MAT361 HOMEWORK ASSIGNMENT SOLUTIONS S2004

This is an optimal Phase I tableau. We now drop the a3 column and eliminate the variable x3 from the Phase II row 0 (z 4x1 4x2 x3 = 0) this yields z 10x1/3 11x2/3 e3/3 = 1, indicating that the current tableau is optimal. Thus the optimal solution to the LP is z = 1, x2 = x1 = 0, x3 = 1. SECTION 4.14. 1. ................
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