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- 1. Instructor’s Solutions Manual for Additional Problems Operations Management EIGHTH EDITION Principles of Operations Management SIXTH EDITION JAY HEIZER Texas Lutheran University BARRY RENDER Upper Saddle River, New Jersey 07458 Rollins College
- 2. VP/Editorial Director: Jeff Shelstad Executive Editor: Mark Pfaltzgraff Senior Managing Editor: Alana Bradley Senior Editorial Assistant: Jane Avery Copyright 2006 by Pearson Education, Inc., Upper Saddle River, New Jersey, 07458. Pearson Prentice Hall. All rights reserved. Printed in the United States of America. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department. Pearson Prentice HallTM is a trademark of Pearson Education, Inc. 10 9 8 7 6 5 4 3 2 1
- 3. Contents Homework Problem Answers Chapter 1 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Supplement 6: Chapter 7 Supplement 7: Chapter 8 Chapter 9 Supplement 10: Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Module A: Module B: Module C: Module D: Module E: Module F: Operations and Productivity ........................................................................... A-1 Project Management ....................................................................................... A-3 Forecasting ...................................................................................................... A-7 Design of Goods and Services ...................................................................... A-11 Managing Quality ......................................................................................... A-15 Statistical Process Control ............................................................................ A-18 Process Strategy ............................................................................................ A-20 Capacity Planning ......................................................................................... A-23 Location Strategies ....................................................................................... A-27 Layout Strategy ............................................................................................. A-30 Work Measurement ...................................................................................... A-34 Inventory Management ................................................................................. A-36 Aggregate Planning ...................................................................................... A-42 Materials Requirements Planning (MRP) & ERP ........................................ A-46 Short-Term Scheduling ................................................................................. A-51 Just-In-Time and Lean Production Systems ................................................. A-55 Maintenance and Reliability ......................................................................... A-57 Decision Making Tools ................................................................................ A-59 Linear Programming ..................................................................................... A-64 Transportation Modeling .............................................................................. A-70 Waiting Line Models .................................................................................... A-75 Learning Curves ........................................................................................... A-79 Simulation ..................................................................................................... A-80 iii
- 4. 1 CHAPTER 1.1 a. Operations and Productivity Last year’s number of units of output total factor = total dollar value of all inputs used productivity 12, 000 units = (12, 000 )( $2.00 ) + (14, 000 )( $10.50 ) + ( 2, 000 )( $8.00 ) + ( 4, 000 )( $0.70 ) + $30, 000 = b. 12, 000 units = 0.0546 units dollar $219,800 This year’s number of units of output total factor = productivity total dollar value of all inputs used = 14, 000 units (14, 000 )( $2.05 ) + (16, 000 )( $11.00 ) + (1,800 )( $7.50 ) + ( 3,800 )( $0.75 ) + $26, 000 = c. 14, 000 units = 0.0567 units dollar $247, 050 This year’s Last year’s total factor − total factor productivity productivity × 100% = 0.0567 − 0.0546 × 100% 0.0546 Last year’s total factor productivity = +3.84% ≈ 3.8% Answer : Total factor productivity increased by 3.798% this year as compared to last year. A-1
- 5. 1.2 57, 600 , where L = number of laborers employed at the plant. (160 )(12 )( L ) 57, 600 So L = = 200 (160 )(12 )( 0.15) 0.15 = Answer : 200 laborers 1.3 Output = 28, 000 customers There are 4 approaches to solving the problem correctly: Input = 7 workers 1. 28, 000 = 4, 000 customers worker Then, 7 2. Input = 7 ( 40 ) labor weeks Then, 3. Input = 7 ( 40 )( 50 ) labor hours Then, 4. 28, 000 = 2 customers labor hour 7 ( 40 )( 50 ) Input = 7 ( 40 )( $250 ) dollars of worker wages Then, 1.4 28, 000 = 100 customers labor week 7 ( 40 ) 28, 000 = 0.40 customers per dollar of labor 7 ( 40 )( $250 ) 6, 600 Cadillacs = 0.10 ( x )( labor hours ) x = 66, 000 labor hours There are 300 laborers. So, 66, 000 labor hours = 220 labor hours laborer 300 laborers 1.5 52 ( $90 ) + 198 ( $80 ) 20,520 $ output = = = $57.00 per labor hour labor hour 8 ( 45 ) 360 A-2
- 6. 3 Project Management CHAPTER Gantt Chart 3.1 A 20 80 B C 120 D 110 E 140 F 150 G 170 H 160 I 50 3.2 150 200 200 100 Hours 60 30 20 B Purchasing D Sawing G Infill AON Network: 20 20 10 30 A Planning E Placement F Assembly I Decoration 100 10 C Excavation H Outfill A-3
- 7. 3.3 AOA Network: G Infill 1 A Plan 2 B Purchase D Saw 3 E Place 4 5 F Assemble 6 7 Dummy Outfill H 8 I Decorate 9 C Excavate 3.4 Path 1–2–3–4–5–6–7–8–9 1–2–3–4–5–6–8–9 1–2–4–5–6–7–8–9 1–2–4–5–6–8–9 Task Times (Hours) 20 + 60 + 30 + 20 + 10 + 20 + 0 + 30 20 + 60 + 30 + 20 + 10 + 10 + 30 20 + 100 + 20 + 10 + 20 + 0 + 30 20 + 100 + 20 + 10 + 10 + 30 Total Hours 190 180 200 190 The longest path clearly is 1 – 2 – 4 – 5 – 6 – 7 – 8 – 9; hence, this is the critical path, and the project will end after 200 hours. 1 A Planning ES = 0 EF = 20 LS = 0 LF = 20 2 ES = 20 EF = 80 LS = 30 LF = 90 Purchasing B 3 ES = 80 EF = 110 LS = 90 LF = 120 Sawing D 4 ES = 120 EF = 140 LS = 120 LF = 140 Placement E C Excavate ES = 20 EF = 120 LS = 20 LF = 120 5 ES = 150 EF = 170 LS = 150 7 LF = 170 Infill Dummy F I G H Assembly Decoration Outfill 6 8 9 ES = 140 ES = 150 ES = 170 EF = 150 EF = 160 EF = 200 LS = 140 LS = 160 LS = 170 LF = 150 LF = 170 LF = 200 Answer : The critical path therefore is A – C – E – F – G – I (200 hours). The activities that can be delayed include ones with slack times > 0. Thus, B (10 hours), D (10 hours), and H (10 hours) can be delayed. A-4
- 8. 3.5 3.6 a + 4m + b Mean: 6 120 A: = 20 6 360 B: = 60 6 600 C: = 100 6 180 D: = 30 6 120 E: = 20 6 60 F: = 10 6 120 G: = 20 6 60 H: = 10 6 180 I: = 30 6 2 b−a Variance 6 2 ( 20 ) = 11.11 A: 36 2 ( 60 ) = 100.00 B: 36 2 120 ) ( C: = 400.00 36 2 (10 ) = 2.78 D: 36 2 (10 ) = 2.78 E: 36 2 0) ( = 0.00 F: 36 2 ( 40 ) = 44.44 G: 36 2 ( 4 ) = 0.44 H: 36 2 40 ) ( I: = 44.44 36 b−a Standard Deviation 6 20 A: = 3.33 6 60 B: = 10.00 6 120 C: = 20.00 6 10 D: = 1.67 6 10 E: = 1.67 6 0 F : = 0.00 6 40 G: = 6.67 6 4 H : = 0.67 6 40 I: = 6.67 6 Since the critical path is A – C – E – F – G – I, only those variances are along the critical path are used. Therefore, the variances along critical path are 11.11, 400, 2.78, 0, 44.44, and 44.44 . So the sum of these variances = 502.77 . Thus, the project completion standard deviation = 502.77 ≅ 22.4 . µ = mean time of critical path = 200 hrs σ = 22.4 hrs 240 − 200 40 The z value = = = 1.8 . Using the cumulative normal distribution table in 22.4 22.4 Appendix I of the text, we observe that 96.4 percent of the distribution lies to the left of 1.8 standard deviations. Hence, there is a 100 − 96.4 = 3.6% chance that it will take more than 240 hrs to build the garden/picnic area. A-5
- 9. 3.7 The critical path is A – C – E – F – G – I. Hence, the project completion variance = 11.11 + 400 + 2.78 + 0 + 44.44 + 44.44 = 502.77. So, the project completion standard deviation = 502.77 ≅ 22.4 . The cumulative normal distribution tells us that 90% of the area lies to the left of 1.29 standard deviations. Therefore, amount of time to build the garden/picnic area should be 200 + 22.4 (1.29 ) = 200 + 29 = 229 hours . 3.8 a. Activity on Nodes Diagram of the project. B 1 E 2 A 1 F 2 C 4 b. c. d. The critical path, listing all critical activities in chronological order: A → B → E → F 1 + 1 + 2 + 2 = 6 ( not CP ) A→C→F 1 + 4 + 2 = 7. This is the CP. The project duration (in weeks): 7 (This is the length of CP.) The slack (in weeks) associated with any and all non-critical paths through the project: Look at the paths that aren’t critical—only 1 here—so from above: A → B → E → F 7 − 6 = 1 week slack. 3.9 We have only 1 activity with probabilistic duration. Due date − µ 8 − (1 + 4 + 2 ) 1 = = = 2 (length of entire path is 7, not 4). For a z = 2 , Z= σ 0.5 0.5 this means P ( Due date < 8) = 97.72% (table lookup) for the path so chance of being OVER 8 weeks is 2.28% (and we know non-CP path will be only 6 weeks) 3.10 Helps to modify the AON with the lowest costs to crash 1. CP is A → C → F ; C is cheapest to crash, so take it to 3 wks at $200. (and $200 < $250) 2. Now both paths through are critical. We would need to shorten A or F, or shorten C and either B/E. This is not worth it, so we would not bother to crash any further. A-6
- 10. 4 CHAPTER 4.1 Present = period ( week ) 6. So: F7 = 4.2 Forecasting t 1 2 3 4 1 1 1 1 1 1 1 1 A6 + A5 + A4 + A3 = ( 52 ) + ( 63) + ( 48 ) + ( 70 ) = 56.75 patients 3 4 4 6 3 4 4 6 At Ft 120 — 136 — 114 128 116 125 120 + 136 256 = = 128 2 2 Checking Data 136 + 114 250 = = 125 2 2 116 + 114 230 F5 = = = 115 = Answer 2 2 4.3 Method 1: Method 2: MAD : MSE : MAD : MSE : 0.20 + 0.05 + 0.05 + 0.20 = 0.5000 ← better 0.04 + 0.0025 + 0.0025 + 0.04 = 0.0850 0.1 + 0.20 + 0.10 + 0.11 = 0.5100 0.01 + 0.04 + 0.01 + 0.0121 = 0.0721 ← better A-7
- 11. 4.4 y = a + bx 4 ∑x y i i = 58,538 i =1 x = 75.75 y = 191.5 n ∑x 2 i i =1 b= = 23, 209 58,538 − 4 ( 75.75)(191.5) 23, 209 − 4 ( 75.75 ) 2 = 513.50 =2 256.75 a = 191.5 − 2 ( 75.75 ) = 40 y ≈ 40 + 2 x x = 85 ⇒ y ≈ 210 4.5 t 1 2 3 4 5 Day Monday Tuesday Wednesday Thursday Friday Actual Forecast Demand Demand 88 88 72 88 68 84 48 80 ← Answer 72 Ft +1 = α At + (1 − α ) Ft . Let α = 1 . Let Monday forecast demand = 88 4 1 3 (88 ) + (88 ) = 88 4 4 1 3 F3 = ( 72 ) + ( 88 ) = 18 + 66 = 84 4 4 1 3 F4 = ( 68 ) + ( 84 ) = 17 + 63 = 80 4 4 1 3 F5 = ( 48 ) + ( 80 ) = 12 + 60 = 72 4 4 F2 = A-8
- 12. 4.6 4.7 Winter Spring Summer Fall 2001 1, 400 1,500 1, 000 600 2002 1, 200 1, 400 2,100 750 2003 1, 000 1,600 2, 000 650 2004 900 1,500 1,900 500 4,500 6,000 7,000 2,500 20, 000 = 1, 250 Average over all seasons: 16 6, 000 = 1,500 Average over spring: 4 1,500 = 1.2 Spring index: 1, 250 5,600 Answer : (1.2 ) = 1, 680 sailboats 4 We need to find the smoothing constant α. We know in general that Ft +1 = α At + (1 − α ) Ft , t = 1, 2, 3 . Choose either t = 2 or t = 3 ( t = 1 won’t let us find α because F2 = 50 = α ( 50 ) + (1 − α ) 50 holds for any α). Let’s pick, e.g., t = 2 . Then F3 = 48 = α ( 42 ) + (1 − α ) 50 . So 48 = 42α + 50 − 50α −2 = −8α 1 =α. 4 Now we can find F5 : F5 = α ( 46 ) + (1 − α ) 50 , with α = F5 = 1 3 ( 46 ) + ( 50 ) = 49 ← Answer 4 4 A-9 1 . So 4
- 13. 4.8 Let X 1 , X 2 , … , X 6 be the prices; Y1 , Y2 , … , Y6 be the number sold. 6 ∑X X= i = Average price = 3.25833 i =1 6 (1) 6 ∑Y i Y = i =1 6 = Average number sold = 550.00 (2) All calculations to the 1 nearest th 100, 000 6 ∑ X Y = 9, 783.00 (3) ∑X (4) i i i =1 6 2 i = 67.1925 i =1 Then y ≈ a + bx , where y = number sold , x = price , and 6 ∑ X Y − n ( X )(Y ) ( 9, 783) − 6 ( 3.25833)( 550 ) −969.489 b= = = = −277.61395 3.49222 67.1925 − 6 ( 3.25833) ∑ X − n( X ) i i i =1 6 2 2 2 i i =1 a = (Y ) − b ( X ) = 1, 454.5578 So at x = 1.80 , y = 1, 454.5578 − 277.61395 (1.80 ) = 954.85270 . Now round to the nearest integer: Answer : 955 dinners n ∑( A − F ) t 4.9 Tracking Signal = Month May June July August September October November December t t =1 At 100 80 110 115 105 110 125 120 MAD Ft 100 104 99 101 104 104 105 109 At − Ft ( At − Ft ) 0 24 11 14 1 6 20 11 Sum: 87 0 –24 11 14 1 6 20 11 Sum: 39 87 = 10.875 8 39 1 Answer : th = 3.586 to the nearest 10.875 1,000 So: MAD : A-10
- 14. 5 Design of Goods and Services CHAPTER (0.80) 90 of 100 non-defect 5.1 $42,500 $27,500 (0.20) 70 of 100 non-defect Use K1 $27,500 –$32,500 (0.85) 90 of 100 non-defect $12,500 (0.15) 75 of 100 non-defect –$43,750 (0.90) 95 of 100 non-defect –$18,750 (0.10) 80 of 100 non-defect Use K2 –$75,000 $4,062.50 Use K3 $24,375 Answer: $27,500—use K1 Outcome Calculations 90 10 ( 500 )( 300 )( $1.20 ) − ( 500 )( 300 )( $1.30 ) = −$100, 000 + 100 100 −$100, 000 + $162, 000 − $19, 500 = $42,500 70 30 (150, 000 )( $1.20 ) − (150, 000 )( $1.30 ) = −$100, 000 + 100 100 −$100, 000 + $126, 000 − $58,500 = −$32, 500 A-11
- 15. 90 10 (150, 000 )( $1.20 ) − (150, 000 )( $1.30 ) = −$130, 000 + 100 100 −$130, 000 + $162, 000 − $19, 500 = $12,500 75 25 (150, 000 )( $1.20 ) − (150, 000 )( $1.30 ) = −$130, 000 + 100 100 −$130, 000 + $35, 000 − $48, 750 = −$43, 750 95 5 (150, 000 )( $1.20 ) − (150, 000 )( $1.30 ) = −$180, 000 + 100 100 −$180, 000 + $171, 000 − $9, 750 = −$18, 750 80 20 (150, 000 )( $1.20 ) − (150, 000 )( $1.30 ) = −$180, 000 + 100 100 −$180, 000 + $144, 000 − $39, 000 = −$75, 000 (0.3) F market 5.2 80.0 66.0 (0.7) U market Use D0 (0.4) F market 84.0 Use D1 60.0 99.0 84.0 (0.6) U market (0.6) F market Use D2 74.0 89.2 80.2 (0.4) U market 66.7 (All $ figures in millions in tree) $ Profits : D0 − F : 1,000 ( 80, 000 ) = $80, 000,000 D0 − U : 750 ( 80, 000 ) = $60, 000,000 D1 − F : 1, 000 (100, 000 ) − 1,000, 000 = $99, 000,000 D1 − U : 750 (100, 000 ) − 1, 000, 000 = $74, 000,000 D2 − F : 1, 000 ( 90, 000 ) − 800, 000 D2 − U : 750 ( 90, 000 ) − 800, 000 = $89, 200, 000 = $66, 700,000 Answer : Answer: Design D1 has an expected profit of $84,000,000. A-12
- 16. (0.3) Demand rises 5.3 $10,000 Purchase overhead hoist $30,000 (0.5) Demand stays $10,000 same –$20,000 (0.4) Demand rises $14,000 (0.2) Demand falls $20,000 Purchase forklift $14,000 Do nothing (0.6) Demand stays $10,000 same $0 Answer : Maximum expected payoff = $14, 000 5.4 $0 Do nothing High demand (0.6) $300,000 Use A 160K Low demand (0.4) High demand (0.6) Use B 180K Use C –$50,000 $300,000 Low demand (0.4) $0 High demand (0.6) $250,000 302K Low demand (0.4) Upgrade to D $380,000 380K No upgrade to D $0 Note: K = $1,000’s Answer : Use Design C. If demands turns out to be low, upgrade to Design D. A-13
- 17. 5.5 Support Bread & Rolls No support Support Bread & Rolls Pies & Cakes No support Support Full Service No support 5.6 Support (p = 0.40) $15,000 Bread & Rolls EMV = $12,000 No support (p = 0.60) Support (p = 0.40) Bread & Rolls Pies & Cakes $10,000 $25,000 EMV = $13,000 No support (p = 0.60) Support (p = 0.40) $5,000 $35,000 Full Service EMV = $7,500 No support (p = 0.60) –$10,000 Based upon this decision tree, Jeff should consider most seriously the medium-sized shop carrying bread, rolls, pies, and cakes. A-14
- 18. 6 Managing Quality CHAPTER 6.1 1. 2. 3. 4. 5. 6. 7. 8. 9. Item Rated 1. 2. 3. 4. 5. 6. 7. 8. 9. Item Rated 1. 2. 3. 4. 5. 6. 7. 8. 9. a. b. Appearance of food Portion size Lighting Speed of service Knowledge of server Quality of service Appearance of room Appropriate amount of space View of stage and audio A 20 4 19 4 0 9 19 0 0 4 80 16 76 16 0 36 76 0 0 B 28 2 20 5 0 30 18 26 0 Overall Grade C 1 30 3 25 27 7 13 24 0 D 1 14 8 5 18 0 0 0 20 E 0 0 0 11 7 4 0 0 30 3 84 6 60 15 0 90 54 78 0 Weights 2 2 60 6 50 54 14 26 48 0 1 1 14 8 5 18 0 0 0 20 0 0 0 0 0 0 0 0 0 0 Highest rated is appearance of food; 2.61. Lowest rated is view of stage; 0.31. A-15 Total 167 96 150 86 72 140 156 126 20 Average 2.61 1.50 2.34 1.34 1.13 2.19 2.44 1.97 0.31
- 19. c. A check sheet will help categorize the comment cards Check Sheet Positive Appearance of food Portion size Lighting Speed of service Knowledge of server Quality of service Appearance of room Appropriate amount of space View of stage and audio Other d. 6.2 ! ! ! !! ! !!!!!! ! ! ! ! chilly ! The written comments are not always consistent: Portion size is highly rated in comments, but 5th in overall grade. View/audio is lowest rated in both. a. y minutes 14 13 12 11 10 9 8 0 b. !!!! ! ! Negative 1 2 3 4 5 6 This is a scatter diagram. A-16 x Trips
- 20. 6.3 a. b. 6.4 40 35 30 25 20 15 10 5 0 34 36 30 24 14 10 6 W R I 4 M 2 O 39% of complaints are W, demeaning towards women. Machines Manpower Incorrect measurement Inadequate cleanup Operator misreads display Technician calculation off Temperature controls off Antiquated scales Variability Equipment in disrepair Inadequate flow controls Incorrect Formulation Jars mislabeled Incorrect weights Lack of clear instructions Damaged raw material Priority miscommunication Incorrect maintenance Inadequate instructions Methods Materials A-17
- 21. 6 Statistical Process Control SUPPLEMENT S6.1 σ 25 We are given a target of X = 420 . So LCL = X − Z = 420 − 4 = 400 . n 25 σ 25 UCL = X + Z = 420 + 4 = 440 . Thus, n 25 Answer : LCL = 400 calories UCL = 440 calories S6.2 p= 7 250 5 9 + 250 + " + 250 7 + 5 + " + 9 300 = = = 0.040 30 7,500 7,500 UCL = p + Z LCL = p − Z S6.3 p (1 − p ) n p (1 − p ) n = 0.040 + 3 ( 0.01239 ) = 0.077 = 0.040 − 3 ( 0.01239 ) = 0.003 We want Z = 2 , since (1 − 0.0455 ) = 0.9545 which implies Z = 2 from the Normal Table. UCL = c + 2 c , where c = average number of breaks = 3 : 3 + 2 3 = 6.46 . S6.4 Z = 3 for x -chart . Here, n = 4 so A2 = 0.729 (from Table S6.1). x = 2.0 , R = 0.1 , UCL x = x + A2 R = 2.0 + 0.729 ( 0.1) = 2.07 S6.5 C chart 0.0027 1.0000 − 0.0027 0.9973 = = 0.49865 ⇒ Z = 3 (see Normal Table) 2 2 UCL = c + 3 c = 1.5 + 3 1.5 = 5.17 A-18
- 22. S6.6 σ x±Z = answers n 384 x= = 16 lbs. 24 σ 0.12 Z = 2 3 = 0.08 n 16.00 + 0.08 = 16.08 = UCL x 16.00 − 0.08 = 15.92 = LCL x S6.7 x = 1.00 , R = 0.10 , A2 = 0.483 (from Table S6.1), LCL = x − A2 R = 1 − ( 0.483)( 0.10 ) = 0.9517 weeks S6.8 R = 3.25 mph, Z = 3 , with n = 8 , from Table S6.1, UCL = 1.864 R = 6.058 LCL = 0.136 R = 0.442 30 S6.9 p= ∑ Number of defects 250 i =1 30 UCL p = p + 2 LCL p = p − 2 S6.10 a. b. = 300 = 0.04 , n = 250 7,500 ( 0.04 )( 0.96 ) 250 ( 0.04 )( 0.96 ) 250 = 0.04 + 2 ( 0.0124 ) = 0.0648 = 0.04 − 2 ( 0.0124 ) = 0.0152 We are counting attributes and we have no idea how many total observations there are (the proportion of drivers who weren’t offended enough to call!) This is a C-chart. 36 = 6 for c , as true c is unknown. Use mean of 6 weeks of observations 6 UCL = c + z c = 6 + 3 ( 2.45 ) = 13.3 LCL = c − z c = 6 − 3 ( 2.45 ) = −1.3 , or 0. c. It is in control because all weeks’ calls fall within interval of [ 0, 13] . d. Instead of using 36 = 6 , we now use c = 4 . UCL = 4 + 3 4 = 4 + 3 ( 2 ) = 10 . 6 LCL = 4 − 3 ( 2 ) = −2 , or 0. Week 4 (11 calls) exceeds UCL. Not in control. A-19
- 23. 7 Process Strategy CHAPTER 7.1 a. Find breakeven points, X p . Mass Customization: 1, 260, 000 + 60 X = 120 X → X p = 21, 000 Intermittent: Repetitive: 1, 625, 000 + 55 X = 120 X → X p = 25, 000 Continuous: b. 1, 000, 000 + 70 X = 120 X → X p = 20, 000 1,960, 000 + 50 X = 120 X → X p = 28, 000 Find least-cost process at X = 24, 000 units . Fixed cost VC Units Mass Customization: 1, 260, 000 + 60 ( 24, 000 ) = 2, 700, 000 Intermittent: Repetitive: 1, 000, 000 + 70 ( 24, 000 ) = 2, 680, 000 1, 625, 000 + 55 ( 24, 000 ) = 2,945, 000 Continuous: 1,960, 000 + 50 ( 24, 000 ) = 3,160, 000 The least-cost process: Intermittent Process. c. 24,000 > 20,000 ? yes! #$% #$% Anticipated Production Volume Intermittent Process Breakeven Point Annual Profit Using Intermittent Process: $ 120 ( 24, 000 ) − 2, 680, 000 = $200, 000 Answer : The intermittent process will maximize annual profit. Annual Profit: $200,000 A-20
- 24. 7.2 Use a crossover chart. First graph. Then solve for breakpoint(s). 3 Cost R (Millions 2 MC of dollars) 1 I 0 V 0 10 P1 15P2 20 5 25 1,000’s of Ovens Finding value of P2: 1, 250, 000 + 50 ( P2 ) = 2, 000, 000 + 5 ( P2 ) . So P2 = 16, 666 (Note: P1 = 12, 500 ). Answer : For volumes of production V such that 16, 666 2 ≤ V ≤ 25,000 . 3 7.3 I R 14 12 C C 10 Cost (millions) R 8 6 4 I 2 0 Volume V 0 5,000 10,000 15,000 20,000 7,500 1, 000, 000 + 1, 650 x = 3, 000, 000 + 1, 250 x 400 x = 2, 000, 000 I&R x = 5, 000 Intersect 1, 000, 000 + 1, 650 ( 5, 000 ) = $9, 250, 000 3, 000, 000 + 1, 250 x = 7,500, 000 + 650 x 600 x = 4,500, 000 R&C x = 7,500 Intersect 3, 000, 000 + 1, 250 ( 7,500 ) = $12,375, 000 For all V between 5, 000 ≤ V ≤ 7,500 A-21 2 3 units .
- 25. 7.4 Breakeven points R : 21, 000, 000 + 450 x = 750 x ⇒ x = 70, 000 a. C : 26, 250, 000 + 400 x = 750 x ⇒ x = 75, 000 M : 15, 000, 000 + 500 x = 750 x ⇒ x = 60, 000 b. Least cost process at x = 65, 000 Cost R: $50,250,000 C: $52,250,000 M: $47,500,000 ← lowest cost with Mass Customization c. 65,000 demand > 60,000 breakeven for M 7.5 Breakeven points Continuous : 2, 400, 000 + 20 x = 80 x ⇒ x = 40, 000 a. Repetitive : 1,950, 000 + 30 x = 80 x ⇒ x = 39, 000 Mass Customization : 1, 480, 000 + 40 x = 80 x ⇒ x = 37, 000 Intermittent : 1,800, 000 + 40 x = 80 x ⇒ x = 45, 000 b. Least cost process at x = 48, 000 Continuous: $3,360,000 ← least cost Repetitive: $3,390,000 Mass Customization: $3,400,000 Intermittent: $3,720,000 c. Is 48,000 > 40,000? Yes, so we use continuous process. Annual profit = $480, 000 (2,000; 300,000) (4,000; 860,000) (11,000; 1,350,000) 7.6 M $ R C I 2,000 4,000 11,000 Volume 15,000 widest Repetitive has the widest production volume range over which it is a least-cost process. 7.7 Total profit now: Profit = 40, 000 × 2.00 − 20, 000 − 40, 000 × 0.75 = 80, 000 − 20, 000 − 30, 000 = 30, 000 Total profit with new machine: Profit = 50, 000 × 2.00 − 2, 000 − 50, 000 × 1.25 = 100, 000 − 25, 000 − 62,500 = 12,500 Since profit decreases with the new piece of equipment added to the line, purchase of the machine probably would not be a good investment. A-22
- 26. 7 Capacity Planning SUPPLEMENT S7.1 Problem is under risk and has two decisions, so use a decision tree: Medium demand 70 (0.4) Small expansion No additional expansion 90 109 High demand (0.6) 148 135 Additional minor expansion 135 Large expansion Medium demand 40 (0.4) 148 High demand (0.6) 220 (Payoffs and Expected Payoffs are in $1,000’s) Answer : Ralph should undertake a large expansion. Then the annual expected profit will equal $148,000. A-23
- 27. No additional minor expansion S7.2 $90,000 High demand 140,000 (0.3) Small expansion Additional minor expansion 70,000 Medium demand (0.7) $40,000 $70,000 High demand (0.3) Large expansion $105,000 14,000 Medium demand (0.7) –$25,000 Maximum value = $70,000 S7.3 No expand $0 Demand up small (0.4) $18,000 Small expand 16,000 Demand up medium (0.6) Large expand $10,000 Demand up medium (0.3) $20,000 –$10,000 18,000 Demand up large (0.7) Answer : $18,000 A-24 $34,000 $140,000
- 28. S7.4 (3) (1) 300,000 (2) 200,000 (2) (1) (2) (1) 100,000 50,000 (3) x 250 400 1,000 2,000 Cap level (2) is lowest for all x so 1, 000 ≤ x ≤ 2, 000 S7.5 Effective Capacity (text Equation S7-3) Efficiency 4.8 cars = 5.5 cars × 0.880. Therefore in one 8 hour day one bay accommodates 38.4 cars = (8 hrs × 4.8 cars per hr ) and to do 200 cars per day requires 5.25 bays or 6 bays = Actual (or expected) Output = 200 cars 38.4 cars per bay S7.6 a. b. F 450 = = 878.05 ∑ 1 − (Vi Pi ) Wi 0.5125 Breakeven ( $ ) = $878.05 Number of pizzas required at breakeven: Whole pizzas = ( 878.05 × 0.30 ) 5.00 = 52.7 → 53 BEP ( $ ) = Slices = ( 878.05 × 0.05 ) 0.75 = 58.5 → 59 Whole pizzas to make slices = 59 6 = 9.8 → 10 Therefore, he needs a total of 63 pizzas. He does not have sufficient capacity. S7.7 a. Remember that Yr 0 has no discounting. Initial coat $1,000,0000 yearly maint Salvage cost $50,000 yearly dues Discount rate 0.100 Year 0 1 2 3 4 5 Cost $1,075,000 75,000 75,000 75,000 75,000 75,000 Revenues $300,000 300,000 300,000 300,000 300,000 350,000 undisc. Profit A-25 75,000 members $300,000 500 Profit $–775,000 225,000 225,000 225,000 225,000 275,000 400,000 dues/member $600 PV Mult PV Profit 1 –$775,000 0.9 $202,500 0.81 $182,250 0.729 $164,025 0.6561 $147,623 0.59049 $162,385 PV Profit $83,782
- 29. b. Assume dues are collected at the beginning of each year. This is a simplification—in reality, people are likely to join throughout the year. (Technically, if equipment is sold at the end of year 5, it should probably appear as a final revenue stream in year 6 but the difference is only $2,952.45. Special deal comparison: $3,000 for all 6 years. Compare the PV cash stream of yearly dues from one member to that of the deal. Since we specified the club will always be full, we can make the assumption that the member (or her replacement) will always be paying the annual fee. Initial cost $0 yearly maint $0 Salvage cost $0 yearly dues $600 Discount rate 0.100 Year 0 1 2 3 4 5 Cost $0 0 0 0 0 0 (Membership fee) Revenues Profit $600 $600 600 600 600 600 600 600 600 600 600 600 undisc. Profit 3,600 PV Mult 1 0.9 0.81 0.729 0.6561 0.59049 PV Profit PV Profit $600 $540 $486 $437 $394 $354 $2,811 Since this is less than $3K, the special deal is worth more to the Health Club. Note also: If Health Club member is using same discount rates, it’s better for her to pay yearly. S7.8 Breakeven: Costs = Revenues 500 + 0.50 × b = b × 0.75 where b = number of units at breakeven or b ( 0.75 − 0.50 ) = 500 , 500 = 2,000 units 0.25 breakeven in units = 2,000 units breakeven in dollars = $0.75 × 2, 000 = $1, 500 and b = a. b. A-26
- 30. 8 CHAPTER 8.1 Cx ( 2, 000 )( 2.5) + ( 5, 000 )( 2.5) + (10, 000 )( 5.5 ) + ( 7, 000 )( 5.0 ) + (10, 000 )( 8.0 ) + ( 20, 000 )( 7.0 ) + (14, 000 )( 9.0 ) = = 6.67 2, 000 + 5, 000 + 10, 000 + 7, 000 + 10, 000 + 20, 000 + 14, 000 ( 2, 000 )( 4.5 ) + ( 5,000 )( 2.5 ) + (10, 000 )( 4.5) + ( 7,000 )( 2.0 ) + (10, 000 )( 5.0 ) + ( 20,000 )( 2.0 ) + (14, 000 )( 2.5 ) Cy = 8.2 8.3 Location Strategies 68, 000 Site A B C D Total Weighted Score 174 185 187 165 ∑ Population weights = 5, 000 Cx = Cy = ( 0 )( 2, 050 ) + ( −1)( 550 ) + ( 2 )(1, 025 ) + ( 3)( 775 ) + ( −2 )( 250 ) + ( −2 )( 350 ) = 0.525 5, 000 ( 0 )( 2, 050 ) + (1.5)( 550 ) + ( −1)(1, 025) + ( 3)( 775) + ( −3)( 250 ) + ( −1)( 350 ) 5, 000 Coordinates: ( 0.525, 0.205 ) 8.4 = 3.02 A 150 (0, 25) B 125 (1/2, 30) $ cost 100 (millions) 75 (2, 60) C 50 25 (1, 35) (0, 20) (0, 10) 0 1 2 3 4 5 6 10,000’s of Autos = V For all V such that 10, 000 ≤ V ≤ 60, 000 . A-27 V = 0.205
- 31. 8.5 Site A B C D Score 5w + 320 4 w + 330 3w + 370 5w + 255 Find all w from 1–30 so that: 3w + 370 ≥ 5w + 320 ⇔ 50 ≥ 2w ⇔ w ≤ 25 3w + 370 ≥ 4 w + 330 ⇔ 40 ≥ w ⇔ w ≤ 40 3w + 370 ≥ 5w + 255 ⇔ 115 ≥ 2w ⇔ w ≤ 57.5 Answer : For all w such that 1.0 ≤ w ≤ 25.0 (1, 35) 8.6 12 10 TC (millions $) Char 8 L.A. K.C. Charlotte K.C. (20, 7) (30, 9.5) 6 4 L.A. 2 V 0 5 10 15 20 25 30 35 (thousands) cutoff 4,100, 000 + 180V ≤ 1, 000, 000 + 300V For all V so: 4,100, 000 + 180V ≤ 2, 000, 000 + 250V 0 ≤ V ≤ 35, 000 3,100, 000 ≤ 120V ⇔ 2,100, 000 ≤ 70V 0 ≤ V ≤ 35, 000 1 25,833 ≤ V ⇔ ⇒ Answer : 30, 000 ≤ V ≤ 35,000 3 30, 000 ≤ V A-28
- 32. 8.7 ∑ weights = 14, 000 (or 14 for calculations below) Cx = Cy = ( 2 )( 3.5 ) + ( 8)( 7 ) + ( 6 )( 3.5) = 6.0 14 ( 6 )( 3.5) + (1)( 7 ) + ( 2 )( 3.5) 14 = 2.5 A-29
- 33. 9 CHAPTER Layout Strategy ( 60 )( 60 sec ) = 3,600 = 20 sec per PLA Cycle time = Theoretical number of work stations = c. 9.2 a. b. 9.1 Yes, it is feasible. 180 PLAs 180 1 D A2 3 C B4 Department pair Weekly Cost AB : 1 ( 8)( 800 ) = 2 1 AC : 2 ( 6 )( 700 ) = AD : 1 ( 4 )( 400 ) = 2 BC : 1 (10 )( 300 ) = 2 BD : 1 ( 7 )( 200 ) = 2 CD : 1 ( 9 )( 600 ) = 2 ($) 3, 200 2,100 800 1,500 700 2, 700 $11, 000 A-30 ∑ task time = 60 = 3 cycle time 20
- 34. 9.3 a. n= ∑t Cycle time C.T. = b. i = 274 ( seconds ) Cycle time ( seconds ) 60 ( 60 ) seconds = 60 seconds ( per truck ) so n = 60 trucks Answer : 5 Steps 1 and 2: Sample Answer c = 60 seconds From (a) number of stations is at least 5 Precedence diagram: 274 = 4.5667 → n = 5 . 60 40 30 B D 20 H 6 E 40 A 50 C 30 25 F 15 J 18 I G Step 3 Task A B C D E Number of Successors 9 4 4 2 2 Task F G H I J A-31 Number of Successors 2 2 1 1 0
- 35. Step 4 Station 1 Station 2 Station 3 Station 4 Station 5 Station 6 c. 9.4 a. Available A B, C B, C B, F, G B, F, G D, E, F, G D, E, G D, E, G E, G E, I E, I I, H I J J Available and Fit A — B, C — B, F, G E, F, G — D, E, G E, G — E, I I, H I — J Assigned A — C — B F — D G — E H I — J (Broke a tie) (Broke ties) (Broke ties) (Broke a tie) (Broke a tie) Answer : Station Tasks (Other answers possible, 1 A depending upon how ties 2 C are broken in above 3 B, F procedure) 4 D, G 5 E, H, I 6 J n = 6 work stations are in our answer. ∑ ti = 274 = 0.7611 Efficiency = n ( C.T.) 6 ( 60 ) First assignment costs = ( 8, 000 + 7, 200 + 1, 600 + 4,800 + 8, 000 + 800 ) × 1 2 = $15, 200 b. New layout costs = ( 8, 000 + 9, 600 + 6, 400 + 3, 600 + 2, 000 + 800 ) × No improvement—both yield the same cost. A-32 1 = $15, 200 2
- 36. 9.5 Cost of 3 attempts: Attempt 1 2 3 a. b. c. 1 S D M Work Area 2 D S D 3 M M S Cost $1,088 $1,142 $1,100 ↑ 2 ( 23)(10 ) + ( 32 )( 5) + ( 20 )( 8) = $1,100 9.6 a. Theoretical minimum number of stations = Cycle time = b. ∑ task times cycle time 60 48 = 12 minutes . So minimum number of stations = = 4 stations 5 12 WS #1 WS #2 10 min 12 min A B WS #5 D 8 min F 6 min WS #4 E C 12 min WS #3 c. 9.7 This requires 5 stations—it cannot be done with 4. 48 48 Efficiency = = = 80% for 5 stations. 5 × 12 60 There are three alternatives: Station 1 2 3 4 5 Alternative 1 Tasks A, B, F C, D E G, H I Alternative 2 Tasks A, B C, D G, H E I Each alternative has an efficiency of 86.67%. A-33 Alternative 3 Tasks A, F, G H, B C, D E I
- 37. 10 SUPPLEMENT Work Measurement 2 Zs S10.1 Required sample size = n = where s = 0.15 , x = 0.4 , z = 1.96 (for 95% confidence), hx (1.96 )( 0.15 ) n= = 54.0225 ≈ 54 ( 0.10 )( 0.4 ) 2 h = 10% accuracy level S10.2 a. b. ( ) 2 hx n ( 0.10 )( 0.40 ) 12 Zs = = 0.924 n = Thus, Z = s 0.15 hx Referring to Appendix I (Standard Normal Table), Area = 0.64 = 64% . The confidence level when n = 12 is 64%, as opposed to 95% when n = 54 (in Problem S10.1) 0.331 + 0.243 + " + 0.484 = 0.4484 minutes Average observed time = 12 Normal time = Average time × perf. rating = 0.4484 × 0.90 = 0.4036 minutes Normal time 0.4036 Standard time = = = 0.429 minutes 1 − allowance factor 1 − 0.06 100 hours × 60 minutes × 0.75 = 22.5 minutes 200 units Normal time = 22.5 minutes × 1.1 = 24.75 minutes Normal time for process 24.75 Standard time for job = = = 29.12 minute unit 1 − Allowance fraction 1 − 0.15 S10.3 Average observed time = S10.4 a. b. c. sum of times 1.74 = = 0.10875 minutes = 6.525 seconds number of cycles 16 Normal time = ( Observed time ) × ( Performance rating factor ) = 6.525 × 95% Observed time = = 6.2 seconds normal time 6.2 6.2 Standard time = = = = 6.739 seconds 1 − allowance factor 1 − 8% 92% A-34
- 38. S10.5 Normal time = 10 minutes × 0.90 = 9 minutes Personal + Fatigue + Delay 5 + 3 + 1 9 = = = 0.15 Allowance fraction = 60 minutes 60 60 Normal time 9 Standard time = = = 10.59 minutes 1 − Allowance fraction 1 − 0.15 Observation (Minutes Per Cycle) S10.6 Element 1 2 3 4 Rating 100% 90% 120% 100% 1 1.5 2.3 1.7 3.5 2 1.6 2.5 1.9 3.6 3 1.4 2.1 1.9 3.6 4 1.5 2.2 1.4 3.6 5 1.5 2.4 1.6 3.2 Average Normal Time Time 1.5 1.50 2.3 2.07 1.7 2.04 3.5 3.50 Normal time for lab test = 9.11 Normal time for process 9.11 Standard time for lab test = = = 11.1 minutes 1 − Allowance fraction 1 − 0.18 A-35
- 39. 12 CHAPTER 12.1 Inventory Management An ABC system classifies the top 70% of dollar volume items as A, the next 20% as B, and the remaining 10% as C items. Similarly, A items constitute 20% of total number of items, B items are 30%; and C items are 50%. Item Code Number Average Dollar Volume Percent of Total $ Volume 1289 → 400 × 3.75 = 1, 500.00 44.0% 2347 → 300 × 4.00 = 1, 200.00 36.0% 2349 → 120 × 2.50 = 300.00 9.0% 2363 → 75 × 1.50 = 112.50 3.3% 2394 → 60 ×1.75 = 105.00 3.1% 2395 → 30 × 2.00 = 60.00 1.8% 6782 → 20 ×1.15 = 23.00 0.7% 7844 → 12 × 2.05 = 24.60 0.7% 8210 → 8 × 1.80 = 14.40 0.4% 8310 → 7 × 2.00 = 14.00 0.4% 9111 → 6 × 3.00 = 18.00 0.5% $3, 371.50 100% Answer : The company can make the following classification: A: 1289, 2347 B: 2349, 2363, 2394, 2395 C: 6782, 7844, 8210, 8310, 9111 12.2 D (Annual Demand) = 4,800 units, P (Purchase Price/Unit) = $27, H (Holding Cost) = $2 2 ( 4,800 )( 30 ) 2 DS = . H 2 HQ SD 2 × 240 30 × 4,800 Thus, TC ( Total Annual Cost ) = PD + + = ( 4,800 × 27 ) + + 2 2 240 2 = 129, 600 + 240 + 600 = $130, 440 S (Ordering Cost) = $30. So, Q∗ ( Order Quantity ) = 240 = A-36
- 40. 12.3 D (Annual Demand) = 14,558, P (Purchase Price/Unit) = $5, H (Holding Cost/Unit) = $4, 2 DS 2 × 14,558 × 22 = = 400 tons per order S (Ordering Cost/Order) = $22, Q = H 4 HQ SD 4 × 400 22 × 14,558 + = ( 5 × 14,558 ) + TC = PD + + = 72, 790 + 800 + 800.69 2 2 400 2 = $74,390.69 Answer : The optimal order quantity ( Q ) = 400 tons order ; total annual inventory cost (TC ) = $74,391 . 12.4 D (Annual Demand) = 400 × 12 = 4,800, P (Purchase Price/Unit) = $350 unit , H (Holding Cost/Unit) = $35 unit year , S (Ordering Cost/Order) = $120 order . So, 2 DS 2 × 4,800 × 120 = = 181.42 = 181 units (rounded off). H 35 HQ SD 35 × 181 120 × 4,800 Thus, TC ( Total Cost ) = PD + + = ( 4,800 × 325 ) + + Q 2 181 2 = 1,560, 000 + 3,168 + 3,182 = $1,566,350 However, if Bell Computers orders 200 units, 35 × 200 120 × 4,800 TC = ( 4,800 × 325 ) + + = 1, 440, 000 + 3,500 + 2,880 = $1, 446,380 2 200 Answer : Bell Computers should order 200 units for a minimum total cost of $1,446,380. Q= 12.5 2 DS 2 × 4,800 × 120 = = 181 units H 35 2 DS 2 × 4,800 ×120 Q2 = = = 188 units H 32.5 2 DS 2 × 4,800 × 120 Q3 = = = 196 units H 30 181 units cannot be bought at $350, hence that isn’t feasible. 196 units cannot be bought at $300, hence that isn’t possible either. So, EOQ = 188 units . HQ SD 32.5 × 188 120 × 4,800 Thus, TC (188 units ) = PD + + = ( 325 × 4,800 ) + + Q 2 2 188 = 1,560, 000 + 3, 055 + 3, 064 = $1,566,119 HQ SD 30 × 200 120 × 4,800 + = ( 300 × 4,800 ) + TC ( 200 units ) = PD + + Q 2 2 200 = 1, 440, 000 + 3, 000 + 2,880 = $1, 445,880 Answer : The minimum order quantity is 200 units yet again, since the overall cost of $1,445,880 is less than ordering 188 units which has an overall cost of $1,566,119. Q1 = A-37
- 41. 12.6 D = 12,500 year , so d = 50 day , p = 300 day , S = $30 order , H = $2 unit year a. b. c. d. 12.7 p 2 × 12, 500 × 30 300 = × = 612.37 × 1.095 = 671 p−d 2 300 − 50 D 12,500 Number of production runs ( N ) = = = 18.63 671 Q d 50 1 Maximum inventory level ( I max ) = Q 1 − = 6711 − = 6711 − = 559 p 300 6 250 Days of demand satisfied by each production run = = 13.42 days in demand 18.63 only mode Q 671 Time in production for each order = = = 2.24 days in production for each p 300 order. Total time = 13.42 days per cycle. 2.24 = 16.7% . Thus, percent of time in production = 13.42 Q= 2 DS × H 4, 000 H = $2 unit year , S = $10 order , D = 4, 000 d = = 16 . So, 250 2 DS 2 × 4, 000 ×10 Q= = = 200 . ROP (reorder point) = l × d , l (lead time = 5 days). H 2 Thus, ROP = 5 ×16 = 80 units . Answer : Saveola, Inc. should place an order for 200 frames every time the inventory of frames falls to 80 units. This will be their inventory policy. 12.8 300 = 12.9 a. b. c. d. e. 12.10 a. b. c. d. e. 2 ( 5, 400 ) 34 H 90, 000 = ; Square both sides 367, 200 , H H= 367, 200 = $4.08 90,000 2 ( 6, 000 )( 30 ) 2 DS = = 189.74 units H 10 Average inventory = 94.87 Optimal number of orders/year = 31.62 250 Optimal days between orders = = 7.91 31.62 Total annual inventory cost = 601,897.37 (including the $600,000 cost of goods) EOQ = Holding cost = $530.33 Set up cost = $530.33 Unit costs = $56,250.00 Total costs = $57,310.66 Order quantity = 16,970.56 units Thus, order 10,001 units for a total cost of $57,310.66. A-38
- 42. 12.11 (52 Weeks) Cumulative Inventory $Value #Ordered Total $ Total Percent of Percent of Item per per Value/Week ($*Weeks) Rank Inventory Inventory Case Week Fish Fillets 143 10 $1,430.00 $74,360.00 1 17.54% 34.43% French Fries 43 32 $1,376.00 $71,552.00 2 16.88% 47.31% Chickens 75 14 $1,050.00 $54,600.00 3 12.88% 59.53% Prime Rib 166 6 $996.00 $51,792.00 4 12.22% 69.83% Lettuce (case) 35 24 $840.00 $43,680.00 5 10.31% 78.85% Lobster Tail 245 3 $735.00 $38,220.00 6 9.02% 83.82% Rib Eye Steak 135 3 $405.00 $21,060.00 7 4.97% 87.25% Bacon 56 5 $280.00 $14,560.00 8 3.44% 90.64% Pasta 23 12 $276.00 $14,352.00 9 3.39% 93.74% Tomato Sauce 23 11 $253.00 $13,156.00 10 3.10% 95.71% Table Cloths 32 5 $160.00 $8,320.00 11 1.96% 97.60% Eggs (case) 22 7 $154.00 $8,008.00 12 1.89% 98.28% Oil 28 2 $56.00 $2,912.00 13 0.69% 98.72% Trash Can Liners 12 3 $36.00 $1,872.00 14 0.44% 99.13% Garlic Powder 11 3 $33.00 $1,716.00 15 0.40% 99.42% Napkins 12 2 $24.00 $1,248.00 16 0.29% 99.72% Order Pads 12 2 $24.00 $1,248.00 17 0.29% 99.83% Pepper 3 3 $9.00 $468.00 18 0.11% 99.93% Sugar 4 2 $8.00 $416.00 19 0.10% 99.93% $312.00 20 0.07% 100.0% Salt 3 2 $6.00 $8,151.00 $423,852.00 100.00% a. b. c. Fish filets total $74,360 C items are items 10 through 20 in the above list (although this can be one or two items more or less) Total annual $ volume = $423,852 12.12 Safety Stock 0 100 200 Carrying Cost 0 100 × 15 = 1,500 200 × 15 = 3,000 Incremental Costs Stockout Cost 70(100 × 0.4 + 200 × 0.2) = 5,600 (100 × 0.2) × (70) = 1,400 0 Total Cost 5,600 2,900 3,000 The safety stock which minimizes total incremental cost is 100 kilos. The re-order point then becomes 200 kilos + 100 kilos or, 300 kilos. A-39
- 43. 12.13 S = $10 order , LT = 4 days , d = 80 day , σ = 20 , H = 10% of $250, D ( for term ) = 80 × 100 = 8, 000 a. b. 2 DS Q days = 10 days = 800 calzones . Order every H d ROP for calzone, σ ( demand during lead time ) = LT × σ = 40 , Z = 1 (from Table) Q= for 0.1587, ROP = d × LT + Z × σ dLT = 360 c. d. ROP − d × LT . σ LT Thus, Z of 0.25 gives 0.5987. So there is a 40.13% chance of stockout. Q D = 400 = average inventory , = 10 orders term . 2 Q Q Holding cost term = H = ( 400 )( 0.25 ) = $100 2 D Order cost term = S = (10 )(10 ) = $100 Q On hand = 85, days left = 1. Since ROP = d × LT + Z × σ dalt , Z = 12.14 S = $16 , H = $0.40 calzone term , p = 160 , d = 80 , D = 8, 000 for 100 days of the term a. b. c. Q= 2 DS = 1,131.37 H (1 − d p ) Q 1,131.37 = = 14.14 d 80 Q 1,131 = 7.07 days days = Run production for 160 p Cycle = A-40
- 44. 12.15 Under present price of $6.40 per box: 2 DS 2 × 5000 × 25 Economic Order Quantity: Q∗ = = = 395.3 or 395 boxes where H 0.25 × 6.40 D = annual demand, S = set-up or order cost, H = holding cost DS QH + + CD Total cost = order cost + holding cost + purchase cost = Q 2 5, 000 × 25 395 × 0.25 × 6.40 = + + 6.4 × 5, 000 = 316.46 + 316.00 + 32, 000 395 2 = $32, 632.46 Note: Order and carrying costs are not exactly equal due to rounding of the EOQ to a whole number. Under the quantity discount price of $6.00 per box: DS QH Total cost = order cost + holding cost + purchase cost = + Q 2 5, 000 × 25 5, 000 × 0.25 × 6.00 = + + 5, 000 × 6.00 = 41.67 + 3, 750.00 + 30, 000 3, 000 2 = $33, 791.67 Therefore, the old supplier with whom they would incur a total cost of $32,632.46, is preferable. 12.16 Economic Order Quantity, non-instantaneous delivery: where: D = period demand, S = set-up or order cost, H = holding cost, d = daily demand rate, p = daily production rate 2 DS 2 × 10, 000 × 200 Q∗ = = = 2,309.4 or 2,309 units 50 d 1.00 (1 − 200 ) H 1− p ( ) A-41
- 45. 13 CHAPTER 13.1 Aggregate Planning The total production required over the year is 8,400 units, of 700 per month. Thus, the schedule is to produce 700 per month and have no costs associated with work force variation. The only costs incurred will be the monthly production cost, the inventory cost, and the shortage cost. The costs are calculated as follows. Beginning Month Inventory January 0 February 200 March 300 April 400 May 400 June 400 July 300 August 100 September 0 October 0 November 0 December 0 Production 700 700 700 700 700 700 700 700 700 700 700 700 8,400 Production Cost $49,000 49,000 49,000 49,000 49,000 49,000 49,000 49,000 49,000 49,000 49,000 49,000 $588,000 Demand 500 600 600 700 700 800 900 900 800 700 600 600 8,400 Ending Inventory 200 300 400 400 400 300 100 0 0 0 0 0 2,100 Inventory Shortage Cost 0 $600 0 900 0 1,200 0 1,200 0 1,200 0 900 0 300 100 0 200 0 200 0 100 0 0 $ 0 600 $ 6,300 The total cost of this plan is the sum of the three costs, or $600,300. A-42 Shortage Cost $0 0 0 0 0 0 0 1,000 2,000 2,000 1,000 0 $6,000
- 46. 13.2 a. Total hotel demand for the year = 7,000,000 Total restaurant demand for the year = 2,080,000 Level staffing Quarter Winter Spring Summer Fall Totals Hotel (Room) Demand 800,000 2,200,000 3,300,000 700,000 7,000,000 Personnel Required 33 33 33 33 132 Restaurant Demand Req. 160,000 12 800,000 12 960,000 12 160,000 12 2,080,000 48 Personal cost = 180 quarters of labor × 5,000 = Hiring cost = 45 hires at $1,000 = Termination cost = none = b. Total 45 45 45 45 180 Hire Terminate 45 – – – – – – – 45 $900,000 45,000 0 $945,000 Total hotel demand for the year = 7,000,000 Total restaurant demand for the year = 2,080,000 Chase staffing (staffing to meet the forecasted demand) Personnel Hotel (Room) Quarter Demand Winter 800,000 Spring 2,200,000 Summer 3,300,000 Fall 700,000 Totals 7,000,000 Personnel TermiTermi- Restaurant Demand Req. Hire nate Req. Hire nate 8 8 160,000 2 2 22 14 800,000 10 8 33 11 960,000 12 2 7 0 26 160,000 2 0 10 70 2,080,000 26 Personnel cost = 96 quarters of labor × 5,000 = Hiring cost = 45 hires @ $1,000 = Termination cost = 36 terminations @ $2,000 = c. Totals Quarter Quarter Terminations Hires 10 22 13 36 45 36 $480,000 45,000 72,000 $597,000 The chase plan developed in part b is most economical ($945,000 vs. $597,000) A-43
- 47. 13.3 Total hotel demand for the year = 7,000,000 Total restaurant demand for the year = 2,080,000 Hiring from local staffing agency all personnel above base requirements. Quarter Fall Spring Summer Fall Totals Hotel (Room) Req. Hire Demand 800,000 8 7 2,200,000 22 0 3,300,000 33 0 700,000 7 0 7,000,000 28* Personnel from Agency 1 15 26 0 42 Restaurant Demand 160,000 800,000 960,000 160,000 2,080,000 Req. 2 10 12 2 26 Quarter Personnel Quarter from from Hires Agency Hire Agency 2 0 2 1 0 8 0 23 0 10 0 36 0 0 0 0 8** 18 35 60 *On Hotel Grand payroll (7 each quarter × 4 quarters = 28) ** On Hotel Grand payroll (2 each quarter × 4 quarters = 8) Total on Grand Hotel payroll = 28+ 8 = 36 Personnel cost = 36 quarters of labor × 5,000 = Hiring cost = 9 ( = 7 + 2 ) hires @ $1,000 = Termination cost = 0 terminations @ $2,000 = Staffing agency costs = 60 @ 6,500 = 13.4 Plan A: Month Demand Production Mar 1,000 900 Apr 1,200 1,200 May 1,400 1,400 June 1,200 1,200 July 1,500 1,500 Aug 1,300 1,300 Total extra cost: $112,000 Hire $180,000 9,000 0 390,000 $579,000 Fire 700 Extra Cost 56,000 12,000 8,000 8,000 12,000 16,000 300 200 200 300 200 Plan B: Month Demand Production Inventory Mar 1,000 1,100 200 Apr 1,200 1,100 100 May 1,400 1,100 June 1,200 1,100 July 1,500 1,100 Aug 1,300 1,100 Total extra cost: $39,000 Therefore, Plan B would be preferred. A-44 Sub-Contracting 200 100 400 200 Extra Cost 2,000 1,000 8,000 4,000 16,000 8,000
- 48. 13.5 Plan Number 1: Month Demand Production Inventory 1 1,000 1,200 300 2 1,200 1,200 300 3 1,400 1,200 100 4 1,200 1,200 100 5 1,500 1,200 6 1,300 1,200 Total extra cost: $112,000 Plan Number 2: Month Demand Production Inventory 1 1,000 1,200 300 2 1,200 1,200 300 3 1,400 1,200 100 4 1,200 1,200 100 5 1,500 1,200 6 1,300 1,200 Total extra cost: $39,000 Sub-Contracting 200 100 Overtime 200 100 Extra Cost 3,000 3,000 1,000 1,000 8,000 4,000 Extra Cost 3,000 3,000 1,000 1,000 2,000 1,000 Therefore, Plan Number 2 would be preferred. 13.6 Plan Y: Month Demand Production 1 1,100 1,100 2 1,600 1,600 3 2,200 2,200 4 2,100 2,100 5 1,800 1,800 6 1,900 1,900 Total extra cost: $112,000 Hire Fire 400 Extra Cost 32,000 20,000 24,000 8,000 24,000 4,000 500 600 100 300 100 Plan Z: Month Demand Production Inventory 1 1,100 1,600 600 2 1,600 1,600 600 3 2,200 1,600 4 2,100 1,600 5 1,800 1,600 6 1,900 1,600 Total extra cost: $52,000 Therefore, Plan Z would be preferred. A-45 Sub-Contracting Extra Cost 6,000 6,000 500 200 300 20,000 8,000 12,000
- 49. 14 Materials Requirements Planning (MRP) & ERP CHAPTER 14.1 14.2 Requirement for 3,500 “Get Well” bud vases. 3,500 Vases 3,500 × 8" white ribbons (2,333 ft.) 3,500 × 8" red ribbons (2,333 ft.) 3,500 signature cards 7,000 sprigs of baby’s breath 7,000 pink roses Lot Size Lot for Lot Lot for Lot Lot for Lot Lot for Lot Lot for Lot Lead Time 1 1 1 1 2 On Safety AlloHand Stock cated 1,000 — — — — — — — — — — — — — — LowLevel Code 0 1 1 1 2 Item ID Period (week) CD Case 1 2 5 6 7 Gross Requirements 650 300 Scheduled Receipts CD Projected On Hand 1,000 1,000 1,000 1,000 350 Case Net Requirements Planned Order Receipts Planned Order Releases 500 550 400 500 50 500 500 400 400 400 500 500 500 Gross Requirements Scheduled Receipts CD Projected On Hand Top Net Requirements Planned Order Receipts Planned Order Releases 500 400 500 500 500 400 400 400 500 500 500 500 400 500 500 500 400 400 400 500 500 500 500 400 500 400 400 500 500 500 500 500 500 400 500 400 500 500 500 500 400 500 100 500 500 500 Gross Requirements Scheduled Receipts CD Projected On Hand Bottom Net Requirements Planned Order Receipts Planned Order Releases 500 Gross Requirements Scheduled Receipts CD Projected On Hand Insert Net Requirements Planned Order Receipts Planned Order Releases Gross Requirements Scheduled Receipts Projected On Hand Black Net Requirements Dye Planned Order Receipts Planned Order Releases A-46 3 500 500 4 100 100 8 100
- 50. 14.3 LowLot Lead On Safety AlloLevel Size Time Hand Stock cated Code Lot for Lot Lot for Lot Lot for Lot Lot for Lot Lot for Lot Lot for Lot Lot for Lot Lot for Lot 1 1 3 1 1 1 1 1 — — — — — 3,000 3,000 — — — — — — — — — — — — — — — — — Item ID 1 1 2 3 Gross Requirements Scheduled Receipts Projected On Hand Net Requirements Planned Order Receipts Planned Order Releases 0 1 Period (day) Ball Point Pens 10,000 10,000 10,000 2,000 — — 2,000 2,000 2,000 Gross Requirements Scheduled Receipts Projected On Hand Body Net Requirements Planned Order Receipts Planned Order Releases Gross Requirements Scheduled Receipts Clip Projected On Hand Fine Net Requirements Pt Planned Order Receipts Planned Order Releases Gross Requirements Scheduled Receipts Std. Projected On Hand 3,000 Clip Net Requirements Planned Order Receipts Planned Order Releases Gross Requirements Scheduled Receipts Std. Projected On Hand 3,000 Ball Point Net Requirements Planned Order Receipts Planned Order Releases Fine Point Ball Point Gross Requirements Scheduled Receipts Projected On Hand Net Requirements Planned Order Receipts Planned Order Releases A-47 5 10,000 — — 10,000 10,000 10,000 10,000 Gross Requirements Scheduled Receipts Projected On Hand Cap Net Requirements Planned Order Receipts Planned Order Releases Gross Requirements Scheduled Receipts Projected On Hand Ink CC Net Requirements Planned Order Receipts Planned Order Releases 4 10,000 10,000 10,000 10,000 5,000 5,000 5,000 5,000 5,000 3,000 2,000 2,000 2,000 5,000 3,000 2,000 2,000 2,000 5,000 5,000 5,000 5,000 6 7 8
- 51. 14.4 LowLot Lead On Safety AlloLevel Size Time Hand Stock cated Code Lot for Lot Lot for Lot Lot for Lot Lot for Lot Lot for Lot Lot for Lot Lot for Lot Lot for Lot Lot for Lot 1 1 1 1 1 1 1 1 1 — — — — — — — — — — — — — — — — — — — — — — — — — — — 0 1 1 1 1 2 2 2 3 Item ID Period (week, day) Ball Point Pens 1 2 3 Gross Requirements Scheduled Receipts Glue Projected On Hand 100 gal Net Requirements Planned Order Receipts Planned Order Releases 100 Gross Requirements Scheduled Receipts Projected On Hand Base Net Requirements Planned Order Receipts Planned Order Releases Gross Requirements Scheduled Receipts Long Projected On Hand Braces Net Requirements Planned Order Receipts Planned Order Releases Gross Requirements Scheduled Receipts Short Projected On Hand Braces Net Requirements Planned Order Receipts Planned Order Releases Gross Requirements Scheduled Receipts Projected On Hand Leg Net Requirements Planned Order Receipts Planned Order Releases Gross Requirements Scheduled Receipts Brass Projected On Hand Caps Net Requirements Planned Order Receipts Planned Order Releases A-48 100 640 128 128 640 640 640 640 640 128 128 128 128 128 128 640 128 128 640 640 640 640 640 128 128 128 128 128 128 80 16 16 80 80 80 80 80 16 16 16 16 16 16 40 8 8 100 60 20 12 640 100 640 40 80 8 80 Gross Requirements Scheduled Receipts Stain Projected On Hand gal Net Requirements Planned Order Receipts Planned Order Releases 7 640 640 6 640 Gross Requirements Scheduled Receipts Projected On Hand Top Net Requirements Planned Order Receipts Planned Order Releases 5 640 Gross Requirements Scheduled Receipts Coffee Projected On Hand Table Net Requirements Planned Order Receipts Planned Order Releases 4 640 128 128 640 640 640 640 640 128 128 128 128 128 128 880 880 880 880 4 1,280 1,280 1,280 1,280 1,280 1,280 1,280 1,280 5,120 5,120 5,120 5,120 5,120 880 5,120 6,000 6,000 880
- 52. 14.5 Coffee Table Master Schedule Hours Required Table Assembly 2 Top Preparation 2 Assemble Base 1 Long Braces (2) 0.25 Short Braces (2) 0.25 Legs (4) 0.25 Total Hours Employees needed @ 8 hrs. each 14.6 Lead Time Day 1 Day 2 Day 3 Day 4 320 320 640 1280 160 1,280 640 320 320 640 3,200 400 1,280 1,280 640 64 64 128 3,456 432 1 1 1 1 1 1 0 0 Day 5 640 1,280 256 128 64 64 128 1,920 240 Day 6 640 256 256 128 Day 7 128 256 640 80 Day 8 128 256 32 The following table lists the components used in assembling FG-A. Also included for each component are the following information: the on-hand supply, lead time, and direct components. Item PG-A SA-B SA-C SA-D E F On-Hand 0 0 0 0 10 5 LT (weeks) 1 1 2 2 1 3 Components SA-B, SA-C(2), SA-D(2) SA-D(2) E, F(2) E (3) — — While not required as part of the question, we recommend you make a product structure tree to help you answer the questions on the Bill-of-Materials and lead time. 1 FGA [LT =1] [LT =1] SA-B [LT =2] SA-C(2) [LT =2] SA-D(2) E(3) [LT = 1] a. [LT =2] SA-D(2) E(3) [LT = 1] E [LT = 1] F(2) [LT = 3] Bill-of-Material associated with 1 unit of FG-A 1 SA-B Note: this is just the master recipe. Not subtracting off on on-hand quantities YET 2 SA-C = 2 ×1 + 2 4 SA-D 14 E = 2 × 3 + 2 ×1 + 2 × 3 = 6 + 2 + 6 = 2× 2 4 F A-49
- 53. b. Total lead time (in weeks) associated with making an item of FG-A, assuming we had no starting on-hand for any part? 6 weeks Look at all branches of the product structure tree (or an assembly time chart, if we had one). The most common mistake people tend to make is to forget the LT associated with final assembly. c. Longest branch is 6 weeks = 1wk FGA + 2wks SA-C + 3wks F Yes, if we wanted to make one FG-A, we need to order more of either E or F. We have enough F on-hand, but need 4 more E. Now we look at the on-hand quantities and see we have 10 E and 5 F. Compare that to the BOM calculated above. No subassemblies like SA-B, SA-C or SA-D, so we are going to need all the component parts. Our on-hand records show we have 14 E and 4 F, so we are short 4 E. 14.7 a. b. 17 × 2 (level 2) = 34 [17 × 3 (B’s)] – 15 = 51 – 15 = 36 36 × 5 (D’s) = 180 14.8 a. b. 17 × 2 (level 2) + 17 × 2 × 3 (level 3) = 136 170 – 12 × 2 (level 2 on hand) = 170 – 24 = 146 A-50
- 54. 15 CHAPTER Short-Term Scheduling 15.1 A dummy task is added to balance the problem. The assignment is May—Task 1 Gray—Task 2 Ray—Task 3 Total time = 1 + 1 + 1 = 3 hours 15.2 Convert the minutes into $, Chris Marketing $80 Finance $120 Operations Human Resources $125 $140 Steve $20 $115 $145 $160 Juana $40 $100 $35 $45 Rebecca $65 $35 $25 $75 (Now subtract smallest number in each column from every number.) The Minimum Cost Solution = Chris Steve Juana Rebecca Finance Marketing Human Resources Operation $120 $ 20 $ 45 $ 25 $210 15.3 The best pairs are assigned as follows: Ajay—Jackie Jack—Barbara Gray—Stella Raul—Dana Total compatibility score (overall) = 90 + 70 + 50 + 20 = 230 A-51
- 55. Gantt Chart 15.4 F 6 D 10 13 C B 20 28 E A 37 10 Project A B C D E F a. b. 20 Hour 30 Time 9 7 3 4 8 6 40 Due Date 22 17 16 13 16 9 Late Days 15 3 0 0 12 0 6 + 10 + 13 + 20 + 28 + 37 114 = = 19 days 6 6 total late days 15 + 3 + 12 Average lateness = = = 5 days number of jobs 6 Average flow time = c. 15.5 Maximum lateness = 15 days (for job A—Gantt Table) a. The shortest processing time ( SPT ) =DBACE D Flow Time Due Hours Late Hours 0 0 0 B 1 12 — A 3 4 — 41 = 8.2 hours 5 Number of deliveries late = C and E = 2 5 + 13 18 Average hours late = = = 9 hours 2 2 Average flow time = A-52 C 6 8 — E 11 6 5 20 7 13 = 41 = 18
- 56. b. The EDD schedule = BCEAD B Flow Time Due Hours Late Hours C 0 0 0 E 2 4 — 7 6 1 A 16 7 9 D 19 8 11 20 12 8 64 = 12.8 hours 5 Number of deliveries late = C, E, A, and D = 4 29 = 7.25 days Average hours late = 4 Average flow time = 15.6 1011 1517 20 45 Cut & sew 1 2 25 3 [123 schedule] Deliver 1 2 3 15 18 20 2325 56 56 Cut & sew 10 10 13 1517 20 3 2 25 1 [321 schedule] Deliver 56 1011 15 2022 25 Answer : The 321 schedule finishes in 22 days, 1 day faster than the 123 schedule, which finishes in 23 days. 15.7 a. b. We begin by taking 5 empty slots Next, we find the shortest time in the table. It is product 2678 on Machine B. Since this is on the second machine, this product can be done as late as possible. 2678 c. d. Then we find the shortest again. It is 2800 on B 2800 2678 Then it is 2731 on A. Since it is on the first machine, it is the earliest job. 2731 2800 2678 A-53
- 57. e. Finally, we get 2731 2134 2387 2800 2678 5 12 19 22 A 2731 2134 2387 2800 2678 B 2731 2134 7 14 21 25 33 35 2387 2800 2678 Total time = 33 hours 15.8 a. b. The jobs should be processed in the sequence: 3, 6, 2, 7, 5, 1, 4 Time = 61 hours 15.9 a. Jobs should be processed in the sequence A, B, C, D, E if scheduled by the FCFS scheduling rule. Jobs should be processed in the sequence A, B, C, D, E if scheduled by the EDD scheduling rule. Jobs should be scheduled in the sequence B, E, A, C, D if scheduled by the SPT scheduling rule. Jobs should be processed in the sequence D, C, A, B, E if scheduled by the LPT scheduling rule. b. c. d. 15.10 Job 103 205 309 410 517 Due Date 214 223 217 219 217 Duration (Days) 10 7 11 5 15 Critical Ratio 1.4 3.3 1.5 3.8 1.1 Jobs should be scheduled in the sequence 517, 103, 309, 205, 412 if scheduled by the critical ratio scheduling rule. A-54
- 58. 16 Just-In-Time and Lean Production Systems CHAPTER a. Q= 2× D× S H 1− d p Q2 = 16.1 2× D× S H 1− d p S= ( ( ) ( ) Q2 ( H ) 1 − d p 2D ) = (1, 000 ) (10 ) (1 − ) = (1, 000, 000 )(10 )(.2 ) = 2, 000, 000 = $10 2 400 500 2 × 100, 000 200, 000 200, 000 10 1 ⇒ hr ⇒ 10 min 60 6 Note that the lead time was not needed in this problem. b. 16.2 10 min. – 2 min. = 8 min. improvement required 16.3 How to improve setups (from Figure 16.4 in Heizer/Render text). Step 1: Separate setup into preparation and actual setup, doing as much as possible while the machine/process is operating. Step 2: Move material closer and improve material handling Step 3: Standardize and improve tooling Step 4: Use one-touch system to eliminate adjustments Step 5: Train operators and standardize work procedures 16.4 Q= 2 × (1, 250, 000 ) ×15 2× D× S 37,500, 000 = = = 3, 750, 000 = 1,936 units 5,000 d 10 20 (1 − 10,000 ) H 1− p ( ) 1, 936 per container = 19.36 containers 100 A-55
- 59. 16.5 Note: S = 6 minutes (or Q= 2× D× S ( H 1− d p ) = 1 10 hour) ×$100 shop labor cost = .1× $100 = $10 ( 2 )(12, 500 )(10 ) = 50 50 (1 − 500 ) 250, 000 = 5, 555 = 74 45 1 = EOQ + Safety stock + Lead time = 74 + 10 + ( 500 ) 2 = 334 oil pumps 334 No. of Kanban containers = = 9.27 = 9 or 10 containers required 36 16.6 Where: D = annual demand, S = set-up or order cost, H = holding cost, d = daily demand rate, p = daily production rate. Solving for S (set-up cost): S= ( Q2 × H × 1 − d p ) = 150 2 150 × 10 × (1 − 1,000 ) = 22,500 × 10 × (1 − 0.15 ) 2D 2 × 40, 000 80, 000 191, 250 = = $2.39 80, 000 $2.39 set-up × 60 minute hour Set-up time = = 2.69 minute set-up $50 hour A-56
- 60. 17 CHAPTER 17.1 17.2 Maintenance and Reliability Failure Rate Analysis Station No. 1: N = 1,000 F = 22 22 FR = = 0.022 1, 000 R = 1 − 0.022 = 0.978 = 97.8% Station No. 2: N = 1,000 F = 51 51 FR = = 0.051 1, 000 R = 1 − 0.051 = 0.949 = 94.5% Answer : Unit #1 is clearly better, with a lower failure rate and better reliability. Reliability ( Rs ) = R1 × R2 × R3 ×! × Rn . So, R5 = ( 0.98 )( 0.99 )( 0.96 ) = 0.9304 Answer : The reliability is 93.04% A-57
- 61. 17.3 Task 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Total Errors 1 6 4 2 3 1 0 2 1 2 1 1 0 2 3 29 FR 0.001 0.006 0.004 0.002 0.003 0.001 0.000 0.002 0.001 0.002 0.001 0.001 0.000 0.002 0.003 Reliability 0.999 0.994 0.996 0.998 0.997 0.999 1.000 0.998 0.999 0.998 0.999 0.999 1.000 0.998 0.997 Overall Reliability ( Rs ) = Product of all reliabilities above = 0.971 = 97.1% 17.4 Reliability (stick) = 1 − (1 − 0.99 ) = 0.9999 2 Reliability (cloth) = 1 − (1 − 0.98 ) = 0.999998 2 The reliability of the snake-charmer’s work = 0.9999 × 0.999998 = 0.9998 = 99.98% 17.5 .90 .96 .93 .99 .90 .96 .93 .99 .90 .96 .93 .90 Mountain test’s reliability = 1 − (1 − 0.96 ) = 0.999936 3 Bump test’s reliability = 1 − (1 − 0.99 ) = 0.999900 2 Speed test’s reliability = 1 − (1 − 0.93) = 0.999657 3 Sudden brake test’s reliability = 1 − (1 − 0.90 ) = 0.999900 4 So, overall reliability ( Rs ) = 0.999936 × 0.999900 × 0.999657 × 0.999900 = 0.9994 = 99.94% A-58
- 62. A Decision Making Tools MODULE States of Nature A.1 Prob. 0.35 0.65 Competes Doesn’t Compete Expected Value Assembly Line $30,000 $26,500 Plant Addition –$10,000 $50,000 $29,000 No New Product Decision Alternatives $20,000 $0 $0 $0 Answer : 0.35 ( $20, 000 ) + 0.65 ( $50, 000 ) − $29, 000 = $10,500 States of Nature Slight Major Increase Increase A.2 Fixed N 0 M –4,000 Decision Alternatives L –10,000 O –50,000 Use “maximin: criterion. No floor space (N). A.3 Increasing capacity Using overtime Buying equipment 2,000 8,000 6,000 4,000 Row Average $700,000 $700,000 $733,333 ← Using equally likely, “Buying equipment” is the best option. A-59 3,000 9,000 20,000 40,000 Minimum 0 –4,000 –10,000 –50,000
- 63. 3 2 35 15 ( 50 ) + ( 20 ) + (100 ) + (100 ) 10 10 100 100 = 15 + 4 + 35 + 15 = 69 A.4 Expected value under certainty = A.5 E ( A ) = 0.4 ( 40 ) + 0.2 (100 ) + 0.4 ( 60 ) = 60 E ( B ) = 0.4 ( 85) + 0.2 ( 60 ) + 0.4 ( 70 ) = 74 ← E ( C ) = 0.4 ( 60 ) + 0.2 ( 70 ) + 0.4 ( 70 ) = 66 E ( D ) = 0.4 ( 65 ) + 0.2 ( 75 ) + 0.4 ( 70 ) = 69 E ( E ) = 0.4 ( 70 ) + 0.2 ( 65 ) + 0.4 ( 80 ) = 73 Choose Alternative B. 1 1 3 1 ( 50 ) + ( 92 ) + ( 40 ) + ( 64 ) = 10 + 23 + 12 + 16 = 61 5 4 10 4 A.6 Expected value under certainty = A.7 Solution Approach: Decision tree, since problem is under risk and has more than one decision. Buy now 33 $30,000 Unavailable $0 (.40) Wait 1 day 33 Buy now Avail. (.60) 55 $55,000 Unavailable $0 (.70) Wait 1 day 21 Buy now $70,000 Avail. (.30) 70 Wait 1 day $0 (Numbers in Nodes are in $1,000’s) Answer : Maximum expected profit: $33,000. Manny should wait 1 day. Then, if an XPO2 is available, he should buy it. Otherwise, he should stop pursuing an XPO2 on the wholesale market. A-60
- 64. A.8 We use a decision table, since expected marginal value of perfect information is asked for. Profits (= Payoffs) D1: D2: D3: Good Market: Bad Market: Good Market: Bad Market: Good Market: Bad Market: 80(400) – 25,000 = $7,000 70(375) – 25,000 = $1,250 85(450) – 30,000 = $8,250 80(425) – 30,000 = $4,000 90(475) – 33,000 = $9,750 80(425) – 33,000 = $1,000 Decision Alternatives : D1, D2, D3, N (N = Do Nothing) States of Nature : Good Market, Bad Market Decision Table and Solution States of Nature Good Market Bad Market (0.6) (0.4) Decision Alternatives D1 D2 D3 N 7,000 8,250 9,750 0 The car enthusiast should use design D2. Exp. value under certainty: ( 0.6 )( 9, 750 ) + ( 0.4 )( 4, 000 ) = $7,450 Minus Max Expected Value: –$6,550 Expected (Marginal) Value → $900 A-61 1,250 4,000 1,000 0 EV 4,700 6,550 6,250 0
- 65. A.9 More than one decision is involved, and problem is under risk, so use a decision tree approach. Low (0.4) demand 200 Expand 242 Small Facility High (0.6) demand 270 270 Don’t expand Modest (0.3) response 544 Advertise Low (0.4) demand Large Facility 223 20 160 160 Sizable (0.7) response 544 Don’t Advertise High (0.6) demand 220 40 800 Note: Payoffs and expected payoffs are in $1000’s. a. Build the large facility. If demand proves to be low, then advertise to stimulate demand. If demand proves to be high, no advertising option is available (so don’t advertise). Expected Payoff: $544,000. b. A.10 Approach: Set up and solve via a decision table. $16.50 hr Cost per dozen bagels: Labor: = $1.65 dozen 10 dozen hr + $0.85 hr Ingredients: $2.50 dozen $3.00 if sold during the day∗ Profit per dozen bagels = ∗∗ −$1.00 if not ∗ $3.00 = $5.50 − $2.50 ∗∗ −$1.00 = $1.50 − $2.50 B = dozens baked in A.M. Suppose: D = dozens sold during the day (= dozens demanded) $3D − $1( B − D ) if B ≥ D Then, total profit = if B ≤ D $3B Also, since demand is between 40–43 dozen bagels, inclusive, always more profitable to bake 40 dozen than fewer than 40 dozen, AND always more profitable to bake 43 dozen than more than 43 dozen. From work above, decision table, expected values, and answer are as follows. A-62
- 66. Dozens Demanded D (= States of Nature) 40 41 42 43 (0.20) (0.30) (0.35) (0.15) 40 41 42 43 Dozens Baked B = Decision Alternatives 120 119 118 117 120 123 122 121 120 123 126 125 120 123 126 129 EV 120 122.2 123.2 122.8 Answer : 42 dozen bagels A.11 a. Payoff Rev – Cost 10 – 6 =4 EMV = 0.6(0.4) + 0.4(–1) = 2 P(high) = 0.6 EMV = 0.5(0) + 0.5(–3) = –1.5 P(success) = 0.5 8 – (6 + 2) = 0 Build advertise 2 Build? P(low) = 0.4 Prob (not success) = 0.5 5 – (6 + 2) = –3 Ad? Don’t Advertise 5 – 6 = –1 Don’t Build b. c. A.12 $0 Build studio, but don’t advertise (even if demand is low) Expected value = $2 million Build a. Build P(high) = 0.6 10 – 6 = 4 Don’t Build 0 EMV = 0.6(4) + Demand 0.4(0) = 2.4 Build P(low) = 0.4 Ad? We don’t really care about this part of the tree (which has a –1 EMV anyways) Build? Don’t Build b. EVPI = EV with PI – EMV = 24 – 2 = 0.4 = $400,000 A-63 0
- 67. B MODULE B.1 a. b. Linear Programming Let T = number of trucks to produce per day Let C = number of cars to produce per day max z = 300T + 220C 1 1 S.T. T + C ≤1 40 60 1 1 T + C ≤1 50 50 T, C ≥ 0 Graph feasible region: C 60 A 50 40 B 30 20 10 O C T 10 20 30 40 50 60 70 80 90 (1) (2) c. Point Coordinates O ( 0, 0 ) A ( 0, 50 ) C ( 40, 0 ) B Solve 2 equations in 2 unknowns derived from (1) and ( 2 ) to obtain ( 20, 30 ) d. Produce 20 trucks and 30 cars daily for a profit of $12,600 per day. A-64 z value 0 11, 000 12, 000 12, 600 ←
- 68. B.2 We solve this problem by the isocost line method: x2 18 16 14 12 10 8 (6, 6) 6 4 (0, 4) 2 x1 2 –2 4 6 z=4 8 10 12 14 16 18 z = 11 z = 12 Answer : Unique optimal solution is (0, 4) with z = 4 B.3 Feasible region is a line segment AB, where A = (0,0), B = (3, 5). Solution via isoprofit line method is shown. (5) x2 8 B (3) (1) 6 4 (4) 2 A –2 (2) x1 2 4 6 –4 –6 –8 z=0 z=5 z = 20 Answer : Unique optimal solution is ( x1 , x2 ) = ( 3, 5) , with objective function value 20. A-65
- 69. B.4 Using the isoprofit line method. B 9 8 (5) Opt. sol. 7 6 (2) 5 4 3 2 (3) (4) 1 z=9 (1) z = –4 z = –6 1 2 3 4 5 6 7 8 9 –1 –2 –3 –4 (6) A Answer : Unique optimal solution is ( A, B ) = (1, 5 ) . It has objective function 9.0. B.5 Feasible region is same as in Problem B.4. Use Isoprofit line method. B –1 –2 –3 –4 z=4 z=1 9 8 7 6 5 4 3 2 1 A 1 2 3 4 5 6 7 Answer : Problem is feasible and unbounded. A-66 8 9
- 70. B.6 Let S = number of standard bags to produce per week Let D = number of deluxe bags to produce per week maximum: z = 10 S + 8 D 1 S + D ≤ 300 2 2 S + D ≤ 360 3 S, D ≥ 0 ( A) ( B) D z = $3,840 (0, 540) 400 z = 2,000 z = 1,000 300 (A) (240, 180) 200 100 (B) 100 200 (360, 0) 300 400 Extreme (Corner) Points Point Profit (0, 0) $0 (0, 300) $2,400 (360, 0) $3,600 $3,840 ← optimal solution and answer (240, 180) A-67 S 500 600
- 71. y B.7 (–1, 3 1/2) 5 4 (–2, 4) (–2, 3) 3 (2) (4) (1) (3) 2 (4, 1) 1 –2 –1 (5)1 x 2 3 4 5 6 x≤4 −x ≤ 2 x + 2y ≤ 6 −x + 2 y ≤ 8 y≥0 ( line 1) ( line 2 ) ( line 3) ( line 4 ) ( line 5 ) There are 5 corner (extreme) points. B.8 Foods Apple Sauce Canned Corn Fried Chicken French Fries Mac & Cheese Turkey Breast Garden Salad Cost Servings Cost/ Serving $0.30 $0.40 $0.90 $0.20 $0.50 $1.50 $0.90 Calories/ Serving 100 150 250 400 430 300 100 Percent Protein 0% 20% 55% 5% 20% 67% 15% Percent Carbs 100% 80% 5% 35% 30% 0% 40% Percent Fat 0% 0% 40% 60% 50% 33% 45% AS CC FC FF MC TB GS 0.3 0.4 0.9 0.2 0.5 1.5 0.9 0 1.333333 0.457143 0 1.129568 0 0 Constraints AS Cals min 100 Cals max 100 Protein min 0 Carb min 100 Fat max 0 Fruit/Veg Min 100 CC 150 150 30 120 0 150 FC 250 250 137.5 12.5 100 0 FF 400 400 20 140 240 0 A-68 MC 430 430 86 129 215 0 TB 300 300 201 0 99 0 GS 100 100 15 40 45 100 Fruit/ Vegetable 1 1 0 0 0 0 1 $1.51 LHS RHS 800 500 800 800 200 200 311.4286 200 288.5714 400 200 200
- 72. Target Cell (Min) Answer Report (Relevant Section) Original Value Final Value $2.91 $1.51 Cell $I$14 Name servings Cell $B$14 $C$14 $D$14 $E$14 $F$14 $G$14 $H$14 Name serving A serving C serving FC serving FF serving M serving T serving G Cell $I$17 $I$18 $I$19 $I$20 $I$21 $I$22 Name Cals min LI Cals max L Protein min Carb min L Fat max LI Fruit + Veg I Cell $B$14 $C$14 $D$14 $E$14 $F$14 $G$14 $H$14 Sensitivity Report (Relevant Section) Final Reduced Objective Allowable Name Value Cost Coefficient Increase serving A 0 0.172602 0.3 1E + 30 serving C 1.333333 0 0.4 0.258904 serving FC 0.457143 0 0.9 0.105072 serving FF 0 0.152691 0.2 1E + 30 serving M 1.129568 0 0.5 0.062909 serving T 0 0.169316 1.5 1E + 30 serving G 0 0.666151 0.9 1E + 30 Allowable Decrease 0.172602 0.225581 0.100581 0.152891 0.707833 0.169318 0.688151 Cell $I$17 $I$18 $I$19 $I$20 $I$21 $I$22 Final Shadow Constraint Name Value Price R.H. Side Cals min LI 800 0 500 Cals max L 800 –0.00023 800 Protein min 200 0.008983 200 Carb min L 311.4286 0 200 Fat max LI 288.5714 0 400 Fruit + Veg I 200 0.001504 200 Allowable Decrease 1E + 30 251.6129 40 1E + 30 111.4286 200 Adjustable Cells Original Value 1.50 0.00 1.33 0.00 0.00 0.00 1.40 Final Value 0.00 1.33 0.46 0.00 1.13 0.00 0.00 Cell Value 800 800 200 311.4286 288.5714 200 Formula $I$17 ≥ $J$1 $I$18 ≤ $J$1 $I$19 ≥ $J$1 $I$20 ≥ $J$2 $I$21 ≤ $J$2 $I$22 ≥ $J$2 Constraints Adjustable Cells Status Not Binding Binding Binding Not Binding Not Binding Binding Slack 300 0 0 111.4286 111.4286 0 Constraints A-69 Allowable Increase 300 200 155 111.4285 1E + 30 485.7143
- 73. C Transportation Modeling MODULE C.1 Destination 2 3 10 9 8 5 3 1 A Source 4 B Demand 3 4 7 8 5 0 2 1 1 5 X Supply 6 5 4 3 X 6 X 12 7 1 X 5 X Answer : a. Final Solution: Shown in the final tableau above. b. Total Cost : 5 (10 ) + 3 ( 9 ) + 1( 3) + 6 ( 2 ) + 5 (1) = $97 c. Perform optimality test: Result : Cell A-3 A-4 A-1 Change in Cost if “Opened Up” 0 Since these numbers 0 are all nonnegative, 0 the solution is optimal. Answer : Yes A-70
- 74. C.2 First, apply the optimality test: Cell C-MI C-J H-L B-MI B-D Change in Cost if Route is “Opened Up” +3 +4 +2 –1 +4 Resulting new basic feasible solution: Destination San Miami Denver Lincoln Jose 7 CHI. 2 55 3 HOU. 40 BUF. 70 5 45 1 100 5 30 Demand Source 4 2 35 6 75 9 7 4 50 90 Supply 45 80 50 TOTAL COST: 30 ( 6 ) + 40 ( 3) + 35 (1) + 55 ( 2 ) + 45 ( 4 ) + 50 ( 4 ) = $825 = Answer A-71
- 75. C.3 Total supply = 450 < 500 = Total demand. W Destination A B C Supply 4 9 6 200 200 10 X Source 0 12 5 100 7 Y Demand 175 6 75 0 Dummy = Z 8 75 0 75 0 50 250 50 100 150 Optimal solution and meaning : Ship 200 tons of grain from W to A Ship 100 tons of grain from X to B Ship 75 tons of grain from X to C Ship 75 tons of grain from Y to C. Results in demands being met at destinations B and C, but in a shortfall of 50 tons at destination A. Total cost of optimal shipping plan: $2,750. A C.4 B 4 1 21 8 6 9 2 5 1 4 3 3 17 7 1 9 1 9 4 D 2 2 3 C 25 8 9 1 2 13 Cost = $84 + 16 + 34 + 9 + 75 + 9 + 26 = $253 . This is also optimal. A-72
- 76. C.5 A 12 1 4 2 Destination D B C 8 5 10 4 8 2 Source: 6 11 3 7 E 4 2 9 12 Cost = $48 + 64 + 40 + 8 + 12 + 36 = $208 C.6 The only cell with a negative cost improvement index is Houston–Miami. It achieves a “–1”. Allocate 10 to that cell. The result is: Denver 0 20 0 Houston St. Louis Chicago Yuma 0 0 20 Miami 10 0 10 Total cost = $170 2 C.7 4 5 9 4 6 7 8 1 1 3 3 4 10 1 11 12 3 Cost = $3 ×1 + 4 × 3 + 4 × 5 + 1× 8 + 1×10 + 3 ×12 = $89 A-73
- 77. C.8 Cell B-2 C-1 Improvement Index –9 +6 Result: A Destinations 1 2 2 8 6 6 9 Sources: B 1 3 7 C Cost = $12 + 9 + 18 + 14 = $53 A-74 4 7 2 7 6 5 2
- 78. D MODULE D.1 Waiting Line Models CURRENT MACHINE: λ = 40 , µ = 60 40 1. Utilization ( ρ ) = = 67% 60 2. ( 40 ) 1 Average number of customers waiting ( Lq ) = =1 60 ( 60 − 40 ) 3 3. Average number of customers in system ( Ls ) = 4. Average time waiting (Wq ) = 2 5. customers 40 = 2 customers 60 − 40 40 = 0.033 hours = 2 minutes 60 ( 60 − 40 ) 1 1 = hours = 3 minutes Average time in system (Ws ) = 60 − 40 20 PROPOSED MACHINE: λ = 40 , µ = 90 40 1. Utilization ( ρ ) = = 44% 90 2. ( 40 ) = 0.356 Average number of customers waiting ( Lq ) = 90 ( 90 − 40 ) 3. Average number of customers in system ( Ls ) = 2 Average time waiting (Wq ) = customers 40 = 0.8 customers 90 − 40 40 = 0.0089 hours = 0.533 minutes 90 ( 90 − 40 ) 1 1 = hours = 1.2 minutes 5. Average time in system (Ws ) = 90 − 40 50 Answer : Thus, we observe that the proposed machine will give better results with a decrease in queries as well as time in system. 4. A-75
- 79. D.2 Two machine system λ 40 1 ρ= = = . Each machine is busy, on average, 33 percent of the time. a. M µ 2 × 60 3 b. Lq = average number of customers waiting in line = 0.081 c. d. e. D.3 Ls = Lq + Lq λ = 0.081 + 0.667 = 0.748 customers in the system µ 0.081 = 0.002 hours = 0.1215 minutes λ 40 = 7.3 seconds average waiting time in the queue 1 1 Ws = Wq + = 0.002 + = 0.0187 hours = 1.12 minutes 60 µ = 67.3 seconds average time in the system The 2-machine system seems to be the best overall, with even further reduction in queues and waiting times. Wq = = This model is, again, the M/M/1 model. The service time of 30 seconds means that the service rate, µ, is 120 per hour. λ 40 1 ρ= = = . In this example and in Problem D.2, the system utilization is the a. µ 120 3 same. λ2 402 1 Lq = = = = 0.167 customers. The average number of b. µ ( µ − λ ) 120 (120 − 40 ) 6 customers waiting in line is twice as large as the average number (0.081) found in Problem D.2. λ 40 1 Ls = = = customer. On the average, the number of customers in c. µ − λ 120 − 40 2 1 3 the system is , rather than the (0.748) found in Problem D.2. 2 4 λ 40 1 1 Wq = = = d. hour = minutes = 15 seconds , which 4 µ ( µ − λ ) 120 (120 − 40 ) 240 exceeds Wq in Problem D.2. e. 1 1 1 3 = = hour = minutes = 45 seconds . Thus, average time 4 µ − λ 120 − 40 80 spent in the system is 15 seconds waiting plus 30 seconds service. The 45 seconds calculated here is less than the 67.3 seconds found in Problem D.2. Ws = A-76
- 80. D.4 EMPLOYEE SYSTEM: The current system is actually an M/D/1 system, because service times are constant. λ 10 ρ = = = 0.83 . The toll collector is busy 83 percent of the time. a. µ 12 b. Lq = λ2 100 100 = = = 2.08 drivers. The average number of 2 µ ( µ − λ ) 2 (12 )(12 − 10 ) 48 drivers waiting to pay the toll is 2.08. c. Ls = Lq + λ = 2.08 + 0.83 = 2.91 drivers. The average number of drivers at any one µ toll booth is 2.91. λ 10 10 = = = 0.208 minutes = 12.5 seconds . The 2µ ( µ − λ ) 2 (12 )(12 − 10 ) 48 d. Wq = e. average time drivers spend waiting is 12.5 seconds. Ws = waiting time + service time = 12.5 seconds + 5 seconds = 17.5 seconds AUTOMATED SYSTEM: The proposed system, although automated, is an M/M/1 system. λ 10 ρ = = = 0.83 . The utilization of the new system is the same as that of the old a. µ 12 system. λ2 100 100 Lq = = = = 4.16 drivers. Twice as many drivers are in b. µ ( µ − λ ) 12 (12 − 10 ) 24 line under the new, automated system. c. d. e. Ls = Lq + λ = 5 drivers. On average, 5 customers are in the system, which is 67 µ percent more customers than the current system allows. λ 10 10 Wq = = = = 0.417 minutes = 25 seconds . The waiting time µ ( µ − λ ) 12 (12 − 10 ) 24 will double with the new system. 1 Ws = Wq + = 30 seconds. The time in the system will rise by 67 percent. µ The employee system seems to be better overall. A-77
- 81. D.5 a. b., c. Each server handles 60 registrants hour , so it takes 4 servers to handle 200 arrivals hour λ = 200 , µ = 60 , M = 4 servers yields: Lq = 3.29 , Wq = 0.0164 hours Server cost = 4 × $15 = $60 Wait cost = ( 3.29 people )( 0.0164 hour )( $100 ) = $5.38 ( rounded ) Total cost = $65.30 With 5 servers, Lq = 0.65 , Wq = 0.0033 d. D.6 Wait cost = ( 0.65 )( 0.0033)( $100 ) = $0.21 , Server cost = 5 × 15 = $75 Total cost = $75.21 The system is optimal with 4 servers. Server utilization rate = 83.33% . a. b. The optimal number of servers is again 4. The wait cost is now $50 person hour × Lq × Wq (which are the same as in Problem D.5). Entertainment cost = $15 hour , Wait cost = ( 3.29 )( 0.0164 )( $50 ) = $2.70 ( rounded ) Total cost = $15 + $60 + 2.70 = $77.70 D.7 λ = 15 hour , µ = 20 hour a. b. D.8 Wq = 0.075 hours = 4.5 minutes Lq = 1.125 people λ = 10 hour , µ = 30 hour a. Wq = 0.0083 hours = b. 1 minute 2 Lq = 0.083 people A-78
- 82. E MODULE E.1 Learning Curves In order to estimate the learning curve rate, we take the ratios of the units that have doubled. report report report report report report report report 2 56 = = 0.848 1 66 4 49 = = 0.860 2 56 6 45 = = 0.849 3 53 8 42 = = 0.857 4 49 The learning curve rates are not identical for each paired comparison, but since they are in the range of 84 percent to 86 percent, we can safely use 85 percent. E.2 Since the 6th report took 45 minutes, the 12th unit should take 85 percent of 45 minutes, or 38.25 minutes. Multiplying again by 85 percent yields a time of 32.5 minutes for the 24th report. One more multiplication yields a time of 27.6 minutes for the 48th report. An even more exact answer can be found using Excel OM or POM for Windows software. Or a third approach is the formula TN = T1C where C ≈ 0.402 (from Table E.3); T48 = ( 66 )( 0.402 ) = 26.5 minutes. E.3 TN = T1C where C = 0.400 (from Table E.3), T50 = ( 66 )( 0.400 ) = 26.4 minutes E.4 TN = T1C where C = 25.513 and TN is now cumulative time. T50 = ( 66 )( 25.513) = 1, 683.85 minutes = 28.06 hours A-79
- 83. F Simulation MODULE F.1 We will use the following random number intervals when simulating demand and lead time. We will select column 1 of text Table F.4 to get the random numbers for demand, while we will use column 2 of the same table to find the lead time whenever an order is placed. Probability 0.20 0.40 0.20 0.15 0.05 Cumulative Probability 0.20 0.60 0.80 0.95 1.00 RN Interval 01–20 21–60 61–80 81–95 96–00 Demand 0 1 2 3 4 Probability 0.15 0.35 0.50 Cumulative Probability 0.15 0.50 1.00 RN Interval 01–15 16–50 51–00 Lead Time 1 2 3 The results are: Units Received 10 10 Begin Inv. 5 4 3 0 10 6 2 1 0 10 RN 52 37 82 69 98 96 33 50 88 90 Demand 1 1 3 2 4 4 1 1 3 3 Total End Inv. 4 3 0 0 6 2 1 0 0 7 23 Lost Sales 0 0 0 2 0 0 0 0 3 0 5 Order? RN Lead time Yes 06 1 Yes 63 3 The total stock out cost = 5($40) = $200. The total holding cost = 23($1) = $23. A-80
- 84. F.2 If the reorder point Problem F.1 is changed to 4 units, we have: Units Received 10 10 Begin Inv. 5 4 13 10 8 4 0 0 10 7 RN 52 37 82 69 98 96 33 50 88 90 End Inv. 4 3 10 8 4 0 0 0 7 4 40 Demand 1 1 3 2 4 4 1 1 3 3 Total Lost Sales 0 0 0 0 0 0 1 1 0 0 2 Order? Yes RN 6 Lead time 1 Yes 63 3 Yes 57 3 The total stock out cost = 2($40) = $80. The total holding cost = 40($1) = $40. The total cost is $120 with a reorder point of 4 and $223 with a reorder point of 2. (Same random numbers were used as in Problem F.1). F.3 Since average waiting time is a variable of concern, a next event time increment model should be used. (1) Customer Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 (2) Random Number 50 28 68 36 90 62 27 50 18 36 61 21 46 01 14 (3) Interval to Arrival 2 1 2 1 4 2 1 2 1 1 2 1 2 0 1 (4) Time of Arrival 9:02 9:03 9:05 9:06 9:10 9:12 9:13 9:15 9:16 9:17 9:19 9:20 9:22 9:22 9:23 (5) Random Number 52 37 82 69 98 96 33 50 88 90 50 27 45 81 66 (6) Service Time 3 2 3 3 4 4 2 3 4 4 3 2 2 3 3 (7) Start Service 9:02 9:05 9:07 9:10 9:13 9:17 9:21 9:23 9:26 9:30 9:34 9:37 9:39 9:41 9:44 (8) End Service 9:05 9:07 9:10 9:13 9:17 9:21 9:23 9:26 9:30 9:34 9:37 9:39 9:41 9:44 9:47 (9) Wait Time 0 2 2 4 3 5 8 8 10 13 15 17 17 19 21 (10) Idle Time 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Read the data as in the following example for the first row: Column 1: Number of customer. Column 2: From third column of random number Table F.4. Column 3: Time interval corresponding to random number (random number of 50 implies a 2-minute interval). Column 4: Starting at 9 A.M. the first arrival is at 9:02. Column 5. From the first column of the random number Table F.4. Column 6: Teller time corresponding to random number 52 is 3 minutes. Column 7: Teller is available and can start at 9:02. Column 8: Teller completes work at 9:05 (9:02 + 0:03). Column 9: Wait time for customer is 0 as the teller was available. Column 10: Idle time for the teller was 2 minutes (9:00 to 9:02). The drive-in window clearly does not meet the manager’s criteria for an average wait time of 2 minutes. As a matter of fact, we can observe an increasing queue buildup after only a few customer simulations. This observation can be confirmed by expected value calculations on both arrival and service rates. A-81
- 85. F.4 Time Until Next Arrival (Minutes) 0 1 2 3 Service Time (Minutes) 1 3 5 10 F.5 Cumulative Probability 0.1 0.3 0.6 1.00 Cumulative Probability 0.1 0.6 0.8 1.00 Probability 0.1 0.2 0.3 0.4 Probability 0.1 0.5 0.2 0.2 Random Number Range 00–09 10–29 30–59 60–99 Random Number Range 00–09 10–59 60–79 80–99 Simulating employment of one detailer: * Random Number 52 37 82 69 98 96 33 50 27 45 81 ** IAT 2 3 2 0 3 3 2 2 1 2 3 Random Number 06 63 57 02 94 52 69 33 32 30 48 Service Time 1 3 2 1 4 2 3 2 2 2 2 Arrival Time 2 5 7 7 10 13 15 17 18 20 23 Detailer 1 Time Time In Out 2 3 5 8 8 10 10 11 11 15 15 17 17 20 20 22 22 24 24 26 26 28 Wait Time 1 3 1 2 2 3 4 4 3 Maximum waiting time: 4 minutes. Total waiting time: 23 minutes. 4 minutes will not meet the 2.5 minute criteria. * From Column 1 of Table F.4 in text. From Column 2 of Table F.4 in text. ** A-82
- 86. F.6 Simulating employment of two detailers: Random Number 52 37 82 69 98 96 33 50 27 45 81 IAT 2 3 2 0 3 3 2 2 1 2 3 Random Service Arrival Number Time Time 06 1 2 63 3 5 57 2 7 02 1 7 94 4 10 52 2 13 69 3 15 33 2 17 32 2 18 30 2 20 48 2 23 Detailer 1 Time Time In Out 2 3 5 8 Detailer 2 Time Time In Out 7 8 10 9 9 14 1 13 19 18 18 20 23 15 17 15 20 22 25 Maximum waiting time: 1 minute. Total waiting time: 1 minute. Two detailers with a waiting time of one (1) minute will meet the 2.5 minute criteria. A-83 Wait Time

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