section22

# section22 - 22 The dual simplex method We have seen how...

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22 The dual simplex method We have seen how, when a linear program in standard equality form has a nondegenerate basic optimal solution, the optimal solution of the dual problem provides sensitivity information with respect to small changes to the right-hand sides. We begin this section by revisiting the question of exactly what we mean by “small” in this context. We begin with one of our earliest examples, once again: maximize x 1 + x 2 subject to x 1 2 x 1 + 2 x 2 4 x 1 , x 2 0 . We can picture the optimal solution graphically as shown below. If we introduce slack variables as usual, the problem becomes maximize x 1 + x 2 = z subject to x 1 + x 3 = 2 x 1 + 2 x 2 + x 4 = 4 x 1 , x 2 , x 3 , x 4 0 . After solving the problem by the simplex method, we arrive at the following tableau: z + 1 2 x 3 + 1 2 x 4 = 3 x 1 + x 3 = 2 x 2 - 1 2 x 3 + 1 2 x 4 = 1 . The corresponding basic feasible solution is nondegenerate and optimal. We can easily check, using the revised simplex method calculations, or comple- mentary slackness, that the optimal solution of the dual problem is y * 1 = 119

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1 2 , y * 2 = 1 2 . Hence, if, for example, the right-hand side 4 changes to 4 + for any small number , then by Theorem 18.1 the optimal value increases by y * 2 = 2 . Reviewing the proof of Theorem 18.1, we see the explanation: the new tableau corresponding to the old optimal basis B = [1 , 2] is unchanged except for the right-hand sides: z + 1 2 x 3 + 1 2 x 4 = 3 + 2 x 1 + x 3 = ( A - 1 B b ) 1 x 2 - 1 2 x 3 + 1 2 x 4 = ( A - 1 B b ) 2 where the vector b is the new right-hand side in the linear program: b = [2 , 4 + ] T . After calculating the inverse of the basis matrix A - 1 B , we find A - 1 B b = 1 0 - 1 2 1 2 2 4 + = 2 1 + 2 , so the new tableau is z + 1 2 x 3 + 1 2 x 4 = 3 + 2 x 1 + x 3 = 2 x 2 - 1 2 x 3 + 1 2 x 4 = 1 + 2 . As predicted by Theorem 18.1, this tableau remains feasible, and hence op- timal, providing is small. In fact, we can be more precise: the tableau remains optimal providing ≥ - 2. We can interpret this observation geometrically. If the right-hand side 4 changes slightly to 4+ , then the diagonal line (corresponding to x 1 +2 x 2 = 4) in the picture of the feasible region above moves slightly, but the optimal solution remains at the intersection of this line with the vertical line x 1 = 2.
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