A solution approach for linear optimization with completely positive matrices

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A solution approach for linear optimization with completely positive matrices Franz Rendl http://www.math.uni-klu.ac.at Alpen-Adria-Universität Klagenfurt Austria joint work with M. Bomze (Wien) and F. Jarre (Düsseldorf) Happy Birthday, Henry p.1/26

Overview COP and CP Inner approximations with COP matrices A Random COP generator Computations Happy Birthday, Henry p.2/26

Completely Positive Matrices Let A = (a 1,...,a k ) be a nonnegative n k matrix, then is called completely positive. X = a 1 a T 1 +... + a k a T k = AAT COP = {X : X completely positive} COP is closed, convex cone X COP, t 0 tx COP X = AA T, Y = BB T COP X + Y = [A B][A B] T COP For basics, see the book: A. Berman and N. Shaked - Monderer: Completely Positive Matrices, World Scientific 2003 Happy Birthday, Henry p.3/26

Copositive Matrices Dual cone COP of COP in S n (sym. matrices): Y COP trxy 0 X COP tra T Y A 0 A 0 a T Y a 0 vectors a 0. By definition, this means Y is copositive, Y CP. CP is dual cone to COP! Bad News: X / CP is NP-complete decision problem. Positive semidefinite matrices PSD: Y PSD a T Y a 0 vectors a. Well known facts: PSD = PSD (PSD cone is selfdual. ) COP PSD CP Happy Birthday, Henry p.4/26

Semidefinite and Copositive Programs Problems of the form max trcx s.t. A(X) = b, X PSD are called Semidefinite Programs. Problems of the form or max trcx s.t. A(X) = b, X CP max trcx s.t. A(X) = b, X COP are called Copositive Programs, because the primal or the dual involves copositive matrices. Happy Birthday, Henry p.5/26

Why Copositive Programs? Copositive Programs can be used to solve combinatorial optimization problems. Stable Set Problem: Let A be adjacency matrix of graph, J be all ones matrix. DeKlerk and Pasechnik (SIOPT 2002) show the following: α(g) = max trjx s.t. tr(a + I)X = 1, X COP = miny s.t. y(a + I) J CP. This is a copositive program with only one equation (in the primal problem). This is a simple consequence of the Motzkin-Strauss Theorem. Happy Birthday, Henry p.6/26

A general copositive modeling theorem Burer (2007) shows the following general result for the power of copositive programming: The optimal values of P and C are equal: opt(p) = opt(c) (P) minx T Qx + c T x a T i x = b i, x 0, x i {0, 1} i B. Here x IR n and B {1,...,n}. (C) min tr(qx) + c T x, s.t. a T i x = b i, ( ) a T i Xa i = b 2 1 x T i, X ii = x i i B, COP x X Happy Birthday, Henry p.7/26

Approximating COP We have now seen the power of copositive programming. Since optimizing over CP is NP-Hard, it makes sense to get approximations of CP or COP. To get relaxations, we need supersets of COP, or inner conic approximations of CP (and work on the dual cone). The Parrilo hierarchy uses Sum of Squares and provides such an outer approximation of COP (dual viewpiont!). Now we consider inner approximations of COP. This can be viewed as a method to generate feasible solutions of combinatorial optimization problems ( primal heuristic!). Happy Birthday, Henry p.8/26

Inner approximations of COP We consider min trcx s.t. A(X) = b, X COP Remember: COP= {X : X = V V T, V 0}. Some previous work by: Bomze, DeKlerk, Nesterov, Pasechnik, others: Get stable sets by approximating COP formulation of the stable set problem using optimization of quadratic over standard simplex, or other local methods. Happy Birthday, Henry p.9/26

Incremental version A general feasible descent approach: Let X = V V T with V 0 be feasible. Consider the regularized, and convex descent step problem: min ǫ C, X + (1 ǫ) X 2, such that A( X) = 0, X + X COP. Happy Birthday, Henry p.10/26

Incremental version A general feasible descent approach: Let X = V V T with V 0 be feasible. Consider the regularized, and convex descent step problem: min ǫ C, X + (1 ǫ) X 2, such that A( X) = 0, X + X COP. For small ǫ > 0 we approach the true optimal solution, because we follow the continuous steepest descent path, projected onto COP. Unfortunately, this problem is still not tractable. We approximate it by working in the V -space instead of the X-space. Happy Birthday, Henry p.11/26

Incremental version: V -space X + = (V + V )(V + V ) T hence, X = V V T X = X( V ) = V V T + V V T + ( V )( V T ). Now linearize and make sure V is small. Happy Birthday, Henry p.12/26

Incremental version: V -space X + = (V + V )(V + V ) T hence, X = V V T X = X( V ) = V V T + V V T + ( V )( V T ). Now linearize and make sure V is small. We get min ǫ 2CV, V + (1 ǫ) V 2 such that 2A i, V = b i A i V,V i, V + V 0 This is convex approximation of nonconvex version in X( V ). Happy Birthday, Henry p.13/26

Correcting the linearization The linearization obtained by replacing X( V ) with V V T + V V T introduces an error both in the cost function and in the constraints A(X) = b. We therefore include corrector iterations of the form V = V old + V corr before actually updating V V + V. Happy Birthday, Henry p.14/26

Convex quadratic subproblem The convex subproblem is of the following form, after appropriate redefinition of data and variables x = vec(v + V ),... min c T x + ρx T x such that Rx = r, x 0. Happy Birthday, Henry p.15/26

Convex quadratic subproblem The convex subproblem is of the following form, after appropriate redefinition of data and variables x = vec(v + V ),... min c T x + ρx T x such that Rx = r, x 0. If V is n k (k columns generating V ), x if of dimension nk and there are m equations. Happy Birthday, Henry p.16/26

Convex quadratic subproblem The convex subproblem is of the following form, after appropriate redefinition of data and variables x = vec(v + V ),... min c T x + ρx T x such that Rx = r, x 0. If V is n k (k columns generating V ), x if of dimension nk and there are m equations. Since Hessian of cost function is identity matrix, this problem can be solved efficiently using interior-point methods. (convex quadratic with sign constraints and linear equations) Main effort is solving a linear system of order m, essentially independent of n and k. Happy Birthday, Henry p.17/26

Test data sets COP problems coming from formulations of NP-hard problems are too difficult ( Stable Set, Coloring, Quadratic Assignment) to test new algorithms. Would like to have: Data (A,b,C) and (X,y,Z) such that (X,y,Z) is optimal for primal and dual (no duality gap). COP is nontrivial (optimum not given by optimizing over semidefiniteness plus nonnegativity ) generate instances of varying size both in n and m. Hard part: Z provably copositive!! Happy Birthday, Henry p.18/26

A COP generator Basic idea: Let G be a graph, with Q G = A + I, and (X G,y G,Z G ) be optimal solution of ω(g) = max J,X such that Q G,X = 1, X COP = miny such that yq G J CP. We further assume: G = H K (strong graph product of H and K). K is perfect and H = C 5 (or some other graph with known clique number and ω(g) < ϑ (G)). This implies: ω(g) = ω(h)ω(k) (same for ϑ ) Max cliques in G through Kronecker products from H,K. Happy Birthday, Henry p.19/26

COP generator (2) Then we have: y G := ω(h)ω(k) implies Z G = y G Q G J is CP. Let X H and X K convex combinations of outer products of characteristic vectors of max cliques in H and K. This implies that X G = X H X K is primal optimal. Select m and Matrix A of size m n 2 and set b = A(X G ). Set C := Z G A T (y G ). This implies that (X G,y G,Z G ) is optimal for the data set A,b,C and that there is no duality gap, and the problem is nontrivial, because of ω(g) < ϑ (G). Happy Birthday, Henry p.20/26

Computational results A sample instance with n = 60, m = 100. z sdp = 9600, 82, z sdp+nonneg = 172.19,z cop = 69.75 it secs b-a(x) f(x) 1 5.17 0.166918-54.5084 10 33.45 0.000028-69.4370 20 64.94 0.000003-69.5439 30 96.49 0.000002-69.6109 The number of inner iterations was set to 3, column 1 shows the outer iteration count. Happy Birthday, Henry p.21/26

Computational results A sample instance with n = 60, m = 100. z sdp = 9600, 82, z sdp+nonneg = 172.19,z cop = 69.75 it secs b-a(x) f(x) 1 5.17 0.166918-54.5084 10 33.45 0.000028-69.4370 20 64.94 0.000003-69.5439 30 96.49 0.000002-69.6109 The number of inner iterations was set to 3, column 1 shows the outer iteration count. But starting point: V 0 =.9 Vopt +.1 rand Happy Birthday, Henry p.22/26

Computational results (2) Example (continued). recall n = 60, m = 100. z sdp = 9600, 82, z sdp+nonneg = 172.19,z cop = 69.75 start iter b-a(x) f(x) (a) 30 0.000000-69.67 (b) 30 0.000002-69.61 (c) 30 0.000100-69.03 Different starting points: (a) V =,95 * V opt +.05 * rand (b) V =.90 * V opt +.10 * rand (c) V = rand(n, 2n) Similar results for other instances using the COP generator. Happy Birthday, Henry p.23/26

More results n m opt found b A(X) 50 100 314.48 315.18 4 10 5 60 120-266.99-266.31 4 10 5 70 140-158.74-155.78 3 10 5 Starting point in all cases: rand(n,2n) Inner iterations: 5 Outer iterations: 25 Happy Birthday, Henry p.24/26

Some experiments with Stable Set max J,X such that tr(x) = 1, tr(a G X) = 0, X COP Only two equations but many local optima. We consider a selection of graphs from the DIMACS collection. Computation times in the order of a few minutes. name n ω clique found keller4 171 11 9 brock200-4 200 17 14 c-fat200-1 200 12 12 c-fat200-5 200 58 58 brock400-1 400 27 24 p-hat500-1 500 9 8 Happy Birthday, Henry p.25/26

Last Slide Unfortunately, the subproblem may have local solutions, which are not local minima for the original descent step problem. The number of columns of V does not need to be larger than ( n+1) 2, but for practical purposes, this is too large. Also, there is dependence on the starting point V. Further technical details in a forthcoming paper by I. Bomze, F. Jarre and F. R.: Quadratic factorization heuristics for copositive programming, technical report, (2008). Happy Birthday, Henry p.26/26