secn8 - 8 Phase 1 of the simplex method Let us summarize...

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Unformatted text preview: 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 X j =1 a ij x j = b i ( i = 1 , 2 , . . . , m ) x j ≥ ( 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 ) X 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 n m X i =1 x n + i : ( S ) holds , x ≥ o We transform this Phase 1 problem to standard equality form by multiply- ing the objective function by- 1. (So the Phase I objective variable w is- ∑ m i =1 x n + i .) An initial feasible tableau is readily available for this prob- lem: 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 r = x n + h say, has become nonbasic, then before proceeding to the next iteration we can delete all the appearances of the artificial variable x n + h in ( T ) to obtain a system ( T ). To justify this process, notice that it is equivalent to setting x n + h = 0 throughout ( T ). Consider the effect of editing the system ( S ) by setting the variable x n + h to zero. This new system, ( S ) say, shares the key motivating property with 46 ( S ): the original system of constraints (8.1) has a feasible solution exactly when the the linear program min n X i 6 = r x n + i : ( S ) holds , x j ≥ 0 ( j = 1 , . . . , n ) , x n + i ≥ 0 ( i = 1 , . . . , m, i 6 = h ) o has a feasible solution with objective value zero, and furthermore the system ( T ) is a feasible tableau for this linear program....
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secn8 - 8 Phase 1 of the simplex method Let us summarize...

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