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mthsc810-lecture14-2x2

# mthsc810-lecture14-2x2 - Farkas lemma MthSc 810...

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MthSc 810: Mathematical Programming Lecture 14 Pietro Belotti Dept. of Mathematical Sciences Clemson University October 25, 2011 Reading for today: Section 4.6 and 4.7 Reading for Oct. 27: Section 4.8-4.10 Midterm 2: Next Tuesday, 2pm–3:15pm Farkas’ lemma Lemma Consider A R m × n and b R m . Then (1) x 0 such that A x = b , or (2) p such that p A 0 and p b < 0, but not both. Therefore, either a (non-negative) solution to a system of equations, or a solution to a system of inequalities (one of which strict). Proof (easy) We need to prove that ( 1 ) ⇒ ¬ ( 2 ) and ¬ ( 1 ) ( 2 ) . ( 1 ) ⇒ ¬ ( 2 ) : If x 0 , A x = b , and p A 0 , then p b = p A 0 x 0 , a contradiction. ¬ ( 1 ) ( 2 ) : Consider the primal-dual pair P : max 0 x D : min p b s . t . A x = b s . t . p A 0 x 0 ¬ ( 1 ) ( x 0 : A x = b ) P is infeasible. p = 0 is feasible for D , which must then be unbounded There exists a feasible p such that p b < 0 More on Farkas’ lemma Corollary 4.3 Consider A R m × n and b R m . If, for any p

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