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

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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 + w 3 + 5 x 2 + 4 w 1 x 3 = 1 − 2 w 3 − 2 x 2 + 3 w 1 w 2 = 4 + w 3 + x 2 − 3 w 1 x 1 = 3 − 2 w 3 w 4 = 2 + w 3 − w 1 w 5 = 0 − x 2 + w 1

A pivot is degenerate if the objective function value does not change.

Examples (based on above dictionary):

1. If x 2 enters, then w 5 must leave, pivot is degenerate.
2. If w 1 enters, then w 2 must leave, pivot is not degenerate.

Hope Fades

An example that cycles using the following pivot rules:

• entering variable: largest-coefficient rule.
• leaving variable: smallest-index rule.

ζ = 10 x 1 − 57 x 2 − 9 x 3 − 24 x 4 w 1 = − 0. 5 x 1 + 5. 5 x 2 + 2. 5 x 3 − 9 x 4 w 2 = − 0. 5 x 1 + 1. 5 x 2 + 0. 5 x 3 − x 4 w 3 = 1 − x 1.

Here’s a demo of cycling...

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: x 2 Leaving variable: w 2

Perturbation Method—Example Con’t.

Recall current dictionary:

Entering variable: x 1 Leaving variable: w 3

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 x 1 + 2 x 2 + 3 x
• subject to x 1 + 2 x 3 ≤ - x 2 + 2 x 3 ≤ - x 1 , x 2 , x 3 ≥ - 1 2 x
  • x - x - x 1 +2 x 3 =
    • x 2 =0 x 2 +2 x 3 = - maximize x 1 + 2 x 2 + 3 x - subject to x 1 + 2 x 3 ≤ - x 2 + 2 x 3 ≤ - x 1 , x 2 , x 3 ≥ - 1 x - x - x 1 x^1 +2 x^3 = - x 2 = - x 2 +2 x 3 =