# lect13 - -100 1 10 6 15 1-1 1 7 1-1-1 5 1 1 Lets now change...

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ISE 536–Fall03: Linear Programming and Extensions October 15, 2003 Lecture 13: Dual Simplex, Farkas Lemma Lecturer: Fernando Ord´ o˜nez 1 Dual Simplex Usual Simplex Maintain a BFS and aim for optimality (i.e. dual feasibility 0 c t - c t B B - 1 A = c t - y t A ). This motivates the converse: Maintain dual feasibility (i.e. optimality of primal) and aim for primal feasibility The tableau would maintain all ¯ c j 0, but potentially some b i < 0. Dual Simplex Algorithm 1. Select row r such that ( x B ) r = u r 0 < 0. The r -th basic variable leaves the basis. 2. Find u 0 s u rs = max j : u rj < 0 u 0 j u rj . The s variable enters the basis. 3. We pivot on u rs to update the tableau. Example Consider our inventory problem from last class, and its optimal tableau:

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Unformatted text preview: -100 1 10 6 15 1-1 1 7 1-1-1 5 1 1 Lets now change the warehouse capacity to 4. What is the new optimal solution? 1 How about if we the second period demand is changed to d 2 = 3? 2 Using LP duality as a proof method How can we identify that P = { x | Ax = b, x ≥ } is infeasible? Is there a way to do it without an algorithm? Geometrically , how can we describe the infeasibility of P ? Theorem (Farkas Lemma) Ax = b, x ≥ 0 is infeasible iﬀ y t A ≥ , y t b < 0 is feasible. proof: 2...
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