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### IEOR3608Fall06-degeneracy1

Course: IE 3608, Fall 2002
School: Columbia
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Word Count: 316

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Degenerate A LP Denition: An LP is degenerate if in a basic feasible solution, one of the basic variables takes on a zero value. Degeneracy is a problem in practice, because it makes the simplex algorithm slower. Original LP maximize subject to x1 + x2 + x3 x1 + x2 x2 + x3 x 1 , x2 , (1) 8 0 0. (2) (3) (4) Standard form. z= x1 + x2 + x3 s1 = 8 x 1 x 2 s2 = x2 + x3 (5) (6) (7) Iteration 1 z= x1 + x2 +...

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Degenerate A LP Denition: An LP is degenerate if in a basic feasible solution, one of the basic variables takes on a zero value. Degeneracy is a problem in practice, because it makes the simplex algorithm slower. Original LP maximize subject to x1 + x2 + x3 x1 + x2 x2 + x3 x 1 , x2 , (1) 8 0 0. (2) (3) (4) Standard form. z= x1 + x2 + x3 s1 = 8 x 1 x 2 s2 = x2 + x3 (5) (6) (7) Iteration 1 z= x1 + x2 + x3 s1 = 8 x 1 x 2 s2 = x2 + x3 (8) (9) (10) Note that one of the basic variables is 0. We choose x1 as the entering variable and s1 as the leaving variable. z=8 + x 3 s1 x1 = 8 x2 s1 s2 = x2 x3 (11) (12) (13) Note again that one of the basic variables is 0. The previous pivot did increase the objective function value from 0 to 8 though. Iteration 2 z=8 + x 3 s1 x1 = 8 x2 s1 s2 = x2 x3 (14) (15) (16) We now choose x3 as the entering variable, and s2 as the leaving variable. These were our only choices. z = 8 + x s1 2 s2 x 1 = 8 x 2 s1 x3 = x2 s2 (17) (18) (19) Note that the objective function did not increase. This occurs because of degeneracy. Iteration 3 z = 8 + x 2 s1 s2 x 1 = 8 x 2 s1 x3 = x2 s2 (20) (21) (22) We now choose x2 as the entering variable and x1 as the leaving variable. z = 16 x1 2s1 s2 x 2 = 8 x 1 s1 x 3 = 8 x 1 s1 s2 (23) (24) (25) Since all coecients of variables in the objective function are negative, we now have the optimal solution, (x1, x2, x3, s1, s2) = (0, 8, 8, 0, 0) with objective value 16. Notice that in the nal solution, the basic variables are all nonzero. In a degenerate LP, it is also possible that even in the nal solution, some of the basic variables will be zero. One other thing to note is that x1 was an entering variable in one iteration, and a leaving variable in another. In general, a variable can be an entering and leave the basic many times in the course of the simplex algorithm.
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