A Branch-and-Cut Algorithm for the Stochastic Uncapacitated Lot-Sizing Problem

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1 Yongpei Guan 1 Shabbir Ahmed 1 George L. Nemhauser 1 Andrew J. Miller 2 A Branch-and-Cut Algorithm for the Stochastic Uncapacitated Lot-Sizing Problem December 12, 2004 Abstract. This paper addresses a multi-stage stochastic integer programming formulation of the uncapacitated lot-sizing problem under uncertainty. We show that the classical (l, S) inequalities for the deterministic lot-sizing polytope are also valid for the stochastic lot-sizing polytope. We then extend the (l, S) inequalities to a general class of valid inequalities, called the (Q, S Q ) inequalities, and we establish necessary and sufficient conditions which guarantee that the (Q, S Q ) inequalities are facet-defining. A separation heuristic for (Q, S Q ) inequalities is developed and incorporated into a branch-and-cut algorithm. A computational study verifies the usefulness of the (Q, S Q ) inequalities as cuts. Key words. Stochastic Lot-Sizing Multi-stage Stochastic Integer Programming Polyhedral Study Branch-and-Cut 1. Introduction The deterministic uncapacitated lot-sizing problem is to determine a minimum cost production and inventory holding schedule for a product so as to satisfy its demand over a finite discrete-time planning horizon. A standard mixed-integer programming formulation for the single item, uncapacitated, lot-sizing problem is (cf. [18]): (LS) : min T (α i x i + β i y i + h i s i ) i=0 s.t. s i 1 + x i = d i + s i i = 0,..., T, x i M i y i i = 0,..., T, x i, s i 0, y i {0, 1} i = 0,..., T, s 1 = 0, where x i represents the production in period i, s i represents the inventory at the end of period i, and y i indicates if there is a production set-up in period i. Problem parameters α i, β i, h i, and d i represent the production cost, set-up cost, holding cost, and the demand in period i, respectively. Since there is no 1 School of Industrial & Systems Engineering, Georgia Institute of Technology, GA Department of Industrial Engineering, University of Wisconsin, Madison, WI This research has been supported in part by the National Science Foundation under Award number DMII

2 2 Yongpei Guan et al. restriction on the production level, the parameter M i is a sufficiently large upper bound on x i. In the absence of backlogging, this bound can be set as M i = T j=i d j. We denote the set of feasible solutions of (LS) as X LS. Although (LS) is solvable in strongly polynomial time using specialized dynamic programming algorithms (cf. [1, 10, 23, 24]), such algorithms are not applicable when (LS) is embedded, as it frequently is, in various multi-period production planning problems. This has motivated the polyhedral study of X LS in order to improve integer programming approaches for such production planning problems. Barany, Van Roy and Wolsey [7, 8] proved that a complete polyhedral description of the convex hull of X LS is given by some of the original inequalities together with the (l, S) inequalities d il y i d 0l, i S x i + i S where l {0, 1,..., T }, S {0, 1,..., l}, S = {0, 1,..., l} \ S, and d ij = j k=i d k. The authors reported good computational results for multiple item capacitated lot-sizing problems using the (l, S) inequalities within a branchand-cut scheme. Following Barany et al. s work, polyhedral structures of many variants of (LS) have been investigated. These include variants of (LS) involving sales and safety stocks [15], start-up costs [22], piecewise linear and concave production costs [2], and constant [14,20], as well as dynamic [6,5,17,19] production capacities, only to name a few. The lot-sizing model (LS) assumes that the cost and demand parameters are known with certainty for all periods of the planning horizon. However, in many applications, these parameters are uncertain, and, at best, only some distributional information may be available. In this case, (LS) can be extended to explicitly address uncertainty by adopting a stochastic programming [21] approach. Haugen, Løkketangen and Woodruff [13] proposed a heuristic strategy for such stochastic lot-sizing problems. Ahmed, King and Parija [3] proposed an extended reformulation of the uncapacitated stochastic lot-sizing problem whose LP relaxation is significantly tighter than the standard formulation. They also point out that the Wagner-Whitin optimality conditions for deterministic uncapacitated lot-sizing problems, i.e., no production is undertaken if inventory is available, do not hold in the stochastic case. The stochastic lot-sizing problem has also been considered as subproblems embedded in some classes of stochastic capacity expansion problems [4], stochastic batch-sizing problems [16], and stochastic production planning problems [9]. In this paper, we study the polyhedral structure of the uncapacitated stochastic lot-sizing problem. We show that the (l, S) inequalities are also valid for the stochastic lot-sizing polytope. We generalize the (l, S) inequalities to a new class of valid inequalities for the stochastic lot-sizing polytope. We provide necessary and sufficient conditions that guarantee that the proposed inequalities are facet-defining, and develop separation algorithms. Our computational experiments demonstrate that the proposed inequalities are extremely useful within a branch-and-cut scheme for stochastic lot-sizing problems.

3 A Branch-and-Cut Algorithm for the Stochastic Uncapacitated Lot-Sizing Problem 3 2. The Stochastic Lot-sizing Problem A stochastic programming extension of the deterministic formulation (LS) is presented in [3]. This extension is described next. Each of the problem parameters α i, β i, h i, and d i are assumed to evolve as a discrete time stochastic process with finite probability space. This information structure can be interpreted as a scenario tree with T levels (or stages) where a node i in stage t of the tree gives the state of the system that can be distinguished by information available up to time stage t. Each node i of the scenario tree, except the root node (indexed as i = 0), has a unique parent a(i), and each non-terminal node i is the root of a subtree T (i) = (V(i), E(i)), which contains all descendants of node i. For notational brevity we use T = T (0) and V = V(0) for the whole tree. The set of leaf nodes of T is denoted by L. The probability associated with the state represented by node i is p i. The set of nodes on the path from the root node to node i is denoted by P(i). If i L then P(i) corresponds to a scenario, and represents a joint realization of the problems parameters over all periods 1,..., T. We define P(i, j) = {k : k P(j) V(i)}, thus P(i) = P(0, i). Similarly, we let d ij = k P(i,j) d k. We let C(i) denote the set of nodes that are immediate children of node i, i.e., C(i) = {j : a(j) = i}; t(i) denote the time stage or level of node i in the tree, i.e., t(i) = P(i) ; L(i) denote the leaf nodes of the subtree T (i). Using this notation, a multi-stage stochastic integer programming formulation of the single-item, uncapacitated, stochastic lot-sizing problem is: (SLS1) : min p i (α i x i + β i y i + h i s i ) i V s.t. s a(i) + x i = d i + s i i V, x i M i y i i V, x i, s i 0, y i {0, 1} i V, s a(0) = 0, where x i represents the production in period t(i) corresponding to the state defined by node i, similarly s i represents the inventory at the end of period t(i) and y i is the indicator variable for a production set-up in period t(i). An upper bound on x i is given by M i = max j L(i) d ij. Upon eliminating variables s i from (SLS1), we obtain the reformulation: (SLS) : min i V(ᾱ i x i + β i y i ) (1) s.t. j P(i) x j d 0i i V, (2) 0 x i M i y i i V, (3) y i {0, 1} i V, (4)

4 4 Yongpei Guan et al. where ᾱ i = p i α i + j V(i) p jh j and β i = p i β i. Throughout this paper, we use the formulation (SLS) for the stochastic lot-sizing problem. The set of feasible solutions to (SLS) defined by the constraints (2)-(4) is denoted by X SLS. 3. Valid Inequalities for the Stochastic Lot-Sizing Problem In this section, we provide valid inequalities for the stochastic lot-sizing problem. We first show that the well-known (l, S) inequalities, for the deterministic lotsizing problem, are valid for (SLS). These inequalities are based on a sequence of consecutive time periods that can be thought of as a path in the scenario tree T. Next, we extend the (l, S) inequalities to a general class, called the (Q, S Q ) inequalities, which are derived from subtrees of T The (l, S) inequalities Theorem 1. Given l V and S P(l), the (l, S) inequality d il y i d 0l, i S x i + i S where S = P(l) \ S, is valid for X SLS. Proof. The proof is analogous to that of the deterministic case (cf. [7]). Given a point (x, y) X SLS, we consider two cases: (a) there exists i S such that y i = 1, and (b) y i = 0 for all i S. Case (a): Let k = argmin{t(i) : i S, y i = 1}. Then y i = 0 and x i = 0 for all i S P(a(k)). Hence d il y i x i + d kl d 0a(k) + d kl = d 0l. i S x i + i S i P(a(k)) Case (b): If y i = 0 for all i S, then x i + d il y i = i S i S x i d 0l. i P(l) 3.2. The (Q, S Q ) inequalities In this section, we extend the (l, S) inequalities to a general class called the (Q, S Q ) inequalities. Consider a subset Q V \ {0} satisfying the following properties: (A1) If i, j Q, then d 0i d 0j.

5 A Branch-and-Cut Algorithm for the Stochastic Uncapacitated Lot-Sizing Problem 5 (A2) If i, j Q, then i / P(j) and j / P(i). (A1) allows us to uniquely index the nodes in the set Q as {1, 2,..., Q} where Q = Q, such that d 01 < d 02 < < d 0Q. (A2) simply gives us a convenient way of defining the subtrees over which the (Q, S Q ) inequalities are defined. We will comment on (A1) and (A2) at the end of this section. Define T Q = {V Q, E Q } to be the subtree of T whose leaf nodes are Q, i.e, V Q = i Q P(i). Note that by (A2), all nodes in Q are leaf nodes of T Q. Given i V Q, we denote by T Q (i) = {V Q (i), E Q (i)} the subtree of T Q with i as the root node. Note that V Q (i) = V(i) V Q. We use Q(i) Q to denote the set of leaf nodes of the subtree T Q (i), i.e., Q(i) = V Q (i) Q. In addition to (A1) and (A2), we need the following property on the set Q for the validity of the (Q, S Q ) inequalities: (A3) Given any node k V Q, and nodes i, j Q such that i < j and i, j Q(k), we have that {i, i + 1,..., j 1, j} Q(k). Given a subset Q, define the following quantities for all nodes i V Q : D Q (i) = max{d 0j : j Q(i)} (5) D Q (i) = { 0, if {j : j Q \ Q(i) such that d0j D Q (i)} = max{d 0j : j Q \ Q(i) such that d 0j D Q (i)}, otherwise (6) M Q (i) = max{d ij : j Q(i)} (7) { Q (i) = min D Q (i) D } Q (i), M Q (i). (8) Given k Q, let Q k = {1, 2,..., k 1, k} and T Qk = {V Qk, E Qk } be the subtree of T with leaf nodes Q k. It is easily verified that, if Q satisfies (A1)- (A3) then every subset Q k for k = 1,..., Q satisfies these properties as well. Now, let K Q, and suppose there exists a j V QK such that j P(K) and D QK (j ) > 0. Then there exists r Q such that D QK (j ) = d 0r. Clearly 1 r K. Let u = argmax{t(i) : i V Qr P(K)}. Figure 1 illustrates the relative position of the nodes j, r, and u, and the set V Qr. In this figure Q K = {1, 2, 3, r, K 1, K}, Q r = {1, 2, 3, r }, V QK is the set of all nodes and V Qr is the set of nodes within the dotted area as shown in the graph. From (A3), it follows that u P(r ). If not, then there exists a r < r, r Q r such that u P(r ) since u V Qr. Thus, we have r, K Q(u ) and r < r K. Then according to (A3), we have r Q(u ), which contradicts u / P(r ). For K, j, r and u defined as above, we need the following two lemmas. Lemma 1. QK (i) Qr (i) for all i P(u ). Proof. We have D QK (i) = d 0K d 0r = D Qr (i) for all i P(u ). (9) Furthermore, for all i P(u ), we have r, K V QK (i). It then follows from (A3) that Q K (i) = Q r (i) {r + 1,..., K}. Thus Q K \ Q K (i) = {1,..., K} \ (Q r (i) {r + 1,..., K}) = ({1,..., K} \ {r + 1,..., K}) \ Q r (i) = Q r \ Q r (i). (10)

6 6 Yongpei Guan et al. Fig. 1. Notation for Lemmas 1 and 2 (For example, in Figure 1, consider node i 1 P(u ), then Q K (i 1 ) = {2, 3, r, K 1, K} and Q r (i 1 ) = {2, 3, r }. Thus Q K \ Q K (i 1 ) = Q r \ Q r (i 1 ) = {1}.) Next, note that for all i P(u ), (9) implies that d 0j D QK (i) = d 0K for all j Q K and d 0j D Qr (i) = d 0r for all j Q r. Thus for all nodes i P(u ), D QK (i) = max{d 0j : j Q K \ Q K (i)} and D Qr (i) = max{d 0j : j Q r \ Q r (i)}. It then follows from (10) that Since Q r (i) Q K (i), we also have D QK (i) = D Qr (i) for all i P(u ). (11) M QK (i) M Qr (i) for all i P(u ). (12) The lemma follows from (9), (11), (12) and the definition of. Lemma 2. QK (i) = Qr (i) for all i V Qr \ P(u ). Proof. We first claim that j V Qr. (13) Suppose that j V Qr. Then there exists r j Q such that r j r < K, i.e., r j Q K (j ). Note that by definition r Q K (j ). Since K Q K (j ) and r j r < K, we have a contradiction to (A3). Thus (13) holds.

7 A Branch-and-Cut Algorithm for the Stochastic Uncapacitated Lot-Sizing Problem 7 Next, we show that Q r (i) = Q K (i) for all i V Qr \ P(u ). (14) Clearly Q r (i) Q K (i). Now, suppose there exists a k Q K (i) such that k > r. Note that i V Qr and j V Qr from (13), thus j P(i). Furthermore we also have i P(j ), otherwise by definition of u we would have i P(u ). Thus i V QK (j ) and so k V QK (j ). Thus d 0r = D QK (j ) = max{d 0j : j Q K \ Q K (j ) and d 0j D QK (j ) = d 0K } d 0k, which is a contradiction to k > r. Thus (14) is true. (The claim is clear in Figure 1. Consider the node i 2 V Qr \ P(u ). Here Q r (i 2 ) = Q K (i 2 ) = {2}.) From (14), we have and D QK (i) = D Qr (i) for all i V Qr \ P(u ), (15) M QK (i) = M Qr (i) for all i V Qr \ P(u ). (16) From (14) and (15), we have D QK (i) = max{d 0j : j Q K \ Q r (i) and d 0j D Qr (i)}. Now, consider the set {j : j Q K \ Q r (i) and d 0j D Qr (i)} = {j : j (Q r {r + 1,..., K}) \ Q r (i) and d 0j D Qr (i)} = {j : j Q r \ Q r (i) and d 0j D Qr (i)}, where the last step follows from the fact that D Qr (i) d 0r and d 0j > d 0r for all j {r + 1,..., K}. Thus D QK (i) = D Qr (i) for all i V Qr \ P(u ). (17) The lemma follows from (15), (16), (17) and the definition of. We are now ready to state the (Q, S Q ) inequalities and prove their validity. Theorem 2. Given any Q V satisfying (A1), (A2), and (A3) and any subset S Q V Q, the inequality x i + Q (i)y i M Q (0), i S Q i S Q where S Q = V Q \ S Q, called a (Q, S Q ) inequality, is valid for X SLS. Proof. We show by induction over k {1,..., Q} that any (Q k, S Qk ) inequality is valid for X SLS. The base case (k = 1): Note that D Q1 (i) = d 01, D Q1 (i) = 0, and M Q1 (i) = d i1 for all i V Q1. Given any point (x, y) X SLS, the left-hand-side of the (Q 1, S Q1 ) inequality is given by x i + min{d 01, d i1 }y i = x i + d i1 y i d 01 = M Q1 (0). i S Q1 i S Q1 i S Q1 i S Q1

8 8 Yongpei Guan et al. The first equality follows from the fact that d 01 d i1, the inequality follows from the validity of the (l, S) inequality with l = 1 and S = S Q1, and the last equality follows from the definition of M Q1 (0). The inductive step: We assume that for all k {1,..., K 1} (where K 1 < Q), given any S Qk V Qk, the (Q k, S Qk ) inequality is valid for X SLS. Consider any S QK V QK, we show that the (Q K, S QK ) inequality i S QK x i + i S QK QK (i)y i M QK (0) is also valid for X SLS. Let F K = {i P(K) S QK : D QK (i) D QK (i) < M QK (i)}. Given any solution (x, y) X SLS, we consider two cases: (a) there exists j F K such that y j = 1, and (b) y j = 0 for all j F K. Case (a): Note that D QK (j ) D QK (j ) < M QK (j ) implies D QK (j ) > 0 since D QK (j ) M QK (j ). Thus there exists r Q such that D QK (j ) = d 0r. Let S Qr = S QK V Qr and S Qr = S QK V Qr. The left-hand-side of the (Q K, S QK ) inequality is then equal to x i + (18) i S Qr x i + (19) i S QK \S Qr QK (i)y i + (20) i S Qr QK (i)y i. (21) i S QK \S Qr As before, let u = argmax{t(i) : i V Qr P(K)}. Expression (20) can be further disaggregated into QK (i)y i + (22) i S Qr P(u ) i S Qr \P(u ) From Lemma 1, it follows that (22) Qr (i)y i, i S Qr P(u ) QK (i)y i. (23)

9 A Branch-and-Cut Algorithm for the Stochastic Uncapacitated Lot-Sizing Problem 9 and from Lemma 2, it follows that (23) = Qr y i. i S Qr \P(u ) From the validity of the (Q r, S Qr ) inequality, we then have (18) + (22) + (23) M Qr (0) = d 0r. Now consider the expression (21). Since j S QK \ S Qr non-negative, we have that and all coefficients are (21) D QK (j ) D QK (j ) = d 0K d 0r. Thus which implies (18) + (22) + (23) + (21) d 0K, (18) + (19) + (22) + (23) + (21) d 0K = M QK (0). Therefore the (Q K, S QK ) inequality is valid. Case (b): The left-hand-side of the (Q K, S QK ) inequality equals x i + QK (i)y i i S QK i S QK x i + QK (i)y i = = i S QK P(K) i S QK P(K) i S QK P(K) x i + x i + d 0K = M QK (0), i S QK P(K) i S QK P(K) i S QK P(K) M QK (i)y i d ik y i where the third expression follows from the fact that y j = 0 for all j S QK P(K) such that D QK (j) D QK (j) < M QK (j), the fourth expression follows from the definition of M QK (j), and the fifth expression follows from the validity of the (l, S) inequality with l = K and S = S QK P(K). Therefore the (Q K, S QK ) inequality is valid. We conclude this section with a discussion of properties (A1) and (A2) and an example that illustrates the (Q, S Q ) inequalities. Suppose property (A1) does not hold for some Q. In particular, suppose there exists a pair of nodes q 1, q 2 Q such that d 0q1 = d 0q2. Without loss of generality, we index the nodes in Q such that q 2 > q 1. Let Q = Q \ {q 2 }. Note that Q satisfies (A1). From the fact that

10 10 Yongpei Guan et al. d 0q1 = d 0q2, it can be easily verified that Q (i) = Q (i) for all i V Q and M Q (0) = M Q (0). Now, let S Q = S Q V Q and S Q = S Q V Q. Then i S Q x i + i S Q = i S Q i S Q Q (i)y i x i + i S Q x i + i S Q M Q (0) = M Q (0). Q (i)y i Q (i)y i Thus the (Q, S Q ) inequality is valid. However, this inequality is clearly dominated by the (Q, S Q ) inequality. Consequently, (A1) is without loss of generality. Suppose property (A2) does not hold for some Q and there exists a pair of nodes q 1, q 2 Q such that q 1 P(q 2 ). Then V Q = V Q\{q1} and we only need to consider (Q, S Q ) inequalities corresponding to Q \ {q 1 } instead of Q. Consequently, (A2) is without loss of generality. Example: Consider an instance of (SLS) with 7 nodes as shown in Figure 2. The problem parameters are shown in the columns labelled α i, β i and d i in Table 1. The optimal LP relaxation objective value of (SLS) is and the corresponding optimal solution (x, y) is shown in the columns labelled x 1 and y 1 in Table 1. We augment the LP relaxation with 3 (Q, S Q ) inequalities: 10y 0 10, i.e., Q = {0}, S Q = {0} x 0 + x 2 + 5y 3 35, i.e., Q = {2, 3}, S Q = {3} x 0 + 5y y 2 + 5y 4 35, i.e., Q = {2, 4}, S Q = {1, 2, 4} Then we obtain an integral optimal solution as shown in columns labelled x 2 and y 2 in Table 1 and the corresponding optimal objective value is Fig. 2. Tree for the example

11 A Branch-and-Cut Algorithm for the Stochastic Uncapacitated Lot-Sizing Problem 11 α i β i d i x 1 y 1 x 2 y Table 1. Data for the example 4. Facets for the Stochastic Lot-Sizing Problem In this section we give some classes of facets for the stochastic lot-sizing polyhedron. First, we identify some facets from the original inequalities defining X SLS. Next, we provide necessary and sufficient conditions under which a (Q, S Q ) inequality is facet-defining. We make the assumption (A4) d i > 0 for all i V throughout the remainder of this paper. Given (A4), the following results can be shown by constructing appropriate sets of affinely independent solutions. Recall that V = N. Proposition 1. The dimension of X SLS is 2N 1. Proposition 2. The inequalities (i) x i M i y i for i V \ {0}, (ii) y i 1 for i V \ {0}, (iii) x i 0 for i V \ {0}, are facet-defining for X SLS. Note that, the inequalities y i 0, i V \ {0}, are not facet-defining because y i = 0 implies x i = 0, and therefore we can have no more than 2N 2 affinely independent solutions satisfying y i = 0. We now establish a set of conditions guaranteeing that a (Q, S Q ) inequality is facet-defining. Let F Q = {i S Q : D Q (i) D Q (i) < M Q (i)} and G Q = S Q \F Q. Thus, V Q = F Q G Q S Q. We need the following definitions. Definition 1. Given Q V and S Q V Q, the neighborhood of (Q, S Q ) is N (Q, S Q ) = i V Q \( j SQ V Q (j)) C(i) \ V Q. For example, in Figure 3, let Q = {1, 2, 3, 4} and S Q = {0, 3, 5, 9}, then N (Q, S Q ) contains the two nodes shaded horizontally.

12 12 Yongpei Guan et al. Fig. 3. Partitioning of the node set V used in the proof of Theorem 3 Definition 2. Given j V Q, let q j = max{i : i Q(j)} and W(j) = } argmin {t(m) : m S Q P(i) \ V Qqj. i Q\Q qj For example, in Figure 3, if j = 9 then q j = 2 and W(j) = {4, 7}; and if j = 6 then q j = 3 and W(j) = {4}. Theorem 3. The (Q, S Q ) inequality x i + Q (i)y i M Q (0) i S Q i S Q is facet-defining if and only if (i) 0 S Q, (ii) M Q (0) max i N (Q,SQ ){d 0i }, (iii) For each j V Q, (a) W(j) P(i), i Q \ Q qj, (b) If j F Q, then D Q (j) d 0a(k), k W(j), (c) If j G Q, then d 0a(j) d 0a(k), k W(j), (d) If j S Q, then D Q (j) > d 0a(k), k W(j), (iv) ( i GQ argmax{j : j Q(i)}) L =. Proof. The proof is constructive and the details are given in the Appendix.

13 A Branch-and-Cut Algorithm for the Stochastic Uncapacitated Lot-Sizing Problem 13 Example (continued): Consider the three inequalities added in the example. The first one is not facet-defining since 0 / S Q. The second one is not facet-defining since it does not satisfy condition (ii) of Theorem 3. However, the third inequality is facet-defining. To illustrate the necessity of condition (iii), the inequality x 0 + x 1 + x 4 + x y 6 45, where Q = {4, 6} and S Q = {6}, is not facetdefining since D Q (4) = d 0a(6) and 6 W(4), which contradicts condition (d) of (iii). On the other hand, the inequality x 0 + x 1 + x 4 + x y 5 45, where Q = {4, 5} and S Q = {5}, satisfies all four conditions of Theorem 3 and therefore is facet-defining. Recall that every (l, S) inequality is a (Q, S Q ) inequality with Q = {l} and S Q = S. We then have the following corollary to Theorem 3. Corollary 1. An (l, S) inequality is facet-defining if and only if l and S are such that (i) 0 S Q, (ii) d 0l max i N(l,S) d 0i, (iv) P(l) \ S, l / L or P(l) \ S =, l L. In this case, the neighborhood is simply N(l, S) = {j : j C(i) \ P(l) where i < argmin{t(k) : k S}}, and condition (iii) is redundant since W(j) = for all j V Q. Remark 1. From above, we can see that (Q, S Q ) inequalities suffice to describe the convex hull of the deterministic case of (SLS) since in this case, they are equivalent to the (l, S) inequalities. Moreover, the (Q, S Q ) inequalities are also sufficient to describe the convex hull when (SLS) has two periods [12]. 5. Separation of (Q, S Q ) inequalities Given the set Q, and a fractional solution (x, y ) of (SLS), let S Q = {i V Q : x i Q (i)y i }. (24) If i S x Q i + i S Q (i)y Q i < M Q(0), then the (Q, SQ ) inequality is violated. On the other hand, if (x, y ) satisfies the (Q, SQ ) inequality then there are no violated (Q, S Q ) inequalities corresponding to the node set Q, since min x i + S Q V Q Q (i)yi = x i + Q (i)yi M Q (0). i S Q i S Q i S Q i S Q The difficulty in separating (Q, S Q ) inequalities is how to determine Q. The (Q, S Q ) inequalities with Q = Q can be separated in O(N Q+1 ) time by enumeration since for each such Q, we can check for a violated (Q, S Q ) inequality in O(N) time. Because we don t know a polynomial algorithm for general Q, we only check all of the Q = 1 and Q = 2 inequalities for violations and then

14 14 Yongpei Guan et al. we apply a heuristic (Algorithm 1) to try to find some violated inequalities for larger Q. The basic idea of Algorithm 1 is to add nodes to Q, using a depth-first strategy, such that the right-hand-side of the inequality is not changed while the left-hand-side decreases. The process stops as soon as we find a violated (Q, S Q ) inequality. If no violated inequality is found after exhausting the depth-first search, we re-start the search with a new node. Algorithm 1 Heuristic separation of {Q, S Q } inequalities with Q 3 Input: a fractional solution (x, y ). for l V do Step 0. Set Q = {l} and i = l. Step 1. If Q 3, go to Step 2. Otherwise, go to Step 3. Step 2. Compute SQ as in (24). If the (Q, S Q ) inequality is violated stop. Step 3. For some node j V(a(i)) \ V(i), let Q = Q {j}. If a node k = argmax{d 0j : j V(a(i)) \ V(i), d 0j < d 0i and i S x Q i + i S Q Q (i)y i < i S Q x i + i S Q(i)y Q i } exists, go to Step 5. Otherwise, go to Step 4. Step 4. If i 0, set i a(i) and go to Step 3. If i = 0 end for. Step 5. Set Q Q {k} and i k and go to Step 1. end for 6. Computational Experiments In this section, we report on the computational effectiveness of the (Q, S Q ) inequalities on randomly generated instances of single-item, uncapacitated, stochastic lot-sizing problems Implementation We implemented a branch-and-cut scheme in which complete separation of (Q, S Q ) inequalities is done for Q = 1 and Q = 2 followed by Algorithm 1. We add all violated Q = 1 inequalities if some are found and repeat until no more are found. We do the same for Q = 2 inequalities. When no more of these are found, we apply Algorithm 1 and add inequalities one-at-a-time until no further violation is found. Our implementation was carried out in C using the callable libraries of CPLEX 8.1. Default CPLEX options were used throughout. All computations were carried out on a 2.4GHz Intel Xeon/Linux workstation with 2GB RAM with one hour time limit per run Test problem generation A number of instances of (SLS) were generated corresponding to different structures of the underlying scenario trees, different ratios of the production cost to

15 A Branch-and-Cut Algorithm for the Stochastic Uncapacitated Lot-Sizing Problem 15 the inventory holding cost, and different ratios of the setup cost to the inventory holding cost. We assumed that the underlying scenario tree is balanced with T stages and K branches per stage. We considered 6 different tree structures with K = 2 and T {10, 11}; K = 3 and T {6, 7}; K = 4 and T {5, 6}. We considered three different levels of production to holding cost ratio α/h {50, 100, 200}, and three different levels of setup to holding cost ratio β/h {1750, 3500, 7000}. For each of the 54 combinations of the tree structure, α/h and β/h, we generated three random instances as follows. For each node i of the tree, the holding cost h i U[0.01, 0.05], i.e., a uniform random number in the interval [0.01, 0.05]; α i U[0.8(α/h) h, 1.2(α/h) h] where h = 0.03 is the average holding cost; β i U[0.8(β/h) h, 1.2(β/h) h]; and d i U[10, 100]. Finally, all K children of a node were assigned equal probabilities Results Tables 2, 3, and 4 report on the effectiveness of the (Q, S Q ) inequalities in tightening the LP relaxation gap for the instances corresponding to K = 2, 3 and 4 at the root node. The column labelled LP Gap % gives the relative LP relaxation gap of the original formulation (SLS) with respect to the best feasible solution found with our branch-and-cut scheme. The columns labelled Q = 1, Q = 2 and General Q correspond to the results from adding all violated (Q, S Q ) inequalities for Q = 1 and then those for Q = 2, and then heuristically for some violated inequalities with Q > 2. For each combination of T, β/h and α/h, there are two rows corresponding to the columns labelled Q = 1, Q = 2 and General Q. The first row gives the LP relaxation gap after adding the (Q, S Q ) inequalities, and the second row gives the number of (Q, S Q ) inequalities added. Note that all reported numbers are averages over three instances. Significant tightening of the LP relaxation is achieved via the proposed (Q, S Q ) inequalities. In some cases, the LP relaxation gap is reduced from over 20% to 0.4%. Furthermore, in most cases, the LP relaxation gap is small after adding the inequalities corresponding to Q = 1 and Q = 2. The results from our branch-and-cut scheme are reported in Tables 5, 6, and 7 for the instances corresponding to K = 2, 3 and 4, respectively. For each combination of T, β/h and α/h, there are two rows. The first row gives the performance of the default CPLEX MIP solver and the second row gives the performance of our branch-and-cut scheme. We give the number of cutting planes added by the default CPLEX MIP solver and by our branch-and-cut scheme respectively, the relative optimality gap upon termination, the number of nodes explored (apart from the root node), and the total CPU time. The reported data is averaged over three instances. The numbers in square brackets indicate the number of instances not solved to default CPLEX optimality tolerance within the allotted time limit of one hour. The default CPLEX MIP solver adds several types of cuts including flow covers, Gomory fractional cuts and mixed integer rounding cuts. Our branch-and-cut algorithm adds (Q, S Q ) cuts at each node after the

16 16 Yongpei Guan et al. CPLEX default cuts have been added. For the total CPU time, we report the average CPU time for instances that are solved to default CPLEX optimality tolerance within the allotted time limit of one hour. Otherwise, we use to represent the case that no instance is solved to default CPLEX optimality tolerance within the allotted time. The efficiency of the (Q, S Q ) inequalities within our branch-and-cut is clearly observed. Our branch-and-cut algorithm proves optimality for all instances for K = 2, has only 11 and 25 instances unsolved to optimality for K = 3 and K = 4, respectively. In contrast, the unsolved instances corresponding to default CPLEX are 6, 43 and 52, respectively. For cases where neither algorithm could prove optimality, our algorithm yielded much smaller optimality gaps. Moreover, our cuts dramatically reduced the number of nodes in the tree and, although we added many more cuts, the running times were smaller as well. Because we add so many (Q, S Q ) inequalities, we thought that the running times might be reduced substantially by deleting cuts that were no longer tight. However, experiments using cut management did not yield significant improvement. Table 2. Results for the root node (K = 2) T β/h α/h LP Gap % Q = 1 Q = 2 General Q

17 A Branch-and-Cut Algorithm for the Stochastic Uncapacitated Lot-Sizing Problem 17 Table 3. Results for the root node (K = 3) T β/h α/h LP Gap % Q = 1 Q = 2 General Q Table 4. Results for the root node (K = 4) T β/h α/h LP Gap % Q = 1 Q = 2 General Q

18 18 Yongpei Guan et al. Table 5. Results for branch-and-cut (K = 2) T β/h α/h No. of cuts Optimality gap % Nodes CPU secs [3] *** [3] *** Table 6. Results for branch-and-cut (K = 3) T β/h α/h No. of cuts Optimality gap % Nodes CPU secs [1] [3] *** [3] *** [3] *** [3] *** [2] [3] *** [2] [3] *** [3] *** [3] *** [3] *** [3] *** [3] *** [3] *** [1] [3] *** [3] 2000 *** [3] *** [3] 1768 ***

19 A Branch-and-Cut Algorithm for the Stochastic Uncapacitated Lot-Sizing Problem 19 Table 7. Results for branch-and-cut (K = 4) T β/h α/h No. of cuts Optimality gap % Nodes CPU secs [3] *** [3] *** [2] [3] *** [3] *** [3] *** [2] [3] *** [2] [3] *** [2] [3] *** [3] *** [3] 483 *** [3] *** [3] *** [3] 425 *** [3] *** [3] 143 *** [3] *** [3] 112 *** [3] *** [3] 46 *** [3] *** [3] 0 *** [3] *** [3] 0 ***

20 20 Yongpei Guan et al. Appendix Theorem 3. The (Q, S Q ) inequality i S Q x i + is facet-defining if and only if i S Q Q (i)y i M Q (0) (i) 0 S Q, (ii) M Q (0) max i N (Q,SQ ){d 0i }, (iii) For each j V Q, (a) W(j) P(i), i Q \ Q qj, (b) If j F Q, then D Q (j) d 0a(k), k W(j), (c) If j G Q, then d 0a(j) d 0a(k), k W(j), (d) If j S Q, then D Q (j) > d 0a(k), k W(j), (iv) ( i GQ argmax{j : j Q(i)}) L =. Proof of sufficiency. We first describe the construction of 2N 1 vectors that are in X SLS and satisfy the (Q, S Q ) inequality at equality. Then we show that the vectors are linearly independent. Given the (Q, S Q ) inequality, we partition V into disjoint sets V = {0} A Z B, where A = V Q \ {0}, Z = {j : j V \ V Q and a(j) V Q } and B = V \ (V Q Z). Note that we have N (Q, S Q ) Z. Nodes in the set V \ V Q correspond to a forest, and Z represents the set of root nodes of the subtrees in this forest. This partitioning is illustrated in Figure 3. Here Q = {1, 2, 3, 4}, V Q = {0, 1, 2,..., 9}, S Q = {0, 3, 5, 9}, S Q = {1, 2, 4, 6, 7, 8} (shaded diagonally), and A = {1, 2,..., 9}. The two horizontally shaded nodes in Z represent N (Q, S Q ). Construction: We create one vector u 0 for the root node {0} and two vectors u j and v j for each node j V \ {0}. We let u 0 = M Q (0)e x0 + e y0 + i Z(M i e xi + e yi ), where e xi and e yi are unit vectors in R 2N corresponding to the coordinates x i and y i, respectively. j B: We let u j = u 0 + e yj, and v j = u 0 + M j e xj + e yj. j A:

21 A Branch-and-Cut Algorithm for the Stochastic Uncapacitated Lot-Sizing Problem 21 If j S Q, we let u j = u 0 + (D Q (j) ε M Q (0))e x0 + εe xj + e yj + i W(j) (M Q(i)e xi + e yi ), where ε is a sufficiently small positive number, and v j = u 0 + e yj. If j S Q, we let j Z: If j N (Q, S Q ), we let u j = u 0 + (D Q (j) Q (j) M Q (0))e x0 + Q (j)e xj + e yj + i W(j) (M Q(i)e xi + e yi ) and v j = u j + εe xj. u j = u 0 M j e xj e yj + i B (M ie xi + e yi ) v j = u j + e yj. and If j Z \ N (Q, S Q ), define k j = argmin{t(i) : i S Q P(j)}. Note that k j S Q by definition. We let u j = u kj + (M kj Q (k j ))e x k j Mj e xj e yj and v j = u j + e yj. Feasibility: It is obvious that u 0 X SLS. Consequently, the vectors {u j, v j } j B and {v j } j SQ are also feasible. Now we verify the feasibility of u j for j S Q. Given j S Q, u j satisfies 0 x i M i y i and y i {0, 1} for all i V since x 0 < M Q (0) M 0, Q (j) M Q (j) M j and M Q (k) M k k W(j). Therefore, we just need to check that u j satisfies constraint (2) for all i V = {0} A Z B. Clearly u j satisfies constraint (2) for i = 0. Also, note that if u j satisfies constraint (2) for i {0} A, then it satisfies constraint (2) for i Z B since x i = M i and y i = 1 for all i Z, and the nodes in Z include an ancestor of each node in B. Therefore, we just need to show that u j satisfies constraint (2) for i A = S Q S Q. Note that u j yields x 0 = D Q (j) Q (j) D Q (j) M Q (j) = d 0a(j), (25) where the second line follows from the definition of Q (j) and the third line follows from the definition of D Q (j) and M Q (j). It then follows that u j satisfies constraint (2) for all i P(a(j)).

22 22 Yongpei Guan et al. Next, note that u j yields x 0 = D Q (j) Q (j) D Q (j) (D Q (j) D Q (j)) = D Q (j), (26) where the second line follows from the definition of Q (j). If D Q (j) > 0, then we know that there exists r j Q such that D Q (j) = d 0rj. Thus (26) implies that u j satisfies constraint (2) for all i V Qrj. Also, note that u j yields x 0 + x j = D Q (j). (27) Since 0 P(i) and j P(i) for all i V Q (j), (27) implies that u j satisfies (2) for all i V Q (j). Next, considering (b) and (c) of condition (iii), (25) and (26) imply that u j satisfies x 0 d 0a(k) k W(j). (28) Thus u j satisfies (2) for all i P(a(k)) k W(j). Finally, note that {0} A = V Q = P(j) V Qrj V Q (j) P(a(k)) k W(j) k W(j) V Q (k). So it only remains to check that u j satisfies (2) for all i k W(j) V Q(k). Given any k W(j), note that u j satisfies x 0 + x k d 0a(k) + M Q (k) = D Q (k), (29) where the first line follows from (28) and the second line follows from the definition of D Q (k). Since, for all i V(k) we have 0 P(i), k P(i) and d 0i D Q (k), it follows that u j satisfies constraint (2) for all i V Q (k) and k W(j). v j for j S Q is feasible because v j satisfies constraint (2) since v j u j and condition (iv) ensures that v j satisfies 0 x i M i y i and y i {0, 1}. The feasibility of u j for j S Q can be established using analogous arguments as long as ε Q (j), D Q (j) ε D Q (j) and D Q (j) ε d 0a(k) k W(j). We now verify the feasibility of u j for j N (Q, S Q ). As before, we only need to verify that u j satisfies constraint (2) for all i V. Since the construction of u j only affects nodes i V(j), from the feasibility of u 0, constraint (2) is satisfied for all i V \ V(j). Given any node i V(j), note that u j satisfies k P(i) x k = M Q (0) + k P(i)\P(j) M k (30) d 0j + k P(i)\P(j) d k = d 0i,

23 A Branch-and-Cut Algorithm for the Stochastic Uncapacitated Lot-Sizing Problem 23 where the first line follows from the construction of u j and the second line follows from condition (ii). Thus u j satisfies (2) for all i V. We now verify the feasibility of u j for j Z \ N (Q, S Q ). Since the construction of u j only affects nodes i V(k j ), from the feasibility of u kj (recall that k j S Q ), constraint (2) is satisfied for all i V \ V(k j ). Given any node i V(k j ), note that u j satisfies x 0 + x kj d 0a(kj) + M kj d 0i, (31) where the first line follows from (28) and the construction of u j, and the second line follows the definition of M kj and the fact that k j P(i). Thus u j satisfies constraint (2) for all i V. Finally, v j for j Z is feasible since v j u j. Tightness of the (Q, S Q ) inequality: Here we prove the claim that the (Q, S Q ) inequality is tight or active at each of the solutions vectors u 0 and {u j, v j } j V\{0}. This claim is true for u 0, {u j, v j } j B, {v j } j SQ and {u j, v j } j N (Q,SQ ). Furthermore, for any j Z \N (Q, S Q ), we have that u j and v j satisfy the claim as long as u kj satisfies the claim since k j S Q. Similarly the solutions {v j } j SQ satisfy the claim as long as {u j } j SQ satisfy the claim. Therefore, we just need to prove the claim for {u j } j SQ S Q. Here we prove the claim for {u j } j SQ. The proof for {u j } j SQ is nearly identical, see Guan [11] for details. Since j S Q and W(j) S Q, we have that u j satisfies x j i = { DQ (j) Q (j) if i = 0 0 if i S Q \ {0}, and y j i = { 1 if i {j} W(j) 0 if i S Q \ ({j} W(j)). Thus u j satisfies i S Q x j i + i S Q Q (i)y j i = D Q(j) + i W(j) Q (i). (32) It remains to show that the right-hand side of the above expression is equal to M Q (0). If W(j) = then D Q (j) = M Q (0) by definition of W(j). If W(j), note that for all i W(j), D Q (i) D Q (i) D Q (i) D Q (j) D Q (i) d 0a(i) = M Q (i), where the first inequality follows from the fact that D Q (i) d 0qj = D Q (j) since i / V Qqj, and the second inequality follows from the fact that D Q (j) D Q (j)

24 24 Yongpei Guan et al. d 0a(i) from case (b) of condition (iii) or D Q (j) d 0a(j) d 0a(i) from case (c) of condition (iii). Thus, for any node i W(j), Q (i) = D Q (i) D Q (i). (33) By Property (A2), we index the nodes in W(j) as i 1, i 2,..., i W such that D Q (i 1 ) < D Q (i 2 ) <... < D Q (i W ). From this indexing scheme, the definition of D Q, D Q, and W(j), it follows that D Q (i 1 ) = D Q (j), D Q (i W ) = M Q (0), and Thus D Q (i k ) = D Q (i k+1 ) k = 1, 2,..., W 1. D Q (j) + i W(j) Q (i) = M Q (0), and the (Q, S Q ) inequality is tight for u j, j S Q. Linear Independence: Given the 2N 1 vectors u 0 and {u j, v j } j V\{0}, we perform a sequence of linear combinations to obtain the following (2N V Q 1) unit vectors. j B: j A: If j S Q : If j S Q : j Z: If j N (Q, S Q ): e xj = 1 M j (v j u j ), and e yj = u j u 0. e yj = v j u 0. e xj = 1 ε (vj u j ). e yj = v j u j, and e xj = 1 M j (u 0 u j e yj + i B (M ie xi + e yi )). If j Z \ N (Q, S Q ), let k j = argmin{t(i) : i S Q P(j)}. e yj = v j u j, and e xj = 1 M j (u kj + (M kj Q (k j ))e x k j u j e yj ). An additional sequence of linear combinations gives the following additional V Q vectors. u 0 = u 0 i Z (M ie xi + e yi ). j S Q \ {0}, u j = u j i Z (M ie xi + e yi ) i W(j) M Q(i)e xi e yj = (D Q (j) ε)e x0 + e y0 + i W(j) eyi + εe xj.

25 A Branch-and-Cut Algorithm for the Stochastic Uncapacitated Lot-Sizing Problem 25 j S Q, v j = v j i Z (M ie xi + e yi ) ( Q (j) + ε)e xj i W(j) M Q(i)e xi = (D Q (j) Q (j))e x0 + e y0 + i W(j) eyi + e yj. We now construct a matrix M whose rows are the (2N 1) vectors u 0, {e xj } j B, {e yj } j B, {u j } j SQ \{0}, {e yj } j SQ \{0}, {e xj } j SQ, {v j } j SQ, {e xj } j N (Q,SQ ), {e yj } j N (Q,SQ ), {e xj } j Z\N (Q,SQ ), and {e yj } j Z\N (Q,SQ ). The resulting matrix M has the following form: {0} B S Q \ {0} S Q N (Q, S Q ) Z \ N (Q, S Q ) x 0 y 0 x y x y x y x y x y {0} M Q (0) 1 B I B I S Q \ {0} E 1 εi F S Q \ {0} I S Q I S Q G 1 H N (Q, S Q ) I N (Q, S Q ) I Z \ N (Q, S Q ) I Z \ N (Q, S Q ) I In the matrix M, the submatrices E and F arise from the nonzero elements of the vectors {u j } j SQ \{0}, and the submatrices G and H arise from the nonzero elements of the vectors {v j } j SQ. Consider the S Q S Q submatrix H. This matrix has a column corresponding to each j S Q. We arrange the columns of H such that the column corresponding to i S Q is before the column corresponding to j S Q if D Q (i) > D Q (j) or t(i) < t(j) if D Q (i) = D Q (j). Note that this arrangement is uniquely defined by assumption (A1) on the set Q. This arrangement guarantees that, for any j S Q, the column corresponding to i W(j) is before the the column corresponding to j. Consequently, the matrix H is lower-triangular and then it follows that the matrix M has rank 2N 1. This is observed by exchanging rows representing {u j } j SQ \{0} and representing {v j } j SQ, and exchanging columns labelled x in S Q \ {0} and y in S Q. Since M was obtained by a sequence of elementary row operations on the (2N 1) 2N matrix whose rows are the vectors u 0 and {u j, v j } j V\{0}, it follows that these vectors are affinely independent. Lemma 3. Consider a feasible solution (x, y) satisfying the (Q, S Q ) inequality at equality. Let j V Q be such that y j = 1, and let q j = argmax{i : i Q(j )}. Then, for all q (Q\Q qj ) {q j }, there exists exactly one node j q F Q P(q) such that y jq = 1 and (i) x i = y i = 0 i S Q P(a(j q )), (ii) x i = 0 i P(a(j q )) \ V Qrjq where r jq = {i Q : d 0i = D Q (j q )}, (iii) x i = 0 i S Q V Q (j q ) and y i = 0 i S Q V Q (j q ).

26 26 Yongpei Guan et al. (iv) i S Qrjq x i + i S Qrjq Q (i)y i = i S Qrjq x i + i S Qrjq Qrjq (i)y i = d 0rjq. Proof. For all q Q, define w(q) = argmin{t(i) : i S Q P(q) and y i = 1}. First consider q = Q. For brevity, let w = w(q). Case (a): If w does not exist, then i P(Q) x i M Q (0) and i / S Q i P(Q). Thus, j / P(Q) since j S Q and the left-hand side of the (Q, S Q ) inequality is at least i P(Q) x i + D Q (j ) D Q (j ) > M Q (0), which contradicts the assumption that the feasible solution satisfies the (Q, S Q ) inequality at equality. Case (b): If w G Q, then i P(a(w)) S Q x i +M Q (w) d 0a(w) +M Q (w) = M Q (0) since x i = y i = 0 i P(a(w)) S Q by the definition of w. Also, j w because w G Q and j F Q. Then the left-hand side of the (Q, S Q ) inequality is at least i P(a(w)) S Q x i + M Q (w) + D Q (j ) D Q (j ) > M Q (0), which again gives a contradiction. Case (c): If w F Q, let r w = {i Q : d 0i = D Q (w)}. Then by Lemmas 1 and 2, we have x i + Q (i)y i x i + (i)y Qrw i d 0rw = D Q (w). i S Qrw i S Qrw i S Qrw i S Qrw Thus the left-hand side of the (Q, S Q ) inequality is i S Qrw x i + i S Qrw Q (i)y i + D Q (w) D Q (w) (34) D Q (w) + D Q (w) D Q (w) (35) = D Q (w) = M Q (0) (36) Therefore, when the (Q, S Q ) inequality holds at equality, we have the following four properties: (a) x i = y i = 0 i S Q P(a(w)), (b) x i = 0 i P(a(w)) \ V Qrw, (c) x i = 0 i S Q V Q (w) and y i = 0 i S Q V Q (w), (d) i S Qrw x i + i S Qrw Q (i)y i = i S Qrw x i + i S Qrw Qrw (i)y i = d 0rw,

27 A Branch-and-Cut Algorithm for the Stochastic Uncapacitated Lot-Sizing Problem 27 where (a) follows from the definition of w, (b) and (c) follow from the tightness of the inequality (34), and (d) follows from the tightness of the inequality (35). Thus, by letting j Q = w, we have proved the claim for q = Q. Now, for any q {Q 1,..., r w + 1}, we have that w(q) = w = j Q. Thus the claim holds for all such q. Now consider the case when q = r w. Recall that Q rw = {1, 2,..., r w }. From property (d), we have x i + (i)y Qrw i = d 0rw. i S Qrw i S Qrw Thus the (Q rw, S Qrw ) inequality is tight. By proceeding recursively in the above manner, we can show properties (a)-(d) for Q rw. Note that this recursion terminates when w = j. Since, otherwise, there must exist a w selected at some step such that w P(j ), which contradicts property (c) since y j 0. Since properties (a)-(d) hold at each recursive step and at termination with w = j, the claim is proven. Proof of necessity. We consider in turn the conditions (i)-(iv) and show that if any condition is removed, the (Q, S Q ) inequality is not facet-defining. Condition (i): The proof is by contradiction. Suppose 0 S Q. Since y 0 = 1 and Q (0) = M Q (0), then we have x i = 0 i S Q \ {0} and y i = 0 i S Q in order to satisfy the (Q, S Q ) inequality at equality. Thus, dim(x SLSF ) 2N 2 S Q \{0} S Q < 2N 2, where X SLSF is the set of feasible solutions satisfying the (Q, S Q ) inequality at equality. Condition (ii): The proof is by contradiction. Suppose there is a node j N (Q, S Q ) such that M Q (0) < d 0j. Let w = {i V Q : j C(i)}. Then M Q (0) = x i + Q (i)y i x i i S Q i S Q i P(w) since i S Q for all i P(w) by the definition of N (Q, S Q ). Then, i P(w) x i M Q (0) < d 0j. Thus, we have y j = 1 for all feasible solutions satisfying the (Q, S Q ) inequality at equality and dim(x SLSF ) < 2N 2. Condition (iii): The proof of (a) is by contradiction. Suppose q = argmax{i Q : W(j) P(i) = }. Then we have x i + (i)y Q\Qq 1 i M Q (0) i S Q\Qq 1 i S Q\Qq 1 corresponding to leaf node set Q \ Q q 1 since i S Q for all i P(q ). Thus, x i = 0 for all i S Qq \ P(q ) and y i = 0 for all i S Qq \ P(q ) are required for the (Q, S Q ) inequality to be tight. This implies dim(x SLSF ) < 2N 2.