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section14

# section14 - 14 The revised simplex method maximize cT x = z...

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14 The revised simplex method Consider once again the standard equality-form linear program maximize c T x = z subject to Ax = b x 0 , Corresponding to any basis B is a tableau that is unique except for the order in which we write the equations. Theorem 12.4 gave a formula for that tableau, that we can rewrite slightly as follows: z - k N ( c k - y T A k ) x k = y T b x B + A - 1 B k N x k A k = x * B , where the vector y solves the equation A T B y = c B , and the vector x * B (whose components are the current values of the basic variables) is A - 1 B b . Recall that N is a list of the nonbasic indices, the vector A k is the k th column of the matrix A ; as usual, A B is the basis matrix, and analogously, the vector c B has entries c i as the index i runs through the basis B . Using this notation, consider how an iteration of the simplex method proceeds. We begin by choosing an entering variable x k (where k N ) with positive reduced cost c k - y T A k . To do this, we must first calculate the vector y . Step 1: Solve the equation A T B y = c B for the vector y . Step 2: Choose an entering index k N such that c k > y T A k . The next step is the ratio test, for which we need the column vector whose components are the coefficients of the entering variable x k in the body of the tableau. We can see from the formula for the tableau that this vector, which we call d , is just A - 1 B A k . Step 3: Solve the equation A B d = A k for the vector d . Step 4: Calculate t min x * i d i : i B, d i > 0 , 73

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and choose a leaving index r from among those indices i attaining this minimum.
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section14 - 14 The revised simplex method maximize cT x = z...

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