2. 1 By: Prof. Y. Peter Chiu MRP & JIT ~ HOMEWORK SOLUTION ~

3. 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 Week Gross Requirements Scheduled Receipts On-hand inventory Net Requirement s Time-Phased Net Requirements Planned Order Release (lot for lot) 2 3 4 5 6 7 8 9 10 11 12 13 126 126 96 36 78 336 135 42 228 114 78 63 270 144 96 0 27 0 0 0 0 51 336 135 42 228 114 51 336 135 42 228 114

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

5. 4 #5

6. 5 #6

7. 6 # 9 (b) Week MPS-end item Component B (P.O.R) Component F -Net. Req. Time Phased Net. Req. Ans. →Planned Order Release 27 28 29 30 31 32 33 34 35 165 180 300 220 200 240 330 360 600 440 400 480

8. 7 # 9 (c) Week MPS-end item P.O.R –Comp. B Net Req. –Comp. E Time Phased –Net Req. P.O.R – Comp. E P.O.R –Comp. G Time Phased –Net Req. Net Req. –Comp. G Net Req. –Comp. I Net Req. –Comp. H Time Phased –Net Req. Ans. → P.O.R –Comp. I Ans. → P.O.R –comp. H 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 165 180 300 220 200 240 330 360 600 440 400 480 660 720 1200 880 800 960 1980 2160 3600 2640 2400 2880 (d)

9. 8 Month Demand 1 2 3 4 5 6 7 8 9 10 11 12 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 Net. Demand # 14 1 2 3 4 5 6 7 8 9 10 11 12 2 12 4 8 15 25 20 5 10 20 5 20 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 ∴ (a) Silver-Meal

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

11. 10 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 (b) LUC Starting in Period 1 : Starting in Period 7 : Starting in Period 5 : C(1) = 40 /20 = 2 C(2) = (40+5) /25 = 1.80 C(3) = [40+5+2(10)] /35 = 1.86 C(1) = 40 /15 = 2.67 C(2) = 65 /40 = 1.63 C(3) = [65+2(20)] /60 = 1.75 ∴ ∴ ∴ # 14

12. 11 (C) PPB Starting in Period 1 : PeriodHolding cost 234234 1* (12) = 12 12+2(4) = 20 20+3(8) = 44 Closer to period 4 K = $40 ∴ 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 ∴ Using LUC ; y = [ 26, 0, 0, 0,40, 0, 25, 0, 35, 0, 0, 20 ] ∴ ∴ (b) LUC # 14

13. 12 Starting in Period 5 : PeriodHolding cost 123123 0 25 25+2(20) = 65 Closer to period 2 ∴ Starting in Period 7 : Starting in Period 10 : 2323 PeriodHolding cost 5 5+2(12) = 29 PeriodHolding cost 234234 5 5+2(10) = 25 25+3(20) = 85 Closer to period 3 ∴ ∴ ∴ Using PPB ; y = [ 26, 0, 0, 0,40, 0, 35, 0, 0, 45, 0, 0 ] (C) PPB K = $40

14. 13 SM LUC PPB Demand Inv. SM Inv. LUC Inv. PPB 1 2 3 4 5 6 7 8 9 10 11 12 18 0 0 23 0 50 0 0 35 0 0 20 26 0 0 0 40 0 25 0 35 0 0 20 26 0 0 0 40 0 35 0 0 45 0 0 2 12 4 8 15 25 20 5 10 20 5 20 16 4 0 15 0 25 5 0 25 5 0 0 24 12 8 0 25 0 5 0 25 5 0 0 24 12 8 0 25 0 15 10 0 25 20 0 Σ = 95 Σ = 104 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 (d)

15. 14 #17 K=200; h=0.3

16. 15 #17

17. 16 #17

18. 17 #18

19. 18 #18 (cont’d)

20. 19 h = $0.4 K = $180 Starting Inventory Week 6 is 75 Receiving: 30 & 10 in week 8 & 10 # 24 Week Net Demand Demand 6 7 8 9 10 11 220 165 180 120 75 300 145 165 150 120 65 300 (a) PPB : Starting in Period 1 : Starting in Period 4 : Holding costs Week Period 123123 0 165*0.4= 66 66+2(0.4)(150)= 186 Closer to period 3 ∴ ∴ 123123 0 65*0.4= 26 26+2(300)(0.4)= 266 Closer to period 3 K = $180

21. 20 # 24 MRP - Mother Boards Week Net. Req. Time-Phased Net Req. P.O.R. (lot-for-lot) Ans. → ( PPB ) P.O.R For DRAM Ans. → Gross Req. For DRAM Time-Phased Net Req. For DRAM P.O.R. 1 2 3 4 5 6 7 8 9 10 11 145 165 150 120 65 300 460 0 0 485 0 0 41,400 0 0 43,650 0 0

22. 21 #25

23. 22 #25 (cont’d)

24. 23 #28(a) K=$200 ; h=$0.30

25. 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 %

26. 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 Net Requirements Time-Phased Net Requirements r = 4 5 6 7 8 9 10 11 12 13 14 15 16 17 42 42 32 12 26 112 45 14 76 38 Production c= Capacity 50 50 50 50 50

27. 26 Week Net Requirements Time-Phased Net Requirements r = 4 5 6 7 8 9 10 11 12 13 14 15 16 17 42 42 32 12 26 112 45 14 76 38 Production c= Capacity # CW.3 excess (c-r) = Capacity 8 8 18 38 24 (62) 5 36 (26) 12 (c-r)’ = [2] Lot-shifting technique (back-shift demand from r j > c j ): 50 50 50 50 50 8 8 18 38 24 (62) 5 36 (26) 12 8 8 18 0 0 0 5 10 0 12 (c-r)’ = final r ’ = 42 42 32 50 50 50 45 40 50 38 [1] First test for: It is okay!

28. 27 # CW.4 ( continues on #CW.3) Determine the optimal production plan. Suppose that with overtime work on 2 nd 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. Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 Production c= Capacity 50 50 50 50 50 Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Production c= Capacity (O-T) 120 120 120 120 120

29. 28 # CW.4 ( continues on #CW.3) Determine the optimal production plan. Suppose that with overtime work on 2 nd 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. Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 Production c= Capacity 50 50 50 50 50 final r ’ = 42 42 32 50 50 50 45 40 50 38 (using regular shift) Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 0 0 0 38 62 0 0 26 0 0 Σ= 126 Ending Inventories =

30. 29 Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 Production c= Capacity (O-T) excess (c-r) = Capacity 120 120 120 120 120 Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 # CW.4 [1] First, the cost for using regular shift is $100(10) + $0.65 (126) [ lot for lot ] = $1,081.9 [ lot for lot ] 78 78 88 108 94 8 75 106 44 82 42 42 32 12 26 112 45 14 76 38 r = 78 78 88 108 94 8 75 106 44 excess (c-r)’= Capacity 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 85 0 120 0 0 120 0 0 114 0 final r ’ =

31. 30 # CW.4 Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 85 0 120 0 0 120 0 0 114 0 final r ’ = [2] T he cost for using Overtime shift is $205(4) + $0.65(372) = $1061.8 43 1 89 77 51 59 14 0 38 0 Ending Inventories = Less than the cost for using regular shift $1,081.9,  Saved $ 20.10

32. 31 # CW.4 Ending Inventories = [3] To think about the following solution: Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Suppose r ’= 74 32 0 38 62 0 0 26 0 0 Σ= 232 116 0 0 50 50 50 45 40 50 38 [ One OT, 7 regular ] T he 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 ?

33. 32 # CW.4 Ending Inventories = [4] A Better Solution : Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Suppose r ’= 74 32 0 27 1 9 14 0 38 0 Σ= 195 116 0 0 39 0 120 50 0 114 0 T he 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 !

34. 33 #33

35. 34 The End The End

36. 35