SimplexMethodExample - ORIE 3300/5300 Fall 2011 Example of...

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ORIE 3300/5300 Fall 2011 Example of the Simplex Method Consider the following linear program in standard equality form: maximize x 1 + 3 x 2 subject to x 1 + x 2 + x 3 = 3 - x 1 + x 2 + x 4 = 1 2 x 1 - x 2 + x 5 = 3 x 1 , x 2 , x 3 , x 4 , x 5 0 Let us see how this linear program can be solved using the Simplex Method. Introducing the new variable z = x 1 + 3 x 3 to keep track of the objective value, we get the initial system of equations z - x 1 - 3 x 2 = 0 x 1 + x 2 + x 3 = 3 - x 1 + x 2 + x 4 = 1 2 x 1 - x 2 + x 5 = 3 This system is already a tableau. The basis is { 3 , 4 , 5 } , the corresponding basic feasible solution is x = [0 0 3 1 3] T , and the objective value is z = 0. ( Make sure you understand and agree with these statements!) In particular, Step 0 of the Simplex Method (initialization) is already complete. Let us carry out the other steps. Iteration 1 Step 1 (Check optimality): The reduced costs are ¯ c 1 = 1, ¯ c 2 = 3, so do not stop . Step 2 (Choose entering index): Choose k N with ¯ c k > 0. Currently N = { 1 , 2 } and the reduced costs are both positive, so either index would work. But we have to pick one, so let us choose the smallest index (“smallest subscript rule”), k = 1. 1
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Step 3 (Check for unboundedness): Since k = 1, we have ¯ A k = ¯ a 3 k ¯ a 4 k ¯ a 5 k = 1 - 1 2 , which is not 0. So do not stop . (Note: the rows of the tableau are indexed by the basic indices i B = { 3 , 4 , 5 } , and so are the elements of ¯ A k . This is why we write ¯ a 3 k , ¯ a 4 k , ¯ a 5 k instead of, for instance, ¯ a 1 k , ¯ a 2 k , ¯ a 3 k .) Step 4 (Choose leaving index): Compute the ratios
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SimplexMethodExample - ORIE 3300/5300 Fall 2011 Example of...

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