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By: Prof. Y. Peter Chiu 12 / 2008

Adv-POM: Mid-Term #2 Chaps.4,5,7 ~ SOLUTIONs ~. By: Prof. Y. Peter Chiu 12 / 2008. B:3,4,5. #1. #1-3. B:3. #2. B:4. #3. B:5. If Q=720/2=360 ( units purchased each time; twice a year). B:1,2. #4-5. #4. B:1. #5. B:2. Solution #6. B:8.

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By: Prof. Y. Peter Chiu 12 / 2008

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  1. Adv-POM: Mid-Term #2 Chaps.4,5,7 ~ SOLUTIONs ~ By: Prof. Y. Peter Chiu 12 / 2008

  2. B:3,4,5 #1 #1-3 B:3 #2 B:4

  3. #3 B:5 If Q=720/2=360 ( units purchased each time; twice a year)

  4. B:1,2 #4-5 #4 B:1

  5. #5 B:2

  6. Solution #6 B:8

  7. Solution #7 Supplier - A B:6

  8. Solution #7 Supplier - B B:6

  9. B:6 Solution #7 Supplier - B

  10. Solution #7 Supplier - B B:6

  11. Solution #8 : Newsboy model B:7

  12. # 9 B:11 Total Costs = 3*($320) + ($0.8)*(930) = $1,704 ( Silver-Meal)

  13. #10 B:12 Starting in Week 1 Order Horizon Total Holding cost 1 2 3 0 $144 (0.8*180) $384 [0.8*180+2*(0.8)*150] K=360 ∵ K is closer to period 3 ∴

  14. #10 Starting in Week 4 B:12 Order Horizon Total Holding cost 1 2 3 0 $224 (0.8*280) $560 [224+2*(0.8)*210] K=360 ∵ K is closer to period 2 ∴

  15. #10 Starting in Week 6 B:12 Order Horizon Total Holding cost 1 2 0 $136 (0.8*170) K=360 Total Costs = 3*($320) + ($0.8)*(930) = $1,704 ( P.P.B.)

  16. #11 Yes Week 1 2 3 4 5 6 7 Req. Capacities 290180 150 320 280 210 170 450 500 100 100450100 100 B:9 (C-R) lot-for-lot Initial Solution 160 320(50) (220)170 (110) (70) R’ 290460100 100450100 100 (C-R’) 160 40 0 0 0 0 0 B:10: K=$225 ; h=$0.8 Alternative #1- ( Not optimal ! ) Saves $17 (A) Week 4th lot : by Week 2 2*($0.8)*40=$64 by Week 1 3*($0.8)*60=$144 Total savings $82 & (B) Week 3rd lot : by Week 1 2*($0.8)*100=$160 Saves $65

  17. #12 Week 1 2 3 4 5 6 7 B:10 290180 150 320 280 210 170 Req. Capacities 450 500 100 100450100 100 R’ 290460100 100450100 100 (C-R’) 160 40 0 0 0 0 0 3505000 100450100 100 Optimal Solution Inventories 60 380 230 10 180 70 0 Alternative #2 K=$225 ; h=$0.8 the best  Total Savings $97 X (A) Week 3rd lot : by Week 2 $(0.8)(40)= $32 by Week 1 $(0.8)(2)(60)=$96 (B) Week 4th lot : by Week 1 3($0.8)(100)=$120 Initial Solution : ($225)*7+($0.8)[280+230+10+180+70]= $2,191 Total costs: Optimal Solution : ($225)*6+($0.8)[60+380+230+10+180+70]=$2,094

  18. The End

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