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Unformatted text preview: IEOR 162 Linear Programming Spring 2007 Homework 7 Solutions 4.7.2 (4 points) To find multiple optimal solutions, we perform the simplex algorithm and instead of stopping when all reduced costs are nonpositive, we continue by entering a variable with 0 reduced cost. ↓ z x 1 x 2 x 3 x 4 RHS ratio 1 3 6 5 7 1 35 5 1 2 1 2 1 ↓ z x 1 x 2 x 3 x 4 RHS ratio 1 3 6 8.5 1 7 2 28 56 17 1 2 1 1 2 1 We would normally stop here because all reduced costs are nonpositive, but since we want to find more optimal solutions we enter x 1 into the basis. z x 1 x 2 x 3 x 4 RHS ratio 1 3 6 1 2 17 7 17 56 17 1 1 17 5 17 45 17 So we have found 2 optimal solutions, x 1 = ( 1 28 ) and x 2 = ( 56 17 45 17 ) . We know that all points that lie on the line between x 1 and x 2 must also be optimal, so to find a third optimal solution, we can simply take the average of x 1 and x 2 , x 3 = x 1 + x 2 2 = ( 28 17 31 17 14 ) ....
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This homework help was uploaded on 04/02/2008 for the course IEOR 162 taught by Professor Zhang during the Spring '07 term at Berkeley.
 Spring '07
 Zhang

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