SOLUTIONS MANUAL for Matching Supply with Demand: An Introduction to Operations Management, 5th Edit by StudyGuide

Matching Supply with Demand 5th Edition an Introduction to Operations Management Gerard Cachon

TABLE OF CONTENTS CHAPTER 1: Introduction CHAPTER 2: The Process View of the Organization CHAPTER 3: Understanding the Supply Process: Evaluating Process Capacity CHAPTER 4: Estimating and Reducing Labor Costs CHAPTER 5: Batching and Other Flow Interruptions: Setup Times and the Economic Order Quantity Model CHAPTER 6: The Link between Operations and Finance CHAPTER 7: Quality and Statistical Process Control CHAPTER 8: Lean Operations and the Toyota Production System CHAPTER 9: Variability and Its Impact on Process Performance: Waiting Time Problems CHAPTER 10: The Impact of Variability on Process Performance: Throughput Losses CHAPTER 11: Scheduling to Prioritize Demand CHAPTER 12: Project Management CHAPTER 13: Forecasting CHAPTER 14: Betting on Uncertain Demand: The Newsvendor Model

CHAPTER 15: Assemble-to-Order, Make-to-Order, and Quick Response with Reactive Capacity CHAPTER 16: Service Levels and Lead Times in Supply Chains: The Order-up-to Inventory Model CHAPTER 17: Risk-Pooling Strategies to Reduce and Hedge Uncertainty CHAPTER 18: Revenue Management with Capacity Controls CHAPTER 19: Supply Chain Coordination

Chapter 2 The Process View of the Organization Q2.1 Dell The following steps refer directly to Exhibit 2.1. #1: For 2001, we find in Dell’s 10-k: Inventory = $400 (in million) #2: For 2001, we find in Dell’s 10-k: COGS = $26,442 (in million) 26, 442$/ year #3: Inventory turns = = 66.105 turns per year 400$ 40% per year #4: Per unit Inventory cost = = 0.605% per year 66.105 per year Q2.2. Airline We use Little’s law to compute the flow time, since we know both the flow rate as well as the inventory level: Flow Time = Inventory/ Flow Rate = 35 passengers/ 255 passengers per hour = 0.137 hours = 8.24 minutes Q2.3 Inventory Cost (a) Sales = $60,000,000 per year / $2000 per unit = 30,000 units sold per year Inventory = $20,000,000 / $1000 per unit = 20,000 units in inventory Flow Time = Inventory/ Flow Rate = 20,000 / 30,000 per year = 2 / 3 year = 8 months Turns = 1/ Flow Time = 1/(2 / 3 year) = 1.5 turns per year

Note: we can also get this number directly by writing: Inventory turns = COGS / Inventory (b) Cost of Inventory: 25% per year /1.5 turns = 16.66%. For a $1000 product, this would make an absolute inventory cost of $166.66 . Q2.4. Apparel Retailing (a) Revenue of $100M implies COGS of $50M (because of the 100% markup). Turns = COGS/ Inventory = $50M/ $5M = 10 . (b) The inventory cost, given 10 turns, is 40%/10 = 4% . For a 30$ item, the inventory cost is 0.4 $30 = $1.20 per unit . Q2.5. La Villa (a) Flow Rate = Inventory / Flow Time = 1200 skiers /10 days = 120 skiers per day (b) Last year: on any given day, 10% (1 of 10) of skiers are on their first day of skiing

This year: on any given day, 20% (1 of 5) of skiers are on their first day of skiing Average amount spent in local restaurants (per skier) Last year = 0.1$50 + 0.9$30 = $32 This year = 0.2$50 + 0.8$30 = $34 % change = ($34 −$32) / $32 = 6.25% increase Q2.6. Highway We look at 1 mile of highway as our process. Since the speed is 60 miles per hour, it takes a car 1 minute to travel through the process (flow time). There are 24 cars on ¼ of a mile, i.e. there are 96 cars on the 1 mile stretch (inventory). Inventory = Flow Rate * Flow Time: 96 cars = Flow Rate * 1 minute Thus, the Flow Rate is 96 cars per minute, corresponding to 96*60 = 5760 cars per hour. Q2.7. Strohrmann Baking The bread needs to be in the oven for 12 minutes (flow time). We want to produce at a flow rate of 4000 breads per hour, or 4000/60 = 66.66 breads per minute. Inventory = Flow Rate * Flow Time: Inventory = 66.66 breads per minute* 12 minutes Thus, Inventory = 800 breads, which is the required size of the oven. Q2.8. Mt Kinley Consulting We have the following information available from the question: Level Associate Manager Partner

Inventory (number of consultants at that level) 200 60 20

Flow Time (time spent at that level) 4 years 6 years 10 years

(a) We can use Little’s law to find the flow rate for associate consultants: Inventory = Flow Rate * Flow Time; 200 consultants = Flow Rate * 4 years; thus, the flow rate is 50 consultants per year, which need to be recruited to keep the firm in its current size (note: while there are also 50 consultants leaving the associate level, this says nothing about how many of them are dismissed vs how many of them are promoted to Manager level). (b) We can perform a similar analysis at the manager level, which indicates that the flow rate there is 10 consultants. In order to have 10 consultants as a flow rate at the manager level, we need to promote 10 associates to manager level (remember, the firm is not recruiting to the higher ranks from the outside). Hence, every year, we dismiss 40 associates and promote 10 associates to the manager level (the odds at that level are 20%)

Now, consider the partner level. The flow rate there is 2 consultants per year (obtained via the same calculations as before). Thus, from the 10 manager cases we evaluate every year, 8 are dismissed and 2 are promoted to partner (the odds at that level are thereby also 20%). In order to find the odds of a new hire to become partner, we need to multiply the promotion probabilities: 0.2*0.2 = 0.04. Thus, a new hire has a 4% chance of making it to partner. Q2.9. Major US Retailers a. Product stays on average for 31.9 days in Costco’s inventory b. Costco has for a $5 product an inventory cost of $0.1311 which compares to a $0.2049 at Wal-Mart Q2.10. McDonald’s a. Inventory turns for McDonald’s were 92.3. They were 30.05 for Wendy’s. b. McDonald’s has per unit inventory costs of 0.32%, which for a 3$ meal about $0.00975. That compares to 0.998% at Wendy’s where the cost per meal is $0.0299 . Q2.11. BCH I = 400 associates, T = 2 years. R = I / T = 400 associates / 2 yrs = 200 associates / yr . Q2.12. Kroger Turns = R / I = 76858 / 6244 =12.3

Matching Supply with Demand: An Introduction to Operations Management 5e Solutions to Chapter Problems Chapter 3 Understanding the Supply Process: Evaluating Process Capacity Q3.1 Process Analysis with One Flow Unit (a) Capacity of the three resources in units per hour are 602 /10 =12 , 601/ 6 =10; 603 /16 =11.25 . The bottleneck is the resource with the lowest capacity, which is resource 2. (b) The process capacity is the capacity of the bottleneck, which is 10 units/hr . (c) If demand = 8 units / hr , then the process is demand constrained and the flow rate is 8 units/hr (d) Utilization = Flow Rate / Capacity . For the three resources they are 8 / 12 , 8 / 10 , and 8 /11.25 .

Q3.2 Process Analysis with Multiple Flow Units a) Bottleneck is resource 3 because it has the highest implied utilization of 125%. The demands per hour of the three products are 40 / 8 = 5 , 50 / 8 = 6.25 and 60 / 8 = 7.5. The total minutes of work demanded per hour at resource 1 is 5 × 5 + 6.25 * 5 + 7.5 * 5 = 93.75. Two workers at resource 1 produce 2 * 60 = 120 min of work per hour. So resource 1’s utilization is 93.75 /120 = 0.78. Utilization at the other resources are similarly evaluated. b) The capacity of resource 3 is 60 /15 = 4 units per hour. Given the ratio of units produced must be 4 to 5 to 6, the process can produce 4 units/ hr of A, 5 units / hr of B and 6 units / hr of C. Q3.3. Cranberry Cranberries arrive at a rate of 150 barrels per hour. They get processed at a rate of 100 barrels per hour. Thus, inventory accumulates at a rate of 150-100 = 50 barrels per hour. This happens while trucks arrive, i.e. from 6am to 2pm. The highest inventory level thereby is 8h*50 barrels per hour = 400 barrels. From these 400 barrels, 200 barrels are in the bins, the other 200 barrels are in trucks. (a) 200 barrels (b) From 2pm onwards, no additional cranberries are received. Inventory gets depleted at a rate of 100 barrels per hour. Thus, it will take 2h until the inventory level has dropped to 200 barrels, at which time all waiting cranberries can be stored in the bins (no more truck waiting) (c) It will take another 2 hours until all the bins are empty (d) Since the seasonal workers only start at 10:00am, the first 4 hours of the day we accumulate 4hours * 50barrels per hour = 200 barrels. For the remaining time that we receive incoming cranberries, our processing rate is higher (125 barrels per hour). Thus, inventory only accumulates at a rate of 25 (150-125 barrels per hour). Given that this happens over 4 hours, we get another 100 barrels in inventory. At 2pm, we thereby have 300 barrels in inventory. After 2pm, we receive no further cranberries, yet we initially process cranberries at a rate of 125 barrels per hour. Thus, it only takes 100 barrels /125 barrels/hour = 0.8 hours = 48 minutes until all bins are empty. From then, we need another 2h until the bins are empty. Q3.4. Western Pennsylvania Milk We start the day with 25,000 gallons of milk in inventory. From 8am onwards, we produce 5,000 gallons, yet we ship 10,000 gallons. Thus inventory is depleted at a rate of 5000 gallons per hour, which leaves us without milk after 5 hours (at 1pm). From then onwards, clients will have to wait. This situation gets worse and worse and by 6pm (last client arrives), we are short 25,000 gallons. (a) 1pm (b) Clients will stop waiting when we have worked off our 25,000 gallon backlog that we are facing at 6pm. Since we are doing this at a rate of 5,000 gallon per hour, clients will stop waiting at 11pm (after 5 more hours). (c) At 6pm, we have a backlog of 25,000 gallons, which is equivalent to 20 trucks (d) The waiting time is the area in the triangle © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

- width: beginning of waiting (1pm) to end of waiting (11pm) = 10 hours - height: maximum number of trucks waiting: 20 (see part c above) Hence, we can compute the area in the triangle as: 0.5*10hours*20trucks = 100 truck* hours The cost for this waiting is 50$/ truck hour 100 truck hours = 5000$ Q3.5. Bagel Store (a) The bottleneck is “Veggies on Bagel”, as it has the highest implied utilization. Resource

Cut Grilled stuff Veggies on Bagel Cream cheese Wrap

Total

Implied utilization

3*11 0

Cream cheese bagel 3*4 0

54 30

54 / 60 30/60

3*5

11*5

70 / 60

4*4

16 / 60

3*2

11*2

4*2

36 / 60

Available minutes per hour 60 60

Grilled veggie bagel 3*3 3*10

Veggie bagel

(b) If we want to keep the product mix constant (i.e. keep the ration between the various bagel types at 3:11:4), we need to scale down demand by 60 / 70 . This leads to the following flow rates: - Grilled veggie: 3 60 / 70 bagels / hour - Veggie: 1160 / 70 bagels / hour - Cream cheese: 460 / 70 bagels / hour If we try to fulfill as much demand as possible, we encounter a problem with Veggie bagels and Grilled + Veggie bagels. Given that we have an implied utilization of 70 / 60% at the bottleneck, we can only fulfill 60 / 70% of demand for these two bagel types. We can meet all of demand for the cream cheese bagels. Q3.6. Valley Forge Income Tax Recall that the demand for the 4 groups is: 1: 15% 2: 5% 3: 50% 4:30% There are three resources, the admin, the junior person, and the senior person. We compute the minutes of capacity they have available every month as well as their work-load, given that there are 50 incoming returns to be processed. This leads to the following computations. Resource

Minutes 1 per month

Total work-

Implied utilization

Admin

9600

Senior

9600

Junior

9600

0.15*50 *(20+50) 0.15*50 *(30+20) 0.15*50 *120

0.05*50 *(40+80) 0.05*50 *(90+60) 0.05*50 *300

0.5*50 *(20+30) 0.5*50 *(0+5) 0.5*50 *80

0.3*50 *(40+60) 0.3*50 *(0+30) 0.3*50 *200

load 3575

0.372

1325

0.138

6650

0.693

(a) The junior person is the bottleneck, her implied utilization is the highest (b) Implied utilizations (see table) (c) One should consider: revenue, future revenue (lifetime value of customer), to what extent the request draws on bottleneck capacity (d) not at all, since improvements at non-bottleneck steps don’t increase capacity Q3.7. Car Wash Supply Process a. The implied utilization at wheel cleaning is: 42 minutes of work from package 3 a day, 84 minutes of work from package 4 per day, giving a total of 126 minutes per day. Relative to 720 minutes of available time, that gives an implied utilization of 17.5%. b. The highest implied utilization is at step 1 (automated washing machine) where the implied utilization is 55.55%. c. The new bottleneck will be the interior cleaning employee with an implied utilization of 111% d. Every day, we are 80 minutes of work short at step “interior cleaning”. That corresponds to four customers with package 4. Q3.8. Starbucks a. Time required of the frozen drink maker = 20*2 = 40 minutes. Each worker has 60 minutes available, so the utilization for the frozen drink maker: 40 / 60 b. Highest implied utilization is for the Espresso Drink Maker with 116.66% c. The workload on the cashier is as follows. 25 Drip coffee customer per hour at 1/ 3 min per customer = 25 / 3 min / hr . 5 Ground customers per hour at 1 min per customer = 5 min / hr . 120 customers per hour at 1/ 3 min to pay per customer = 40 min / hr . 30% of 120 customers per hour buying food, which is 36 customers per hour and they require 1/ 3 min per customer, for a total of 12 min / hr . In total the workload on the cashier is 25 / 3 + 5 + 40 +12 = 65.33 min / hr . The implied utilization is 65.33 / 60 = 108.9% . The implied utilization of the other persons don’t change and they are 66.66% and 116.66%. Q3.9. Paris Airport a. The implied utilization levels of the resources are computed as follows Servers

Avail. Min/Hr

Requested

Imp. Ulll.

Security Agents Kiosk

4 6 3

240 360 180

150 440 160

62.50% 122.22% 88.89%

b. The backlog at the bottleneck accumulates at a rate of 80 “requested minutes” per hour. After 4 hours this is 4*80 = 320 “requested minutes” of work from the agents. This takes 6 agents 320 / 360 = 8 / 9 = 0.89 hours to complete, or 53.4 minutes after the last arrival c. When Kim arrives, the backlog is half as long as the final backlog in question 2, e.g. there are 160 “requested minutes” of work in front of Kim. This will take the agents 160 / 360 = 0.45 hours = 26.6 minutes to complete. Kim takes 3 minutes to complete service with the agent and 30 seconds to pass through security (there is no line at security) so the answer is 30 minutes and 10 seconds d. We compute the new utilizations as follows:

Security Agents Kiosk

Servers Avail Min/Hr 4 240 6 360 3 180

Requested 150 376 32

Imp. Ulll. 62.50% 104.44% 17.78%

Extra work accumulates at a rate of 16 minutes per hour for the agents, for a total of 64 minutes after the last arrival. This takes the 6 agents 64 / 360 = 0.178 hours = 10.7 minutes to complete, so 8:10 PM is the closest answer.

Matching Supply with Demand: An Introduction to Operations Management 5e Solutions to Chapter Problems Chapter 4 Estimating and Reducing Labor Costs Q4.1. Empty System Labor Utilization (a) Time to complete 100 units: #1 The process will take 10 + 6 + 16 minutes = 32 minutes to produce the first unit. #2 We know from problem xyz that resource 2 is the bottleneck and the process capacity is 0.1666 units per minute #3 Time to finish 100 units = 32 minutes +

99 units = 626 minutes 0.1666 units / min

(b) + (c) + (d) Use Exhibit for Labor computations

#1 Capacities are: Resource 1: 2 /10 units/ minute = 0.2 units / minute Resource 2: 1/ 6 units/ minute = 0.1666 units / minute Resource 3: 3 /16 units/ minute = 0.1875 units / minute Resource 2 is the bottleneck and the process capacity is 0.1666 units/minute #2 Since there is unlimited demand, the flow rate is determined by the capacity and thereby 0.1666 units/minute ; this corresponds to a cycle time of 6 minutes/unit #3 Cost of direct labor =

610$ / h 60 0.1666 units / h

= 6$ / unit

#4 Compute the idle time of each worker for each unit: Idle time for workers at resource 1 = 6 min / unit  2 −10min / unit = 2 min / unit Idle time for worker at resource 2 = 6 min / unit1− 6 min / unit = 0 min / unit Idle time for workers at resource 3 = 6 min / unit 3 −16 min / unit = 2 min / unit #5 Labor content = 10 + 6 +16 min / unit = 32 min / unit #6 Average labor utilization =

32 = 0.8888 32 + 4

Q4.2. Assign tasks to workers (a) Worker Task(s) Processing Time (sec) Capacity (units per hour) 1 1 30 120 2 2 25 144 3 3,4 75 48 4 5,6 45 80 The capacity of the current line is restricted by the capacity of the step with the longest processing time. Therefore, capacity = 1/ 75 sec = 48 units per hour . (b) Worker Task(s) Processing Time (sec) Capacity (units per hour) 1 1,2 55 65.45 2 3 35 102.86 3 4 40 90 4 5,6 45 80 Therefore, capacity of the revised line =1/ 55 sec = 65.45 units per hour . (a) No matter how you organize the tasks, maximum capacity of the line is 65.45 units per hour, i.e. at a cycle time of 55 seconds.

Q4.3. Power Toys (a) Since every resource has exactly one worker assigned to it, the bottleneck is the assembly station with the highest processing time (#3) (b) Capacity = 1/ 90 sec = 40 units per hour (c) Direct labor cost = Labor cost per hour / flow rate = 915$/ h / 40 trucks per hour = 3.38 $ / truck (d) Direct labor cost in work cell = (75 + 85 + 90 + 65 + 70 + 55 + 80 + 65 + 80)sec/ truck $15 / hr = 2.77$ / truck (e) Utilization = flow rate / capacity 85 sec / 90 sec = 94.4% (f) Worker 1 2 3 4 5 6

Station(s) 1 2 3 4,5 6,7 8,9

Processing Time (sec) 75 85 90 135 135 145

Capacity (units per hour) 48 42.35 40 26.67 26.67 24.83

(g) Capacity = 1/145 units / second = 24.83 toy - trucks per hour Q4.4. 12 tasks to 4 workers (a) Worker Task(s) 1 1,2,3 2 4,5,6 3 7,8,9 4 10,11,12

Processing Time (sec) 70 55 85 60

Capacity (units per hour) 51.43 65.45 42.35 60

(a) Capacity = 1/ 85 sec = 42.35 units per hour (b) Direct labor content = (70 + 55 + 85 + 60) sec = 270 sec/unit or 4.5 min / unit (c) Labor utilization = labor content / ( labor content + total idle time) = 270sec / (270 +15 + 30 + 0 + 25 sec) = 79.41% (d) Note that we are facing a machine paced line, thus the first unit will take 4*85 seconds top go through the empty system. Flow Time = 4 85 sec + 99 /(1/ 85 sec) = 8755 sec or 145.92 min or 2.43 hrs (e) There are multiple ways to achieve this capacity. This table shows only one example. © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Worker Task(s) Processing Time (sec) 1 1,2 55 2 3,4,5 50 3 6,7 70 4 8,9 35 5 10,11,12 60 Capacity = 1/ 70 units / sec = 51.43 units per hour

Capacity (units per hour) 65.45 51.43 51.43 102.86 60

(f) There are multiple ways to achieve this capacity. This table shows only one example. Worker Task(s) Processing Time (sec) Capacity (units per hour) 1 1,2 55 65.45 2 3,4,6 55 65.45 3 5,8,10 55 65.45 4 7 50 72 5 9,11,12 55 65.45 Capacity = 1/ 55 units / sec = 65.45 units per hour (g) We have to achieve a cycle time of 3600 / 72 = 50 seconds/unit. The following task allocation includes a lot of idle time, but is the only way to achieve the cycle time, given the constraints we face. Worker Task(s) Processing Time (sec) 1 1 30 2 2,3 40 3 4,5 35 4 6 20 5 7 50 6 8,9 40 7 10,11 40 8 12 20 Therefore, a minimum of 8 workers are required to achieve a capacity of 72 units per hour. Q4.5. Geneva Watch (a) Station E is the bottleneck with a process capacity of 1 unit every 75 seconds. (b) Capacity = 1/ 75 sec = 48 watches per hour (c) Direct labor content = 68 + 60 + 70 + 58 + 75 + 64 = 395 sec (d) Utilization = 60 sec / 75 sec = 80% (e) Idle time = (75 − 70)sec/ 75 sec60 min per hour = 4 min per hour ; as an alternative computation, we can observe that the worker has 5 seconds idle time per cycle (i.e. per unit) and that there are 48 cycles (units) per hour. Thus, the idle time over the course of an hour is 240 seconds = 4 minutes.

(f) Time to produce 193 watches = time for the first watch + time for the remaining 192 watches = 6*75 seconds + 192*75 seconds=14,850 seconds=4h7min30sec. Production begins at 8:00, so 193 watches will be completed by 12:07:30 Q4.6. Yoggo Soft Drink a. Bottling machine capacity: 1 bottles / second Lid machine capacity: 0.333 bottles / sec Two labeling machines capacity: 10 / 25 = 0.4 bottles / sec Packaging machine capacity: 0.25 bottles/sec So the process capacity is going to be 0.25 bottles / sec = 0.253600 = 900 bottles / hour b. The packaging machine c. It has no effect on the capacity since it is not the bottleneck d. Process capacity = 90 boxes/ hour , implied utilization = demand / capacity = 0.666 = 66% Q4.7. Atlas Inc a. The bottleneck is the worker with the highest processing time (across activities), which is Worker 2 (60 seconds) b. Capacity of the line is decided by the processing time of the bottleneck step. Hence we have Capacity = 1/60 sec = 60 units/hour c. Utilization is given by Flow Rate/ Capacity . Hence, we have Utilization = 45 sec / 60 sec = 75% . d. As we are facing an empty system, the first unit would take (50+60+30+45+40)=225 seconds to go through the system. Hence, Flow time = 225 + (100-1)*60 = 102.75 minutes e. Labor Utilization is given by Labor Content / ( Labor Content + Idle Time) . Total Labor Content can be calculated as (50+60+30+45+40)=225 seconds. Idle time for each worker can be calculated as processing time of bottleneck - processing time of worker. Hence, we have Labor Utilization = 225 / (225 +10 + 30 +15 + 20) = 75%. f. Direct Labor Cost = Labor Cost per Hour / Flow Rate = 5$15 / 60 = $1.25 / unit g. Again, there are multiple configurations that minimize completion time, but in all of these the processing time of the bottleneck resource is 55 seconds. Hence maximum achievable capacity is 1/55 sec = 65.5 units/hour . h. Direct Labor Cost = (30 + 20 + 35 + 25 + 30 + 45 + 40)sec$15 / hour = $0.9375 / unit i. The bottleneck is worker 3, and process capacity is given by 1/75 sec = 48 units/hour

Q4.8. Glove Design a. Cutting has a process capacity of 1 glove / 2 minutes*60 minutes = 30 gloves / hour . Dyeing has a process capacity of 1 glove / 4 minutes*60 minutes = 15 gloves / hour . Stitching has a process capacity of 1 glove / 3 minutes 60 minutes = 20 gloves / hour . Packaging has a process capacity of 1 glove / 5 minutes 60 minutes = 12 gloves / hour . Therefore, the capacity is a. 12 gloves/hour . b. The first statement is incorrect because packaging is the bottleneck. The second statement is incorrect because in a machine-paced line or conveyor belt, the unit spends the same amount of time at each station as the bottleneck. The fourth statement is incorrect because cutting is not the bottleneck. c. By reducing packaging time the process capacity increases is correct because packaging is the bottleneck. c. If the demand is 10 gloves/hour , then the implied utilization at packaging = 10 /12 = d . 83.3%. d. A glove spends 5 minutes in each of 4 stations, so the flow time = 5*4 = c. 20 minutes.

Q4.9. Worker Paced a. We know that Step 4 is the bottleneck and has a process capacity = to the capacity of the entire process because the utilization = 100%. We are given the fact that the process capacity = 36 units/hour , and Step 5 has a utilization of 40%. Therefore, the capacity of Step 5 = 36/ 0.4 = d . 90 units per hour. b. The step with the highest utilization is the bottleneck, or d. Step 4. c. The step with the highest utilization has the highest process capacity. Step 1 has a process capacity of 36 / (4 / 30) = 270 units/hour . Step 2 has a process capacity of 36 /(4 /15) =135 units/hour . Step 3 has a process capacity of 36 / (4 / 5) = 45 units/hour . Step 4 has a process capacity of 36 units/hour . Step 5 has a process capacity of 36 / (2 / 5) = 90 units/hour . Therefore, the step with the highest capacity is a. Step 1. d. There are 5 workers per hour to make 36 units. The wages per hour, then = 5*$36 =$180 in labor costs to make 36 units. Therefore, the direct cost of labor per unit = $180 / 36 = a . $5 unit.

Matching Supply with Demand: An Introduction to Operations Management 5e Solutions to Chapter Problems Chapter 5 Batching and Other Flow Interruptions: © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Setup Times and the Economic Order Quantity Model Q5.1 Window Boxes a) The production cycle consists of setup to produce part A (120 minutes), produce A (1 minute per part), setup to produce B (120 minutes), and produce B (1 minute for two sides). The total setup time in this production cycle is 240 minutes. The processing time for the two components is 2 minutes. There are 360 component sets produced in each production cycle. So the capacity of the stamping machine is 360 /(240 + 2  360) = 0.375 units per minute . b) We want to choose a batch size so that the capacity of the stamping process is the same as the capacity of the assembly process. The capacity of the assembly process is (1/ 27 units per minute) 12 workers =12 / 27 units per minute . The desired batch size is found using the following equation: batch size = (flow rate  setup time) (1- flow rate  processing time). Hence, batch size = (12 / 27 240) / (1−12 / 27 2) = 960 . c) The batch size is 1260 part A. Because this batch is larger than 960 (from question 5.1b), the bottleneck is assembly. Hence, the flow rate of part A is 12/27 part units per minute (the capacity of assembly). The processing time of part A is 1 min. Average inventory = 12 1260 units (1−12 / 27 units per minute 1 minute per part) = 350 part As. Q5.2 Two-step a) Capacity of step B is 5 / (9 + 0.15) = 0.53 units per minute . The first activity makes 1 unit per minute, so the bottleneck is the second step. b) The flow rate is determined by the bottleneck, which is step A, and equals 1 unit per minute. The processing time at step B is 0.1 min per unit. The average inventory with a batch size of 15 is 12 15 (1−1 unit per min  0.1 min per unit) = 6.75 units . c) The desired flow rate is 1 unit per minute because that is the capacity of step A. Recommended batch size = 19 / (1−1 0.1) = 10 Q5.3 Simple Set-Up a) First, we calculate the process capacity at each step using the formula for process capacity with batching: B / S + Tb where B = batch size, S = set-up time, and Tb = time to process the entire batch. So, the first step has a process capacity of50 units / (20 minutes

(

)

+ 150 minutes = .714 units / minute = 42.86 units / hour . The second step has a

process capacity of 50 units /(250) minutes = 30 units/hour . The third step has a process capacity of 50 units /(1.550) minutes = 40 units/hour. . Therefore, the process capacity of the entire process = 30 units/hour .

b) The only step that is dependent on the batch size is Step 1. With a batch size of 10 units, the process capacity of the first step becomes 10 units /(20 minutes + (110) minutes) =10 units / 30 minutes = 20 units / hour . Therefore, Step 1 becomes the bottleneck. c) In order to determine an optimal batch size, we set the process capacity of Step 1 (the only step dependent on batch size) equal to the process capacity of the bottleneck capacity. The bottleneck capacity is 30 units/hour , or 1 unit every 2 minutes. So, we solve for B using the following formula: 1 unit / 2 minutes = B / (20 minutes + B minutes) B = 20 units. If B < 20 units, then Step 1 becomes the bottleneck and the capacity of the entire process decreases. d) Based on the answer to part (c), with a batch size of 40 the bottleneck is step 2 and the Flow rate is 30 units/hour . The processing time at step 1 is 1 min/unit , which converted to hours is 1/ 60 hour/unit . Average inventory after step 1 is therefore = 12  40 units  (1− 30 units / hour 1/ 60 hour / unit) = 10 units .

Q5.4 Set-Up Everywhere a) Capacity = B / (S+ B Processing Time) = 35 parts / (30 minutes + (35 0.25) minutes) = .903 units / minute = 54.19 units / hour b) Step 2 is never the bottleneck because its processing time and setup time are always less than Step 1’s processing time and setup time. So the task can be simplified to determining for what values Steps 1 and 3 are the bottleneck. The processing time for the two steps are equal when 30 + 0.25B = 45 + 0.15B, or when B =150. For batches smaller than 150 parts, Step 3 is the bottleneck. For batches larger than 150 parts, Step 1 is the bottleneck.

Q5.5 JCL a) Capacity of a step is given by Batch Size/ (Set-Up-Time + Batch  Processing Time) . Using this formula, the capacity of the three steps can be calculated as: Deposition: 100 / (45 + 0.15100) = 1.67 units / min = 100 units / hour Patterning: 100 /(30 + 0.25100) = 1.82 units / min = 109 units / hour Etching: 100 /(20 + 0.2100) = 2.5 units / min =150 units / hour The bottleneck is the “deposition” step and it determines process capacity. b) If batch size is B, then processing time of step 3 is 20 + 0.20B which is always lesser than the processing time of step 2, 30 + 0.25B, for a positive value of B. Hence step 3 can never be the bottleneck.

c) Under the new technology, the process capacity of step 1 is no longer dependent on batch size and is (1/ 0.45min / unit) = 2.22 units / min . This is clearly the maximum overall process capacity that can be targeted. Since the capacity of step 2 can never exceed the capacity of step 3 (since step 2’s run time and setup time are both higher), we set the batch size so the capacity of step 2 matches 2.22 units / min . Thus, the target batch size = Target capacity  Setup time / (1− ( Target capacity  Processing time)) = 2.22 units / min30 min /(1− (2.22 units / min 0.25 min / unit)) =150 units . Q5.6 Kinga Doll a) The process capacity of molding = 500 / (15 + (.25500)) = 3.57 dolls / minute = 214.9 dolls/hour . The process capacity of painting = 500 / (30 + (.15500)) = 4.76dolls / minute = 285.7 dolls/hour . The process capacity of dressing = 1/ .3 = 3.33 dolls / minute = 200 dolls / hour . Therefore, the process capacity = 200 dolls / hour . b) For molding the target batch size = Target capacity  Setup time / (1− ( Target capacity x Processing time)) = 3.33 dolls / min15min / (1− (3.33 dolls / min 0.25min / doll)) = 300 dolls For painting the target batch size = Target capacity  Setup time /(1− ( Target capacity Processing time)) = 3.33 dolls / min30 min /(1− (3.33 dolls / min0.15 min / doll)) = 200 dolls Therefore, the optimal batch size = 300 units. (If the smaller batch size were selected, then molding’s capacity would be too low.) c)

The flow rate is the minimum of the process capacity, which is 3.33 dolls / minute or 200 dolls/hour , and demand. We know demand = 4000 dolls / week , with a 40-hour work week. Therefore, the demand = 100 dolls/hour, which is less than the process capacity. Therefore, current flow rate =100 dolls/hour or 1.67 dolls/minute . For molding the target batch size = Target capacity  Setup time /(1− ( Target capacity  Processing time)) = 1.67 dolls / min15min / (1− (1.67 dolls / min 0.25min / doll)) = 42.9 dolls For painting the target batch size = Target capacity  Setup time /(1− ( Target capacity  Processing time)) =1.67 dolls / min30 min /(1− (1.67 dolls / min 0.15 min / doll)) = 66.7 dolls Again, choose the larger batch size so that the flow rate isn’t decreased 0 - the optimal batch size is 66.7. Q5.7 PTest a) PTest can test 300 samples in 12300 / 60 + 30 = 90 min =1.5 hour . The capacity is 300 samples/1.5 hours = 200 samples/hour . © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

b) The smallest batch size that achieves a flow rate of 2.5 samples per minute is 2.530 / (1−12 / 60 2.5) = 150 samples . c) The number of basic tests per minute = 70 / (15 / 60 70 +1.530 + 20) = 70 / 82.5 = 0.848 Q5.8 Gelato a) Each batch consists of a set of three flavors. The total setup time is 3 + 1 +1/ 6 = 17 /12 hours . The desired flow rate is 10 +15 + 5 = 30kgs / hr . The 4 2 processing time is 1/ 50 hrs / kg . The desired batch size for the set of three is 3017 /12 (1− 301/ 50) = 106 kgs . b) Fragola is 10/30ths of demand, so produce (10 / 30) 106 = 35.33 kgs of fragola c) Chocolato is 15/30ths of demand, so the chocolato batch is (15 / 30) 106 = 53 kgs . The flow rate of chocolato is 15 kgs / hour . The Processing time for chocolato is 1/ 50hr / kg . So average inventory = 12  Batch size(1− Flow rate  Processing time) = 12 53(1−15 1/ 50) = 18.55 . Q5.9 Carpet a) The flow unit is a yard of carpet. Total demand = 100 + 80 + 70 + 50 = 300 yrds / hr . Setup time = 4 carpets × 3 hr per carpet = 12hr. Processiong time =1 hr /(350 yrds / hr) = 0.002857 hr . Target batch size = = 12hr 300 yrds / hr /(1− 300 yrds/hr*0.002857 hr)=25200 yrds . Of the carpet produced per production cycle, 100 / 300th of it is carpet A. Hence, batch size of carpet A = 25, 200100 / 300 = 8400 b) Batch size for carpet A is 16,800. Flow rate for carpet A is 100 yards/hr and the Processing time is 1/350 hr /yard . Thus, Average inventory = 12  Batch size(1 - Flow rate  Processing time) = 12 16,800 (1−1001/ 350) = 6, 000 yards . Q5.10 Catfood a) Holding costs are $0.5015%/ 50 = 0.0015 per can per week. Note, each can is purchased for $0.50 , so that is the value tied up in inventory and therefore determines the 2  7  500 = 2160 T . holding cost. The EOQ is then 0.0015 b) The ordering cost is $7 per order. The number of orders per year is 500 / EOQ . Thus, 7  500 = 1.62 $ / week = 81$ / year . order cost = EOQ c) The average inventory level is EOQ / 2 . Inventory costs per week are thus 0.5*EOQ*0.0015 = $1.62 . Given 50 weeks per year, the inventory cost per year is $81. d) Inventory turns = Flow rate / Inventory © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Flow Rate = 500 cans per week Inventory = 0.5 * EOQ

Thus, Inventory Turns = R / (0.5 EOQ) = 0.462 turns per week = 23.14 turns per year

Q5.11 Beer Distributor The holding costs are 25% per year = 0.5% per week = 8*0.005 = $0.04 per week 210010 = 223.6 0.04 b) Inventory turns = Flow Rate / Inventory = 10050 / (0.5EOQ) = 5000 / EOQ = 44.7 turns per year. 2  0.04 10 = 0.089$ / unit c) Per unit inventory cost = 100 d) You would never order more than Q = 600 For Q = 600, we would get the following costs: 0.56000.040.95 +10100 / 600 =13.1The cost per unit would be 13.1/100 = $0.131 a) EOQ =

The quantity discount would save us 5%, which is $0.40 per case. However, our operating costs increase by $0.131 - 0.089 = $0.042 . Hence, the savings outweigh the cost increase and it is better to order 600 units at a time.

Q5.12 Millenium Liquors The fixed cost of refrigeration can be ignored because that cost does not change as we vary our order quantity. a)

Weekly holding cost 15%/ 50 per week 120$/ case = 0.36$/ week .

2  300  45 = 273.9 cases per order. 0.36 c) We would get slightly lower ordering costs, which results in more frequent orders and lower inventory b) The ordering cost is $290 + $10 = $300 . EOQ =

Q5.13 Powered by Koffee The holding costs are $1.50 per month ( $1 storage and $0.50 capital) a) EOQ =

28550

= 75.27 1.5 b) Order frequency: (1250) / EOQ = 8 times per year c) Average inventory is EOQ / 2 . So months of supply = (EOQ / 2) / 50 = 0.75 months d) inventory costs per month = (EOQ/ 2) 1.50 = 56.46$/ month

e) The monthly holding cost per bag is $1+ 0.02 20 =1.4 . Annual purchase quantity is 12 x 50 = 600 bags. The average inventory will be 600 / 2 = 300 , and so the monthly holding cost is 300$1.4 = $420 . The yearly holding cost is 12$420 = $5040 . The annual purchase cost is 600$20 = $12, 000 . The total annual cost of this option is $12, 000 +$500 +$5040 = $17,540 . The current system operates at costs of 12 285501.5 + 600 25 =16,355 Thus, the original system is cheaper.

Matching Supply with Demand: An Introduction to Operations Management 5e Solutions to Chapter Problems Chapter 6 The Link between Operations and Finance Q6.1. Crazy Cab a) see tree below b) see tree below c) Value drivers include the % of distance driven empty, the number of trips per day, and the distance of the trip. This is a high fix cost business with lots of capital, thus the more revenue you can squeeze out of the cabs, the more money you make. And interesting issue would be to see if by reducing the time the cab drives empty, one could increase the number of trips further d) Similar to the airline ratios discussed previously. We can look at labor efficiency as: Revenue / labor cost = Revenue / mile  miles / trip  trips / day  days / labor cost

The first ratio is the pricing power, the second the length of a trip, the third how many trips we get out of a cab, and the fourth is a measure of wage rates.

Q6.2. Penne Pesto a) The restaurant will serve 200 guests on an evening b) The ROIC is 18.25% (based on $100 net profit per day) c) Making the clean-up faster allows to turn over the table quicker and leads to an increase in the number of guests served. We get 218 guests served and a 66% ROIC d) A reduction in OH cost has a less dramatic effect. New ROIC would be 36.5% Q6.3 Philly Air PA1. Draw an ROIC (return on invested capital) tree for the company that incorporates all of the above information.

PA2. What is the current ROIC? Revenues per year = load factor × 50 × 2 × 24 × 365 × 100 = 61,320,000 Costs per year = 60,000,000 Returns = Revenues per year – Costs per year = 1,320,000 ROIC = Returns per year / Invested Capital = 1, 320,000 / 5,000,000 = 26.4%

PA3. What is the minimum load factor at which the company breaks even?

The load factor at which the company breaks even satisfies Returns = load factor × 50 × 2 × 24 × 365 × 100 − 60,000,000 = 0. This gives load factor = 60,000,000 / (50  2  24  365  100) = 68.49%

PA4. What load factor would the company have to achieve so that it obtained a ten percentage point increase in the ROIC (e.g. an ROIC increasing from 5% to 15%)? One needs to solve for (load factor  50  2  24  365 100 − 60,000,000) / 5,000,000 = 36.4% ,

which gives load factor = 70.57 %

Q6.4 Oscar’s office building

Q6.5 OPIM Bus Inc OB1. Draw an ROIC (return on invested capital) tree for the company. This is a very top heavy tree, in which only the revenue branch is broken up in detail. Costs and invested capital are given in the question. Revenue: 2 vehicles * 24 trips * 35 seats * $10 * 365 days = 6,132,000 Profit: 6,132,000−6,000,000 = 132,000 OB2. What is the current ROIC? © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

ROIC = 3,132, 000 / 500, 000 = 0.264

OB3. What is the minimum load factor at which the company breaks even? 365*2*24*50*load factor * 10=6,000,000 Load factor = 68.49% OB4. By how much will the firm have to increase the load factor to achieve a ten percentage point increase in the ROIC (e.g. from 5% to 15%)? ROIC increases to 0.364 365*2*24*50*load factor * 10−6,000,000 = 500,000*0.364 Load factor = 0.7057 (load factor needs to increase from 0.70 to 0.7057) Corresponding

Matching Supply with Demand: An Introduction to Operations Management 5e Solutions to Chapter Problems Chapter 7 Quality and Statistical Process Control Q7.1 Pizza The time that a customer leaves the pizza in the refrigerator is outside the control of the firm and hence is more of an environmental variable than an input variable.

Q7.2 MakeStuff Answer C. The process capability increases because we are lowering our standards by allowing for a broader specification interval and hence a larger capability score. But, nothing has changed other than the measurement, so this will not improve the true quality. Q7.3 Precision Machining a) We compute the capability score as C_p = (1.01− 0.99) /(60.005) = 0.667 b) To move to a six sigma process, the capability score has to be increased to C_p=2. To find the sigma that accomplishes this, we solve C_p = (1.01− 0.99) /(6  sigma) = 2 , which yields Sigma = 0.02 /12 = 0.00167 Q7.4 Vetro In © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

a) We compute the capability score as C_p = (5.04 − 0.4.96) /(60.01) = 1.333 b) To move to a six sigma process, the capability score has to be increased to C_p=2. To find the sigma that accomplishes this, we solve C_p = (5.04 − 0.4.96) /(6  sigma) = 2 , which yields Sigma = 0.0066667 Q7.5 Cyclo Cross a) We compute the capability score as C_p = (23.2 − 22.8) /(60.25) = 0.26667 b) To move to a six sigma process, the capability score has to be increased to C_p=2. To find the sigma that accomplishes this, we solve C _ p = (23.2 − 22.8) /(6  sigma) = 2 , which yields Sigma = 0.03333

Q7.6 Quality Check Answer B. Because of the high costs of paint, inspection makes sense before the first step. Step 2 is the bottleneck and we should not waste scarce capacity on defects there, so another inspection step is warranted at that point.

Matching Supply with Demand: An Introduction to Operations Management 5e Solutions to Chapter Problems Chapter 8 Lean Operations and the Toyota Production System Q8.1 Waste Consider each of the activities in the work of the waiter. Again, note that we can have arguments about any one of them: – Waiting for a customer order: this is wasteful as it leaves the waiter idle Taking the order: taking orders is probably what waiters are supposed to do and hence is value add work. You might argue that this could be done cheaper with an app; that is entirely a matter what the customer values. Most likely depends on the customer – Waiting for the kitchen to confirm: again, this is idle time. Also, confirmation does not help add value to the customer – Bringing the food: this really is transportation and hence waste, though unless the customer is eating in the kitchen, this will be required. – Serving the customer: most likely value add – Collecting payment: again, most likely value add, though some of us would prefer to pay through an app Q8.2 Push a) In a push process, the first machine would run at capacity and thus u_1=100% © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

b) The pull system would slow down the first machine to the bottleneck to avoid unnecessary inventory; the utilization would drop to u_1=60% Q8.3 Three Step a) As a push system, inventory would pile up in front of steps 2 and 3 as both of them have a faster (higher capacity) step supplying them b) In a pull system, we would slow down the production rate of steps 1 and 2 to the rate of step 3 or the demand rate, whichever is lower Q8.4 Heijunka Answer b To achieve a mixed model production requires shorter set-ups. Other actions, including Kaizen and Jidoka and Poka Yoke might be welcome, but are less critical at this point. Q8.5 ITAT For process 1, the defective flow unit needs to journey through the remaining 11 stations. The flow rate is 1 unit per minute and the inventory is 300 units. So, it will take 300 minutes until the defective unit is detected at inspection. For process 2, the defective unit will be detected at station 10. There are about 5 units of inventory between step 9 and 10 (50 units over 10 buffers). So the time to detection is about 5 minutes. Q8.6 Jidoka Answer g None of these descriptions captures the idea of Jidoka. Q8.7 Kanban Answer A. In a Kanban system, work needs to be authorized by Kanban cards.

Matching Supply with Demand: An Introduction to Operations Management 5e Solutions to Chapter Problems Chapter 9 Variability and Its Impact on Process Performance: Waiting Time Problems Q9.1 Online retailer a) m = 3, a = 2 min, p = 4 min, CVa = 2 / 2 =1 , CVp = 2 / 4 = 0.5 . Utilization = p / (a m) = 4 / (23) = 0.6667 . Tq = (4/3) × [0.6667 𝖠 sqrt((2 × (3 + 1)) − 1)/(1 − 0.6667)] × [(1𝖠2 + 0. 5𝖠2) / 2] = 1.19 min . © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

b) Iq = (1/ a) Tq = (1/ 2) 1.19 = 0.60 . Ip = (1/ a) p = 4 / 2 = 2 . Total emails in the system that have not been processed is Iq + Ip = 2.59 Q9.2. My Law a) inter-arrival time: 10 emails per hour = 1 email / 6 min a = 6 min, CVa = 1 processing time: p = 5 min, CVp = 4 min / 5 min = 0.8

(

)

waiting time = 5 min[5 / 6 /(1− 5 / 6)]   12 + 0.82 / 2 = 20.5 min total response time = waiting time + processing time = 25.5 min

b) emails per day = 10 emails per hour * 10 hours = 100 emails per day c) idle time = (1- utilization) * 10 hours = 1.66 hours d) The average amount of time would not change, since utilization is not dependent on the variance or standard deviation, but on the average processing and inter-arrival time. e) processing time: p = 5 min, CVp = 0.5 min / 5 min = 0.1

(

)

waiting time = 5 min [5 / 6 /(1− 5 / 6)]   12 + 0.12 / 2 = 12.63 min total response time = waiting time + processing time = 17.63 min

Q9.3. Car Rental Company a) We approach this problem as though the rental car is the “server”. We know that a = 2.4 hours, p = 72 hours, CVa = (2.4 / 2.4) = 1, CVp = (24 / 72) = 0.33 , and m = 50 cars. To determine the number of cars on the lot, we can look at the utilization rate of our "servers" = (1/ a) /(m/ p) = (1/ 2.4) /(50 / 72) = 60% . Therefore, on average 60% of the cars are in use or 30 cars, so on average 20 cars are in the lot. b) We assume that the standard deviations DO NOT change. If the average demand is increased to 12 rentals per day, then a = 2 hours. If the average rental duration increases to 4 days, then p = 96 hours. These values raise the utilization rate to (1/ 2) / (50 / 96) = 96% . This means that 48 cars are rented on average. With the initial rate average revenue per day = $8030 cars = $2400 . With the proposed rate average revenue per day = $55 48 = $2640 . Therefore, the company should make the proposed changes. c) Using the wait time formula, the average wait time = 0.019 hours or 1.15 minutes. d) Using the wait time formula the average wait time is computed as 0.0462 hours. 2 Time in queue = Tq =

 Activity time    utilization 2(m+1) −1    CVa + CV p     1− utilization   m  2       

Activity time = 96 hours Utilization = p / (ma) = 96 / (50 3) = 64%

CVp = 0! Plugging it all in we get Tq = .0459 hours Q9.4. Tom Opim a) idle time = 1 min / 3 min 8 hours =160 min pages read = 160 min 1 page / min = 160 pages b)

inter-arrival time:1 call / 3 min a = 3 min, CVa = 3 min / 3 min =1 processing time: p = 2 min, CVp =1 min / 2 min = 0.5

waiting time = 2 min[2 / 3 / (1− 2 / 3)][(12 + 0.52) / 2] = 2.5 min c) The average total time a line is used per customer = average wait time + average processing time. In this case, the average total time per customer = 2.5 + 2 = 4.5 minutes per customer. There are an average of 20 customers per hour, so the average number of minutes per hour = 20*4.5 = 90 minutes. Thus, the total per hour charge = (90 / 60) $5 = $7.50 per hour or $60 for 8 hours. Another way to approach the same problem is to look at the average number of callers at any given time = average number of callers on hold at any given time ( Iq ) + average number of callers talking to Tom at any given time. We can calculate Iq = R * Tq where the flow rate =1 call / 3 minutes and Tq = 2.5 minutes. Thus, Iq = 0.83 calls. The average number of callers talking to Tom must be a number between 0 and 1, and is equal to Tom’s utilization = 0.67. So, the average number of callers at any given time = 0.83 + 0.67 = 1.5 callers. The line charge for 8 hours = $58 = $40 per line . Therefore the total cost over an 8-hour shift =1.540 = $60 .

Q9.5. Atlantic Video a) inter-arrival time: 30 customers per hour =1 customer / 2 min a = 2 min, CVa = 1 process time: p = 1.7 min, CVp = 3 min /1.7 min =1.765 waiting time = 1.7 min [(1.7 / 2) /(1−1.7 / 2)][(12 +1.7652 ) / 2] = 19.82 min b)

utilization = 1.7 min / 2 min = 0.85 idle time = 0.158 hrs = 72 min /1.5 min per sort = 48 sorts

c) Using R = minimum of 1/ a and 1/ p , we calculate R = 0.5. Thus, the average number of customers in line waiting I = R *Tq = 0.5*19.82 = 9.9 customers. In addition, an average of .85 customers (equal to the utilization u) are being served at any given time. So the average number of customers at the check-out desk = 9.9 + .85 = 10.75. © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

d) Because only 90% of customers go through checkout, the inter-arrival time of paying customers changes: 27 customers per hour = 1 customer / 2.22 min

(

)

waiting time = 1.7 min [(1.7 / 2.22) /(1−1.7 / 2.22)]   12 +1.7652 / 2 = 11.38 min

e) The average person waits 19.81 minutes, and 30 customers arrive in one hour, so there are approximately 19.82*30 = 594.6 minutes of wait time per hour. This costs the store .75594.6 = $445.95 for wait time. If 2 servers are used, we can apply the formula, and using the same methodology, calculate a cost of .75(0.8830) = $19.79. Finally, if 3 servers are used, the cost is .75 (0.16230) = $3.65 . Adding any more employees would not be cost effective. The most cost-effective number of employees is 3. Q9.6. Rent-a-Phone a) To answer this question, we must first set up the problem where we consider the phones as “servers”. Thus m = 80, a = 24 hours / 25 customers = 0.96 hours average interarrival time, and p = 72 hours. So, utilization = p / ( m  a ) = 72 / (80.96) = 94% . This means that .94*80 phones = 75 phones are in use, and 5 phones are available on average. b) We are given that CVa = 1 and CVp =100 / 72 =1.39 . Using the wait time formula, we calculate an average wait time of 9.89 hours. c) We first need to calculate the average number of people in the queue. We know that the flow rate R = demand rate =1/.96 =1.04 customers / hour . Time in the queue Tq = 9.89 hours, so Iq = Tq * R = 9.89 * 1.04 = 10.31 people in the queue on average. So we can multiply $1 24 hours/day 30 days 10.31 people in the queue = $7419.87 . d) We can repeat the same calculations for m = 81 and obtain Iq = 7.379 . The new expenses would be 5312, thus it would pay to buy at least one additional phone. e) We repeat the waiting time calculations with Cp = 0 and a = 24 / 20 ; we now get a utilization of u = 0.75 and a waiting time of Tq = 0.06166 hours Q9.7. Webflux a) Demand interarrival time = a = 10, service time = p = 7. Utilization = ( flow rate) / ( capacity) = (1/ a) / (1/ p) = p / a = 0.7 . Time in queue = ( service time) ( utilization / (1-utilization)) 

((CV + CV ) / 2) = 7  2

(0.7 / 0.3) ((1+1) / 2) = 16.33 days . Adding the shipping time, the answer is 17.33 days. b) Queue length = ( time in queue) / (1/ (flow rate)) = 16.33 / a = 16.33 /10 = 1.63 . c) Number of servers = m = 2. r = ( implied utilization)( number of servers) = p / a = 7 / 3 = 2.333 . From the Erlang Loss Table, Pm = Prob{all 2 servers are busy = all 2 DVDs are rented out} = 0.4495.

Q9.8. Security Walking Escorts a) We know that a =5, p = 25, and m=8. Utilization = (1/ a) / (m / p) = (1/ 5) / (8 / 25) =.625 . Therefore, 8*(1-.625) = 3 officers available. b) The average wait time = 1.8 minutes. The average time to walk to the destination = 25 minutes. Therefore, the total time between calling and arriving = 25+1.8 = e. 26.8 minutes. c) Again, we know that a = 3.125, m = 8, and p = 25. We also know that if the coefficient of variation = 1 then interarrivals follow an exponential distribution. We can also calculate r = p / a = 25 / 3.125 = 8 . So we use the Erlang Loss table to calculate P8(8) = approximately 0.2356. Since 19.2 students/hr request an escort, 19.20.2356 = 4.5 students / hour find no escort available. d) Again, using the Erlang Loss table we see that to have .20 or less, we need m=9. Therefore we need 9 officers. Q9.9. Mango Electronics a) We know that a = 7 months, p = 28 months, and m = 5 centers. Therefore, utilization = (1/ a) / (m / p) = (1/ 7) / (5 / 28) =.80 or 80%. b) Average wait time = 40.4 months. The average development time = 28 months. Therefore the total time = 40.4 + 28 = 68.4 months. c) The patent is given for 20 years. If the entire development process = 68.4 months, or 5.7 years, then there will be 14.3 years left.

Matching Supply with Demand: An Introduction to Operations Management 45e Solutions to Chapter Problems Chapter 10 The Impact of Variability on Process Performance: Throughput Losses Q10.1 Loss System a) The interarrival time is 60 minutes per hour divided by 55 units arriving per hour, which is an interarrival time of a =1.0909 minutes / unit . The processing time is p = 6 minutes/unit ; this allows us to compute r = p / a = 6 /1.0909 = 5.5 . With r = 5.5 and m = 7, we can use the Erlang Loss Formula Table to look up Pm(r) = P7(5.5) as 0.1525. b) Compute the flow rate: R = 1/ a (1− Pm) = 1/1.0909 (1− 0.153) = 0.77 unit per minute or 46.585 units per hour. c) Compute lost customers: Customers lost 1/ aPm =1/1.09090.153 = 0.14 unit per minute which corresponds to 8.415 units per hour. Q10.2. Home security We have m = 5, an interarrival time of a = 5 minutes and a service time of p = 90 minutes. a) We compute r = p / a = 18 ; we then use the Erlang loss table to find that the probability that all 5 servers are utilized, P5 (18) = 0.7402 .

b) We compute the loss rate of the system, since this corresponds to the calls that will be handled by the police. The loss rate is 0.1480 calls per minute, which is 6395 calls per month. Since each of the calls would lead to a 500$ charge by the police, we would have monthly payments of 3.2 million $. What a dull business idea! Q10.3. Video Store a) We have m = 9, a = 24 /15 hours and p = 36 hours. r = p / a = 22.5 b) We use the Erlang loss table to find that all servers will be utilized with probability P9 (22.5) = 62.44% . That means that there will be a movie available with probability 37.56%. c) The flow rate is 0.235 movies per hour. Over the course of a day (and at 5$ per movie), this brings in 28.166 $/ day revenue d) The loss rate is 0.39 customers per hour; since each of them gets $1, that means a loss of $9.36 a day. e) We repeat all calculations with m = 10. We find that we get new (higher) revenues of 31.19 $/day and new (lower) losses of 8.76 $/ day . Thus, we increase profits by $3.62 per day. In less than 14 days, we have then recovered our investment for the new copy of the movie. Q10.4. Gas station a) 20 cars want to fill in per hour. This leaves us with an interarrival time of a = 60 / 20 minutes = 3 minutes . The service time is p = 5 minutes (r = 1.667) and we have m = 6 servers. All servers will be busy with a probability of 0.0056 b) The flow rate is 19.89 cars per hour Q10.5. Two work stations a) For both, scenario 1 and scenario 3, we obtain a flow rate of 12 units per hour (one unit every 5 minutes). Scenario 3 will not be able to run faster than this, since the bottleneck is at the first resource and has a capacity of 12 units per hour. Scenario 1 and 3 will flow at one unit every 5 minutes on average. In both cases the flow is constrained by step 1. In scenario 2, it is the 50% of the time that the processing time at step 2 is 6 minutes that causes the problem. Since there is no buffer, sometimes step 1 will be blocked. b) Scenario 3 has the lowest flow time. Given that it is never above 10 minutes, but sometimes (if station B takes 3 minutes) below, the average flow time will be below 10 minutes. In scenario 1, the flow time is always exactly 10 minutes. In scenario 2, it will be longer than 10 minutes, as occasionally (when B takes 6 minutes), station A will be blocked, leading to longer than 5 minutes flow time at A. At station B, the flow time in this scenario will – on average – be 5 minutes. There is no buffer and the flow time is driven only by the processing times. The shortest expected processing time is 5 + 0.5 * (3 + 5) = 9 minutes in scenario 3. Q10.6. Xtremely Fast Service a) r = p / a = 5 / 3 . Given that number of servers, m = 4, the Erlang Loss Table yields a probability of 0.0624. b) Flow Rate = demand rate * probability that not all servers are busy * price = © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

1 customer /3 minutes * (1-0.0624) * 60 min/hr * 8 hr/day * $1 / call = $150 . c) Flow rate = demand rate * probability that all servers are busy * cost = 1/3 * (0.0624) * 60 * 8 * $5/day = $49.92 , which is closest to $50.

Q10.7. Gotham City Ambulance Services a) M = 8 ambulances; processing time p = 90 minutes; interarrival time a = 15 minutes; r = p / a = 6 ; fraction of emergencies handled by neighboring community: 12.19% b) There are 3.514 emergencies served per hour, i.e. 84.2976 per day. c) Reducing the coefficient of variation has no effect whatsoever on the number of emergencies they can attend. d) They would have to buy 2 additional ambulances.

Matching Supply with Demand: An Introduction to Operations Management 5e Solutions to Chapter Problems Chapter 11 Scheduling to Prioritize Demand Q11.1 Answer = 3. 10 jobs in 200 minutes yields a flow rate of 10 jobs / 200min = 0.05 jobs per minute, which is 60 min per hr × 0.05 jobs per min = 3 jobs per hour Q11.2 Answer = 0.5 projects / week . 5 projects are completed over 3 × 1 + 3 + 4 = 10 weeks, so the flow rate is 5 projects /10 weeks = 0.5 projects / week . Q11.3 a) Answer = C. Jobs are sequenced E, D, A, C, F, B b) Answer = 25. F is the 5th job processed because it has the 5th largest processing time. Its flow time is then the sum of the five smallest processing times, 1+2+5+8+9 c) Answer =1/ 6 . 6 jobs are completed in a total of 36 minutes, so the flow rate is 6 / 36 jobs per minute. d) Answer = 14.8. The flow time of the 6 jobs are 1, 1+2 = 3, 1+2+5=8, 1+2+5+8 = 16, 1+2+5+8+9 = 25, and 1+2+5+8+9+11 = 36. The average flow time is then (1+ 3 + 8 +16 + 25 + 36) / 6 = 89 / 6 = 14.8 minutes e) Answer = 2.47. To evaluate inventory we need the flow rate and flow time, which we evaluated in the previous questions. Hence, Inventory = Flow Rate  Flow Timp = 1/ 6 jobs per min14.8 minutes = 2.47 jobs . Q11.4 a) Answer = 0.3556. Total processing time is 4 +1 +2 … + 0.5 = 22.5hrs. 8 jobs completed in 22.5hrs yields a flow rate of 8 jobs / 22.5hrs = 0.3556 jobs / hr b) Answer = 13.4. Flow times are 4, 4+1 = 5, 4+1+2 = 7, etc. The average flow time is (4 + 5 + 7 ++ 22.5) / 8 = 13.4 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

c) Answer = 4.78. I = R × T = 0.3556 × 13.4 = 4.78 Q11.5 a) Answer = C. The weight/processing time ratios are 1.5 /1 =1.5 , ¼ = 0.25 , 1.25 / 2 = 0.625 , 0.5 / 3 =1/ 6 , 2 / 5 = 0.4 . The 2nd highest weight is job C’s. b) Answer = 0.333 jobs /min. The total of the processing times is 15 min. 𝑅 = 5 jobs/15 min = 0.333 jobs/min c) Answer = 7.8 min. Sequence the jobs in WSPT yields the sequence A, C, E, B, D. The flow times are 1, 1+2 = 3, 1+2+5 = 8, etc. The average flow time is T = (1 + 3 + 8 + 12 + 15)/5 = 7.8 min. jobs d) Answer = 2.60 jobs. 𝐼 = 𝑅 × 𝑇 = 0.333 × 7.8 min = 2.60 jobs . min

Q11.6 Answer = A. The ratios of weights to processing times are 5 /15 =1/ 3, 10 / 5 = 2 , 1/25 , and rd 15 / 30 = 12 . The 3 largest ratio is 1/ 3 , from job A.

Plane A B C D E Passengers 150 50 400 200 250 TABLE TAKEOFF: Number of passengers on 5 planes ready for departure. Q11.7 Answer = d. They should use Weighted Shortest Processing Time. Because they all have the same processing times, they should depart in decreasing order of the number of passengers on the plane. Q11.8. a) Answer = 15. Using EDD, job C is processed 4th because it has the fourth due date. Thus, it is completed at time 20+25+5+40 = 90. It is due on day 75, so its lateness is 9075 = 15. b) Answer = -10. Using EDD, job E is processed 1st because it has the earliest due date. Thus, it is completed at time 20 but it is due at time 30, meaning that it’s lateness is 20-30 = -10. c) Answer = 0. Using EDD, job D is processed 3rd because it has the third due date. It is completed at time 20+25+5 = 50. It is due at time 55, so it is completed before its due date. Hence its tardiness is zero. d) Answer = 15. Given EDD, the jobs are processed in the following sequence E, B, D, C, A. The jobs are completed at times, 20, 20+25=45, 20+25+5 = 50, 20+25+5+40 = 90, 20+25+5+40+10 = 100. The due dates are 30, 40, 55, 75 and 100. Thus, only jobs B and C are tardy. Job B is tardy by 5 days and job C is tardy by 15 days, so the maximum tardiness is 15. Q11.9 a) Answer = 2 b) Answer = 0. © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

c) Answer = 2.5 d) Answer = 5

Matching Supply with Demand: An Introduction to Operations Management 5e Solutions to Chapter Problems Chapter 12 Project Management Q12.1 Venture Fair a) The dependency matrix consists of 11 rows and 11 columns.

b) The critical path is given by the longest path through the network. As can be seen in the table below, path 4 (A1-A2-A4-A6-A7-A9-A10-A11) is the longest and will take 46 days to complete.

Since project completion will take 46 days, once the project has been started, we have to take a calendar, pick April 18, and go 46 days forward. That gets us to March 3. Q12.2 10 Activities a) The project graph is as follows:

b) There are three paths through the network of the 10 activities. The three paths are as follows: • A1-A2-A4-A10 • A1-A2-A5-A7-A9-A10 • A1-A3-A6-A8-A9-A10 The durations of each of these three paths are computed in the table below.

The third path (A1-A3-A6-A8-A9-A10) is the critical path. c) From this, we can compute the slack times as follows:

Q12.3 Graduation Party a) The project graph looks like this:

We can also display this information in a dependency matrix: Finals Music Evaluate Guest Visit Hotel Sites list sites Finals Music X Evaluate X sites Guest X list Visit X sites Hotel X T-shirt X Event X planner Beach X X

T-shirt

Event planner

Beach

b) The critical path is Final Exams – Evaluate Sites – Visit Sites – Event Planner – go to Beach. Thus, they can arrive at the beach 19 days after the beginning of their final exams. c) The slack time calculations are as follows: LCT Finals 5 Music 19 Evaluate 11 sites Guest 13 list

LST 0 14 5

Slack 0 9 0

Visit sites Hotel T-shirt Event planner Beach

19 19 19

16 13 15

4 1 0

Q12.4 Three Activities with Uncertainty It takes an expected 5.5 days. Note that this is longer than one would expect based on the average completion time of activities A and B (both 4 days) followed by activity C (one day). The reason for this is that in order for Activity C to start after 3 days, both A and B have to be early. This only happens in 25% of the cases. In 75% of the cases, Activity C is starting late (after 5 days) leading to a completion time of 6 days. The overall expected completion time is thus .25*4 + 0.75*6 = 5.5 days.

Matching Supply with Demand: An Introduction to Operations Management 5e Solutions to Chapter Problems Chapter 13 Forecasting Q13.1 Jim and John The calculations in the table below show that Jim has an average forecast error of ˗3 while John has one of 0. So, John is biased, while Jim is not. Jim has an MSE of 35.5 and an MAE of 4.5. John has an MSE of 52 and an MAE of 6. Day Jim John Actual Jim: FE Jim: FE2 Jim: |FE| John: FE John: FE2 John: |FE| 1 56 47 45 -11 121 11 -2 4 2 2 50 49 51 1 1 1 2 4 2 3 45 51 41 -4 16 4 -10 100 10 4 59 51 61 2 4 2 10 100 10 Average -3.0 35.5 4.5 0.0 52.0 6.0 Q13.2 Tom’s Towing 1 The naïve forecast would be 15, as this corresponds to the last observed demand. Q13.3 Tom’s Towing 2 With a four-day moving average, we would forecast the demand for Friday as the average of the demand for Monday, Tuesday, Wednesday and Thursday. That would be: Forecast for Friday = Average(27, 18, 21, 15) = 20.25 Q13.4 Tom’s Towing 3 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 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

for the next day. This gives us forecast for Friday of 18.2976.

Q13.5 Online-MBA We forecast the demand for the next year as follows:

So, the forecast for the next period is 776.8. Q13.6 GoPro We forecast the demand for the next year as follows: Season Quarter average SI 1 128 2 117 3 350 4 315 Overall 227.5 alpha

0.2

0.563 0.514 1.538 1.385

Deseasonalized Smoothed demand demand 197.29 143.46 233.33 161.43 227.50 174.65 240.50 187.82 231.05 196.46 211.94 199.56 194.35 198.52 220.28 202.87 254.16 213.13 237.22 217.95 260.65 226.49 221.72 225.53

Forecast 2015Q1 2015Q2 2015Q3 2015Q4

126.89 115.99 346.98 312.28

Q13.7 Four Partners The forecast combination is simply the average of the four forecasts. In this case 37.

Matching Supply with Demand: An Introduction to Operations Management 5e Solutions to Chapter Problems Chapter 14 Betting on Uncertain Demand: The Newsvendor Model Q14.1 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

a) First find the z-statistic for 400 (Dan’s blockbuster threshold): z = (400 − 200) / 80 = 2.50 . From the Standard Normal Distribution Function Table we see that (2.50) = 0.9938 . So there is a 99.38% chance demand is less than 400 or fewer. It is greater than 400 with probability 1 − (2.50) = 0.0062 , i.e., there is only a 0.62% chance this is a blockbuster. b) First find the z-statistic for 100 units (Dan’s dog threshold): z = (100 − 200) / 80 = −1.25 . From the Standard Normal Distribution Function Table we see that (−1.25) = 0.1056 . So there is a 10.56% chance demand is less than 100 or fewer, i.e., a 10.56% chance this book is a dog. c) Demand is within 20% of the mean if it is between 1.2 200 = 240 and 0.8 200 = 160 . First find the z-statistic for 240 units (the upper limit on that range): z = (240 − 200) / 80 = 0.5 . From the Standard Normal Distribution Function Table we see that (0.5) = 0.6915 . Repeat the process for the lower limit on the range: z = (160 − 200) / 80 = −0.5 and (−0.5) = 0.3085 . The probability demand is between 160 and 240 is (0.5) − (−0.50) = 0.6915 − 0.3085 = 0.3830 , i.e., 38.3% d) The underage cost is Cu = 20 −12 = 8 . The salvage value is 12 – 4 = 8, because Dan can return left over books for a full refund (12) but incurs a 4 cost of shipping and handling. Thus, the overage cost is cost minus salvage value: Co = 12 − 8 = 4 . The critical ratio is Cu / (Co + Cu ) = 8 / 12 = 0.6667 . In the Standard Normal Distribution Function Table we see that (0.43) = 0.6664 and (0.44) = 0.6700 , so use the round-up rule and choose z = 0.44 . Now convert z into the order quantity for the actual demand distribution: Q =  + z  = 200 + 0.44  80 = 235.2 . e) We want to find a z such that (z) = 0.95 . In the Standard Normal Distribution Function Table we see that (1.64) = 0.9495 and (1.65) = 0.9505 , so use the round-up rule and choose z = 1.65 . Now convert z into the order quantity for the actual demand distribution: Q =  + z  = 200 +1.65  80 = 332 . f) If the in-stock probability is 95%, then the stockout probability (which is what we are looking for) is 1 minus the in-stock, i.e., 1 – 95% = 5%. g) The z-statistic for 300 units is z = (300 − 200) / 80 = 1.25 . From the Standard Normal Inventory Function Table we see that I (1.25) =1.3006. Expected leftover inventory is I (1.25) =104.0. Expected sales = 300 – 104.0 = 195.95 Expected profit = ( Price - Cost )  Expected sales −(Cost − Salvage value) Expected left over inventory = (20 −12)195.95 − (12 − 8)104.05 = 1151.4

Q14.2 a) We need to evaluate the stock-out probability with Q = 3. From the Poisson Distribution Function Table, F(3) = 0.34230. The stock-out probability is 1 – F(3) = 65.8%. b) They will need to markdown 3 or more baskets if demand is 7 or fewer. From the © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Poisson Distribution Function Table, F(7) = 0.91341, so there is a 91.3% probability this will occur. c) First evaluate their critical ratio. The underage cost (or, cost of a lost sale), is $55 −$32 = $23. The overage cost (or, the cost of having a unit left in inventory) is $32 −$20 = $12 . The critical ratio is Cu / (Co + Cu ) = 0.6571 . From the Poisson Distribution Function Table, with a mean of 4.5, we see that F(4) = 0.53210 and F(5) = 0.70293, so we apply the round-up rule and order 5 baskets. d) With 4 baskets Expected Leftover Inventory is 0.58808. Expected Sales = 4 – 0.58808 = 3.4. e) With 6 baskets Expected Leftover Inventory is 1.823. f) From the Poisson Distribution Function Table F(6) = 0.83105 and F(7) = 0.91314. Hence, order 7 baskets to achieve at least a 90% in-stock probability (in fact, the instock probability will be 91.3%) g) If they order 8 baskets then Expected Sales is 4.5 – 0.06758 = 4.43242 and Expected Left Over Inventory is 3.56758. Profit is then $23 4.43242 − $123.56758 = $59.13. Q14.3 a) If they purchase 40,000 units, then they need to liquate 10,000 or more units if demand is 30,000 units or lower. From the table provided, F(30000) = 0.7852 , so there is a 78.52% chance they need to liquidate 10,000 or more units. b) The underage cost is Cu = 12 − 6 = 6 , the overage cost is Co = 6 − 2.5 = 3.5 and the critical ratio is 6 /(3.5 + 6) = 0.6316 . Looking in the demand forecast table we see that F(25000) = 0.6289 and F(30000) = 0.7852 , so use the round-up rule and order 30,000 Elvis wigs. c) We want to find a Q such that F(Q) = 0.90 . From the demand forecast table we see that F(35000) = 0.8894 and F(40000) = 0.9489 , so use the round-up rule and order 40,000 Elvis wigs. The actual in-stock probability is then 94.89% d) From the table I(Q) = 25061 e) A 100% in-stock probability requires an order quantity of 75,000 units. Expected left over inventory is 75000 – 25000 = 50000. Evaluate expected sales = 75,000 – 50,000 = 25,000. Expected profit =( Price - Cost ) Expected sales

− (Cost − Salvage value) Expected left over inventory

=(12 − 6) 25000 − ( 6 − 2.5) 50000 =−25,000

So a 100% in-stock probability is a money losing proposition. Q14.4 a) It is within 25% of the forecast if it is greater than 750 and less than 1250. The zstatistic for 750 is z = (750 −1000) / 600 = −0.42 and the z-statistic for 1250 is z = (1250 −1000) / 600 = 0.42 . From the Standard Normal Distribution Function

Table we see that (−0.42) = 0.3372 and (0.42) = 0.6628 . So there is a 33.72% chance demand is less than 750 and a 66.28% chance it is less than 1250. The chance it is between 750 and 1250 is the difference in those probabilities: 0.6628 − 0.3372 = 0.3256 . b) The forecast is for 1000 units. Demand is greater than 40% of the forecast if demand exceeds 1400 units. Find the z-statistic that corresponds to 1400 units: Q −  1400 −1000 z= = = 0.67  600 From the Standard Normal Distribution Function Table, (0.67) = 0.7486 . Therefore, there is almost a 75% probability that demand is less than 1400 units. The probability demand is greater than 1400 units is 1 − (0.67) = 0.2514 , or about 25%. c) Find the expected profit maximizing order quantity, first identify the underage and overage costs. The underage cost is Cu = 121 − 72 = 49 , because each lost sale costs Flextrola its gross margin. The overage cost is Co = 72 − 50 = 22 , because each unit of left over inventory can only be sold for 50. Now evaluate the critical ratio Cu = 49 = 0.6901 . Co + Cu 22 + 49 Lookup the critical ratio in the Standard Normal Distribution Function Table: (0.49) = 0.6879 and (0.50) = 0.6915 , so choose z = 0.50. Now convert the zstatistic into an order quantity: Q =  + z  = 1000 + 0.5 600 = 1300 . d) If Q = 1200 , then the corresponding z-statistic is z = (Q −  ) /  = (1200 − 1000) / 600 = 0.33 . From the Standard Normal Distribution Inventory Table we see that I(0.33) = 0.5876. Expected leftover inventory is 600 × 0.5876 = 352.5. Expected sales = Q – Expected leftover inventory = 1200 – 352.5 = 846.7. e) Flextrola sells its left over inventory in the secondary market, which from the previous question is 352.5. f) To evaluate the expected gross margin percentage we begin with Expected revenue = (Price  Expected sales ) +(Salvage value  Expected left over inventory ) = (121 846.7) + (50  353.3) = 120,116 Then we evaluate expected cost = Q  c = 1200 72 = 86, 400 . Finally, expected gross margin percentage =1−86, 400 /120,116 = 28.1% g) Use the results from parts d and e to evaluate expected profit: Expected profit = ( Price - Cost )  Expected sales −(Cost − Salvage value) Expected left over inventory = (121− 72) 846.7 − (72 − 50) 353.3 = 33716 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

h) Solectric’s expected profit is 1200  (72 − 52) = 24,000 because units are sold to Flextrola for 72 and each unit has a production cost of 52. i) Flextrola incurs 400 or more units of lost sales if demand exceeds the order quantity by 400 or more units, i.e., if demand is 1600 units or greater. The z-statistic that corresponds to 1600, z = (Q −  )/ = (1600 − 1000)/ 600 = 1 . In the Standard Normal Distribution Function Table, (1) = 0.8413 . Demand exceeds 1600 with the probability 1 − (1) = 15.9% . j) The critical ratio is 0.6901. From the graph of the distribution function we see that the probability demand is less than 1150 with the Log Normal distribution is about 0.70. Hence, the optimal order quantity with the Log Normal distribution is about 1150 units.

Q14.5 a) The underage cost is Cu = 70 − 40 = 30 and the overage cost is Co = 40 − 20 = 20 . The critical ratio is Cu /(Co + Cu ) = 30 / 50 = 0.6 . From the Standard Normal Distribution Function Table, (0.25) = 0.5987 and (0.26) = 0.6026 , so we choose z = 0.26 . Convert that z-statistic into an order quantity Q =  + z  = 500 + 0.26  200 = 552 . Note that the cost of a truckload has no impact on the profit maximizing order quantity. b) We need to find the z in the Standard Normal Distribution Function Table such that (z) = 0.9750 because (z) is the in-stock probability. We see that (1.96) = 0.9750 , so we choose z = 1.96 . Convert to Q =  + z  = 500 +1.96  200 = 892 . c) If 725 units are ordered, then the corresponding z-statistic is z = (Q −  )/ = (725 − 500) / 200 = 1.13 . Expected leftover inventory is 200 × I(z) = 238.1. Expected sales = Q – Expected leftover inventory = 725 – 238.1 = 486.9. Expected profit is (70-40) × 486.9 + (40-20) × 238.1 = 9,847. The total expected profit is 5 × 9,847 – 2000 = $47,235 d) The stockout probability is the probability demand exceeds the order quantity 725, which is 1 − (1.13) = 12.9% . e) If we order the expected profit maximizing order quantity for each sweater, then that equals 5  552 = 2,760 sweaters. With an order quantity of 552 sweaters expected left over inventory is 108.5 and expected sales is 552 – 108.5 = 443.5 and Expected profit per sweater is Expected profit = (70 − 40) 443.5 − (40 − 20)108.5 = 11,135 Because two truckloads are required, the total profit is then 511,136 − 2 2000 = 51, 675 . If we order only 500 units per sweater type, then we can evaluate the expected profit per sweater to be 11,010. Total profit is then 511,010 − 2000 = 53,050. Therefore, we are better off just ordering one truckload © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

with 500 sweaters of each type. Q14.6

a) The parka sells less than half of the forecast if demand is 2100 / 2 =1050 or fewer units. Normalize the quantity 1050: z = (1050 − 2100 )/1200 = −0.88. From the Standard Normal Distribution Function Table, (−0.88) = 0.1894 , which implies there is a 18.9% probability that the parka will be a dog. b) To determine the profit maximizing order quantity, begin with the underage cost, Cu = 22 −10 = 12 , and the overage cost, Co = 10 − 0 = 10. The critical ratio is 12 /(10 +12) = 0.5455 . We see from the Standard Normal Distribution Function Table that (0.11) = 0.5438 and (0.12) = 0.5478 , so we choose z = 0.12 . Convert that z-statistic back into an order quantity, Q =  + z  = 2100 + 0.12 1200 = 2,244. To hit the target in-stock probability of 98.5%, we need to find the z-statistic such that (z) = 0.9850 . We see from the Standard Normal Distribution Function Table that (2.17) = 0.9850 , so we choose z = 2.17 . Convert to Q: Q = 2100 + 2.17 1200 = 4704 . d) If 3000 parkas are ordered then the corresponding z-statistic is (3000 − 2100)/1200 = 0.75 . Expected leftover inventory = 1200 × I(z) = 1057.4. Expected sales = 3000 – 1057.4 = 1942.5 Expected profit = (22 −10)1942.6 − (10 − 0)1057.4 = 12, 737 e) The stock out probability is 1− (z) = 1− (0.75) = 22.7%

Q14.7 a) The underage cost is Cu = 54 − 40 = 14 , and the overage cost, Co = 40 − 54 / 2 = 13. The critical ratio is 14 /(13 +14) = 0.5185 . We see from the Standard Normal Distribution Function Table that (0.04) = 0.5160 and (0.05) = 0.5199 , so we choose z = 0.05 . Convert that z-statistic back into an order quantity, Q =  + z  = 400 + 0.05  300 = 415. b) Expected left over inventory is 110. Expected sales is 380 – 110 = 270. Expected profit is (54-40) × 270 – (40-27) × 110 = $2,351 c) We first have to evaluate Teddy Bower’s expected profit with the optimal order quantity, 415 boots. Expected leftover inventory = 127.3, expected sales = 287.7 and expected profit = $2,372.4 Given that the optimal number of boots to order is 415, if Teddy Bowers is going to get the quantity discount then they should only order 800 boots. With an order quantity of 800 the z-statistic is z = (800 − 400) / 300 = 1.33 . Expected profit is $3, 257 . Therefore, it is in Teddy Bower’s interest to order 800 units to get the quantity discount. Q14.8 a) With option 1 Land’s End sales price is $100 , purchase cost is $65 and salvage value is $53 (because Geoff buys back unsold glasses for $53 ). So the underage cost is © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Cu = 100 − 65 = 35 and the overage cost is Co = 65 − 53 = 12 . The critical ratio is Cu / (Co + Cu ) = 35 / 47 = 0.7422 . From the Standard Normal Distribution Function Table we see (0.65) = 0.7422 and (0.66) = 0.7454 , so we choose z = 0.66. The optimal order quantity is then Q =  + z  = 200 + 0.66 125 = 282.5. b) With option 1 Land’s End sales price is $100 , purchase cost is $55 and the salvage value is $0. So the underage cost is Cu = 100 − 55 = 45 and the overage cost is Co = 55 . The critical ratio is Cu / (Co + Cu ) = 45 / 100 = 0.4500 . From the Standard

Normal Distribution Function Table we see (−0.13) = 0.4483 and (−0.12) = 0.4522 , so we choose z = −0.12. The optimal order quantity is then Q =  + z  = 200 − 0.12 125 = 185 . c) We need to evaluate expected profit in each case to determine which option Lands End should choose. With option 1, Geoff sells 282.5 units at $65 for total revenue of 18,363 and production cost of 282.5 2 = 7063. Geoff also credits Lands End for each returned sunglass, so we need to evaluate how many sunglasses Lands End will return. Expected left over inventory is 282.5 – 180.9 = 101.6. Expected sales is 282.5 – 101.6 = 180.9. Expected profit is then Expected profit = (100 − 65)180.9 − (65 − 53)101.6 = 5112 With option 2, expected profit is Expected profit = (100 − 55)142.28 − (55 − 0 )  42.72 = 4053 So Land’s End prefers option 1. d) If Land’s End chooses option 1 and orders 275 units, then Geoff’s earns $65  275 = $17,875 in initial revenue and incurs a production cost of $25  275 = $6875 . But Geoff also has to buy back left over glasses from Land’s End. With an order quantity of 275, the z-statistic is (275 − 200) /125 = 0.60 . Expected lost sales is then 125  L(0.60) = 21.09 . Expected left over inventory is Q −  + Expected lost sales , which is 275 − 200 + 21.09 = 96.09 . So Geoff’s buy back cost is 53 96.09 = 5093 . Geoff’s expected profit is then 17875 − 6875 − 5093 = 5907 . Q14.9 a) We first need to evaluate the overage and underage costs. The underage cost is Cu = 0.6 − 0.20 = 0.4 , i.e., it is the gross margin on each bagel. The overage cost is slightly more complex to evaluate. Assume Day Old bagels are sold for $0.165 each, but only about 2/3rds of them are sold. Hence, the average salvage value received on Day Old bagels is $0.165  (2 / 3) = $0.11. The overage cost is then Co = 0.20 − 0.11 = 0.09 because the average loss on each unsold bagel is 9 cents. The critical ratio is Cu / (Co + Cu ) = 0.40 / 0.51 = 0.8163 . In the Standard Normal

Distribution Function Table we see that (0.90) = 0.8159 and (0.91) = 0.8186 , so we choose z = 0.91. Convert that z-statistic to a quantity, © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Q =  + z  = 54 + 0.91 21 = 73.11. Hence, the store should have approximately 73 bagels to maximize its expected profit. b) If there is an additional $5 cost to a lost sales, then the underage cost is now Cu = 0.4 + 5 = 5.4 , i.e., it is the gross margin plus the additional penalty. The critical

ratio is now Cu / (C o + Cu ) = 5.40 / 5.51 = 0.9836 . In the Standard Normal Distribution Function Table we see that (2.13) = 0.9834 and (2.14) = 0.9838 , so we choose z = 2.14 . Convert that z-statistic to a quantity, Q =  + z  = 54 + 2.14  21 = 98.94 . c) If 101 bagels are in stock at 3pm., then the z-statistic is (101 − 54) / 21 = 2.24 . Expected left over inventory = 21 × I(z) = 47.09

Q14.10 a) The overage cost is Co = $2 because left over burritos are disposed. The underage cost is Cu = $2.55 , because not having a burrito means the gross margin is lost on a burrito plus a soda. The critical ratio is 2.55 / (2 + 2.55) = 0.5604 . From the table, F (22) = 0.5564 and F (23) = 0.6374 so the optimal number of burritos to make is 23. b) The overage cost remains the same, Co = $2 because left over burritos still are disposed. The underage cost is now Cu = $2 − 0.5 = $1.5 , because not having a burrito means the gross margin is lost on a burrito, but at least the gross margin on a Pop Tart is earned. The gross margin on the soda is captured either way, so it no longer figures into the analysis. The critical ratio is 1.5 / (2 +1.5) = 0.4286 . From the table, F (20) = 0.3869 and F(21) = 0.4716 so the optimal number of burritos to make is now 21.

Matching Supply with Demand: An Introduction to Operations Management 5e Solutions to Chapter Problems Chapter 15 Assemble-to-order, Make-to-order, and Quick Response with Reactive Capacity Q15.1 a) Teddy will order from the American supplier if demand exceeds 1500 units. With for Q = 1500 the z-statistic is z = (1500 − 2100) /1200 = −0.5 . From the Std Norm Dist Func Table we see that (−0.50) = 0.3085 , which is the probability demand is 1500 or fewer. The probability demand exceeds 1500 is 1−(−0.50) = 0.6915 , or about 69%. b) The supplier’s expected demand equals Teddy’s expected lost sales with an order quantity of 1500 parkas. From the Std Norm Loss Func Table, L(−0.50) = 0.6978 . Expected lost sales is   L(z) = 1200  0.6978 = 837.4. c) The overage cost is C0 = 10 − 0 = 10 , because left over parkas must have been purchased in the 1st order at a cost of $10 and they have no value at the end of the © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

season. The underage cost is Cu = 15 −10 = 5 because there is a $5 premium on units ordered from the American vendor. The critical ratio is 5 / (10 + 5) = 0.3333 . From the Std Norm Dist Func Table we see that (−0.44) = 0.3300 and (−0.43) = 0.3336 , so choose z = -0.43. Convert to Q : Q = 2100 – 0.43  1200 = 1584. d) First evaluate some performance measures. We already know that with Q = 1584 the corresponding z is –0.43. From Std Norm Loss Func Table, L(−0.43) = 0.6503 . Expected left over inventory is 1200 × I(z) = 264.4. Expected sales = Q – expected leftover inventory = 1584 – 264.4 = 1319.6. Expected lost sales is then 2100 – 1319.6 = 780.4. Now evaluate expected profit with the American vendor option available. Expected revenue is 2100 22 = $46, 200 . The cost of the 1st order is 158410 = $15,840 . Salvage revenue from left over inventory is 264.4  0 = 0. Finally, the cost of the 2nd order is 780.415 = $11, 706 . Thus, profit is 46200 −15840 −11706 = $18, 654 . e) If Teddy only sources from the American supplier, then expected profit would be ($22 − $15)  2100 = $14, 700 , because expected sales would be 2100 units and the gross margin on each unit is $7 = $22 − $15 . Q15.2

a) Expected sales = 1000 and the gross margin per sale is 121−83.5 = $37.5. Expected profit is then 1000$37.5 = $37,500 . b) Co = 72 − 50 = 22 . CU = 83.5 − 72 = 11.5 : the premium on orders from XE is $11.5 . The critical ratio is 11.5 / (22 +11.5) = 0.3433. From the Std Norm Dist Func Table (−0.41) = 0.3409 and (−0.40) = 0.3446 , so z = -0.40. Convert to Q = 1000 – 0.4  600 = 760. c) The underage cost on an option is the change in profit if one additional option had been purchased that could be exercised. For example, if 700 options are purchased, but demand is 701, then 1 additional option could have been purchased. The cost of the option plus exercising it is $25 +$50 = $75 . The cost of obtaining the unit without the option is $83.5 , so purchasing the option would have saved Cu = $83.5 −$75 = $8.5 The overage cost on an option is the extra profit that could have been earned if the option were not purchased assuming it isn’t needed. For example, if demand were 699, then the last option was not necessary. The cost of that unnecessary option is C0 = $25 . The critical ratio is 8.5 / (25 + 8.5) = 0.2537 . From the Std Norm Dist Func Table (−0.67) = 0.2514 and (−0.66) = 0.2546 , so z = 0.66. Convert to Q = 1000 – 0.66  600 = 604. d) Evaluate some performance measures. Expected left over inventory = 600 × I(z) = 91.7. Expected number of options exercised (expected sales) is 604 – 91.7 = 512.3. Expected number of units ordered beyond the purchased options (expected lost sales) is 1000 – 512.3 = 487.7. Expected revenue is 1000$121 = $121, 000 . So profit is revenue, minus the cost of purchasing options (604$25 = $15,100) , minus the cost of exercising options (512.3$50 = $25, 615) , minus the cost of units purchased without options (487.7$83.5 = $40723) :121000 −15100 − 25615 − 40723 = $39,562

. Q15.3 a) The underage cost is Cu = 0.4 − 0.05 = $0.35 : if her usage exceeds the minutes she purchases then she could have lowered her cost by $0.35 per minute if she had purchased more minutes. The overage cost is Co = 0.05 because each minute purchase but not used provides no value. The critical ratio is 0.35 / (0.05 + 0.35) = 0.8749 . From the Std Norm Dist Func Table (1.15) = 0.8749 and (1.16) = 0.8770 , so z = 1.16. Convert to Q, Q = 250 + 1.16  24 = 278. b) We need to evaluate the number of minutes used beyond the quantity purchased (expected lost sales). z = (240 – 250) / 24 = -0.42, Expected leftover inventory = 24 x I(z) = 5.4. Expected sales = 240 – 5.4 = 234.6. Expected lost sales = 250 – 234.6 = 15.4 minutes. Each minute costs $0.4 , so the total surcharge is 15.4$0.4 = $6.16 . c) Find the corresponding z-statistic: z = (280 − 250) / 24 =1.25 . Now evaluate performance measures. L(1.25) = 0.0506 and expected lost sales = 24  0.0506 = 1.2 minutes, i.e., only 1.2 minutes are needed on average beyond the 280 purchased. The minutes used out of the 280 (expected sales) is 250 – 1.2 = 248.8. The unused minutes (expected left over inventory) is 280 – 248.8 = 31.2. d) Find the corresponding z-statistic: z = (260 − 250) / 24 = 0.42 . Expected leftover inventory = 24 × I(z) = 15.4. Expected sales = 260 – 15.4 = 244.6. Expected lost sales = 250 – 244.6 = 5.4 minutes. Total bill is 2600.05 + 5.40.4 = $15.16 e) From the Std Norm Dist Func Table (1.64) = 0.9495 and (1.65) = 0.9505 , so with z = 1.65 there is a 95.05% chance the outcome of a Standard Normal is less than z. Convert to Q, Q = 250 + 1.65  24 = 290 f) With “Pick your minutes” the optimal number of minutes is 278. The expected bill is then $14.46 : z = (278 − 250) / 24 =1.17; L(1.17) = 0.0596 ; expected surcharge minutes = 24  0.0596 = 1.4; expected surcharge = $0.41.4 = $0.56 ; purchase cost is 278 0.05 = $13.9 ; so the total is $13.9 + 0.56 . With “No Minimum” is total bill is $22.5 : minutes cost $0.07 250 = $17.5 ; plus the fixed fee, $5 . So she should stick with the original plan. Q15.4 a) The overage cost is Co = $60 because ordering a plate for a guest that doesn’t show up costs $60 . The underage cost is Cu = $85 − 60 = $25 , because not committing to a guest that does show up costs an extra $25 . The critical ratio is 25 / (60 + 25) = 0.2941. From the table, F (92) = 0.2727 and F(93) = 0.3030 , so the optimal number of guests to commit to is 93. b) Expected number of extra guests is L(105) = 2 , so her bill is $60105 +$85 2 = $6, 470 . c) The overage cost is Co = $45 because committing to a guest that doesn’t show only costs $45 . The underage cost is still Cu = $85 − 60 = $25 , because not committing to a guest that does show up costs an extra $25. The critical ratio is

25 / (45 + 25) = 0.3571. From the table, F(94) = 0.3333 and F(95) = 0.3636, so the optimal number of guests to commit to is 95. d) With the original plan we need to evaluate the expected bill. Expected number of extra guests is L(93) = 8.36 , so her bill is $6093 +$858.36 = $6, 291 . At $70 per guest, her expected bill is $70100 = $7000 . So go with the original option contract.

Q15.5

a) With 22 workers there is 22  40 = 880 hours of work available during the week. The z-statistic for Q = 880 is (880 − 793) /111 = 0.78 . 1− (z) is the probability the workload demand is greater than Q : 1− (0.78) = 1 – 0.7823 = 0.2177 If there are 52 weeks in a year, then you will use overtime in 0.2177  52 = 11.32 of them. b) With 18 workers we have 18 40 = 720 hours available per week. The z-statistic for 720 is (720 − 793) /111 = −0.66 . The probability demand is less than 720 is (z) , which is (−0.66) = 0.2546 . With 52 weeks a year we can expect to be underutilized in 0.2546  52 = 13.24 of them. c) The overage cost is the hourly cost of labor, because if we hire workers and they are idle, then that costs us their hourly wages. The underage cost is the cost of the overtime premium: every hour of overtime costs us the premium over the regular cost of labor. Because the overtime premium is 50% of the hourly wage, the underage cost is 50% of the overage cost: Cu = Co / 2 . (Note, the underage cost is not 150% of the hourly wage. Similarly, if we have a lost sale of a product then the underage cost is not the sales price but rather it is the gross margin.) Hence, the critical ratio is Cu / (Cu + Co ) == (0.5 Co ) / (1.5 Co ) = 1/ 3 = 0.3333 . From the Std Norm Dist Func Table, (−0.44) = 0.3300 and (−0.43) = 0.3336 , so choose z = −0.43. Convert to Q : Q = 793 − 0.43 111 = 745.27 . That translates into 745.27 / 40 =18.6 workers . d) The critical ratio evaluated in part c still holds even if you are using the empirical distribution function. We need to convert the given histogram, which is a density function, into a distribution function. That is done in the following table:

Hours Observations 480 1 520 0 560 1 600 3 640 8 680 8 720 12 760 14 800 13 840 16 880 11 920 8 960 4 1000 3 1040 1 1080 1

Cumulative observations 1 1 2 5 13 21 33 47 60 76 87 95 99 102 103 104

Percentile 0.0096 0.0096 0.0192 0.0481 0.1250 0.2019 0.3173 0.4519 0.5769 0.7308 0.8365 0.9135 0.9519 0.9808 0.9904 1.0000

We are looking for 0.3333 in the percentile column. From the table we see that 31.73% of the time the workload is 720 hours and 45.19% of the time the workload is 760 hours. Based on the round up rule we should hire 19 workers so that we have 760 hours of regular time available each week.

Q15.6 a) This is a newsvendor problem because you have one opportunity to satisfy your “demand” for shillings at the favorable rates in the capital. Buying a shilling in the capital costs $0.50 but buying one in the town costs $1/1.6 = $0.625 . The underage cost of buying too few shillings is Cu = 0.625 − 0.50 = 0.125 : each shilling that you could use but need to substitute an American dollar costs you an additional $0.125 . The overage cost of buying too few shillings is what you lose converting them back to dollars: Co = 0.5 −1/ 2.5 = 0.1. The critical ratio is 0.125 / (0.125 + 0.1) = 0.5556 . From the Std Norm Dist Func Table we see that (0.13) = 0.5517 and (0.14) = 0.5557 , we choose z = 0.14 . Convert the z-statistic to Q = 400 + 0.14 100 = 414 . Because you want 414 shillings, you should convert $207 dollars . b) A 1 in 200 chance of stocking out means a 0.5% stockout probability, which in turn means a 99.5% in-stock probability. From the Standard Normal Distribution Function Table we see that (2.57) = 0.9949 and (2.58) = 0.9951, so choose z = 2.58 . Convert the z-statistic to Q = 400 + 2.58  100 = 658. If you need 658 shillings, then you must convert 658 / 2 = $329 .

Q15.7 a) TEC’s gross margin is $1100.25 = $27.5.4101 units are produced, so total profit is 4101$27.5 = $112, 778. b) TEC’s regular production cost is 0.75$110 = $82.5. TEC’s expensive production cost is 2$82.5 =165 . TEC earns $27.5 on each of the 3263 units in O’Neill’s 1st order. TEC charges 1.2$110 = $132 for units in the 2nd replenishment. O’Neill’s expected 2nd order quantity is 437, and TEC “earns” $132 − $165 = −$33 on those units despite the premium charged of 20%. Hence, TEC’s profit is $27.53263 −$33 437 = $75,312 c) Now TEC can produce more than 3263 units. The overage cost is $82.5 − 30 = $52.5 . If a unit is not produced in the 1st production run but could be sold, TEC “earns” −$33 on that unit. If the unit were produced in the 1st production run, TEC earns $132 −$82.5 = $49.5 . Hence, the underage cost is $49.5 −(−$33) = $82.5 . In other words, every unit produced in the 1st production run that O’Neill eventually orders saves TEC $82.5 in profit relative to producing that unit in the 2nd production run. The critical ratio is 82.5 / (52.5 + 82.5) = 0.6111. We find that (0.28) = 0.6103 and (0.29) = 0.6141, so choose z = 0.29. Convert the z-statistic to Q = 3192 + 0.29  1181 = 3534. Because that quantity is greater than O’Neill’s initial order of 3263, TEC should produce 3534 units in the 1st production run. d) If TEC produces 3534, then its expected 2nd production run is 320 units: L(0.29) = 0.2706 ,   L(z) = 1181 0.2706 = 320 . Expected left over inventory is tricky to evaluate. Expected left over inventory with Q = 3534 is 662 units. Expected left over inventory with Q = 3263 is 508 units. Hence, if TEC produces 3534 units and O’Neill’s 1st order is 3263 units, then among the 271 units (3534-3263) TEC produces above O’Neill’s order, TEC can expect to have 662 – 508 = 154 remaining at the end of the season. (To explain, suppose demand is only 3200 units. Then TEC has 271 units left over, not 3534-3200 = 334, i.e., left over inventory is the amount that is left over if the order quantity is 3534 minus the amount that would be left over if the order quantity is 3263. This is probably not obvious, which is why this is labeled as a hard question.) TEC’s revenue is then revenue from the 1st order $1103263 = $358,930 , plus revenue from the 2nd order $132 437 = $57, 684 , plus revenue from left over inventory $30154 = $4620 , for total revenue of $421, 234 . Costs include the 1st production run = 3534$82.5 = $291,555 and 2nd production run costs = 320$165 = $52,800 . Expected profit is then $421, 234 −$291555 −$52800 = $76,878 .

Q15.8 The overage cost is Co = $5 because any part remaining in inventory is charged $5 . The underage cost is Cu = $50 −$32 = $18 , because any emergency shipment requires an extra $18 shipping cost. The critical ratio is 18 / (5 +18) = 0.7826 . From the table, F(7) = 0.6728 and F (8) = 0.7916 , so the optimal number of parts to have on hand is 8.

There are three parts in inventory, so 5 parts should be ordered and arrive on Feb 15. Q15.9 There are two approaches to answering this question. To begin, look in the Poisson Inventory Function Table for mean 5.5 and Q = 5 and we see that expected leftover inventory is 0.67824. That means that expected sales is 5 – 0.67824 = 4.32. Hence, among the first 5 cars, he can expect to sell 4.32 of them. He earns $350 for each of those cars, for a total earning of $350 4.32 = $1512 . He might also sell some cars above the 5 car threshold. Expected lost sales = 5.5 – 4.32 = 1.18 cars. Among those cars he earns $400 , for a total commission of $4001.18 = $472 . Combining the commissions from the two sets (any of the first 5 cars and any of the cars sold above 5) yields a total commission of $1512 +$472 = $1984 . For the second approach, notice that for every car Smith sells he gets $350 and an additional $50 for every car sold over 5 cars. He expects to sell 5.5 cars in total and 1.18 cars above 5 (i.e., the cars that he earns the additional commission). Hence, his total commission is (3505.5) + (501.18) = $1984 .

Matching Supply with Demand: An Introduction to Operations Management 5e Solutions to Chapter Problems Chapter 16 Service Levels and Lead Times in Supply Chains: The Order-up-to Inventory Model

Q16.1 a) Inventory position = inventory level + on-order = 100 + 85 = 185. Order enough to raise the inventory position to the order upto level, in this case 220 – 185 = 35 desks. b) As in part a, inventory position = 160 + 65 = 225. Because the inventory position is above the order upto level, 220, you do not order additional inventory. c) From the Std Norm Dist Func Table: (2.05) = 0.9798 and (2.06) = 0.9803 , so choose z = 2.06. The lead time, l, is 2, so  = (2 + 1) 40 = 120 and  = 2 +1  20 = 34.64 . S =  + z  = 120 + 2.0634.64 = 191.36. d) The z-statistic that corresponds to S = 120 is z = (120 −120) / 34.64 = 0 . Expected onhand inventory is 34.64 x I(0) = 13.82 e) From part d, on-hand inventory is 13.82 units, which equals 13.82$200 = $2764 . Cost of capital is 15%, so the cost of holding inventory is 0.15$2764 = $414.6 . Q16.2

a) Mean demand over (l +1) periods is 0.5  (4+1) = 2.5 units. From the Poisson Distribution Function Table, with mean 2.5 we have F(6) = 0.9858 and F(7) = 0.9958 , so choose S = 7 to achieve a 99% in-stock. b) Pipeline inventory is l × expected demand in one period = 4 × 0.5 = 2 units. The

order upto level has no influence on the pipeline inventory. c) From the Poisson Inventory Function Table with mean 2.5 and S = 5, expected onhand inventory = 2.56 units d) A stockout occurs if demand is 7 or more units over (l +1) periods, which is 1 minus the probability demand is 6 or fewer in that interval. From the Poisson Distribution Function Table with mean 2.5 we see that F(6) = 0.9858 and 1− F(6) = 0.0142 , i.e., about a 1.4% chance of a stockout occurring. e) The store is out of stock if demand is 6 or more units over (l +1) periods, which is 1 minus the probability demand is 5 or fewer in that interval. From the Poisson Distribution Function Table with mean 2.5 we see that F (5) = 0.9580 and 1− F(5) = 0.0420 , i.e., about a 4.2% chance of being out of inventory at the end of any given week. f) The store has one or more units of inventory if demand is 5 or fewer over (l +1) periods. From part g, F (5) = 0.9580, i.e., about a 96% chance of having one or more units at the end of any given week. g) Now the lead time is 2 periods (each period is 2 weeks and the total lead time is 4 weeks, or two periods). Demand over one period is 1.0 units. Demand over (l +1) periods is (2+1) × 1 = 3.0 units. From the Poisson Distribution Function Table with mean 3.0 we have F (7) = 0.9881 and F (8) = 0.9962 , so choose S = 8 to achieve a 99% in-stock. h) Pipeline inventory is average demand over l periods = 2  1 = 2.0 units. Q16.3 a) If S = 700 and the inventory position is 523 + 180 = 703, then 0 units should be ordered because the inventory position exceeds the order up-to level. b) From the Std Norm Dist Function Table, (2.32) = 0.9898 and (2.33) = 0.9901, so choose z = 2.33. Convert to S =  + z  = 600 + 2.33159.22 = 971. Q16.4

a) The critical ratio is $25 / ($0.5 + $25) = 0.98039 . The lead time is l = 0, so demand over (l +1) periods is Poisson with mean 1.5. From the Poisson Distribution Function Table with mean 1.5 we see F (3) = 0.9344 and F(4) = 0.9814 , so choose S = 4. There is currently no units on-order or on-hand, so order to raise the inventory position to 4: order 4 units. b) The in-stock probability is the probability demand is satisfied during the week. With S = 3 the in-stock is F (3) = 0.9344 , i.e., a 93% probability. c) Demand is not satisfied if demand is 5 or more units, which is 1− F(4) = 0.9814 = 1− 0.9814 = 0.0186 = 1, or about 1.9%. d) From the Poisson Distribution Function Table with mean 1.5 F(4) = 0.9814 and F (5) = 0.9955 so choose S = 5 to achieve a 99.5% in-stock probability. e) If S = 5, then from the Poisson Inventory Function Table with mean 1.5 we see

expected on-hand inventory is 3.51 units. The holding cost is 3.51$0.5 = $1.76

Q16.5

a) From the Std Norm Dist Function Table, (2.43) = 0.9925 , so choose z = 2.43. Convert to S =  + z  = 159.62 + 2.4395.51 = 392 . b) The holding cost is h = 0.75 and the backorder penalty cost is 50. The critical ratio is 50 / (0.75 + 50) = 0.9852. From the Std Norm Dist Function Table, (2.17) = 0.9850 and (2.18) = 0.9854 , so choose z = 2.18. Convert to S =  + z  = 159.62 + 2.1895.51 = 368 . c) The holding cost is h = 0.05 and the backorder penalty cost is 5. The critical ratio is 5 / (0.05 + 5) = 0.9901. Lead time plus one demand is Poisson with mean 1 × 3 = 3. From the Poisson Distribution Function Table, with  = 3 , F (7) = 0.9881 and F (8) = 0.9962 , so S = 8 is optimal.

Q16.6 a) His inventory position is 200 + 73 = 273. His order-up-to level is 285, so he orders 285 – 273 = 12 pints. b) From the Normal Distribution Function Table, () = 0.9898 and () = 0.9901, so choose z = 2.33. Convert to an order-up-to level: 200 + 2.33 × 48.08 = 312. c) The average order quantity equals average demand during a single period. In this case average daily demand will be 100 / 7 = 14.3 pints . Q16.7

a)  =  (demand in one period) and L = 3 (lead time). On order or pipeline inventory =  × L = 1.5 × 3 = 4.5. b) Compute the probability demand over lead time plus one period exceeds 8. Demand over (l + 1) periods is Poisson with mean = 4 ×  = 4 × 1.5 = 6. From the Poisson Distribution Function Table, Probability{D > 8}= 1 – F(8) = 1 – 0.84724 = 0.15276. c) Overage cost, Co = $0.01 (daily holding cost). Underage cost, Cu = $6 (stockout cost). Critical Ratio = Cu / (Cu + Co ) = 6 / (6 + 0.01) = 0.9983 . Expected demand over l + 1 days is Poisson with mean = (3 + 1) × 1.5 = 6. From the Poisson Distribution Function Table with mean 6, F(13) = 0.99637 and F(14) = 0.99860. Thus the order upto level should be S = 14.

Q16.8

a) Mean demand over (l +1) periods is  = 2000 and the standard deviation is  = 555. Note, is we convert daily demand into weekly demand we get the same mean (which makes sense) but a different standard deviation. Our conversion procedure assumes demand across periods is independent, hence we have data to indicate the independence assumption is not valid. Therefore, we should use the mean and standard deviation evaluated directly from the demand data for (l +1) periods. Use

equation 14.3. The critical ratio is 0.45 / (0.01+ 0.45) = 0.9783 . From the Std Norm Dist Function Table, (2.02) = 0.9783 , so choose z = 2.02. Convert to S =  + z  = 2000 + 2.02 × 555 = 3121. b) The base stock level has not influence over the amount of inventory on order, which equals the lead time (4) times one period demand (400) = 1600 units. c) Evaluate expected on-hand inventory = 555 × I(1.44) = 818.65. Annual holding cost is $0.01 260818.65 = $2129 . d) From the Std Norm Dist Function Table, (1.88) = 0.9699 and (1.89) = 0.9706 so choose z = 1.89. Convert to S =  + z  = 2000 +1.89555 = 3049 .

Q16.9

a) From the Std Norm Dist Function Table, (3.08) = 0.9990 , (3.09) = 0.9990 and (3.10) = 0.9990 , so choose z = 3.10. The lead time is 0 weeks, so (l +1) periods is one week. Weekly demand has mean  = 5 × 178 = 890 and standard deviation

 = 5  45 = 100.62 . (We are not evaluating demand over 6 days because we want the demand over one period first, and then we worry about demand over l +1 periods.) Convert to S =  + z  = 890 + 3.10  100.62 = 1202. On-hand inventory (at the end of the week) is 312.03. The average order quantity is 890, because that is the average weekly demand. (An order equals last period’s demand, so the average order equals the average of last period’s demand, which is just the average demand in a period.) So the average inventory is 312.03 + 890 / 2 = 757, and the weekly holding cost is 757$0.08 = $60.56 . Annual holding cost is 52$60.56 = $3149 . b) An order is placed within a week if there is positive demand. With mean 890 and standard deviation 100.62, demand is greater than 0 with essentially 100% probability. If an order is placed every week, the annual ordering cost is $58 × 52 = $3016. c) If orders are placed every 2 weeks, the ordering cost is cut in half, 3016 / 2 = 1508. But holding costs are potentially increased. Repeat the process in part a with the assumption that one period is 2 weeks and the lead time is still 0. Now one period’s mean demand is 1780 and standard deviation is 142.30

( 10  45). z = 3.10 is still

optimal. The order up-to level is S = = 1780 + 3.10  142.30 = 2221. On-hand inventory is 441.04. The annual holding cost is $0.08 × (441.04 + 1780/2)  52 = $5,537. So the annual order cost decreases by $1508, but the annual holding cost increases by $5537 – $3149 = $2388. The holding costs dominate the ordering cost savings, so order weekly. Q16.10 a) From the Std Norm Dist Function Table, (2.24) = 0.9875 , so choose z = 2.24. Convert to S =  + z  = 165 + 2.24  51.96 = 281. But 20 bags are included in each facing, so S = 281 requires 15 facings. b) 11 facings translates into S = 11 × 20 = 220. With S = 220, z = (220 −165) / 51.96 = 1.06 . Expected on-hand inventory is 51.96 × I(1.06) = 59. c) There will be an empty facing if on-hand inventory is 10 or fewer units. If S = 220, then there will be 10 or fewer units if demand over (l +1) periods is 210 or greater, which equals 1 minus the probability demand is 209 or fewer. The z that corresponds to 209 is z = (209 −165) / 51.96 = 0.85 . From the Std Norm Dist Function Table, (0.85) = 0.8023 . Finally, 1 − 0.8023 = 0.1977. So there is about a 20% chance there will be an empty facing.

Solutions to Chapter Problems Chapter 17 Risk-Pooling Strategies to Reduce and Hedge Uncertainty

Q17.1 a) New standard deviation is 30x 50 = 212 . b) Pipeline inventory = expected demand per week × lead time = 200×50×10 = 100,000. Q17.2 The coefficient of total demand (pooled demand) is the coefficient of the product’s demand times the square root of (1+ Correlation) / 2 . Therefore, sqrt((1− 0.7) / 2)  0.6 = 0.23 Q17.3 a) Assume Fancy Paints implements the order up-to inventory model. Find the appropriate order up-to level. With a lead time of 4 weeks, the relevant demand is demand over 4+1 = 5 weeks, which is 5 × 1.25 = 6.25. From the Poisson Distribution Function Table, F(10) = 0.946 and F(11) = 0.974, a base stock level S = 11 is needed to achieve at least a 95% in-stock. From the Poisson Inventory Function Table, the Expected Inventory is I(11) =4.8. There are 200 SKUs, so total inventory is 200 × 4.8 = 960. b) The standard deviation over (4 + 1) weeks is  = 5 8 = 17.89 and  = 5 × 50 = 250. From the Std Norm Distribution Function Table, we see that (1.64) = 0.9495 and (1.65) = 0.9505, so we choose z = 1.65 to achieve the 95% in-stock probability. The base stock level is then S =  + z ×  = 250 + 1.65 × 17.89 = 279.5. On hand inventory is 17.89 × I(1.65) = 29.9. There are 5 basic SKUs, so total inventory in the store is 29.9 × 5 = 149.5. c) The original inventory investment is 960$14 = $13, 440 , which incurs holding cost $13, 440 0.20 = $2, 688 . Repeat part b but now the target in-stock probability is 98%. From the Std Norm Distribution Function Table, we see that (2.05) = 0.9798 and (2.06) = 0.9803, so we choose z = 2.06 to achieve the 98% in-stock probability. The base stock level is then S =  + z ×  = 250 + 2.06 × 17.89 = 286.9. On-hand inventory for one product is 17.89 × I(2.06) = 37.0. There are 5 basic SKUs, so total inventory in the store is 37.0 × 5 = 185. With the mixing machine the total inventory investment is 185$14 = $2590 . Holding cost is $25900.2 = $518 , which is only 19% (518 / 2688) of the original inventory holding cost. Q17.4 a) Use the newsvendor model to decide an order quantity. From the table we see that F(3500) = 0.8480 and F(4000) = 0.8911, so order 4000 for each store. b) Evaluate expected lost sales and the expected left over inventory. Expected lost sales comes from the table, L(4000) = 185.3 . Expected sales is © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

 −185.3 = 2251−185.3 = 2065.7 . Expected left over is Q minus expected sales, 4000 – 2065.7 = 1934.3. Across 200 stores there will be 200  1934.3 = 386,860 units left over. c) The mean is 450200. The coefficient of variation of individual stores is 0.7108 =1600 / 2251. The coefficient of variation of total demand, we are told, is ½ of that, 0.3554 = 0.7108 / 2 . Hence, the standard deviation of total demand is 160001 = 450200  0.3554. From the Std Norm Dist Function Table we see (1.03) = 0.8485 and (1.04) = 0.8508 , so choose z = 1.04. Covert to Q = 450200 + 1.04  160001 = 616601. d) Expected left over inventory = 160001  I(z) = 178,753, which is only 46% of what would be left over if individual stores held their own inventory. e) The total order quantity is 4000  200 = 800000. With a mean of 450200 and standard deviation of 160001 (from part c), the corresponding z is (800000 − 450200) /160001 = 2.19 . From the Std Norm Dist Function Table we see (2.19) = 0.9857, so the in-stock would be 98.57% instead of 89.11% if the inventory were held centrally. Q17.5 a) With a lead time of 3 weeks,  = (3 + 1)  5200 = 20800 and  = 3 +1  3800 = 7600 . The target expected backorders is (5200 / 7600)  (1− 0.999) = 0.0007 . From the Std Norm Distribution Function Table, we see that (3.10) = 0.9990, so we choose z = 3.10 to achieve the 99.9% in-stock probability. Convert to S = 20800 + 3.10  7600 = 44,360. Expected on-hand inventory for each product is 7600 × I(z) = 23,562. The total inventory for the two is 2 × 23,562 = 47,124. b) Weekly demand for the two products is 5200  2 = 10400. The standard deviation of the two products is 2 (1− Correlation )  Standard deviation of one product = 2 (1− 0.20)  3800 = 4806.66 . Lead time plus one expected demand is 10400  4 = 41600. Standard deviation over (l +1) periods is (3 +1)  4806.66 = 9613 . Now repeat the process in part a with the new demand parameters. Convert to S = 41,600 + 3.10  9,613 = 71,401. Expected on-hand inventory is 9,613 × I(z) = 29,804. The inventory investment is reduced by (47,124 − 29,804) / 47,124 = 37%. Q17.6

a) Demand over (l +1) periods has mean  = 1.25  13 = 16.25. Use Excel to construct the Poisson Distribution Function Table for a Poisson with mean 16.25. We find that F(26) = 0.9910 and F(27) = 0.9949, so the appropriate order up-to level is S = 27. Expected on-hand from the Poisson Inventory Function Table is I(27) = 10.76. Annual turns is 521.25 /10.76 = 6.04 b) Repeat the process in part a, but now the lead time is 1 week. Demand over (l +1) periods has mean  = 1.25  2 = 2.5. From the Poisson Distribution Function Table in the Appendix we find that F(6) = 0.9858 and F (7) = 0.9927, so the appropriate

order up-to level is S = 7. From the Poisson Inventory Function Table I(7) = 4.51. Annual turns is 521.25 / 4.51 =14.4 . Q17.7

a) Demand over (l +1) periods has mean  = 3100 = 300 and standard deviation

 = 3  65 = 112.58 . From the Std Norm Dist Function Table we see (1.88) = 0.9699 and (1.89) = 0.9706, so choose z = 1.89. Convert to S = 300 + 1.89  112.58 = 513. Expected on-hand inventory for each desk is 112.58 × I(z) = 214. Total inventory for the two desks is 2 × 214 = 428. b) Expected on-hand inventory is 214 units because the demand distribution, lead time and the target in-stock is the same. c) Zero. The order up-to level of the gray bases has to be the sum of the order up-to levels of the tops. If it were lower, then there is a chance that demand is less than the order up-to level of each top but more than the order up-to level of the gray base, in which case the in-stock probability would fall below 97%. Hence, to ensure that the in-stock probability remains 97%, there must be one gray base for each top. If the instock probability on the tops were raised, then the number of gray bases could be reduced and still achieve a 97% in-stock probability. Q 17.8 a) Use the newsvendor model. If the order quantity is 5 and mean demand is Poisson with mean 0.9, then expected lost sales from the table is L(5) = 0.00039 . Expected sales is 0.9 – 0.00039 = 0.89961 and expected left over inventory is 5 – 0.89961 = 4.10. Production cost is $2695 = $1345 . Revenue from sales is $3500.89961 = $314.86 and revenue from salvaging is $100 4.10 = $410 . Profit is $314.86 +$410 − $1345 = −620 . b) Expected demand is reduced by 12.5% to 0.875  2.0 = 1.75 units. Repeat the process in part a with the new demand parameter. Expected lost sales from the table is L(5) = 0.01191. Expected sales is 1.75 – 0.01191 = 1.738 and expected left over inventory is 5 – 1.738 = 3.26. Production cost is $2695 = $1345 . Revenue from sales is $3501.738 = $608.33 and revenue from salvaging is $1003.26 = $326 . Profit is $608.33+$326 − $1345 = −411. Hence, this product is not profitable even with only one color. Q17.9 Option a) provides the longest chain, covering all four areas. This gives the maximum flexibility value to the firm, so that should be the chosen configuration. To see that it forms a long chain, Alice can do Regulations, as well as Bob. Bob can do Taxes, as well as Doug. Doug can do Strategy, as well as Cathy. Cathy can do Quota, as well as Alice. Hence, there is a single chain among all four consultants. The other options do not form a single chain.

Matching Supply with Demand: An Introduction to Operations Management 5e Solutions to Chapter Problems © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

Chapter 18 Revenue Management with Capacity Controls Q18.1 a) The booking limit is capacity minus the protection level, which is 150 – 50 = 100, i.e., allow up to 100 bookings at the low fare. b) The underage cost is Cu = 200 −120 = 80 and the overage cost is Co = 120 . The critical ratio is 80 / (120 + 80) = 0.4 . From the Std Normal Dist Function Table we see  (-0.26) = 0.3974 and  (-0.25) = 0.4013, so choose z = -0.25. Evaluate Q = 70 – 0.25  29 = 63. c) Decreases. The lower price for business travelers leads to a lower critical ratio and hence to lower protection level, i.e., it is less valuable to protect rooms for the full fare. d) The number of unfilled rooms with a protection level of 61 is the same as expected left over inventory. Evaluate the critical ratio, z = (61− 70) / 29 = −0.31. Expected left over inventory is 29 × I(z) = 7.62. So we can expect 7.62 rooms to remain empty. e) 70$200 + (150 − 70)$120 = $23, 600 , because on average 70 rooms are sold at the high fare and 150 – 70 = 80 are sold at the low fare. f) 150$120 = $18, 000 . g) If 50 are protected we need to determine the number of rooms that are sold at the high fare. The z statistic is (50 − 70) / 29 = −0.69 . Expected inventory is 29 × I(z) = 4.2 Expected sales is 50 – 4.2 = 45.8. Revenue is then (150 − 50)$120 + 45.8$200 = $21,155 . Q18.2 a) The underage cost is $120 , the discount fare. The overage cost is $325 . The critical ratio is 120 / (325 +120) = 0.2697 . From the table F(12) = 0.2283 and F(13) = 0.3171, so the optimal overbook quantity is 13. b) A reservation cannot be honored if there are 9 or fewer no-shows. F(9) = 0.0552, so there is a 5.5% chance the hotel will be overbooked. c) It is fully occupied if there are 15 or fewer no-shows, which has probability F(15) = 0.5170. d) Bumped customers equals 20 – number of no-shows, so it is equivalent to left over inventory. Expected left over inventory/bumped customers = I(20) = 4.78. Each one costs $325 , so the total cost is $325 4.78 = $1554 . Q18.3 a) First evaluate the distribution function from the density function provided in the table: F(8) =0, F(9) = F(9) + 0.05 = 0.05, F(10) = F(9) + 0.10 = 0.15, etc. Let Q denote the number of slots to be protected for sale later and let D be the demand for slots at $10,000 each. If D > Q, we reserved too few slots and the underage penalty is Cu = $10, 000 −$4, 000 = $6, 000 . If D < Q, we reserved too many slots and overage penalty is Co = $4, 000 . The critical ratio is 6000 / (4000 + 6000) = 0.6 . From the table we find F(13) = 0.6, so the optimal protection quantity is 13. Therefore, WAMB © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

should sell 25-13 = 12 slots in advance. b) The underage penalty remains the same. The overage penalty is now Co = $4000 −$2500 = $1500 : setting the protection level too high before meant lost revenue on the slot, but not at least $2500 can be gained from the slot, so the loss is only $1500 . The critical ratio is 6000 / (1500 + 6000) = 0.8 . From the table, F(15) = 0.8, so protect 15 slots and sell 25 – 15 = 10 in advance. c) If the booking limit is 10, there are 15 slots for last-minute sales. There will be standby messages if there are 14 or fewer last-minute sales, which has probability F(14) = 0.70. d) Over overbooking means the company is hit with a $10000 penalty, so Co = 10000 . Under overbooking means slots that could have sold for $4000 are actually sold at the standby price of $2500 , so Cu = 4000 − 2500 = 1500 . The critical ratio is 1500 / (10000 + 1500) = 0.1304. From the Poisson Distribution Function Table with mean 9.0, F(5) = 0.1157 and F(6) = 0.2068, so the optimal overbooking quantity is 6, i.e., sell up to 31 slots. e) The overage cost remains the same: we incur a penalty of $10000 for each bumped customer (and we refund the $1000 deposit of that customer too). The underage cost also remains the same. To explain, suppose they overbooked by 2 slots but there are 3 withdrawals. Because they have one empty slot, they sell it for $2, 500 . Had they overbooked by one more (3 slots), then they would have collected $4,000 on that last slot instead of the $2, 500 , so the difference is Cu = $4, 000 −$2,500 = $1500 . Note, the non-refundable amount of $1000 is collected from the 3 withdrawals in either scenario, so it doesn’t figure into the change in profit by overbooking one more unit. The critical ratio is 1500 / (10000 +1500) = 0.1304 . From the Poisson Distribution Function Table with mean 4.5, F(1) = 0.0611 and F(2) = 0.17358, so the optimal overbooking quantity is 2, i.e., sell up to 27 slots. Q18.4

a) The z-statistic is (100 − 70) / 40 = 0.75. Expected left over inventory is 40 × I(z) = 35.248. b) Expected revenue is $10000 64.752 = $647,520 . c) The underage cost is $10000 −$6000 = $4000 , because under protecting boutique sales means a loss of $4000 in revenue. Over protecting means a loss of $6000 in revenue. The critical ratio is 4000 / (6000 + 4000) = 0.4 . From the Std Normal Dist Function Table we see  (-0.26) = 0.3974 and  (-0.25) = 0.4013, so choose z = 0.25. Evaluate Q = 40 – 0.25  25 = 33.75. So protect 34 dresses for sales at the boutique, which means sell 100 – 34 = 66 dresses at the show. d) If 34 dresses are sent to the boutique then expected inventory is 25  I(z) = 7.26. Expected sales = 34 – 7.26 = 26.74. So revenue is 26.74$10000 + (100 − 34) 6000 = $661,900 . e) From part d, expected left over inventory is 7.26 dresses.

a) The overage cost is $800 (over overbooking means a bumped passenger, which costs $800). The underage cost is $475 (an empty seat). The critical ratio is 475 / (800 + 475) = 0.3725 . From the Std Normal Dist Function Table we see (−0.33) = 0.3707 and (−0.32) = 0.3745 , so choose z = -0.32. Evaluate Y = 30 – 0.32  15 = 25.2. So the max number of reservations to accept is 200 + 25 = 225. b) 220 – 200 = 20 seats are overbooked. The number of bumped passengers equals 20 minus the number of no-shows, which is equivalent to left over inventory with an order quantity of 20. The z-statistic is (20 − 30) /15 = −0.67 . Expected left over inventory is 15  I(z) = 2.3. If 2.3 customers are bumped, then the payout is $800 2.3 = $1840. c) You will have bumped passengers if there are 19 or fewer no-shows. The z-statistic is (19 − 30) /15 = −0.73. (−0.73) = 0.2317 , so there is about a 23% chance there will be bumped passengers. Q18.6 a) The underage cost is $675 −$375 = $300 and the overage cost is $375 . The critical ratio is 300 / (375 + 300) = 0.4444 . All reservations are sold if the high fare demand exceeds the protection level, which has probability 1 – 0.4444 = 0.5556. b) With Q = 85 the z statistic is (85 − 80) / 35 = 0.14 and L(0.14) = 0.3328, so expected lost sales is 35  0.3328 = 11.65 c) Empty seats equals 85 minus high fare demand, which is equivalent to expected left over inventory. Expected left over inventory is 35  I(z) = 16.65. d) Revenue is (200 − 85)$375 + 68.35$675 = $89, 261. Q18.7 a) The high-fare protection level = Capacity – Low-fare booking limit = 100 – 50 = 50. Next, z = (Q −μ) / σ = (50 − 40) / 30 = 10 / 30 = 0.333 . Expected lost sales =  × L(z) = 30 × 0.2555 = 7.665. Expect sales (high fare) = µ - Expected lost sales = 40 - 7.665 = 32.335. b) Critical Ratio = Cu / (Cu + Co ) = (rh − rl ) / rh = (80 − 0.580) / 80 = 0.5 . F(0) = 0.5, i.e. z = 0. Q =  + z   =  = 60. c) An increase in the discount leads to lower student ticket prices which increases the critical ratio. Critical Ratio = Cu / (Cu + Co ) = (rh − rl ) / rh = (80 − 0.4580) / 80 = 0.55. Thus, the optimal protection level, Q will increase. This means fewer seats available for students, demand from whom is abundant and more seats that could potentially go empty if not enough full-price demand materialize. Thus, the expected number of empty seats will increase. d) Co = 10 Cu . (Overbooking) Critical Ratio = Cu / (Cu + Co ) = Cu / (10 Cu + Cu ) = 1/11 = 0.091. Look in the Poisson Distribution Function Table with a mean of 8: F(3) = 0.04238, F(4) = 0.099963, so use the round-up rule and overbook by 4 seats. Q18.8 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

a) The overage cost is the cost of protecting too many rooms, which means a room goes empty that could have been sold at the discount. The incremental profit from selling that room is $159 − $45 = $114 = Co . The underage cost remains $225 − $159 = Cu = $66 . The critical ratio is 66 / (114 + 66) = 0.3667 . From Table 16.2 we find F(24) = 0.3040 and F(25) = 0.3760, so it is optimal to protect 25 rooms. Q18.9 a) The class size will be at least 720 if there are 30 or fewer students who decline the offer. The corresponding z statistic is (30 − 50) / 21 = −0.95 and  (-0.95) = 0.1711. Hence about a 17% probability the class size will be at least 720. b) The cost of admitting too many (the overage cost) is twice the cost of under admitting, so the critical ratio is Cu / (Co + Cu ) = Cu / ( 2 Cu + Cu ) = 0.3333 . From the Std Norm Distribution Function Table,  (-0.44) = 0.3300 and  (-0.43) = 0.3336, so choose z = -0.43. The overbooking quantity is then Y = 50 – 0.43  21 = 40.97. Admit 720 + 41 = 761 students. c) The cost of admitting too many (the overage cost) is five times the cost of under admitting, so the critical ratio is Cu / (Co + Cu ) = Cu / (5 Cu + Cu ) = 0.16667 . From the Std Norm Distribution Function Table,  (-0.97) = 0.1660 and  (-0.96) = 0.1685, so choose z = -0.96. The overbooking quantity is then Y = 50 – 0.96  21 = 29.84. Admit 720 + 30 = 750 students. Q18.10 a) With 58 units the z statistic is (58 − 65) / 45 = −0.16.L(−0.16) = 0.4840 , so expected lost sales is 45  0.4840 = 21.78. Expected sales is 65 – 21.78 = 43.33. Profit is ($2100 −$330)  43.33 = $76, 499 . b) With long term contracts the company sells 58 units at $1875 each for a total profit of 58($1875 −$330) = $89, 610 . c) The overage cost is $1875 (over protect and lose the opportunity to collect $1875 in revenue) while the underage cost is $2100 − $1875 = $225 . The critical ratio is 225 / (1875 + 225) = 0.1071 . From the Std Norm Distribution Function Table, (1.25) = 0.1056 and (−1.24) = 0.1075 , so choose z = -1.24. The optimal protection level is 65 – 1.24  45 = 9.2. Hence, the optimal booking limit for long term contracts is 58 – 9.2 = 48.8. d) Now the overage cost is $1875 −$330 = $1545 because the incremental profit on a sale is only $1545 . The underage cost remains $2100 − $1875 = $225 . The critical ratio is 225 / (1545 + 225) = 0.1271 . From the Std Norm Distribution Function Table,  (-1.15) = 0.1251 and  (-1.14) = 0.1271, so choose z = -1.14. The optimal protection level is 65 – 1.14  45 = 13.7. Hence, the optimal booking limit for long term contracts is 58 – 13.7 = 44.3.

Matching Supply with Demand: An Introduction to Operations Management 5e

Solutions to Chapter Problems Chapter 19 Supply Chain Coordination Q19.1 a) If orders are made every week, then the average order quantity equals one week’s worth of demand, which is 25 cases. If at the end of the week there is one week’s worth of inventory, then the average inventory is 25 / 2 + 25 = 37.5. (In this case inventory “saw-toothes” from a high of two week’s worth of inventory down to one week, with an average of 1.5 weeks.) On average the inventory value is 37.59.25 = $346.9. The holding cost per year is 52 x 0.4% = 20.8%. Hence, the inventory holding cost with the first plan is 20.8%$346.9 = $72 . Purchase cost is 52 25$9.25 = $12, 025 . Total cost is $12, 025 +$72 = $12, 097 . b) Four orders are made each year, each order on average is for (52 / 4)  25 = 325 units . Average inventory is then 325 / 2 + 25 =187.5 . The price paid per unit is $9.40 0.95 = $8.93. The value of that inventory is 187.5$8.93 = $1674 . Annual holding costs are $1674 20.8% = $348 . Purchase cost is 52 25$8.93 = $11, 609 . Total cost is $348 + $11, 609 = $11,957 . c) P&G prefers our third plan as long as the price is higher than in the second plan, $8.93 . But the retailer needs a low enough price so that its total cost with the third plan is not greater than in the second plan, $11,957 (from part b). In part a we determined that the annual holding cost with a weekly ordering plan is approximately $72 . If we lower the price, the annual holding cost will be a bit lower, but $72 is a conservative approximation of the holding cost. So the retailer’s purchase cost should not exceed $11,957 −$72 = $11,885 . Total purchase quantity is 25 x 52 = 1300 units. So if the price is $11,885 /1300 = $9.14 , then the retailer will be slightly better off (relative to the second plan) and P&G is much better off (revenue of $11,885 instead of $11, 609 ). Q19.2 a) Use the newsvendor model. The overage cost is Co = cost − salvage value = $20 −$28 / 4 = $13 . The underage cost is Cu = price − cost = $28 −$20 = $8 . The critical ratio is 8 / (13 + 8) = 0.3810 . Look up the critical ratio in the Standard Normal Distribution Function table to find the appropriate z statistic = -0.30. The optimal order quantity is Q =  + z x  = 100 – 0.30 x 42 = 87. b) Expected left over inventory = 42 x I(z) = 11.2. Expected sales = Q - Expected left over inventory = 100 – 11.2 = 75.8. Profit = price x Expected sales + salvage value x Expected leftover inventory − Q cost = $2875.8 +$7 11.2 −87 $20 = $459 . c) The publisher's profit = Q(wholesale price − cost) = 87($20 −$7.5) = $1087.5 . d) The underage cost remains the same because a lost sale still costs Dan the gross margin, Cu = $8 . However, the overage cost has changed because Dan can now return © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

books to the publisher. He buys each book for $20 and then returns leftover books for a net salvage value of $15 − $1 (due to the shipping cost) = $14 . So his overage cost is now Co = cost − salvage value = $20 − $14 = $6 . The critical ratio is 8 / (6 + 8) = 0.5714 . Look up the critical ratio in the Standard Normal Distribution Function table to find the appropriate z statistic = 0.18. The optimal order quantity is Q =  + z x  = 100 + 0.18 x 42 = 108. e) Expected left over inventory = sigma x I(z) = 20.8. Expected sales = Q – 20.8 = 87.2.. Profit = price x Expected sales + salvage value x Expected leftover inventory − Q cost = $2887.2 + $14 20.8 −108$20 = $573 . f) The publisher’s sales revenue is $20108 = $2160 . Production cost is $7.5108 = $810 . The publisher pays Dan $15 21.2 = $318 . The publisher’s total salvages revenue on returned books is $6 20.8 = $124.8 . Profit is then $2160 −$810 −$318 + $124.8 = $1157 . Note that both the publisher and Dan are better off with this buy back arrangement. g) Equation 19.1 in the text gives the buyback price that coordinates the supply chain (that is, maximizes the supply chain’s profit). That buyback price is $1+ $28 −($28 −$20) ($28 −$6) / ($28 −$7.5) = $20.41. Note, the publisher’s buyback price is actually higher than the wholesale price because the publisher needs to subsidize Dan’s shipping cost to return books: Dan’s net loss on each book returned is $20 − (20.41−1) = $0.59 . Q19.3 a) The number of options exercised is the same as expected sales (if demand is less than Q, then sales/options exercised equals demand, but if demand is greater than Q, then sales/options exercised equals Q). Evaluate the z-statistic for Q = 30000 : z = (30000 − 24000) / 8000 = 0.75 . Expected leftover inventory is 8000 x I(z) = 7050. Expected sales is 30000 – 7050 = 22950. b) The expected number of displays purchased on the spot market equals expected lost sales = 24000 – 22950 = 1050. c) Handi’s procurement cost includes the cost of buying 30000 options (30000$4.5 = $135, 000) plus the cost of exercising options (22950.4$3.5 = $80,326.4) plus the cost of purchases on the spot market (1050$9 = 9450) , for a grand total of $224, 776 . d) Use the newsvendor model to determine the optimal number of options to purchase. If Handi buys one option too few, then Handi must make a spot market purchase of $9 . If Handi had purchased an additional option for $4.5 , then that option would have been exercised at a cost of $3.5 , for a total cost for the unit of 4.5 + 3.5 = 8. So Handi’s cost of under purchasing options is Cu = 9 − 8 = 1 . If Handi purchases an option that is not used (over buys), the cost to Handi is Co = 4.5 , because there is no residual value to the option. The critical ratio is 1/ (4.5 +1) = 0.1818 . Look up in the Standard Normal Distribution Function Table and find that (-0.91) = 0.1814 and (-0.90) = 0.1841, so choose z = -0.90. The optimal number of options is then Q = © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

24000 – 0.90 x 8000 = 16,800. e) Following the processes in parts a to c, the expected number of spot market purchases is 8003 = 8000 x L(-0.91) = 8000 x 1.0004. Expected sales (number of options exercised) is 24000 – 8003 = 15997. The total procurement cost is then $4.516,800 +$3.515997 +$98003 = $203, 618 . Q19.4 a) Cusano’s discount price is (1-0.75)x 250 = 62.5. Cusano’s overage cost is Co = 185 − 62.5 = 122.5 . Cusano’s underage cost is Cu = 250 −185 = 65 . The critical ratio is 65 / (122.5 + 65) = 0.34667 . From the Poisson Distribution Function Table F(6) = 0.23051 and F(7) = 0.35398, so the optimal order quantity is Q = 7. b) Expected left over inventory from the Poisson Inventory Function Table is 0.46. Expected sales is 7 – 0.46 = 6.54. Expected profit is then 6.54 x 250 + 0.46 x 62.5 – 7 x 185 = 369. c) Profit is 7 x (185 – 100) = 595. d) The supply chain’s overage cost is Co = 100 − 62.5 = 37.5 . The supply chain’s underage cost is Cu = 250 −100 = 150 . The critical ratio is 150 / (37.5 +150) = 0.8000 From the Poisson Distribution Function Table F(10) = 0.73519 and F(11) = 0.82657, so the optimal order quantity is Q = 11. e) The supply chain’s underage cost remains the same, Cu = 250 −100 = 150 . The supply chain’s overage cost is the consequence of having one grill left at the end of the season. If that grill had not been ordered, then the $15 shipping cost could have been avoided. (One might argue that damaged grills should not be sent back to SJ, but let’s assume that all grills are sent back to SJ and SJ determines whether they are damaged or not.) Fortunately, that grill may be sold next year, but if it is not sold because it is damaged, then an additional cost is incurred. There is a 45% chance it is not sold, so the expected loss on a left over grill is 100 0.45 = $45 : the cost of the grill times the chance the grill is disposed for no value. Hence, the total overage cost is 15 + 45 = $60 . But if left over grills were disposed at Cusano’s, then the overage cost would be $100 −$62.5 = $37.5 . Hence, grills should NOT be sent back to SJ, they should be disposed of at Cusano’s because the supply chain loses only 37.5 per grill salvaged at Cusano’s while it loses 62.5 per grill sent back to SJ. Hence, the analysis is exactly as in part d, send 11 grills to Cusano’s. f) The buy back price is 0.9 x 185 = 166.5. Cusano’s underage remains Cu = 250 −185 = 65 . Cusano’s overage cost is 185 – 166.5 + 15 = 33.5. The critical ratio is 65 / (33.5 + 65) = 0.65990 . From the Poisson Distribution Function Table F(9) = 0.6203 and F(10) = 0.7352, so the optimal order quantity is Q = 10. g) Expected left over inventory is I(10) = 1.92. Expected Sales is 10 = 1.92 = 8.08. Expected profit is then $2508.08 + ($166.5 −15) 1.92 −10185 = $461. h) From part g we know that Cusano will return 1.92 grills on average. SJ must pay Cusano 166.5 per grill returned. SJ earns the full wholesale price on the grills that are not returned and 55% of the returned grills. So the full wholesale price is earned on 10 – 1.92 = 8.08 grills that are not returned and 0.55 x 1.92 = 1.056 grills that are © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC.

returned in good condition, for a total of 8.08 + 1.056 = 9.136 grills. Profit is then $1859.136 − $10010 − $166.51.92 = 370.48. i) In part e we determined that returning grills to SJ is not optimal for the supply chain. But we can now evaluate what the supply chain’s optimal decision would be assuming grills are returned to SJ. In part e we determined that the underage cost is Cu = 250 −100 = 150 and the overage cost is Co = 15 + 45 = $60 . So the critical ratio is 150 / (60 +150) = 0.7143 . From the Poisson Distribution Function Table F(9) = 0.6203 and F(10) = 0.7352, so the optimal order quantity is Q = 10. Equation 16.1 in the text gives us the optimal buy-back price. But to use the equation we need to know the supply chain’s salvage value. If each grill costs $100 to make and Co = $60 , then the salvage value is $100 −$60 = $40 . So, from equation 16.1, the buy-back price we seek is 15 + 250 −(250 −185) (250 − 40) / (250 −100) =174 . A buy-back of $174 implies a return credit of 174 /185 = 94% . j) If Cusano does not pre-book enough, then Cusano must purchase a grill at the regular price of $185 instead of the discount pre-book price = 0.9$185 = $166.5 . Hence, Cusano’s underage cost is Cu = 10% $185 = $18.5 . If Cusano pre-books too many grills, then the left-over grills are salvaged for $62.5 each (see part a), so Co = 166.5 − 62.5 = 104 . The critical ratio is 18.5 / (104 +18.5) = 0.15102 . From the Poisson Distribution Function Table, F(5) = 0.1317 and F(6) = 0.23051, so the optimal pre-book quantity is 6 grills. k) With Q = 6, expected left over inventory is I(6) = 0.23 and expected sales is 6 - 0.23 = 5.77. Expected lost sales is 8.75 – 5.77 = 2.98. The lost sales are not actually lost: if demand exceeds the pre-book quantity of 6 units, then Cusano purchases the needed extra grills at the regular wholesale price of $185 and sells them for $250. So expected profit on non pre-book grills is 2.98($250 − $185) = $193.7 . Expected profit on pre-book grills is $2505.77 +$62.50.23 − $166.56 = $457.9 . Total profit is 193.7 + 457.9 = $652 . l) SJ’s profit on non-prebook grills is 2.98 (185 −100) = $253.3. SJ’s profit on prebook grills is 6 (166.5 −100) = $399 . Total profit is then $652 . m) The advanced purchase contract has a higher profit for all firms. However, the advanced purchase contract requires SJ to hold a sufficient stock of grills to satisfy Cusano’s requests (and other retailers’ requests) during the season. If that is too costly (e.g., due to expedited shipping, or additional warehouse space, or left over inventory at SJ, etc), then maybe the advanced purchase contract is not as attractive as this analysis suggests.

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