Unformatted text preview: 15.083J/6.859J Integer Optimization Lecture 4: Ideal formulations I 1 Outline
Total unimodularity Dual Methods Slide 1 2 Total unimodularity n S = fx 2 Z+ j Ax bg, A 2 Z m�n and b 2 Z m . P = fx 2 <n j Ax bg. + When P = conv(S ) for all integral vectors b? Slide 2 2.1 Cramer's rule 2.2 De nition A 2 Z m�n of full row rank is unimodular if the determinant of each basis of A is 1, or 1. A matrix A 2 Z m�m of full row rank is unimodular if det(A) = 1: A matrix A 2 Z m�n is totally unimodular if the determinant of each square submatrix of A is 0, 1, or 1. 2.3 Examples
21 1 61 0 A=6 0 1 6 4 11 "1 1 0#
101 is TU. 1 0 1 1 0 0 1 0 1 3 02 1 7 7 B6 7 5 is not TU: det @4 1 0 10 01 11 31 7C = ;2 5A : A 2 <n�n nonsingular. et( Ax = b () x = A;1b () 8 i : xi = ddet(A )) : A Ai: Aij = Aj for all j 2 f1 : : : ng n fig and Aii = b. Slide 3 i Slide 4 Slide 5 2.4 Proposition A is TU if and only if A I ] is unimodular. 2 A3 67 A is TU if and only if 6 A 7 is TU. 4 I5 I A is TU if and only if A0 is TU. Slide 6 2.5 Theorem 2.5.1 Proof Conversely, P (b) integral for all b 2 Z m . B f1 : : : ng with AB nonsingular. b = AB z + ei , where z integral: z + A;1 ei 0 for all i. B A;1 b = z + A;1 ei 2 Z m for all i. B B ;1 ith column of A B is integral for all i. A;1 is an integer matrix, and thus, since AB is also an integer matrix, and B det(AB )det(A;1 ) = 1, we obtain that det(AB ) =1 or 1. B For second part: A is TU if and only if A I ] is unimodular. For any b 2 Z m the extreme points of fx 2 <n j Ax bg are integral if and only if the extreme + points of f(x y) 2 <n+m j Ax + I y = bg are integral. + 2.6 Corollary 2 Let A be an integral matrix. A is TU if and only if fx j Ax = b 0 x ug is integral for all integral vectors b and u. A is TU if and only if fx j a Ax b l x ug is integral for all integral vectors a b l u. Assume that A is unimodular. b 2 Z m and P (b) = . 6 x = (xB xN ) extreme point of P (b), xB = A;1 b and xN = 0. B Since A unimodular det(AB ) = 1. By Cramer's rule and the integrality of AB and b, xB is integral. P (b) is integral. P Aninteger matrix of full row rank. A is unimodular if and only if (b) = fx 2 6 <+ j Ax = bg is integral for all b 2 Z m for which (b) = . A integer matrix. A is TU if and only if (b) = fx 2 <n j Ax bg is integral + 6 for all b 2 Z m for which (b) = .
P P P Slide 7 Slide 8 Slide 9 2.7 Theorems 2.8 Corollary The following matrices are TU: The nodearc incidence matrix of a directed graph. The nodeedge incidence matrix of an undirected bipartite graph. A matrix of zeroone elements, in which each column has its ones consecutively. 2.9 Example 21 A = 6 ;0 41 0 1 ;1 00 ;1 0 0 ;1 10 01 0 0 1 ;1 0 0 ;1 1 3 7 5 2 1 4 1 2 4 3 5 6 3 3 into two parts so that the sum of the columns in one part minus the sum of the columns in the other part is a vector with entries 0, +1, and 1. A is TU if and only if each collection Q of rows of A can be partitioned into two parts so that the sum of the rows in one part minus the sum of the rows in the other part is a vector with entries only 0, +1, and 1. A is TU if and only if each collection J of columns of A can be partitioned Slide 10 Slide 11 Slide 12 2.10 Implications Following problems can be solves as LOs: Network ows Matching in biparite graphs Stable set in biparite graphs. Slide 13 3 Dual methods
Z Slide 14 LP = min Let P be a nonempty polyhedron with at least one extreme point. The polyhedron P is integral if and only if ZLP is integer for all c 2 Z n . For converse, assume x 2 P , extreme point with xj fractional. c 2 Z n : x unique optimum. There exist a 2 Z : x optimum for c = c + (1=a)ej . ac0 x ; ac0 x = xj , either 0 0 ac x or ac x is fractional. Contradiction. c0 x st x2
:: P Construct a solution to the dual of the LP relaxation and an integer solution, feasible to IO with ZH = ZD . Since ZD ZLP ZIP ZH if ZH = ZD , ZLP = ZIP . 3.1 Key idea 3.2 Submodular functions
f Slide 15 : 2N 7! <+ is submodular if
fS Slide 16
T ( ) + f (T ) ( ) + f (T ) fS ( \ T ) + f (S ) )
T: 8 ST N: f : 2N 7! <+ is supermodular if
fS fS ( \ T ) + f (S T 8 ST N: It is nondecreasing, if
fS () fT () 8 S 4 3.3 Polynatroids
maximize subject to n X X xj j 2S xj 2 Z+ P (f ) = x 2 <n + X
j 2S If the function f is submodular, nondecreasing, integer valued, and f ( ) = 0, then P (f ) = conv(F ) F set of feasible integer solutions. 3.3.1 Theorem 3.4 Proof
maximize subject to dual n X X X j=1 xj j2S xj 0 minimize subject to SN S X fSy ( )S
y fSjj2Sg
y S 0 x is integer, x j 0 5 S x y 8 j; < S= : 0k
c c j= 0 fS ( j ) ; f (S j;1 )
c j+1 for 1 j k for j > k: for S = S j 1 j < k for S = S k otherwise : c 1 c2 0= . c k > 0 c k+1 ::: cx jj j =1 cj xj Slide 17 f (S ) xj f (S ) SN j 2 N:
8 S N: Slide 18 Slide 19 fS () S N: j 2N c j j 2N
N: S j = f1 cn . S Slide 20
j ::: g for j 2 N , and X
j2T x j= X
fj j j2T j kg ; ; fS = f (S k \ T ) ; f ( ) f (T ) ; f ( ) = f (T ): fj j j2T j kg y is dual feasible because X y S 0 and + ySk = cj
c if j k fS j j2Sg y S = ySj +
y X fS j j2Sg S =0 Primal objective value: Dual objective value:
j=1 Pk j j;1 j=1 cj f (S ) ; f (S ) k X; j=1
c ; k;1 X (cj ; cj+1 )f (S j ) + ck f (S k ) = 3.5 Matroids (N I ) independence system, r(T ) = maxfjS j : X maximize cj xj subject to
xj r (S ) j2S xj 2 f0 1g: j2N X 8 S 3.6 Greedy algorithm 6 n 1. Given a matroid (N I ), and weights cj for j 2 N , sort all elements of N
cj . Let J = k = 1. in decreasing order of cj : cj1 cj2 2. For k = 1 : : : m, if J fjk g is an independent set, let J = J fjk g 3. An optimum solution is given by the set J . Theorem: (N I ) independence system. It is a matroid if and only if its rank function r(S ) = maxfjS j : S 2 I S T g is submodular. X ( j ) ; f (S j;1 ) fS ( j \ T ) ; f (S j;1 \ T ) j if j > k : j f (S j ) ; f (S j;1 ) : S 2I S T g: Slide 21 N Slide 22 ...
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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.
 Spring '04
 DimitrisBertsimas

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