# sec7_1 - 7 Finding an initial feasible tableau To begin the...

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7 Finding an initial feasible tableau To begin the simplex method for solving a linear program in standard equal- ity 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 0 . 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 0 . Our initial system of equations is then z - x 1 - x 2 = 0 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 0 , 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. 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 0 . 40

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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 - x 4 - x 5 : ( * ) holds and x 1 , . . . , x 5 0 . 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. To see this, notice that, given any feasible solution for (7.1), setting x 4 = x 5 = 0 gives a feasible solution for (7.2) with value zero, and clearly no feasi- ble solution for (7.2) can have greater objective value than zero. Conversely, if a feasible solution for (7.2) has objective value zero, then both x 4 and x 5
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• TODD
• Optimization, linear program, initial feasible tableau

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