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Unformatted text preview: 7 Finding an initial feasible tableau To begin the simplex method for solving a linear program in standard equality form, we need to find an initial feasible tableau. Sometimes, this is easy. For example, the first linear program we studied was the problem maximize x 1 + x 2 subject to x 1 ≤ 2 x 1 + 2 x 2 ≤ 4 x 1 , x 2 ≥ . We transformed this problem to standard equality form by introducing slack variables: 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 ≥ . Our initial system of equations is then z- x 1- x 2 = x 1 + x 3 = 2 x 1 + 2 x 2 + x 4 = 4 , and, as we observed, this system is already a feasible tableau, corresponding to the basis [3 , 4]. More generally, given any linear program in standard inequality form, maximize c T x subject to Ax ≤ b x ≥ , if we transform to standard equality form by introducing slack variables, then the original system of constraints constitute a tableau for the trans- formed problem, where the basic variables are exactly the slack variables. Furthermore, this tableau is feasible, provided b ≥ 0. In general, however, it may not so easy to find an initial feasible tableau, or even to decide whether a system of constraints has a feasible solution. We next introduce a technique to resolve this difficulty. We illustrate with an example. 35 Consider the linear program (7.1) maximize 2 x 1 + x 2 subject to x 1 + x 2 = 3- x 1 + x 2- x 3 = 1 x 1 , x 2 , x 3 ≥ . For the moment, we ignore the objective function, and simply try to decide whether or not the problem has a feasible solution, and, if so, to find one. With this in mind, we introduce two new artificial variables that measure the error in each of the equality constraints: ( * ) x 4 = 3- ( x 1 + x 2 ) x 5 = 1- (- x 1 + x 2- x 3 ) . We will try to force these errors to zero by solving the linear program (7.2) max n- x 4- x 5 : ( * ) holds and x 1 , . . . , x 5 ≥ o . Central to the success of this approach is the following property relationship between the linear programs (7.1) and (7.2): (7.1) is feasible ⇔ (7.2) has an optimal solution with value zero....
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This note was uploaded on 02/20/2009 for the course ORIE 3300 taught by Professor Todd during the Fall '08 term at Cornell.
- Fall '08