IEOR162_SampleMidterm2_sol

# IEOR162_SampleMidterm2_sol - IEOR 162, Fall 2011 Suggested...

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IEOR 162, Fall 2011 Suggested Solution to Sample Midterm 2 1. The standard form is max x 1 - 2 x 2 s.t. x 1 + 2 x 2 - x 3 - x 4 = 2 x 1 - x 2 + x 5 = 3 x i 0 i = 1 ,..., 5 . The following table lists all the nine basic solutions. The last column shows whether a solution is feasible or not. There are four basic feasible solutions. x 1 x 2 x 3 x 4 x 5 Basic solution? Feasible? 0 0 0 - 2 3 Yes No 0 0 - 2 0 3 Yes No 0 0 - - 0 No No 0 1 0 0 4 Yes Yes 0 - 3 0 - 8 0 Yes No 0 - 3 - 8 0 0 Yes No 2 0 0 0 1 Yes Yes 3 0 0 1 0 Yes Yes 3 0 1 0 0 Yes Yes 8 3 - 1 3 0 0 0 Yes No 2. The augmented standard form LP is max s.t. x 1 + 3 x 2 + x 3 + + x 4 - Mx 7 - Mx 8 x 1 + x 2 - 2 x 3 + x 5 = 6 - 2 x 1 + 3 x 2 - x 4 - x 6 + x 7 = 9 3 x 2 - 2 x 3 - x 4 + x 8 = 3 x i 0 i = 1 ,..., 8 . An initial basic feasible solution for the augmented standard form is (0 , 0 , 0 , 0 , 6 , 0 , 9 , 3). We start from this solution and try to optimize the augmented form LP. Because - 1 - 3 - 1 - 1 0 0 M M 0 1 1 - 2 0 1 0 0 0 6 - 2 3 0 - 1 0 - 1 1 0 9 0 3 - 2 - 1 0 0 0 1 3 1

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is not a valid tableau, we ﬁx the objective row and then run two simplex iterations: - 1 + 2 M - 3 - 6 M - 1 + 2 M - 1 + 2 M 0 M 0 0 - 12 M 1 1 - 2 0 1 0 0 0 x 5 = 6 - 2
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## This note was uploaded on 02/25/2012 for the course IEOR 162 taught by Professor Zhang during the Fall '07 term at Berkeley.

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IEOR162_SampleMidterm2_sol - IEOR 162, Fall 2011 Suggested...

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