Optimization Exercise Set n. 4 :

Optimization Exercise Set n. 4 : Prepared by S. Coniglio and E. Amaldi translated by O. Jabali 2018/2019 1

4.1 Airport location In air transportation, usually there is not a direct connection between every pair of airports. Indeed, passengers of intercontinental flights leaving from small airports usually travel on small aircrafts toward the so-called hub airports, where they stop over and are boarded on higher capacity aircrafts for the intercontinental flight. If their final destination is a small airport, they stop over in a second hub from which they reach their final destination. This allows the airline company to fully exploit the high capacity of intercontinental flights by aggregating several passengers with different origins and destinations. An airline company serves a set of cities C. Suppose the number t ij of passengers traveling from city i C to city j C is known. The company must decide where to open its p hubs among the airports in set C. The distance d ij between airports i and j is given. The cost of one passenger traveling from i to j is proportional to the distance d ij. However, due to the economy of scale, the cost per kilometer of an inter-hub flight is lower than the one of a flight between a non-hub and a hub airport: the cost of the travel between i and j is βd ij if either i or j is not a hub, and it is equal to αd ij if i and j are both hubs, where α < β. Give an Integer Linear Programming formulation for the problem of minimizing the overall traveling cost in the following case: a) multiple assignment: each non-hub airport can send (and receive) its passengers to (and from) more than one hub. b) single assignment: each non-hub airport can send (and receive) its passengers to (and from) a single hub. Solution a) Multiple assignment version Variables: x kl ij 0: number of passengers travelling from city i to city j via the hubs in cities k and then l y i {0, 1}: y i = 1 if a hub is open in city i and y i = 0 otherwise Model: min s.t. i C j C k C l C (βd ik + βd lj + αd kl )x kl ij (transportation cost) i C y i = p (number of hubs) k C l C xkl ij = t ij i, j C (transport all passengers) x kl ij y k k C, i C, j C, l C (consistency) x kl ij y l l C, i C, j C, k C (consistency) y i {0, 1} i C x kl ij Z+ i C, j C, k C, l C. Page 2

b) Single assignment version Variables: x ik {0, 1}: x ik = 1 if all passengers flying out of/in city i travel via the hub k, and x ik = 0 otherwise y i {0, 1}: y i = 1 if a hub is open in city i and y i = 0 otherwise Model: min s.t. β i C +α i C k C (d ik j C t ij + d ki j C t ji)x ik j C l C d klt ij x ik x jl (transportation cost) i C y i = p (number of hubs) k C x ik = 1 i C (assignments) x ik y k k C, i C (consistency) y i {0, 1} i C k C x kl ij Z+ i C, j C, k C, l C. The objective function can be linearized by introducing the variables zij kl = x ikx jl, and the constraints zij kl x ik, zij kl x jl and zij kl x ik + x jl 1, i, j, k, l C. Page 3

4.2 Chvátal-Gomory cuts for the maximum clique problem Given an undirected graph G = (V, E), consider the problem of finding a clique, i.e., a subset of nodes K V such that {i, j} E for every pair of nodes i, j K, of maximum cardinality. Consider the following natural Integer Linear Programming formulation: max x i (1) i V s.t. x i + x j 1 {i, j} / E (2) x i {0, 1} i V, (3) where S denotes the set of all the stable sets (or independent sets) in G, i.e., the subsets of nodes such that no two nodes are adjacent. Consider the stable set inequalities x i 1 S S. (4) i S a) Verify that these inequalities are valid. Are they tighter than those in (2)? b) Which stable set inequalities can be obtained by applying a single round of the Chvátal-Gomory procedure? c) Show that all the stable set inequalities can be obtained applying the Chvátal- Gomory procedure. Which is the Chvátal-Gomory rank of a stable set inequality of cardinality k? Solution a) A stable set is a subset S of nodes that are not pairwise adjacent, whereas a clique is a completely connected subgraph. Since a clique can contain at most one node of any stable set, the inequalities (4) are valid. Constraints (2) are x i + x j 1 α t x α 0. For S = 3, we can write Constraints (4) as x i + x j + x k 1 β t x β 0. The latter inequality is stronger since α 0 = β 0 and β t is the same or contains more nonzero components than α t. b) Starting from the above natural ILP formulation and applying one round of the Chvátal-Gomory procedure, we can obtain all the stable set inequalities associated to stable sets of cardinality 3. Indeed, for any stable set S V with S = 3 it is sufficient to consider the three nodes i, j and k in S (which are not connected in G) and the associated inequalities x i + x j 1 x i + x k 1 x j + x k 1. Page 4

Multiplying them by 1/2 and adding them we get: which is equivalent to x i + x j + x k 3 2 x i + x j + x k 1. Thus, for any stable set S with S = 3, the stable set inequality associated to S is a Chvàtal-Gomory cut of rank 1. c) It is easy to show by induction that the stable set inequalities associated to stable sets S = k with k 3 are Chvátal-Gomory cuts of rank k 2 with respect to the above natural ILP formulation. In part b) we have verified that this is true for k = 3. Let us now verify that if the property holds for all stable sets with S = k 1 then it also holds for all stable sets with S = k. Consider any stable set S with S = k, the corresponding stable set inequality can be derived by considering all stable subsets of S of cardinality k 1. The latter k subsets are easily obtained by iterating on each node i S and building, for each i, a stable subset containing all nodes in S \ {i}. The aggregated inequality (with unit weight) is therefore i S (k 1)x i k. By dividing each coefficient by k 1 and rounding down to the largest integer less the right hand side, we obtain i S x i 1. Thus the property holds for all stable sets of cardinality k 3. Page 5

4.3 Cover inequalities for the 0-1 knapsack problem Consider the 0-1 knapsack problem with the feasible region X = {x {0, 1} 6 : 12x 1 + 9x 2 + 7x 3 + 5x 4 + 5x 5 + 3x 6 14}. a) List all the minimal cover inequalities for X. b) Assuming x 1 = x 2 = x 4 = 0, consider the minimal cover inequality x 3 + x 5 + x 6 2, which is valid for X = X {x {0, 1} 6 : x 1 = x 2 = x 4 = 0}. Show that this minimal cover inequality defines a facet of conv(x ). c) Apply the sequential lifting procedure to the minimal cover inequality x 3 + x 5 + x 6 2 according to the order x 1, x 2, x 4 to obtain an inequality α 1 x 1 + α 2 x 2 + x 3 + α 4 x 4 + x 5 + x 6 2 valid for X. Show that this lifted minimal cover inequality defines a facet of conv(x). d) Describe the separation problem for the cover inequalities and propose a simple heuristic to identify violated cover inequalities. Given the optimal solution of the current linear relaxation with the cover inequalities generated so far, how do we need to tackle the above separation problem to make sure that no other violated cover inequality exist? Solution a) The minimal covers are {1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {2, 3}, {2, 4, 5}, {2, 4, 6}, {2, 5, 6}, {3, 4, 5}, {3, 4, 6}, {3, 5, 6}. b) It suffices to verify that the three points (1 1 0), (1 0 1), (0 1 1), which satisfy the minimal cover inequality x 3 + x 5 + x 6 2 with equality, are affinely independent. This is obvious since they are linearly independent. c) If x 1 = 0 then the inequality α 1 x 1 + x 3 + x 5 + x 6 2 is valid for all values of α 1. If x 1 = 1, the maximum of α 1 for α 1 x 1 + x 3 + x 5 + x 6 2 to be a valid inequality must satisfy α 1 + max{x 3 + x 5 + x 6 : 7x 3 + 5x 5 + 3x 6 14 12 = 2} 2 namely α 1 + 0 2, thus α 1 = 2. The maximum of α 2 for 2x 1 + α 2 x 2 + x 3 + x 5 + x 6 2 to be a valid inequality must satisfy α 2 + max{2x 1 + x 3 + x 5 + x 6 : 12x 1 + 7x 3 + 5x 5 + 3x 6 14 9 = 5} 2 namely α 2 + 1 2, thus α 2 = 1. The maximum of α 4 for 2x 1 + x 2 + x 3 + α 4 x 4 + x 5 + x 6 2 to be a valid inequality must satisfy α 4 +max{2x 1 +x 2 +x 3 +x 5 +x 6 : 12x 1 +9x 2 +7x 3 +5x 5 +3x 6 14 5 = 9} 2 namely α 4 + 2 2, thus α 4 = 0. The resulting lifted minimal cover inequality is 2x 1 + x 2 + x 3 + x 5 + x 6 2. This inequality defines a facet of conv(x). First note that dim(conv(x)) = 6: 0 and the 6 unit vectors e i, 1 i 6, are 7 affinely independent vectors Page 6

belonging to X. Considering the components in the order x 1, x 2, x 3, x 5, x 6, x 4, the following 6 points satisfy the lifted minimal cover inequality with equality: (0, 0, 1, 1, 0, 0) (0, 0, 1, 0, 1, 0) (0, 0, 0, 1, 1, 0) (1, 0, 0, 0, 0, 0) (0, 1, 0, 1, 0, 0) (0, 0, 0, 1, 1, 1) where the first three points are derived by expanding the points from part b). Clearly the 6 above points are linearly independent and hence affinely independent. d) The separation problem: given a fractional solution x with 0 x 1 where 1 j n, decide whether x satisfies all the cover inequalities or determine one violated by x. The problem consists in finding C N such that j C a j > b and j C (1 x j ) < 1. Defining the incidence vector z {0, 1} n corresponding to C N, the separation problem is equivalent to verifying if the optimal solution to the following problem (binary knapsack variant) has an objective function value strictly less than one: min s.t. j C (1 x j )z j j C a jz j > b z j {0, 1} j C. The problem can be solved heuristically. One greedy heuristic consists in inserting items in a non-decreasing order of 1 x j a j until a cover is obtained. In order to ensure that no cover inequalities are violated we must solve the separation problem exactly. Page 7

4.4 Valid inequalities for the maximum stable set problem Consider the maximum stable set problem where, given an undirected graph G = (V, E), we wish to find a stable set, i.e., a subset of nodes that are not pairwise adjacent, of maximum cardinality. This NP-hard problem arises in many fields of application where the edges represent incompatibilities between the entities associated to the corresponding nodes. If, for every node i V, the binary variable x i is equal to 1 if node i is selected in the stable set and 0 otherwise, a natural ILP formulation is as follows: max i v x i s.t. x i + x j 1 {i, j} E x i {0, 1} i V. Exercise session a) Which 5 valid inequality can be Optimization associated to every clique K V of G? Prof. Such E. Amaldi valid inequalities are referred to as clique inequalities. Consider the b) instance Considerinthe theinstance figure: of the maximum stable set problem defined by the following undirected graph: Add to the above natural ILP formulation all the 1 2 3 4 5 7 6 Add to the natural ILP formulation all the clique inequalities of cardinality 3. clique inequalities of cardinality 3, that is, containing 3 nodes. Show, c) by Show, exhibiting by exhibiting a fractional a fractional feasible optimal soluzione solution x LPx of the corresponding linear relaxation, that ming therelaxation above inequalities of the ILPare formulation not sufficient obtained to describe in (b), the thatconvex the clique hullin- of all integer LP of the linear program- solutions equalities (incidence arevectors not sufficient of thetostable describe set). the convex hull of all integer solutions (incidence vectors of the stable sets in G). Construct the auxiliary bipartite graph to separate odd hole inequalities w.r.t. x LP and use itd) to Determine derive the the inequalities. dimension of the stable set polytope P Stab, i.e., the convex hull of the incidence vectors of all stable sets in G. Show that the clique inequality associated to a clique K defines a facet of P Stab if and only if K is a maximal clique (in terms of inclusion). e) State the separation problem for the clique inequalities and mention how it can be solved? Page 8

f) A subset H V is a hole of G if the induced subgraph is a simple cycle, i.e., the nodes of H are connected via a cycle and for each pair of non adjacent nodes i and j we have {i, j} / E. Which valid inequality can be associated to any odd hole H of G, that is, a hole of G with an odd number of nodes? Explain why such odd hole inequalities are valid. g) What about the odd hole inequalities of cardinality 3 and the clique inequalities of cardinality 3? List all the odd hole inequalities of cardinality 5 for the instance in (b). h) Consider the instance in (b) and the odd hole inequality associated to the odd hole H = {2, 3, 7, 6, 5} and lift the coefficient of the variable x 4 associated to node 4. Solution a) A clique K V is a completely connected subgraph, whereas a stable set is a subset of nodes that are not pairwise adjacent. Clearly, given any clique K of G, at most one node of K can be contained in any stable set. Thus, the clique inequality x i 1 is valid. i K b) The cliques of cardinality 3 are {1, 2, 3}, {2, 3, 4}, {2, 4, 5}, {4, 5, 6}, {1, 3, 7} and the associated clique inequalities are: x 1 + x 2 + x 3 1 x 2 + x 3 + x 4 1 x 2 + x 4 + x 5 1 x 4 + x 5 + x 6 1 x 1 + x 3 + x 7 1. c) We observe that the clique inequality x i + x j + x k 1 associated to any clique of cardinality 3 with K = {i, j, k}, dominates the three corresponding edge inequalities x i + x j 1 {i, j} E with i, j K. The ILP formulation for the instance of point b) including all the clique inequalities of cardinality 3 is as follows max i v x i s.t. x i + x j 1 {i, j} E x 1 + x 2 + x 3 1 x 2 + x 3 + x 4 1 x 2 + x 4 + x 5 1 x 4 + x 5 + x 6 1 x 1 + x 3 + x 7 1 x i {0, 1} i V. Page 9

The unique optimal solution to the linear programming relaxation of the above formulation is (1/3, 1/3, 1/3, 1/3, 1/3, 1/3, 1/3). Since this optimal solution is not integer, we conclude that the clique inequalities are not sufficient to describe the convex hull of all integer feasible solutions. d) Consider the bounded polyhedron P Stab = conv({x {0, 1} V : x i + x j 1, {i, j} E}) which is known as the stable set polytope. P Stab = conv(x) where X = {x {0, 1} V : x incidence vector of a stable set of G}. First note that dim(p Stab ) = n = V. Since 0 and the n unit vectors e i, 1 i n, belong to X, there are n + 1 affinely independent vectors in X. A clique K V is a maximal clique if K {i} is not a clique for every i V \S. Clearly the undominated clique inequalities are those corresponding to the maximal cliques. Indeed, consider two cliques K and K with K K, then i K x i 1 clearly dominates i K x i 1. Proposition: A clique inequality x i 1 (5) i K defines a facet of P Stab if and only if K is a maximal clique in G. It is easy to establish this necessary and sufficient condition: Suppose the maximal clique is K = {1,..., k}. Clearly the binary vectors e i with i = 1,..., k, satisfy (5) 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 other components equal to 0. Then x i X and it satisfies (5) with equality for i = k + 1,..., n. Since e 1,..., e k, x k+1,..., x n are linearly (affinely) independent, the inequality (5) is facet defining. Conversely, if a clique K is not maximal, there exists i K such that K {i} is a clique and thus l K {i} x i 1 is valid for P Stab. Since (5) is the sum of x i 0 and l K {i} x i 1, it is not facet defining. e) Separation problem for clique inequalities. Given a fractional optimal solution x of the current partial linear programming relaxation max i V x i s.t. x i + x j 1 {i, j} E constraints (5) 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 x i 1 for all i V as above, assign a weight x i to each node i V and look for a clique K V of maximum total weight. Page 10

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 this separation problem is NP-hard, heuristics are used. f) For any hole H V of G, no more than half the nodes of H, i.e., at most H 2 of them, can be contained in any stable set, otherwise two of these nodes would be adjacent. Since starting from any node in H and going through the hole we can select every other node, H 2 is a valid bound. Therefore, for any hole H V, the hole inequality i H x i H 2 is valid, i.e., satisfied by the incidence vectors of all the stable sets. If H is an even hole (with an even number of nodes) or an odd hole with 3 nodes, the corresponding hole inequality is clearly implied by the edge inequalities s.t. x i + x j 1 {i, j} E associated to all the edges of the hole (it can be obtained by aggregating these edge inequalities with multipliers 1/2). However, if H is an odd hole (with an odd number of nodes) then H 2 = H 1 2 and the valid odd hole inequality i H x i H 1 2 is not implied by the edge inequalities associated to the edges of H. g) An odd hole inequality with cardinality 3 is a clique. The odd holes of cardinality 5 for the instance in (b) are {1, 2, 5, 6, 7} and {3, 2, 5, 6, 7, }, the corresponding (valid) odd hole inequalities are x 1 + x 2 + x 5 + x 6 + x 7 2 x 3 + x 2 + x 5 + x 6 + x 7 2. h) The inequality associated to the hole {3, 2, 5, 6, 7} is x 3 +x 2 +x 5 +x 6 +x 7 2. We consider the inequality αx 4 + x 3 + x 2 + x 5 + x 6 + x 7 2. For x 4 = 0, the inequality is valid for all α. If x 4 = 1, then α 2 x 3 x 2 x 5 x 6 x 7. Since x 4 = 1 implies that x 3 = x 2 = x 5 = x 6 = 0, then α 2 x 7. Therefore α = 1. Thus we obtain the stronger lifted odd hole inequality x 4 + x 3 + x 2 + x 5 + x 6 + x 7 2. Page 11