princeton univ. F 17 cos 521: Advanced Algorithm Design Lecture 7: LP Duality Lecturer: Matt Weinberg
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1 prnceton unv. F 17 cos 521: Advanced Algorthm Desgn Lecture 7: LP Dualty Lecturer: Matt Wenberg Scrbe: LP Dualty s an extremely useful tool for analyzng structural propertes of lnear programs. Whle there are ndeed applcatons of LP dualty to drectly desgn algorthms, t s often more useful to gan structural nsght (such as approxmaton guarantees, etc.). In ths lecture, we ll see statements of LP dualty. We ll practce applyng t n the homeworks. 1 Weak LP Dualty Consder a lnear program of the form: max c x A x b, x 0,. We ll call ths the prmal LP. x s called a prmal soluton, and our goal s to fnd a prmal soluton that maxmzes our obectve, subect to the feasblty constrants. On the other hand, nstead of thnkng about drectly searchng for good prmal solutons, we could alternatvely thnk about searchng for good upper bounds on how good a prmal can possbly be. Ths s called the dual problem. How can we derve an upper bound on how good a prmal can possbly be? Consder the followng: f we have weghts w 0 for each nequalty, and take a lnear combnaton of the feasblty constrants, we may drectly conclude that any feasble x must satsfy: w A x w b. Okay, so we can upper bound some lnear functon of any feasble x, so what? Well, f we happen to have chosen our w s so that w A = c for all, now we re n busness! We ll have drectly shown that c x w b. In fact, because x 0, even f we only have w A c we re n busness, as we d have: c x w A x w b. 1
2 2 Note that the frst nequalty s only true because x 0. So now we can thnk of the followng dual approach: search over all weghts w to fnd the ones that nduce the best upper bound. Note that our search s constraned to fnd weghts such that c w A, so ths tself s a lnear program: mn w b w A c, w 0,. Ths s called the dual LP. As an exercse, verfy that the dual of the dual LP s tself the prmal. Note that we have already proved that every feasble soluton of the dual provdes an upper bound on how good any prmal soluton can possbly be. Therefore, we have establshed what s called weak LP dualty: Theorem 1 (Weak LP Dualty) Let LP1 be any maxmzaton LP and LP2 be ts dual (a mnmzaton LP). Then f: The optmum of LP1 s unbounded (+ ), then the feasble regon of LP2 s empty. The optmum of LP1 fnte, t s less than or equal to the optmum of LP2, or the feasble regon of LP2 s empty. Proof: We have already proven the second bullet. To see the frst bullet, observe that f the feasble regon of LP2 s non-empty, then we have drectly found a fnte upper bound on LP1. So f LP1 s unbounded, LP2 must be empty. In fact, we wll see a stronger clam later. Weak Dualty s easy to prove, and t s good to remember ths ntuton. Strong Dualty (later) s good to know, but the ntuton s largely captured by the proof of Weak Dualty. 1.1 Complementary Slackness We ll also want to dscuss propertes of optmal prmal/dual pars. One useful property s called complementary slackness. A x and w are sad to satsfy complementary slackness f they satsfy condton 1) n the theorem statement below. Theorem 2 Consder a prmal LP, LP1 and ts dual LP, LP2, and feasble (not necessarly optmal) solutons x for the prmal and w for the dual. Then the followng are equvalent: 1. (w = 0 OR ( A x = b for all ) AND x = 0 OR ) A w = c for all. 2. c x = w b (and therefore both x s an optmal prmal and w s an optmal dual).
3 3 Proof: Note that we can wrte: c x w b ( A w ) x w b = ( ) w A x b. The nequalty s because w s a feasble soluton to LP2. The equalty s ust rearrangng the order of sums. Let s now analyze the RHS. Observe that A x b 0 for all as x s feasble for LP1. Observe also that w 0 for all, as w s feasble for LP2. So every term n the summand multples a non-negatve number by a non-postve number and s therefore non-postve. Ths means that the RHS s zero f and only f for all, w = 0 or A x b = 0. Now we turn our attenton to the nequalty. Note that because c A w for all, the nequalty s strct f and only f there exsts an for whch x > 0 and c < A w. So the LHS s equal to the mddle term f and only f for all, x = 0 or c = A w. Takng the two bold-font clams together, ths means that the LHS s equal to zero f and only f 1) holds. If 1) does not hold, then ether the RHS s < 0, or the LHS s less than the mddle term (whch s 0). Fnally, observe that 2) holds f and only f the LHS above s equal to zero. 2 Weak Partal Dualty We ll dscuss a slghtly more general dualty (t s not obvous that the prevous dualty s a specal case of ths, but t s a good exercse to show so). We ll agan only prove the weak case for now. Defnton 1 Consder an LP of the form: max c x A x b, x 0,. Then a Lagrangan relaxaton of the above LP for a subset S of constrants and Lagrangan multples λ 0 for all S s the followng (whch we ll refer to as LPS λ: max c x + ( λ b ) A x S A x b, / S x 0,.
4 4 Theorem 3 (Weak Partal Dualty ) For all S, λ, and any LP, the value of LPS λ upper bounds the value of LP. Proof: Let x optmze LP. Then because x s feasble for LP, t s also feasble for LPS λ (as the feasblty constrants n LPS λ are a proper subset of those n LP). Also, because x s feasble for LP, we have b A x 0 for all. As we also have λ 0, ths means that S λ (b A x ) 0. Ths drectly mples that x s feasble for LPS λ, and also that x acheves a greater obectve value when evaluated by LPS λ than LP. So every settng of λ agan nduces an upper bound on how good the soluton to LP can possbly be. We can also thnk about searchng for the best bound of ths form (for a fxed S). We ll agan call λ a canddate dual soluton snce t helps wtness an upper bound on how good a prmal soluton can be. The problem below can be wrtten as an LP n terms of the varables λ (by ntroducng a varable t constraned so that t c x + S λ (b A x ) and mnmzng t, we saw ths trck n Lecture 5 to mnmze an absolute value). We ll refer to the followng program as the partal Lagrangan w.r.t. S. mn {max c x + ( λ b ) A x {λ 0, S} S A x b, / S x 0, }. 2.1 Complementary Slackness There s a smlar defnton of Complementary Slackness for ths noton of dualty. Property 1) below captures ths defnton. Theorem 4 (Complementary Slackness for Partal Lagrangan) Let LP1 be a lnear program and LP2 ts Partal Lagrangan w.r.t. S. Let x be a canddate prmal soluton to LP1, and λ a canddate dual soluton LP2. Then the followng are equvalent: 1. For all S, λ = 0 OR A x = b, AND x = arg max x A x b / S,x 0 { c x + S λ (b A x )}. 2. c x = max x A x b / S,x 0 { c x + S λ (b A x )} (and therefore, x s optmal for LP1, and λ s optmal for LP2). Proof omtted, but smlar to that n Secton Strong Dualty (Proof adapted from Anupam Gupta s scrbed lecture notes here:
5 5 The prevous sectons dscussed weak dualty: usng dual solutons as upper bounds on how good a prmal soluton could be. In fact, somethng qute strong s true: there s always a dual wtnessng that the optmal prmal s optmal. We ll gve a proof, but note that most of the ntuton (asde from geometry/lnear algebra) s provded by Weak Dualty. We ll ust dscuss the classc case, the partal case s smlar and omtted. Theorem 5 (Strong LP Dualty) Let LP1 be any maxmzaton LP and LP2 be ts dual (a mnmzaton LP). Then: If the optmum of LP1 s unbounded (+ ), the feasble regon of LP2 s empty. If the feasble regon of LP1 s empty, the optmum of LP2 s ether unbounded ( ), or also nfeasble. If optmum of LP1 fnte, then the optmum of LP2 s also fnte, and they are equal. The key ngredent n the proof wll be what s called the Separatng Hyperplane Theorem. Theorem 6 (Separatng Hyperplane Theorem) Let P be a closed, convex regon n R n, and x be a pont not n P. Then there exsts a w such that x w > max y P { y w}. Proof: Consder the pont y P closest to x (that s, mnmzng x y 2 over all y P. As dstance s a postve contnuous functon, and P s a closed regon, such a y exsts. Now consder the vector w = x y. We clam that the chosen w s the desred wtness. Observe frst that ( x y) w = w 2 2 > 0, so ndeed x w > y w. We ust need to confrm that y = arg max z P { z w} and then we re done. Assume for contradcton that z w > y w and z P. Then as P s convex, z ε = (1 ε) y + ε z P as well for all ε > 0. Observe that x z ε 2 2 = x y + ε( y z) 2 2 = x y 2 2 2ε( x y) ( y z) + ε2 y z 2 2 = x y 2 2 2ε( w) ( y z) + ε2 y z 2 2. By hypothess, w ( y z) < 0, and y z 2 2 s fnte, so for suffcently small ε, we get x z ε 2 2 < x y 2 2, a contradcton. Now, consder the optmum x of LP1. Let S denote the for whch A x = b, and S the constrants for whch A x < b. We clam that c can be wrtten as a convex combnaton of the vectors A, S (up to possble scalng). Lemma 1 Let x be the optmum of LP1, and let S denote the for whch A x = b. Then there exst {λ 0} S such that c = S λ A for all. Proof: Assume for contradcton that ths were not the case. As the space X of all vectors for whch there exsts {λ 0} S such that c = S λ A for all s clearly closed and convex, we can apply the separatng hyperplane theorem. So there would exst some γ such that c γ > max y X { y γ}. Now consder the vector x + ε γ. We know that for all S, A γ = 0. So for all S, A (x + εγ ) = b. Moreover, for all / S, A x < b, and A γ s fnte. So there exsts a suffcently small ε so that x + ε γ s feasble for LP1.
6 6 Fnally, observe that max y X { y γ} 0, as 0 X. So c γ > 0, and we have shown that x was not optmal. Now wth the lemma n hand, we want to show a dual whose value matches c x. Let c = S λ A wth λ 0 as guaranteed by the lemma. Set w = λ for all S, and w = 0 for all / S. Frst, s t clear that w s feasble for LP2, as we have explctly set w so that c = w A for all. Now we ust need to evaluate ts value: b w = b w + b 0 = ( S / S S A x )w = So ts obectve value s exactly the same as LP1. A w x = S c x.
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