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Question: i will rate the answer asap as the answer seems...

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Here is a variant of Farkas lemma: 3x : Ax<b + Ay20,y A-0,y b<0. In this exercise we will use this Farkas lemma to show that every bounded LP has an optimal solution. Let P be the polyhedron that is the feasible region of the LP. Since the LP is bounded, there is some δ such that δ sup {cTx : x E P). To argue that there is an optimal solution x to the LP, we will show that there is an x P such that CTX 〉 δ. Then it will follow that CTx-δ and this x would be the optimal solution of the LP. Use the above variant of Farkas lemma to argue that there is an x : Ax b and CTX > .

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