15.083J/6.859J Integer Optimization. Lecture 3: Methods to enhance formulations

Transcription

1 15.083J/6.859J Iteger Optimizatio Lecture 3: Methods to ehace formulatios

2 1 Outlie Polyhedral review Slide 1 Methods to geerate valid iequalities Methods to geerate facet defiig iequalities Polyhedral review.1 Dimesio of polyhedra Defiitio: The vectors x 1,...,x k R uique solutio of the liear system are affiely idepedet if the Slide k a i x i = 0, i=1 k a i =0, i=1 is a i =0 for all i =1,...,k. Propositio: The vectors x 1,...,x k R are affiely idepedet if ad oly if the vectors x x 1,...,x k x 1 are liearly idepedet. Slide 3 Defiitio: Let P = {x R Ax b}. The, the polyhedro P has dimesio k, deoted dim(p )= k, if the maximum umber of affiely idepedet poits i P is k +1. P = {(x 1,x ) x 1 x =0, 0 x 1 1, 0 x 1}, dim(p )=?. Valid Iequalities..1 Defiitios a x b is called a valid iequality for a set P if it is satisfied by all poits i P. Slide 4 Let f x g be a valid iequality for a polyhedro P,ad let F = {x P f x = g}. The, F is called a face of P ad we say that f x g represets F. A face is called proper if F =Ø,P. Aface F of P represeted by the iequality f x g, is called a facet of P if dim(f )= dim(p ) 1. We say that the iequality f x g is facet defiig. 1

3 x 3 cov(s) x 1.. Theorem For each facet F of P, at least oe of the iequalities represetig F is ecessary i ay descriptio of P. Slide 5 Every iequality represetig a face of P of dimesio less tha dim(p ) 1 is ot ecessary i the descriptio of P, ad ca be dropped...3 Example S = {(x 1,x ) Z x 1 3,x 1 1, x 1 +x 4, x 1 + x 8, x 1 +x 3}. Facets for cov(s): x 1 3, x 1 1, x 1 +x 3, x 1 + x 5, x 1 + x 1. Faces of dimesio oe: x 1 +x 4, ad x 1 + x 8. Slide 6 3 Methods to geerate valid iequalities 3.1 Roudig Choose u = (u 1,...,u m) 0; Multiply ith costrait with u i ad sum: (u A j )x j u b. Slide 7 Sice u A j u A j ad x j 0: ( ) u A j x j u b. ( ) As x Z + : u A j x j u b.

4 3.1.1 Matchig { } Slide 8 S = x {0, 1} E xe 1, i V. e δ({i}) U V, U = k +1. For each i U, multiply add: 1 1 x e + x e U. e E(U) Sice x e 0, x e 1 U. e E(U) Roud to ( U is odd) e δ(u) 1 U 1 x e U =, e E(U) e δ({i}) xe 1 by 1/, ad 3. Superadditivity 3..1 Defiitio A fuctio F : D R Ris superadditive if for a 1, a D,: a 1 + a D : Slide 9 F (a 1 )+ F (a ) F (a 1 + a ), It is odecreasig if F (a 1 ) F (a ), if a 1 a for all a 1, a D. 3.. Theorem Slide 10 If F : R m R is superadditive ad odecreasig with F (0) = 0, { } F (A j )x j F (b) is valid for the set S = x Z + Ax b Proof By iductio o x j,we show: F (A j )x j F (A j x j ). For x j = 0 it is clearly true. Assumig it is true for x j = k 1, the Slide 11 F (A j )k = F (A j )+ F (A j )(k 1) F (A j )+ F (A j (k 1)) F (A j + A j (k 1)), by superadditivity, ad the iductio is complete. Slide 1 Therefore, F (A j )x j F (A j x j ). 3

5 By superadditivity, F (A j x j ) F A j x j = F (Ax). Sice Ax b ad F is odecreasig F (Ax) F (b). 3.3 Modular arithmetic { } S = x Z + a j x j = a 0, d Z +.We write a j = b j + u j d, where b j (0 b j <d, b j Z +). The, b j x j = b 0 + rd, for some iteger r. Sice bj xj 0ad b0 <d,weobtai r 0. The, bj xj b0 is valid for S Examples For d = 1, ad a j are ot itegers. I this case, sice x 0, we obtai a j x j a 0. Sice x Z, a j x j a 0, ad thus the followig iequality is valid for S (a j a j )x j a 0 a 0. If the iequality b is valid for S a j x 1 R j +, ad the iequality c j x j d is valid for S R +, the the iequality 4 S = {x Z + 7x 1 +17x 64x 3 + x 4 = 03}. For d = 13, iequality x 1 +4x + x 3 + x 4 8 is valid for S. 3.4 Disjuctios Propositio Slide 13 Slide 14 Slide 15 mi(a j,c j )x j max(b, d) is valid for S 1 S. 4

6 3.4. Theorem If the iequality aj xj d(x k α) b is valid for S for some d 0, ad the iequality aj xj + c(x k α 1) b is valid for S for some c 0, the the iequality aj xj b is valid for S. Example: I previous example, we write x 1 +x 4ad x 1 1 as follows: Slide 16 α =, x 1 + x 1 is valid. ( x 1 + x )+(x 3) 1, ( x 1 + x ) (x ) Mixed iteger roudig Propositio For v R, f(v) = v v, v + =max {0,v}. Slide 17 X = {(x, y) Z R + x y b} y 3 5 x y = 3 x y = x The iequality x 1 1 f (b) y b is valid for cov(x) Proof P 1 = X {(x, y) x b }, P = X {(x, y) x b +1}. Add 1 f(b) times the iequality x b 0ad 0 y: (x b ) (1 f(b)) y is valid for P 1. For P we combie (x b ) 1ad x y b with multipliers f(b) ad 1: (x b )(1 f(b)) y. By disjuctio, (x b )(1 f(b)) y is valid for cov(p 1 P )=cov(x). Slide 18 5

7 3.5.3 Theorem { } Slide 19 S = x Z + A j x j b, j =1,...,. For every u Q+ m the ( iequality u A j + [f(u Aj ) f ( u ) b)] + x 1 f ( u b j u b is valid for cov(s). ) 4 Facets 4.1 By the defiitio Stable set max w i x i i V s. t. x i + x j 1, {i, j} E, x i 1, x i {0, 1}, i V. A collectio of odes U, such that for all i, j U, {i, j} E is called a clique. i U for ay clique U ( ) Coversely, sice U is ot maximal, there is a ode i / U such that U {i} is a clique, ad thus xj 1 ( ) is valid for cov(s). Sice iequality (*) is the sum of j U {i} x i 0 ad iequality (**), the (*) is ot facet defiig. Slide 0 is valid. Slide 1 A clique U is maximal if for all i V \ U, U {i} is ot a clique. (*) is facet defiig if ad oly if U is a maximal clique. U = {1,...,k}. The, e i, i = 1,...,k satisfy (*) with equality. For each i U, there is aode j = r(i) U, such that (i, r(i)) E. x i with x i i =1, x i r(i) = 1, ad zero elsewhere is i S, ad satisfies iequality (*) with equality. e 1,...,e k, x k+1,...,x are liearly idepedet; hece, affiely idepedet. 4. Liftig Slide Slide 3 6 Slide 4 max i=1 x 1 + x + x 3 1 x 1 + x 3 + x 4 1 x 1 + x 4 + x 5 1 x 1 + x 5 + x 6 1 x 1 + x + x 6 1. x i uique optimal solutio x 0 =(1/)(0, 1, 1, 1, 1, 1). Do maximal clique iequalities describe covex hull? Slide 5 6

8 x 0 does ot satisfy x + x 3 + x 4 + x 5 + x 6. (1) Stable sets {, 4}, {, 5}, {3, 5}, {3, 6}, {4, 6} satisfy it with equality. Not facet, sice there are o other stable sets that satisfy (1) with equality. { } (1) is facet defiig for S x {0, 1} 6 x 1 =0. Cosider ax 1 + x + x 3 + x 4 + x 5 + x 6, a>0. Select a i order for (1) to be still valid, ad to defie a facet for S. For x 1 = 0, (1) is valid for all a. If x 1 =1, a x x 3 x 4 x 5 x 6.Sice x 1 = 1 implies x = = x 6 =0, the a. Therefore, if 0 a, (1) is valid. For a =, {, 4}, {, 5}, {3, 5}, {3, 6}, {4, 6}, ad {1} satisfy it with equality. x 1 + x + x 3 + x 4 + x 5 + x 6, is valid ad defies a facet cov(s). Slide Geeral priciple { } Slide 7 Suppose S {0, 1},S i = S x {0, 1} x1 = i, i =0, 1, ad a j= j x j a 0 () is valid for S 0. If S 1 =Ø, the x 1 0 is valid for S. j= If S 1 =Ø, the a1x 1 + aj xj a0 (3) is valid for S for ay a1 a 0 Z, Z = j= aj xj s.t. x S1. If a 1 = a 0 Z ad () defies a face of dimesio k of cov(s 0 ), the (3) gives a face of dimesio k +1 of cov(s). 4.. Geometry Slide 8 7

9 x 3 a 1 x 1 + a x + a 3 x 3 a 0 a x + a 3 x 3 a 0 x x Order of liftig P =cov {x {0, 1} 6 5x 1 +5x +5x 3 +5x 4 +3x 5 +8x 6 17}. Slide 9 x 1 + x + x 3 + x 4 3 is valid for P {x 5 = x 6 =0}. Liftig o x 5 ad the o x 6, yields x 1 + x + x 3 + x 4 + x 5 + x 6 3. Liftig o x 6 ad the o x 5, yields x 1 + x + x 3 + x 4 +x

10 15.083J/6.859J Iteger Optimizatio Lecture 13: Lattices I

11 1 Outlie Iteger poits i lattices. Slide 1 Is {x Z Ax = b} oempty? Iteger poits i lattices B = [b 1,...,b d ] R d, b 1,...,b d are liearly idepedet. Slide L = L(B) = {y R y = Bv, v Z d } is called the lattice geerated by B. B is called a basis of L(B). b i = e i, i =1,..., e i is the i-th uit vector, the L(e 1,...,e )= Z. x, y L(B) ad λ, μ Z, λx + μy L(B)..1 Multiple bases b 1 =(1, ), b =(, 1), b 3 =(1, 1). The, L(b 1, b )= L(b, b 3 ). Slide 3 x b b b x

12 . Alterative bases Let B = [b 1,...,b d ] be a basis of the lattice L. If U R d d is uimodular, the B = BU is a basis of the lattice L. If B ad B are bases of L, the there exists a uimodular matrix U such that B = BU. If B ad B are bases of L, the det(b) = det(b)..3 Proof For all x L: x = Bv with v Z d. det(u )= ±1, ad det(u 1 )=1/ det(u) = ±1. x = BUU 1 v. From Cramer s rule, U 1 has itegral coordiates, ad thus w = U 1 v is itegral. B = BU. The, x = Bw,with w Z d, which implies that B is a basis of L. B = [b 1,...,b d ]ad B = [b 1,...,b d ] be bases of L. The, the vectors b 1,...,b d 1 d ad the vectors b,...,b are both liearly idepedet. V = {By y R } = {By y R }. There exists a ivertible d d matrix U such that Slide 4 Slide 5 B = BU ad B = BU 1. b i = BU i, U d 1, U d i Z ad b i = BU 1 Z. i U ad U 1 are both itegral, ad thus both det(u) ad det(u 1 ) are itegral, leadig to det(u )= ±1. det(b) = det(b) det(u ) = det(b). i.4 Covex Body Theorem Let L be a lattice i R ad let A R be a covex set such that vol(a) > det(l)ad A is symmetric aroud the origi, i.e., z A if ad oly if z A. The A cotais a o-zero lattice poit..5 Iteger ormal form A Z m of full row rak is i iteger ormal form, if it is of the form [B, 0], where B Z m m is ivertible, has itegral elemets ad is lower triagular. Elemetary operatios: (a) Exchagig two colums; (b) Multiplyig a colum by 1. (c) Addig a itegral multiple of oe colum to aother. Theorem: (a) A full row rak A Z m ca be brought ito the iteger ormal form [B, 0] usig elemetary colum operatios; (b) There is a uimodular matrix U such that [B, 0] = AU. Slide 6 Slide 7

13 .6 Proof We show by iductio that by applyig elemetary colum operatios (a)-(c), we ca trasform A to [ ] α 0 v C, (1) m 1 where α Z + \{0}, v Z ad C Z (m 1) ( 1) is of full row rak. By proceedig iductively o the matrix C we prove part (a). By iteratively exchagig two colums of A (Operatio (a)) ad possibly multi plyig colums by 1 (Operatio(b)), we catrasform A (ad reumber the colum idices) such that Slide 8 a 1,1 a 1,... a 1, 0. Sice A is of full row rak, a 1,1 > 0. Let k = max{i : a 1,i > 0}. If k = 1, the we have trasformed A ito a matrix of the form (1). Otherwise, k ad by applyig k 1 operatios (c) we trasform A to [ ] a 1,1 a 1,k 1 A = A 1 A,...,A k 1 A k, A k, A k+1,...,a. a 1, a 1,k Repeat the process to A, ad exchage two colums of A such that max{i : a 1,i > 0} k a 1,1 a 1,... a 1, 0. k k 1 k a 1,i (a 1,i a 1,i+1)+ a 1,k = a 1,1 < a 1,i, i=1 i=1 i=1 which implies that after a fiite umber of iteratios A is trasformed by elemetary colum operatios (a)-(c) ito a matrix of the form (1). Each of the elemetary colum operatios correspods to multiplyig matrix A by a uimodular matrix as follows: (i) Exchagig colums k ad j of matrix A correspods to multiplyig matrix A by a uimodular matrix U 1 = I + I k,j + I j,k I k,k I j,j. det(u 1)= 1. (ii) Multiplyig colum j by 1 correspods to multiplyig matrix A by a uimodular matrix U = I I j,j, that is a idetity matrix except that elemet (j, j) is 1. det(u )= 1. (iii) Addig f Z times colum k to colum j, correspods to multiplyig matrix A by a uimodular matrix U 3 = I + fi k,j. Sice det(u 3)= 1, U 3 is uimodular. Performig two elemetary colum operatios correspods to multiplyig the correspodig uimodular matrices resultig i aother uimodular matrix. 3

14 .7 Example [ ] [ ] [ ] Slide 9 Reorderig the colums [ ] Replacig colums oe ad two by the differece of the first ad twice the secod colum ad the secod ad third colum, respectively, yields [ ] Reorderig the colums [ ] Cotiuig with the matrix C =[19, 5], we obtai successively, the matrices [19, 5], [4, 5], [5, 4], [1, 4], [4, 1], [0, 1], ad [1, 0]. The iteger ormal form is: [ ] Characterizatio A Z m,fullrow rak;[b, 0] = AU. Let b Z m ad S = {x Z Ax = b}. m (a) The set S is oempty if ad oly if B 1 b Z. (b) If S, every solutio of S is of the form Slide 10 where U 1, U : U =[U 1, U ]. x = U 1B 1 b + U z, z Z m, (c) L = {x Z Ax = 0} is a lattice ad the colum vectors of U costitute a basis of L..9 Proof y = U 1 x. Sice U is uimodular, y Z if ad oly if x Z. Thus, S is oempty if ad oly if there exists a y Z such that [B, 0]y = b. Sice B is m ivertible, the latter is true if ad oly B 1 b Z. Slide 11 4

15 We ca express the set S as follows: S = {x Z Ax = b} = {x Z x = Uy, [B, 0]y = b, y Z } = {x Z x = U 1w + U z, Bw = b, w Z m, z Z m }. m Thus, if S, the B 1 b Z from part (a) ad hece, S = {x Z x = U 1B 1 b + U z, z Z m }. Let L = {x Z Ax = 0}. By settig b = 0 i part (b) we obtai that L = {x Z x = U z, z Z m }. Thus, by defiitio, L is a lattice with basis U..10 Example Is S = {x Z 3 Ax = b} is oempty [ ] [ ] A = ad b = Slide 1 Iteger ormal form: [B, 0] = AU, with [ ] [ ] [B, 0] = ad U = Note that det(u )= 1. Sice B 1 b = (3, 13) Z, S =. All iteger solutios of S are give by [ ] [ ] [ ] [ ] z 3 x = z = z, z Z z.11 Itegral Farkas lemma Let A Z m, b Z m ad S = {x Z Ax = b}. The set S = if ad oly if there exists a y Q m, such that y A Z ad y b Z. / The set S = if ad oly if there exists a y Q m, such that y 0, m y A Z ad y b Z /..1 Proof m Assume that S =. If there exists y Q m, such that y A Z ad y b Z, / the y Ax = y b with y Ax Z ad y b Z. / Coversely, if S =, the by previous theorem, u = B 1 b / Z, that is there exists a i such that u i Z /. Takig y to be the ith row of B 1 proves the theorem. m m Slide 13 Slide 14 5

16 .13 Reformulatios max c x, x S = {x Z + Ax = b}. [B, 0] = AU. There exists x 0 Z : Ax 0 = b iff B 1 b / Z. Let Let U be a basis of L, i.e., x S x = x 0 + y : Ay = 0, x 0 y. L = {y Z Ay = 0}. m Slide 15 L = {y Z y = U z, z Z m }. max c U z s.t U z x 0 z Z m. Differet bases give rise to alterative reformulatios max c Bz s.t. Bz x 0 z Z m. 6

17 MIT OpeCourseWare J / 6.859J Iteger Programmig ad Combiatorial Optimizatio Fall 009 For iformatio about citig these materials or our Terms of Use, visit: