Lifted Inequalities for 0 1 Mixed-Integer Bilinear Covering Sets

Transcription

1 1 2 3 Lifted Inequalities for 0 1 Mixed-Integer Bilinear Covering Sets Kwanghun Chung 1, Jean-Philippe P. Richard 2, Mohit Tawarmalani 3 March 1, Abstract In this paper, we study 0 1 mixed-integer bilinear covering sets. We derive several families of facetdefining inequalities via sequence-independent lifting techniques. We then show that these sets have polyhedral structures that are similar to those of certain fixed-charge single-node flow sets. As a result, we obtain new facet-defining inequalities for these sets that generalize well-known lifted flow cover inequalities from the integer programming literature Introduction and motivation Nonlinear branch-and-bound is a method to solve mixed-integer nonlinear programming (MINLP problems to global optimality; see [9, 17]. This method has been implemented in commercial solvers such as BARON [27] and LINDO Global [18]. It requires that convex relaxations of the problem be recursively solved over smaller and smaller subsets of the feasible region obtained by branching on variables. Most existing commercial software use a method proposed by McCormick [21] to obtain these convex relaxations for factorable problems. McCormick s relaxation is an instantiation of a more general technique that relaxes (nonconvex constraints of the form g(x r into (convex constraints of the form ḡ(x r where ḡ(x is a concave overestimator of g(x. This technique does not use the right-hand-side of the inequality in the process. As a result, the relaxation obtained is typically not the strongest possible. Some of the functional forms that appear most frequently in the formulation of nonlinear programs are probably multilinear inequalities and equalities. In particular, bilinear inequalities of the covering type a j x j y j d, (1 j N where a j > 0, x j S R +, and y j S R + appear in the formulation of various practical problems (including trimloss applications; see Harjunkoski et al. [16] for an example, and are among the simplest nonconvex inequalities that can be studied. Therefore, sets of the form (1 provide an important test bed for the derivation of new, stronger convexification methods that use right-hand-side information. When variables do not have upper bounds, we have derived in [30] closed-form expressions for the convex hull of feasible solutions of (1 over various subsets of the nonnegative orthant. For problems where variables are continuous and have finite upper bounds, we also derived in [29] convex relaxations of (1 that are stronger than McCormick s. In this paper, we study further the convex hull of feasible solutions to (1 when variables are bounded. In particular, we consider 0 1 mixed-integer bilinear covering sets of the form B = (x, y {0, n 1}n [0, 1] n a j x j y j d, This work was supported by NSF CMMI grants and Center for Operations Research and Econometrics, Belgium. 2 Department of Industrial and Systems Engineering, University of Florida, corresponding author. 3 Krannert School of Management, Purdue University. 1

2 where n Z ++, a j > 0 j N := {1,..., n}, and d > 0. Results similar to those derived in this paper can also be obtained for sets defined through constraints of the form k (a jx j y j + b j x j + n j=k+1 a jy j d. This generalization allows us to extend the applicability of our study to problems where the bounds on y are not 0 and 1 and, in addition, to problems where some of the x variables are fixed. Our proofs extend easily to such a setup because the two sets share strong relationships that are described in Proposition 5.1 and the discussion following it. In order to guarantee that B is not empty, we impose Assumption 1. n a j d. 42 On the theoretical side, we are interested in studying relaxation techniques for B that will take both 43 the right-hand-side d and upper bounds on the variables into account. On the one hand, it follows from the separability of j N a jx j y j over j that conv { (x, y, z {0, 1} n [0, 1] n R z j N a } 44 jx j y j is 45 described by the McCormick constraints that overestimate each bilinear term separately [1]. Therefore, the tightest relaxation of the type ḡ(x d, where ḡ(x is a concave overestimator of 46 j N a jx j y j restricted to {0, 1} n [0, 1] n over [0, 1] 2n 47, is the relaxation described above that uses McCormick constraints. On the 48 other hand, if upper bounds on the variables are absent, the convex hull of the bilinear covering set can be 49 obtained explicitly [30]. Yet, as we will see in Proposition 1.1, it is difficult to optimize linear functions over B 50 and therefore the study of PB will help us understand better the difficulties that arise from the simultaneous 51 presence of a right-hand-side and upper-bounds on the variables. 52 On the practical side, we are interested in deriving convex relaxations of B since they directly yield convex relaxations for problems with constraints of the form n f j(zx j d, where z R p 53, by replacing f j (z with 54 a j y j + b j where y j [0, 1]. We are also interested in studying B because of its relations to some important 55 mixed-integer linear sets. In particular, since the set B is a relaxation of the fixed-charge single-node flow 56 set without inflows F = (x, y {0, n 1}n [0, 1] n a j y j d, x j y j j N, see Lemma 4.1 for a proof, valid inequalities for B will also be valid for F. Further, we will show in Section 4 that facets of either F or B can be easily identified if facet-defining inequalities for the other set are known. As a result, the inequalities we derive for B also reveal new families of facet-defining inequalities for the convex hull of F. We next argue that it is typically difficult to find globally optimal solutions to problems containing B as a constraint by showing that it is NP-hard to optimize a linear function over B. To this end, consider the following optimization problem (Q that seeks to minimize a linear objective function over the bilinear set B: n n (Q min b j x j + c j y j (x, y B where b R n and c R n. Proposition 1.1. Problem (Q is NP-hard. Proof. The proof is by reduction from the 0 1 knapsack problem, which is proven to be NP-hard in [10]. Consider the following 0 1 knapsack instance: n n (K z K = min b j x j a j x j d, x j {0, 1} j N. We define a corresponding instance of (Q by setting c j = 1 for all j N, i.e. n n n (P z P = min b j x j y j a j x j y j d, x j {0, 1}, y j [0, 1] j N. 2

3 The reduction from (K to (P is clearly polynomial. Observe further that if x is a feasible solution to (K, then (x, 1 is feasible to (P, therefore showing that z P z K n. Similarly, if (x, y is an optimal solution to (P, then x is feasible to (K as n a jx j n a jx j y j d. Therefore zk z P + 1 y z P + n. We conclude that z P = z K n and that x is an optimal solution to (K if and only if (x, 1 is an optimal solution to (P. In this paper, we are interested in studying the convex hull of B, conv(b, that we denote by PB. Since B is a finite union of polytopes, PB is polyhedral. Proposition 1.2. PB is a polytope. It follows that, when studying PB, it is sufficient to consider linear inequalities. Proposition 1.1 suggests that finding a complete closed-form expression for the convex hull of B is difficult. As a result, we will focus our efforts on constructing families of strong cutting planes for optimization problems containing the constraints of B by studying the convex hull of B. To construct these inequalities, we will use lifting. Lifting is a well-known integer programming technique that generates strong inequalities for a given set by transforming an inequality valid for a restricted subset of the feasible region into a globally valid constraint. Early work on lifting in integer programming can be found in Wolsey [32, 33]. A generalization to nonlinear programming is given in Richard and Tawarmalani [24]. In particular, lifting is said to be sequence-independent if the order in which the restrictions are removed does not change the derived inequality. Subadditivity of a certain perturbation function, called the lifting function, is a sufficient condition for lifting to be sequence-independent when the restrictions involve fixing the variables at their bounds; see Proposition 3.2 and [24]. In this paper, we derive new tools to verify that functions are subadditive that we exploit to derive large families of facetdefining inequalities for PB. These results illustrate that lifting can successfully use bounds on variables in the generation of cuts for MINLPs. Further, the results have implications for fixed-charge flow models, a family of problems both theoretically and practically important in mixed-integer linear programming. The paper is structured as follows. In Section 2, we derive basic polyhedral results about PB. We provide necessary and sufficient conditions for trivial inequalities to be facet-defining. Then, we derive a linear description of PB for the special case where n = 2. This result is used to identify the seed inequalities that will be used in lifting procedures. In Section 3, we show that for a general class of multi-dimensional functions, it suffices to check the subadditivity condition at certain points to establish the subadditivity of the function everywhere. Then, using this result, we derive, in closed-form, three families of facet-defining inequalities for PB using sequence-independent lifting techniques. One requires the use of a subadditive approximation of the lifting function. In Section 4, we prove that there are some tight connections between the facets of PB and those of PF. In particular, we show that the lifted inequalities developed for PB generalize certain families of flow cover cuts and yield new facet-defining inequalities for the fixed-charge single-node flow set without inflows F. We summarize the contributions of our work and conclude with directions of future research in Section 5. 2 Basic polyhedral results In this section, we derive basic results about the polyhedral structure of PB. First, we provide necessary and sufficient conditions for PB to be full-dimensional. Proposition 2.1. PB is a full-dimensional polytope if and only if n a j a i d for all i N. Proof. First, we show that if n a j a i d for all i N, then PB is full-dimensional. For all i N, construct p i = (1 e i, 1 and q i = (1, 1 e i. Also define r = (1, 1. The points p i, q i, and r belong to B. These points are affinely independent because r p i and r q i for all i N are linearly independent. Since we have described 2n+1 affinely independent points in PB, we have shown that PB is full-dimensional. Next, we prove that if PB is a full-dimensional polyhedron, then n a j a i d for all i N. Assume by contradiction that n a j a i < d for some i N. Since n a j d from Assumption 1, B is nonempty and so x i = 1 in every feasible solution of B, showing that PB is not full-dimensional. This is the desired contradiction. 3

4 In the remainder of this paper, we will assume that PB is full-dimensional. Assumption 2. n a j a i d for all i N. Observe that Assumption 2 strictly dominates Assumption 1 and implies that n 2. We next identify some basic characteristics of the facet-defining inequalities of PB. Proposition 2.2. Let n α j x j + n β j y j δ (2 be a facet-defining inequality for PB that is not a scalar multiple of x i 1 for i N or y i 1 for i N. Then, (i α i 0, i N, (ii β i 0, i N, and (iii δ 0. Proof. Select i N. Since (2 is a facet-defining inequality for PB that is not a scalar multiple of x i 1, there exists (x, y B with x i < 1 such that n α j x j + n β j yj = δ. (3 Consider now ( x, ȳ = (x, y + (1 x i (e i, 0. This point belongs to B and therefore satisfies (2, i.e., n α j x j + n β j ȳ j δ. (4 Subtracting (3 from (4, we obtain that α i 0. The proof that β i 0 for all i N is similar. The fact that δ 0 then follows from (3 after noting that all terms in the left-hand-side are nonnegative. The following proposition further studies facet-defining inequalities whose right-hand-sides are zero. Proposition 2.3. Let n α j x j + n β j y j 0 (5 be a facet-defining inequality for PB. Then, (5 is a scalar multiple of x j 0 for j N or of y j 0 for j N. Proof. Assume for a contradiction that (5 is not a scalar multiple of x j 0 for j N or of y j 0 for j N. Then, for each i N, there exists (x i, y i B such that x i i > 0 and for which n α j x i j + n β j yj i = 0. (6 Since we know from Proposition 2.2 that α j 0 and β j 0 for all j N, we obtain from (6 that 0 = n α j x i j + n β j yj i α i x i i 0. (7 We conclude that, for each i N, α i = 0 since x i i > 0. Similarly, we can establish that β i = 0 i N. This is a contradiction to the fact that (5 is facet-defining for PB. We now focus on these inequalities that play a special role in Propositions 2.2 and 2.3 and characterize when they are facet-defining for PB. We refer to these inequalities as bound inequalities. Proposition 2.4. The upper bound inequalities x i 1, y i 1 are facet-defining for PB for all i N. Further, for i N, the lower bound inequalities x i 0, y i 0 are facet-defining for PB if and only if n a j a i a l(i d where l(i argmax{a j j N \ {i}}. 4

5 Proof. The validity of all these inequalities is trivial since they belong to the description of B. Assume for a contradiction that x i 1 is not facet-defining for PB. Then, it follows from Proposition 2.2 that (e i, 0 is a recession direction of PB, a contradiction to the fact that PB is a polytope; see Proposition 1.2. The proof that y i 1 is facet-defining for PB is similar. Now, we show that x i 0 is facet-defining if n a j a i a l(i d by describing 2n affinely independent points in B that satisfy x i = 0. For k N \{i}, we construct the 2(n 1 points, p k = (1 e i e k, 1 e i e k and q k = (1 e i e k, 1 e i. We also define r 1 = (1 e i, 1 e i and r 2 = (1 e i, 1. Clearly, the points r 1, r 2 and p k, q k for k N \ {i} satisfy x i 0 at equality and are feasible for B since n a j a i n a j a i a k n a j a i a l(i d. These points are affinely independent since p k r 1, q k r 1, and r 2 r 1 can be easily verified to be linearly independent. To prove the reverse direction, assume now that x i 0 is facet-defining for PB. We claim that n a j a i a l(i d. Assume for a contradiction that n a j a i a l(i < d. This condition implies that every feasible solution (x, y of PB with x i = 0 also must satisfy x l(i = 1. As a result, the dimension of the face defined by x i = 0 is less or equal to 2n 2, which is a contradiction. Similarly, it can be proven that y i 0 is facet-defining for PB if and only if n a j a i a l(i d. Observe that the above proofs are also valid when y i {0, 1} instead of y i [0, 1] for some subset J N. We next study another simple facet-defining inequality for PB. Proposition 2.5. The inequality n a jy j d is facet-defining for PB. Proof. Validity is easily verified since n a jy j n a jx j y j d. To prove that n 172 a jy j d is facetdefining, we present 2n points (x i, y i in B that satisfy n a jyj i d at equality and such that the system 173 αxi + βyi 174 = δ for i = 1,..., 2n only has solutions (α, β, δ that are scalar multiples of (0, a, d. Consider the n points p k = (1, k (1 e k and q k = (1 e k, k (1 e k where k = d n a j a k for k N. Note that because of Assumption 2, 0 < k 1 for all k N. Clearly, p k and q k belong to B and satisfy n a jy j d at equality. These 2n points yield the system: n n α j + k β j β k = δ k N, (8 n n α j α k + k β j β k = δ k N. ( By subtracting (8 from (9, we obtain that α k = 0 for k N. From (8 and the definition of k, we then conclude that, for all k, l N, n β j β k = δ n n a j a k and β j β l = δ n a j a l. d d Subtracting these expressions yields β k δ d a k = β l δ d a l. After defining β k δ d a k = θ for k N and using these relations in (8, we obtain that θ = 0, which implies β k = δ d a k for k N. Therefore, we conclude that all solutions (α, β, δ to the system (8 and (9 are scalar multiples of (0, a, d. In the remainder of this paper, we will often use the term facet to refer to a facet-defining inequality. We will also refer to inequalities x i 1, y i 1, and n a jy j d as trivial facets of PB. To illustrate the richness of the polyhedral structure of PB, we present an example next. The linear inequalities describing the convex hull of this set were obtained using PORTA; see Christof and Löbel [6]. Example 2.6. Consider the 0 1 mixed-integer bilinear covering set { } B = (x, y {0, 1} 4 [0, 1] 4 19x 1 y x 2 y x 3 y x 4 y

6 The linear description of PB has 58 inequalities that are presented in the Appendix. They include: 50x x x y y (10 70x x x y y (11 19x x y y 4 20 (12 17x x y y 4 20 (13 19y y y y 4 20 (14 14x x 3 + 5x y 2 15 (15 12x x 3 + 5x y 1 15 (16 10x 3 + 5x y y 2 15 (17 x 1 + x 2 + x y 4 2 (18 x 1 + x 2 + x 3 + x 4 2 (19 x 1 0 (20 y 1 0 (21 x 1 1 (22 y 1 1 (23 Among the inequalities in Example 2.6, we recognize the upper bound inequalities (22 and (23 that are shown to be facet-defining for PB in Proposition 2.4. In this example, the lower bound inequalities (20 and (21 are also facet-defining, as can be established from Proposition 2.4. Further, (14 is the trivial facet-defining inequality studied in Proposition 2.5. Our goal is now to discover families of valid inequalities for PB that would explain (10 (13 and (15 (19. To derive these nontrivial facet-defining inequalities, we first study the convex hull of B when n = 2 with the goal of identifying seed inequalities for subsequent lifting procedures. We show that the linear description of PB has at most three nontrivial inequalities. In this study, Assumption 2 requires that a 1 d and a 2 d. Proposition 2.7. Let B 2 = { } (x, y {0, 1} 2 [0, 1] 2 a 1 x 1 y 1 + a 2 x 2 y 2 d, where a 1 d, a 2 d and d > 0. Then, conv(b 2 = X := (x, y [0, 1] 2 [0, 1] 2 x 1 + x 2 1 dx 1 + a 2 y 2 d a 1 y 1 + dx 2 d a 1 y 1 + a 2 y 2 d Proof. We prove the result using disjunctive programming techniques; see [5]. We define X 10 := B 2 d {x 1 = 1, x 2 = 0} = {(1, y 1, 0, y 2 a 1 y 1 1, 0 y 2 1}, X 01 := B 2 d {x 1 = 0, x 2 = 1} = {(0, y 1, 1, y 2 0 y 1 1, a 2 y 2 1}, X 11 := B 2 {x 1 = 1, x 2 = 1} = {(1, y 1, 1, y 2 a 1 y 1 + a 2 y 2 d, 0 y 1 1, 0 y 2 1}. It is easily verified that conv(b 2 = conv(x 10 X 01 X 11 = conv(x 2 X 11 where X 2 := conv(x 10 X 01. We first use disjunctive programming techniques to obtain a linear description of X 2 and then compute conv(b 2 as conv(x 2 X 11. Using Theorem 2.1 in Balas [5], we write (x 1, y 1, x 2, y 2 = (λ, z 1 + ẑ 1, 1 λ, z 2 + ẑ 2, d λ z 1 λ, 0 z 2 λ, X 2 = proj (x 1, y 1, x 2, y 2, z 1, z 2, ẑ 1, ẑ 2, λ a 1 (x,y d. 0 ẑ 1 1 λ, (1 λ ẑ 2 1 λ, a 2 0 λ 1 We then use Fourier-Motzkin elimination [34] to compute the projection. We first eliminate the variables λ, ẑ 1 and ẑ 2 using the equations λ = x 1, ẑ 1 = y 1 z 1, and ẑ 2 = y 2 z 2. We obtain x 1 + x 2 = 1, 0 x 1 1, 6

7 and d a 1 x 1 z 1 x 1, x 1 + y 1 1 z 1 y 1, 0 z 2 1 x 2, y 2 x 2 z 2 y 2 d a 2 x 2, from which we project variables z 1 and z 2 to obtain X 2 = conv(x 10 X 01 = (x x 1 + x 2 = 1, x 1 0, x 2 0, 1, y 1, x 2, y 2 d d x 1 y 1 1, x 2 y 2 1 a 1 a 2 since x 1 1 and x 2 1 are implied by x 1 + x 2 = 1, x 1 0 and x 2 0. Now, compute conv(x 2 X 11 as proj (x,y (x 1, y 1, x 2, y 2, ū 1, ū 2, v 1, v 2, ˆv 1, ˆv 2, λ (x 1, y 1, x 2, y 2 = (ū 1 + (1 λ, v 1 + ˆv 1, ū 2 + (1 λ, v 2 + ˆv 2, ū 1 + ū 2 = λ, ū 1 0, ū 2 0, d d ū 1 v 1 λ, ū 2 v 2 λ, a 1 a 2. a 1ˆv 1 + a 2ˆv 2 d(1 λ, 0 ˆv 1 1 λ, 0 ˆv 2 1 λ, 0 λ 1 We again obtain the projection using Fourier-Motzkin elimination. Using the equations x 1 = ū λ, x 2 = ū λ, and ū 1 + ū 2 = λ, we obtain that λ = 2 (x 1 + x 2, ū 1 = 1 x 2, and ū 2 = 1 x 1. Using these relations together with v 1 = y 1 ˆv 1 and v 2 = y 2 ˆv 2 to eliminate the corresponding variables, we obtain and x 1 1, x 2 1, 1 x 1 + x 2 2, y 1 + x 1 + x 2 2 ˆv 1 y 1 d a 1 (1 x 2, a 2 a 1 ˆv 2 + d a 1 (x 1 + x 2 1 ˆv 1, 0 ˆv 1 x 1 + x 2 1, y 2 + x 1 + x 2 2 ˆv 2 y 2 d a 2 (1 x 1, 0 ˆv 2 x 1 + x 2 1, where inequality is clearly redundant. Projecting ˆv 1, we obtain and d x 1 1, x 2 1, 1 x 1 + x 2, y 1 1, a 1 (1 x 2 y 1, a 1 x 1 + (a 1 dx 2 2a 1 d d a 2 x 1 a1 a 2 y 1 ˆv 2, (d a 1 (x 1 +x 2 1 a 2 ˆv 2, y 2 + x 1 + x 2 2 ˆv 2 y 2 d a 2 (1 x 1, 0 ˆv 2 x 1 + x 2 1, where obviously redundant inequalities have been omitted. Again, inequalities are redundant since x 1 1, x 2 1, x 1 + x 2 1, a 1 d and a 2 d > 0. Projecting ˆv 2, we obtain the system x 1 1, x 2 1, 1 x 1 + x 2, y 1 1, d a 1 (1 x 2 y 1, and d a 1 y 1 + a 2 y 2, a 2 (a 2 dx 1 + a 1 y 1 + a 2 x 2, (a 2 dx 1 + a 2 x 2 2a 2 d, y 2 1, d a 2 (1 x 1 y 2, 1 x 1 + x 2, (R 7

8 where inequalities are either repeated or redundant. In particular, (R is redundant since it can be obtained as a conic combination with weights (a 2 d and 1 of valid inequalities x 1 + x 2 1 and a 1 y 1 + dx 2 d. Therefore, conv(x 2 X 11 is defined by bounds and the four inequalities given in the description of X, concluding the proof. Next, we give generalizations of the nontrivial facets of conv(b 2 that we prove are facet-defining for more general instances of conv(b. In particular, we give a generalization of inequalities dx 1 + a 2 y 2 d and a 1 y 1 + dx 2 d in Proposition 2.9 and of inequality x 1 + x 2 1 in Proposition We will use these generalizations as seed inequalities for lifting procedures in Section Lemma 2.8. Inequality is valid for PB. j N min{dx j, a j x j, a j y j } d ( Proof. We first show that j N min{dx j, a j x j y j } d is valid for B. Consider (x, y B. If there exists j N such that dx j < a j x j y j then x j = 1 and, consequently, the inequality is satisfied. Otherwise, the inequality reduces to the defining inequality of B. Since (x j, y j [0, 1] 2 implies that x j y j min{x j, y j } and a j 0, it follows that min{dx j, a j x j y j } min{dx j, a j x j, a j y j } and, therefore, (24 is valid for PB. The set of solutions in [0, 1] 2n that satisfy (24 is a subset of the convex relaxations of B discussed in Section 1. In particular, when each bilinear term is outer-approximated using McCormick envelopes, we obtain the inequality j N a j min{x j, y j } d, which is clearly implied by (24. Further, using orthogonal disjunctions, see [30], it can be shown that { O := conv (x, y R 2n + } { a j x j y j d = (x, y R 2n + aj x j y j } d. j N j N This convex relaxation is obtained without making use of the bounds or the integrality of the variables x. It follows from the inequality relating elementary means (see Theorem 5 in [15] that da j x j y j min{dx j, a j y j }. Therefore, the feasible solutions to (24 are contained in O. However, when (x, y C R n +, a procedure described in [29] permits strengthening relaxation O by restricting attention to C. When one exploits the fact that (x, y C = {0, 1} n [0, 1] n, this construction yields (24. Proposition 2.9. Let L N be such that j N\L a j > d. Define ā = j N\L a j max i N\L a i and assume that S = {(x, ȳ {0, 1} L [0, 1] i L min{a i, d}x i + āȳ = d} =. Then, min{a j, d}x j + a j y j d (25 j L j N\L is facet-defining for PB. In particular, (25 is facet-defining for PB if (i L L >, or (ii L =, or (iii ā max i L min{a i, d}, or as a special case (iv ā d where L > := {j N a j > d}. Proof. Validity of (25 for PB follows from Lemma 2.8. We now prove that (25 is facet-defining for PB by providing 2n affinely independent points (x i, y i in B that satisfy (25 at equality. Assume without loss of generality that L = {1,..., l}. Define n = N \ L and denote the points as (x L, x N\L, y L, y N\L. Let (x, ȳ S and define a = j N\L a j. Let p 0 = (0, 1, 0, d a 1 and p j = p 0 + ɛ(0, 0, 0, 1 a j e j 1 a j+1 e j+1 for j = 1,..., n 1. For i L, define q i = (e i, 1, e i, d min{a i,d} a min{a 1, i,d} ri = p 0 + (0, 0, e i, 0 if a i d, and r i = (e i, 1, d a i e i, 0 if a i > d. For j {1,..., n }, s j = (x L, 1 e j, 1, ȳ i N\(L {j} ā a (1 e j. It can be i easily verified that p 0, q i, s j and r i belong to B and that p j belongs to B when ɛ is sufficiently small. We now show that the above points are affinely independent. Clearly, for j {0,..., n 2}, p 0,..., p j satisfy j+1 i=1 a i( d a y l+i = 0, whereas p j+1 does not. Therefore, p j are affinely independent. Further, for i L, q i are affinely independent of p j since the latter satisfy (x i, y i = (0, 0. For i L, r i are independent of p j and q j since the latter satisfy y i = x i. Finally, s i are affinely independent of p j, q j, r j since the latter satisfy x i = 1. 8

9 The family of inequalities described in Proposition 2.9 is typically exponential in size. In the case of Example 2.6, it contains multiple inequalities including ( More generally, it can be verified that inequalities (10-(19 in the Appendix are of the form (25. In the remainder of the paper, we use the following notation extensively. For N 0, N 1 N such that N 0 N 1 = and Ñ0, Ñ1 N such that Ñ0 Ñ1 =, we let { } B(N 0, N 1, Ñ0, Ñ1 := (x, y B x j = 0 for j N 0, x j = 1 for j N 1, y j = 0 for j Ñ0,. y j = 1 for j Ñ1 We also define PB(N 0, N 1, Ñ0, Ñ1 := conv(b(n 0, N 1, Ñ0, Ñ1. In particular, B(,,, N is equivalent to the classical 0 1 knapsack set { n } x {0, 1} n a j x j d, whose polyhedral structure was first studied by Balas [4], Hammer et al. [14] and Wolsey [31]. The following proposition describes, among other things, relations between the bilinear set B and the 0 1 knapsack set B(,,, N. 303 Proposition Let j N α j x j + j I β j y j δ (26 be an inequality for PB(,,, N \ I that is not a scalar multiple of a bound inequality. facet-defining for PB(,,, N \ I if and only if (26 is facet-defining for PB. Then, (26 is 307 Proof. We first prove that if (26 is facet-defining for PB(,,, N \ I, then (26 is facet-defining for PB. To show that (26 is valid for B, we assume for a contradiction that there exists a point (x, y 308 B with j N α jx j + j I β jy j < δ. Since (x, y B, we have that j N a jx j y j d. Next, we define ( x, ȳ as x = x, ȳ j = y j 310 for j I, and ȳ j = 1 for j N \ I. Observe that ( x, ȳ B(,,, N \ I as j I a j x j ȳ j + j N\I a j x j j N a jx j y j d. Since (26 is valid for B(,,, N \ I, ( x, ȳ satisfies j N α jx j + j I β jy j = j N α j x j + j I β 312 jȳ j δ. This is the desired contradiction. 313 Next, we show that (26 is facet-defining for PB. Since (26 is facet-defining for PB(,,, N \ I and 314 δ 0 as (26 is not a bound, there exist n + I linearly independent points in B(,,, N \ I, call them (x k, y k 315, that satisfy (26 at equality. Clearly, these points belong to B and satisfy (26 at equality. Now, for 316 each j N \ I, we construct one new point in B \ B(,,, N \ I that satisfies (26 at equality. Choose j 317 arbitrarily in N \ I. Since (26 is not a scalar multiple of x j 1, there exists k j {1,..., n + I } such that 318 x kj j = 0. Now define ( x kj, ȳ kj such that x kj i = x kj i i N, ȳ kj i = y kj i i N \ {j} and y kj j = 0. Clearly, the point ( x k j, ȳ k j belongs to B and satisfies (26 at equality. Further, it is easily seen that the points (x k, y k 319 and ( x k j, ȳ k j 320 for j N\I are linearly independent and therefore show that (26 is facet-defining for PB. 321 To prove the reverse implication, we assume that (26 is a facet-defining inequality for PB that is not 322 a scalar multiple of a bound. Validity is trivial since for B(,,, N \ I B. Now, we show that (26 is 323 facet-defining for PB(,,, N \ I. Since δ 0 as (26 is not a bound, the set of 2n affinely independent points (x k, y k 324 in B for k = 1,..., 2n that satisfy (26 at equality are also linearly independent. Therefore, x x 1 n y y 1 n x x 2 n y yn x 2n 1... x 2n n y1 2n... yn 2n Therefore, there must exist n + I rows i 1,..., i n+ I where I = {j 1,..., j I } such that x i x i1 n y i 1 j1... y i 1 j I x i x i 2 n y i2 j 1... y i2 j I x i n+ I 1... x i n+ I n y i n+ I j 1... y i n+ I j I 0. 9

10 Hence, we see that the n+ I points (x i k 1,..., x i k n ; y i k j1,..., y i k j I for k = 1,..., n+ I are linearly independent. Now, define the points ( x i k, ỹ i k for k = 1,..., n + I such that x i k = x i k, ỹ i k j = y i k j for j I, and ỹ i k j = 1 for j N \ I. The points ( x i k, ỹ i k are feasible to B(,,, N \ I and satisfy (26 at equality. Therefore, we conclude that (26 is facet-defining for PB(,,, N \ I. Observe that Proposition 2.10 implies that all nontrivial facets of the 0 1 knapsack polytope can be found in B and that it is sufficient to study the facets of B to obtain the facets of the 0 1 knapsack polytope. Next, we use Proposition 2.10 to generalize the inequality x 1 + x 2 1 of Proposition 2.7 into an inequality that will be used as a seed for lifting procedures in Section 3.4. Proposition Assume that j N a j a k a m < d for all k, m N with k m. The clique inequality is facet-defining for PB. x j N 1 (27 j N Proof. Because of Proposition it is sufficient to prove that (27 is facet-defining for PB(,,, N. To prove validity, assume for a contradiction that there exists x B(,,, N such that j N a jx j d and j N x j N 2. Since j N x j N 2, there exist k, m N with k m such that x k = 0 and x m = 0. Therefore, j N a j a k a m j N a jx j d. This contradicts the assumption that j N a j a k a m < d for all k, m N with k m. We next show that (27 is facet-defining for PB(,,, N. It can be easily verified using Assumption 2 that the points p k = (1 e k, 1 for k N belong to B(,,, N. Since these points are linearly independent and satisfy (27 at equality, we conclude that (27 is facet-defining for PB(,,, N Lifted inequalities In this section, we derive three families of strong valid inequalities for PB via lifting. The first two families are obtained using sequence-independent lifting from (25 and are facet-defining for PB. In this case, lifting is simple since the lifting function is subadditive. The third inequality is obtained by lifting (27. Although the lifting function associated with this seed inequality is not subadditive, we obtain lifted inequalities using approximate lifting. We then identify conditions under which these inequalities are facet-defining for PB. 3.1 Sequence-independent lifting for bilinear covering sets Sequence-independent lifting is a well-known technique to construct strong valid inequalities for mixed-integer linear programs; see Wolsey [33] and Gu et al. [13]. We next give a brief description of how this technique can be used to derive strong valid inequalities for PB. A more general treatment of lifting in nonlinear programming is given in Richard and Tawarmalani [24]. Given = S N, we consider B(S,, S,, which is the restriction of B obtained when all variables (x j, y j for j S are fixed to (0, 0. Let S = {s,..., n} for some s 2 and define S i = {i + 1,..., n} for i S. Assume that the inequality s 1 s 1 α j x j + β j y j δ (28 is facet-defining for PB(S,, S,. In sequential lifting, we reintroduce the variables (x j, y j for j S one at the time in (28. Assuming that variables (x j, y j have already been lifted in the order j = s,..., i 1, we next review how to lift variables (x i, y i in the inequality i 1 i 1 α j x j + β j y j δ, (29 10

11 which is assumed to be facet-defining for PB(S i 1,, S i 1,. To perform this lifting, we first compute the lifting function i 1 i 1 P i (w = max δ α j x j + β j y j s.t. i 1 a j x j y j d w x j {0, 1}, y j [0, 1] j = 1,..., i 1. Once the lifting function P i (w is computed, the lifting coefficients (α i, β i are obtained from P i (w as follows. Proposition 3.1 (Richard and Tawarmalani [24]. Let (29 be a valid inequality for B(S i 1,, S i 1,. Assume that there exist (α i, β i R 2 such that α i x i + β i y i P i (a i x i y i for (x i, y i {0, 1} [0, 1] \ {0, 0}. ( Then, the inequality is valid for B(S i,, S i,. i i α j x j + β j y j δ ( The result of Proposition 3.1 can be applied recursively to construct a valid inequality for PB from (28. Note that, at each step, the lifting function P i 380 (w must be recomputed to account for the changes 381 in the lifted inequality. Further, if B(S,, S, is full-dimensional, the seed inequality (28 is facet-defining 382 for B(S,, S,, and for each i S, the lifting coefficients (α i, β i of the variables (x i, y i are chosen so that (30 is satisfied at equality by two points (x i, y1 i and (x2 i, y2 i such that (0, 0, (x1 i, y1 i and (x2 i, y2 i 384 are affinely independent (a feature we refer to as maximal lifting, then the final lifted inequality will be facet-defining for PB. In this scheme, (recomputing the lifting functions P i 385 (w for each i S is often the 386 most computationally demanding task. However, this computational work is unnecessary when the lifting function P s 387 (w is subadditive. This observation, first made by Wolsey [33], leads to the following result Proposition 3.2 (Richard and Tawarmalani [24]. Assume that (28 is valid for B(S,, S,. Assume also that (i P s (w is subadditive over R +, i.e, P s (w 1 + P s (w 2 P s (w 1 + w 2 w 1, w 2 R +, and (ii there exist (α i, β i R 2 for all i S such that α i x i + β i y i P s (a i x i y i for (x i, y i {0, 1} [0, 1] \ {0, 0}. (32 Then, the inequality n n α j x j + β j y j δ (33 is valid for PB. Further, if (i Inequality (28 is facet-defining for B(S,, S,, (ii B(S,, S, is fulldimensional and (iii coefficients (α i, β i are chosen in a way that two linearly independent points satisfy (32 at equality, then (33 is facet-defining for PB. The fundamental difference between Proposition 3.1 and Proposition 3.2 lies in equations (30 and (32. In the latter, the lifting coefficients of all variables (x i, y i are obtained from the same lifting function P s (w while in the former, they are obtained from P i (w for i S. Although this difference might seem minor, it has important practical implications. In particular, the subadditivity of lifting functions typically permits the derivation of closed-form expressions for lifting coefficients that would otherwise be difficult to obtain. Observe also that in Proposition 3.2, the subadditivity of P s (w is required only over R + since all coefficients a i in PB are assumed to be nonnegative. 11

12 Proposition 3.1 describes how to perform lifting when then variables (x j, y j for j S are fixed at (0, 0. When variables (x j, y j are fixed at (1, 1, similar results can be obtained. In this case, condition (30 must be changed to α i (x i 1 + β i (y i 1 P i (a i x i y i a i for (x i, y i {0, 1} [0, 1] \ {1, 1}. ( Similarly, Proposition 3.2 can be adapted to allow sequence-independent lifting for variables (x j, y j fixed at (1, 1 by replacing P i (w with P s (w in (34 and by requiring that the lifting function P s 409 (w is subadditive 410 over R. Subadditive lifting can also be used to generate facets of PB if B(, S,, S is full-dimensional, the 411 seed inequality (28 is facet-defining for B(, S,, S, and for each i S, the lifting coefficients (α i, β i of the variables (x i, y i are chosen so that (34 is satisfied at equality by two points (x i, y1 i and (x2 i, y2 i such that (1, 1, (x i, y1 i and (x2 i, y2 i are affinely independent. 414 We next show in the following proposition that all interesting lifted inequalities that can be obtained by 415 fixing variables (x i, y i at (0, 1 or (1, 0 can also be obtained by fixing variables (x i, y i at (0, Proposition 3.3. Assume that (29 defines a nonempty face of PB(S i 1,, S i 1, = PB(S i 1,,, S i 1 = PB(, S i 1, S i 1,. Then any inequality obtained from maximally lifting (29 in PB(S i 1,,, S i 1 or PB(, S i 1, S i 1, could have been obtained by maximally lifting (29 in PB(S i 1,, S i 1,. Proof. First, we consider the case when (x i, y i is fixed at (1, 0. In this situation, valid lifting coefficients must satisfy α i (x i 1 + β i y i P i (a i x i y i for (x i, y i {0, 1} [0, 1]. (35 We next show that maximal lifting coefficients (α i, β i in (35 must also satisfy α i x i + β i y i P i (a i x i y i for (x i, y i {0, 1} [0, 1] (36 and be maximal for (36. This is sufficient to prove the result since restricting (x i, y i = (0, 0 instead of (1, 0 does not change the restricted set and, therefore, the seed inequality is still a face of same dimension. Let (0, yi satisfy (35 at equality. Such a point exists since lifting is assumed to be maximal. Then, 0 α i = β i yi P i (a i yi 0, where the first inequality follows from (35 by setting (x i, y i = (0, 0, the equality holds since (0, yi 428 satisfies (35 at equality, the second inequality is satisfied from (35 with (x i, y i = (1, yi 429 and the last inequality is verified since a i yi Therefore, equality holds throughout and, in particular, α i = 0. It follows that 431 α i (x i 1 + β i y i = α i x i + β i y i and, consequently, (α i, β i is valid and maximal to ( Now, we fix (x i, y i at (0, 1. Then, we show that any (α i, β i that is valid and maximal to 433 α i x i + β i (y i 1 P i (a i x i y i (37 is also valid and maximal to (36. Let yi = min{y i [0, 1] α i + β i (y i 1 = P i (a i y i }, i.e., (1, yi 434 satisfies 435 (37 at equality. It follows that β i (y i 1 = P i (a i y i α i P i (a i y i P i (a i 0, where the first inequality follows from (37 by substituting (x i, y i = (0, yi 437, the equality is satisfied since (1, yi 438 satisfies (37 at equality, the second inequality is verified by substituting (1, 1 in (37, and the last inequality holds since P i ( is non-decreasing and a i y i 439 a i. Therefore, the equality holds throughout and, in particular, β i (yi 1 = 0. It follows that either β i = 0 or yi 440 = 1. We show that β i = 0 in the latter case as well. If yi 441 = 1, because lifting is assumed to be maximal and because of the definition of y i, there is a y i [0, 1 such that (0, y i satisfies (37 at equality. Therefore, β i(y i = 0 and so β i = 0. It follows that 443 α i x i + β i (y i 1 = α i x i + β i y i and, consequently, (α i, β i is valid and maximal for (36. 12

13 Subadditivity of lifting functions In this section, we provide a general result that helps in proving subadditivity of functions. For specific functions, it has been observed (see Proposition 3.10 in [23], Theorem 7 in [11], Proposition 4.2 in [8], and Lemma 21 in [25] that subadditivity of a function over R n can often be established by checking it at a small subset of points. The corresponding proofs are often detailed and are the key step in proving subadditivity. In Theorem 3.4, we identify a fairly large class of functions for which a similar result holds. We use this result to prove subadditivity of functions that arise during lifting of inequalities for mixed-integer 0 1 bilinear covering set. The scope of applications of Theorem 3.4 is, however, much larger and we provide this general result with the hope that it may be useful in other applications. Theorem 3.4. For i N = {1,..., n}, let b i R m, f i R, and h i (x : R m R be subadditive functions. Let h(x : R m R be a function with h(0 = 0 that majorizes h i (x for all i. Let B i = {x R m h i (x b i = h(x b i }. Define f(x : R m R as f(x = min n i=1{ fi + h i (x b i }. Then, f(x + h(y x f(y for each (x, y R m R m. For x R m, let i(x N be such that f(x = f i(x + h i(x (x b i(x. If x B i(x then f(b i(x f(b i(x + y f(x f(x + y. Further, if y B i(y as well, then f(b i(x + f(b i(y f(b i(x + b i(y f(x + f(y f(x + y. Proof. Let (x, y R m R m. Then, f(x+h(y x f(x+h i(x (y x = f i(x +h i(x (x b i(x +h i(x (y x f i(x +h i(x (y b i(x f(y, (38 where the first inequality follows since h( h i(x (, the second inequality holds since h i(x is a subadditive function, and the third since the minimum defining f(y includes a term that equals f i(x + h i(x (y b i(x. Note that 0 h i(x (0 h(0 = 0 where the first inequality follows from subadditivity of h i(x. Therefore, h i(x (0 = 0. Further, if x B i(x : f(x = f i(x + h i(x (x b i(x f(b i(x + h i(x (x b i(x = f(b i(x + h(x b i(x f(x, where the first inequality holds since h i(x (0 = 0 implies f(b i(x f i(x as f i(x is one of the terms in minimum defining f(b i(x, the second equality since x B i(x and the last inequality by (38. Therefore, equality holds throughout and, in particular, f i(x = f(b i(x. Now, consider (x, y R m R m with x B i(x. Then, f(b i(x f(b i(x + y = f(x h(x b i(x f(b i(x + y f(x f(x + y, (39 where the equality follows from the definition of i(x, x B i(x, and f i(x = f(b i(x, and the inequality follows from (38 since f(b i(x + y + h(x b i(x f(x + y. Further, if y B i(y, f(b i(x + f(b i(y f(b i(x + b i(y f(b i(x + f(y f(b i(x + y f(x + f(y f(x + y, where each of the inequalities follows from (39. Any function, say g(x, that is subadditive and satisfies g(0 = 0 can be expressed as f(x, defined in Theorem 3.4, by setting n = 1, b 1 = 0, f 1 = 0, and h(x = h 1 (x = g(x. Observe that lifting functions derived from seed inequalities that are tight on the restricted set always satisfy the condition g(0 = 0. In other words, Theorem 3.4 can be interpreted as a recursive tool for proving subadditivity of such lifting functions, f(x, that exploits the subadditivity of the constituent simpler functions, h i (. For example, Theorem 3.4 shows that it suffices to check the subadditivity of f(x at a small subset of points if f(x can be expressed as a minimum of finitely many translates of a subadditive function. Since positively-homogenous convex functions belong to the class of subadditive functions, see Theorem 4.7 in [26], they can be used as building blocks in the application of Theorem 3.4. In Corollaries 3.5 and 3.6, we apply Theorem 3.4 to prove subadditivity of two functions that will be used in Sections 3.3 and 3.4 to derive inequalities for the 0 1 mixed-integer bilinear covering set. In both cases, functions h i are univariate positively-homogenous convex functions. 13

14 Corollary 3.5. Let ν and D i for i = 0, 1,..., r be nonnegative integers that satisfy ν > 0, D 0 = 0, and D i D i 1 + ν for i = 1,..., r. Then the function 0 if w < D 0 w iν if D g(w := i w < D i+1 ν, i = 0,..., r 1, D i iν if D i ν w < D i, i = 1,..., r 1, D r rν if D r ν w is subadditive over R if and only if D i + D j D i+j for 0 i j r with i + j r. Proof. First note that g(w = min r i=0{ Di iν +h i (w D i }, where h i (w = max{0, w} for i = 0,..., r 1 and h r (w = 0. We observe that i(x = 0 for x < D 1 ν, i(x = i for x [D i ν, D i+1 ν and i = 1,..., r 1, and i(x = r for x D r ν. Let h(w = max{0, w}. We also see that B 0 = B 1 =... = B r 1 = R and B r = R. Assume that D i + D j D i+j for 0 i j r with i + j r. Consider x, y R with x y. We argue next that g(x + g(y g(x + y. Define i = i(x and j = i(y. Clearly, i j. We consider two cases. Assume first that j = r. Then, g(y = D r rν g(x + y and therefore g(x + g(y g(x + y as g(x 0. Assume next that j r 1. Since x y < D r ν, it follows that x B i and y B j and, therefore, from Theorem 3.4 that g(d i + g(d j g(d i + D j g(x + g(y g(x + y We next argue that the left-hand-side of the above expression is nonnegative, which proves the result. Let t = min{j, r i}. Then, g(d i +D j = g(d i +D t = g(d i+t +D i +D t D i+t g(d i+t +D i +D t D i+t = g(d i +g(d t g(d i +g(d j, where the first equality holds since t = r i implies D i + D j D i + D t D r, the first inequality follows from (38 and D i+t D i + D t, the second equality since g(d i = D i iν, and the last inequality from (38 since D t D j. We now prove the reverse implication. For w > 0, g(d k w g(d k ν max{0, w ν} = g(d k max{0, w ν} > g(d k w, (40 where the first inequality follows from (38 and the last inequality since ν > 0 and w > 0. If i + j r and D i + D j < D i+j then g(d i + g(d j g(d i + D j < g(d i + g(d j g(d i+j D i D j + D i+j = 0, yields a contradiction to subadditivity of g, where the strict inequality follows from (40 where k = i + j and w = D i+j D i D j since D i +D j < D i+j and the equality holds since g(d k = D k kν for k {i, j, i+j}. Corollary 3.5 equivalently shows the superadditivity of w g(w, generalizing prior similar results in the literature. In particular, see Lemmas 6 and 7 in [3] and Definition 4 in [19]. Corollary 3.6. Let λ and C i for i = 0, 1,..., s be nonnegative integers that satisfy λ > 0, C 0 = 0 and C i 1 + λ C i for i = 1,..., s. Then the function 0 if w < C 0 i + g(w = w C i λ if C i w < C i + λ, i = 0,..., s, i if C i 1 + λ w < C i, i = 1,..., s, s + 1 if C s + λ w. is subadditive over R if and only if C i + C j C i+j for 0 i j s with i + j s. Proof. Let C s+1 = max{c i + C j i + j = s + 1}. Note that g(w = min s+1 { i=0 i + hi (w C i } 520 where h i (w = max{0, w λ } for i = 0,..., s and h s+1(w = 0. Let h(w = max{0, w 521 λ }. 522 Assume that C i + C j C i+j for 0 i j s with i + j s. Consider x, y R with x y. We 523 argue next that g(x + g(y g(x + y. Define i = i(x and j = i(y. Assume first that j = s + 1. Then 14

15 g(y = s + 1 g(x + y and therefore, g(x + g(y g(x + y as g(x 0. Next assume that j s. Since x y < C s + λ, it follows that x B i and y B j and therefore, from Theorem 3.4 that g(c i + g(c j g(c i + C j g(x + g(y g(x + y We next argue that the left-hand-side of the above expression is nonnegative, which proves the result. t = min{j, s + 1 i}. Then, g(c i + C j g(c i+t = i + t i + j = g(c i + g(c j, Let where the first inequality follows from (38 and C i+t C i + C j when t = j and from h s+1 (w = 0 when t < j, and the last equality holds because g(c k = k for k {i, j}. We now prove the reverse implication. For w > 0 and i s, { g(c k + w g(c k + max 0, w } > g(c k, (41 λ where the first inequality follows from (38 and the strict inequality from w > 0 and λ > 0. If i + j s and C i + C j > C i+j then g(c i + g(c j g(c i + C j < g(c i + g(c j g(c i+j = 0, yields a contradiction to subadditivity of g( where the strict inequality follows from (41 where k = i + j and w = C i + C j C i+j and the equality holds since g(c k = k for k {i, j, i + j}. 3.3 Lifted inequalities by sequence-independent lifting In this section, we derive strong inequalities for PB through lifting using (25 as seed inequality. To describe the general form of these inequalities, we use the notion of a cover, which is adapted from the definition of a cover for the 0 1 knapsack polytope; see Balas [4], Hammer et al. [14], and Wolsey [31]. Definition 3.7. Let C N. We say that C is a cover for B if j C a j > d. Further, we define the excess of the cover as µ = j C a j d > 0. We create lifted inequalities by first partitioning the set of variables N into (C, {l}, M, T in such a way that: (A1 C := C {l} is a cover for B with excess µ, (A2 a l a j, j C, (A3 a l > µ, (A4 j C T a j > d + a l, i.e., j T a j > a l µ. Note that (A1 and (A3 might be reminiscent of conditions that make a cover minimal for the 0 1 knapsack polytope. We note however that minimal covers require a j > µ for all j C and not simply a l > µ. Note also that (A4 implies that T. To obtain lifted inequalities from (C, {l}, M, T, we first fix the variables (x j, y j for j M to (0, 0 and the variables (x j, y j for j C to (1, 1. The resulting (full-dimensional set B(M, C, M, C is then defined by the inequality a l x l y l + a j x j y j d a j = a l µ. j T j C Since a l > µ and j T a j > a l µ from Conditions (A3 and (A4, we conclude from Proposition 2.9(i that (a l µx l + j T a j y j a l µ (42 15

16 is facet-defining for PB(M, C, M, C. We will create two different families of lifted inequalities for PB by reintroducing the variables (x j, y j for j M C in different orders. To derive both families, we use the lifting function P (w := max (a l µ (a l µx l + a j y j j T s.t. a l x l y l + a j x j y j a l µ w (43 j T x j {0, 1}, y j [0, 1] j {l} T. 565 We next derive a closed-form expression for P (w. 566 Proposition 3.8. P (w = if w < j T a j µ, w + µ if j T a j µ w < µ, 0 if µ w < 0, w if 0 w < a l µ, a l µ if a l µ w Further, P (w is subadditive over R and R + respectively. Proof. We first derive a closed-form expression for P (w. Observe that, if (43 is feasible, there exists an optimal solution (x, y to (43 for which x j = 1 for j T and y l = 1 since the coefficients of x j for j T and y l in the objective are equal to 0. Defining ā = j T a j and ȳ = formulation of P (w in (43 as: P (w = max (a l µ {(a l µx l + āȳ} j T a jy j ā, we can simplify the s.t. a l x l + āȳ a l µ w (44 x l {0, 1}, ȳ [0, 1]. When w < ā µ, (44 is infeasible and so P (w =. When w a l µ, the optimal solution is x 575 l = 0 and ȳ 576 = 0 with P (w = a l µ. For ā µ w < a l µ, there are two cases. When ā µ w < a l ā µ, then every feasible solution (x l, ȳ has x l = 1. Further, the optimal solution has ȳ = max{ µ w 577 ā, 0}. It 578 follows that P (w = min{w + µ, 0}. When a l ā µ w a l µ, an optimal solution must be found among the solutions (1, ( µ w+ ā and (0, a l µ w ā. It follows that P (w = max{(w + µ 579, w} from which we obtain 580 the desired expression for P (w after considering both the cases where a l ā < 0 and a l ā Subadditivity of P (w over R and R + follows from Karamata/Hardy-Littlewood-Polya inequality [15], concavity of P (w over these domains and P (0 = We note that, although P (w is subadditive over R + and over R, P (w is not subadditive over R as P (2a l µ + P ( a l = (a l µ + ( a l + µ = 0 < a l µ = P (a l µ Lifted bilinear cover inequalities To obtain lifted bilinear cover inequalities, we will lift first the variables (x i, y i for i C from (1, 1 and then lift the variables (x i, y i for i M from (0, 0. Since P (w is subadditive over R, we can apply sequence-independent lifting for the variables (x i, y i for i C. Proposition 3.9. Under Conditions (A1, (A2, (A3 and (A4, j µ j C(a + x j + a j y j j µ j T j C(a + (45 is facet-defining for PB(M,, M,. 16