lecture05

lecture05 - 15.083J/6.859J Integer Optimization Lecture 5:...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 15.083J/6.859J Integer Optimization Lecture 5: Ideal formulations II 1 Outline Randomized rounding methods Slide 1 2 Randomized rounding Solve c x subject to x 2 P for arbitrary c. x be optimal solution. From x create a new random integer solution x, feasible in P : E c x] = ZLP = c x . ZLP ZIP E ZH ] = ZLP : Hence, P integral. 0 0 0 Slide 2 2.1 Minimum s ; t cutX minimize subject to f u vg2E xuv yu ; yv xuv yv ; yu ys = 1 yt = 0 yu xuv 2 f0 1g: cuv xuv Slide 3 fu vg 2 E fu vg 2 E 2.1.1 Algorithm Solve linear relaxation. Position the nodes in the interval (0 1) according to the value of yu . Generate a random variable U uniformly in the interval 0 1]. Round all nodes u with yu U to yu = 0, and all nodes u with yu > U to yu = 1. Set xuv = jyu ; yv j for all fu vg 2 E . Slide 4 2.2 Theorem For every nonnegative cost vector c, E ZH ] = ZIP = ZLP : Slide 5 1 U yu is rounded to 0 yv is rounded to 1 yt yu yv ys Proof: = = fu vg2E X = ZLP ZIP 2.3 Stable matching 2.3.1 Formulation w1 >m w2 if man m prefers w1 to w2 . m1 >w m2 if woman w prefers m1 to m2 . Decision variables xij = 1 0 if mi is matched to wj , otherwise. 2 n men fm1 : : : mn g and n women fw1 : : : wng, with each person having a list of strict preference order. Find a stable perfect matching M of the men to women: There does not exist a man m and a woman w who are not matched under M , but prefer each other to their assigned mates under M . fu vg2E cuv jyu ; yv j X ; cuv P min yu yv ZIP 2 3 X cuv xuv 5 E ZH ] = E 4 fu vg2E ; U < max yu yv Slide 6 Slide 7 N = f1 : : : ng n X j=1 n X xij = 1 i2N j2N xkj 1 i j2N i j 2 N: Slide 8 xij + X xij = 1 i=1 xij 2 f0 1g xik + X fkjwk <mi wj g fkjmk <wj mi g 2.3.2 Proposition x 2 PSM . If xij > 0, then X xij + fj k wk<mi wj g xik + X fj k mk <wj mi g xkj = 1: 2.3.3 Proof min Dual max s.t. x 2 PSM. nn XX i=1 j=1 xij Slide 9 s.t. n X i=1 i+ n X j=1 j; x 2 PSM nn XX i=1 j=1 ij i+ j; X Set ij i= 0: n X j=1 fkjwk mi wj g ik ; X fkjmk >wj mi g kj 1 i j2N ij j = n X i=1 ij and ij = xij for all i j 2 N: Dual: ij + X fkjwk <mi wj g ik + X fkjmk <wj mi g kj 1 8i j2N feasible if ij = xij and x 2 PSM . Objective n n nn nn X X XX XX xij : i+ j; ij = Complementary slackness of optimal primal and dual solutions. i=1 j=1 i=1 j=1 i=1 j=1 3 mi X fj wj 2.4 Key Theorem PSM = conv(S ): 2.4.1 Randomization The matching is stable: wk who is preferred by mi to his mate wj under the assignment, i.e., the interval spanned by xik is on the right of the interval spanned by xij in the row corresponding to mi , is assigned a mate whom she strictly prefers to mi , since in the row corresponding to wk the random number U lies strictly to the left of the interval xik . xU = 1 if mi and wj are matched. ij R1 u x du: x can be written as a convex combination of stable matchings 0 ij xu as u varies over the interval 0 1]. E xU ] = P(U lies in the interval spanned by xij ) = xij : ij xij = 4 Generate a random number U uniformly in 0,1]. Match mi to wj if xij > 0 and in the row corresponding to mi , U lies in the interval spanned by xij in 0 1]. Accordingly, match wj to mi if in the row corresponding to wj , U lies in the interval spanned by xij in 0 1]. Key property: xij > 0, then the intervals spanned by xij in rows corresponding to mi and wj coincide in 0 1]. 0 1 11 00 k wk<mi wj g xik xij fj xkj k mk >wj mi g X 11 00 11 00 Slide 10 Slide 11 Slide 12 ...
View Full Document

This note was uploaded on 10/13/2010 for the course CS 15.083J taught by Professor Dimitrisbertsimas during the Spring '04 term at MIT.

Ask a homework question - tutors are online