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Lecture 8 - Linear Programming with Equality Constraints

Lecture 8 - Linear Programming with Equality Constraints -...

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Lecture 8 Linear Programming with Equality Constraints UCSD Math 171A: Numerical Optimization Philip E. Gill ( [email protected] ) Friday, January 23, 2009 Linear programming First, we consider a special class of LP’s with equality constraints ELP minimize x R n c T x subject to Ax = b , The feasible region is IF = { x R n : Ax = b } UCSD Center for Computational Mathematics Slide 2/29, Friday, January 23, 2009 Reasonable assumptions on the problem: A1. A has full rank A2. b range( A ) (i.e., IF is not empty). Definition x * is a minimizer of ELP if c T x * c T x for all x IF = { x R n : Ax = b } (i.e., ( x * ) ( x ) for all feasible x ). UCSD Center for Computational Mathematics Slide 3/29, Friday, January 23, 2009 An obvious case: If m n , then any x * such that Ax * = b is the unique minimizer of ( x ) because IF contains only one point. Without loss of generality, we assume that m < n , i.e., A = IF contains infinitely many points for all vectors b . UCSD Center for Computational Mathematics Slide 4/29, Friday, January 23, 2009
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The standard strategy Given a non-optimal point ¯ x , find a direction p such that positive steps α along p give points ¯ x + α p that are feasible and give x + α p ) < ‘ x ) The direction p is known as a feasible direction . A precise definition follows: UCSD Center for Computational Mathematics Slide 5/29, Friday, January 23, 2009 Definition Given a feasible ¯ x , p is a feasible direction if p 6 = 0 and there exists a step σ > 0 such that ¯ x + α p is feasible for all α such that 0 < α σ (i.e., positive steps that cannot be longer than σ .) The step σ is called the maximum feasible step in the direction p . UCSD Center for Computational Mathematics Slide 6/29, Friday, January 23, 2009 ¯ x p ¯ x + σp feasible points How can we characterize the set of all such feasible directions? UCSD Center for Computational Mathematics Slide 7/29, Friday, January 23, 2009 If ¯ x is feasible, then A ¯ x = b If ¯ x + α p is feasible for all α > 0, then A x + α p ) = b A ¯ x + α Ap = b A ¯ x - b + α Ap = 0 α Ap = 0 As α is a positive step, it must hold that Ap = 0 p null( A ) If p null( A
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