15.083J/6.859J Integer Optimization Lecture 4: Methods to enhance formulations II1 Outline Slide 1 • Independence set systems and Matroids • Strength of valid inequalities • Nonlinear formulations 2 Independence set systems 2.1 Definition Slide 2 • N finite set, I collection of subsets of N . • (N, I)is an independence system if: (a) ∅∈I; (b) if A ⊆ B and B ∈I,then A ∈I. • Combinatorial structures that exhibit hereditary properties 2.2 Examples Slide 3 • Node disjoint paths; G =(V,E), I1 collection of node disjoint paths in G. (E,I1)is IS. Why? • Acyclic subgraphs. I2 collection of acyclic subgraphs (forests) in G =(V,E). (E,I2)is IS. Why? • Linear independence; A matrix; N index set of columns of A; I3 collection of linearly independent columns of A.(N, I3)is IS. Why? • Feasible solutions to pac king problems. S = {x ∈{0, 1}n | Ax ≤ b}, A ≥ 0, N = {1, 2,...,n}.For x ∈ S, A(x)= {i | xi =1}. I4 = ∪x∈S A(x). (N, I4)is IS. Why? 2.3 Rank Slide 4 • (N, I) independence system • An independent set of maximal cardinality contained in T ⊆ N is called abasis of T . The maximum cardinality of a basis of T , denoted by r(T ), is called the rank of T . • S ⊆ T ; |A| = r(T ).A ∩ S and A ∩ (T \ S) are independent sets contained in S and T \ S • r(S)+ r(T \ S) ≥|A ∩ S| + |A ∩ (T \ S)| = |A| = r(T ). 1 2.4 Matroids Slide 5 • (N, I) is a matroid if: Every maximal independent set contained in F has the same cardinality r(F ) for all F ⊂ N. • (E, I1) (node disjoint paths in G). Is (E, I1)a matroid? • F = {(1, 2), (2, 3), (2, 4), (4, 5), (4, 6)}. Maximal indep endent sets in F : {(1, 2), (2, 4), (4, 5)} and {(1, 2), (2, 3), (4, 5), (4, 6)}. • Is (E,I2) of forests a matroid? • (N, I3) of linearly independent columns of A is a matroid. T ⊂ N index of columns of A , AT =[Aj ]j∈T . r(T )=rank(AT ). • Is (N, I4) of feasible solutions to packing problems a matroid? 2.5 Valid Inequalities Slide 6 • C ⊆ N a circuit in (N, I). maximize c x subject to xi ≤|C|−1 for all C ∈C i∈C x ∈{0, 1}n . • Rank inequality i∈T xi ≤ r(T ) • BW contains conditions for rank inequalities to be facet defining. For matroids, rank inequalities completely characterize convex hull. 3 Strength of valid inequalities Slide 7 • S set of integer feasible vectors. n • Pi = {x ∈+ | Aix ≥ bi},i =1, 2, Ai, bi ≥ 0; covering type polyhedra. • The strength of P1 with respect to P2 denoted by t(P1,P2) is the minimum value of α> 0 such that αP1 ⊂ P2. • P1 = {x ∈R|x ≥ 0}, P2 = {x ∈R|x ≥ 1}.Strength ? 3.1 Characterization 3.1.1 Theorem Slide 8 • αP1 ⊂ P2 if and only if for all c ≥ 0, Z2 ≤ αZ1, where Zi =min c x: x ∈ Pi. • Proof If αP1 ⊂ P2,then Z2 ≤ αZ1 for all c ≥ 0. • For converse, assume Z2 ≤ αZ1 , for all c ≥ 0,and there exists x0 ∈ αP1, but x0 ∈/ P2. • By the separating hyperplane theorem, there exists c: c x0 < c x for all x ∈ P2, i.e., c x0 <Z2. • x0 ∈ αP1, x0 = αy0, y0 ∈ P1. Z1 ≤ c y0, i.e., αZ1 ≤ c x0, and thus αZ1 <Z2. Contradiction. Z2• t(P1,P2)= supc≥0 Z1 . 2  3.1.2 Computation Slide 9 Pi = { x ∈n | ai� x ≥ bi,i=1,...,m},and ai ≥ 0,bi ≥ 0 for all i =1,...,m.+ Then, t(P1,P2)= max bi , i=1,...,m di where di =min ai� x: x ∈ P1.(If di =0, then t(P1,P2) is defined to be +∞. 3.2 Strength of an inequality Slide 10 • The strength off x ≥ g, f ≥ 0,g> 0with respect to P = {x ∈n | Ax ≥ b} of covering type is defined as g/d,where d =minx∈P f x. + • By strong duality, d =max b p s.t. A p ≤ f p ≥ 0. • p feasible dual solution. b p ≤ d. Then, the strength of inequality f x ≥ g with respect to P is at most g/(b p). 4 Nonlinear formulations Slide 11 n ZIP =min cj xj j=1 n s.t. Aj xj = b j=1 xj ∈{0, 1}. 4.1 SDP relaxation Slide 12 n • Multiply each constraint by xi: j=1 Aj xj xi = bxi. • Introduce zij = xixj . zii = xi 2 = xi ∀ i =1,...,n. xixj ≥ 0 ⇐⇒ zij ≥ 0 ∀ i, j; i = j. xi(1 − xj ) ≥ 0 ⇐⇒ zij ≤ zii ∀ i, j, i = j. (1 − xi)(1 − xj ) ≥ 0 ⇐⇒ zii + zjj − zij ≤ 1 ∀ i, j, i = j. • Matrix Z = xx is positive semidefinite, Z  0, i.e., for u ∈n , u Zu = ||u x||2 ≥ 0. 3       4.2 SDP relaxation Slide 13 n ZSD =min cj zjj j=1 n s.t. Aj zij − bzii = 0, i =1,...,n, j=1 n Aj zjj = b, j=1 0 ≤ zij ≤ zii, i,j =1 ,...,n,i = j, 0 ≤ zij ≤ zjj , i,j =1,...,n,i = j, 0 ≤ zii ≤ 1, j =1,...,n, zii + zjj − zij ≤ 1 i, j =1 ,...,n,i = j, Z  0. ZLP ≤ ZSD ≤ ZIP . Why? 4.3 Stable set Slide 14 n ZIP =max wixi i=1 s.t.xi + xj ≤ 1, ∀{i, j}∈E, xi ∈{0, 1}, i ∈ V. n ZSD =max wizii i=1 s.t.zij =0, ∀{i, j}∈E, zii + zjj ≤ 1, ∀{i, j}∈E, zik + zkj ≤ zkk, ∀{i, j}∈E, zii + zjj + zkk ≤ 1+zik + zjk , ∀{i, j}∈E, Z  0. 4.4 Max-Cut Slide 15 max wij (xi + xj − 2xixj ) {i,j}∈E s.t.xs =1, xt =0, xi ∈{0, 1}, ∀ i ∈ V. ZSD =max wij (zii + zjj − 2zij ) {i,j}∈E s.t.zss =1, ztt =0, zst =0 Z  0. 4   •        Also 0 ≤ zii ≤ 1,zij ≤ zii,zij ≤ zjj ,zii + zjj − zij ≤ 1. 4.5 Scheduling Slide 16 • Jobs J = {1,...,n} and m machines. • pij processing time of job j on machine i. • Completion time Cj . Objective: assign jobs to machines, and schedule each machine to minimize j∈J wj Cj . • If jobs j and k are assigned to machine i,thenjob j is scheduled b efore job k on machine i, denoted by j ≺i k if and only if wk wj> . pik pij 4.5.1 Formulation Slide 17 • xij is one, if job j is assigned …