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Unformatted text preview: 5. a. b. It would merely serve to reduce his net profit by \$1,000. Since he must incur this cost regardless of what kind of hot tub he makes there is not reason to include this cost in the formulation of the model. MAX 350 X1 + 300 X2 - 900 Y1 - 800 Y2 ST 1 X1 + 1 X2 d 200 9 X1 + 6 X2 d 1,520 12 X1 + 16 X2 d 2,650 X1 - M Y1 d 0 X2 - M Y2 d 0 X1, X2 must be nonnegative integers Y1,Y2 must be binary MIN ST X1 + X2 + X3 + X4 + X5 + X6 + X7 X2 + X3 + X4 + X5 + X6 t 18 X3 + X4 + X5 + X6 + X7 t 17 X1 + X4 + X5 + X6 + X7 t 16 X1 + X2 + X5 + X6 + X7 t 16 X1 + X2 + X3 + X6 + X7 t 16 X1 + X2 + X3 + X4 + X7 t 14 X1 + X2 + X3 + X4 + X5 t 19 Xi t 0 & integer 6. a. b. c. d. 7. a. See file Prb6_6.xls X1 =5, X2 =2, X3 =6, X4 =2, X5 =6, X6 =2, X7 =1 (alternate optimal solutions exist). 3 MIN: ST 32X1 + 80X2 + 32X3 + 80X4 + 32X5 + 80X6 + 32X7 X1 + X2 t 11 X1 + X2 t 11 X1 + X2 t 11 X1 + X2 t 11 X2 + X3 + X4 t 24 X2 + X3 + X4 t 24 X2 + X3 + X4 t 16 Chapter 6 - Integer Linear Programming : S-2 ---------------------------------------------------------------------------------------- X2 + X3 + X4 t 16 X4 + X5 + X6 t 10 X4 + X5 + X6 t 10 X4 + X5 + X6 t 22 X4 + X5 + X6 t 22 X6 + X7 t 17 X6 + X7 t 17 X 6 + X7 t 6 X 6 + X7 t 6 X2 + X4 t 0.3*(X2+X3+X4) X4 + X6 t 0.3*(X4+X5+X6) X2 t 1 X6 t 1 Xi t 0 & integer b. c. See file Prb6_7.xls X1=9, X2=2, X3=16, X4=6, X5=15, X6=1, X7=16 Total cost = \$2,512 8. 1 Minimum Capacity in MW a. 780 2 860 Year 3 950 4 1060 5 1180 Let Xij = # of generators of size j purchased in year i. MIN 300 X1,10 + 460 X1,25 + 670 X1,50 + 950 X1,100 + 250 X2,10 + 375 X2,25 + 558 X2,50 + 790 X2,100 + 200 X3,10 + 350 X3,25 + 465 X3,50 + 670 X3,100 + 170 X4,10 + 280 X4,25 + 380 X4,50 + 550 X4,100 + 145 X5,10 + 235 X5,25 + 320 X5,50 + 460 X5,100 b. c. 9. a. 750 + 10 X1,10 + 25 X1,25 + 50 X1,50 + 100 X1,100 t 780 750 + 10 (X1,10 + X2,10 ) + 25 (X1,25 + X2,25) + 50 (X1,50 + X2,50 ) + 100 (X1,100 + X2,100) t 860 750 + 10 (X1,10 + X2,10 + X3,10 ) + 25 (X1,25 + X2,25 + X3,25) + 50 (X1,50 + X2,50 + X3,50) + 100 (X1,100 + X2,100 + X3,100) t 950 750 + 10 (X1,10 + X2,10 + X3,10 + X4,10) + 25 (X1,25 + X2,25 + X3,25 + X4,25) + 50 (X1,50 + X2,50 + X3,50 + X4,50) + 100 (X1,100 + X2,100 + X3,100 + X4,100) t 1060 750 + 10 (X1,10 + X2,10 + X3,10 + X4,10 + X5,10) + 25 (X1,25 + X2,25 + X3,25 + X4,25 + X5,25) + 50(X1,50 + X2,50 + X3,50+ X4,50 + X5,50)+100(X1,100 + X2,100 + X3,100 + X4,100 + X5,100)t1060 Xij t 0 and integer See file: Prb6_8.xls X1,100 = X2,10 = X3,100 = X4,100 = X5,25 = X5,100 = 1, Total Cost = \$3,115,000 MIN ST 450 X1 + 650 X2 + 550 X3 + 500 X4 + 525 X5 X 1 + X3 t 1 X 1 + X2 + X4 + X5 t 1 X 2 + X4 t 1 X 3 + X5 t 1 X 1 + X2 t 1 X 3 + X5 t 1 X 4 + X5 t 1 ST ------------------------------------------------------------------------------------ All Xi are binary X1 = 1 if Sanford is selected, 0 otherwise X2 = 1 if Altamonte is selected, 0 otherwise X3 = 1 if Apopka is selected, 0 otherwise X4 = 1 if Casselberry is selected, 0 otherwise X5 = 1 if Maitland is selected, 0 otherwise b. c. See file: Prb6_9.xls X1= X4X5 =1 (Build at Sanford and Casselberry & Maitland) Minimum total cost = \$1,475,000 X1 = batches of CD players to produce X2 = batches of tape decks to produce X3 = batches of stereo tuners to produce MAX ST (75*150) X1 + (50*150) X2 + (40*150) X3 (3*150) X1 + (2*150) X2 + (1*150) X3 d 400,000 50,000/150 d X1 d 150,000/150 50,000/150 d X2 d 100,000/150 50,000/150 d X3 d 90,000/150 where: 10. a. b. c. See file: Prb6_10.xls X1=334, X2=532, X3=600 Maximum profit = \$11,347,500 MIN ST 21X1 + 23X2 + 25X3 + 24X4 + 20X5 + 26X6 + 1000Y1 + 950Y2 + 875Y3 + 850Y4 + 800Y5 + 700Y6 X1 + X2 + X3 + X4 + X5 + X6 = 1800 X1 - 500 Y1 d 0 X2 - 600 Y2 d 0 X3 - 750 Y3 d 0 X4 - 400 Y4 d 0 X5 - 600 Y5 d 0 X6 - 800 Y6 d 0 Xi t 0 All Yi are binary 11. a. b. c. See file: Prb6_11.xls X1=500, X2=600, X4=100, X5=600 Total cost = \$42,300 13. a. Let Xi = 1 if project i is selected, 0 otherwise MAX ST 650 X1 + 550 X2 + 600 X3 + 450 X4 + 375 X5 + 525 X6 + 750 X7 7 X1 + 6 X2 + 9 X3 + 5 X4 + 6 X5 + 4 X6 + 8 X7 d 20 250 X1 + 175 X2 + 300 X3 + 150 X4 + 145 X5 + 160 X6 + 325 X7 d 950 X2 + X6d 1 (implemented as X2 d 1-X6 ) b. c. 14. a. See file: Prb6_13.xls X1=X6=X7=1, Total NPV = \$1,925,000 Let Xij = bushels (in 1000s) shipped from grove i to processing plant j Yij = 1 if Xij t 0, 0 otherwise \$168 Y14 + \$400 Y15 + \$320 Y16 \$280 Y24 + \$240 Y25 + \$176 Y26 \$440 Y34 + \$160 Y35 + \$200 Y36 ST X14 + X15 + X16 = 275 X24 + X25 + X26 = 400 X34 + X35 + X36 = 300 X14 + X24 + X34 d 200 X15 + X25 + X35 d 600 X16 + X26 + X36 d 225 Xij - MijYij d 0 Xij t 0 Yij binary Note: Mij = MIN(supply of i, capacity of j) MIN b. c. See file: Prb6_14.xls The solution is: X15= 275, X24= 200, X26= 200, X35= 300, Y15= 1, Y24= 1, Y26= 1, Y35= 1. Minimum trucking cost = \$1,016. X0 = number of efficiency apartments to build X1 = number of 1-bedroom apartments to build X2 = number of 2-bedroom apartments to build X3 = number of 3-bedroom apartments to build MAX ST 350 X0 + 450 X1 + 550 X2 + 750 X3 X0 + X1 + X2 + X3 d 40 500X0 + 700X1 + 800X2 + 1,000X3 d 40,000 0 d X0 d 40 5 d X1 d 15 8 d X2 d 22 0 d X3 d 10 15. a. 18. a. Let Pij = production at plant i allocated to meet demand for customer j Yi= 1 if any Xij t 0, 0 otherwise MIN 35 P1X + 30 P1Y + 45 P1Z + 45 P2X + 40 P2Y + 50 P2Z + 70 P3X + 65 P3Y + 50 P3Z + 20 P4X + 45 P4Y + 25 P4Z + 65 P5X + 45 P5Y + 45 P5Z + 1000 (1,325 Y1 + 1,100 Y2 + 1,500 Y3 + 1,200 Y4 + 1,400 Y5) P1X + P1Y + P1Z d 40,000 Y1 P2X + P2Y + P2Z d 30,000 Y2 P3X + P3Y + P3Z d 50,000 Y3 P4X + P4Y + P4Z d 20,000 Y4 P5X + P5Y + P5Z d 40,000 Y5 P1X + P2X + P3X + P4X + P5X t 40,000 P1Y + P2Y + P3Y + P4Y + P5Y t 25,000 P1Z + P2Z + P3Z + P4Z + P5Z t 35,000 Pij t 0 Yi binary ST b. c See file: Prb6_18.xls The solution is: Build plants at locations 1, 4 and 5 (i.e., Y1=Y4=Y5 = 1). P1X=P1Y=P4X=20,000, P5Y=5,000, P5Z=35,000. Total cost = \$7,425,000. Additional Constraints: Y1+Y2 d 1 and Y4+Y5 d 1 (implement as Y1d 1-Y2 and Y4d 1-Y5) See file: Prb6_19.xls The solution is: Build plants at locations 1, 3 and 4 (i.e., Y1=Y3=Y4 = 1). P1X=15,000, P1Y=25,000, P3X=5,000, P3Z=35,000, P4X=20,000. Total cost = \$7,800,000. MIN ST 13 A1+ 9 B1+ 10 C1+ 11 A1+ 12 B2+ 8 C2 + 55 YA1+ 93 YB1+ 60 YC1+ 65 YA2+ 58 YB2+ 75 YC2 A1+ A2 = 3 B1+ B2 = 7 C1+ C2 = 4 0.4 A1+ 1.1 B1+ 0.9 C1d 8 0.5 A2+ 1.2 B2+ 1.3 C2d 6 A1- 3YA1 d 0 B1- 7YB1 d 0 C1- 4YC1 d 0 A2- 3YA2 d 0 B2- 7YB2 d 0 C2- 4YC2 d 0 Ai, Bi, Ci t and integer Yij binary 19. a. b. c. 20. a. b. c. See file: Prb6_20.xls The optimal solution is: A2=3, B1=4, B2=3,C1=4, YA2=YB1=YB2=YC1=1 Total cost = \$421 ...
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This note was uploaded on 05/29/2010 for the course BMS 5487 taught by Professor Askjdakjb during the Fall '97 term at Universität Zürich.

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