BUAD311/311T – Operations Management | Illustrative Exam 11 Solutions

BUAD311/311T – Operations Management | Illustrative Exam 11 Solutions 1 Illustrative Exam 11 Solutions BUAD311/311T – Operations Management Section A: Multiple Choices (circle only one) 1. Which of the following s ...

BUAD311/311T – Operations Management | Illustrative Exam 11 Solutions 1 Illustrative Exam 11 Solutions BUAD311/311T – Operations Management Section A: Multiple Choices (circle only one) 1. Which of the following statement is correct? a. Make-to-order systems are better than make-to-stock systems. b. Make-to-stock systems are better than make-to-order systems. c. Hybrid systems are perfect and better than both MTO and MTS systems. d. Hybrid systems are worse than both MTO and MTS systems. e. None of the above e): It is hard to tell which one is better between MTO and MTS. Each has its pros and cons. The answer really depends on the settings, i.e. customer demands and operating cost. 2. According to Little’s Law, a restaurant owner may increase the revenue by a. reducing the throughput rate b. having larger space and higher WIP c. having smaller space and lower WIP d. increasing the time customers staying in the restaurant e. None of the above b): The manager actually wants to increase the Flow Rate (Throughput Rate). 3. To improve the utilization rate, we can I: Cross-train the workers II: Adopt flexibility equipment III: Shift from MTS systems to MTO systems Choose the most appropriate. a. I b. II c. III d. I and II e. I, II, and III d): MTO and MTS have nothing to do with the utilization, since utilization is a concept for AVERAGE. Cross-trained workers and flexible equipment belong to the same concept—convertible resource, which can improve utilization. 4. Comparing make-to-order systems and make-to-stock systems, which of the following statement is false? a. Make-to-order systems are more responsive to the customer needs. b. Make-to-order systems have lesser or no finished-goods inventory. c. Make-to-stock systems can fulfill customer demand faster d. Make-to-stock systems usually have a lower utilization rate than made-to-order systems. e. IN-N-OUT burger is a make-to-order restaurant. d) 1 These are all actual questions that appeared in past exams. Note we may be covering slightly different materials this semester.2 5. Consider a business process with multiple stages. The utilization rate of the bottleneck resource is currently 90%. Which of the following will increase the bottleneck utilization? a. Increase in capacity of the bottleneck resource b. Increase in capacity of non-bottleneck resources c. Decrease in capacity of the bottleneck resource d. Decrease in capacity of non-bottleneck resources e. Decrease in throughput rate c): Bottleneck process has nothing to do with non-bottleneck process. 6. Bank XYZ is hiring, and they would like to determine the number of tellers they should hire. Each teller is paid $15 per hour. Bank XYZ has assessed a cost on customer waiting at $10 per customer in line per hour. (In other words, each customer standing in line at the bank and not being helped by a teller costs the bank $10 per hour.) How many tellers must the Bank hire to minimize their total hourly cost? Assume customers arrive according to a Poisson process at rate 20 per hour, and that each customer spends an exponentially distributed amount of time with a teller that has mean 8 minutes. a. 2 b. 3 c. 4 d. 5 e. 6 c): a = 3 mins; p = 8 mins; both a and p are exponentially distributed, resulting to CVa = 1, CVp = 1, and . For m = 2, , so m > 2. For m = 3, , , mins, , waiting cost = Iq * 10 = 64.5, tellers’ salary = m * 15 = 45, total cost = waiting cost + tellers’ salary = 109.5. For m = 4, , , mins, , waiting cost = Iq * 10 = 8.3, tellers’ salary = m * 15 = 60, total cost = waiting cost + tellers’ salary = 68.3. For m = 5, , , mins, , waiting cost = Iq * 10 = 2.4, tellers’ salary = m * 15 = 75, total cost = waiting cost + tellers’ salary = 77.4. For m = 6, , , mins, , waiting cost = Iq * 10 = 0.9, tellers’ salary = m * 15 = 90, total cost = waiting cost + tellers’ salary = 90.9. 7. The following figure shows the production process flow for HappyDoll and CryingDoll. Station D and E are flexible and can handle either product. No matter the type of the product, station D can finish 90 units per day and station E can finish 100 units per day. Station A works only for HappyDoll and have a capacity of 60 units per day. Station B and C are only for CryingDoll and have capacity of 80 and 70 units per day, respectively. The demands for both products are 50 units3 per day. The capacities (units/day) of the machines are marked in the graph. Which station is the bottleneck? a. A b. B c. C d. D e. E d): The demand for crying babies and happy babies are half and half. 8. Vicky was always amazed Disneyland accurately forecasts waiting times. She was fortunate enough to secure an internship position at Disneyland last summer, and learned firsthand how they estimate waiting times at each attraction. What is the information that they use to estimate waiting times? Choose the most appropriate. a. Attraction’s duration and #people waiting b. #people waiting and #people that the attraction can accommodate at one time c. #people that the attraction can accommodate and the attraction’s duration d. #people that the attraction can accommodate and the attraction’s duration and #people waiting e. None of the above. d): Vicky was looking for the FT (or TT) though the queue. # of people waiting is the WIP in the queue. is the FR (or TR) of attraction or ride which is also=FR of queue. 9. Hotel guests usually complain about the long waiting times for the elevators. Which of the following may help to improve customers’ waiting experience? I: Install more elevators II: Upgrade to faster elevators III: Install mirrors near the elevators a. Only I may. b. Only II may. c. Only III may. d. I and II may, but not III. e. All the three may. e): Both more elevators and faster elevators increase the FR (or TR). Mirrors increase occupied time. 10. The operator of a coffee stand has noticed that over the past week the utilization at the stand has increased from 0.6 to 0.75, and as a result the average total amount of time a customer spends there (in line and in service) has increased by one minute and a half. If the utilization increases next week to 0.9, the average total time will: a. Decrease. b. Increase by less than one minute and a half. c. Increase by exactly one minute and a half. B: 80 A: 60 D: 90 C: 70 E: 1004 d. Increase by more than one minute and a half. e. None of the above. d): m = 1, , . The average total amount of time a customer spends here is Tq + p, in which only Tq depends on u. When u = 0.6, . When u = 0.75, . When u = 0.9, . Therefore, the increase of Tq from 0.75 to 0.9 is more than that from 0.6 to 0.75. 11. Consider an M/M/1 queue with an arrival rate of 4 customers per hour, an average service time of 12 minutes. What is the percentage of time an arriving customer does not have to wait? a. 33.33% b. 80.00% c. 66.66% d. An arriving customer always has to wait. e. None of the above e): Customers don’t have to wait means the system is empty and the server is idle. Hence, the percentage of time an arriving customer does not have to wait is 1 - u = 20%. 12. TinyBank has only one ATM machine. A previous statistical study indicates that customers arrive to this ATM machine according to a Poisson process, and that the average amount of time a customer spends at the ATM is 6 minutes. Which is the following is NOT true? a. If, on average, 12 customers per hour come to the ATM, the bank needs to buy and install one more ATM machine. b. If the average arrival rate is 8 customers per hour and the service time is constant (exactly 6min for every customer), no customer needs to wait for the ATM. c. The average waiting time in line when service time is a constant 6 minutes will be half of the average waiting time in line when service time is exponentially distributed with a mean of 6 minutes. (everything else equal) d. This is an M/G/1 system. e. None of the above b): p = 6, m = 1. (a) For a = 5, , so not enough capacity. (b) Although u is less than 100% and p is a constant, a is random. (c) Since arrival is a Poisson process, CVa = 1. If the service time is a constant, CVp = 0, and ; if the service time is exponentially distributed, CVp = 1, and . (d) As the question doesn’t tell us any information about the distribution of the service time, it is a M/G/1 queue. 13. For a DMV office, which the following tends to result in longer waiting time in line? a. Hire more service representative b. Require more customers to make appointment c. Reduce variability in service time. d. Combine multiple waiting lines and make all walk-in customers wait in one line e. None of the above5 e): From queuing theory, . Hence, either increase service time, or reduce # of servers (which eliminates (a)), or increase variability in service time (which eliminates (c)), or increase utilization (decrease inter-arrival time, which eliminates (b)). For (d), risk pooling doesn’t change the average waiting time if customers don’t have preference. 14. Which of the following may NOT help improve the customer’s waiting experience? a. Provide menu and snacks for customers waiting to be served in a restaurant b. Give reasonable explanations to airline passengers when flights are delayed c. Play soothing music while customers are being held on a phone d. Put mirrors next to the elevator switch in a high-rise hotel c): ARES Reading: The Psychology of Waiting Lines 15. Bank of Lancaster has two ATM machines. According to a statistical study, on average 20 customers arrive at the ATM station every hour. Each customer spends on average 4 minutes using one of the ATM machines. Which of the following is true? a. An ATM machine can serve only 15 customers per hour and one more ATM machine should be purchased and installed. b. If the inter-arrival time of customers is constant, no customers need to wait to use an ATM machine. c. If customers wait in two lines, the utilization rate of the machines will be higher than if they wait in one line. d. None of the above is true. d): p = 4, m = 2, a = 3. (a) If a = 4, , enough capacity. (b) Although u is less than 100% and a is a constant, p is random. (c) Risk pooling doesn’t change the utilization if customers don’t have preference. 16. Located in the University Village Food Court, Tube is a make-to-order sandwich store specialized in exotic custom-made sandwiches. Currently, customers first line up to place an order. There are two workers accepting orders, and three workers preparing sandwiches. After receiving the order, the customer pays for it. Tube has only one cashier accepting payments. Currently, the sandwich preparation process is the bottleneck. Which of the following will NOT improve the throughput at Tube? a. Become a make-to-stock sandwich store. b. Cross-train the cashier and sandwich preparers c. Cross-train the order takers and sandwich preparers. d. Install state-of-the-art order entry terminals. d): Since preparation is the bottleneck, cross-training workers at the bottleneck with workers at the nonbottleneck improves the throughput. 17. Recall the Kristen’s Cookie Company case. Kristen realizes that the bottleneck resource is the oven, which makes 6 dozen cookies per hour. A friend of Kristen’s would like to give her an old oven, which produces 3 dozen cookies per hour. If Kristen accepts the old oven, the new hourly capacity for Kristen’s Cookie Company will be: a. 6 dozen b. Between 6 and 9 c. 9 dozen d. Greater than 9 dozen6 b): The oven works for 10 mins per batch and Kristen works for 8 mins per batch. That is, the oven makes 6 batches per hour and Kristen makes 7.5 batches per hour. With the additional oven, the ovens make 9 batches per hour but Kristen still makes 7.5 batches per hour. 18. At Caffe Vita Coffee House, one cashier takes orders, and one skilled worker prepares coffee. It takes a total of 5 minutes to finish both order-taking and preparing stages for a customer. The capacity of Caffe Vita Coffee House is a. 12 customers per hour b. Greater than 12 customers per hour c. Less than 12 customers per hour d. Depends on the actual demand b): It takes in total 5 mins for both order-taking and preparing, so for each step, the throughput time is less than 5 mins. That is, each one’s capacity is greater than 12 customers per hour, no matter which is the bottleneck. 19. Holding other quantities a constant, which of the following increases the average waiting time in a queue? a. Increasing the service rate b. Decreasing the average inter-arrival time of customers c. Increasing the number of servers d. Risk pooling e. None of the above b) 20. Consider a queue with an arrival rate of 5 customers per hour and an average service time of 10 minutes. There is one server. What is the percentage of time an arriving customer does not have to wait? a. 100% b. 50% c. 16.67% d. 0% e. None of the above. c) 21. After receiving complaints about excessive waiting times for teller service, a bank manager has installed TV monitors in the bank lobby and in waiting areas. Which of the following principles is used in this instance? a. Unoccupied time feels longer than occupied time b. Pre-process waits feel longer than in-process waits c. Unexplained waits are longer than explained waits d. The more valuable the service, the longer a person will be willing to wait e. Solo waiting feels longer than group waiting a): ARES Reading: The Psychology of Waiting Lines. 22. Which of the following is false? (The Psychology of Waiting Lines) a. A customer’s satisfaction with his service depends on both his perception of his service encounter and his expectation for the service. b. The beginning of the service encounter is very important in order that the customer rates his service well.7 c. A customer that has nothing to do while waiting for his service (except to wait) is likely to perceive his wait for service to be longer than it actually is. d. Customers tend to be more tolerant of waiting when they perceive order and fairness when assessing the waiting situation. e. None of the above. e) For questions 23-25 (Kristen’s Cookie Case): Recall Kristen’s Cookie Company. To refresh your memory, it takes Kristen 6 minutes to mix, and 2 minutes to load. It takes the roommate 1 minute to set the oven, 2 minutes to pack, and 1 minute to collect payment. The oven can bake one dozen cookies at one time, and it takes 9 minutes to bake. After baking, it takes 5 minutes to cool the cookies. There are many trays, and the mixer can mix up to 3 dozen at one time. The company makes many kinds of different cookies, depending on what the customer requests. 23. If all customers order one dozen cookies, what is the capacity of Kristen’s Cookie Company? That is, how many dozen cookies can Kristen produce in one hour if we assume that the company operates 24 hours a day, 7 days a week? a. 6 dozen per hour b. 7.5 dozen per hour c. 15 dozen per hour d. 20 dozen per hour e. None of the above a): For each dozen, the working time of Kristen is 6+2 = 8 mins; of the roommate is 1+2+1 = 4 mins; of the oven is 1+9 = 10 mins. Therefore, the oven is the bottleneck. The whole process’s capacity is the bottleneck’s capacity, which is . 24. If all customers order one dozen cookies and Kristen installs a second oven, what is the capacity of Kristen’s Cookie Company? a. Same as the answer to Q23. b. Greater than the answer to Q23. c. Less than the answer to Q23. b): For each dozen, the working times of Kristen and the roommate are the same, 8 mins and 4 mins; of each oven of the two is (1+9)/2 = 5 mins. Therefore, Kristen becomes the bottleneck. The whole process’s capacity increases: . 25. If all customers order three dozen cookies and Kristen installs a second oven, what is the capacity of Kristen’s Cookie company? a. Same as the answer to Q24. b. Greater than the answer to Q24. c. Less than the answer to Q24.8 b): For each dozen of the three, the working time the roommate is still 4 mins; of each oven of the two is (1+9)/2 = 5 mins; of Kristen is 6/3 + 2 = 4 mins. Therefore, the oven is still the bottleneck. The whole process’s capacity increases: . 26. What is the minimum number of servers needed for an M/M/m queue with an arrival rate of 5 customers per hour, and an average service time of 36 minutes? a. 1 b. 2 c. 3 d. 4 e. 5 or more d): p = 36 mins, a = 60/5 = 12 mins, u=p/ma=(36/12m) < 1  m >3. 27. Consider an M/M/1 queue with an arrival rate of 12 customers per hour, and an average service time of 4 minutes. What is the percentage of time that the server is idle (i.e., is not helping customers)? a. 0% b. 10% c. 20% d. 30% e. None of the above. c): p = 4 mins, a = 60/12 = 5 mins, u = 0.8, the server is idle 1 - u = 20%. 28. Consider a M/G/1 queue in which the arrival rate is 15 customers per hour, and the service time is 6 minutes with probability 0.25 and 1 minute with probability 0.75. What is the coefficient of variation of the service times, CVp? a. 1 or less b. More than 1 and less than or equal to 2 c. More than 2 and less than or equal to 3 d. More than 3 and less than or equal to 4 e. More than 4 a): the mean of the service time is p = 0.25*6 + 0.75*1 = 2.25; the variance of the service time is Var p = 0.25*(6- p)2 + 0.75*(1- p)2 = 4.6875; the standard deviation of the service time is S p = Varp 0.5 = 2.165; the coefficient of variation of the service time is CV p = Sp / p = 0.96. 29. Consider a serial sequence of 3 tasks - Station 1, Station 2 and Station 3. Station 1 is the bottleneck. Further assume that the capacity of Station 2 is lower than Station 3. Assume demand is greater than system capacity. Suppose the processing time of Station 2 is decreased, and Station 1 still remains the bottleneck. Consider the following statements: I. System capacity or flow rate will increase II. Flow time will decrease III. WIP in the system will increase a. Only I is true b. Only II is true c. Only III is true d. Both I and II are true. III is false Station 1 Station 2 Station 39 b): I. Since the bottleneck didn’t change, and the bottleneck’s capacity remained the same, the system capacity will remain the same. II. the system flow time is the summation of the flow times of the three stations III. WIP = FT * FR, when FR remains the same and FT decreases, WIP decreases. 30. Drive-through customers at Trojan Tacos line up in the drive-through lane, and wait until they reach the order/payment/delivery counter. They then place an order, pay for it, and wait at the counter until the order is delivered. The drive-through lane holds an average of 10 cars. On average, customers spend 12 minutes from the time they enter the drive-through lane until they receive their order. How many orders (cars) per hour, on average, does the counter serve? a. 120 b. 72 c. 60 d. 50 d): WIP = 10 cars, FT = 12 mins, FR = 10/12 cars/min = 10/12*60 = 50 cars/hour.10 Section B: Problems 1. [20 pts] Eastern Coffee follows the flow chart below to serve its customers. It takes a worker two minutes to take order and receive payment, two minutes to prepare coffee, and three minutes to clean equipment. Eastern Coffee has two workers: worker A takes order and prepares coffee, while worker B handles the cleaning. a. [6 pts] On average, 10 customers per hour show up and order coffee. What is the utilization rate of worker A? And what is the utilization rate of worker B? Capacity of worker A: 60/(2+2) = 15 customers/hour Capacity of worker B: 60/3 = 20 customers/hour Utilization of A: 10/15 = 0.666 or 66.6…% Utilization of B: 10/20 = 0.5 or 50% b. [6 pts] How many customers can Eastern Coffee serve per hour? From the above, Capacity of Eastern Coffee is 15 customers/hour Western Coffee follows the same flow chart above, and each activity takes the same amount of time as Eastern. Western Coffee also has two workers: worker C only takes order and payment, while worker D handles the coffee preparation and cleaning. c. [6 pts] How many customers can Western Coffee serve per hour? Capacity of worker C: 60/2 = 30 customers/hour Capacity of worker D: 60/(2+3) = 12 customers/hour Capacity of Western Coffee is 12 customers/hour d. [2 pts] The manager of Western Coffee notices that cleaning is not a critical activity in the sense that it can be delayed and be finished when there are fewer customers. Therefore, during the peak hour when many customers come in, workers can focus on serving customers and temporarily ignore the cleaning activity. Then how many customers can Western Coffee serve during the peak hour? Capacity of worker D during the peak hour: 60/2 = 30 customers/hour Capacity of Wewstern Coffee is 30 customers/hour 2. [20 pts] A resort manager observes that every day, on average, 1000 guests arrive at the resort. On average there are 5700 guests in the resort. a. [4 pts] How long, on average, does a visitor stay in the resort? 5700/1000 = 5.7 days b. [4 pts] The resort manager knows that there are two types of guests: Premier and Traveler. After some research the manager found that, on average, 700 Premier guests arrive at the resort every day and that there are 4200 Premier guests in the resort. How long, on average, does a Premier guest stay at the resort? 4200/700 = 6 days c. [4 pts] On average how many Traveler guests are there in the resort? 5700-4200 = 1500 guests. Take order and receive payment Prepare the coffee Clean equipment11 d. [4 pts] How long, on average, does a Traveler guest stay at the resort? 1000-700 = 300 Traveler guests per day 1500/300 = 5 days. Due to the large number of empty rooms in the resort, the manager decided to encourage Premier guests to stay longer in the resort. After a promotional campaign, the manager observed that Premier guests are now staying for 9 days on average. e. [4 pts] How many Premier guests are staying in the hotel now, on average? Assume that the number of Premier guests arriving every day did not change. 9*700 = 6300 Premier guests. 3. [15 pts] Bank of San Pedro has only one teller. On average, one customer comes every 6 minutes, and it takes the teller an average of 3 minutes to serve a customer. To improve customer satisfaction, the bank is going to implement a unique policy called, “We Pay While You Wait.” Once implemented, the bank will pay each customer $3 per minute2 while she or he waits in line. (So the clock starts when a customer comes to the end of the line, and stops when he or she begins to talk to the teller.) Bank of San Pedro hired you as a consultant and you are responsible for estimating how much the “We Pay While You Wait” program will cost. Your preliminary study indicates there are, on average, 0.5 customers waiting in line.3 a. [3 pts] Calculate the capacity of the teller. State the unit. The service time is p = 3 mins. The service rate is 1/p cus per min = 60/p cus per hour = 20 cus per hour. b. [3 pts] Calculate proportion of the time the teller is busy. Since a = 6 mins and m = 1, u = p / ma = 50%. c. [4 pts] How long, on average, does a customer wait in line? State the unit. Since I q = 0.5, Tq = Iq * a = 3 mins. d. [5 pts] Calculate the expected hourly cost of the “We Pay While You Wait” program. Solution 1: each customer waits, on average, 3 minutes. So each customer receives, on average, 3*3=$9. There are 10 customers arriving per hour, so the overall hourly cost is 9*10=$90. Solution 2: at any a moment, there are Iq customers waiting, so for each min the bank pays to its waiting customers Iq * $3 = $1.5. In 60 mins, the overall hour cost the bank pays to its customers is $1.5*60 = $90 4. [15 pts] Time Travelers Insurance Company (TTIS) processes 10,000 claims per year. The average processing time is 3 weeks. Assume 50 weeks per year. a. [5 pts] What is the average number of claims that are in process? 10,000 claims per year = 200 claims per week. WIP=TR*TT=200claims/week * 3weeks = 600 claims. 50% of all the claims that TTIS receives are car insurance claims, 10% motorcycle, 10% boat, and the remaining 30% house insurance claims. There are, on average, 150 car, 150 motorcycle, and 100 boat claims being processed. b. [5 pts] How long, on average, does it take to process a car insurance claim? 2 Assume linear cost. If a customer waits for 20 seconds in line, Bank of San Pedro will pay $1. 3 This does not include any customer being served.12 TR=200*0.5=100claims/week TT=150/100=1.5 weeks c. [5 pts] How long, on average, does it take to process a house insurance claim? WIP for house is 200. TR=200*0.3=60 claims/week. TT=200/60=3.33 weeks 5. [20 pts] Macrina Bakery is a small bakery that makes cupcakes and muffins. The bakery never stops working. They work 24 hours a day, 365 days a year. Making of a cupcake involves four stages: prepare the batter, fill the cupcake pan, bake, and put icing on the cupcake. Batter for 50 cupcakes can be mixed in a bowl – therefore cupcakes flow through the system in batches of 50. There are four skilled workers at the batter preparation stage, each working to prepare their own batch of batter. Each worker can prepare a bowl of batter in 20 minutes. They then pass the bowl of batter to the filling stage. (There are plenty of extra bowls for mixing batter.) At the filling stage, a single worker can pour the batter into the cupcake pans at a rate of 700 cupcakes per hour. The oven can bake 3 pans, or 150 cupcakes, at a time and baking takes 12 minutes. Assume that loading and unloading the oven takes very little time. A worker at the final stage puts the icing on the cupcakes and it takes 8 seconds per cupcake. Muffin flow in the system is the same as for cupcakes with the following exceptions. i) Muffins need to be baked for 24 minutes. ii) Muffins don’t need icing; hence they leave the bakery right after they are taken out of the oven. Note that cupcakes and muffins share the other resources in the system (4 workers at batter preparation, one worker at fill the pan, and one oven). Assume Macrina Bakery can sell as many cupcakes and muffins as they produce. a. [4 pts] Draw the process flowchart diagram clearly indicating processes and inventories. Use at least three inverted triangles. Avoid redundancy. Use arrows, not lines. b. [4 pts] Suppose for now that Macrina Bakery makes only cupcakes. What is the bottleneck? What is bakery capacity? Clearly state the unit. What is the utilization of the “Batter Preparation” stage? Batter: 60/20=3 batches/hour  3*4*50=600 cupcakes/hour Filling 700 cupcakes/hour Baking (60/12)*150=750 cupcakes/hour Icing (60/8)*60=450 cupcakes/hour Bakery Capacity: bottleneck capacity=Icing capacing: 450 cupcakes/hour Utilization of “Batter Preparation” stage: 450/600 = 75% c. [4 pts] Suppose that Macrina Bakery makes only muffins. What is bakery capacity? Clearly state the unit. What is the utilization of the “Filling Pan” stage? Batter: 600 muffins/hour Filling: 700 muffins/hour Baking: (60/24)*150 = 375 muffins/hour Bakery Capacity: 375 muffins/hour Utilization of the “Filling Pan” stage? 375/700 = 53.5% d. [4 pts] Suppose that Macrina Bakery makes both muffins and cupcakes; the product mix is 60% muffins and 40% cupcakes. What is bakery capacity for cupcakes? State the unit clearly. Prepare batter Fill Pan Bake Put Icing13 Batter: 600 per hour (cupcakes and muffins combined) Filling: 700 per hour (cupcakes and muffins combined) Baking: Average baked goods (both combined) takes on average 0.4*12+0.6*23=19.2 (60/19.2)*150 = 468.75 units/hour. 468.75*0.4=187.5 Bakery Capacity for cupcakes: 187.5 units/hour 6. [13 pts] Detroit and Temecula (DT) has in total 100 specialists to handle clients’ tax issues. On average, DT needs to hire 20 specialists every year to replace those who leave the company for any reason. DT’s hiring process follows a Poisson distribution. a. [4 pts] How long does a specialist stay in DT? State the unit clearly. TT=WIP/TR=100/20=5 years There are two types of specialists, personal income tax specialists and corporate tax specialists. No specialists can handle both personal income and corporate tax issues. 60% of those hired each year are personal income tax specialists. On average, a personal income tax specialist stays with DT for 3 years. b. [4 pts] How many personal income tax specialists are there in DT on average? TR=20*0.6=12 WIP=TR*TT=12*3=36 personal income tax specialists c. [5 pts] How long does a corporate tax specialist stay in DT on average? Clearly state the unit. WIP=100-36=64 TR=20-12=8 TT=WIP/TR=64/8=8 years. 7. [18 pts] Sausage Kitchen is a very popular make-to-order gourmet sausage place in downtown Los Angeles. It is a pay-first restaurant in which customers first order and pay before they get seated. At the order-taking counter, there is one cashier. In the kitchen there is one employee and one grill that can cook up to 10 sausages at one time. Each sausage cooks in 15 minutes. In the spacious eating area, there are 80 seats. On a typical day, a customer’s experience is as follows. First she waits, on average, 30 minutes before she places an order. It takes 2 minutes on average to order. She then walks to the eating area. (Walking time is negligible.) After she finds a seat, it takes on average 20 minutes to receive the meal. Finally, it takes 25 minutes to finish the meal. Assume each customer orders exactly one sausage. Recently Sausage Kitchen found that the customers are complaining about long waits. It is not clear, however, whether customers are complaining about waiting before placing orders or waiting before they eat. Sausage Kitchen is willing to address this issue either by increasing the number of cashiers, increasing the number of grills, or increasing the number of seats. a. [3 pts] What is the capacity of the order-taking counter? State the unit. It takes 2 minutes per customer. Capacity=30 customers/hour b. [3 pts] What is the capacity of the kitchen (or the grill)? State the unit. 15 minutes to cook each sausage and the grill can cook 10 sausages at one time. Capacity = (60/15)*10 = 40 sausages/hour, or 40 customers/hour c. [4 pts] What is the capacity of the eating area? State the unit. (Hint: “80 seats” is probably not an appropriate capacity here.) A customer spends 45 minutes (20 minutes wait + 25 minutes eat). Using Little’s Law, Capacity = 80/(45/60) = 106.6 customers/hour14 d. [3 pts] What is the capacity of the restaurant? State the unit. Also identify the bottleneck. Is it order taking, kitchen or eating area? Briefly explain. Capacity of the entire restaurant is 30 customers/hour which is the least of the above three. The bottleneck is “order-taking.” To respond to customer complaints, Sausage Kitchen is wondering if they want to: I. Hire a second cashier. II. Install a second grill. III. Increase the number of seats to 100. e. [2 pts] Which of the above three options should Sausage Kitchen pursue? Briefly Explain. I, because the cashier is the bottleneck. f. [3 pts] After they implement your recommendation above, what is the resulting capacity of Sausage Kitchen? Now the capacity of ordertaking is 60. Sausage Kitchen’s capacity is determined by “grilling” -> 40 customers/hour. 8. [16 pts] You operate an online specialty toy store that takes orders from customers 7 days a week. In an average week, you receive 350 orders for toys. In general, you can fulfill 50% of those orders immediately from your stock. Of the remaining 50%, you order a half from manufacturer XYZ, and you custom-make the other half. All orders must be packaged before being mailed, and on average an order spends one day in your packaging facility before being shipped. There are on average 40 orders at manufacturer XYZ, and 20 orders that you are in the process of custom-making. a. [4 pts] How many days on average does manufacturer XYZ take to complete an order for you? WIP = TR x TT WIP = 40 orders TR = 350 * 0.25 = 87.5 orders per week TT = 40 / 87.5 = 0.457 weeks or 3.2 days b. [4 pts] How many orders are there on average in the packaging facility? WIP = TR x TT TR = 350 orders per week, or 50 orders per day TT = 1 day WIP = 50 * 1 = 50 orders c. [4 pts] What is the average time in days between when you receive an order and when it is shipped? WIP = TR * TT WIP = 50 + 20 + 40 = 110 orders TR = 350 orders per week TT = 110 / 350 = 0.314 weeks, or 2.2 days. d. [4 pts] What is the average time in days between when you receive an order that must be custommade and when it is shipped? 1 day + (20 / 87.5)*7 = 2.6 days15 9. [20 pts] Grocery store XYZ has 5 cashiers. Each hour, an average of 30 customers arrive to checkout, and pay for their groceries. It takes an average of 8 minutes to check out a customer, with a standard deviation of 4 minutes. Assume that customer inter-arrival times are exponential. a. [3 pts] What is the average arrival rate of customers to check-out each hour ()? State the unit. The arrival rate is 30 cus per hour. b. [3 pts] What is the average service rate? State the unit. The service time is 8 mins The service rate is 1/p cus per min = 60/p cus per hour = 7.5 cus per hour c. [6 pts] What is the average time a customer spends in the system, including the wait time and the time checking out? State the unit. Since a is exponentially distributed, the arrival type is M and CVa = 1. Since we know p’s mean and standard deviation, the service type is G and CVp = 0.5 We also know that a = 60/30 = 2mins, m = 5, . The total time is T q + p = 10.8852 mins. d. [4 pts] What is the average number of customers waiting in line to check out? e. [4 pts] Suppose that the inter-arrival times have mean 2 minutes, and standard deviation 1 minute. Do you expect the average number of customers waiting to check out to increase, to decrease, or to stay the same? Explain. The new CVa = ½ = 0.5 mins, decreased from 1. Therefore, both Tq and Iq will decrease. 10. Wells Fargo operates one ATM machine in a certain Trader Joe’s. There is on average 8 customers that use the ATM machine every hour, and each customer spends on average 6 minutes at the ATM machine. Assume customer arrivals follow a Poisson process, and the amount of time each customer spends at the ATM follows an exponential distribution. a. What is the percentage of time the ATM is in use? p = 6 mins, a = 7.5 mins and m = 1. b. What is the average time a customer must wait to use the ATM? State your answer in minutes. Since arrivals follow Poisson process, and p is exponentially distributed, it is an M/M/1 queue with CVa2 = 1, CVp2 = 1. c. On average, how many customers are in line waiting to use the ATM? d. What is the probability that there is nobody in front of the ATM machine? Nobody in front of the ATM machine means the system is empty and the server is idle. Hence, the probability is 1 – u = 20%.16 11. BOA operates 3 ATM machines (next to each other) in a shopping mall. There, an average of 15 customers arrive to use the ATM every hour, and the time between customer arrivals follows an exponential distribution. Each customer is equally likely to spend 3, 5, or 10 minutes at the ATM machine. What is the average time a customer must wait to use the ATM machine? a = 4 mins and m = 3. Since arrivals follow Poisson process, and we know p is neither deterministic nor exponentially distributed, it is an M/G/3 queue with CVa2 = 1. First let’s calculate : Mean of p = 3*33.3% + 5*33.3% + 10*33.3% = 6 mins; Variance of p = (3-6)2*33.3% + (5-6)2*33.3% + (10-6)2*33.3% = 8.67; . Therefore, CVp = 0.49. . 12. Price Waterhouse Anderson Coopers (PWAC) is one of the major accounting firms. As a free service (benefit) to its employees, PWAC has an in-house medical clinic so that employees can consult a doctor at work. The clinic is staffed by one doctor. A Marshall intern has estimated that patients will arrive throughout the day at an average rate of 5 patients per hour, with the interarrival times being exponentially distributed. She has also determined that the average time the doctor spends with a patient is 10 minutes, and that the doctor’s service times are exponentially distributed. The clinic is open 8 hours every working day. (In your analysis, you may assume that the clinic always starts and ends on time and patients who are not seen in the afternoon will be carried over to next morning.) a. What is the utilization rate? p = 10 mins, a = 12 mins and m = 1. b. How many hours will the doctor be idle every day? The doctor is idle with the probability of 1 - u. Consider the doctor is working for 8 hours per day, they are idle for 8 * (1 - u) = 1.33 hours. c. How much time, on average, will a patient spend at the clinic (in the system)? Since arrivals follow Poisson process, and p is exponentially distributed, it is an M/M/1 queue with CVa2 = 1, CVp2 = 1. They stay for 1 hour in total (the waiting time is 50 mins, and the service time is 10 mins) d. Suppose the average salary of an employee (patient) is $50,000 per year. What is the implicit cost per day that the company has to bear for the total employee time spent at the clinic (i.e., forgone productive time)? Each patient spends 1 hour in the clinic. There are 5*8 = 40 patients coming to the clinic every day. Therefore, 40 working hours are wasted per day. Assuming working time 50 weeks per year and 5 days per week, and 8 hours per day, then for each working hour, the company is wasting $50000/50/5/8 = $25. Therefore, the company has to bear lost of $25*40 = $1000 per hour.17 13. After a detailed study on peak hour operations, Eastern Coffee cross-trains its two workers, worker A and worker B. After the cross-training, it takes a constant time of 3 minutes per customer to take the order and receive payment, and a constant time of 3 minutes per customer for coffee preparation. Each worker can handle both activities. Therefore, one customer order can be served by only one of the two workers. If worker A takes your order, for instance, worker A can prepare your coffee. a. How many customers can an individual worker serve during an hour? 60/(3+3) = 10 customers/hour b. During the peak hours, on average, 15 customers show up per hour. What is the utilization rate of workers A and B? A and B are two identical servers: a = 60/15 = 4 mins, p = 60/10 = 6 mins, m = 2 u = p / ma = 6 / (2*4) = 0.75 c. For the arrival rate in b), on average, how many customers are waiting in line? (Assume exponential inter-arrival time) Since inter-arrival time is exponential, the arrival type is M and CVa = 1. Since both processes take constant times, the service type is D and CVp = 0. Western Coffee follows the same flow chart above and each activity takes exactly the same amount of time (a constant 3 minutes for taking order/receiving payment, and a constant 3 minutes for preparing coffee). Western Coffee also has two workers: worker C only takes orders and payments, while worker D only handles the coffee preparation. d. During the peak hours, on average, 15 customers show up per hour. What is the utilization rate of worker C, and of worker D? For C: a = 60/15 = 4 mins, p = 3 mins, m = 1, u = p / ma = 3 / (1*4) = 0.75 For D: a = 60/15 = 4 mins, p = 3 mins, m = 1, u = p / ma = 3 / (1*4) = 0.75 e. For the arrival rate in d) above, on average, how many customers wait in line for worker C? (Assume exponential inter-arrival time) For worker C, the inter-arrival time is exponential: CVa = 1 f. For the arrival rate in d) above, how many customers, on average, wait in line for worker D? Since both workers have the same constant processing time, customers don’t have to wait for the service from worker D. So the answer is ZERO. g. Which store has more customers waiting? Explain briefly. Western has more customers waiting, because Eastern did cross training. Take order and receive payment Prepare the coffee18 14. UNDERGROUND is a small local sandwich store. Customers first line up to place an order. After receiving their orders, customers enter the line for payment. UNDERGROUND has one worker who can prepare a sandwich in exactly 3 minutes, and one cashier who can process payment in exactly 2.4 minutes. During a typical hour, 10 customers visit the store and place an order. Assume the interarrival time follows an exponential distribution. a. What is the utilization rate for the cashier? a = 60/10 = 6 mins, p = 2.4 mins, m = 1, u = 2.4/(1*6) = 0.4 b. What is the utilization rate for the worker? a = 60/10 = 6 mins, p = 3 mins, m = 1, u = 3/(1*6) = 0.5 c. What is the average waiting time in line (before placing an order)? State the unit. CVa = 1 and CVp = 0, p = 3 mins, m = 1, u = 0.5 d. What is the average number of customer waiting in line (before placing an order)? e. The store manager understands that the inter-arrival time may not exactly follow exponential distribution. As a result, he handcounts the number of people waiting in line for placing the order, and find, on average, 1.5 people waiting in line. How would your answer to part d) change? WIP = 1.5 customers, FR = 10 cus/hour, FT = 1.5 / 10 = 0.15 hours = 9 mins f. After receiving the order, how long will the customer wait in line for payment? State the unit. Since both workers have constant processing time, and the worker in step 2 (cashier) has shorter processing time than the worker in step 1, customers don’t have to wait for the service from the cashier. So the answer is ZERO. 15. [24 pts] University Airport, a small airport operated by a well-known university is required to screen checked baggage for all of its passengers. Before they can check-in, all passengers must take their bags to a screening station. At the screening station, the current method for screening checked baggage is for one employee to search each piece by hand. This takes an average of 4 minutes per piece of baggage, and the hand-search time is exponentially distributed. The airport is considering the acquisition of an automatic baggage-screening machine to replace hand screening. The new baggage-screening machine will only take an average of 3 minutes per piece of baggage, with a standard deviation of 2 minutes. The airport has an average of 4 flights per day, and an average of 30 passengers per flight, and on average only 50% of the passengers have one checked piece of luggage (the other 50% have no checked-in luggage). The airport is required to screen all passengers. The airport operates 5 hours per day, and passenger arrivals are Poisson, and the average rate of passenger arrivals is constant over the time the airport is open. a. [3 pts] At what rate do pieces of baggage requiring screening arrive (in bags per hour)? 30 passengers/flight * 4 flights/day = 120 passengers/day 120 passengers/day * 0.5 bag/passenger = 60 bags/day Place the order Pay the cashier19 60 bags/day / 5 hours/day = 12 bags/hr FOR THE REST OF THE PROBLEM ASSUME AN ARRIVAL RATE AT SCREENING OF 12.5 BAGS PER HOUR b. [6 pts] What are the utilizations for the current system and the proposed automatic system? Arrival Rate (1/a) = 12.5 bags/hr  a = (1/12.5) hr/bag = (1/12.5)*60 = 4.8 minutes Current Hand-Screening: p = 4 min, u = (p/ma) = (4/4.8) = 5/6 = 0.833 Proposed Automatic-screening: p = 3 min, u = (3/4.8) = 0.625 c. [8 pts] What is the average amount of time an average passenger spends in the system (waiting and being screened) under the current hand-screening system and the proposed automaticscreening system? Current Hand-Screening: CVa = 1, CVp = 1, m = 1, u = 5/6 mins mins             2(m+1)-1 2 2 2 2 a p q q p u 4 (5 / 6) 1 + 1 CV + CV T = = = 20 m 1- u 2 1 (1 / 6) 2 T = T + p = 20 + 4 = 24 Proposed Automatic-Screening: CVa =1, m = 1, u = 0.625 Sa=2 min and p = 3 min  CVp = 2/3 mins mins             2 2 2(m+1)-1 2 2 a p q q p u 3 (0.625 CV + CV 1 +( ) 23 T = = = 3.16 m 1- u 2 1 (0.375) 2 T = T + p = 3.16 + 3 = 6.16 d. [2 pts] If the airport acquires the automatic-screening system, what is the average number of passengers that will be waiting in line to be screened? Iq = (Tq/a) = 3.16/4.8 = 0.752 passenger e. [5 pts] If the airport is willing to spend up to $2000 more per month (30 days) for the new system (versus what the old system costs, including labor), what value are they implicitly placing on a minute of passenger time spent in the screening process (waiting plus screening time)? Assume that the average amount of time an average passenger spends in the system (waiting and being screened) under the current hand-screening system is 35 minutes, and is 20 minutes under the proposed automatic-screening system. (This may or may not be the same as your answer to (c)) Difference in avg. wait time (T) of a screened customer = 35-20 = 15 minutes. Total number of customers screened per month = (12.5 customers per hour)*(5 open hours per day) *(30 days per month) = 1875 customers Total expected savings in wait time in screening system = 1875*15 =28,125 minutes Value of one minute of wait = $2000 / (28,125 min) = $0.0711/min ≈ 7.1 cents per min 16. You are managing a restaurant. You do not accept reservations, and so all tables are walk-in. Customers that arrive and request a table are divided as follows: 50% require a table for 2, 40% require a table for 4, and 10% require a table for 6. The waiting time for a table for 2 is on average 10 minutes, the waiting time for a table for 4 is on average 30 minutes, and the waiting time for a table for 6 is on average 40 minutes. The number of parties waiting for a table for 2 is on average 2. a. [5 pts] What is the arrival rate of customers requesting tables (of any size) per hour into your restaurant? WIP = (flow rate)*(flow time) 2 = (0.5x)*(10/60)20 x = 24 24 customers per hour b. [5 pts] What is the average number of parties waiting for a table for 4? WIP = (flow rate)*(flow time) = (24*0.4)*(30/60) = 4.8 parties Suppose that after a table is seated, regardless of the number of customers seated at the table, 1/3 of the tables order from the prix fixe menu and 2/3 of the tables order from the a la carte menu. Suppose also that there are on average 15 tables ordering from the prix fixe menu and 10 tables ordering from the a la carte menu. Assume that the rate at which customers are seated is 30 tables per hour. (This may or may not be consistent with your answer in part 1 above.) c. [5 pts] How long on average does a table ordering from the prix fixe menu occupy the table? WIP = (flow rate)*(flow time)  15 = (30*1/3)*(x)  x = 1.5 hours d. [5 pts] Suppose that the percentage of tables that order from the prix fixe menu increases (so that it is now greater than 1/3), but that the rate at which customers arrive remains 30 tables per hour. What will happen to the average number of occupied tables: will it be more than 10+15=25 tables or less, and why? A table that orders from the a la carte menu on average occupies the table for 10 / (30*2/3) = 0.5 hours. Therefore, if the percentage of tables that order from the prix fixe menu increases, then the average flow time will increase. Since WIP = (flow rate)*(flow time), the WIP (that is, the average number of occupied tables) must increase since the flow rate remains constant at 30 tables per hour. 17. The smartphone repair shop Winc had a bad year and they made major changes to their processes, including stocking all spare parts needed in their own shop and eliminating all sources of variability. The revised process is as follows. The first step performed on any broken phone is Diagnosis which takes 50 minutes per phone. This is followed by a Prepare activity in which the phone is prepared for the actual repair. This activity includes locating the spare part and takes 30 minutes. The final step is Repair which takes 40 minutes per phone. Winc currently has four employees of which two employees work on Diagnosis, one employee works on Prepare and one employee on Repair. Based on this information, please answer the following questions. a. [4 pts] Draw a process flow diagram depicting this process. b. [6 pts] What is the capacity of the diagnosis, prepare, and repair stations in phones per hour? Capacity of Diagnosis station = 2 * 60/50 = 2.4 phones/hr Capacity of Prepare station = 60/30 = 2 phones/hr Capacity of Repair station = 60/40 = 1.5 phones/hr c. [4 pts] What is the capacity of Winc in phones per hour? Which of the three stations (Diagnosis/Prepare/Repair) is the bottleneck? So, capacity of the process is 1.5 phones per hour and the Repair station is the bottleneck. d. [6 pts] Suppose Winc hires an additional employee who can perform any activity. Which single activity should Winc have this employee perform? What is the new process capacity? Is the bottleneck the same station as in your answer to (c)? Diagnosis Prepare Repair21 The Bottleneck is Repair station so the hired employee should do this. The revised capacity of Repair station=2x1.5=3 phones/hr. Now the bottleneck is Prepare station (which is different from (c)), and the new process capacity is 2 phones/hr. e. [4 pts] Disregard part (d). Winc realizes that one of the employees currently performing Diagnosis can perform Repairs as well. So, Winc decides to make this employee do the Repair activity instead of Diagnosis. Is this a good idea? Please explain why or why not. This is not a good idea because with only one employee Diagnosis station’s capacity drops to 1.2 phones per hour. So, this change lowers the process capacity to 1.2 phones per hour. f. [4 pts] Again disregard part (d). Winc decides to cross-train its employees, so that all four employees can diagnose, prepare, and repair phones. Then, will the process capacity be more or less than your answer to part (c)? What is the new process capacity? Each employee requires 50+30+40=120 minutes with one phone, and so each employee can process ½ phone per hour. There are 4 employees, and so the new process capacity is 2 phones per hour. 18. [25 pts] An automated call center has 3 computer servers, which allow customers to hear recorded messages and, if necessary, can direct a call to an appropriate agent. You are hired by the manager to evaluate the performance of these computer servers. With the knowledge you learned from BUAD311, you collect the following information in order to answer questions from the manager.  Currently there is one telephone line for each server  On average, each computer server receives 30 calls per hour.  Arrivals follow a Poisson process.  The time a customer spends with the computer server is exponentially distributed with a mean of 90 seconds. a. [4 pts] What is the utilization rate of the servers? For each line, 1/a = 30 calls/hour. => a = 2 min p = 90 sec = 1.5 min, utilization rate u = 1.5/2 = 75% b. [5 pts] On average, how long will a customer wait before being serviced by one of the servers? 4.5min 0.25 0.75 1.5* 1 2 2( 1) 1 2 2          a p m q CV CV u u p m T For part c) – f) The manager wants you to help with the cost. Currently there are two types of variable costs in consideration. First, it is estimated the cost of waiting is $0.02 per second per call. In other words, the call center will lose $0.02 for a customer waiting in line for a second. Secondly, the call center incurs server idle cost as $0.05 per server per second. In other words, the call center will lose $0.05 (you may think of it as opportunity cost) for a server being empty for a second. c. [4 pts] What is the total cost of waiting per hour for the call center? 90 calls/hour * 4.5 min * 60 * $0.02/sec/call = $486 d. [4 pts] What is the expected total idle cost per hour for the call center? For a server, P0 = 1 – u =0.25 Total expected idle cost is 3 * 60 min * 60 sec * 0.25 * $0.05/server/second = $135 For part e) - f)22 You also recommend pooling all incoming telephone calls into one telephone line and using new software that will let the first available server to pick the call that is first in line. Please answer question e) through f) given this new approach. e. [4 pts] On average, how long will a customer wait before being serviced by one of the servers? m=3, p = 1.5 min, a = 2/3 min, CV = 1 for both 1.182min 0.25 0.75 * 5. 3 1 1 2 2( 1) 1 2 2 8 1           a p m q CV CV u u p m T f. [4 pts] On average, how many customers will be in the system, including waiting and being served? I = Tq/a + mu = 1.182 / (2/3) + 3 * 0.75 = 4.02 19. [25 pts] Trojan Bank is a small community bank that offers loans to residential customers. Each loan application goes through 3 steps. First, the loan originator conducts a basic screening process by gathering information about the applicant such as credit scores. Then, the application is reviewed by an underwriter, who provides a recommendation. The VP of Lending, taking into account the information gathered by the loan originator and the underwriter’s recommendation, makes the final approval decision. Trojan Bank divides their loan applications into 3 groups: Group 1 (New Customers with Excellent Credit Record): 40% of the applications Group 2 (New Customers with Poor Credit Record): 30% of the applications Group 3 (Existing Customers): 30% of the applications The required processing time for each group by each person is given in the following table. All times are in minutes. The expected profit for each application is also provided. Group Basic Screening by Loan Originator Review by Underwriter Decision by VP Expected Profit 1 40 60 30 $600 2 50 90 60 $360 3 30 20 20 $340 All three persons work 40 hours per week, and Trojan Bank receives about 40 loan applications each week. a. [5 pts] Which of the three persons (Loan Originator, Underwriter, or VP) is the bottleneck? Avg time spent by Loan Originator per app = (0.4 * 40) + (0.3 * 50) + (0.3 * 30) = 40 min Avg time spent by Underwriter per app = (0.4 * 60) + (0.3 * 90) + (0.3 * 20) = 57 min Avg time spent by VP per app = (0.4 * 30) + (0.3 * 60) + (0.3 * 20) = 36 min The Underwriter is the bottleneck because s/he is the slowest person. b. [5 pts] What is the utilization of the loan originator? Underwriter? VP? The number of hours per week available = 40hr * 60 min/hr = 2400 min Utilization of Loan Originator = (40 app * 40 min/app) / (2400 min) = 66.67% Utilization of Underwriter = (40 app * 57 min/app) / (2400 min) = 95% Utilization of VP = (40 app * 36 min/app) / (2400 min) = 60% Suppose that Trojan Bank decides to run a promotion that is expected to generate a 20% increase in loan applications. c. [4 pts] Based on the expected profit information in the table, how much is the expected profit increase per week if Trojan Bank can capture all the incremental demand? The original expected net profit per job is = (40%*$600) + (30%*$360) + (30%*$340) = $45023 The original expected net profit per week is = 450*40 = $18,000 The expected increase per week is $18,000 * 20% = $3,600 d. [4 pts] Can Trojan Bank process all of the new loan applications? If the loan volume increases by 20%, it means that the bank receives 48 applications per week. The Underwriter spends an average of 57 min per application, and thus, reviewing all 48 applications require a total of (57 * 48) = 2,736 min. This exceeds the number of available minutes in on week, so the bank does not have the capacity to process all of the new loan applications. e. [4 pts] Suppose Trojan Bank decides to hire an additional underwriter, who can process the loan at the same speed of the existing underwriter. Can Trojan Bank process all of the new loan applications? Which person is the bottleneck? Yes, in this case, we double the capacity of the underwriter, so the avg time spent by per Underwriter per app is now (57 min / 2) = 28.5 min. So, the bottleneck has now shifted to the Loan Originator. Screening 48 applications takes a total of 32 hours, so the bank can process all of the new loan applications in one week. f. [3 pts] After a cost-benefit analysis, Trojan Bank finds hiring an additional underwriter unprofitable. Instead of hiring an additional underwriter, Trojan Bank decides to outsource some of the applications, or some of the applications from a specific group, to an affiliate. As outsourcing the applications means that the affiliate instead of Trojan Bank would capture the profit, Trojan Bank needs to decide which group of application should be outsourced. What is your recommendation and what is your rationale? Trojan Bank should outsource Group 2, New Customers with Poor Credit Record. Because the Underwriter is the only bottleneck, we need to prioritize jobs based on the profit of the job and the bottleneck resource time. For group 1, each minute of bottle resource generates 600/60 = $10. For group 2, each minute of bottle resource generates 360/90 = $4. For group 3, each minute of bottle resource generates 340/20 = $17 By comparing the effective usage of bottleneck resource, we conclude that Trojan Bank should outsource group 2. 20. [30 pts] Ricky Shaw is a small shop that specializes in custom-made hand-bags. To make a handbag, materials are first cut and then weaved together. Currently, two cutting machines are available and each can cut enough materials for 1 bag in exactly 30 minutes. Ricky Shaw recently retired two old weaving machines and installed a new weaving machine NCX-10, which can weave 1 bag in exactly 20 minutes. With the growth of the business, Ricky Shaw expects customers to arrive randomly with a mean demand of 3.5 bags per hour. a. [5 pts] Draw the flow chart of Ricky Shaw using two triangles and two rectangles. b. [5 pts] What is the capacity of Ricky Shaw? Does Ricky Shaw have enough capacity? The capacity of cutting machines 2*(60/30)=4 jobs/hour The capacity of weaving machine 1*(60/20)=3 jobs/hour The capacity of Ricky Shaw is 3 jobs/hour. The capacity of Ricky Shaw is smaller than the mean demand of 3.5 bags per hour; therefore, Ricky Shaw does not have enough capacity. Cut Weave24 c. [4 pts] The workers for the cutting machines get their jobs done as quickly as possible, and Ricky Shaw finds on average there are 5 bags (either waiting or being cut) in the cutting stage. What will be the flow rate of the cutting stage? What is the average time a bag stays in the cutting stage? As the workers for the cutting machines get their jobs done as quickly as possible, the flow rate of cutting stage would be the demand rate 3.5 jobs/hour because the cutting stage has enough capacity to handle the demand. By Little’s Law, Flow time = WIP / Flow Rate = 5 / 3.5 = 1.43 hours. d. [4 pts] Suppose, like the workers at the cutting machine, the workers at the weaving machine get their jobs done as quickly as possible, what will happen to the WIP and flow time for the weaving stage? The arrival rate of the weaving stage is higher than the capacity of the weaving machine. The WIP and flow time for the weaving stage will grow to infinity. e. [4 pts] Ricky Shaw understands that the weaving machine is the “Herbie” and decides to slow down the production at the cutting stage so that while the weaving machine is working all the time, at most 3 bags are in work-in-process (either waiting or being weaved) in the weaving stage. What will be the flow rate of weaving stage? What will be the maximum time a bag stays in the weaving stage? The flow rate in this case would be the system flow rate, which is limited by the system capacity (weaving machine capacity) of 3 jobs. As the weaving machine is working at all the time, the flow rate is 3 jobs / hour. As the cutting stage has at most 3 bags and each bag takes exactly 20 minutes, the maximum flow time can be calculated by, FT = WIP / FR = 3 / 3 = 1 hours. f. [4 pts] Rather than slowing down the production of the cutting stage, Ricky Shaw decides to bring back one of the retired machine which can weave 1 bag in exactly 45 minutes. Will Ricky Shaw have enough capacity? The capacity of old machine is 1*(60/45) = 1.33 jobs / hour. As both old machine and new machine can weave, now Ricky Shaw can weave 3+1.33 = 4.33 jobs in an hour, which is greater than the demand rate of 3.5 jobs/hour. Ricky Shaw will have enough capacity. g. [4 pts] With the help of the old machine that can weave 1 bag in exactly 45 minutes, the process stabilizes and Rick Shaw finds on average there are 5 bags (either waiting or being cut) in the cutting stage and 2 bags (either waiting or being weaved) in the weaving stage. What will be the system flow rate? What is the average total processing time for a custom-made hand-bag order? As the process stabilizes the system flow rate is the demand rate, which is 3.5 jobs/hour. The average total processing time for a custom-made hand-bag order will be Flow time = WIP / Flow Rate = (5 + 2) / 3.5 = 2 hours. 21. (25 points) In apparel manufacturing, capacity is one of the most important factors used by buyers for vendor selection. Consider the following process flowchart for Trojan Apparel. There are two products—Basic and Fashion. First, the raw material (fabric) is sent to the Cutting Workshop. The average time to cut the fabric is 30 seconds per piece for a Basic item, and 40 seconds per piece for a Fashion item. There is 1 worker in the Cutting Workshop. Next, the cut fabric goes to the Sewing Workshop for sewing. Sewing workshop B sews only Basic products and it takes an average processing time of 5 minutes per piece per worker. There are 8 workers in workshop B. All fashion25 items are sewn in workshop F where it takes an average processing time of 8 minutes per piece per worker. There are 10 workers in Sewing workshop F. The (Quality Assurance) QA workstation monitors the quality of Fashion items by have screening (checking the quality) of fashion items finished at workshop F. There is 1 QA worker whose average inspection time is 20 seconds per piece. Any items not meeting quality standards are sent back by QA will to workshop F for rework. The Packing Workshop packs both Basic and fashion products. It takes a worker an average processing time of 1 minute per piece. The packing workshop has 2 workers. For part a) to part e), please assume i. Trojan Apparel makes ONLY fashion apparel ii. On average, 5% of the units fail the QA inspection and are sent back to Workshop F for rework. iii. For simplicity, no item will be sent for rework twice. a) (2pts) What is the capacity (pieces/hour) of the Cutting Workshop? 60*60/40 = 90 / hour b) (3pts) What is the capacity (pieces/hour) of the Sewing Workshop F? 10 * 60/(8*1.05) = 71.4/ hour c) (2pts) What is the capacity (pieces/hour) of QA? 60*60/20 = 180 / hour d) (2pts) What is the capacity (piece/hour) of the Packing Workshop? 2*60/1 = 120 / hour e) (2pts) Which workshop is the bottleneck? Sewing Workshop F For part f) to part k), please assume Cutting Workshop Sewing Workshop B Sewing Workshop F QA Packing Workshop Basic Fashion26 i. Trojan Apparel makes both Basic and Fashion ii. On average, 5% of the units fail the QA inspection are sent back to Workshop F for rework. iii. For simplicity, no item is sent for rework twice. iv. The average hourly demand is 20 pieces/hour for Basic and 60 pieces/hour for Fashion. f) (2pts) What is the utilization of Sewing Workshop B? 8*60/5 = 96 /hour 20 / 96 = 20.83% g) (3pts) What is the utilization of the Cutting Workshop? Utilization = (20 *30 + 60 * 40)/(60*60) = 3000/3600= 83.33% h) (2pts) What is the utilization of Sewing Workshop F? 60 / 71.4 = 84.03% i) (2pts) What is the utilization of the Packing Workshop? 80 / 120 = 66.67% j) (2pts) Which workshop is the bottleneck? Sewing workshop F since it has the highest utilization k) (3pts) Can Trojan Apparel produce more Basic product besides the stated demand in (iv) above? If yes, how many more (show your work)? If no, explain. Yes. Workshop B can make additional 96 – 20 = 76 pieces. Cutting station can make (3600 – 3000)/30 = 20 more pieces. Packing workshop can make additional 120 – 80 = 40 pieces. Thus Trojan can make min{76, 20, 40} = 20 more pieces of Basic.27 23.(24 points, 4pts per part) Currently, 50 people per hour arrive at the emergency room (ER) of Trojan Hospital. On arrival, they take a number and wait in the Check-In area. When their number is called, they move to the Registration area where they are registered by nurses. The registration process takes about 2 minutes. After Registration, the patients move to the Waiting Room area, where they wait until called to the doctors’ examination rooms. When their turn comes, each patient is seen by a doctor. After being examined by a doctor, the patient exits the process, either with a prescription or with admission to the hospital. About 10% of patients are admitted to the hospital. On average, 30 people are waiting in the Check-In area (waiting to be registered) and 40 patients are registered and waiting (in the Waiting Room area) to see a doctor. Among patients who receive prescriptions, average time spent with a doctor is 5 minutes. Among those admitted to the hospital, average time spent with a doctor is 30 minutes. a. On average, how many minutes do patients wait in the Check-In area? FR=50 patients/hr=50/60 patients/min=5/6 patients/min WIP=30 patients FT=30/(5/6)=36 minutes b. On average, how many patients are being registered at any one time? FT=2 mins FR=5/6 p/mi WIP=2*(5/6)=1.67 patients c. On average, how many minutes does a patient spend in the Waiting Room? WIP=40 patients FR=5/6 p/min FT=40/(5/6) =48 minutes d. On average, how many patients are being examined by the doctors? FT=5(0.9) + 30(0.1)=7.5 mins FR=5/6 p/min WIP=7.5*(5/6)=6.25 patients e. On average, how many patients are there in the ER at any one time? 30+1.67+40+6.25=77.92 patients28 f. On average, how many minutes does a patient spend in the ER? 36+2+48+7.5=93.5 mins 22. (27 points) Trojan café serves made-to-order sandwiches. First, customers line-up at the cashier to choose and pay for their sandwich. Second, customers wait while their sandwich is being made. Finally, customers pick up their sandwich and go to the eating area. Customers arrive to Trojan café in accordance with a Poisson process having rate 15 customers per hour. The amount of time a customer spends choosing and paying for his or her sandwich is exponentially distributed, with mean 3 minutes. Then, each sandwich takes 5 minutes, with standard deviation 2 minutes, to make. Trojan café has 3 employees. There is 1 employee that is the cashier, and 2 employees that make sandwiches. a. What is the percentage of time the cashier is idle? (5 points) Customer mean inter-arrival time: a=60/15=4 minutes Cashier mean service time: p = 3 minutes Number of cashiers: m = 1 Utilization: u=p/(a*m) = 0.75 The cashier is idle 1-0.75 = 0.25 = 25% of the time. b. What is the average amount of time a customer spends waiting to be served by the cashier? (3 points) Tq = p m æçè ö÷ø ´ u 2(m+1)-1 1- u æçè ö÷ø ´ CV 2 a + CV 2 p 2 æçè ö÷ø = 3´ 0.75 1- 0.75 æçè ö÷ø ´1 = 9 minutes c. What is the average amount of time a customer spends waiting for his or her sandwich to be made, after having ordered and paid? (Hint: Assume that the arrival process to the sandwichmaking process is Poisson.) (8 points) Customer departure rate from cashier = customer arrival rate to the sandwich maker Customer mean inter-arrival times: a = 4 minutes29 Customer mean service time: p = 5 minutes Number of servers: m = 2 Utilization: u=p/(a*m) = 5/8 = 0.625 CVa = 1; CVp = 2/5 = 0.4 Tq = p m æçè ö÷ø ´ u 2(m+1)-1 1- u æçè ö÷ø ´ CV 2 a + CV 2 p 2 æçè ö÷ø = 5 2 æçè ö÷ø ´ 0.625 6-1 1- 0.625 æçè ö÷ø ´ 1+ 0.42 2 æçè ö÷ø = 1.96 minutes d. Suppose Trojan Café cross-trains its workers so that all 3 employees can perform all duties. In this setting, each arriving customer is served by one employee, that both takes the order and accepts payment, and makes the sandwich. You may assume that the time it takes to take the order and accept payment is independent of the time it takes to make the sandwich, and the service times are exponentially distributed for taking order/accepting payment and for making the sandwich. i. Now, what is the average service time? (2 points) p = 3+5 = 8 minutes ii. Now, what is the coefficient of variation of the service time? (Hint: the standard deviation of an exponential random variable is equal to its mean.(3 points) Service time variance = (order and paying time variance) +(sandwich making time variance) = 32 + 22 = 13 minutes Service time standard deviation = Sqrt(13) Coefficient of variation = Sqrt(13)/8 = 0.45 iii. Now, what is the average amount of time a customer spends waiting to be served by the cashier? Is this more or less than the total waiting time (the sum of your answers to parts 2 and 3) when 1 employee was the cashier and 2 employees made sandwiches? (3 points) Customer mean inter-arrival time: a = 4 minutes30 Cashier mean service time: p = 8 minutes Number of servers: m = 3 Utilization: u=p/(a*m) = 8/12 = 2/3 CVa = 1; CVp = 0.45 Tq = p m æçè ö÷ø ´ u 2(m+1)-1 1- u æçè ö÷ø ´ CV 2 a + CV 2 p 2 æçè ö÷ø = 8 3 æçè ö÷ø ´ 0.667 8-1 1- 0.667 æçè ö÷ø ´ 1+ 0.452 2 æçè ö÷ø = 2.3 minutes The time is much less, since 2.3 minutes < 9+1.96=10.96 minutes. iv. In this setting, suppose the customer arrival rate increases to 35 customers per hour. What is the minimum number of (cross-trained) employees that Trojan café must staff? (3 points) The service time is 8 minutes. Therefore, one server can handle 60/8 = 7.5 customers per hour. In order that the utilization is less than one, we must have: (arrival rate) / (service capacity) < 1; i.e., 35 / (7.5*m) < 1. Hence m > 35/7.5 = 4.667. The minimum number of employees is 5.