Part 2 Designing and Managing Processes C WAITING LINE ? Part 2 Designing and Managing Processes WAITING

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172 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. Supplement C Part 2 Designing and Managing Processes WAITING LINE MODELS PROBLEMS 1. Paula Caplin. Littles Law. a. = 120 jobs/day W = 4 days Current work-in-process = L = W = (120 jobs/day)(4 days) = 480 jobs. b. L must be reduced to 240 jobs. Therefore, either the average number of repairs, , or the time in the system, W, must be cut in half (or some combination). Paula has little or no control over the number of repairs, but has several options for reducing the time in the system. First, she can identify the bottleneck in the total repair process and apply the theory of constraints to utilize the bottleneck to its maximum performance. Second, she can do a process analysis and improve the work methods at the bottleneck as well as all other processes feeding the bottleneck to improve overall throughput in the repair process. Finally, if all else fails, she can add capacity until the goal has been met. 2. Banco Mexicali. Littles Law. = 20 customers/hour L = 4 customers L = W , or W = L/ W = (4 customers)/(20 customers/hour) = 0.20 hour, or 12 minutes. 3. Hasty Burgers. Single-server model, 20 a. Find resulting in L = 4. 204204 80 204 10025 LThe required service rate is 25 customers per hour. b. Find the probability that more than four customers are in the system. This is one minus the probability of four or fewer customers in the system. First we calculate average utilization of the drive-in window. Waiting Lines SUPPLEMENT C 173 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 20 0.825 The probability that more than four customers are in line and being served is: 0 1 2 3 41P P P P P P where 0 1 23 40 1 2 3 411 1 1 11 11 1nnPPP when 0.8 1 0.2 1 0.8 0.64 0.512 0.40960.3277 PP Consequently, there is about a 33 percent chance of more than four customers in the system. c. Find the average time in line. 110.825 20qW W 0.16qW hours or 9.6 minutes Ten minutes borders on being unbearable, particularly in the atmosphere of exhaust fumes. Keep in mind that this is an average, and some people must wait longer. 4. Precision Machine Shop. Single-server model. With the junior attendant, the average number of idle machinists, L 8 410 8L Average hourly idle machinist cost = $20(L) = $20(4) = $80 With the senior attendant, average number of idle machinists, L 8 116 8 L Average hourly cost of idle machinists drops to $20(L) = $20(l) = $20 Adding the attendant pay gives a total cost of $85 per hour ($80 + $5) for the junior attendant and $32 per hour ($20 + $12) for the senior attendant. The best choice is the senior attendant. 174 PART 2 Managing Processes Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 5. KRAN radio. Single-server model. 60 min hr 6 calls hr10 min call 60 min hr 2.4 calls hr25 min call 2.4 0.46 A caller will not receive a busy signal when there are zero, one, or two callers in the system. Therefore, the probability of receiving a busy signal is one minus the probability of two or fewer callers in the system. 22110011 0.4 0.4 0.0961 0.4 0.4 0.2401 0.4 0.4 0.6000.936nnPPPP 1 0.936 0.064 Jakes callers will get busy signals 6.4 percent of the time. 6. Local Bank Service rate 60 3 min. per customer = 20 customers/hour. a. Average utilization, 50 3 20 0.8333 s . b. 110132011! ! 150 20 50 20 1! 3! 1 0.83331 2.5 3.125 15.6220.0449n ssonnnPn sn b. 320.0449 50 20 0.83333.50633! 1 0.8333! 1soqPLs c. 3.5063 0.070150qqLW hours, or 4.2 minutes d. 1 50 0.0701 1 20 6.005qL W W or 6 customers Waiting Lines SUPPLEMENT C 175 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 7. Fantastic Styling Salon a. A common queue, two-server model. 8 , 5 , s = 2 8 0.82 2 5 Customers average waiting time in queue = qqLW 2! 1soqPLs 11010 1 21211! ! 11 0! 1! 2! 18 5 11 8 52 1 0.801 1.6 6.41 9 0.1111n sonPn s 220.1111 8 5 0.80 0.2275 2.8440.082! 1 0.80qL 2.844 0.368qW hours or 21.6 minutes b. Two separate, single-server models. The results for Jenny and Jill are identical. 0.5 8 4 , 5 , 4 0.805 Waiting in line is given by 1qW W 0.8 0.805 4qW hours or 48 minutes c. On average, a customers waiting will be more than twice that of the multiple server design. The reason is that a server will serve any customer as they enter the system in the multiple-server system, thereby reducing the wait time on average. 176 PART 2 Managing Processes Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 8. Moore, Akin, and Payne (dental clinic). Multiple-server model. 3s , 5 , 2 , 5 0.83333 2 sa. Probability of no patients, 0 P 1 11 20 50 0 610 1 2 35615 2 5 21 1! ! 1 ! 3! 15 2 5 2 5 2 5 2 10! 1! 2! 3! 11 2.5 3.125 2.604 61 0.044946.625 15.625n s n ssn nPn s n b. The probability of 6 or more customers in the clinic is: 0 0.04494P (from part a), s = 3 for n s 015212522!0.04494 0.112351!0.04494 0.140442!nnP PnPP for n s 03523 04524 15525 2!0.04494 0.117033!30.04494 0.097533!30.04494 0.081273!3nn n sP Ps sPPP 0 1 2 3 4 511 0.04494 0.11235 0.14044 0.11703 0.09753 0.08127 0.40644P P P P P P c. The average number of patients waiting in the lobby, 35 50 2 62 2 15660.04494 0.585163.5109! 1 3! 1sqPLs d. The average time spent in the clinic, W = 1 1 3.5109 1 1.20225 2qqLw hours Waiting Lines SUPPLEMENT C 177 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 9. Bennys Arcade Because there are only six machines, we must use the finite source model. a. To calculate the Jimmys utilization, we need to compute the probability that he will have no machines to repair. 1/50 0.02 machines per hour 1/15 0.0667 machines per hour 16 6! 0.026 ! 0.0667non oPn 113.92 0.0718 Jimmys utilization 1 0.0718 0.9282 b. Average number of machines out of service. 0.06676 1 0.0718 6 3.095 2.9050.02 L machines c. Average time a machine is out of service. 2.905 6 2.905 0.02 46.93 W hours 10. Solomon, Smith and Samson a. Single-server model, average utilization rate. 8 0.810 or 80% utilization b. The probability of four or fewer documents in the system is 0.6723 as shown following. Therefore, the probability of more than four documents in the system is 1 0.6723 = 0.3277. 443322110011 0.8 0.8 0.08191 0.8 0.8 0.10241 0.8 0.8 0.12801 0.8 0.8 0.16001 0.8 0.8 0.20000.6723nnPPPPPP c. The average number of pages of documents waiting to be typed, 8 8 3.2 pages10 10 8qL L 178 PART 2 Managing Processes Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. ADVANCED PROBLEMS 11. Quarry a. Current System: Single-server model 9 hour ; 10 hour Average waiting line in the system 1 1 110 9 W hour or 60 minutes b. First Alternative: Improved single-server model 6 hour ; 15 hour Average time in the system 1 1 0.111115 6 9 W hour or 6.67 minutes c. Second Alternative: Two-server model 211012129 hour ; 10 hour; 22 9 2 10 9 20 0.451 1! 11! ! 10.90 11 0.902! 1 0.451 0.90 0.73640.37930.3793 9 10 0.452! 1q qsq on ssonqsW W LL P sPn sL 2 0.22850.450.2285 9 1 100.1254 hours or 7.52 minutesW The second alternative results in an 87 percent reduction in waiting time relative to the current system, however, the first alternative dominates the second alternative in time and cost. Nonetheless, to make a final determination between the current system and alternative 1, the cost of the waiting time of the trucks must be considered. Waiting Lines SUPPLEMENT C 179 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 12. Benton University, Finite Source Model a. To calculate the utilization in a finite source waiting line situation, we must first compute the probability that the maintenance person will have no customers. 10011 2 34 5!!5! 0.4 5! 0.4 5! 0.415 1 ! 2.5 5 2 ! 2.5 5 3 ! 2.55! 0.4 5! 0.45 4 ! 2.5 5 5 ! 2.51 0.nNnNPN n 108 0.512 0.246 0.079 0.0131 0.37742.65P 01 1 0.3774 0.6226 P b. Copy machines in repair system 012.55 1 0.37740.45 3.8911.109 L N Pc. Time spent in repair system 11.1095 1.109 0.40.712 days5.7 hours, assuming an 8-hour dayW L N L 13. Pinball Wizard. Multiple-server model. In this analysis we determine the expected total labor and machine failure costs for the existing complement of three employees and then compare it to larger maintenance complements until costs begin to rise. Three maintenance people: s = 3, = 0. 333, = 0. 125, Average utilization 0.333 0.8883 0.125 s180 PART 2 Managing Processes Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. Probability of an empty system 1 130.333 0.3331 20.125 0.12500 010 1 2 30.333 0.333 0.333 0.3330.125 0.125 0.125 0.1251 1! ! 1 ! 3! 1 0.88810! 1! 2! 3! 1 0.8881n s nsn nPn s n 102.664 3.548 3.151 8.9291 0.02837.212 28.135P Average number of machines in waiting line 30.3330 0.1252 20.0283 0.888! 1 3! 1 0.8880.4751 6.3100.0753sqqPLsL The average time waiting in line 6.310 18.9490.333qqLW hrs. Average time in system 1118.9490.12526.949 hrsqW W Average number of machines in system 0.333 26.9498.974 machines L WL The total expected hourly costs for the crew size of three employees is: Labor: 3 ($8 per hour) $ 24.00 Machine downtime: 8.974 ($10 per hour) 89.74 TOTAL $113.74 Four maintenance people: Average utilization 0.333 0.6664 0.125 sProbability of an empty system Waiting Lines SUPPLEMENT C 181 Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 1 140.333 0.3331 30.125 0.12500 010 1 2 3 41 1! ! 1 ! 4! 1 0.6662.664 2.664 2.664 2.664 2.664 10! 1! 2! 3! 4! 1 0.6661 2.664 3.54n s nsn nPn s n 108 3.151 2.099 2.99410.060110.363 6.284P Average number of machines in waiting line 402 20.0601 2.664 0.666! 1 4! 1 0.6662.016 0.7532.677sqqPLsL The average time waiting in line 0.753 2.2610.333qqLW hrs. Average time in system 112.2610.12510.261qW W Average number of machines in system 0.333 10.2613.417 L WLThe total expected hourly costs for the crew size of four employees is: Labor: 4 ($8 per hour) $ 32.00 Machine downtime: 3.417 ($10 per hour) 34.17 TOTAL $ 66.17 Five maintenance people: Average utilization 0.333 0.53285 0.125 sProbability of an empty system 182 PART 2 Managing Processes Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. 1 150.333 0.3331 40.125 0.12500 010 1 2 3 4 51 1! ! 1 ! 5! 1 0.53282.664 2.664 2.664 2.664 2.664 2.664 10! 1! 2! 3! 4! 5! 0.46721 2n s nsn nPn s n 10.664 3.548 3.151 2.099 1.118 2.1401 0.067312.462 2.393P Average number of machines in waiting line 402 20.0673 2.664 0.5328! 1 5! 1 0.53284.811 0.18426.193sqqPLsL The average time waiting in line 0.184 0.5530.333qqLW Average time in system 110.5530.1258.553 qW WAverage number of machines in system 0.333 8.5532.848 L WLThe total expected hourly costs for the crew size of five employees is: Labor: 5 ($8 per hour) $ 40.00 Machine downtime: 2.848 ($10 per hour) 28.48 TOTAL $ 68.48 This total is higher than that for employing four maintenance people. Therefore, the manager of the Pinball Wizard should add only one more maintenance person.