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Unformatted text preview: CO350 Linear Programming Chapter 8: Degeneracy and Finite Termination 24th June 2005 Chapter 8: Finite Termination 1 Recap The perturbation method ( P ) max c T x s.t. Ax = b x ≥ Assumption: B is a feasible basis with A B = I . Perturb the right hand side to b = b + [ ε,ε 2 ,...,ε m ] T to get ( P ) max c T x s.t. Ax = b x ≥ We showed that B is also a feasible basis of ( P ) . Tableaux for ( P ) and ( P ) differ in right hand side only = ⇒ choices of leaving variables are affected. Lemma 8.2 If ε is positive and sufficiently small, then α + α 1 ε + α 2 ε 2 + ··· + α m ε m < β + β 1 ε + β 2 ε 2 + ··· + β m ε m ⇐⇒ ( α ,α 1 ,α 2 ,...,α m ) L < ( β ,β 1 ,β 2 ,...,β m ) Chapter 8: Finite Termination 2 Example (cycling example on pg 107) Initial tableau: z 2 x 1 3 x 2 + x 3 + 12 x 4 = 0 2 x 1 9 x 2 + x 3 + 9 x 4 + x 5 = 0 1 3 x 1 + x 2 1 3 x 3 2 x 4 + x 6 = 0 Tableau for perturbed problem: z 2 x 1 3 x 2 + x 3 + 12 x 4 = 0 2 x 1 9 x 2 + x 3 + 9 x 4 + x 5 = ε 1 3 x 1 + x 2 1 3 x 3 2 x 4 + x 6 = ε 2 ¯ c 2 is largest positive reduced cost, so x 2 enters. min { ,ε 2 / 1 } = ε 2 , so x 6 leaves. Pivot on (6 , 2) : z x 1 + 6 x 4 + 3 x 6 = 3 ε 2 x 1 2 x 3 9 x 4 + x 5 + 9 x 6 = ε + 9 ε 2 1 3 x 1 + x 2 1 3 x 3 2 x 4 + x 6 = ε 2 ¯ c 1 is only positive reduced costs, so x 1 enters. min { ( ε + 9 ε 2 ) / 1 ,ε 2 / 1 3 } = 3 ε 2 , so x 2 leaves. Pivot on (2 , 1) : z + 3 x 2 x 3 + 6 x 6 = 6 ε 2 3 x 2 x 3 3 x 4 + x 5 + 6 x 6 = ε + 6 ε 2 x 1 + 3 x 2 x 3 6 x 4 + 3 x 6 = 3 ε 2 The perturbed problem is unbounded. Same pivots on original problem gives same conclusion. Chapter 8: Finite Termination 3 Theorem 8.3 (pg 111) (a) ( P ) is nondegenerate....
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This note was uploaded on 06/11/2011 for the course C 350 taught by Professor Wolkowicz during the Fall '97 term at Waterloo.
 Fall '97
 Wolkowicz

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