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Unformatted text preview: 15750 — Graduate Algorithms — Spring 2009Miller and Dinitz and TangwongsanAssignment 3 Solutions1Duality Theory(25 pts.)Given a linear program min{c>x:Ax≥b,x≥}and a solutionx, how can one decide whetherxis an optimal solution? More generally, how can one calculate agoodlower bound on such linearprograms? Duality is a key concept in linear programming that can help answer these questions.Since we did not have time in class to cover such an important topic, you will learn about dualitytheory firsthand in this problem.More concretely, suppose we are given a linear program (which we will call the primal LP)minimizec>x,subject toAx≥bandx≥.The following LP is called theduallinear program:maximizey>b,subject toA>y≤candy≥.Several useful properties about the dual program can be derived. For our discussion, defineP={x∈Rd:Ax≥bandx≥}andD={A>y≤candy≥}.(a)Weak duality.Show that ifx∈Pandy∈D, theny>b≤c>x.Note that the dual program therefore gives a lower bound for a minimization LP. In fact,what you have just proven here implies the following theorem, known asweak duality.Theorem:(Weak duality) If the primal is a minimization linear program with optimum valuez, then it has a dual, which is a maximization problem with optimumvaluewandw≤z.Solution:We will be working with two types of inequalities. To tell them apart, we will write≤cwfor the componentwise version and≤for the scalar one. Letx∈Pandy∈D. Webegin by observing that since≤cwyandb≤cwAx(becausexis feasble), we know thaty>b≤y>(Ax). By a similar reasoning, we have(A>y)>x≤c>x. Therefore, we establishy>b≤y>(Ax) = (A>y)>x≤c>x,where the equality follows from the matrix identity(AB)>=B>A>. This concludes the proof.(b)Strong duality.The strong duality theorem states that if both the primal and the dual arefeasible (i.e.,P6=∅), thenmin{c>x:x∈P}= max{y>b:y∈D}.In this part, you will prove the strong duality theorem. For simplicity, assume thatPandDare bounded. You can use the following lemma without providing a proof:15750HW 3 Solutions2Farkas’s Lemma:LetA∈Rm×n, andb∈Rm. Exactly one of the followingholds:1. There existsx∈Rnsuch thatAx≤b.2. There existsy∈Rm≥such thatA>y=andb>y<0.Solution:Letz= min{c>x:x∈P}andw= max{y>b:y∈D}. Assume without loss ofgenerality thatAx≤balready contains the constraintx≥cw(otherwise, construct a new LPwith this constraint added in). We know from weak duality thatw≤z. To show thatw=z, itremains to show thatw≥z. Indeed, we are looking for a vectorysuch thatA>y≤cwcandb>y≥z. We will show that such a vector exists.Suppose for a contradiction that no suchyexists. Use Farkas’s lemma withA=A>b>andb=cz....
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This document was uploaded on 11/03/2009.
 Spring '09
 Algorithms

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