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Unformatted text preview: CO350 Linear Programming Chapter 7: The TwoPhase Method 17th June 2005 Chapter 7: The TwoPhase Method 1 Recap ( P ) max ( z =) x 1 s.t. 3 x 1 + 5 x 2 + 2 x 3 x 4 = 7 2 x 1 + 5 x 2 + 3 x 3 + x 4 = 3 x 1 , x 2 , x 3 , x 4 So far this week, we learned to construct the auxiliary problem ( A ) , to detect infeasibility by solving ( A ) , to obtain a feasible basis of ( P ) from optimal tableau for ( A ) , two observations that may simplify the construction and solution of ( A ) . Chapter 7: The TwoPhase Method 2 An Exceptional Case ( 7.4) It may happen that ( A ) has optimal value , and yet the optimal tableau has basic artificial variables . BUT we want a feasible basis for ( P ) ! I.e., no artificial variables should be basic . Good news: Optimal value is 0 = all artificial variables = 0 . So, every basic artificial variable has value . (This, by definition, means that the bfs is degenerate.) To force the artificial variable out of basis, we pivot! Chapter 7: The TwoPhase Method 3 Example (Pg 96) Suppose that only x 5 is artificial . w + 2 x 4 = 0 x 2 + 4 x 3 + 3 x 4 = 2 2 x 3 2 x 4 + x 5 = 0 x 1 + 3 x 3 + 2 x 4 = 1 We want x 5 to leave. We choose either x 3 or x 4 to enter. BUT this violates the rules of the simplex method . Neither c 3 nor c 4 are positive. Neither a 53 nor a 54 are positive. HOWEVER, b 5 = 0 = r.h.s. values will not change . i.e., obj. value remains , and tableau remains feasible....
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 Fall '97
 Wolkowicz

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