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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 : AK b + Zy 0, y A-0, y l b < 0. In this exercise we will use this Farkas lemma to show that every feasible bounded LP has an optimal solution. Let P = {IE Rn : Ax-b} 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€ P). To argue that there is optimal solution X to the LP. we will show that there is an x E 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 c | x > δ Hints: l. Suppose Ar : Ax < b, and cTx 〉 δ. Then the above version of the Farkas lemma says that there is a y such that something happens. Write down what this is. 2. Argue that the component of y that multiples cT can be assumed to be l 3. Derive a contradiction to one of the duality theorems.I will rate the answer asap as the answer seems correct. Please follow the hints. Thx

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