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Unformatted text preview: Lecture 13 Implications of Farkas Lemma UCSD Math 171A: Numerical Optimization Philip E. Gill http://ccom.ucsd.edu/~peg/math171 Monday, February 9th, 2009 Recap: Optimality conditions for LP LP minimize x c T x subject to Ax b with A an m n matrix, b an mvector and c an nvector. The vector x * is a solution of LP if and only if: (a) Ax * b (b) c = A T a * a for some vector * a 0, where A a is the active constraint matrix at x * UCSD Center for Computational Mathematics Slide 2/35, Monday, February 9th, 2009 Example: minimize 2 x 1 + x 2 subject to the constraints: constraint #1: x 1 + x 2 1 constraint #2: x 2 constraint #3: x 1 Written in the form min c T x subject to Ax b , we get c = 2 1 ! , A = 1 1 1 1 , b = 1 UCSD Center for Computational Mathematics Slide 3/35, Monday, February 9th, 2009 x 1 x 2 a 1 a 3 a 2 a 1 c c x * x c a 1 c a 1 a 3 a 2 x 1 x 2 x * x At the point x * = 1 ! , the active set is A = { 1 , 3 } with A a = 1 1 1 ! , b a = 1 ! Solving for the Lagrange multipliers gives A T a a = c 1 1 1 ! a = 2 1 ! a = 1 1 ! x * is optimal. UCSD Center for Computational Mathematics Slide 6/35, Monday, February 9th, 2009 Recap: Farkas Lemma Result (Farkas Lemma) (A) c T p 0 for all p such that A a p if and only if (B) c = A T a * a for some * a UCSD Center for Computational Mathematics Slide 7/35, Monday, February 9th, 2009 In the proof we defined a feasible descent direction p such that A a p = e s where s is an index such that ( * a ) s < The direction p satisfies: a T j p = ( 0 for j 6 = s , with a T j the j th row of A a 1 for j = s , with a T s the s th row of A a A step x + p keeps all active constraint residuals fixed at zero except the s th, which increases....
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 Winter '08
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