Lecture 19 - Solving LPs in Standard Form

# Lecture 19 - Solving LPs in Standard Form - Recap: Linear...

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Lecture 19 Solving LPs in Standard Form UCSD Math 171A: Numerical Optimization Philip E. Gill http://ccom.ucsd.edu/~peg/math171 Wednesday, February 25th, 2009 Recap: Linear programs in standard form minimize x R n c T x subject to Ax = b , x 0 The working set for the mixed-constraint simplex method has the form: A k = A I k ! all rows of A n - m rows of In and b k = b 0 ! all elements of b n - m components of 0 n UCSD Center for Computational Mathematics Slide 2/31, Wednesday, February 25th, 2009 Recap: Linear programs in standard form We can deﬁne a column permutation matrix P such that A I k ! P = A k P = B N 0 I n - m ! where: B is the matrix of basic columns (nonsingular) N is the matrix of nonbasic columns The permutation moves the basic variables to the front. UCSD Center for Computational Mathematics Slide 3/31, Wednesday, February 25th, 2009 The column permutation applied to a row vector produces a row vector with basic part ﬁrst. c T P = ( c T B c T N ) P T c = ± c B c N ² x T P = ( x T B x T N ) P T x = ± x B x N ² p T P = ( p T B p T N ) P T p = ± p B p N ² UCSD Center for Computational Mathematics Slide 4/31, Wednesday, February 25th, 2009

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AP = ( B N ) and I k P = ( 0 I n - m ) 4 = ( 0 I N ) so that ± A I k ² P = ± B N 0 I N ² UCSD Center for Computational Mathematics Slide 5/31, Wednesday, February 25th, 2009 Result P is an orthogonal matrix , i.e., P T P = I = PP T and det( P ) = ± 1. Applying the column permutation P T to ( c T B c T N ) returns c T : ( c T B c T N ) = c T P ( c T B c T N ) P T = c T PP T = c T Also P ± c B c N ² = PP T c = c UCSD Center for Computational Mathematics Slide 6/31, Wednesday, February 25th, 2009 The vertex x k deﬁned by the simplex working set satisﬁes A k x k = b k A k PP T x k = b k ( A k P )( P T x k ) = b k ± B N 0 I N ²± x B x N ² = ± b 0 ² Bx B + Nx N = b x N = 0 Bx B = b Note: to avoid clutter, we don’t put a suﬃx on B or N . UCSD Center for Computational Mathematics Slide 7/31, Wednesday, February 25th, 2009 Result The matrix A k is nonsingular if and only if B is nonsingular. Proof: From the deﬁnition of A k : det( A k ) = ± det( A k ) det( P ) = ± det( A k P ) = ± det ± B N 0 I N ² = ± det( B ) det( I N ) = ± det( B ) . UCSD Center for Computational Mathematics
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## This note was uploaded on 10/23/2010 for the course MATH 171a taught by Professor Staff during the Winter '08 term at UCSD.

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Lecture 19 - Solving LPs in Standard Form - Recap: Linear...

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