Two-Term Disjunctions on the Second-Order Cone

Transcription

1 Noname manuscript No. (will be inserted by the editor) Two-Term Disjunctions on the Second-Order Cone Fatma Kılınç-Karzan Sercan Yıldız the date of receipt and acceptance should be inserted later Abstract Balas introduced disjunctive cuts in the 970s for mixed-integer linear programs. Several recent papers have attempted to extend this work to mixedinteger conic programs. In this paper we study the structure of the convex hull of a two-term disjunction applied to the second-order cone, and develop a methodology to derive closed-form expressions for convex inequalities describing the resulting convex hull. Our approach is based on first characterizing the structure of undominated valid linear inequalities for the disjunction and then using conic duality to derive a family of convex, possibly nonlinear, valid inequalities that correspond to these linear inequalities. We identify and study the cases where these valid inequalities can equivalently be expressed in conic quadratic form and where a single inequality from this family is sufficient to describe the convex hull. In particular, our results on two-term disjunctions on the second-order cone generalize related results on split cuts by Modaresi, Kılınç, and Vielma, and by Andersen and Jensen. Keywords Mixed-integer conic programming, second-order cone programming, cutting planes, disjunctive cuts Mathematics Subject Classification (00) 90C, 90C6 Introduction. Motivation and Related Work A mixed-integer conic program is a problem of the form supd x : Ax = b, x K, x j Z j J} where K R n is a regular (full-dimensional, closed, convex, and pointed) cone, A is an m n real matrix, d and b are real vectors of appropriate dimensions, and Tepper School of Business Carnegie Mellon University, Pittsburgh, PA fkilinc,syildiz}@andrew.cmu.edu

2 Fatma Kılınç-Karzan, Sercan Yıldız J,..., n}. Mixed-integer conic programming (MICP) models arise naturally as robust versions of mixed-integer linear programming (MILP) models in finance, management, and engineering [,5]. MILP is the special case of MICP where K is the nonnegative orthant, and it has itself numerous applications. A successful approach to solving MILP problems has been to first solve the natural continuous relaxation obtained by dropping the integrality conditions, then add cuts to eliminate regions that do not contain any feasible integer points, and finally perform branch-and-bound using this strengthened formulation. A powerful way of generating such cuts is to impose a valid disjunction on the continuous relaxation, and derive tight convex inequalities for the resulting disjunctive set. Such inequalities are known as disjunctive cuts. Specifically, the integrality conditions on the variables x j, j J, imply split disjunctions of the form π x π 0 π x π 0 + where π 0 Z, π j Z, j J, and π j = 0, j J. Following this approach, the feasible region for MICP problems can be relaxed to x K : Ax = b, π x π 0 π x π 0 + }. More general two-term disjunctions arise in complementarity [4, 34] and other non-convex optimization [8, 6] problems. Therefore, it is interesting to study relaxations of MICP problems of the form supd x : x C C } where C i := x K : Ax = b, c i x c i,0 } for i, }. () Note that the feasible region C C of the problem above is still possibly nonconvex, but using its particular disjunctive structure, one can build convex relaxations by deriving valid inequalities. The tightest convex relaxations of C C that one can obtain through this procedure are its convex hull conv(c C ) and closed convex hull conv(c C ). When K is the nonnegative orthant, Bonami et al. [4] characterize conv(c C ) by a finite set of linear inequalities. The purpose of this paper is to study the structure of conv(c C ) for other cones such as the second-order (Lorentz) cone L n := x R n : (x ;... ; x n ) x n} and provide the explicit description of conv(c C ) with convex inequalities in the space of the original variables. We first review related results from the literature. Disjunctive cuts were introduced by Balas [4] for MILP in the early 970s. Since then, disjunctive cuts have been studied extensively in mixed integer linear and nonlinear optimization [5, 3, 7, 9, 3, 7, 5, 6]. Chvátal-Gomory, lift-andproject, mixed-integer rounding (MIR), and split cuts are all special types of disjunctive cuts. Recent efforts on extending the cutting plane theory for MILP to the MICP setting include the work of Çezik and Iyengar [8] for Chvatal-Gomory cuts, Stubbs and Mehrotra [33], Drewes [], Drewes and Pokutta [], and Bonami [3] for lift-and-project cuts, and Atamtürk and Narayanan [, 3] for MIR cuts. Kılınç- Karzan [6] analyzed properties of minimal valid linear inequalities for general conic sets with a disjunctive structure and showed that these are sufficient to describe their closed convex hulls. Such general sets from [6] include two-term disjunctions on the cone K considered in this paper. Bienstock and Michalka [] studied the characterization and separation of valid linear inequalities that convexify the epigraph of a convex, differentiable function restricted to a non-convex domain. While the second-order cone can be viewed naturally as the epigraph of the function, we note here that this function is not differentiable and a two-term disjunction on the domain of would require c,n = c,n = 0 in our setting. In the last few years, there has been growing interest in developing

3 Two-Term Disjunctions on the Second-Order Cone 3 closed-form expressions for convex inequalities that fully describe the convex hull of a disjunctive conic set. Dadush et al. [0] and Andersen and Jensen [] derived split cuts for ellipsoids and the second-order cone, respectively. Modaresi et al. [8] extended this work on split disjunctions to essentially all cross-sections of the second-order cone, and studied their theoretical and computational relations with extended formulations and conic MIR inequalities in [9]. Belotti et al. [0] studied the families of quadratic surfaces that have fixed intersections with two given hyperplanes and showed that these families can be described by a single parameter. Building on this, in [9] they identified a procedure for constructing two-term disjunctive cuts under the assumptions that C C = and the sets x K : Ax = b, c x = c,0 } and x K : Ax = b, c x = c,0 } are bounded. In this paper we study general two-term disjunctions c x c,0 c x c,0 on conic sets and give closed-form expressions for the disjunctive cuts that can be obtained from these disjunctions in a large class of instances. We focus on the case where C and C in () above have an empty set of equations Ax = b. That is to say, we consider C := x K : c x c,0 } and C := x K : c x c,0 }. () Our results can, however, easily be extended to two-term disjunctions on sets x R n : Ax b K} where A has full row rank through the affine transformation discussed in []. Our main contribution in this paper is to give an explicit outer description of conv(c C ) when K is the second-order cone. Similar results have previously appeared in [], [8], and [9]. Nevertheless, our work is set apart from [] and [8] by the fact that we study two-term disjunctions c x c,0 c x c,0 on the cone K in full generality and do not restrict our attention to split disjunctions, which are defined by parallel hyperplanes. Furthermore, unlike [9], we do not assume that C C = and the sets x K : c x = c,0 } and x K : c x = c,0 } are bounded. Our analysis shows that the resulting convex hulls can turn out to be significantly more complex in the absence of such assumptions. We also stress that our proof techniques originate from a conic duality perspective and are completely different from what is employed in the aforementioned papers; in particular, we believe that they are intuitive in terms of their derivation and transparent in explaining the structure of the resulting convex hulls. Therefore, we hope that they have the potential to be instrumental in extending several important results in this growing area of research. A preliminary version of this paper has appeared in [7]. Yıldız and Cornuéjols [35] have recently extended our convex hull characterizations to two-term disjunctions on cross-sections of the second-order cone. They have shown that a single valid inequality of the type derived in this paper can be used to describe conv(c C ) explicitly even when the definition of C and C in () includes a nonempty set of equations Ax = b.. Outline of the Paper We would like to give a brief outline of our results before we proceed. In Section we introduce the tools that are useful in our analysis. In Section. we set out our notation and identify the cases that are of main interest to us with Conditions and. These conditions are indeed mild requirements and mainly introduced

4 4 Fatma Kılınç-Karzan, Sercan Yıldız to avoid trivial pathological cases. We discuss the pathologies that arise in the absence of these conditions in Section.3. It is a well-known fact from convex analysis that conv(c C ) can be described by linear inequalities alone. However, the set of linear inequalities that are valid for conv(c C ) will typically be very large, and only a small subset of these will be needed besides the cone constraint x K in a minimal description of conv(c C ). In Section., for a general regular cone K, we identify and characterize the structure of a subset of strong valid linear inequalities which, along with the cone constraint x K, are sufficient to describe conv(c C ). These inequalities are tight on conv(c C ) and K-minimal in the sense defined in [6]. We term such linear inequalities undominated. In Section 3 we focus on the case where K is the second-order cone, L n. In Section 3. we prove our main result, Theorem 3, which we state below in a slightly modified form. The proof of Theorem 3 uses conic duality, along with the characterization of undominated valid linear inequalities from Section., to derive a family of convex, possibly linear or conic, valid inequalities (3) for conv(c C ). Theorem Let C and C be defined as in () where K = L n and c,0, c,0 0, ±}. Suppose Conditions and are satisfied. For any β, β > 0 such that β c β c / ± int L n, let N (β, β ) := β c β c (β c,n β c,n ). Then the inequality minβ c,0, β c,0 } (β c + β c ) x ((β c β c ) x) + N (β, β ) ( x n x ) (3) is valid for conv(c C ). Furthermore, inequalities (3) are sufficient to describe conv(c C ) along with the cone constraint x L n. In Section 3. we explore when (3) can equivalently be expressed in conic quadratic form (Propositions 3 and 4). In Section 4 we identify and study the cases where only one inequality of the form (3) is sufficient to describe conv(c C ) (Theorem 4). In the special case where C C is defined by a split disjunction on the second-order cone, the convex hull is closed and our results show that a single additional valid inequality in conic quadratic form is always sufficient to describe conv(c C ). We formulate this conclusion into Theorem, which we state below. This recovers the related results of [] and [8] on split disjunctions on the second-order cone. Theorem Let C and C be defined by a split disjunction in () where K = L n and c,0, c,0 0, ±}. Suppose Condition is satisfied. If c,0 = c,0 =, then ( ) } conv(c C ) = x L n : N x (c c c x ) L n c,n + c,n where N := c c (c,n c,n ) ; otherwise, conv(c C ) = L n. Example As an application of Theorem, consider the split disjunction 4x x on the second-order cone L 3. Let e i R n be the i th standard unit vector, i,..., n}. Theorem states that in this case conv(c C ) is the set of points x L 3 that satisfy the conic quadratic inequality 5x (4x )e L 3.

5 Two-Term Disjunctions on the Second-Order Cone 5 Figures (a) and (b) show the disjunctive set C C and the conic quadratic inequality which is introduced to convexify C C, respectively. (a) C C (b) Conic quadratic inequality describing conv(c C ) Fig. : Sets associated with the split disjunction 4x x on L 3. In spite of these good news, there are cases where it is not possible to obtain conv(c C ) with a single inequality of the form (3). In Section 5 we study these cases and outline a technique to characterize conv(c C ) with closed-form formulas in general. Preliminaries The main purpose of this section is to characterize the structure of undominated valid linear inequalities for conv(c C ) when K is a regular cone and C and C are defined as in (). First, we present our notation and assumptions.. Notation and Assumptions Given a set S R n, we let span S, int S, and bd S denote the linear span, interior, and boundary of S, respectively. We use rec S to refer to the recession cone of a convex set S. The dual cone of K R n is K := y R n : y x 0 x K}. Recall that the dual cone K of a regular cone K is also regular and the dual of K is K itself. We can always scale the inequalities c x c,0 and c x c,0 defining the disjunction so that their right-hand sides are 0 or ±. Therefore, from now on we assume that c,0, c,0 0, ±} for notational convenience. When C C, we have conv(c C ) = C. Similarly, when C C, we have conv(c C ) = C. In the remainder we focus on the case where C C and C C.

6 6 Fatma Kılınç-Karzan, Sercan Yıldız Condition C C and C C. In particular, Condition implies C, C and C, C K. Hence, c i K when c i,0 = + and c i K when c i,0 =. We also need the following technical condition in our analysis. Condition C and C are strictly feasible sets. That is, C int K and C int K. Under Condition, the sets C and C always have nonempty interior. Note that the set C i is always strictly feasible when it is nonempty and c i,0 ±}. Therefore, we need Condition to supplement Condition only when c,0 = 0 or c,0 = 0. We note that Condition is essentially used in proving the sufficiency results, that is, explicit closed convex hull characterizations, and even in the absence of Condition, our techniques yield convex valid inequalities. We evaluate the necessity of Condition for our sufficiency results with an example in Section.3. Throughout the paper, we consider sets C and C which are defined as in () with c,0, c,0 0, ±} and which satisfy Conditions and. We say that such sets C and C satisfy the basic disjunctive setup. Conditions and have several simple implications, which we state next. The first lemma extends ideas from Balas [6] to disjunctions on more general convex sets. Its proof is left to the appendix. Lemma Let S R n be a closed, convex, pointed set, S := x S : c x c,0 }, and S := x S : c x c,0 } for c, c R n and c,0, c,0 R. Suppose S S and S S. Then (i) S S is not convex unless S S = S, (ii) conv(s S ) = conv(s + S+ ) where S+ := S +rec S and S + := S +rec S. Clearly, when conv(c C ) = K, we do not need to derive any new inequalities to get a description of the closed convex hull. The next lemma obtains a natural consequence of Condition through conic duality. Lemma Consider C, C defined as in (). Suppose Condition holds. Then the following system of inequalities in the variable β is inconsistent: β 0, β c,0 c,0, c β c K. (4) Similarly, the following system of inequalities in the variable β is inconsistent: β 0, β c,0 c,0, c β c K. (5) Proof Suppose there exists β satisfying (4). For all x K, this implies (c βc ) x 0 c,0 βc,0. Then any point x C satisfies βc x βc,0 and therefore, c x c,0. Hence, C C which contradicts Condition. The proof for the inconsistency of (5) is similar.

7 Two-Term Disjunctions on the Second-Order Cone 7. Properties of Undominated Valid Linear Inequalities It is well-known that the closed convex hull of any set can be described by valid linear inequalities alone (see, e.g., [3, Theorem 4..3]). In this section, using the special structure of the set C C, we show that a subset of strong valid linear inequalities is sufficient to describe conv(c C ). Besides being smaller in size, this subset of linear inequalities also has a particular structure which is instrumental in the derivation of the nonlinear valid inequality of Theorem 3 in Section 3. A valid linear inequality µ x µ 0 for a feasible set S K is said to be tight if inf xµ x : x S} = µ 0 and strongly tight if there exists x S such that µ x = µ 0. A valid linear inequality ν x ν 0 for a strictly feasible set S K is said to dominate another valid linear inequality µ x µ 0 if it is not a positive multiple of µ x µ 0 and implies µ x µ 0 together with the cone constraint x K. Furthermore, a valid linear inequality µ x µ 0 is said to be undominated if there does not exist another valid linear inequality ν x ν 0 such that (µ ν, µ 0 ν 0 ) K R + \ (0, 0)}. This notion of domination is closely tied with the K-minimality definition of [6] which says that a valid linear inequality µ x µ 0 is K-minimal if there does not exist another valid linear inequality ν x ν 0 such that (µ ν, µ 0 ν 0 ) (K \ 0}) R +. In particular, a valid linear inequality for conv(c C ) is undominated in the sense considered here if and only if it is K-minimal and tight on conv(c C ). In [6], Kılınç-Karzan defines and studies K-minimal inequalities for disjunctive conic sets of the form x R n : Ax H, x K }, where H is an arbitrary set and K is a regular cone. Our set C C represented in this form as ( ) ( ) ( ) } x R n c : c x =,0 } + R + R, x K. R c,0 } + R + c can be Because C C is full-dimensional under Condition, we can use Proposition 3. of [6] to conclude that the extreme rays of the convex cone of valid linear inequalities } (µ, µ 0 ) R n R : µ x µ 0 x conv(c C ) are either tight, K-minimal inequalities or implied by the cone constraint x K. Hence, one needs to add only undominated valid linear inequalities to the cone constraint x K to obtain an outer description of conv(c C ). Because C and C are strictly feasible sets by Condition, conic duality implies that a linear inequality µ x µ 0 is valid for conv(c C ) if and only if there exist α, α, β, β such that (µ, µ 0, α, α, β, β ) satisfies µ = α + β c, µ = α + β c, β c,0 µ 0, β c,0 µ 0, α, α K, β, β R +. This system can be strengthened significantly when we consider undominated valid linear inequalities. (6)

8 8 Fatma Kılınç-Karzan, Sercan Yıldız Proposition Let C, C satisfy the basic disjunctive setup. Then, up to positive scaling, any undominated valid linear inequality for conv(c C ) has the form µ x µ 0 with (µ, µ 0, α, α, β, β ) satisfying µ = α + β c, µ = α + β c, minβ c,0, β c,0 } = µ 0, α, α bd K, β, β R + \ 0}. (7) Proof Let ν x ν 0 be a valid inequality for conv(c C ). Then there exist α, α, β, β such that (ν, ν 0, α, α, β, β ) satisfies (6). If β = 0 or β = 0, then ν x ν 0 is implied by the cone constraint x K. If minβ c,0, β c,0 } > ν 0, then ν x ν 0 is implied by the valid inequality ν x minβ c,0, β c,0 }. Hence, we can assume without any loss of generality that any undominated valid linear inequality for conv(c C ) has the form ν x ν 0 with (ν, ν 0, α, α, β, β ) satisfying ν = α + β c, ν = α + β c, minβ c,0, β c,0 } = ν 0, α, α K, β, β R + \ 0}. We are now going to show that when α int K or α int K, any such inequality is either dominated or equivalent to a valid inequality µ x ν 0 that satisfies (7). Assume without any loss of generality that α int K. There are two cases that we need to consider: α = 0 and α 0. First suppose α = 0. We have α = β c β c int K. By Lemma and taking β, β > 0 into account, we conclude β c,0 < β c,0. Hence, ν 0 = β c,0. If ν 0 > 0, let 0 < ɛ < β be such that α := α ɛ c K and β c,0 β c,0 ɛ c,0 and define β := β ɛ and µ := ν ɛ c. If ν 0 0, let ɛ > 0 be such that α := α + ɛ c K and β c,0 β c,0 + ɛ c,0 and define β := β + ɛ and µ := ν + ɛ c. In either case, the inequality µ x ν 0 is valid for conv(c C ) because (µ, ν 0, α, α, β, β ) satisfies (6). Furthermore, it dominates (or, in the case of ν 0 = 0, is equivalent to) ν x ν 0 because µ = β β ν and β < β when ν 0 > 0 and β > β when ν 0 0. Now suppose α 0. Let 0 < ɛ be such that α := α ɛ α K, and define α := ( ɛ )α and µ := ν ɛ α. The inequality µ x ν 0 is valid for conv(c C ) because (µ, ν 0, α, α, β, β ) satisfies (6). Furthermore, µ x ν 0 dominates ν x ν 0 since ν µ = ɛ α K \ 0}. The system (7) is homogeneous in the tuple (µ, µ 0, α, α, β, β ). Therefore, in an undominated valid linear inequality µ x µ 0, we can assume without any loss of generality that the whole tuple has been scaled by a positive real number so that β = or β =. Proposition Let C, C satisfy the basic disjunctive setup. Then, up to positive scaling, any undominated valid linear inequality for conv(c C ) has the form µ x

9 Two-Term Disjunctions on the Second-Order Cone 9 µ 0 with (µ, µ 0, α, α, β, β ) satisfying one of the following systems: (i) µ = α + β c, µ = α + c, β c,0 c,0 = µ 0, α, α bd K, β R + \ 0}, β =, (ii) µ = α + c, µ = α + β c, β c,0 c,0 = µ 0, α, α bd K, β R + \ 0}, β =. Keeping c,0, c,0 0, ±} in mind, observe that the first of the two systems in (8) is infeasible when c,0 > c,0 and the second is infeasible when c,0 > c,0. Therefore, in these cases it suffices to consider only one of these systems. When c,0 = c,0 however, one may need valid linear inequalities that are associated with either of the two systems in (8) to be able to describe conv(c C ). Still, for this case Proposition implies that any undominated valid linear inequality for conv(c C ) can be written in the form µ x µ 0 with µ 0 = c,0 = c,0. Note that in any tuple (µ, µ 0, α, α, β, β ) satisfying the first system in (8), we must have c β c / ± int K since having c β c ± int K would contradict either α = α + (c β c ) bd K or α = α (c β c ) bd K. Similarly, in any tuple (µ, µ 0, α, α, β, β ) satisfying the second system in (8), we must have c β c / ± int K. Proposition and the discussion above lead us to Corollary. For ease of exposition, we define the sets B C,C := β > 0 : β c,0 c,0, c β c ± int K }, B C,C := β > 0 : β c,0 c,0, c β c ± int K }, M C,C (β, β ) := µ R n : α, α bd K, µ = α + β c = α + β c }. Corollary Let C, C satisfy the basic disjunctive setup. The closed convex hull of C C is given by } conv(c C ) = x R n µ x c :,0, µ M C,C (β, ), β B C,C, µ. x c,0, µ M C,C (, β ), β B C,C (8).3 Revisiting Condition When C i = x K : c i x c i,0} is nonempty and c i,0 ±}, it is not difficult to show that C i has to be strictly feasible. Therefore, Condition is not needed when, for instance, C and C are nonempty sets defined by a split disjunction which excludes the origin. Indeed, the only situation where Condition may be needed in addition to Condition occurs when c,0 = 0 or c,0 = 0. Note that in such a case, linear inequalities that satisfy system (6) (or (7)) are still valid for conv(c C ); they may just not be sufficient to define it completely. We next give an example which shows that Condition is necessary to establish the sufficiency of the linear inequalities that satisfy (6) (or (7)) when c,0 = c,0 = 0. Consider the second-order cone L 3 and the disjunction x x 3 0 x x 3 0 (c := e e 3, c := e e 3, c,0 = c,0 = 0). Note that c, c bd L 3, and C and C are the rays generated by e + e 3 and e + e 3, respectively. Therefore, conv(c C ) = x L 3 : x = 0} and x 0 is a valid inequality for conv(c C ). However, letting µ = e in (6), we see that any α which satisfies µ = α + β c for some β R cannot be in L 3 because α = β e + e + β e 3 / L 3.

10 0 Fatma Kılınç-Karzan, Sercan Yıldız 3 Deriving the Disjunctive Cut In this section we focus on the case where K is the second-order cone L n = x R n : x x n} and x := (x ;... ; x n ). Recall that the dual cone of L n is again L n. As in the previous section, we consider sets C and C which satisfy the basic disjunctive setup. When in addition K = L n, we say that C and C satisfy the second-order cone disjunctive setup. 3. A Convex Valid Inequality Proposition gives a nice characterization of the form of undominated linear inequalities valid for conv(c C ). In the following we use this characterization and show that, for a given pair (β, β ) satisfying the conditions of Proposition, one can group all of the corresponding linear inequalities µ x µ 0, where µ M C,C (β, β ) and µ 0 = minβ c,0, β c,0 }, into a single convex, possibly nonlinear, inequality (3) valid for conv(c C ). Then Theorem immediately follows from the fact that the inequalities (3) that are associated with all such pairs (β, β ) yields conv(c C ). By Corollary, without any loss of generality, we focus on the case c,0 c,0 and the linear inequalities µ x c,0 where µ M C,C (β, ) and β B C,C. For any fixed β B C,C, β c c / ± int L n. Furthermore, β c c / L n by Lemma. This leaves us two distinct cases to consider: β c c bd L n and β c c / ±L n. Remark Let C, C satisfy the second-order cone disjunctive setup. For any β B C,C such that β c c bd L n, the inequality β c x c,0 (9) is valid for conv(c C ) and dominates all valid linear inequalities µ x c,0 such that µ M C,C (β, ). Proof Since β B C,C, we have β c,0 c,0. The validity of (9) follows easily from β c,0 c,0 for C and β c c L n for C. Let µ M C,C (β, ). Then µ β c =: α L n, and since β c x c,0 is valid as well, we have that µ x c,0 is dominated unless α = 0. We note that Remark remains true in the general case where K is an arbitrary regular cone. Theorem 3 Let C, C satisfy the second-order cone disjunctive setup. For any β B C,C such that β c c / ±L n, the inequality c,0 (β c + c ) x ((β c c ) x) + N (β ) ( x n x ) (0) with N (β ) := β c c (β c,n c,n ) () is valid for conv(c C ) and implies all valid linear inequalities µ x c,0 such that µ M C,C (β, ).

11 Two-Term Disjunctions on the Second-Order Cone Proof Consider the set M C,C (β, ). Because β c c / ±L n, Moreau s decomposition theorem implies that there exist µ, α 0, α 0 such that α α, α, α bd L n, and µ = β c + α = c + α. Hence, M C,C (β, ) is indeed nonempty. We can write M C,C (β, ) = µ R n } : µ β c = µ n β c,n, µ c = µ n c,n = µ R n : µ c = µ β c + β c,n c,n, µ β c = µ n β c,n After taking the square of both sides of the first equation in M C,C (β, ), noting β c c / L n, and replacing the term µ β c with µ n β c,n, we arrive at M C,C (β, ) = µ R n µ : (β c c ) µ n(β c,n c,n ) = M, } µ β c = µ n β c,n where M := β( c c,n) ( c c,n). Note that, by Corollary, x conv(c C ) and β B C,C (thus β c,0 c,0 ) imply x L n and µ x c,0 µ M C,C (β, ). } x L n and inf µ x : µ M C,C (β, ) c,0. µ Unfortunately, the optimization problem stated above is non-convex due to the second equality constraint in the description of M C,C (β, ). We show below that the natural convex relaxation for this problem is tight. Indeed, consider the relaxation inf µ x : µ µ (β c c ) µ n(β c,n c,n ) = M, } µ β c µ n β c,n The feasible region of this relaxation is the intersection of a hyperplane with a closed, convex cone shifted by the vector β c. Any solution which is feasible to the relaxation but not the original problem can be expressed as a convex combination of solutions feasible to the original problem. Because we are optimizing a linear function, this shows that the relaxation is equivalent to the original problem. Thus, we have }. x conv(c C ) x L n and inf µ x : µ µ (β c c ) µ n(β c,n c,n ) = M, } c,0 µ β c µ n β c,n which is exactly the same as x conv(c C ) x L n and inf µ x : µ µ (β c c ) µ n(β c,n c,n ) = M, } µ β c L n c,0. () The minimization problem in the last line above is feasible since µ, defined at the beginning of the proof, is a feasible solution. Indeed, it is strictly feasible since

12 Fatma Kılınç-Karzan, Sercan Yıldız α + α is a recession direction of the feasible region and belongs to int L n. Hence, its dual problem is solvable whenever it is feasible, strong duality applies, and we can replace the problem in the last line with its dual without any loss of generality. Considering the definition of N (β ) = β c c (β c,n c,n ) and the hypothesis that β c c / ±L n, we get N (β ) > 0. Then x conv(c C ) ( ) x L n and max ρ,τ β c ρ + M τ : ρ + τ β c c = x, β c,n + c,n ρ L n c,0. x L n and max β c x N ( ) } (β ) β c τ : x + + c τ L n c,0, τ β c,n c,n and since the optimum solution will be on the boundary of feasible region, x L n and minτ, τ + } (β c x c,0 ) N (β ) where τ ± := (β c c ) x ± ((β c c ) x) + N (β )(x n x ). N (β ) x L n and τ (β c x c,0 ). N (β ) x L n and N (β )τ (β c x c,0 ). Rearranging the terms of the inequality in the last expression above yields (0). The next two observations follow directly from the proof of Theorem 3. Remark Under the hypotheses of Theorem 3, the set of points that satisfy (0) in L n is convex. Proof The inequality (0) is equivalent to () by construction. The left-hand side of () is a concave function of x written as the pointwise-infimum of linear functions, while the right-hand side is a constant. We note here that one can convert (0) into an inequality that is convex on all of R n by rewriting it as g(x) 0 where g : R n R + } is c g(x) :=,0 (β c + c ) x ((β c c ) x) + N (β ) ( x n x ), x L n, +, x / L n. However, we avoid using this construction to keep the notation simple in what follows. Remark 3 Inequality (0) reduces to the linear inequality (9) in L n when β c c bd L n. Proof When β c c bd L n, N (β ) = 0. Together with x L n, this also implies (β c c ) x 0, and hence, (0) of Theorem 3 becomes c,0 (β c + c ) x (β c c ) x. This is equivalent to (9).

13 Two-Term Disjunctions on the Second-Order Cone 3 Some comments about Theorem 3 are in order. Remark 4 For β B C,C, the inequality (0) can in fact be considered to be derived for the relaxed disjunction β c x c,0 c x c,0. Indeed, under the hypothesis β B C,C of Theorem 3, we have β c x β c,0 c,0 for all x C. However, somewhat contrary to intuition, inequalities (0) obtained from such weaker disjunctions are sometimes necessary for a complete description of conv(c C ). We will study these cases in more detail in Section 5. Remark 5 Inequality (0) has a simple geometrical meaning when the two sides of the relaxed disjunction β c x c,0 c x c,0 on L n do not intersect, except possibly at their boundary. Consider a point x R n which is on the hyperplane defined by β c x = c,0. Then disjointness of the two sides of the disjunction implies c x c,0. Replacing β c x with c,0 on both sides of (0), we can see that when β c c / ±L n (N (β ) > 0), such a point x satisfies (0) if and only if x ±L n. Similarly, a point x which is on the hyperplane defined by c x = c,0 satisfies (0) if and only if x ±L n. Thus, the region defined by (0) has the same cross-section as ±L n at the hyperplanes defined by the equalities β c x = c,0 and c x = c,0. Suppose C and C have been chosen so that c,0 c,0. When c,0 > c,0, B C,C =, and by Corollary, the family of inequalities given in Remark and Theorem 3 is sufficient to describe conv(c C ). On the other hand, when c,0 = c,0, we also need to consider valid linear inequalities µ x c,0 such that µ M C,C (, β ) and β B C,C. Following Remark and Theorem 3, such linear inequalities can be condensed into a single inequality for every fixed β B C,C. In particular, letting N (β ) := c β c (c,n β c,n ), one can derive the valid inequalities β c x c,0, and (3) c,0 (c + β c ) x ((c β c ) x) + N (β ) ( x n x ) (4) for β B C,C such that β c c bd L n and β c c / ±L n, respectively. Using these inequalities and recalling Remark 3, conv(c C ) can be written as } conv(c C ) = x R n : x satisfies (0) β B C,C and (4) β B C,C. Note that for any β B C,C and β B C,C, we have N (β ) 0 and N (β ) 0; hence, the right hand sides of the inequalities (0) and (4) above are well-defined for any x L n. Thus, this proves Theorem of Section.. In the remainder of this section, we continue to focus on the case where β = and β B C,C with the understanding that our results are also applicable to the symmetric situation. 3. A Conic Quadratic Form While having a convex valid inequality is nice in general, there are certain cases where (0) can be expressed in conic quadratic form. In the following, we first

14 4 Fatma Kılınç-Karzan, Sercan Yıldız identify a symmetry condition that guarantees the existence of an equivalent conic quadratic form for (0) and then show that such a symmetry condition holds under a disjointness condition. Proposition 3 Let C, C satisfy the second-order cone disjunctive setup, and let β B C,C be such that β c c / ±L n. Let x L n be a point for which c,0 + (β c + c ) x ((β c c ) x) + N (β ) ( x n x ) (5) holds with N (β ) defined as in (). Then x satisfies (0) if and only if it satisfies the conic quadratic inequality ( ) N (β )x (β c β c x c,0 ) c L n. (6) β c,n + c,n Furthermore, if (5) holds for all x conv(c C ), then (6) is valid for conv(c C ) and implies (0). Proof Let x L n be a point for which (5) holds. Then x satisfies (0) if and only if it satisfies c,0 (β c + c ) x ((β c c ) x) + N (β ) ( x n x ). We can take the square of both sides without any loss of generality and rewrite this inequality as ( ) ( ( ) c,0 (β c + c ) x (β c c ) x) + N (β ) x n x. ( ) 4(β c x c,0 )(c x c,0 ) N (β ) x n x. Because β c c / ±L n, we have N (β ) > 0, and the above inequality is equivalent to ( ) 0 N (β ) x n x 4N (β )(β c x c,0 )(c x c,0 ). The right-hand side of this inequality is identical to the following quadratic form which has a single negative eigenvalue: ( N (β )x n+(β c N x c,0 )(β c,n c,n )) (β ) x (β c x c,0 )(β c c ). Therefore, we arrive at N (β ) x (β c x c,0 )(β c c ) (. N (β )x n+(β c x c,0 )(β c,n c,n )) Let A(x) := N (β ) x (β c x c,0 )(β c c ) and B(x) := N (β )x n + (β c x c,0 )(β c,n c,n ). We have just proved that x satisfies (0) if and only if it satisfies A(x) B(x). In order to finish the proof, all we need to show is that A(u) B(u) is equivalent to

15 Two-Term Disjunctions on the Second-Order Cone 5 A(u) B(u) for all u L n. It will be enough to show that either A(u) + B(u) > 0 or A(u) = B(u) = 0 holds for all u L n. Suppose A(u)+B(u) 0 for some u L n. Using the triangle inequality, we can write 0 A(u) + B(u) = N (β )ũ (β c u c,0 )(β c c ) + N (β )u n + (β c u c,0 )(β c,n c,n ) N (β ) ũ + β c u c,0 β c c + N (β )u n β c u c,0 β c,n c,n = N (β )(u n ũ ) + β c u c,0 ( β c c β c,n c,n ). Because u L n and β c c / ±L n, we have u n ũ 0 and β c c β c,n c,n > 0. Hence, β c u = c,0. This implies A(u) + B(u) = N (β )(u n + ũ ) which is strictly positive unless u = 0, but then A(u) = B(u) = 0. The second claim of the proposition follows immediately from the first under the hypothesis that (5) holds for all x conv(c C ). Note that by changing the roles of β c and c, the proof of Proposition 3 can be repeated to show that a point x L n for which (5) holds satisfies (0) if and only if it satisfies ( ) N (β )x + (c β c x c,0 ) c L n. β c,n + c,n We next give a sufficient condition, based on a disjointness property of the intersection of C and C, under which (5) is satisfied by every point in L n. Note that this condition thus allows our convex inequality (0) to be represented in an equivalent conic quadratic form (6). Proposition 4 Let C, C satisfy the second-order cone disjunctive setup. Let β B C,C be such that βc c / ±L n. Then (5) holds for all x L n that satisfy β c x c,0 or c x c,0. Furthermore, if x L n : β c x > c,0, c x > c,0 } =, (7) then (5) holds for all x L n and (6) is equivalent to (0) on L n. Proof Let x L n satisfy β c x c,0 or c x c,0. Using Theorem 3 on the disjunction β c u c,0 or c u c,0 shows that x satisfies (5). The second claim of the proposition now follows immediately from Proposition 3 and (7). Condition (7) of Proposition 4 says that the two sides of the relaxed disjunction β c x c,0 c x c,0 on the cone L n have to be almost disjoint. This condition, together with the results of Proposition 3 and Theorem 3, identifies some cases in which (0) can be expressed in an equivalent conic quadratic form. For instance, in a split disjunction on the cone L n, it is easy to see using Lemma that C and C are both nonempty and conv(c C ) L n if and only if c, c / ±L n and c,0 = c,0 =. Furthermore, C C = ; hence, for a proper

16 6 Fatma Kılınç-Karzan, Sercan Yıldız two-sided split disjunction on L n, (7) is trivially satisfied with β =. Moreover, in the next section, we will see that conv(c C ) can be described completely with linear inequalities µ x such that µ M C,C (, ) (β = β = in (7)) when C C is defined by such a split disjunction. 4 When does a Single Inequality Suffice? In this section we give two conditions under which a single convex inequality of the type derived in Theorem 3, together with the cone constraint x L n, describes conv(c C ) completely. The main result of this section is Theorem 4 which we state below. Theorem 4 Let C, C satisfy the second-order cone disjunctive setup with c c / ±L n and c,0 c,0. Then the inequality c,0 (c + c ) x ((c c ) x) + N ( x n x ) (8) is valid for conv(c C ) with N = c c (c,n c,n ). Furthermore, conv(c C ) = x L n : x satisfies (8)}, when, in addition, (i) c L n, or c L n, or (ii) c,0 = c,0 ±} and undominated valid linear inequalities that are tight on both C and C are sufficient to describe conv(c C ). Theorem 4 shows that, under certain conditions, conv(c C ) is completely described by the cone constraint x L n and the inequality obtained by setting β = in (0). A recent result of Yıldız and Cornuéjols [35] complements this sufficiency result by showing that a similar statement is true when c,0 = c,0 = 0. Theorem 5 [35, Theorem ] Let C and C satisfy the second-order cone disjunctive setup with c,0 = c,0 = 0. Then conv(c C )= x L n : (β c + c ) x ((β c c ) x) +N (β ) ( x n x )} where β = / c c,n c c,n. The proof of Theorem 4 will require additional results on the structure of undominated valid linear inequalities. These are the subject of the next section.

17 Two-Term Disjunctions on the Second-Order Cone 7 4. Further Properties of Undominated Valid Linear Inequalities In this section we continue studying the disjunction c x c,0 c x c,0 on a regular cone K and refine the results of Section. on the structure of undominated valid linear inequalities. The results that we are going to present in this section hold for any regular cone K. The lemma below shows that the statement of Proposition can be strengthened substantially when c K or c K. Lemma 3 Let C, C satisfy the basic disjunctive setup with c,0 c,0. Suppose c K or c K. Then, up to positive scaling, any undominated valid linear inequality for conv(c C ) has the form µ x c,0 where µ M C,C (, ). Proof First note that having c i K implies rec C i = K. Therefore, when c,0 0, we can use Lemma to conclude conv(c C ) = K. In this case all valid inequalities for conv(c C ) = K are implied by the cone constraint x K, and the claim holds trivially because there are no undominated valid inequalities. Thus, we only need to consider the situation in which c,0 = c,0 =. Assume without any loss of generality that c K. Let ν x ν 0 be a valid inequality of the form given in Proposition. Then ν 0 = c,0 = c,0 =, and there exist α, α, β, β such that (ν,, α, α, β, β ) satisfies one of the two systems in (8). In particular, ν = α + β c = α + β c K and minβ, β } =. We are going to show that either ν M C,C (, ) or ν x is dominated. If β = β =, ν M C,C (, ) and we are done. We divide the rest of the proof into the following two cases: β > β and β < β. First suppose β > β. Then β = and ν = α +β c = α +c. Having α = 0 contradicts Condition through Lemma ; therefore, α 0. Let ɛ be such that 0 < ɛ β β, and define α := ( ɛ )α + ɛ c, β := ( ɛ )β, α := ( ɛ )α and µ := ν ɛ α. The inequality µ x is valid for conv(c C ) because (µ,, α, α, β, ) satisfies (6). Furthermore, µ x dominates ν x since ν µ = ɛ α K \ 0}. Now suppose β > β =. Observe that (ν,, α, α + (β )c,, ) is also a solution satisfying (6). Having α = 0 contradicts Condition through Lemma ; therefore, α 0. If α + (β )c int K, we can find a valid inequality that dominates ν x by subtracting a positive multiple of α from µ as in the proof of Proposition. Otherwise, α + (β )c bd K and ν M C,C (, ) since ν = α + c = (α + (β )c ) + c. When c,0 = c,0 ±}, a similar result holds for undominated valid linear inequalities that are tight on both C and C. Lemma 4 Let C, C satisfy the basic disjunctive setup with c,0 = c,0 ±}. Then, up to positive scaling, any undominated valid linear inequality for conv(c C ) that is tight on both C and C has the form µ x c,0 where µ M C,C (, ). Proof Let µ x µ 0 be an undominated valid inequality for conv(c C ) that is tight on both C and C. Using Proposition, we can assume that µ 0 = c,0 = c,0 and there exist α, α, β, β such that (µ, µ 0, α, α, β, β ) satisfies one of the two systems in (8). In particular, either β = and β µ 0 µ 0, or β = and β µ 0 µ 0. In any case, minβ µ 0, β µ 0 } = µ 0. We are going to show β = β =.

18 8 Fatma Kılınç-Karzan, Sercan Yıldız Consider the following pair of minimization problems and their duals inf x µ x : x C } and inf x µ x : x C }, supδµ 0 : µ = γ + δc, γ K, δ 0} and δ,γ supδµ 0 : µ = γ + δc, γ K, δ 0}. δ,γ The pairs (α, β ) and (α, β ) are feasible solutions to the first and second dual problems, respectively. Because µ x µ 0 is tight on both C and C, the optimal values of both minimization problems are µ 0 and we must have β µ 0 µ 0 and β µ 0 µ 0 by duality. This implies β µ 0 = β µ 0 = µ 0 and β = β =. Proof of Theorem 4 The validity of (8) follows from Theorem 3 by setting β =. Lemmas 3 and 4 show that we can limit ourselves to valid linear inequalities µ x c,0 where µ M C,C (, ) in Corollary. When this is the case, the implication in () in the proof of Theorem 3 is actually an equivalence. 4. A Topological Condition: Closedness of the Convex Hull Next, we identify an important case where the family of tight inequalities specified in Lemma 4 is rich enough to describe conv(c C ) completely. The key ingredient is the closedness of conv(c C ). Proposition 5 Let C, C satisfy the basic disjunctive setup. Suppose conv(c C ) is closed. Then undominated valid linear inequalities that are strongly tight on both C and C are sufficient to describe conv(c C ), together with the cone constraint x K. Proof Suppose conv(c C ) is closed. When conv(c C ) = K, no new inequalities are needed for a description of conv(c C ) and the claim holds trivially. Therefore, assume conv(c C ) K. We prove that given u K \ conv(c C ), there exists an undominated valid inequality that separates u from conv(c C ) and is strongly tight on both C and C. Let v int(conv(c C )) \ (C C ). Note that such a point exists since otherwise, we have int conv(c C ) C C which implies conv(c C ) C C through the closedness of C C. By Lemma, this is possible only if C C = K which we have already ruled out. Let 0 < λ < be such that w := ( λ)u+λv bd(conv(c C )). Then w K\(C C ) by the convexity of K \ (C C ) = x K : c x < c,0, c x < c,0 }. Because w conv(c C ), there exist x C, x C, and 0 < κ < such that w = κx + ( κ)x. Furthermore, by Corollary, the fact that w bd(conv(c C )) implies that there exists an undominated valid inequality µ x µ 0 for conv(c C ) such that µ w = µ 0. Because µ w = κµ x + ( κ)µ x = µ 0, µ x µ 0, and µ x µ 0, it must be the case that µ x = µ x = µ 0. Thus, the inequality µ x µ 0 is strongly tight on both C and C. The only thing that remains is to show that µ x µ 0

19 Two-Term Disjunctions on the Second-Order Cone 9 separates u from conv(c C ). To see this, observe that u = λ (w λv) and that µ v > µ 0 since v int conv(c C ). Hence, we conclude µ u = λ (µ w λµ v) < µ 0. Proposition 5 demonstrates the close relationship between the closedness of conv(c C ) and the sufficiency of valid linear inequalities that are tight on both C and C. This motivates us to investigate the cases where conv(c C ) is closed. The set conv(c C ) is always closed when c,0 = c,0 = 0 (see, e.g., Rockafellar [30, Corollary 9..3]) or when C and C are defined by a split disjunction (see Dadush et al. [0, Lemma.3]). In Proposition 6 below, we generalize the result of Dadush et al.: We give a sufficient condition for conv(c C ) to be closed and show that this condition is almost necessary. In Corollary, we show that the sufficient condition of Proposition 6 can be rewritten in a more specialized form using conic duality when the base set is the regular cone K. The proofs of these results are left to the appendix. Proposition 6 Let S R n be a closed, convex, pointed set, S := x S : c x c,0 }, and S := x S : c x c,0 } for c, c R n and c,0, c,0 R. Suppose S S and S S. If r rec S : c r = 0} r rec S : c r 0} and r rec S : c r = 0} r rec S : c r 0}, (9) then conv(s S ) is closed. Conversely, if (i) there exists r rec S such that c r < 0 = c r and the problem inf xc x : x S } is solvable, or (ii) there exists r rec S such that c r < 0 = c r and the problem inf xc x : x S } is solvable, then conv(s S ) is not closed. Corollary Let C, C satisfy the basic disjunctive setup. If there exist β, β R such that c β c K and c β c K, then conv(c C ) is closed. Theorem 4, Proposition 5, and Corollary imply that (8) is sufficient to describe conv(c C ) when there exist β, β R such that c β c L n and c β c L n and c,0 = c,0 ±}. Nevertheless, it is easy to construct instances where these hypotheses are not satisfied. We explore these cases further in Section 5. Consider the case of c,0 = c,0 0, ±}. Then by Lemma, c c / L n. Suppose also that (a) condition (i) or (ii) of Theorem 4 is satisfied, and (b) x L n : c x > c,0, c x > c,0 } =.

20 0 Fatma Kılınç-Karzan, Sercan Yıldız We note that statement (a) holds, for instance, when the sets C and C are defined by a split disjunction which excludes the origin because in this case c,0 = c,0 = and conv(c C ) is closed by Corollary. Moreover, statement (b) simply means that the two sets C and C defined by the disjunction do not meet, except possibly at their boundaries. This also holds for split disjunctions. Then by Theorem 4, conv(c C ) is completely described by (8) together with the cone constraint x L n. Furthermore, by Proposition 4, (5) is satisfied by every point in L n with β =, and by statement (b), we have that (8) can be expressed in an equivalent conic quadratic form (6). Thus, we conclude ( ) } conv(c C ) = x L n : N x (c c c x c,0 ) L n c,n + c,n where N = c c (c,n c,n ). Proof of Theorem By Lemma, conv(c C ) = L n unless c,0 = c,0 = because 0 C C and rec(conv(c C )) = L n. Therefore, suppose c,0 = c,0 =. Because C and C are both nonempty by hypothesis and c is a negative multiple of c, we have c, c / ±L n and c c / ±L n. Moreover, C and C are also both strictly feasible in this case; so Condition is satisfied as well. After verifying statements (a) and (b) as above, Theorem follows from Theorem 4 and Propositions 3 and 4. Theorem 4 and Proposition 3, together with Proposition 4, recover the results of [8] and [] for split disjunctions on the cone L n and extend them significantly to more general two-term disjunctions. 4.3 Example where a Single Inequality Suffices Consider the cone L 3 and the disjunction x 3 x + x 3 (c := e 3, c := e + e 3, c,0 = c,0 = ). Note that c, c L 3 in this example. Hence, we can use Theorem 4 to characterize the closed convex hull: conv(c C ) = x L 3 : (x + x 3 ) } x 3 x. Figures (a) and (b) depict the disjunctive set C C and the associated closed convex hull, respectively. In order to give a better sense of the convexification operation, we plot the points added to C C to generate the closed convex hull in Figure (c). We note that in this example the condition on the disjointness of the interiors of C and C that was required in Proposition 4 is violated. Nevertheless, the inequality that we provide is still intrinsically related to the conic quadratic inequality (6) of Proposition 3: The sets described by the two inequalities coincide in the region conv(c C )\(C C ) as a consequence of Proposition 4. We display the corresponding cone for this example in Figure (d). Note that the resulting conic quadratic inequality is in fact not valid for some points in conv(c C ).

21 Two-Term Disjunctions on the Second-Order Cone (a) C C (b) conv(c C ) (c) conv(c C ) \ C C (d) Underlying cone generating the convex inequality Fig. : Sets associated with the disjunction x 3 x + x 3 on L 3. 5 When are Multiple Convex Inequalities Needed? Lemma 4 allows us to simplify the characterization (7) of undominated valid linear inequalities which are tight on both C and C in the case c,0 = c,0 ±}. The next proposition shows the necessity the condition c,0 = c,0 in the statement of this lemma. Unfortunately, when c,0 c,0, undominated valid linear inequalities are tight on exactly one of the two sets C and C. The proof of this result is left to the appendix. Proposition 7 Let C, C satisfy the basic disjunctive setup. If c,0 > c,0, then any undominated valid linear inequality for conv(c C ) is tight on C but not on C. This result, when combined with Proposition 5, yields the following corollary. Corollary 3 Let C, C satisfy the basic disjunctive setup with c,0 > c,0. If conv(c C ) K, then conv(c C ) is not closed. Proof Suppose conv(c C ) is closed, and let x K \ conv(c C ). By Proposition 5, there exists an undominated valid linear inequality which cuts off x from conv(c C ) and is tight on both C and C. This contradicts Proposition 7.

22 Fatma Kılınç-Karzan, Sercan Yıldız 5. Describing the Closed Convex Hull As Proposition 7 hints, there are cases where linear inequalities µ x µ 0 such that µ M C,C (, ) and µ 0 = minc,0, c,0 } (β = β = in (7)) may not be sufficient to define conv(c C ). In this section, we study these cases when K = L n and outline a procedure to find closed-form expressions describing conv(c C ). Recall that, by Theorem 3 and Remark 3, conv(c C ) = x L n : x satisfies (0) β B C,C and (4) β B C,C }. For ease of notation, let R := R(c, c, x) = (c x) + ( c c,n)(x n x ), P := P(c, c, x) = (c x)(c x) + ( c c c,n c,n )(x n x ), Q := Q(c, c, x) = (c x) + ( c c,n)(x n x ), and f c,c,x (β ) := β c x + Rβ Pβ + Q. Then Rβ Pβ + Q = ( + ( ) (β c c ) x) N (β ) x n x. Similarly, define f c,c,x (β ) := β c x + Qβ Pβ + R and note Qβ Pβ + R = ( + ( ) (c β c ) x) N (β ) x n x. Through these definitions, we reach conv(c C ) = = } x L n : c,0 c x f c,c,x (β ) β B C,C, c,0 c x f c,c,x (β ) β B C,C } x L n : c,0 c x inf β B C,C c,0 c x inf β B C,C f c,c,x (β ), f c,c,x (β ). (0) It follows that, for any given x L n, we can check whether x conv(c C ) by calculating the optimal values of the problems on the right-hand side of the inequalities in (0). Furthermore, whenever the minimizer β := β(c, c, x) of inf β B C,C f c,c,x (β ) exists and can be expressed parametrically in terms of c, c, and x, one can replace the inequality c,0 c x inf β B C,C in (0) above with c,0 c x f c,c,x f c,c,x (β ) (β) for all points x L n such that β(c, c, x) B C,C. Similarly, one can define β := β(c, c, x) and replace c,0 c x inf β B C,C f c,c,x (β ) with c,0 c x f c,c,x (β) for all points x L n such that β (c, c, x) B C,C. We illustrate this procedure on an example in the next section.

23 Two-Term Disjunctions on the Second-Order Cone 3 5. Example where Multiple Inequalities are Needed Consider the cone L 3 and the disjunction x 0 x 3 (c := e, c,0 = 0, c := e 3, c,0 = ). Since c,0 > c,0, by Proposition 7, we know that any undominated valid linear inequality for conv(c C ) will be tight on C but not on C. Therefore, we follow the approach outlined in Section 5.. By noting that c β c int L 3 for 0 β < and c β c ± int L 3 for β, we obtain B C,C = [, ). Furthermore, for β =, using (9) in Remark, we obtain x as a valid linear inequality for all x conv(c C ). It is also clear that B C,C =. Since we are interested in cutting off only points x L 3 such that x and x / conv(c C ), consider x L 3 such that 0 < x and x 3 >. Note that x L 3 and x > 0 imply x 3 x > 0. In this setup we have R(c, c, x) = x 3 x, P(c, c, x) = x x 3, Q(c, c, x) = x + x. The resulting f c,c,x is a convex function of β and has a critical point at ˆβ := ˆβ (c, c, x) = P R c x P QR R (c x) R = x x 3 x 3 + x x x 3 (x + x )(x 3 x ) x x 3 x ( )(x 3 x x ) = x x 3 + x x x 3 x = x x 3 x where in the last equation we used the fact that x L 3 and thus x 3 >. For any x L 3 such that x x 3 x, we have ˆβ. } By the convexity of f c,c,x, the minimum of f c,c,x occurs at β = max ˆβ, =. Moreover, for any x L 3 such that x x 3 x, we have ˆβ. For such points, β = ˆβ and f c,c,x (β) = x x (x3+ x ) = x x x 3 x x. Therefore, for all x 3 x L3 such that 0 < x, x 3 >, and x x 3 x, we can enforce c,0 c x f c,c,x (ˆβ ) which translates to + x 3 x x x in this example. Using 0 < x 3 x and x 3 x > 0, we can rewrite this inequality as + x x 3 max0, x }. Putting this together with x, we arrive at } conv(c C ) = x L 3 : + x 3 f c,c,x (β ) β [, ) = x L 3 : x, + x x 3 } max0, x }, where both inequalities are convex on R 3. In fact, both inequalities are conic quadratic representable in a lifted space as expected. In Figures 3(a) and (b), we plot the disjunctive set C C and the resulting conv(c C ), respectively. The closed convex hull is obtained by imposing various convex inequalities (0), each corresponding to a different value of β B C,C, on

24 4 Fatma Kılınç-Karzan, Sercan Yıldız L 3. In Figure 3(c) we show the conic quadratic counterparts (6) of the inequalities. Note that these inequalities are not necessarily valid for all points in C C because Condition (7) is not satisfied, but they describe how the boundary of conv(c C ) is formed outside C C. In Figure 3(d) we show the cross-section of C C and the regions defined by the conic quadratic inequalities (6) at x 3 = 4. (a) C C (b) conv(c C ) (c) Underlying cones generating the convex inequalities (d) Cross-section at x 3 = 4 Fig. 3: Sets associated with the disjunction x 0 x 3 on L 3. 6 Appendix Proof of Lemma To prove the first claim, suppose S S S and pick x 0 S \ (S S ). Also, pick x S \ S and x S \ S. Let x be the point on the line segment between x 0 and x such that c x = c,0. Similarly, let x be the