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HW8_soln

# HW8_soln - IEOR 162 4.11.1 Z 1 0 0 0 step 1 1 0 0 0 x1 5 4...

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IEOR 162 Spring 2009 Homework 8 4.11.1 Z x1 x2 s1 s2 s3 rhs ratio 1 5 3 0 0 0 0 0 4 2 1 0 0 12 3 0 4 1 0 1 0 10 2.5 0 1 1 0 0 1 4 step 1 1 0 1.75 0 1.25 0 12.5 0 0 1 1 1 0 2 2 0 1 0.25 0 0.25 0 2.5 10 0 0 0.75 0 0.25 1 1.5 2 The tie in the ration test indicates that the next b.f.s will be degenerate. step 2 Z x1 x2 s1 s2 s3 rhs ratio 1 0 0 1.75 0.5 0 16 0 0 1 1 1 0 2 0 1 0 0.25 0.5 0 2 4 0 0 0 0.75 0.5 1 0 0 step 3 1 0 0 1 0 1 16 0 0 1 0.5 0 2 2 0 1 0 0.5 0 1 2 0 0 0 1.5 1 2 0 This is an optimal tableau with the optimal solution being z = 16, x 1 = x 2 = 2, s 2 = s 3 = s 1 = 0. This bfs corresponds to the following three sets of basic variables: {x 1 , x 2 , s 1 }, {x 1 , x 2 , s 2 }, and {x 1 , x 2 , s 3 }. The degeneracy is due to the fact that the 3 = 2 + 1 (problem has two decision variables!) constraints meet in a single point (point A = (2, 2) in the figure).

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IEOR 162 Spring 2009 Homework 8 4.12.1 After adding slack, excess and artificial variables we obtain min z = 4x 1 + 4x 2 + x 3 + Ma 3 s.t. x 1 + x 2 + x 3 + s 1 = 2 2x 1 + x 2 + s 2 = 3 2x 1 + x 2 + 3x 3 - e 3 + a 3 = 3 Eliminating the basic variable a 3 from z - 4x 1 - 4x 2 - x 3 - Ma 3 = 0 we obtain z + (2M - 4)x 1 + (M - 4)x 2 + (3M - 1)x 3 - Me 3 = 3M.
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HW8_soln - IEOR 162 4.11.1 Z 1 0 0 0 step 1 1 0 0 0 x1 5 4...

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