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....
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 Spring '08
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 Math, Logic, Englishlanguage films, Farkas

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