postoptonotes(1).pdf - Section 3.3 4(a x1 = 0 x2 = 4 s1 = 2...

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Section 3.3 4.(a) x 1 = 0 , x 2 = 4 , s 1 = 2 , s 2 = 0 , s 3 = 9. 4.(b) Pivot in column 1 and row 3: x 1 x 2 s 1 s 2 s 3 0 0 1 - 7 6 2 3 8 0 1 0 1 3 - 1 3 1 1 0 0 1 6 1 3 3 0 0 0 2 3 1 3 7 4.(c) Yes, the solution is optimal because the objective row contains only nonnegative entries. Section 3.4 4.(a) Substitute x = [20 , 0 , 10 , 5 , 0] T into the constraints: 3 · 20 + 2 · 0 - 10 + 5 = 55 2 · 20 + 0 + 10 + 0 = 50 The solution x is not basic because it has 3 positive components while a basic solution can only have 2 – one for each constraint. 4. If we choose y = [0 , 1 , - 1 , - 1 , - 10] T , then setting t = 30 in x + ty yields the basic solution x b = [0 , 80 , 20 , 70 , 0] T . 12. The choice of solution to the homogeneous case y T = [ - 2 , 4 , - 12 , 0 , 30] T leads to the basic solution x T = [40 , 80 , 0 , 0 , 0] T . Section 3.5 4. Let S , M and G denote the numbers of the respective lines to produce. Then the linear program, with the objective function in $100’s is: Maximize : ( . 9 · 1 . + . 1 · . 85)145 S + ( . 9 · 1 . + . 1 · . 85)190 M + ( . 8 · 1 . 00 + . 2 · . 8)230 G Subject to : S + M + G 100 , 000 G 20 , 000 S + M 90 , 000 7 S - M - 8 G 0 S 0 , M 0 , G 0
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