This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 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 X k N ( c k y T A k ) x k = y T b x B + A 1 B X 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 ....
View
Full
Document
 Fall '07
 BLAND

Click to edit the document details