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

3. 2 #4 (a) MPS for the computers   

4. 3 #5

5. 4 #6

6. 5 # 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

7. 6 # 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)

8. 7 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

9. 8 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

10. 9 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

11. 10 (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

12. 11 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

13. 12 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)

14. 13 #17 K=200; h=0.3

15. 14 #17

16. 15 #17

17. 16 #18

18. 17 #18 (cont’d)

19. 18 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

20. 19 # 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

21. 20 The End The End