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IEOR162_hw09_sol

# IEOR162_hw09_sol - IEOR 162 Fall 2011 Suggested Solution to...

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IEOR 162, Fall 2011 Suggested Solution to Homework 09 Problem 1 (Modified from Problem 6.10.1) (a) The dual problem is min 600 y 1 + 400 y 2 + 500 y 3 s.t. 4 y 1 + y 2 + 3 y 3 6 9 y 1 + y 2 + 4 y 3 10 7 y 1 + 3 y 2 + 2 y 3 9 10 y 1 + 40 y 2 + y 3 20 y i 0 i = 1 , 2 , 3 . (b) First we convert the dual problem to its standard form min 600 y 1 + 400 y 2 + 500 y 3 s.t. 4 y 1 + y 2 + 3 y 3 - t 1 = 6 9 y 1 + y 2 + 4 y 3 - t 2 = 10 7 y 1 + 3 y 2 + 2 y 3 - t 3 = 9 10 y 1 + 40 y 2 + y 3 - t 4 = 20 with all variables nonnegative. Since x 1 , x 4 , and s 3 are positive, by complementary slackness we know t 1 , t 4 , and y 3 must be 0. This information then reduces the first and the fourth constraints to 4 y 1 + y 2 = 6 10 y 1 + 40 y 2 = 20 . Solving this system then gives us y 1 = 22 15 and y 2 = 2 15 . If we plug in these values back to the dual objective function, we will get 2800 3 , which agrees with strong duality. The values of y 1 and y 2 also allow us to derive t 2 = 10 3 and t 3 = 5 3 . (c) The interpretations for the four conditions are listed below. (i) s 3 > 0 y 3 = 0. Since we still have unused glass in the optimal production plan (primal solution), the shadow price of glass, y 3 , should be 0. That is, it is worthless to obtain more glass even for free. (ii) y 1 > 0 s 1 = 0. For the shadow price of molding time to be positive, the only reason is that there is no unused molding time in the optimal production plan (primal solution), i.e., s 1 = 0. Problem 2 (a) We can verify that (6 , 0 , 0) is the optimal solution and 30 is the objective value in one iteration: - 5 - 4 - 3 0 0 0 1 2 3 1 0 6 2 1 2 0 1 18 0 6 12 5 0 30 1 2 3 1 0 6 0 - 3 - 4 - 2 1 6 1

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(b) Because the objective value is c B B - 1 b , we know if b 0 1 = b 1 + 1, then the improvement is c B B - 1 b 0 - c B B - 1 b = c B B - 1 ( b 0 - b ) = c B B - 1 1 0 = ( c B B - 1 ) 1 , which means the first component of c B B - 1 is the shadow price for constraint 1. In general, the i th component of c B B - 1
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IEOR162_hw09_sol - IEOR 162 Fall 2011 Suggested Solution to...

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