lecture 11 - (This is relevant to HW5 Q3...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 482 (Lecture 11): Duality IV: Farkas' lemma This lecture discusses Farkas Lemma, which we introduced in the in class exercises to lecture 10. In class exercises for lecture 10 Proof of Farkas lemma During the above proof of Farkas Lemma, the following was used: In class exercise: Prove that if (LP) is min c'x subject to Ax=b, x≥ 0 (notice the "=") Then the dual (DLP) is max y'b subject to y'A ≥0 (and y is unconstrained) Brief solution notes: Take the (LP) and put it in canonical form, where you know the dual, using 2n dual variables (y1',y2',. ..,yn', y1'', y2'',. ..,yn''). Now you want to compress this to just n variables using y1=y1'-y1'', y2=y2'-y2'', etc. (Why can you do this?) In class exercise:
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (This is relevant to HW5 Q3 "Gordan's theorem"). The hint to the problem asks that you show Ax<0 is solvable iff Ax 1 is solvable. One direction is clear (which one?). Show the other one. Solution notes: The LP max z=0 s.t. Ax<=-1 has a feasible solution iff Ax<0 is solvable. Of course, if the LP is solvable, Ax<0 is solvable. Now, suppose Ax<0 holds for some X. We can use this fact to show the LP is feasible. Find some positive numbers a1,a2,. ..,an such that AX<=(-a1,. ..,-an)^T, now rescale to find Y such that AY<=-1. To finish the homework problem itself: The proof has a similar form to the Farkas' lemma proof we just gave. Consider the Dual LP to the LP above....
View Full Document

This note was uploaded on 02/19/2012 for the course MATH 482 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

Ask a homework question - tutors are online