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### Final Exam Review Problems

Course: MGT 302, Spring 2012
School: Miami University
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Word Count: 645

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302 MGT Section E Final Exam Review Problems Chapter 10: 1, 3, 5 Chapter 11: 1, 2, 3, 12, 16, 17 Chapter 13: 1, 3, 5, 9, 10, 13, 24 Chapter 14: 4, 5, in-class exercise Chapter 10 1.) D = 10 gauges per hour L = 2 hours S = .20 C = 5 gauges K = DL(1+S)/C K = 10(2)(1+0.20) / 5 = 4.8 5 Kanban card sets 3.) D = 2,400 bottles/2 hours = 1200/60 minutes = 20 per minute L = 40 minutes S = .10 C = 120 bottles K =...

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302 MGT Section E Final Exam Review Problems Chapter 10: 1, 3, 5 Chapter 11: 1, 2, 3, 12, 16, 17 Chapter 13: 1, 3, 5, 9, 10, 13, 24 Chapter 14: 4, 5, in-class exercise Chapter 10 1.) D = 10 gauges per hour L = 2 hours S = .20 C = 5 gauges K = DL(1+S)/C K = 10(2)(1+0.20) / 5 = 4.8 5 Kanban card sets 3.) D = 2,400 bottles/2 hours = 1200/60 minutes = 20 per minute L = 40 minutes S = .10 C = 120 bottles K = DL(1+S)/C K = 2(40)(1+0.10) / 120 = 7.33 8 Kanban cards 5.) D = 32 catalytic converters per hour L = 1 hour S =.125 C = 8 catalytic converters K = DL(1+S)/C K = 32(1)(1+0.125) / 8 = 4.5 5 Kanban cards Chapter 11 1. x 1 2 3 4 5 6 7 8 9 10 11 12 Total 78 Y 4200 4300 4000 4400 5000 4700 5300 4900 5400 5700 6300 6000 60200 = = b== a= = xy 4200 8600 12000 17600 25000 28200 37100 39200 48600 57000 69300 72000 418800 x2 y2 1 17640000 4 18490000 9 16000000 16 19360000 25 25000000 36 22090000 49 28090000 64 24010000 81 29160000 100 32490000 121 39690000 144 36000000 650 308020000 Y (y-Y)2 3958.97 58093.360 4151.28 22117.028 4343.59 118053.912 4535.90 18468.113 4728.21 73872.452 4920.51 48625.904 5112.82 35036.160 5305.13 164128.863 5497.44 9493.754 5689.74 105.194 5882.05 174681.131 6074.36 5529.257 728205.128 6.5 5016.667 192.3077 3766.667 LINEST (Microsoft Excel function) will produce the same result: b = 192.3007. Now calculate Y with 3766.67 + 192.3077 (Month) to get: Month Forecast 13 14 15 16 17 18 19 20 21 22 23 24 b. 6266.67 6458.97 6651.28 6843.59 7035.90 7228.21 7420.51 7612.82 7805.13 7997.43 8189.74 8382.05 = 269.85 Therefore, 3 standard errors of the estimate would be 3(269.85) = 809.55 or 810 2. a. FJuly = .60(15) + .30(16) + .10(12) = 15.0 b. FJuly = (15 + 16 + 12) / 3 = 14.3 c. FJuly = FJune + (AJune FJune) = 13 + .2(15-13) = 13.4 d. Total x 1 2 3 4 5 6 21 = = b== a= = y 12 11 15 12 16 15 81 xy 12 22 45 48 80 90 297 x2 1 4 9 16 25 36 91 3.5 13.5 0.77 10.8 Y = a + bx = 10.8 + .77x e. FJuly, where July is the 7th month. Y = a + bx = 10.8 + .77(7) = 16.2 3. a. Ft+1 = Ft + (At Ft), = .20 Month Demand Forecast Absolute Deviation January 100 80 20 February 94 84 10 March 106 86 20 April 80 90 10 May 68 88 20 June 94 84 10 Total 90 b. MAD = 90/6 = 15 12. Month April May June July August September Absolute Forecast Actual Deviation RSFE deviation 250 200 -50 -50 50 325 250 -75 -125 75 400 325 -75 -200 75 350 300 -50 -250 50 375 325 -50 -300 50 450 400 -50 -350 50 Sum of absolute deviations 50 125 200 250 300 350 MAD 50.0 62.5 66.7 62.5 60.0 58.3 TS -1 -2 -3 -4 -5 -6 For September, the MAD is 58.3 and the TS is 6. The model is performing poorly since the tracking signal is 6 and moving in a downward direction. 16. a. FSeptember = (170 + 180 + 140)/3 = 163.3 b. FSeptember = .50(170) + .30(180) + .20(140) = 167.0 a. FJuly = FJune + (AJune FJune) = 130 + .3(140 - 130) = 133.00 FAugust = FJuly + (AJuly FJuly) = 133.00 + .3(180 133.00) = 147.10 FSeptember = FAugust + (AAugust FAugust) = 147.10 + .3(170 147.10) = 153.97 17. a. FOctober = (75 + 80 + 60 + 75)/4 = 72.5 b. FOctober = FSeptember + (ASeptember FSeptember) = 65 + .2(75 65) = 67.0 c. = 405/6 = 67.5 = 21/6 = 3.5 b = = 3.86 a = = 67.5 3.86(3.5) = 54.00 Y = a + bx = 54.0 + 3.86x d. FOctober = 54.00 + 3.86(7) = 81.01 Chapter Cu 13 1. = \$10 - \$4 = \$6 Co = \$4 - \$1.50 = \$2.50 , NORMSINV(.7059)=0.541446 Should purchase 250 + .541446 (34) = 268.4 or 268 boxes of lettuce. 3. = 22.36 22 5. Service level P = .98, = 150, T = 4 weeks, L = 3 weeks, = 30 per week, and I =500 pounds. = 79.4 From Standard normal distribution, z = 2.05 150(4+3) + 2.05(79.4) 500 = 712.77 713pounds 9. a. = 100 units 10. = 3120 units = 180 units From Standard normal distribution, z = 2.05 = 300(4) + (2.05)180 = 1200.0 + 369 = 1569 If safety stock is reduced by 50 percent, then ss = 185 units. , = 1.03, so the service probability is 84.8% 13. = 2,000 units = 80 units From Standard normal distribution, z = 2.05 = 250(4) + (2.05)80 = 1000 + 164 = 1164 If safety stock is reduced by 100 units, then ss = 64 units. , = .80 From Standard normal distribution, z = .80, service probability is 79% 24. = 31.62 32 refrigerators 10 refrigerators From Standard normal distribution, z = 1.88 = (500/365)(7) + (1.88)10 = 9.59 + 18.8 = 28.39 28 refrigerators Order 32 refrigerators when the on-hand inventory level reaches 28 refrigerators. Chapter 14 4. Level 0 Z A(2) B(4) 1 D(4) Period Item Z LT= 2 Q= L4L 1 Gross requirements Scheduled receipts On hand from prior period Net requirements Planned order receipts Planned order releases Item A LT= 1 Q= L4L Gross requirements Scheduled receipts On hand from prior period Net requirements Planned order receipts Planned order releases Item B LT= 1 Q= L4L Gross requirements Scheduled receipts On hand from prior period Net requirements Planned order receipts Planned order releases Item C LT= 1 Q= L4L Gross requirements Scheduled receipts On hand from prior period Net requirements Planned order receipts Planned order releases Item D Gross requirements Scheduled receipts 2 2 E(2) C(3) 3 3 4 5 6 LT= 1 Q= L4L On hand from prior period Net requirements Planned order receipts Planned order releases Item E LT= 3 Q= L4L Gross requirements Scheduled receipts On hand from prior period Net requirements Planned order receipts Planned order releases 800 5. Level A C B(3) E(2) Period 1 1 2 B D E(2) F Item A LT= 2 Q= L4L 0 D E(2) F 2 3 D(2) 4 F 3 4 5 6 7 Gross requirements Scheduled receipts On hand from prior period 0 10 1 0 10 Net requirements Planned order receipts Planned order releases Item B LT= 1 Q= L4L Gross requirements 50 Scheduled receipts On hand from prior period Net requirements Planned order receipts Planned order releases Item C LT= 1 Q= 50 5 0 0 50 50 60 Gross requirements Scheduled receipts On hand from prior period 10 10 1 0 10 Net requirements Planned order receipts Planned order releases Item D LT= 2 Q= 50 Gross requirements 5 0 Scheduled receipts On hand from prior period 0 0 5 0 5 0 Net requirements Planned order receipts Planned order releases Item E LT= 1 Q= 50 60 100 100 Gross requirements Scheduled receipts 60 1 0 0 50 220 200 On hand from prior period 50 100 1 0 0 Net requirements 220 400 Planned order receipts Planned order releases Item F LT= 1 Q= L4L 4 0 0 4 0 0 Gross requirements Scheduled receipts On hand from prior period 50 150 200 Net requirements Planned order receipts Planned order releases 0 200 2 0 0 2 0 0 2 0 0
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