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HW5_soln-1 - HW-5 Solution IEOR162 Spring 09 SECTION 4.1 1...

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HW-5 Solution, IEOR162, Spring 09 SECTION 4.1 1. max z = 3x 1 + 2x 2 s.t. 2x 1 + x 2 + s 1 = 100 x 1 + x 2 + s 2 = 80 x 1 + s 3 = 40 2. min z = 50x 1 + 100x 2 s.t. 7x 1 + 2x 2 - e 1 = 28 2x 1 + 12x 2 - e 2 = 243. 3. min z = 3x 1 + x 2 s.t. x 1 - e 1 = 3 x 1 + x 2 + s 2 = 4 2x 1 - x 2 = 3 SECTION 4.4 1. From Figure 2 of Chapter 3 we see that the extreme points of the feasible region are Basic Feasible Solution H = (0, 0) s 1 = 100, s 2 = 80, s 3 = 40 x 1 = x 2 = x 3 = 0 E = (40, 0) x 1 = 40, s 1 = 20, s 2 = 40 x 2 = x 3 = s 3 = 0 F = (40, 20) x 1 = 40, x 2 = 20, s 2 = 20 x
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