Linear Programming Lecture Notes - Penn State Personal Web Server
Linear Programming Lecture Notes - Penn State Personal Web Server
Linear Programming Lecture Notes - Penn State Personal Web Server
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CHAPTER 7<br />
Degeneracy and Convergence<br />
In this section, we will consider the problem of degeneracy and prove (at last) that there<br />
is an implementation of the Simplex Algorithm that is guaranteed to converge to an optimal<br />
solution, assuming one exists.<br />
1. Degeneracy Revisited<br />
We’ve already discussed degeneracy. Recall the following theorem from Chapter 5 that<br />
defines degeneracy in terms of the simplex tableau:<br />
Theorem 5.16. Consider Problem P (our linear programming problem). Let B ∈ R m×m be<br />
a basis matrix corresponding to some set of basic variables xB. Let b = B −1 b. If bj = 0 for<br />
some j = 1, . . . , m, then xB = b and xN = 0 is a degenerate extreme point of the feasible<br />
region of Problem P .<br />
We have seen in Example 5.15 that degeneracy can cause us to take extra steps on<br />
our way from an initial basic feasible solution to an optimal solution. When the simplex<br />
algorithm takes extra steps while remaining at the same degenerate extreme point, this is<br />
called stalling. The problem can become much worse; for certain entering variable rules,<br />
the simplex algorithm can become locked in a cycle of pivots each one moving from one<br />
characterization of a degenerate extreme point to the next. The following example from<br />
Beale and illustrated in Chapter 4 of [BJS04] demonstrates the point.<br />
Example 7.1. Consider the following linear programming problem:<br />
min −<br />
(7.1)<br />
3<br />
4 x4 + 20x5 − 1<br />
2 x6 + 6x7<br />
s.t x1 + 1<br />
4 x4 − 8x5 − x6 + 9x7 = 0<br />
x2 + 1<br />
2 x4 − 12x5 − 1<br />
2 x6 + 3x7 = 0<br />
x3 + x6 = 1<br />
xi ≥ 0 i = 1, . . . , 7<br />
It is conducive to analyze the A matrix of the constraints of this problem. We have:<br />
⎡<br />
⎤<br />
1 0 0 1/4 −8 −1 9<br />
(7.2) A = ⎣0 1 0 1/2 −12 −1/2 3⎦<br />
0 0 1 0 0 1 0<br />
The fact that the A matrix contains an identity matrix embedded within it suggests that<br />
an initial basic feasible solution with basic variables x1, x2 and x3 would be a good choice.<br />
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