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Linear Prog COP4355_Part_18

# Linear Prog COP4355_Part_18 - Chapter 7 Integer Linear...

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76 Chapter 7 Integer Linear Programming 7.15 UB = 380.5 (X 1 = 22.167, X 2 = 0.167) LB = 374 (X 1 = 22, X 2 = 0) 1 Z=380.5 X 1 < 22 X 1 > 23 2 X 1 = 22 X 2 = 0.29 3 Infeasible Solution UB = 380.5 LB = 380.29 5 X 1 = 21 X 2 = 1 Stop 4 X 1 = 22 X 2 = 0 Stop *Inferior* X 2 < 0 X 2 > 1 UB = 380.5 LB = 379 UB=380.5 LB = 374 Stop *Optimal* 7.16 X 1 = 2; X 2 = 3; X 3 = 2; Z MAX = 1260. 7.17 X 1 = 2; X 2 = 4; Z MAX = 48 7.18 X 1 = 0; X 2 = 1; X 3 = 1; X 4 = 1; X 5 = 1; X 6 = 0; Z MAX = 72000. 7.19 X 1 = 1; X 2 = 0; X 3 = 0; X 4 = 1; X 5 = 1; X 6 = 1: Z MAX = 500. 7.20 X 1 = 15; X 2 = 15; X 3 = 25: Z MAX = 822500. 7.21 X 1 = 6; X 2 = 3; X 3 = 118; Z MAX = 18625. 7.22 X 1 = 7; X 2 = 1; X 3 = 1; Z MAX = 85. 7.23 X 1 = 6; X 2 = 0.667; Z MAX = 32. 7.24 X 1 = 6; X 2 = 1.6; Z MIN = 55.2. 7.25 Z = 360000. 7.26 Z = 314. 7.27 Z = 318.

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Solutions Manual and Workbook 77 Chapter 8 The Transportation Model 8.1 (Initial Solution) 4 Note: Using the Northwest corner method, we do not need the costs to obtain an initial feasible solution. If we use them, we can compute the total cost as: 35(10)+35(8)+15(8)+15(3)+65(6)+10(4)+ 55(3) = \$1,390 8.2 Note: At one point there is a tie between Row A, Row B, and Column X. It was broken arbitrarily in favor of Row A. Total Cost by VAM: 35(4)+50(5)+70(4)+10(3)+ 20(5)+25(4)+20(3) = \$960 .
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