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mt.sol

# mt.sol - CO 350 Winter 2010 Midterm Exam Solutions 1(a The...

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CO 350 Winter 2010 Midterm Exam Solutions 1. (a) The dual is min 8 y 1 + 12 y 2 + 20 y 3 subj. to 2 y 1 y 3 = 1 y 1 + 4 y 2 + 5 y 3 2 y 1 + 2 y 2 0 y 1 , y 3 0 ( y 2 free) (b) Replace min x 1 + 2 x 2 by max x 1 2 x 2 . Let x 3 = x 3 , and write x 1 = u 1 v 1 . The problem in SEF is max u 1 v 1 2 x 2 subj. to 2 u 1 2 v 1 + x 2 + x 3 + s 1 = 8 4 x 2 2 x 3 = 12 u 1 + v 1 + 5 x 2 + s 2 = 20 2 u 1 + 2 v 1 + 3 x 2 x 3 s 3 = 3 u 1 , v 1 , x 2 , x 3 , s 1 , s 2 , s 3 0 2. (a) x = [0 , 0 , 1 , 4 , 0 , 12] T . (b) z = 2. (c) x 2 entering and x 3 leaving, x 5 entering and x 6 leaving, x 5 entering and x 4 leaving. (d) (Add 1 / 2 times the x 3 -row to the z -row to obtain:) z = 2 . 5. 3. (a) ¯ c 2 = 2 0. ¯ c 3 = 2 > 0 . x 3 enters. Compute t = min( 6 4 , 2 1 , ) = 1, so x 4 leaves. We pivot on the position (4,3). The new tableau is z + 2 x 2 + x 4 + x 5 = 6 x 1 x 2 2 x 4 3 x 5 = 2 x 3 + 1 2 x 4 + x 5 = 1 2 x 2 + x 4 + 3 x 5 + x 6 = 3 Now ¯ c 2 = 2 0. ¯ c 4 = 1 0. ¯ c 5 = 1 0. The current basic solution x = (2 , 0 , 1 , 0 , 0 , 3) T is optimal.

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