tut5 - Degeneracy in Linear Programming Sorry, Tim. But the...

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1 Degeneracy in Linear Programming Tim, the turkey I heard that today’s tutorial is all about Ellen DeGeneres Ellen DeGeneres Sorry, Tim. But the topic is just as interesting. It’s about degeneracy in Linear Programming. Degeneracy? Students at MIT shouldn’t learn about degeneracy. And I heard that 15.053 students have already studied convicts’ sex. Reverend Jerry Falwell Actually, they studied convex sets.
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2 What is degeneracy? As you know, the simplex algorithm starts at a corner point and moves to an adjacent corner point by increasing the value of a non-basic variable x s with a negative value in the z-row (objective function). Typically, the entering variable x s does increase in value, and the objective value z improves. But it is possible that that x s does not increase at all. It will happen when one of the RHS coefficients is 0. In this case, the objective value and solution does not change, but there is an exiting variable. This situation is called degeneracy. x 1 -3 3 1 0 -4 2 0 1 = = 2 6 x 2 x 4 x 3 3 -2 0 0 z 0 0 1 = 2 0 A basic feasible solution is called degenerate if one of its RHS coefficients (excluding the objective value) is 0. This bfs is degenerate.
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3 Nooz, the most trusted name in fox. Great. I now know what degeneracy is. Now we can move on to other matters. Ollie, the computationally wise owl. No, Nooz**. This tutorial has many more slides. Degeneracy adds complications to the simplex algorithm. And if you understand what occurs under degeneracy, you really understand what is going on with the simplex algorithm. ** As you know, “No, Nooz” is good news.” Cleaver Professor Orlin apologizes for this bad pun, but feels that he could not resist the temptation.
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4 Degeneracy and Basic Feasible Solutions We may think that every two distinct bases lead to two different solutions. This would be true if there was no degeneracy. But with degeneracy, we can have two different bases, and the same feasible solution. We now pivot on the “2” in Constraint 2 and obtain a second tableau. x 1 -3 3 1 0 -4 2 0 1 = = 2 6 x 2 x 4 x 3 3 -2 0 0 z 0 0 1 = 2 0 x 1 -3 3 1 0 -1 2 0 1 = = 2 6 x 2 x 4 x 3 3 -2 0 0 z 0 0 1 = 2 0 2 2 0 1 -1/2 1 1/2 -3/2 0 -3/2 Both tableus correspond to the same feasible solution with z = 2, x 1 = x 2 = x 4 = 0; x 3 = 6. But the basic variables and the coefficients of the two tableaus are different.
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This note was uploaded on 01/18/2012 for the course MGMT 15.053 taught by Professor Jamesorli during the Spring '07 term at MIT.

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tut5 - Degeneracy in Linear Programming Sorry, Tim. But the...

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