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sample-final-solutions

# sample-final-solutions - IEOR 162 Linear Programming Fall...

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IEOR 162 Linear Programming Fall 2006 Sample Final Solutions 1. Starting from x = (1 , 1 . 5 , 0 , 0), implies that we are starting from the basis with basic variables x B = ( x 1 , x 2 , s 1 ), since s 1 = 5 - 3 x 2 - x 3 - x 4 = . 5. We can calculate A - 1 B = 0 0 1 3 0 1 2 - 1 6 1 - 3 2 1 2 , which we then use to create the tableau for this basis: z x 1 x 2 x 3 x 4 s 1 s 2 s 3 RHS ratio 1 0 0 - 3 - 1 0 0 0 0 0 1 0 2 3 1 3 0 0 1 3 1 1 . 5 0 0 1 1 6 - 1 6 0 1 2 - 1 6 1 . 5 9 0 0 0 .5 3 2 1 - 3 2 1 2 . 5 1 * Performing one simplex iteration we get to the following tableau: z x 1 x 2 x 3 x 4 s 1 s 2 s 3 RHS 1 0 0 0 8 6 - 9 3 3 0 1 0 0 - 5 3 - 4 3 2 - 1 3 1 3 0 0 1 0 - 2 3 - 1 3 1 - 1 3 4 3 0 0 0 1 3 2 - 3 1 1 We see that this tableau is not optimal because there is a negative entry in row 0 for s 2 . 2. a. Dual: max 6 y 1 + 4 y 2 + 3 y 3 s.t. 4 y 1 + y 2 + 3 y 3 4 3 y 1 + 2 y 2 + y 3 1 y 1 + y 2 + y 3 0 y 1 0 , y 2 0 , y 3 urs b. Complementary Slackness conditions: (4 x 1 + 3 x 2 + x 3 - 6) y 1 = 0 ( x 1 + 2 x 2 + x 3 - 4) y 2 = 0 (3 x 1 + x 2 + x 3 - 3) y 3 = 0 (4 y 1 + y 2 + 3 y 3 - 4) x 1 = 0 (3 y 1 + 2 y 3 + y 3 - 1) x 2 = 0 ( y 1 + y 2 + y 3 ) x 3 = 0 c. Given the solution ( x 1 , x 2 , x 3 ) = (0 . 2 , 1 . 4 , 1 . 0) and the complementary slackness conditions of part (b), we can deduce what values of ( y 1 , y 2 , y 3 ) will satisfy the complementary slackness conditions. If these values of ( y 1 , y 2 , y 3 ) also satisfy the dual constraints, then we know that both x and y are optimal for the primal and dual respectively. First note that ( x 1 , x 2 , x 3 ) = (0 . 2 , 1 . 4 , 1 . 0) satisfies all primal constraints at equality, so the first three complimentary slackness equations are already satisfied. Since all values of x are positive, we must also 1

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satisfy all dual constraints at equality (because of the last 3 complimentary slackness conditions). This is
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