# hw3solns - 15-750 Graduate Algorithms Spring 2009 Miller...

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15-750 — Graduate Algorithms — Spring 2009 Miller and Dinitz and Tangwongsan Assignment 3 Solutions 1 Duality Theory (25 pts.) Given a linear program min { c > x : A x b , x 0 } and a solution x 0 , how can one decide whether x 0 is an optimal solution? More generally, how can one calculate a good lower bound on such linear programs? 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 duality theory first-hand in this problem. More concretely, suppose we are given a linear program (which we will call the primal LP) minimize c > x , subject to A x b and x 0 . The following LP is called the dual linear program: maximize y > b , subject to A > y c and y 0 . Several useful properties about the dual program can be derived. For our discussion, define P = { x R d : A x b and x 0 } and D = { A > y c and y 0 } . (a) Weak duality. Show that if x 0 P and y 0 D , then y > 0 b c > x 0 . 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 as weak duality. Theorem: (Weak duality) If the primal is a minimization linear program with op- timum value z , then it has a dual, which is a maximization problem with optimum value w and w z . Solution: We will be working with two types of inequalities. To tell them apart, we will write cw for the component-wise version and for the scalar one. Let x 0 P and y 0 D . We begin by observing that since 0 cw y 0 and b cw A x 0 (because x 0 is feasble), we know that y > 0 b y > 0 ( A x 0 ) . By a similar reasoning, we have ( A > y 0 ) > x 0 c > x 0 . Therefore, we establish y > 0 b y > 0 ( A x 0 ) = ( A > y 0 ) > x 0 c > x 0 , 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 are feasible (i.e., P 6 = ), then min { c > x : x P } = max { y > b : y D } . In this part, you will prove the strong duality theorem. For simplicity, assume that P and D are bounded. You can use the following lemma without providing a proof:

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15-750 HW 3 Solutions 2 Farkas’s Lemma: Let A R m × n , and b R m . Exactly one of the following holds: 1. There exists x R n such that A x b . 2. There exists y R m 0 such that A > y = 0 and b > y < 0. Solution: Let z = min { c > x : x P } and w = max { y > b : y D } . Assume without loss of generality that A x b already contains the constraint x cw 0 (otherwise, construct a new LP with this constraint added in). We know from weak duality that w z . To show that w = z , it remains to show that w z . Indeed, we are looking for a vector y such that A > y cw c and b > y z . We will show that such a vector exists.
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