3.8 Strong valid inequalities

Transcription

1 3.8 Strong valid inequalities By studying the problem structure, we can derive strong valid inequalities which lead to better approximations of the ideal formulation conv(x ) and hence to tighter bounds. Two definitions related to polyhedra and their description: Consider a polyhedron P = {x R n + : Ax b}. Definition: Given two valid inequalities π t x π 0 and µ t x µ 0 for P, π t x π 0 dominates µ t x µ 0 if u > 0 such that uµ π and π 0 uµ 0 with (π, π 0) (uµ, uµ 0). Poiché uµ t x π t x π 0 uµ 0, chiaramente {x R n + : π t x π 0} {x R n + : µ t x µ 0} Example: x 1 + 3x 2 4 dominates 2x 1 + 4x 2 9 since for (π, π 0) = (1, 3, 4) and (µ, µ 0) = (2, 4, 9) we have 1 2 µ π and π0 1 2 µ0. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

2 Definition: A valid inequality π t x π 0 is redundant in the description of P if there exist k 2 valid inequalities π i x π0 i for P with u i > 0, 1 i k, such that k k ( u i π i )x u i π0 i dominates π t x π 0. i=1 i=1 Example: P = {(x 1, x 2) R 2 + : x 1 + 2x 2 4, x 1 2x 2 3, x 1 + x 2 5/3, 1 x 1 3} x 1 + x 2 5/3 is redundant because it is dominated by x 1 + x 2 3/2, which is implied by x 1 + 2x 2 4 and x 1 1 (with u 1 = u 2 = 1 2 ) Observation: When P = conv(x ) is not known explicitly it can be very difficult to check redundancy. In pratice, we should avoid inequalities that are dominated by other already available inequalities. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

3 3.8.1 Faces and facets of polyhedra Consider a polyhedron P = {x R n : Ax b} Definitions The vectors x 1,..., x k are affinely independent if the k 1 vectors x 2 x 1,..., x k x 1 are linearly independent. The dimension of P, dim(p), is equal to the maximum number of affinely linearly independent points of P minus 1. P is full-dimensional if dim(p) = n, i.e., no equation a t x = b is satisfied with equality by all the points x P. Illustrations: Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

4 For the sake of simplicity, we assume that P is full dimensional Theorem: If P is of full dimension, P admits a unique minimal description P = {x R n : a t i x b i, i = 1,..., m} where each inequality is unique within a positive multiple. Each inequality is necessary: deleting anyone of them we obtain a polyhedron that differs from P. Moreover, each valid inequality for P which is not a positive multiple of one of the a t i x b i is redundant (can be obtained as linear combination with non negative coefficients of two or more valid inequalities). Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

5 Alternative characterization of necessary inequalities Definitions Let F = {x P : π t x = π 0} for any valid inequality π t x π 0 for P. Then F is a face of P and the inequality π t x π 0 represents or defines F. If F is a face of P and dim(f ) = dim(p) 1, then F is a facet of P. Consequences: The faces of a polyhedron are polyhedra and a polyhedron has a finite number of faces. Theorem: If P is full dimensional, a valid inequality is necessary for the description of P if and only if it defines a facet of P. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

6 Example Consider the polyhedron P R 2 described by: x 1 + 2x 2 4 (1) x 1 2x 2 3 (2) x 1 + x (3) x 1 3 (4) x 1 1 (5) Verify that P is full dimensional. Establish which inequalities define facets of P or are redundant. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

7 How to show that a valid inequality is facet defining Consider X Z n + and a valid inequality π t x π 0 for X Assumption: conv(x ) is bounded and full dimensional Two simple approaches to show that π t x π 0 defines a facet of P = conv(x ): 1) Direct approach (definition): Find n points x 1,..., x n X that satisfy the inequality with equality (π t x = π 0) and are affinely independent. 2) Indirect approach: (i) Select t points x 1,..., x t X, with t n, that satisfy π x = π 0. Suppose that they all belong to a generic hyperplane µ t x = µ 0. (ii) Solve the linear system n µ j xj k = µ 0 for k = 1,..., t j=1 with the n + 1 unknowns µ 0, µ 1,..., µ n. (iii) If the only solution is (µ, µ 0) = λ(π, π 0) with λ 0, then the inequality π x = π 0 defines a facet of conv(x ). Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

8 Example: Consider X = {(x, y) R m {0, 1} : m i=1 x i my, 0 x i 1 i} i) Verify that dim(conv(x )) = m + 1. ii) Show with the indirect approach that the valid inequality x i y defines a facet of conv(x ). Consider the points (0, 0), (e i, 1) and (e i + e j, 1) for j i which are feasible and satisfy x i = y. Since (0, 0) belongs to the hyperplane defined by m k=1 µ kx k + µ m+1y = µ 0, then µ 0 = 0. Since (e i, 1) belongs to the hyperplane defined by m k=1 µ kx k + µ m+1y = µ 0, then µ i = µ m+1. Since (e i + e j, 1) belongs to the hyperplane defined by m k=1 µ kx k µ i y = µ 0, then µ j = 0 for j i. Thus the hyperplane is µ i x i µ i y = 0 and the inequality x i y defines a facet of conv(x ). Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

9 3.8.2 Cover inequalities for the binary knapsack problem Consider X = {x {0, 1} n : n j=1 a jx j b} with b > 0 and N = {1,..., n}. Assumptions: For each j with 1 j n, a j > 0 (if a j < 0 set x j = 1 x j ) and a j b. Definition: A subset C N is a cover for X if j C a j > b. A cover is minimal if, for each j C, C \ {j} is not a cover. Example: For X = {x {0, 1} 7 : 11x 1 + 6x 2 + 6x 3 + 5x 4 + 5x 5 + 4x 6 + x 7 19} two minimal covers are: {1, 2, 3} and {3, 4, 5, 6} Proposition: If C is a cover for X, the inequality x j C 1 j C is valid for X, and is called a cover inequality. Example: For X = {x {0, 1} 7 : 11x 1 + 6x 2 + 6x 3 + 5x 4 + 5x 5 + 4x 6 + x 7 19} some minimal cover inequalities: x 1 + x 2 + x 3 2, x 1 + x 2 + x 6 2, x 1 + x 5 + x 6 2 Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

10 Can such cover inequalities be strengthened? Proposition: If C is a cover for X, the extended cover inequality x j C 1 j E(C) is valid for X, where E(C) = C {j N : a j a i for all i C}. Example: Consider X = {x {0, 1} 7 : 11x 1 + 6x 2 + 6x 3 + 5x 4 + 5x 5 + 4x 6 + x 7 19}. The extended cover inequality for C = {3, 4, 5, 6} is x 1 + x 2 + x 3 + x 4 + x 5 + x 6 3 which clearly dominates x 3 + x 4 + x 5 + x 6 3. (6) Observation: Since a 1 = 11, a i 5 for i {3, 4, 5}, a 6 = 4 and b = 19, if x 1 = 1 at most one of the other variables in (6) can take value 1 and the inequality is valid and in turn dominates (6). 2x 1 + x 3 + x 4 + x 5 + x 6 3 Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

11 Lifting procedure: to strengthen such valid inequalities to obtain facet-defining ones Let j 1,..., j r be the indices of N \ C and set t = 1. Let t 1 i=1 α j i x ji + j C x j C 1 be the inequality obtained at iteration t 1. At iteration t: Determine the maximum value of α jt such that t 1 α jt x jt + α ji x ji + x j C 1 i=1 j C is valid for X by solving the (binary knapsack) problem σ t = max s.t. and by setting α t = C 1 σ t. Terminate when t = r. t 1 i=1 α j i x ji + j C x j t 1 i=1 a j i x ji + j C a jx j b a jt x {0, 1} C +t 1 N.B.: σ t = maximum amount of space used up by the variables of indices C {j 1,..., j t 1} when x jt = 1. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

12 Example: Consider X = {x {0, 1} 7 : 11x 1 + 6x 2 + 6x 3 + 5x 4 + 5x 5 + 4x 6 + x 7 19}. Applying the lifting procedure to x 3 + x 4 + x 5 + x 6 3 considering in the order x 1, x 2 and x 7, we obtain the valid inequality 2x 1 + x 2 + x 3 + x 4 + x 5 + x 6 3. Proposition: If C is a minimal cover and a i b i N, the lifting procedure is guaranteed to yield a facet-defining inequality of conv(x ). Example cont.: The valid inequality 2x 1 + x 2 + x 3 + x 4 + x 5 + x 6 3 defines a facet of conv(x ). Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

13 Separation of cover inequalities Separation problem: Given a fractional solution x with 0 x j 1, 1 j n, decide whether x satisfies all the cover inequalities or determine a one violated by x. Since j C x j C 1 can be written as j C (1 x j) 1, we have to answer the question: Does there exist a subset C N such that j C a j > b and j C (1 x j ) < 1? If z {0, 1} n is the incidence vector (binary characteristic vector) of the subset C N, it is equivalent to the question: ζ = min{ j N (1 x j )z j : j N a jz j > b, z {0, 1} n } < 1? Proposition: (i) If ζ 1, x satisfies all the cover inequalities. (ii) If ζ < 1 with optimal solution z, then j C x j C 1 with C = {j : zj = 1, 1 j n}, cuts x by a quantity 1 ζ. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

14 Example: Consider max z = 5x 1 + 2x 2 + x 3 + 8x 4 s.t. 4x 1 + 2x 2 + 2x 3 + 3x 4 4 x 1, x 2, x 3, x 4 {0, 1} Optimal solution of the LP relaxation x LP = (1/4, 0, 0, 1) with z LP = The separation problem amounts to the following binary knapsack problem: ζ 3 = min z1 + z2 + z3 4 s.t. 4z 1 + 2z 2 + 2z 3 + 3z 4 > 4 z 1, z 2, z 3, z 4 {0, 1} where the > constraint can be replaced with 4z 1 + 2z 2 + 2z 3 + 3z 4 5. Optimal solution z = (1, 0, 0, 1) with ζ = 3 4. Thus the cover inequality x 1 + x 4 1 cuts away the current LP optimal solution x LP by 1 ζ = 1 4. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

15 3.8.3 Strong valid inequalities for the Maximum Independent Set problem Definitions - A subset S V is an independent/stable set if {i, j} / E for each pair of nodes i, j S. - A subset K V is a clique if {i, j} E for each pair of nodes i, j K,. - A subset S V is a maximal independent/stable set (in the sense of inclusion) if S {i} is not independent/stable for each i V \ S. - A subset K V is a maximal clique if K {i} is not a clique for each i V \ S. Illustrations: Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

16 Maximum Independent Set problem (MIS): (maximum stable set) Given an undirected G = (V, E), determine an independent set S V of maximum size. Proposition: MIS is NP-hard. MIS has a wide range of interesting applications. For each i V, let the binary variable x i indicate whether i S. ILP formulation: max i V x i (7) s.t. x i + x j 1 {i, j} E (8) x i {0, 1} i V (9) Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

17 Definition: The bounded polyhedron P IS = conv({x {0, 1} V : x i + x j 1, {i, j} E}) is the independent set polytope. P IS = conv(x ) where X = {x {0, 1} V : x incidence vector of an independent set of G}. Observation: dim(p IS ) = n = V Indeed: since 0 and the n unit vectors e i, 1 i n, belong to X, there are n + 1 affinely independent vectors in X. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

18 1) Clique inequalities Property: For any clique K V of G, the clique inequality x i 1 is valid for X. i K Observation: The undominated clique inequalities are those corresponding to the maximal cliques. Consider two cliques K and K with K K, then i K x i 1 clearly dominates i K x i 1. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

19 Proposition: A clique inequality defines a facet of P IS if and only if K is a maximal clique in G. Proof: x i 1 (10) i K Suppose K = {1,..., k}. Clearly the vectors e i with i = 1,..., k, satisfy (10) with equality. Since K is a maximal clique, for each i K, there is a node j(i) K such that {i, j(i)} E. Denote by x i the binary vector with components i and j(i) equal to 1 and all others to 0. Then x i X and it satisfies (10) with equality for i = k + 1,..., n. Since e 1,..., e k, x k+1,..., x n are linearly (affinely) independent, inequality (10) is facet defining. Conversely, if K is not maximal i K such that K {i} is a clique and thus l K {i} x i 1 is valid for P IS. Since (10) is the sum of x i 0 and l K {i} x i 1, it is not facet defining. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

20 Separation problem for clique inequalities: Given a fractional optimal solution x of the current linear relaxation of (7)-(9) max i V x i s.t. x i + x j 1 {i, j} E constraints (10) generated so far x i 0 i V, find a clique inequality that is violated by x or establish that no such inequality exists. Separation procedure: Given x with 0 xi 1 for all i V as above, assign a weight to each node i V and look for a clique K V of maximum total weight. x i Indeed, if i K x i > 1, the clique inequality i K x i 1 is violated by x, otherwise all those not yet introduced in the partial ILP formulation are satisfied by x. Since the this separation problem is NP-hard, heuristics are used. N.B.: The larger class of so-called orthonormal representation (OR) valid inequalities, includes the clique inequalities and can be separated in polynomial time. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

21 2) Odd hole inequalities Definition: A subset H V is a hole of G if the induced subgraph H is a simple cycle, i.e., the nodes in H are connected via a cycle and for each pair of non adjacent nodes i, j we have {i, j} / E. Example: outer cycle of a wheel with n = 6 nodes Proposition: For each hole with an odd number of nodes, the odd hole inequality is valid for X. i H x i H 2 = H 1 2 We cannot select more than half of the nodes of each hole; for odd holes the upper bound is H 1 2. Observation: The inequalities corresponding to even holes and odd holes with 3 nodes, are implied by (8) and (10). An odd hole with 3 nodes is a clique; every even hole inequality can be obtained by aggregating the inequalities (8) associated to all the edges of the hole with multipliers 1 2. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

22 Separation problem: Given x optimal solution of the current linear relaxation, find an odd hole inequality that is violated by x or establish that no such inequality exists. Proposition: Odd hole inequalities can be separated by computing a shortest path between n = V pairs of nodes in an ad hoc bipartite graph with appropriate edge costs. Separation algorithm: Given an optimal solution x as above, we look for a violated odd hole inequality. Construct an auxiliary bipartite graph G = (V, E ), including each node i V of G in V 1 and a copy i in V 2. For each edge {i, j} E, include in E (G ) two edge {i, j } and {i, j}, where i, j V 1 and i, j V 2. To each edge {i, j } E we assign a cost 1 x i x j. This way, a cycle C in G has cost {i,j} C (1 xi xj ) = C 2 where V (C) is the set of nodes of C in G. If such cost < 1, then i V (C) x i i V (C) xi = V (C) 2 V (C) 1 2 is violated by x. i V (C) x i, Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

23 To consider all possible odd holes, for each i V 1 it suffices to look for a minimum cost path p i from node i V 1 to its copy i V 2. - If p i is a hole and its cost is < 1, we have a violated inequality. - If pi is not a hole and its cost < 1, then it contains an odd subcycle whose cost is necessarily smaller than 1 (since the costs xi 0), which corresponds to a violated odd hole inequality. - If all the paths found have a costs 1, x satisfy all the odd hole inequalities. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

24 Example: G = (V, E) is the wheel with 6 nodes and 10 edges (5 nodes in the outer cycle/hole). i) List all the maximal clique inequalities. ii) Indicate the unique optimal solution x LP of the linear programming relaxation including all the maximal cliques inequalities. iii) Determine the violated odd hole inequality. iv) Lift this odd hole inequality so that it defines a facet of the Independent Set polytope P IS. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

25 3.8.4 Equivalence between separation and optimization Consider a family of LPs min{ c t x : x P o} with o O, where P o = { x R no : A ox b o } polytope with rational (integer) coefficients and a very large (e.g., exponential) number of constraints. Examples: 1) Linear relaxation of asymmetric TSP with cut-set inequalities (O set of all graphs) 2) Maximum Matching problem: For each G = (V, E), the matching polytope conv({x {0, 1} E : x e 1, i V }) coincides (Edmonds) with {x R E + : x e 1, i V, e δ(i) e E(S) e δ(i) x e S 1, S V with S 3 odd} 2 We adopt a cutting plane approach where constraints are only generated if needed. Assumption: Even though the number of constraints m o of P o is exponential in n o (e.g., O(2 no )), A o and b o are specified in a concise way (as a function of a polynomial number of parameters w.r.t. n o). Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

26 Optimization problem: Given a rational polytope P R n and a rational objective vector c R n, find a x P minimizing c t x over x P or establish that P is empty. N.B.: we assume that P is bounded (polytope) just to avoid unbounded problems. Separation problem: Given a rational polytope P R n and a rational vector x R n, establish that x P or determine a rational vector π R n such that πx < πx for each x P (indentify a cut that separate x from P). Theorem: (consequence of Grötschel, Lovász, Schriver 1988 theorem) The separation problem for a family of polyhedra can be solved in polynomial time in n e log U if and only if the optimization for that family can be solved in polynomial time in n and log U, where U is an upper bound on all a ij and b i. Proof based on the Ellipsoid method, first LP polynomial algorithm (Khachiyan 1979). For now theoretical tool: the resulting algorithm is not efficient but the equivalence may guide the search for more practical polynomial-time algorithms. Corollary: The linear relaxation of the ILP formulation for ATSP with cut-set inequalities can be solved in polynomial time in spite of the exponential number of constraints. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27

27 3.8.5 Remarks on cutting plane methods Consider a generic discrete optimization problem with rational coefficients c i. min{ c t x : x X R n +} When designing a cutting plane method, we should be aware that: It can be difficult to describe one or more families of strong (possibly facet defining) valid inequalities for conv(x ). The separation problem for a given family F may require a considerable computational effort (if NP-hard we devise heuristics). Even when finite convergence is guaranteed (e.g., with Gomory cuts), pure cutting plane methods tend to be very slow. To solve hard/large problems optimally, it is often necessary to embed strong valid inequalities into a Branch-and-Bound framework, leading to a so-called Branch and Cut method. The subfield of Discrete Optimization studying the polyhedral structure of the ideal formulations (conv(x )) is known as Polyhedral Combinatorics. Edoardo Amaldi (PoliMI) Ottimizzazione A.A / 27