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By: Prof. Y. Peter Chiu

MRP & JIT ~ HOMEWORK SOLUTION ~. By: Prof. Y. Peter Chiu. #4 (a) MPS for the computers. . . . #5. #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).

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By: Prof. Y. Peter Chiu

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

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

  3. #5

  4. #6

  5. # 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

  6. # 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

  7. # 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> ∴

  8. # 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 ]

  9. # 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> ∴

  10. # 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 ∴

  11. (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 ]

  12. # 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 !

  13. K=200; h=0.3 #17

  14. #17

  15. #17

  16. #18

  17. #18 (cont’d)

  18. # 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 ∴

  19. # 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

  20. The End

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