Dual bounds: can t get any better than...

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1 Bounds, relaxations and duality Given an optimization problem z max{c(x) x 2 }, how does one find z, or prove that a feasible solution x? is optimal or close to optimal? I Search for a lower and upper bound on the optimal objective value z apple z apple z. I If z z, then the optimal objective value is found. I If x? 2 satisfies z apple c(x? ) apple z, and z z apple, where > 0 is an optimality tolerance parameter, then x? is guaranteed to satisfy c(x? ) z. Primal bounds can you beat this? Lower(upper) bounds on the optimal objective function value of a max(min) problem Any feasible solution can be used to obtain a primal bound. IOE 51 Introduction to IP, Winter 2012 Bounds, relaxations, duality Page 37 c Marina A. Epelman Dual bounds can t get any better than... Upper(lower) bounds on the optimal objective function value of a max(min) problem are often obtained via relaxations of the problem Relaxation (IP) z max{c(x) x 2 R n } A problem (RP) z R max{f (x) x 2 R n } is a relaxation of (IP) if (i),and(ii) f (x) c(x) forallx 2. Proposition 2.1 If (RP) is a relaxation of (IP), then z R z. Proposition 2.3 (i) If a relaxation (RP) is infeasible, the original problem (IP) is infeasible. (ii) Let x? be an optimal solution of (RP). If x? 2 and f (x? )c(x? ), then x? is an optimal solution of (IP). IOE 51 Introduction to IP, Winter 2012 Bounds, relaxations, duality Page 3 c Marina A. Epelman

2 Linear Programming Relaxations LP realaxation For the integer program max{c x x 2 P \ Z n } with formulation P {x 2 R n Ax apple b}, thelinear programming relaxation is the linear program z LP max{c x x 2 P}. Clearly a relaxation P \ Z n P and objective function is unchanged. Proposition 2.2 Suppose, P 1, P 2 are two formulations for the IP max{c x x 2 Z n } with P 1 P 2.If max{c x x 2 P i }, i 1, 2, then z1 LP apple z2 LP for all c. z LP i If P is the ideal formulation, then z LP z?, and any extreme point optimal solution of the LP is an optimal solution of the IP. IOE 51 Introduction to IP, Winter 2012 Bounds, relaxations, duality Page 3 c Marina A. Epelman Combinatorial relaxations SP SP a SP tour is an assignment (or permutation) containing no subtours; hence z SP min c ij is a tour ; min c ij is an assignment ; zass IOE 51 Introduction to IP, Winter 2012 Bounds, relaxations, duality Page 40 c Marina A. Epelman

3 Combinatorial relaxations symmetric SP SSP aspwithc ij c ji ree and 1-tree A tree on a set of k nodes is an undirected graph that satisfies any two of the following three properties (i) It has k 1 edges (ii) It contains no cycles (iii) It is connected. A 1-tree on n nodes is a graph consisting of two edges adjacent to node 1, plus the edges of a tree on nodes {2,...,n}. Every tour is a 1-tree, so z SSP min min c ij is a tour ; c ij is a 1-tree ; z1-tree IOE 51 Introduction to IP, Winter 2012 Bounds, relaxations, duality Page 41 c Marina A. Epelman Lagrangian relaxation (IP) z max{c x Ax apple b, x 2 Z n }. Proposition 2.4 Let z(u) max{c x + u (b all u 0. Ax) x 2 }. henz apple z(u) for Above is called the Lagrangian relaxation of (IP). It is, indeed a relaxation larger feasible region, and for any x feasible for (IP), z(u) c x + u (b Ax) c x. IOE 51 Introduction to IP, Winter 2012 Bounds, relaxations, duality Page 42 c Marina A. Epelman

4 Duality a dual problem Duality he two problems (IP) z max{c x x 2 Z n },and (D) w min{w(u) u 2 U} form a (weak)-dual pair if c(x) apple w(u) forallx 2 and all u 2 U. Whenz w, theyform a strong-dual pair. Advantage any feasible solution of the dual provides a dual bound (whereas a relaxation needs to be solved to optimality ). Proposition 2.6 (i) If (D) is unbounded, the original problem (IP) is infeasible. (ii) If x? 2 and u? 2 U satisfy c(x? )w(u? ), the x? is optimal for (IP) and w? is optimal for (D). Proposition 2.5 he integer program z max{c x Ax apple b, x 2 Z n +} and the linear program w LP min{u b u A c, u 2 R m +} form a weak dual pair. IOE 51 Introduction to IP, Winter 2012 Bounds, relaxations, duality Page 43 c Marina A. Epelman A dual pair combinatorial perspective Given a graph G (V, E), a matching M E is a set of disjoint edges; a covering by nodes is a set R V of nodes such that every edge has at least one endpoint in R. Proposition 2.7 he problem of finding a maximum cardinality matching and the problem of finding a minimum cardinality covering by nodes form a weak-dual pair. Proof If M {(i 1, j 1 ),...,(i k, j k )} then the nodes i 1, j 1,...,i k, j k are distinct, and any covering by nodes R must contain at least one node from each pair. Hence R k M. IOE 51 Introduction to IP, Winter 2012 Bounds, relaxations, duality Page 44 c Marina A. Epelman

5 A dual pair LP perspective Node-edge incidence matrix Let n V and m E. Defineann by m node-edge incidence matrix A as follows a j,e 1ifnodej is an endpoint of edge e, and 0 otherwise. Maximum cardinality matching z max{e x Ax apple e, x 2 Z m +} Minimum cardinality cover by nodes w min{e y A y e, y 2 Z n +} Let z LP and w LP be the values of the corresponding LP relaxations. hen z apple z LP w LP apple w, establishing weak-duality. 2 Not a strong-dual pair! Consider A IOE 51 Introduction to IP, Winter 2012 Bounds, relaxations, duality Page 45 c Marina A. Epelman

to work with) can be solved by solving their LP relaxations with the Simplex method I Cutting plane algorithms, e.g., Gomory s fractional cutting

$to work with) can be solved by solving their LP relaxations with the Simplex method I Cutting plane algorithms, e.g., Gomory s fractional cutting$ Summary so far z =max{c T x : Ax apple b, x 2 Z n +} I Modeling with IP (and MIP, and BIP) problems I Formulation for a discrete set that is a feasible region of an IP I Alternative formulations for the