# sol6 - y 1 = 0 the objective coeFcient of s 1 y 2 = 1 the...

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AMS 341 Spring, 2011 Homework Set # 6 — Solution Notes #4, 6.5: The dual of the LP is: max z = 6 x 1 + 8 x 2 s . t . x 1 + x 2 4 2 x 1 - x 2 2 2 x 2 = - 1 x 1 0 , x 2 unrestricted #2, 6.7: (a). The dual of the LP is: min z = 3 y 1 + 2 y 2 + y 3 s . t . y 1 + y 3 ≥ - 2 y 1 + y 2 ≥ - 1 y 1 + y 2 + y 3 1 y 1 0 , y 2 0 , y 3 unrestricted (b). Using the formulas on page 310, we get y 1 = 0, the objective coefficient of s 1 , y 2 = - 1 the negative of the coefficient of e 2 , and y 3 = 2, the coefficient of a 3 minus M . #5, 6.7: Using the formulas on page 310, we get
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Unformatted text preview: y 1 = 0, the objective coeFcient of s 1 , y 2 = 1 the coeFcient of s 2 . So the optimal to the dual is 6 y 1 +10 y 2 = 10, but the optimal to the primal is 20 / 3. This would contradict the strong duality theorem, which says that at the optimal solutions, opt of primal is equal to opt of dual....
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