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Linear Programming Degeneracy, Lecture Notes - Mathematics, Study notes of Linear Programming

Linear Programming Degeneracy, Lecture Notes - Mathematics - Prof. J Vanderbei.pdf, Prof. J Vanderbei, Mathematics, Linear Programming, Degeneracy, Cycling, Pivot Rules, Perturbation Method

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2010/2011

Uploaded on 11/02/2011

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Download Linear Programming Degeneracy, Lecture Notes - Mathematics and more Study notes Linear Programming in PDF only on Docsity! Linear Programming: Chapter 3 Degeneracy Robert J. Vanderbei October 17, 2007 Operations Research and Financial Engineering Princeton University Princeton, NJ 08544 http://www.princeton.edu/∼rvdb Degeneracy Definitions. A dictionary is degenerate if one or more “rhs”-value vanishes. Example: ζ = 6 + w3 + 5x2 + 4w1 x3 = 1 − 2w3 − 2x2 + 3w1 w2 = 4 + w3 + x2 − 3w1 x1 = 3 − 2w3 w4 = 2 + w3 − w1 w5 = 0 − x2 + w1 A pivot is degenerate if the objective function value does not change. Examples (based on above dictionary): 1. If x2 enters, then w5 must leave, pivot is degenerate. 2. If w1 enters, then w2 must leave, pivot is not degenerate. wi w3 x1 ¥3 x x2 w3 x3 x2 wa 0.0 -6.75 0.75 0.25 -13.25 2.75 0.25 -2.75 x3 x4 w3 wi x4 wa wl w3 Ei * x2 Bi x2 AEE Perturbation Method Whenever a vanishing “rhs” appears perturb it. If there are lots of them, say k, perturb them all. Make the perturbations at different scales: other nonzero data  1  2  · · ·  k > 0. An Example. Entering variable: x2 Leaving variable: w2 Theoretical Results Cycling Theorem. If the simplex method fails to terminate, then it must cycle. Why? Fundamental Theorem of Linear Programming. For an arbitrary linear program in standard form, the following statements are true: 1. If there is no optimal solution, then the problem is either infeasible or un- bounded. 2. If a feasible solution exists, then a basic feasible solution exists. 3. If an optimal solution exists, then a basic optimal solution exists. Geometry maximize x1 + 2x2 + 3x3 subject to x1 + 2x3 ≤ 3 x2 + 2x3 ≤ 2 x1, x2, x3 ≥ 0. 1 2 x2 x3 x1 x1+2x3=3 x2+2x3= x2=0 maximize x1 + 2x2 + 3x3 subject to x1 + 2x3 ≤ 2 x2 + 2x3 ≤ 2 x1, x2, x3 ≥ 0. 1 x2 x3 x1 x1+2x3=2 x2=0 x2+2x3=2
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