CO350 Linear Programming Chapter 8: Degeneracy and Finite Termination

Transcription

1 CO350 Linear Programming Chapter 8: Degeneracy and Finite Termination 22th June 2005

2 Chapter 8: Finite Termination Recap On Monday, we established In the absence of degeneracy, the simplex method will terminate after a nite number of iterations. The following observations. Iteration degenerate =) old tableau degenerate. The converse is not true. 2. Iteration degenerate =) new tableau degenerate. The converse is not true. 3. More than one choice of leaving variable =) new basis degenerate. The converse is not true. 4. Iteration is degenerate () basic solution remains the same.

3 Chapter 8: Finite Termination 2 Example of degeneracy (pg 05) x 2 6 x A (; 2) AU (0; ) x 0 - x + x 2 Feasible region H HHHH (0; 0) CO C H (; 0) HHHH H H x + x 2 = z x 2x 2-6 x x 2 0 LP problem: max z = x + x 2 s.t. x + x 2 x 2x 2 x x ; x 2 0

4 Chapter 8: Finite Termination 3 Geometry of degeneracy (bottom of pg 06) From the example: degeneracy in 2-dimension is represented by having more than two lines intersecting at an extreme point. =) there is a redundant constraint. In general: degeneracy in d-dimension is represented by having more than d hyperplanes intersecting at an extreme point. However, this does not necessarily mean that there is a redundant constraint. Example degenerate b.f.s

5 Chapter 8: Finite Termination 4 Cycling In some extreme cases, degeneracy may stall the simplex method indenitely: We may visit the same sequence of bases over and over again. This phenomenon is called cycling. Example of cycling (pg 07) Using largest coecient rule for entering variable. Initial tableau: z 2x 3x 2 + x 3 + 2x 4 = 0 Pivot on (6; 2): 2x 9x 2 + x 3 + 9x 4 + x 5 = 0 x 3 + x 2 x 3 3 2x 4 + x 6 = 0 z x + 6x 4 + 3x 6 = 0 Pivot on (5; ): x 2x 3 9x 4 + x 5 + 9x 6 = 0 x 3 + x 2 x 3 3 2x 4 + x 6 = 0 z 2x 3 3x 4 + x 5 + 2x 6 = 0 x 2x 3 9x 4 + x 5 + 9x 6 = 0 x x 3 + x 4 3 x 5 2x 6 = 0

6 Chapter 8: Finite Termination 5 Example of cycling (cont'd) Pivot on (2; 4): z + 3x 2 x 3 + 6x 6 = 0 Pivot on (; 3): x + 9x 2 + x 3 2x 5 9x 6 = 0 x x 3 + x 4 3 x 5 2x 6 = 0 z + x + 2x 2 2x 5 3x 6 = 0 Pivot on (4; 6): x + 9x 2 + x 3 2x 5 9x 6 = 0 3 x 2x 2 + x x 5 + x 6 = 0 z + 6x 2 + 3x 4 x 5 = 0 Pivot on (3; 5): 2x 9x 2 + x 3 + 9x 4 + x 5 = 0 3 x 2x 2 + x x 5 + x 6 = 0 z 2x 3x 2 + x 3 + 2x 4 = 0 2x 9x 2 + x 3 + 9x 4 + x 5 = 0 3 x + x 2 3 x 3 2x 4 + x 6 = 0 This is the same as the initial tableau!

7 Chapter 8: Finite Termination 6 The Perturbation Method (x8.3) In this chapter, we consider the LP problem (P ) max c T x s.t. Ax = b x 0 and assume that we are given a feasible basis B. Motivation Degeneracy occurs because too many hyperplanes intersect at an extreme point. Solution: nudge each hyperplane slightly so that such intersection disappear. More precisely, we change b to b 0 = b + " b 2 + " 2. b m + " m where " is a very small but positive real number, and consider the LP problem (P 0 ) max c T x s.t. Ax = b 0 x 0

8 Chapter 8: Finite Termination 7 We need to ensure that. the new LP problem (P 0 ) is feasible, and 2. the new LP problem (P 0 ) is nondegenerate.. (P 0 ) is feasible If A B = I, then the b.f.s. of fax = b 0 ; x 0g determined by B is x B = b 0, x N = 0, which is feasible. If A B 6= I, then we rst change the equality constraints of (P ) to (A B A)x = A B b Feasible region remains unchanged. The new equality constraints b Ax = b b satises b A B = I. Then we change b b to b b 0 by adding "; " 2 ; : : : ; " m. In both cases, (P 0 ) is feasible.

9 Chapter 8: Finite Termination 8 Before showing that (P 0 ) is nondegenerate, consider Question: How can we apply the simplex method without knowing the value "? Answer: " only appears on the r.h.s. of each tableau =) it only aects the choice of leaving variable =) we only need to compare numbers of the form d 0 + d " + d 2 " d m " m Good news: Such numbers are comparable if " is small enough. Lemma 8.2 (pg 0). Consider the numbers and = 0 + " + 2 " m " m = 0 + " + 2 " m " m For all " suciently small and positive,. = if and only if i = i for 0 i m. 2. > if and only if k > k where k is the least integer i for which i 6= i. 3. < if and only if k < k where k is the least integer i for which i 6= i.

10 Chapter 8: Finite Termination 9 Example (Not in notes) Arrange the following numbers in ascending order, assuming that " is very small and positive. p = 3" 3" 2 p 2 = 3" + 0" 2 p 3 = 00" + 5" 2 p 4 = 2 + 7" 2 p 5 = 2 7" Comparing constants: p ; p 2 ; p 3 < p 4 ; p 5. Comparing " terms: p 5 < p 4 and p 3 < p ; p 2. Comparing " 2 terms: p < p 2. Answer: p 3 < p < p 2 < p 5 < p 4 Lexicographic ordering: Suppose we represent p by (0; 3; 3) and p 2 by (0; 3; 0). Then p < p 2 corresponds to the fact that (0; 3; 3) is lexicographically less than (0; 3; 0). We write (0; 3; 3) L < (0; 3; 0) This is also precisely how we order words in dictionaries: cap cat go golf to

11 Chapter 8: Finite Termination 0 We need to ensure that. the new LP problem (P 0 ) is feasible, and 2. the new LP problem (P 0 ) is nondegenerate. 2. (P 0 ) is nondegenerate Theorem 8.3 (pg ) (a) (P 0 ) is nondegenerate. (b) B is a feasible basis of (P 0 ) =) B is a feasible basis of (P ). (c) B is an optimal basis of (P 0 ) =) B is an optimal basis of (P ). (d) x k can enter and x r can leave in tableau for (P 0 ) corresponding to B =) same for tableau for (P ) corresponding to B. (e) Tableau for (P 0 ) corresponding to B detects unboundedness =) same for tableau for (P ) corresponding to B. Corollary 8.3 (pg 2) The simplex method applied to the perturbed problem (P 0 ) starting from a feasible basis B with A B = I will terminate after a nite number of iterations. Moreover, B 0 optimal for (P 0 ) =) B 0 optimal for (P ), and (P 0 ) unbounded =) (P ) unbounded.