sec8_Notes

# sec8_Notes - 8 Phase 1 of the simplex method Let us...

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8 Phase 1 of the simplex method Let us summarize the method we sketched in the previous section, for finding a feasible basis for the constraint system (8.1) n j =1 a ij x j = b i ( i = 1 , 2 , . . . , m ) x j 0 ( j = 1 , 2 , . . . , n ) . We can assume each right-hand side b i is nonnegative: otherwise we multiply the corresponding equation by - 1. The idea of “Phase 1” of the simplex method is as follows: correspond- ing to each constraint, introduce an “artificial variable” x n + i 0 (for i = 1 , 2 , . . . , m ), which we think of as the error in that constraint. We therefore consider the system ( S ) j a ij x j + x n + i = b i ( i = 1 , 2 , . . . , m ) . We then try to force the values of these artificial variables x j , j > n , to zero, by using the simplex method to minimize their sum: (8.2) min m i =1 x n + i : ( S ) holds , x 0 We transform this Phase 1 problem to standard equality form by multiplying the objective function by - 1. An initial feasible tableau is readily available for this problem: we choose x n +1 , x n +2 , . . . , x n + m as the initial list of basic variables, and eliminate them from the top equation defining the Phase 1 objective function using row operations. We then begin the simplex method. If, after some iteration of the Phase 1 simplex method, we arrive at a tableau ( T ) where one of the new variables, x n + r say, has become nonbasic, then before proceeding to the next iteration we can delete all the appearances of the artificial variable x n + r in ( T ) to obtain a system ( T ) . To justify this process, notice that it is equivalent to setting x n + r = 0 throughout ( T ). Consider the effect of editing the system ( S ) by setting the variable x n + r to zero. This new system, ( S ) say, shares the key motivating property with ( S ): the original system of constraints (8.1) has a feasible solution exactly 42

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when the the linear program min i = r x n + i : ( S ) holds , x j 0 ( j = 1 , /ldots, n ) , x n + i 0 ( i = 1 , /ldots, m, i = r ) has a feasible solution with objective value zero, and furthermore the system ( T ) is a feasible tableau for this linear program. We repeat this deletion process throughout Phase 1: each time an artifi- cial variable becomes nonbasic, we delete it. Assuming Phase 1 terminates with an optimal tableau for the auxiliary problem (8.2), we face three possi- bilities: The optimal value is strictly positive. In this case, the original system
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