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By: Prof. Y. Peter Chiu 12 / 2010

Adv-POM: Mid-Term #2 Chaps.4,5,7 ~ SOLUTIONs ~. By: Prof. Y. Peter Chiu 12 / 2010. #1. (a). (b). τ = 6 months = 0.5 years > T * R = λ * τ eff = (1200)*(0.5-0.361) = 334 pounds (  reorder point ). #2. All Unit Quantity discount Problem. #3. • Compare.

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By: Prof. Y. Peter Chiu 12 / 2010

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  1. Adv-POM: Mid-Term #2 Chaps.4,5,7 ~ SOLUTIONs ~ By: Prof. Y. Peter Chiu 12 / 2010

  2. #1 (a) (b) τ= 6 months =0.5 years > T* R = λ * τeff =(1200)*(0.5-0.361) = 334 pounds (reorder point)

  3. #2

  4. All Unit Quantity discount Problem #3 • Compare

  5. Solution # 4 J-55R 1500 33600 $102 $20 $4 3.8214 5732 0.1887 283 0.0084 2.02 K-18R 540 24000 187 12 2.3460 1267 0.1887 102 0.0043 1.03 2.4 Z-344 2880 39600 263.5 45 9 8.3455 24035 0.1887 543 0.0137 3.29 

  6. Solution # 4 (a) (b)

  7. Solution #5

  8. Solution # 6

  9. Solution # 6

  10. Solution # 6

  11. Solution # 7 (a)

  12. Solution # 7 (a) Total Costs = Setup costs + holding costs = 2*($500) + $2*(690) = $ 2380 ( Silver-Meal) Solution # 7 (b) Starting in Week 1 Order Horizon Total Holding cost 1 2 3 0 200 (2*100) 540 [200+2*(2)*85] K=500 ∵ K is closer to period 3 ∴

  13. Solution # 7 (b) Starting in Week 4 Order Horizon Total Holding cost 1 2 3 0 240 (2*120) 620 [240+2*(2)*95] K=500 ∵ K is closer to period 3 ∴

  14. Solution # 7 (b) Total Costs = Setup costs + holding costs = 3*($500) + $2*(580) = $ 2660 ( P.P.B.) Total Costs = Setup costs + holding costs = 2*($500) + $2*(690) = $ 2380 ( Silver-Meal) = 3*($500) + $2*(580) = $ 2660 ( P.P.B.) Silver-Meal has the lower costs. (12% lower)

  15. Solution # 8 (a) Solution # 8 (b)

  16. Solution # 8 (b)

  17. Solution # 9 (a) Yes Week 1 2 3 4 5 6 7 Req. Capacities 275 55 170 300 90 240 180 400 400 120 120 400 120 120 (C-R) lot-for-lot Initial Solution 125 345 (50) (180) 310 (120) (60) r’ 275 285 120 120 270 120 120 (C-R’) 125 115 0 0 130 0 0 Ending-Inv. 0 230 180 0 180 60 0 Costs for Initial Solution : ($300)*7+($1)[230+180+180+60]=$2,750

  18. Solution # 9 (b) Week 1 2 3 4 5 6 7 Req. Capacities 275 55 170 300 90 240 180 400 400 120 120 400 120 120 (C-R) lot-for-lot Initial Solution 125 345 (50) (180) 310 (120) (60) r’ 275 285 120 120 270 120 120 (C-R’) 125 115 0 0 130 0 0 (b) 400 400 0 0 390 120 0 One Solution Ending-Inv. 125 470 300 0 300 180 0 Costs for this Solution : ($300)*4+($1)[125+470+300+300+180]=$2,575 But not Optimal !! Total Savings = $175

  19. Solution # 9 (b) Week 1 2 3 4 5 6 7 Req. Capacities 275 55 170 300 90 240 180 400 400 120 120 400 120 120 (C-R) lot-for-lot Initial Solution 125 345 (50) (180) 310 (120) (60) r’ 275 285 120 120 270 120 120 (C-R’) 125 115 0 0 130 0 0 (b) 280 400 0 120 390 0 120 Better Solution Ending-Inv. 5 350 180 0 300 60 0 Costs for this Solution : ($300)*5+($1)[5+350+180+300+60]= $2,395 Optimal !! Total Savings = $355

  20. The End

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