Lecture10

# Lecture10 - C&O 355 Mathematical Programming Fall 2010...

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Unformatted text preview: C&O 355 Mathematical Programming Fall 2010 Lecture 10 N. Harvey • How should we define corner points? • Under any reasonable definition, point x should be considered a corner point x What is a corner point? • Attempt #1: “x is the ‘farthest point’ in some direction” • Let P = { feasible region } • There exists c 2 R n s.t. c T x>c T y for all y 2 P n {x} • “For some objective function, x is the unique optimal point when maximizing over P” • Such a point x is called a “ vertex ” c x is unique optimal point What is a corner point? • Attempt #2: “There is no feasible line-segment that goes through x in both directions” • Whenever x= ® y+(1- ® )z with y,z ≠ x and ® 2 (0,1), then either y or z must be infeasible. • “If you write x as a convex combination of two feasible points y and z, the only possibility is x=y=z” • Such a point x is called an “ extreme point ” y z (infeasible) x What is a corner point? • Attempt #3: “x lies on the boundary of many constraints” x lies on boundary of two constraints x 4x 1- x 2 · 10 x 1 + 6x 2 · 15 What is a corner point? • Attempt #3: “x lies on the boundary of many constraints” • What if I introduce redundant constraints? y also lies on boundary of two constraints y Not the right condition x 1 + 6x 2 · 15 2x 1 + 12x 2 · 30 What is a corner point? • Revised Attempt #3: “x lies on the boundary of many linearly independent constraints” • Feasible region: P = { x : a i T x · b i 8 i } ½ R n • Let I x ={ i : a i T x=b i } and A x ={ a i : i 2I x }. (“ Tight constraints ”) • x is a “ basic feasible solution (BFS) ” if rank A x = n y x 1 + 6x 2 · 15 2x 1 + 12x 2 · 30 x y’s constraints are linearly dependent 4x 1- x 2 · 10 x’s constraints are linearly independent x 1 + 6x 2 · 15 What is a corner point? Proof of (i) ) (ii): x is a vertex ) 9 c s.t. x is unique maximizer of c T x over P Suppose x = ® y + (1- ® )z where y,z 2 P and ® 2 (0,1). Suppose y ≠ x. Then c T x = ® c T y + (1- ® ) c T z ) c T x < ® c T x + (1- ® ) c T x = c T x Contradiction! So y=x. Symmetrically, z=x. So x is an extreme point of P. ¥ · c T x (since c T x is optimal value) < c T x (since x is unique optimizer) Lemma : Let P be a polyhedron. The following are equivalent. i. x is a vertex (unique maximizer) ii. x is an extreme point (not convex combination of other points) iii. x is a basic feasible solution (BFS) (tight constraints have rank n) Proof Idea of (ii) ) (iii): x not a BFS ) rank A x · n-1 Lemma : Let P={ x : a i T x · b i 8 i } ½ R n . The following are equivalent....
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## This note was uploaded on 06/16/2011 for the course CO 355 taught by Professor Harvey during the Winter '10 term at Waterloo.

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Lecture10 - C&O 355 Mathematical Programming Fall 2010...

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