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Unformatted text preview: 14 The revised simplex method Consider once again the standard equalityform linear program maximize c T x = z subject to Ax = b x , Corresponding to any basis B is a unique tableau. Theorem 12.2 gave a formula for that tableau, which we can rewrite slightly as follows: z X j N ( c j y T A j ) x j = y T b x B + X j N A 1 B A j x j = 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 j is the j 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 . (If none, the current solution is optimal.) 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 n x i d i : i B, d i > o , and choose a leaving index r from among those indices i attaining this minimum. (If d 0, the problem is unbounded.) 70 Since the ratio test at the next iteration will involve the new values of the basic variables, it is helpful to calculate these values. Step 5: Update the current values of the variables, x k t, x B x B td, and then the basis: Replace r by k in the ordered list B. We write B B { k } \ { r } ....
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