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By: Prof. Y. Peter Chiu

MRP & JIT ~ HOMEWORK SOLUTION ~. By: Prof. Y. Peter Chiu. What is the MRP Calculations for the slide assemblies ? ( 3 slide assemblies for each valve casing ). Lead Time = 2 weeks. Assume On-hand inventory of 270 slide assemblies at the end of week

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By: Prof. Y. Peter Chiu

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  1. MRP & JIT ~ HOMEWORK SOLUTION ~ By: Prof. Y. Peter Chiu

  2. What is the MRP Calculations for the slide assemblies ? ( 3 slide assemblies for each valve casing ) • Lead Time = 2 weeks • Assume On-hand inventory of 270 slide assemblies at the end of week 3 & Scheduled receipts of 78 & 63 at the beginning of week 5 & 7 • Solution to MRP Calculations for the slide assemblies 2 3 4 5 6 7 8 9 10 11 12 13 Week Gross Requirements 126 126 96 36 78 336 135 42 228 114 78 63 Scheduled Receipts On-hand inventory 270 144 96 0 27 Net Requirements 0 0 0 051336 135 42 228 114 Time-Phased Net Requirements 51 336 135 42 228 114 Planned Order Release (lot for lot) 51 336 135 42 228 114

  3. #4(a) MPS for the computers   

  4. #5

  5. #6

  6. # 9 (b) Week 27 28 29 30 31 32 33 34 35 MPS-end item 165 180 300 220 200 240 Component B (P.O.R) 330 360 600 440 400 480 Component F 330 360 600 440 400 480 -Net. Req. Time Phased Net. Req. 330 360 600 440 400 480 Ans. →Planned Order Release 330 360 600 440 400 480

  7. # 9 (c) Week 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 165 180 300 220 200 240 MPS-end item P.O.R –Comp. B 330 360 600 440 400 480 330 360 600 440 400 480 Net Req. –Comp. E 330 360 600 440 400 480 Time Phased –Net Req. P.O.R – Comp. E 330 360 600 440 400 480 660 720 1200 880 800 960 Net Req. –Comp. G 660 720 1200 880 800 960 Time Phased –Net Req. P.O.R –Comp. G 660 720 1200 880 800 960 660 720 1200 880 800 960 Net Req. –Comp. I 660 720 1200 880 800 960 Time Phased –Net Req. Ans. → P.O.R –Comp. I 660 720 1200 880 800 960 (d) 1980 2160 3600 2640 2400 2880 Net Req. –Comp. H Time Phased –Net Req. 1980 2160 36002640 2400 2880 Ans. → P.O.R –comp. H 1980 2160 36002640 2400 2880

  8. # 14 Month 1 2 3 4 5 6 7 8 9 10 11 12 Demand 6 12 4 8 15 25 20 5 10 20 5 12 • Current Inventory : 4 • An ending Inventory should be : 8 • h = $ 1 • k = $ 40 Month 1 2 3 4 5 6 7 8 9 10 11 12 Net. Demand 212 4 8 15 25 20 5 10 20 5 20 (a) Silver-Meal • Starting in Period 1 : C(1) = 40 C(2) = (40+12)/2 = 26 C(3) = [40+12+2(4)] /3 = 20 C(4) = [40+12+2(4)+3(8)] /4= 21 <stop> ∴

  9. # 14 (a) Silver-Meal C(1) = 40 C(2) = [40+15]/2 = 27.5 C(3) = [40+15+2(25)] /3 = 35 <stop> • Starting in Period 4 : ∴ • Starting in Period 6 : C(1) = 40 C(2) = [40+20]/2 = 30 C(3) = [40+20+2(5)] /3 = 23.3 C(4) = [40+20+2(5)+3(10)] /4 = 25 <stop> ∴ • Starting in Period 9 : C(1) = 40 C(2) = [40+20]/2 = 30 C(3) = [40+20+2(5)] /3 =23.3 C(4) = [40+20+2(5)+3(20)] /4 = 32.5 <stop> ∴ ∴ ∴ Using Silver- Meal ; y = [ 18 , 0 , 0 , 23 , 0 , 50 , 0 , 0 , 35 , 0 , 0 , 20 ]

  10. # 14 (b) LUC • Starting in Period 1 : C(1) = 40 /2 = 20 C(2) = 52 /14 = 3.71 C(3) = 60 /18 = 3.33 C(4) = 84 /26 = 3.23 C(5) = (84+60) /41 = 3.51 <stop> ∴ • Starting in Period 5 : C(1) = 40 /15 = 2.67 C(2) = 65 /40 = 1.63 C(3) = [65+2(20)] /60 = 1.75 <stop> ∴ • Starting in Period 7 : C(1) = 40 /20 = 2 C(2) = (40+5) /25 = 1.80 C(3) = [40+5+2(10)] /35 = 1.86 <stop> ∴

  11. # 14 (b) LUC • Starting in Period 9 : C(1) = 40 /10 = 4 C(2) = 60 /30 = 2 C(3) = [60+2(5)] /35 = 2 C(4) = [70+60] /55 = 2.36 <stop> ∴ ∴ ∴ Using LUC ; y = [ 26 , 0 , 0 , 0 ,40 , 0 , 25 , 0 , 35 , 0 , 0 , 20 ] (C) PPB Period Holding cost • Starting in Period 1 : 1* (12) = 12 12+2(4) = 20 20+3(8) = 44 2 3 4 K = $40 Closer to period 4 ∴

  12. (C) PPB Period Holding cost • Starting in Period 5 : 1 2 3 0 25 25+2(20) = 65 K = $40 Closer to period 2 ∴ Period Holding cost • Starting in Period 7 : 2 3 4 5 5+2(10) = 25 25+3(20) = 85 K = $40 ∴ Closer to period 3 Period Holding cost • Starting in Period 10 : 2 3 5 5+2(12) = 29 K = $40 ∴ ∴ Using PPB ; y = [ 26 , 0 , 0 , 0 ,40 , 0 , 35 , 0 , 0 , 45 , 0 , 0 ]

  13. # 14 (d) 1 2 3 4 5 6 7 8 9 10 11 12 SM 18 0 0 23 0 50 0 0 35 0 0 20 LUC 26 0 0 0 40 0 25 0 35 0 0 20 PPB 26 0 0 0 40 0 35 0 0 45 0 0 Demand 2 12 4 8 15 25 20 5 10 20 5 20 Inv. SM Σ = 95 16 4 0 15 0 25 5 0 25 5 0 0 24 12 8 0 25 0 5 0 25 5 0 0 Inv. LUC Σ = 104 Inv. PPB 24 12 8 0 25 0 15 10 0 25 20 0 Σ = 139 Cost of S.M. ($40*5)+($1*95) = $295 Cost of LUC ($40*5)+($1*104) = $304 Cost of PPB ($40*4)+($1*139) = $299 ∴ Silver Meal Method is the least expensive one !

  14. K=200; h=0.3 #17

  15. #17

  16. #17

  17. #18

  18. #18 (cont’d)

  19. # 24 • h = $0.4 • Starting Inventory Week 6 is 75 • K = $180 • Receiving: 30 & 10 in week 8 & 10 Week 6 7 8 9 10 11 Demand 220 165 180 120 75 300 Net Demand 145 165 150 120 65 300 (a) PPB : Holding costs Week • Starting in Period 1 : 1 2 3 0 165*0.4= 66 66+2(0.4)(150)= 186 K = $180 Closer to period 3 ∴ Holding costs • Starting in Period 4 : Period K = $180 1 2 3 0 65*0.4= 26 26+2(300)(0.4)= 266 Closer to period 3 ∴

  20. # 24 MRP - Mother Boards Week 1 2 3 4 5 6 7 8 9 10 11 Net. Req. 145 165 150 120 65 300 Time-Phased Net Req. P.O.R. (lot-for-lot) 145 165 150 120 65 300 145 165 150 120 65 300 Ans. →( PPB ) P.O.R 460 0 0 485 0 0 For DRAM Ans. →Gross Req. 41,400 0 0 43,650 0 0 For DRAM Time-Phased Net Req. 41,400 0 0 43,650 0 0 For DRAM P.O.R. 41,400 0 0 43,650 0 0

  21. #25

  22. #25 (cont’d)

  23. #28(a) K=$200 ; h=$0.30

  24. #28(b) 17(b) S-M: ($1052 - $754) / ($754) = 40 % 17(c) LUC: ($1052 - $852) / ($852) = 23 % 17(d) PPB: ($1052 - $754) / ($754) = 40 %

  25. # CW.3 Consider the example presented previously for the scheduling of the valve casing assembly. Suppose that the production capacity in any week is 50 valve casings. Determine the feasible planned order release for the valve casings. Recall that the time phased net requirements for the valve casings as followed: Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Net Requirements 42 42 32 12 26 112 45 14 76 38 Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 Production c= Capacity 50 50 50 50 50 50 50 50 50 50

  26. # CW.3 [1] First test for: It is okay! [2] Lot-shifting technique (back-shift demand from rj > cj): Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Net Requirements 42 42 32 12 26 112 45 14 76 38 Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 Production c= Capacity 50 50 50 50 50 50 50 50 50 50 excess (c-r) = Capacity 8 8 18 38 24 (62) 5 36 (26) 12 (c-r)’ = 8 8 18 38 24 (62) 5 36 (26) 12 (c-r)’ = 8 8 18 0 0 0 5 10 0 12 final r ’ = 42 42 32 50 50 50 45 40 50 38

  27. # CW.4 ( continues on #CW.3) Suppose that with overtime work on 2nd shifts, the company could increase the weekly production capacity to 120 valve casings, however, extra cost per week = $105. Where K=$100, is the regular setup cost. The holding cost per valve casing per week is estimated to be $0.65. Determine the optimal production plan. Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 Production c= Capacity 50 50 50 50 50 50 50 50 50 50 Production c= Capacity (O-T) 120 120 120 120 120 120 120 120 120 120

  28. # CW.4 ( continues on #CW.3) Suppose that with overtime work on 2nd shifts, the company could increase the weekly production capacity to 120 valve casings, however, extra cost per week = $105. Where K=$100, is the regular setup cost. The holding cost per valve casing per week is estimated to be $0.65. Determine the optimal production plan. Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 Production c= Capacity 50 50 50 50 50 50 50 50 50 50 final r ’ = 42 42 32 50 50 50 45 40 50 38 (using regular shift) Ending Inventories = 0 0 0 38 62 0 0 26 0 0 Σ= 126

  29. # CW.4 [1] First, the cost for using regular shift is$100(10) + $0.65 (126) = $1,081.9 [ lot for lot ] Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 Production c= Capacity (O-T) 120 120 120 120 120 120 120 120 120 120 excess (c-r) = Capacity 78 78 88 108 94 8 75 106 44 82 r = 42 42 32 12 26 112 45 14 76 38 excess (c-r)’= Capacity 78 78 88 108 94 8 75 106 44 35 77 0 31 43 0 6 42 42 32 12 26 112 45 14 76 38 85 43 120 89 77 120 59 0 114 0 0 0 0 0 final r ’ = 85 0 120 0 0 120 0 0 114 0

  30. # CW.4 Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 final r ’ = 85 0 120 0 0 120 0 0 114 0 Ending Inventories = 43 1 89 77 51 59 14 0 38 0 [2] The cost for using Overtime shift is$205(4) + $0.65(372) = $1061.8 Less than the cost for using regular shift $1,081.9, Saved $ 20.10

  31. # CW.4 [3] To think about the following solution: Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 Suppose r ’= 116 0 050 50 50 45 40 50 38[ One OT, 7 regular ] Ending Inventories = 74 32 0 38 62 0 0 26 0 0 Σ= 232 The cost for using only one Overtime shift on week 4 is$205(1) + $100(7) + $0.65(232) = $1055.8 Less than the cost for using regular shift $1,081.9, Saved $ 26.1 Less than the cost for using all Overtime shift $1061.8 Saved $ 6.0 WHY ?

  32. # CW.4 [4] A Better Solution : Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 Suppose r ’= 116 0 0 39 0 120 50 0 114 0 Ending Inventories = 74 32 0 27 1 9 14 0 38 0 Σ= 195 The cost for using the above solution is$205 (3) + $100 (2) + $0.65(195) = $ 941.75 Less than the cost for using regular shift $1,081.9, Saved $ 140.05 Wow !

  33. #33

  34. The End

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