests served per attraction change? For which types of attractions does it change the most (e.g., the busiest or the least busy ones)?

This question requires students to think through how Fastpass could influence the utilization of attractions in the park. It might be tempting to say that Fastpass wouldn’t change the number of guests served per attraction because the number of guests remains the same (and clearly the number of attractions remains the same). However, that would be true only if all of the attraction have 100% utilization, which isn’t likely.

Say there are two attractions, one named ―Pop‖ and the other ―OK‖. Attraction Pop is very popular and attraction ―OK‖ is liked, but not as much as Pop. With FCFS, Pop has 100% utilization and a long queue, while OK has less than 100% utilization and a short queue. So guests do essentially three things in the park – wait in line for the Pop, experience Pop, and experience OK. The time waiting for Pop is time that can’t be used to experience OK even though guests would rather experience OK than spending time in line for Pop. Now suppose Fastpass is used. Guests no longer need to spend as much time waiting for Pop. That time can be used for other activities, such as experiencing OK. Thus, the introduction of Fastpass, by reducing time in lines, increases the time guests can spend experiencing attractions.

This cannot increase the number of guests that experience the most popular attractions, because they were already 100% utilized. But it can increase the number of guests that experience the 175

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less popular attractions. Thus, overall, the number of guests served per attraction should increase, and the increase should be greatest for attractions that would not be fully utilized with FCFS. To get these ideas across to students, it might help to draw a simple diagram with two boxes representing attractions and one queue in front of one of the boxes. With FCFS all of the people are in one of those three places. But with Fastpass, some of the people that were in the queue, now can be at the OK attraction.

Q2. Disney needs to decide how many Fastpass tickets to issue for each time window. What are some of the trade-offs associated with this decision? For example, what are good and bad issues about increasing the number of tickets available?

As the previous discussion reveals, the more Fastpass tickets issued, the less time guests spend on physical lines. That allows them to do other things in the park that increase the quality of their experience (other attractions, resting, shopping, etc.) However, there is a downside to Fastpass tickets. Fastpass is like an appointment system. And with appointment systems, there can be ―no- shows‖. If only Fastpass tickets were issued, then it is possible that a popular attraction might lose some capacity due to no-shows. That would not be beneficial. Hence, while 0% Fastpass tickets (i.e., just FCFS) is not best, 100% Fastpass tickets is probably not ideal either.

Q3. Is Fastpass fair?

This is an open-ended question that raises the issue of what is fair in a service system. Most people believe FCFS is fair because everyone is perceived to have equal access. This is often the case, but not always. For example, suppose a university were to assign seats in courses starting at 8am but athletes are required to be at practice until 9am. A FCFS system might not be perceived to be fair in that case because the athletes don’t have equal access.

Although fairness is desirable, it shouldn’t be confused with total welfare. For example, imagine everyone working for a company doing the same job is paid $15 per hour. This might be considered fair because there is equal pay for equal work. However, now suppose some are paid $15 per hour while others are paid $20. Total welfare is higher in this case because no one is worse off, but perceived fairness might be lower.

Let’s return to the issue of fairness and Fastpass. It seems clear that Fastpass can only increase total value because less time is spent on a relatively undesirable task (waiting). If all guests have equal access to Fastpass, then it might be perceived to be a fair system. While in theory Fastpass might offer equal access, it might not achieve that perception in practice. For example, some guests might not trust that Fastpass actually works or is beneficial. If they choose to wait in the physical line, they might resent some guests ―cutting in line‖, thereby experiencing the attraction without the cost of waiting. Or some guests might not be aware of Fastpass. To them, they merely see a violation of the cultural norm of FCFS. Or, some guests might not be physically able to enjoy the full benefits of Fastpass. For example, say a guest requires the use of a wheelchair for mobility. That guest might not be able to enjoy as much of an advantage from the virtual wait via Fastpass because they are not able to easily move between attractions.

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Finally, while it may be apparent after reading this chapter and careful thought that Fastpass can enhance total welfare (by allowing guests to experience more attractions or to spend their time on other activities that are more desirable than waiting in line), this benefit might not be obvious to many guests. They may have the belief that guests who use Fastpass get something desirable (the use of a popular attraction) without incurring a cost (waiting in line). That feeling that other received ―something for nothing‖ may leave some guests jealous and annoyed.

Given the potential concerns with perceptions of fairness, it seems prudent that Disney should (a) educate as many guests as possible as to the benefits of Fastpass and (b) attempt to have as many guests as possible take advantage of the program.

Q4. Should Disney charge for Fastpass?

Students might have different opinions on this. Some will say that Disney is providing value to guests, so they should charge for the service. Others might say that Disney already charges a park entrance fee, so a Fastpass fee would be like charging twice. Furthermore, if they charge, then they would create two types of customers – Fastpass customer and non-Fastpass customers.

That could raise further concerns regarding fairness.

Instead of charging for Fastpass explicitly, it might be possible to charge implicitly. To be specific, Disney could raise the park entrance fee. In effect, Disney would charge all guests for Fastpass without telling each guest how much they are paying for it. Economists refer to this as ―bundle‖ pricing. In fact, Disney is quite familiar with this approach – they long ago abandoned the practice of charging for each attraction separately, choosing instead to bundle all attractions into a single part entrance fee.

CHAPTER 19 PROJECT MANAGEMENT CONCEPTUAL QUESTIONS 1. A project has four activities that take 4, 3, 6, and 7 days, respectively. What is the total completion time for the project?

Answer: D. It is not possible to determine the total completion time from the given information.

Feedback: This really depends on the precedence relationships between the activities which determines the critical path which determines the total completion time for the project.

2. Which of the statements is correct?

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Answer: B. A1 is a predecessor activity to A2.

3. Consider a project with two activities that are interdependent. Which of the following statements is correct?

Answer: C. Dynamic coordination of the two activities implies that the two activities are carried out in an iterative manner.

4. Which of the following statements is correct with respect to an AON graph of a project?

Answer: B. In an AON graph, the activities are on the nodes of a graph.

5. In an AON graph of a project, there only exists one path from the first to the last activity. True or false?

Answer: B. False 6. The critical path in a project is the sequence of activities that: Answer: D. None of these statements is correct.

7. What is the relationship between the earliest start time (EST) of an activity and the earliest completion time (ECT) of the activity?

Answer: A. EST + Activity time = ECT 8. Which of the following statements is correct with respect to slack time?

Answer: D. All of these statements are correct.

9. What is the late start schedule for a project?

Answer: A. It is the latest start and the latest completion time of the project activities so that the overall project is still completed on time.

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10. Which of the following statements is correct with respect to a Gantt chart?

Answer: E. All of these statements are correct.

11. Consider two projects that have the same activities and the same dependencies. In the first project, the activity times are expected outcomes. The actual times will vary. In the second project, the activity times are always as expected. Assuming the expected activity times are identical across the two projects, which project will be completed first?

Answer: B. The second 12. A project manager is concerned about rework, specifically about having to iterate on two activities that are on the critical path. A consultant tells the project manager not to worry, because the effect of uncertain activity times is likely to dominate the effect of iteration. The project manager disagrees. Who is right?

Answer: A. The project manager 13. They key objectives of a project can be summarized in the form of a: Answer: B. triangle.

14. The key objectives of a project are given by: Answer: A. time, scope, and budget.

15. What is the effect of overlapping activities that are not on the critical path?

Answer: C. No effect 16. Which of the following descriptions best captures the project manager’s responsibilities?

Answer: A. Define the project, plan the project, control the project.

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PROBLEMS AND APPLICATIONS 1. Which of the following statements is correct with respect to a dependency matrix?

Answer: D. All of these statements are correct.

2. Consider the AON graph of a project shown in Figure 19.17. Which of the following statements is correct?

Answer: D. All of these statements are correct.

3. Consider the AON graph of a project shown in Figure 19.18. The activity times are 3, 4, 2, 5, 7, 2, 2, and 3 days for activities A1 through A8, respectively. What is the earliest completion time of the project?

Answer: 19 days Feedback: The Critical Path is A1 (3 days), A2 (4 days), A5 (7 days), A7 (2 days), A8 (3 days) for a total of 19 days.

4. Consider the AON graph of a project shown in Figure 19.19. The activity times are 3, 4, 2, 5, 7, 2, 2, and 3 days for activities A1 through A8, respectively. What is the critical path?

Answer: C. A1, A2, A5, A7, A8 5. A project activity has an earliest completion time of 5, an activity time of 3, and a latest completion time of 6. What is the slack time for the activity?

Answer: A. 1 Feedback: Latest completion time – earliest completion time = 6 – 5 = 1.

6. What is the definition of the latest completion time (LCT) of a project activity?

Answer: B. LCT(Ai) = Min{LST(Aj) for which Aj is a successor of Ai} 7. Which of the following statements is correct with respect to describing a Gantt chart?

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Answer: A. In a Gantt chart, activities with longer times are captured by bars that are longer along the x-dimension of the chart.

Feedback: In a Gantt chart, activities with longer times are captured by bars that are longer along the x-dimension of the chart.

8. A project consists of three tasks: A, B, and C. Task A is known to take 5 days for sure. Tasks B and C are each dependent on A. B takes 5 days. C takes 7 days with a probability of 0.5 and 3 days with a probability of 0.5. What is the expected completion time of the project?

Answer: C. 11 days Feedback: Critical Path #1: A 5 days + C 7 days = 12 days Critical Path #2: A 5 days + B 5 days = 10 Days (Critical Path #1 12 days × 50%) + (Critical Path #2 10 days × 50%) = 11 days 9. Which of the following statements best defines the scope of a project?

Answer: A. What must be accomplished in order to complete the project 10. What happens as you increase the number of people working on a project from three to six?

Answer: F. The project might or might not be accelerated and the communication requirements for the project increase far more than by 2 times.

11. You manage a project with 10 activities. Activities A1, A3, A5, and A9 form the critical path.

Because you have a large budget for the project, you consider crashing activity A2, which has the potential to shorten the time of A2 by 3 days. What do you think about this opportunity?

Answer: B. It is a bad idea because A2 is not on the critical path.

12. Which of the following activities is not part of defining a project?

Answer: D. The critical path 13. You have been asked by the CEO of your company to manage a really important project. The CEO suggested five employees to be on your project team. You now think about the right organizational structure. Under what project management structure would these five employees face the least trade-offs with their current responsibilities?

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Answer: A. A dedicated project team

CASE Building a House in Three Hours Q.1 How is it possible to build a house so quickly? Or, asked differently, why are other residential construction projects taking 100 times longer?

A couple of factors contributed to the short completion time: • • • •

With 200 workers, the construction crew was much larger than for a typical residential construction project.

In typical construction projects, a lot of the time is lost waiting for contractors to show up. In this project, resources were waiting ―shovel ready‖ so that they could immediately start working when the predecessor task was completed.

Since the foundation was pre-built, a number of tasks could be carried out in parallel, especially after the house was framed (the roof can go on at the same time as the plumbing is done) The scope of the project was clear from the beginning

Q.2 What lessons can be learned from this record-time construction for other projects such as the construction of new airports, ocean liners, or campus buildings?

In any project, we face a trade-off between the productivity of the resources (the contractors) and the project completion time. As long as there exists some variability, it will not be possible to schedule the contractors such that they immediately show up when the predecessor activity is completed. The job was done quickly, yet most contractors spend a lot of their time idle.

The idea of working in parallel and having a well-defined scope apply to other projects as well.

CHAPTER 20 NEW PRODUCT DEVELOPMENT CONCEPTUAL QUESTIONS 1. Your friend Tony has many ideas. Five times a week, he calls you and explains to you a new device he invented. However, few of these devices address a user need. Are these ideas innovation?

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Solution Manual for Operation Management 3rd Edition by Gerard Cachon, Christian Terwiesch

2. BMW launches a new 3-series model every 4 to 6 years. Such launches are which type of innovation?

Answer: A. Horizon 1 innovations 3. Walter is vice president of product development at a big computer maker. At a strategic planning retreat, he talks with the vice president of purchasing and the vice president of production. The computer maker spends about 80 percent of its cost on purchasing, 10 percent on production, and only 2 percent on product development. Which of the following statements is correct?

Answer: B. The costs for production and purchasing are mostly determined during product development and that is why, to reduce costs, product development is critical.

4. Which of the following activities is not part of the product development process?

Answer: A. Strategic planning 5. A manager of a product development project reviews the results of the concept validation performed by the team. By showing some concepts to consumers, the team discovered an additional need and now wants to add it to the needs document. However, the manager argues that the identification of customer needs is long over and thus should not be changed.

Which statement is correct?

Answer: B. Product development is an iterative process and it is common that activities are repeated.

6. The Kano model classifies needs into three categories. Which of the following is not a category in the Kano model?

Answer: D. Important needs 7. Latent needs are typically associated with which category in the Kano model?

Answer: B. Delighters

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8. Which of the following statements about contextual observation is correct?

Answer: D. All of these statements are correct.

9. What is the difference between primary needs and secondary needs?

Answer: A. Primary needs are more important than secondary needs.

10. The product concept is concerned with: Answer: C. both the consumer needs and solution approach.

11. Consider two prototypes of a new race car. One is a two-dimensional sketch; the other one is a physical model that almost looks like the new race car, except it has no engine inside. Which of the two is a higher-fidelity prototype?

Answer: B. The physical model 12. Which of the following techniques is not a form of decomposing a product for the sake of generating product concepts?

Answer: C. Expert-based decomposition 13. A value curve shows what type of information for a product concept?

Answer: B. The degree by which various needs are fulfilled with the product concept.

14. Which of the following statements is correct as far as the definition of a service blueprint is concerned?

Answer: C. Both of these statements are correct.

15. A minimum viable product is: 185

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Answer: E. A and C 16. What is the purpose of building a minimum viable product?

Answer: C. To test if consumers want to buy the product 17. The purchase probability is the percentage of customers on the purchase intent survey who indicated they would want to buy the product. True or false?

Answer: B. False. The fraction of customers that on the purchase intent survey indicated that they would want to buy the product still ought to be adjusted for the fact that far fewer customers buy a product relative to those customers who say that they would buy.

18. To get to the sales forecast, we just multiply the market size by the purchase probability.

True or false?

Answer: B. False 19. Consider two products, A and B, both of which have the same purchase probability. Product A has a population size of 1000 and a 50 percent awareness probability. Product B has a population size of 10,000 and a 5 percent awareness probability. Which product do you expect to sell more?

Answer: C. They are expected to sell the same amount.

PROBLEMS AND APPLICATIONS 1. Which of the following activities is not part of the product development process?

Answer: E. New product strategy

2. Dishwasher 2020 is a new product under development at a large appliance company looking to develop the next generation of products in the dishwasher market. The team has done some 186

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market research and has listed a set of user needs. Water conservation/reducing energy consumption emerges as a need expressed by many consumers. The team debates how that need fits into the Kano model. In your opinion, which type of need is captured by the conservation of water?

Answer: A. Linear satisfier – the more water saved the better

3. A need related to a product triggers the following reaction by consumers. Consumers are happy if the need is fulfilled but not really dissatisfied if the need is not fulfilled. What type of need in the Kano model fits this description?

Answer: C. Delighter Feedback: Delighters are those needs that will lead to customer satisfaction when fulfilled, but will not lead to dissatisfaction when not fulfilled.

4. Which of the following statements is correct as far as a service blueprint is concerned?

Answer: B. A service blue print distinguishes between customer actions, onstage actions, backstage actions, and support processes

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5. Which of the following statements correctly captures the idea of a minimum viable product?

Answer: D. None of these statements is correct.

6. Which of the following statements best describes the idea of a minimum viable product?

Answer: C. A prototype of a new product or service that allows the team to collect the maximum amount of learning with the least amount of effort Feedback: 7. You are given the following customer responses to a purchase intent survey for a new electric bicycle that has all its batteries built into the frame and hence almost looks exactly like a regular bike: • • • • •

Definitely would use: 12 Probably would use: 22 Might or might not use: 31 Probably would not use: 111 Definitely would not use: 201

Out of a target market of 1,500,000 potential customers, how many do you expect to adopt the product, assuming you can reach a 5 percent awareness in the population? Assume Cprobably = 0.2 and Cdefinite = 0.4. Select the closest option.

Answer: C. 1830 CASE Innovation at Toyota Q.1 In your opinion, are the Prius and the FCV horizon 1, 2, or 3 innovations?

This is an opinion question, but the Prius most likely corresponds to a horizon 2 innovation. It uses new technology (the hybrid engine) to please customers that are already used to driving a car. The same can be said about the FCV technology—as radical as a fuel cell might appear at first sight, it is still a car.

Q.2 From the perspective of today, what would a horizon 3 innovation in the automotive industry look like?

Horizon 3 innovations could include an autonomous vehicle (a car that drives itself), personal transporters such as the Segway, or even more revolutionary modes of transportation such as Elon Musk’s high-speed transportation system Hyperloop, which puts humans in a capsule that then is moved forward via induction and air pressure.

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Q.3 Would you as an executive in the automotive industry invest in a project like this?

This answer could be both yes or no. It would depend in the risk-taking ability of the executive as well as the innovation capacity of the firm.

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