CH-9-11 Service Processes – Operations and Supply Chain Management : Book Guide

This Document Contains Chapters 9 to 11 CHAPTER 9 SERVICE PROCESSES Discussion Questions 1. What is the service package of your college or university? The service package for your school could be extended beyond the degree you will earn. If you are a resident on the campus, you are using the housing that is offered by your school. Additionally, you are able to buy all your books at the book store on the campus or your school ships them to you. Dining facilities are often contracted out to food service companies so you can eat on campus. The library is another service that is provided. Less obvious services include the maintenance of the building, clearing the parking lot of snow, cleaning the building after hours. 2. Relative to the behavioral science discussion, what practical advice do you have for a hotel manager to enhance the ending of a guest's stay in the hotel? How about putting a couple of Starbuck's coffee coupons in the envelope along with the bill that is slipped under the door in the early morning of check-out day? For guests staying several days, arrange to have the manager say a personal good bye. 3. List some occupations or sporting events where the ending is a dominant element in evaluating success. Attorney's closing argument at trial; the professor's final lecture in the course; the comedian saving his best joke for last, the magician saving best trick for last; the "dismount" in all gymnastic exercises; the entry into the water in diving competitions. 4. Behavioral scientists suggest that we remember events as snap shots, not movies. How would you apply this to designing a service? What this means is that not all parts of the service are of equal impact in one's memory, and therefore place your resources on those areas that give you the most bang for the "memory buck." That is, think about which snap shots you want to clearly "frame" in the customer's memory-- these might be thought of as the sweet spots of the service. You might even want to create them. For example, a Lexus dealership in Los Angeles hands out a long stemmed rose when a customer comes to the service counter to pay for car servicing. While the general car servicing process involves mainly inconvenience, the "movie" experience was dominated by a positive snapshot of receiving a rose. Another example: At Splash Mountain at Disneyland, photographers regularly take pictures of people coming to the end of their ride. These become the keepsakes long after the details of the day are forgotten. 5. Some suggest that customer expectation is the key to service success. Give an example from your own experience to support or refute this assertion. Most responses will probably support rather than refute this assertion. 6. Where would you place a drive-in church, a campus food vending machine, and a bar’s automatic mixed drink machine on the service-system design matrix? Referring to the exhibit 9.3, the church would likely be “face-to-face tight specs.” The most logical placement for the latter two would appear to be “Internet and on-site technology”. 7. Can a manufacturer have a service guarantee in addition to a product guarantee? Certainly. With some makes of automobiles, a rental car is guaranteed if the product fails. Many products are sold with warranties promising free and prompt service. 8. Suppose you were the manager of a restaurant and you were told honestly that a couple eating dinner had just seen a mouse. What would you say to them? How would you recover from this service crisis? The only thing you can do is to try to overcompensate for the event. Make dinner free. Promise to contract with an exterminator. Offer them another free dinner after a “cooling off” period. The Tylenol response is a good model to following in this situation. 9. What strategy do the following organizations seem to use to manage customer-introduced variability? a. EBay Low Cost Accommodation b. Ritz-Carlton Hotels Classic Accommodation c. New airline check-in procedures Low Cost Accommodation 10. How have price and variety competition changed McDonald’s basic formula for success? McDonald’s originally emphasized quick delivery of a limited menu. In response to competition, McDonald’s has continuously expanded its menu and instituted customized ordering. Some evidence suggests that service, including quick delivery, has suffered as a result. 11. Could a service firm use production line approach or self-serve design and still keep a high customer focus (personal attention)? Explain and support your answer with examples. Yes, this is possible. In many instances, the customer actually desires more technology and feels that the service will not have as high a quality without it. For example, dental care with its mechanized X-ray techniques requires far less customer time and less exposure to radiation. The self-service design can also support customer needs, for example the ATM provides customer access to funds at a variety of locations and 24 hours a day. 12. Why should a manager of a bank home office be evaluated differently than a manager of a bank branch? Since a bank home office typifies low contact quasi-manufacturing, while a bank branch typifies medium contact mixed service, the problems faced by management differ considerably. 13. Identify the high-contact and low-contact operations of the following services: a. A dental office. Dental office high contact includes waiting rooms, receptionists, dentist(s), hygienist(s), x-ray, etc., while labs would be low contact. b. An airline. Airline high contact includes reservations desk, loading concourse, plane with crew and attendants, etc. Low contact includes maintenance, baggage handling, tower operations, etc. c. An accounting office. In an accounting office, high contact includes reception and CPAs, while low contact includes records, computer, library, etc. d. An automobile agency Automobile agency high contact includes showroom and offices. Low contact includes maintenance, preparation, records-files, etc. e. Amazon.com High contact include the, customer service representatives. Low contact includes order pickers, QA checkers, order packers, buyers who purchase the product for Amazon and the computer technicians who keep the servers running. 14. At first glance, asking the customer to provide their own service in the self-service approach may not seem very consumer friendly. What are some of the characteristics of self-service operations that have led to their acceptance and significant popularity? There are many benefits, both for the consumer and the company. For the consumer, selfservice operations can often be faster (self-checkout at grocery store), less expensive, more convenient (24-hr ATM service), and the customer can have the service done his/her own way. For the company it can cut down on labor costs, attract repeat customers, keep the customer occupied during service so they are not as bored, and can cut down on complaints about the quality of service – will the customers complain about themselves? A good example of this is a steak restaurant where the customers grill their own steaks over a long charcoal grill at the center of the restaurant. Steaks never go back because they were overcooked. For the self-service approach to work, it must be well-designed, easy to use, and dependable. If the self-scanner at the grocery store 15. Do you think a service operation can be successful by developing a system that combines characteristics from the three contrasting service designs presented in the chapter? Why or why not? Please provide examples. Generally this would not be a good idea – it is akin to the concept of straddling discussed in the strategy chapter. It is s fairly strong argument that the “personal attention” approach is not compatible with the other two service designs presented. However, there are examples of successfully merging the “production line” and “self-service” approaches to a degree. For example, fast food operations often have the customer fill their own drink after placing their order. That reduces the workload on the service staff, occupies the customers while they are waiting for the order to be ready, and allows them to fix the drink exactly as they would like it. Objective Questions 1. What is the term used for the bundle of goods and services that are provided in some environment by every service operation? Service package 2. Are service operations with a high degree of customer contact more or less difficult to control than those with a low degree of customer contact? More difficult 3. List at least three significant ways in which service systems differ from manufacturing systems. You cannot inventory service, the process is part of the product, patents and copyrights cannot be obtained, service is not tangible, specific training and certifications are often provided. 4. A ride at an amusement park is an example of a service operation where there is direct contact between the customer and server, but little variation in the service process -- neither the customer nor server has much discretion in how the service will be provided. As shown on the Service System Design Matrix, which type of service is being delivered? Face-to-face, tight specs 5. As the degree of customer contact increases in a service operation, what generally happens to the efficiency of the operation? It decreases 6. As the degree of customer contact increases in a service system, what worker skills would be more important, clerical skills or diagnostic skills? Diagnostic skills 7. An important difference between service and manufacturing operations is that customers induce far more variability into the operation in a service system. Name at least three of the basic types of variation that customers bring to a service system. Arrival variability, request variability, capability variability, effort variability, subjective preference variability. 8. Flowcharts are a common process design and analysis tool used in both manufacturing and services. What is a key feature on flowcharts used in service operations that differentiates between the front-office and back-office aspects of the system? Line of visibility 9. What are the “Three Ts” relevant to poke-yokes in service systems? Task to be done, Treatment of the customer, Tangible features of the service. 10. List at least four characteristics of a well-designed service system. 1. Each element of the system is consistent with the operating focus of the firm. 2. It is user-friendly. 3. It is robust. 4. It is structured so that consistent performance by its people and system is easily maintained. 5. It provides effective links between the front-office and back-office parts of the system. 6. It manages the evidence of service quality so that customers see the value of the service. 7. It is cost-effective. 11. A psychological therapist treats patients according to their individual needs. Patients are all treated in the same office on a scheduled basis. Each patient’s treatment is customized for the individual according to the therapist’s professional training. Which of the contrasting service designs in the chapter would best describe the therapist’s service operation? Personal attention approach CASE: Pizza USA - Teaching Note This exercise is used to illustrate the ideas and issues involved in designing a service product and the processes to deliver the service. It is highly unlikely that any student has never had a pizza delivered. Therefore, they should all have personal experience as customers, thus making part 1 very straight-forward. Ahead of class, ask that the students to prepare a list that would describe their personal definition of very good to excellent pizza delivery service. Remind them that they can assume that the pizza restaurant can make the pizza any way that they want; the objective is to focus on the delivery service. This focus could include concerns about what happens to the pizza while it is being delivered. Start the students with a review/discussion of the issues in designing service products and services. Then break them into groups to compare their individual lists (about five minutes or so). Next, go to the board and ask each group to give two or three requirements. The discussion around these requirements often results in students thinking up more requirements or debating the meaning of the requirements. Next ask the groups to organize these items under major headings to make the next step somewhat easier to do. Once the list is divided under major headings, take one of the headings (usually one with relatively few items such as “timeliness”) and ask them how they would do this if they ran the restaurant. Then ask them how they would measure this and would there be any qualifications to a normal target. For example, regarding “fast delivery”, most students quickly figure out that this will be affected by the delivery area (service radius). Most also recognize that the target should be adjusted in case of bad weather or special situations (like Super Bowl Sunday). If someone says, “we’ll get a computer system to do this”, ask them how they would evaluate such a system. What factors should the system consider? How would you determine that such a system is giving you reasonably accurate output? Challenge them to think of a non-computer method. Work through the list of requirements, given the time available for this exercise. Many thanks to Mark Ippolito of Indiana University – Purdue University Indianapolis for contributing this exercise. CHAPTER 10 WAITING LINE ANALYSIS AND SIMULATION Discussion Questions 1. Distinguish between a channel and a phase. A channel is the initial service point of a queuing system. A phase refers to the number of stages that the service points provide. It is possible to have single to multiple service channels and single to multiple service phases. 2. In what way might the first-come, first-served rule be unfair to the customer waiting for service in a bank or hospital? In a bank, FCFS may be perceived to be unfair by customers who have large accounts, but who must wait while the less "important” customers obtain service. In a hospital, especially in an emergency room, FCFS is probably the exception rather than the rule. FCFS would be unfair when a patient with a minor problem is treated before another experiencing severe pain. 3. Define, in a practical sense, what is meant by an exponential service time. An exponential service time means that most of the time, the service requirements are of short time duration, but there are occasional long ones. Exponential distribution also means that the probability that a service will be completed in the next instant of time is not dependent on the time at which it entered the system. We can see that a barber, for example, does not fit an exponential distribution in either case. The barber has an average time for cutting hair, and a person who has been sitting in the chair getting a haircut for the past 15 minutes has a higher probability of being completed in the next minute than a person who just walked in and sat down. 4. Would you expect the exponential distribution to be a good approximation of service times for a. Buying an airline ticket at the airport? Yes. Although certain customers will require special routings and payment methods, most ticketing is pretty straightforward and entails a short service time. b. Riding a merry-go-round at a carnival? No. The Merry-go-round has a fixed cycle time and hence constant services rate. c. Checking out of a hotel? No. Probably a better approximation would be a normal distribution since fast and slow checkouts are likely to be fairly equally balanced. d. Completing a midterm exam in your OM class? No. From our experience, students require the entire class period (and then some) to finish the typical mid-term. Thus, the service rate is close to constant. 5. Would you expect the Poisson distribution to be a good approximation of a. Runners crossing the finish line in the Boston Marathon? Yes. The arrival pattern typically shows a few runners arriving “early” and the majority arriving in a bunch and the remainder spread out along the tail of the distribution. b. Arrival times of the student in your OM class? No. Arrivals are not random since there is a schedule to be met. c. Arrival times of the bus to you stop at school? No. Again, arrivals are not random since the bus follows a set schedule. 6. What it the major cost trade-off that must be made in managing waiting line situations? The classic trade-off is between the cost of waiting for service versus the cost of providing additional service capacity, e.g., the cost of idle WIP versus the cost of adding more workers and machines to process the inventory. 7. Which assumptions are necessary to employ the formulas given for Model 1? Poisson arrival rates, exponential service rates, which imply a purely random process, but with a known mean (and hence known variance). Also assumed is that the process has reached a point of stochastic equilibrium. In other words, steady state conditions prevail. Infinite calling population and unlimited queue length. 8. Why is simulation often called a technique of last resort? Simulation is called a technique of last resort because simulation models are time consuming to build (flow charting, coding, etc.) and do not “guarantee” an optimal solution or indeed any solution. Therefore, it makes sense to investigate other problem solving methods such as linear programming or waiting line theory before embarking on simulation. 9. Must you use a computer to get good information from a simulation? Explain. A computer is a must for any but the most simple simulation problems. Because simulation is a sampling process, it stands to reason that a large number of observations is desirable, and the computer is the only practical way of providing them. Of course, computerization is no guarantee of “good” information. Simulating an invalid model on the computer will only provide a larger volume of questionable data. 10. What methods are used to increment time in a simulation model? How do they work? Time incrementing methods include fixed time increments and variable time increments. With fixed time increments, uniform clock times are specified (minutes, hours, days, etc.) and the simulation proceeds by fixed intervals from one time period to the next. At each point in clock time, the system is scanned to determine if any events have occurred and time is advanced; if none have, time is still advanced by one unit. With variable time increments, the clock time is advanced by the amount required to initiate the next event. It is interesting to note that variable time incrementing generally is more difficult to program unless one is using a special simulation language such as GPSS. 11. What are the pros and cons of starting a simulation with the system empty? With the system in equilibrium? The pros of starting a simulation with the system empty are that this enables evaluation of the transient period in terms of time to reach steady-state and the activities which are peculiar to the transient period. One con is that it takes a longer period of time to perform the simulation. A second is that the model will be biased by the set of initial values selected, since the time to achieve steadystate and the activities which take place during the transient period will be affected by the initial values. Steps must be taken to remove these initial values if steady-state results are needed. The advantages of starting the system in equilibrium are that the run time may be greatly reduced, and that the aforementioned bias may be eliminated. The disadvantage of starting the simulation in equilibrium is, in essence, that it assumes that the analyst has some idea of the range of output he is looking for. This, in a sense, constitutes “beating” the model and may lead to incorrect conclusions from the simulation run. 12. Distinguish between known mathematical distributions and empirical distributions. What information is needed to simulate using a known mathematical distribution? A “known mathematical distribution” is one that can be generated mathematically and is amenable to the laws of statistical probability. Examples of such distributions are the normal, binomial, Poisson, Gamma, and hypergeometric. An empirical distribution is one that is obtained from observing the probability of occurrence of phenomena relating to a specific situation. While it may be possible to define the moment generating function for such distributions, their applicability to other situations is likely to be small. The information required to simulate using a known distribution, of course, depends on the known distribution selected. Generally speaking, however, at a minimum, the analyst must be able to estimate the mean and standard deviation of the population to be sampled since they are parameters of distributions. Alternatively, the analyst can use a discrete probability table to represent the empirical distribution. Objective Questions 1. The exponential distribution is often used to model what in a queuing system? The time between customer arrivals and/or service times. 2. If the average time between customer arrivals is 8 minutes, what is the hourly arrival rate? 7.5 customers per hour 3. How much time on average would a server need to spend on a customer to achieve a service rate of 20 customers per hour? Three minutes 4. What is the term used for the situation where a potential customer arrives at a service operation and upon seeing a long line decides to leave? Balking 5. What is the most commonly used priority rule for setting queue discipline, likely because it is seen as most fair? First-come, first-served (FCFS) 6. Use model 1. = 4/hour = 6/hour a. 1−=1−=1− 4 = .3333 or 33.33%  6 2 b. Lq =  = 42 =1.33, Wq =Lq = 1.33 = 1/3 hour or 20 minutes (−) 6(6−4)  4 c. Lq = 2 = 42 = 1.33 students (−) 6(6−4) d. At least one other student waiting in line is the same as at least two in the system. This probability is 1-(P0+P1). n Pn =1− P0 =1−440 = .3333  66 1 P1 =1−44 = .2222  66 Probability of at least one in line is 1-(.3333 + .2222) = .4444 7. Use model 2. = 60/50 per minute = 60/45 per minute minutes. Lq = 2 = (60/50)2 = 4.05 cars 2(−) 2(60/ 45)(60/ 45− 60/50) Ls =Lq + = 4.05 + (60/50)/(60/45) = 4.95 cars Lq 4.05 Wq = = = 3.375 minutes  (60/50) Ls 4.95 = 4.125 minutes Ws = =  (60/50) 8. Use Model 1. = 100 per hour = 120 per hour a. Ls =  = 100 = 5 customers − 120−100 Ls 5 = .05 hours or 3 minutes Ws = =  100 b. Now, = 180 per hour Ls =  = 100 = 1.25 customers − 180−100 Ls 1.25 = .0125 hours or .75 minutes or 45 seconds Ws = =  100 c. Using model 3, = 100 per hour = 120 per hour S =2 , and== 100 = .8333, from spreadsheet, Lq = .1756  120 Ls =Lq + =.1756+100= 1.01 customers  120 Ls 1.01 = .0101 hours or .605 minutes or 36.3 seconds Ws = =  100 9. Use model 2. = 10 per hour = 12 per hour 2 102 a. Lq =  = = 2.083 people 2(−) 2(12)(12−10) b. Ls =Lq + = 2.083 + 10/12 = 2.917 people c. Wq =Lq = 2.083 =.2083 hours.  10 L 2.917 d. Ws = s = = .2917 hours  10 2 e. It will cause it to increase, at = 12 per hour, L  122 q = 2(−) = 2(12)(12−12) → 10. Use model 1 = 3 per minute = 4 per minute  3 a. Ls = = = 3 customers − 4−3 b. Ws = Ls = 3 = 1 minute  3  3 c. = = = .75 or 75%  4 d. Probability of 3 or more is equal to 1 – probability of 0, 1, 2 0 1 2 P0 =1−33 = .2500, P1 =1− 33 = .1875, P2 =1−33 = .1406  44  44  44 Total of P0 + P1 + P2 = (.2500 + .1875 + .1406) = .5781 Therefore, the probability of three or more is 1 - .5781 = .4219 e. If a automatic vendor is installed, use model 2. (a. revisited) 2 32 Lq = = 2(−) 2(4)(4−3) =1.125 customers Ls =Lq += 1.125 + ¾ =1.875 customers (b. revisited) Lq 1.125 Wq = = =.375 minutes Ws = = =.625 minutes 3 By converting to constant service time, the number in line is reduce from 3 to 1.875 people (a reduction of 1.125, and time in system is reduced from 1 minute to .625 minutes (a reduction of .375 minutes or 22.5 seconds). 11. Use model 4. N = 4, population of 4 engineers, S = 1, one technical specialist, T = 1, average time to help engineer, U = 7, time between requests for help T 1 a. X = = = .125, look up value of F in Exhibit 10.10 T+U 1+7 F = .945, therefore, L = N(1-F) = 4(1-.945) = .22 engineers waiting b. W=L(T+U) = .22(1+7) = .466 hours or 28 minutes N−L 4−.22 c. From Exhibit 10.10 at X=.125, and S=1, D = .362. In other words, 36.2% of the time an engineer will have to wait for the specialist. 12. Use model 1. = 20 per hour = 30 per hour a. Ls =  = 20 = 2 people in the system − 30− 20 Ls 2 = .10 hours or 6 minutes b. Ws = =  20 c. Probability of 3 or more is equal to 1 – probability of 0, 1, 2 n Pn =1− 0 P0 =1− 2020 = .3333  3030  20201 P1 =1−   = .2222  3030 P2 =1−20202 = .1481  3030 Total of P0 + P1 + P2 = (.3333 + .2222 + .1481) = .7036 Therefore, the probability of three or more is 1 - .7036 = .2964  20 d. = = = .67 or 67%  30 e. Use model 3.  20 = = =.6667  30 From Exhibit 10.9, Lq = .0093 Ls =Lq += .0093 + 20/30 = .676 Ls .676 = .0338 hours or 2.03 minutes Ws = =  20 13. Use model 4. N = 4, population of 4 pieces of equipment, S = 1, one repairperson, T = .0833, average time for repair, U = .5, time between requests for repair (hours) T .0833 a. X = = = .1428, look up value of F in Exhibit 10.10 T+U .0833+.5 F = .928, therefore, L = N(1-F) = 4(1-.928) = .288 machines waiting b. J=NF(1-X) = 4(.928)(1-.1428) = 3.18 machines operating c. H = FNX =.928(4).1428 = .53 machines being serviced d. n =L + H = .288 + .53 = .818 machines in the system Cost of downtime is .818 times $20 per hour = $16.36 per hour Cost of one serviceperson = $ 6.00 per hour Total cost per hour = $22.36 With 2 repairpersons, S =2, T .0833 X = = = .1428, look up value of F in Exhibit 10.10 T+U .0833+.5 F = .995, therefore, L = N(1-F) = 4(1-.995) = .020 machines waiting H = FNX =.995(4).1428 = .568 machines being serviced n = L + H = .020 + .568 = .588 Cost of downtime is .588 times $20 per hour = $11.76 per hour Cost of two serviceperson = $12.00 per hour Total cost per hour = $23.76 No, added cost of $1.42 per hour would be added for second repairperson 14. Use model 1. = 2 per hour = 3 per hour 2 22 a. Lq = = = 1.333 customers waiting (−) 3(3−2) L b. Wq = q = 1.333 = .667 hours or 40 minutes  2 c. Ls =  = 2 = 2, Ws =Ls = 2 = 1 hour − 3− 2  2  2 d. = = = .67 or 67% of the time  3 15. Use model 1. = .55 =1.5/hour 1.5 .55 =  = 2.727 .85 =  2.727 = 2.318/hour 16. Use model 1. = 6 per hour = 10 per hour  6 a. Ls = = = 1.5 people − 10−6 Ls 1.5 = .25 hours or 15 minutes Ws = =  6  6 b. = = = .60 or 60%  10 c. Probability of more than 2 people is equal to 1 – probability of 0, 1, or 2 n Pn =1− 0 P0 =1− 6  6  = .4000  1010  6  6 1 P1 =1−   = .2400  1010 P2 =1− 6  6 2 = .1440  1010 Total of P0 + P1 + P2 = (.4000 + .2400 + .1440) = .7840 Therefore, the probability of three or more is 1 - .7840 = .2160 d. Use model 3.  6 = = = .60, from Exhibit 10.9, Lq = .0593  10 Ls =Lq +=.0593+6/10 =.6593 Ws =Ls = .6593/6 = .1099 hours or 6.6 minutes 17. Use model 1. = 25 per hour = 30 per hour  25 a. = = = .833 or 83.3%  30 b. Ls =  = 25 = 5.00 document in the system − 30− 25 c. Ws =Ls = 5.00 = .20 hours or 12 minutes  25 d. Probability of 4 or more is equal to 1 – probability of 0, 1, 2, 3 n Pn =1−  0 P0 =1− 2525 = .1667  3030 1 P1 =1− 2525 = .1389  3030  25252 P2 =1−   = .1157  3030  25253 P3 =1−   = .0965  3030 Total of P0 + P1 + P2 + P3 = (.1667 + .1389 + .1157 + .0965) = .5178 Therefore, the probability of three or more is 1 - .5178 = .4822 or 48.22% 2 302 e. Lq = = → (−) 30(30−30) 18. Use model 1. = 4 per hour = 6 per hour  4 a. = = = .667 or 66.7%  6 b. Ls =  = 4 = 2.00 students in the system − 6− 2 Ls 2.00 = .50 hours or 30 minutes c. Ws = =  4 d. Probability of 4 or more is equal to 1 – probability of 0, 1, 2, or 3 n Pn =1−  440 P0 =1−   = .3333  66 1 P1 =1−44 = .2222  66 2 P2 =1− 44 = .1481  66 3 P3 =1−44 = .0988  66 Total of P0 + P1 + P2 + P3 = (.3333 + .2222 + .1481 + .0988) = .8024 Therefore, the probability of four or more is 1 - .8024 = .1976 or 19.76% 2 62 e. Lq = = → (−) 6(6−6) 19. Use model 1. = 10 per hour = 12 per hour 2 102 a. Lq = = = 4.17 people (−) 12(12−10) b. Ls =  = 10 = 5, Ws =Ls = 5 = .5 hours or 30 minutes − 12−10  10  10 c. = = = .833 or 83.3%  12 d. Probability of 3 or more is equal to 1 – probability of 0, 1, 2 n Pn =1−  10100 P0 =1−   = .1667  1212 P1 =1−10101 = .1389  1212 2 P2 =1−1010 = .1157  1212 Total of P0 + P1 + P2 + P3 = (.1667 + .1389 + .1157) = .4213 Therefore, the probability of three or more is 1 - .4213 = .5787 or 57.87% 20. Use model 1. [Students may want to use model 4 because of the limited number of cars in such an attraction, but there is insufficient information in the problem to solve it under model 4. Kudos to students who note this.] = 2 per hour With one repair person: = 2 Ls =  = 2 → − 2− 2 With two repair people: = 3 Ls =  = 2 = 2cars − 3− 2 With three repair people: = 4 Ls =  = 2 =1car − 4− 2 Service rate 1 2  $  $ 20 $  2 3 2 80 40 120 3 4 1 40 60 100 Number of repair per hour Cost of waiting Cost of service Total cost personnel () n s per hour1 per hour2 per hour Note: 1 = cost of waiting is number in system times downtime cost of $40 per hour. 2 = cost of service is number of repair personnel times wage rate ($20 per hour). We should use three repair persons. 21. Use model 2. = 750 per hour = 900 per hour a. Ws =Ls = 2.92/750 = .003889 hours or .2333 minutes or 14 seconds 2 7502 b. Lq = = = 2.083 cars 2(−) 2(900)(900−750) Ls =Lq + = 2.083 + 750/900 = 2.92 cars 22. Use waiting line approximation because the service times do not follow an exponential distribution. [Note: Student answers may vary slightly due to rounding differences.] Xa =3, Sa = 3, Xs =15, Ss = 7, S = 6 Sa 3 1, Cs = Ss = 7 = .4667, Ca = = = Xa 3 Xs 15 = 1 = 1 = .3333, = 1 = 1 = .06667, =  = .3333 =.8332 Xa 3 Xs 15 S 6(.06667) a. On average how many customers would be waiting in line? Lq = 2(S+1) XCa2 +Cs2 = .8332 2(6+1) X 12 +.46672 =1.8443 1− 2 1−.8332 2 b. On average how long would a customer spend in the bank? Ls =Lq +Sp =1.848+ (6)(.8332) = 6.8435 Ls 6.8468 20.5325 Ws = = =  .3333 c. If a customer arrived, saw the line and decided not to get in line that customer has ________________________________. Balked d. A customer who enters the line but decides to leave the line before getting service is said to have ________________________________. Reneged 23. Use waiting line approximation because the service times do not follow an exponential distribution. Xa = 4, Sa = 4, Xs =7, Ss = 3, S = 2 Sa 4 1, Cs = Ss = 3 = .4286, Ca = = = Xa 4 Xs 7 = 1 = 1 = .2500, = 1 = 1 = .1429, =  = .2500 = .875 Xa 4 Xs 7 S 2(.1429) [Note: Student answers may vary slightly due to rounding differences.] a. On average, how long will each line be at each of the cashier windows? Lq = 2(S+1) Ca2 +Cs2 = .8750 2(2+1) X 12 +.42862 = 3.5015 X 1− 2 1−.8750 2 b. On average how long will a customer spend in the bank (assume they enter, go directly to one line and leave as soon as service is complete). Ls =Lq +Sp = 3.4138+ (2)(.8750) = 5.2509 Ls 5.1638 21.0036 Ws = = =  .2500 You decide to consolidate all the cashiers so they can handle all types of customers without increasing the service times. c. What will happen to the amount of time each cashier spends idle? (increase, decrease, stay the same, depends on _____ ) Stay the same d. What will happen to the average amount of time a customer spends in the bank? (increase, decrease, stay the same, depends on _____ ) Decrease 24. Use model 1. Avg number in system (Ls) = 4 Avg time in system (Ws) = 1.176 4 1.176= = 3.4  4 = = 4.25 25. Time between arrivals (minutes) Probability RN assignment 1 .08 00-07 2 .35 08-42 3 .34 43-76 4 .17 77-93 5 .06 94-99 Service Time (minutes) Probability RN assignment 1.0 .12 00-11 1.5 .21 12-32 2.0 .36 33-68 2.5 .19 69-87 3.0 .07 88-94 3.5 .05 95-99 Customer number RN Interarrival time Arrival time Service begins RN Service time Service ends Waiting time Idle time 1 08 2 2 2.0 74 2.5 4.5 0.0 2.0 2 24 2 4 4.5 34 2.0 6.5 0.5 0.0 3 45 3 7 7.0 86 2.5 9.5 0.0 0.5 4 31 2 9 9.5 32 1.5 11.0 0.5 0.0 5 45 3 12 12.0 21 1.5 13.5 0.0 1.0 6 10 2 14 14.0 67 2.0 16.0 0.0 0.5 Average waiting time = 1.0/6 = 1/6 minute, and average teller idle time between customers = 4.0/6 = 4/6 minute or 40 seconds. Average teller idle time percent = 4.0/16 = .25 or 25%. 26. Machine Interarrival Breakdown Service Repairman 1 Repairman 2 Down breakdown time time time time number RN (hours) (hours) RN (hours) begin end begin end (hours) 1 30 1.0 1.0 81 3.0 1.0 4.0 3.0 2 02 0.5 1.5 91 4.0 1.5 5.5 4.0 3 51 1.0 2.5 08 0.5 4.0 4.5 2.0 4 28 0.5 3.0 44 1.0 4.5 5.5 2.5 5 86 3.0 6.0 84 3.0 6.0 9.0 3.0 Average down time is 14.5/5 = 2.9 hours Case: Community Hospital Evening Operations Room 1. Calculate the average customer arrival rate and service rate per hour. The customer arrival rate lambda = 0.0212 patients per hour. This is calculated 62/(8 x 365) = 0.0212. The service rate mu = 0.7427 patients per hour. This calculated 60 min/hour divided by 80.79 minute/patient = 0.7427 patients per hour. 2. Calculate the probability of zero patients in the system (P0), probability of one patient (P1), and the probability of two or more patients simultaneously arriving during the night shift. P(0) = (1 – lambda/mu) = (1 - .0212/.7427) = 0.9714 the probability of no patients in the system is over 97 percent. P(1) = (1 – lambda/mu)(lambda/mu)1 = (1 - .0212/.7427)(.0212/.7427) = 0.0278 the probability of exactly 1 patient in the system is 2.78 percent. The probability that 2 or more patients are in the system is 1 – (P(1) + P(0)) = 1 – (.0278 + 0.9714) = 0.0008. The probability of two or more patients occurring simultaneously on the night shift is less than 0.1% (less than one chance in 1,000). 3. Using a criterion that if the probability is greater than 1 percent, a backup OR team should be employed, make a recommendation to hospital administration. A second OR is not needed at this hospital. ANALYTICS EXERCISE: Processing Customer Orders – Analyzing a Taco Bell Restaurant 1. Draw a diagram of the process using the format in Exhibit 9.5. 2. Consider a base case where a customer arrives every 40 seconds and the Customer Service Champion can handle 120 customers per hour. There are two Food Champions each capable of handling 100 orders per hour. How long should it take to be served by the restaurant (from the time a customer enters the kiosk queue until her food is delivered)? Use queuing models to estimate this.  = 90, order = 120, prep = 100. For the Customer Service Champion, the average service time is .0083 hours (40 seconds) and the average time in line is .0250 hours (from queuing formulas) for a total of .0333 hrs. (120 seconds). This is calculated using Model 1. For preparing the food for the two Food Champions, the average service time is .01111 hours (40 seconds). Using model 3, and with lambda/mu = .9, Lq from Exhibit 10.9 is 0.2285 customers. The time a customer spends in the system for food to be prepared is .01254 hours or 45.1 seconds. The total time is 120 + 45.1 = 165.1 seconds or 2.75 minutes. 3. On average, how busy are the Service Champion and the two Food Champions? The Service Champion is busy  = 90/120 = 75% of the time and the Food Champions are busy () = 90/200 = 45% of the time. 4. On average, how many cars do you expect to have in the drive-thru line? (Include those waiting to place order and waiting for food.) Lq due to the order taking = λ2/µ(µ - λ) = 8100/3600 = 2.25 customers Lq due to the food preparation = .2285 customers Average total customers waiting in line = 2.4785 customers Ls due to the order taking = 3 customers Ls due to the food preparation = 1.1285 customers Average number of customers in the system = 4.1285 5. If the restaurant runs a sale and the customer arrival rate increases by 20%, how would this change the total time expected to serve a customer? How would this change the average number of cars in the drive-thru line? The current customer arrival rate is 90 per hour, this would go up to 90(1.2) = 108 per hour. Ls due to order taking = (λ/µ – λ) = 108/(120-108) = 9.0 cars Ws = Ls/λ = 9.0/108 = .08333 hours = 5.0 minutes Ls due to order preparation = Lq + λ/µ = .445 + 108/100 = 1.525 cars (note λ/µ = 108/100 = 1.08, Lq = .445 from the spreadsheet) Ws = Ls/λ = 1.525/108 = .01412 hours = .8472 minutes Average total time in the system = 5.0 + .8742 = 5.8472 minutes Average total number of cars in the drive thru = 9.0 + 1.525 = 10.525 cars 6. Currently, relatively few customers (less than ½ percent) order the Crunchwrap Supreme. What would happen if we ran the sale and demand jumped on the Crunchwrap Supreme and 30% of our orders were for this item? Take a quantitative approach to answering this question. Assume that the two processes remain independent. Just to make this interesting, let’s assume that the customer arrival rate went to 108 customers per hour. Assume the order taking rate stays the same at 120/hour. For the order preparation, assume that 70% of the orders still take (60*60)/100 = 36 seconds on average to prepare, but 30% of them now take 72 seconds. The average time to make an order would be .7(36) + .3(72) = 46.8 seconds So they could prepare (60*60)/46.8 = 76.92 orders per hour (each Food Champ). Ls due to order taking = λ/(µ – λ) = 108/(120-108) = 9.0 cars Ws = Ls/λ = 9.0/108 = .08333 hours = 5.0 minutes → this is the same as before. Ls due to order preparation = Lq + λ/µ = 1.3449 + 108/76.92 = 2.749 cars (note λ/µ = 108/76.92 = 1.4, Lq = 1.3449 from exhibit 10.9.) Ws = Ls/λ = 2.749/108 = .02545 hours = 1.527 minutes Average total number of cars in the drive thru = 11.749 cars Average total time in the system = 6.527 minutes Not much of an impact. 7. For the type of analysis done in this case, what are the key assumptions? What would be the impact on our analysis if these assumptions were not true? A big assumption is that there is no interference between the two processes. This would occur if one process was causing a delay in the other process for some reason. It does not appear that this would be true in this process. We are also assuming the Food Champs are working independent, not as a team. Other major assumptions are the distribution associated with arrival and service rates, and that these rates are valid during the peak noon period. 8. Could this type of analysis be used for other service type businesses? Give examples to support your answer. Yes, many examples can be given. CHAPTER 11 PROCESS DESIGN AND ANALYSIS Discussion Questions 1. Define a process in general. Apply this definition in detail to a university, a grocery store, and a beer brewing company. A process is any part of an organization that takes inputs and transforms them into outputs, adding value in the process. Students’ specific responses to the three examples will vary widely – some of the more common expected answers follow. University Inputs: Book, other printed material, chalk, supplies, students Transformation: Education and training Output: Educated graduates Grocery Store Inputs: Shelf-stable food, frozen and perishable foods, grocery bags, promotional displays, shoppers Transformation: Local exchange for cash Output: Satisfied shoppers Brewer Inputs: Grain, hops, water, yeast, bottles/cans, packaging Transformation: Conversion of raw materials via brewing Output: Beer Expect that parts of the process itself (personnel, equipment, facilities) will often show up as inputs in student answers. Whether they should be considered as inputs to or parts of the process is left to the instructor’s discretion. 2. Consider your favorite fast food restaurant. The next time you are there, pay special attention to how the food preparation and delivery process works. Develop a flowchart of the process like that in Exhibit 11.1. Try to include as much detail as is needed to explain the process, remembering to differentiate between your activities as a customer and those of the service process. Student responses will vary widely based on their experiences. Have students develop PowerPoint slides of their flowcharts so the entire class can discuss the charts. This can be a very good exercise for teaching the nuances of flowcharting with student-instructor interaction. 3. Describe cycle time as it relates to business processes. Why is it important to the management of business processes? How does it relate to concepts like productivity and capacity utilization? Cycle time is the average time between completions of successive units in a process. It is directly related to the output capacity of the system – as cycle time is reduced, the process can produce more units in the same period of time, increasing the best operating level used in capacity analysis. Without an actual increase in production volume, capacity utilization will decrease as cycle time decreases but productivity remains the same. As the process produces more output per time period, productivity and capacity utilization will then increase. 4. Compare McDonald's old and new processes for making hamburgers. How valid is McDonald's claim that the new process will produce fresher hamburgers for the customers? Comparing McDonald's new process to the processes used by Burger King and Wendy's, which process would appear to produce the freshest hamburgers? Exhibit 11.2 illustrates the various processes. McDonald's old process was a make-to-stock, where orders were pulled from finished goods. However, McDonald's new process will assemble-to-order. Therefore, McDonald's claim of a fresher hamburger should hold. Burger King's process is a combination of McDonald's old and new processes. The best Burger King can hope to do is match McDonald's with their orders that are assembled-toorder. The ones that are taken from finished goods will generally not be as fresh. Wendy's, on the other hand, should beat both McDonald's and Burger King on freshness, since they cook-to-order (Make-to-order)! 5. State in your own words what Little's Law means. Think of an example that you have observed where Little's Law applies. Little's Law shows the relationship between throughput rate, throughput time, and the amount of work-in-process inventory. Specifically, it is throughput time equals amount of work-in-process inventory divided by the throughput rate. Little's Law is useful for examining the performance of a process. Example 11.1 illustrates an application of Little’s Law. 6. Explain how having more work-in-process inventory can improve the efficiency of a process? How can this ever be bad? More work-in-process inventory can be used to buffer multiple stage processes. Specifically, it can help with blocking or starving. Blocking is when the activities in the stage must stop because there is no place to deposit the item just completed. Starving is when the activities in a stage must stop because there is no work. Buffer inventories between operations can help relieve these problems, and improve the efficiency of the overall process. Increasing work-in-process inventory can be bad in that it involves more investment in inventory, as well as taking-up valuable floor space. Also, the JIT philosophy views work-in-process as being negative for a variety of reasons (more on JIT in a later chapter). 7. Recently some Operations Management experts have begun insisting that simple maximizing process velocity, which actually means minimizing the time that it takes to process something through the system, is the single most important measure for improving a process. Can you think of a situation when this might not be true? The problem with focusing exclusively on process velocity is that other dimensions might be ignored, such as quality or safety. There are many examples. One would be if drying time was reduced, this might impact the quality of the process. Another example would be whiskey, reducing the aging time would probably impact its quality. 8. What is job enrichment and what has led to its importance in job design? Job enrichment is the modification of a specialized job to make it more interesting to an employee. It has become important to counteract the negative aspects of job specialization. To increase the efficiency of business processes, many companies have turned to job specialization. While this has increased the productivity of these systems it can leader to worker boredom with the job and a resultant lack of understanding and concern for the overall process. Enrichment can give the worker more of a feeling of ownership of the process, leading to quality improvements and even productivity gains. 9. Why are work measurement and time standards important for a firm? Are there any negatives to the implementation of these standards? Are there ways to achieve the same objectives without setting firm standards? The reasons for work measurement and time standards are outlined in the chapter: 1. To schedule work and allocate capacity 2. To provide an objective basis for motivating the workforce and measuring workers’ performance 3. To bid for new contracts and evaluate performance on existing ones 4. To provide benchmarks for performance. Essentially they are important to have a baseline measure against which to evaluate and set goals for the system. They can have a negative effect on workers’ attitudes towards the job and the firm, especially if they are unreasonable or unachievable. Various managerial approaches can be used to try and increase workers’ motivation for the job and improved performance. Job enrichment as discussed in the chapter is one such example. 10. From your own experiences, compare the processes of your favorite bricks and mortar department store and a comparable online retailer. What advantages does each have over the other for the company? How about for you, the customer? Answers will vary widely based on student experiences. They should identify immediate delivery, immediate face-to-face advice from salespeople, and the possibility of visual inspection of the goods as advantages at a B&M store. For online retailers: wide variety of products and superior convenience. Advanced answers might include the advantages of pooled inventory and the lower infrastructure cost for the online retailer. 11. What is the effect of waiting time on a manufacturing process? Why is it good to reduce waiting time? Can it be eliminated altogether? Waiting time is non-productive time for material in the process. As waiting time increases, average WIP inventory increases. Decreasing wait time reduces the investment in inventory and improves the flow of material through the system. Completely eliminating waiting time would be very difficult if not impossible to eliminate in most systems. Variability in any process is likely to induce wait time at some point in the system. 12. How does seasonal variability in demand affect the flow and waiting time through a process? How might a company respond to reduce the effect of this variability? As demand peaks during busy season the process can be overwhelmed if there is insufficient capacity in the system to meet demand when it occurs. Orders can backup, creating increased WIP inventory and extending the wait time for orders in the system. In a make-tostock system the firm can manage the varying demand by building up inventory during the slow periods to augment production during peak periods. In a make-to-order system the company can try to level demand by offering promotions and discount pricing during the traditionally slow periods to shift some of the peak demand to the off-peak periods. Where feasible, another option is postponement, where the firm might build up inventory of partially completed products, and perform the final customizing steps as orders come in. Of course there is always the option of increasing the capacity of the system, but that can be quite expensive. Objective Questions 1. A manufacturing company has a small production line dedicated to the production of a particular product. The line has four stations in serial. Inputs arrive at station 1 and the output from station 1 becomes the input to station 2. The output from station 2 is the input to station 3 and so on. The output from station 4 is the finished product. Station 1 can process 2,700 units per month, station 2 can process 2,500/month, station 3 can process 2,300/month, and station 4 2,100/month. What station sets the maximum possible output from this system? What is that maximum output number? Station 4, 2,100/month 2. In a flowchart, what is used to represent a storage activity in the process? Inverted Triangle 3. A process is part of an organization that takes _________________, turns them into _______________, and adds ___________________ while doing so. Inputs, outputs, value 4. Wait Time = 50 Minutes 10 Cars * 5 minutes per car 5. Using Little’s Law we know that Inventory = Throughput x Flow Time Before change: Inventory = 1 per hour × 15 hours = 15 × $1500 = $22,500 After change: Inventory = 1 per hour × 10 hours = 10 × $1500 = $15,000 Reduction in WIP = $22,500 - $15,000 = $7,500 6. a. The throughput rate is 30 buses per hour. The throughput time is 20/60 =1/3 hours. Inventory = Throughput * Flowtime = 30 * 1/3 = 10 buses are traveling to and from the airport at any one time. b. This reduces the average waiting time, but has no effect on the average traveling time. The average traveling time does not change and will remain 20 min. This would, however, result in 40 buses being in transit between the airport and the office at any given time. Does Avis have that many buses? 7. a. TR = 60 per week, TT = 2/7 weeks. WIP = TT *TR = 60*2/7 = 120/7 = 17.1 On average there are 17.1 new mothers in the Children’s Hospital. b. TR = 210 per week, TT = 2/7 weeks. WIP = TT * TR = 210*2/7 = 60. On average there are 60 new mothers in Swedish Hospital. c. TR = 270 per week, TT = 2/7 weeks. WIP = TT * TR = 270*2/7 = 540/7 = 77.1 On average there will be 77.1 new mothers in the unified hospital. No, it will not decrease the total number of mothers in the hospital. 8. What are the four basic techniques for measuring work and setting time standards? Two direct methods: time study, work sampling; and two indirect methods: predetermined motion-time data systems (PMTS), elemental data. 9. Which work measurement technique is most appropriate for tasks that are infrequent or have a long cycle time? Work sampling 10. If 10 companies are selected: Traditional method: 20 minutes setup plus 10 companies times 2 minutes per company. 20 + (10*2) = 40 minutes. Alternative method: 1 minute setup plus 10 companies times 5 minutes per company. 1 + (10*5) = 51 minutes. Therefore, the traditional method is best. This can also be thought of more generally as a breakeven point question. The breakeven point occurs when 20 + 2x = 1 + 5x, or x = 6 1/3. If an applicant selects 6 or fewer firms, use the alternative method. If the applicant selects 7 or more firms, use the traditional method. 11. The longest process on this "assembly line" will govern the output. Therefore, the maximum output from this line will be: Output = available time/cycle time = (40 hours per week)*(60 minutes per hour)/1.5 minutes per student = 1,600 students per week. Therefore, this line cannot produce the 2,000 students per week. 12. a. Hour Water In Water Out Water Remaining 1 4 3 1 2 4 3 2 3 4 3 3 4 4 3 4 5 4 3 5 6 4 3 6 7 4 3 7 8 4 3 8 As you can see, every hour you’re adding exactly 1 unit of water (inventory) to your tub (system). At the completion of hour 50, the bathtub will be at maximum capacity, if the flow of water is not stopped, storage limitations will begin to cause problems. b. Bucket Hour Water In Water Out Water Remaining 1 1 & 2 5 5 0 2 3&4 5 5 0 3 5&6 5 5 0 With this scenario, essentially we have cut the inbound flow rate to 2.5 gallon per hour. Although we are receiving loads in batches, we can empty (consume) all 5 gallons before the next batch is received. Every batch results in 20 minutes of downtime. 13. a. One operator per project: 10 projects per day/8 hours per day = 1.25 projects/hour. The productivity of this option is also 1.25 projects/hour. For the two operator approach, the second operator will limit the system to a rate of 2 projects/hour (this assumes 30 minutes per project). The first operator would be idle for an average of 10 minutes each project. The productivity for the two operator approach is 2 projects per hour/2 hours of labor = 1 project/hour. b. With the one operator, 1000 projects would take 1000 projects/1.25 project per hour = 800 hours or 100 days. With two operators, it would take 1000 projects/2 projects per hour = 500 hours or 62.5 days. The labor content for the first option is 800 hours. The second option requires 1000 hours of labor. 14. Current plans are to make 100 units of component A, then 100 units of component B, then 100 units of component A, then 100 units of component B, etc, where the setup and run times for each component are given below. Component Setup / Changeover Time Run Time/unit A 5 minutes 0.2 minutes B 10 minutes 0.1 minutes 5 + 10 + .2(100) + .1(100) = 15 + 30 = 45 minutes/100 units 45/100 = 60/X X = 133.3 units/hr. 15. Y Z a. (8 X 60 X 60)/38 = 757.9 b. Efficiency: 263/ (9 X 38) = 76.9% c. 6 x 38 = 228 seconds 16. a. Take Order = 100 per hour * 12 hours = 1200 Pick Order = 80 per hour * 24 hours = 1920 Pack Order = 60 per hour * 24 hours = 1440 Maximum output is determined by order taking (1200) since the pick and pack operations can work up to 24 hours to clear out their order backlog. b. If we take the maximum of 1200 orders then: Pick Order = 1200 orders/80 per hour = 15 hours Pack Order = 1200 orders/60 per hour = 20 hours c. Orders can be taken at a rate of 100/hours and can be picked at the rate of 80/hour so they build at the rate of 20/hour. Orders are taken for 12 hours. Maximum orders waiting for picking = 20/hour * 12 hours = 240 d. Orders can be picked at a rate of 80/hours and can be packed at the rate of 60/hour so they build at the rate of 20/hour. Orders are picked for 15 hours. Maximum orders waiting for packing= 20/hour * 15 hours = 300 e. (b. revisited) If we take the maximum of 1200 orders then: Pick Order = 1200 orders/80 per hour = 15 hours Pack Order = 1200 orders/120 per hour = 10 hours However, Packing has to wait for the orders to be picked so it would be 15 hours (c. revisited) This answer does not change. Orders can be taken at a rate of 100/hours and can be picked at the rate of 80/hour so they build at the rate of 20/hour. Orders are taken for 12 hours. Maximum orders waiting for picking = 20/hour * 12 hours = 240 (d. revisited) Orders can be picked at a rate of 80/hours and can be packed at the rate of 120/hour so they build at the rate of 0/hour. Orders are picked for 15 hours. Maximum orders waiting for packing= 0/hour * 15 hours = 0 17. a. The maximum capacity at National State would be (8 tellers * 60 minutes)/5 minutes per customer or 96 customers per hour. b. We cannot handle all the customers by 5:00 pm (see table below). The last customers are processed after National State closes their doors at 5:00 but they will be done at 5:05. c. The maximum waiting time is 12.5 minutes (see table below) and it occurs from 4:40 – 4:45. Customers Customers Customers Customers Arriving Departing at Waiting During During Teller at Expected Period Period or end of Waiting Time (Cumulative) (Cumulative) Waiting Period Time* 4:00 - 4:05 2 2(2) 2 0 0 4:05 - 4:10 5(7) 5(7) 5 0 0 4:10 - 4:15 6(13) 6(13) 6 0 0 4:15 - 4:20 8(21) 8(21) 8 0 0 4:20 - 4:25 10(31) 8(29) 10 2 1.25 4:25 - 4:30 12(43) 8(37) 14 6 3.75 4:30 - 4:35 16(59) 8(45) 22 14 8.75 4:35 - 4:40 12(71) 8(53) 26 18 11.25 4:40 - 4:45 10(81) 8(61) 28 20 12.50 4:45 - 4:50 6(87) 8(69) 26 18 11.25 4:50 - 4:55 4(91) 8(77) 22 14 8.75 4:55 - 5:00 2(93) 8(85) 16 8 5.00 5:00 - 5:05 0(93) 8(93) 8 0 0 * Waiting time is customers waiting * .625 minutes. A customer should complete service every .625 minutes (5 minutes service/8 tellers) 18. a. Maximum output is 21.6 patients a day A. Greet/Register the patient (60 min. per hr/2 min per patient)* 10 hours = 300 patients/ day B. Optometrist (60 min. per hr/25 per patient) * 9 hours (one hour for lunch) = 21.6 patients/day C. Frame/lenses selection (60 min. per hr/20 min per patient)* 10 hours = 30 patients/ day D. Wait for glasses to be made (60 min per hr/ 60 min per set of glasses * 6) * 10 hours/day = 60 pairs E. Final fitting (60 min. per hr/5 min per patient)* 10 hours = 120 patients/ day The bottleneck limiting out is task B. b. Adding another optometrist would allow more patients to be processed. c. Capacity would be the same, but the time to fill an order would increase to 5-7 days. 19. Lead Time = 20 quotes/4 quotes/day = 5 days 20. The average number of customers in the barbershop at any time to be 5. (.5 X10) 21. Assuming sufficient work-in-process inventory, the expected output per hour will initially be the average output per hour, or 3.5 faculty members per hour. With the variability in the process, shifting bottlenecks are likely though, so we could expect the average output per hour to be somewhat less than that in the long run. CASE: Analyzing Casino Honey-Handling Processes Case Questions: 1. Draw a diagram of the drop process. How long should it take to empty 300 silver dollar slot machines? Get slot Deliver filled Get drop cart(20 buckets) cabinet bucket to hard count keys (10 min/slot) Room (30 min/cart) (15 min) Getting the slot cabinet keys only needs to be done one time and takes 15 minutes. Getting the drop buckets will take 3,000 minutes (300 x 10). To empty 300 slot machines, 15 carts need to be delivered (300/20), and delivering the 15 carts will take 450 minutes (15 x 30). Total time to complete the process will be 15 + 3,000 + 450 = 3,465 minutes. This 57 ¾ hours of work. This assumes that only 1 team is doing the work. 2. Draw a diagram of the hard count process. How long should this process take to complete for 300 silver dollar slot machines? Assume that each slot machine has an average of 750 silver dollars when it is emptied. Let’s assume that the weigh scale test is ok. This will take 10 minutes. To weigh and record 300 buckets takes 2,100 minutes (300 x 7). An average of 30 rolls need to be wrapped for each bucket processed (750/25). This means that a total of 9,000 rolls will need to be wrapped (30 x 300). At 10 rolls/minute this will take 900 minutes (9,000/10). Rolls are then placed in cans. To can 9,000 rolls 225 cans need to be filled (9000/40). At 5 minutes/can this will take 1,125 minutes. The summary report is then run that takes 5 minutes. Finally, can counts are manually verified. At 2 minutes/can this will take 450 minutes. The total time to complete the process is 10 + 2,100 + 900 + 1,125 + 5 + 450 = 4,500 minutes or 76.5 hours. 3. The casino is considering the purchase of a second coin-wrapping machine. What impact would this have on the hard count process? Is this the most desirable machine to purchase? Actually, the slowest process that involves a machine is the weighing process. The company should look into a second scale. 4. What would be the impact of purchasing “electronic” slot machines that do not use coin? Of course, this would radically change the way business is done at the casino. All of this manual labor would be eliminated assuming that the machines used a credit card to track the status of the customer’s account. Solution Manual for Operations and Supply Chain Management F. Robert Jacobs, Richard B. Chase 9780078024023, 9780077824921, 9781260238907, 9780077228934, 9781259666100