Two-term disjunctions on the second-order cone

Abstract

Balas introduced disjunctive cuts in the 1970s for mixed-integer linear programs. Several recent papers have attempted to extend this work to mixed-integer 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.

Appendix

Proof of Lemma 1

To prove the first claim, suppose $S_1\cup S_2\subsetneq S$ and pick $x_0\in S{\setminus }(S_1\cup S_2)$. Also, pick $x_1\in S_1{\setminus } S_2$ and $x_2\in S_2{\setminus } S_1$. Let $x'$ be the point on the line segment between $x_0$ and $x_1$ such that $c_1^\top x'=c_{1,0}$. Similarly, let $x''$ be the point between $x_0$ and $x_2$ such that $c_2^\top x''=c_{2,0}$. Note that $x'\notin S_2$ and $x''\notin S_1$ by the convexity of $S{\setminus } S_1$ and $S{\setminus } S_2$. Then a point that is a strict convex combination of $x'$ and $x''$ is in ${{\mathrm{conv}}}(S_1\cup S_2)$ but not in $S_1\cup S_2$.

Corollary 9.1.2 in [31] implies $S_1^+$ and $S_2^+$ are closed and ${{\mathrm{rec}}}S_1^+={{\mathrm{rec}}}S_2^+={{\mathrm{rec}}}S_1+{{\mathrm{rec}}}S_2$ because $S$ is pointed. The inclusions $S_1\subseteq S_1^+$ and $S_2\subseteq S_2^+$ imply that ${{\mathrm{conv}}}(S_1\cup S_2)\subseteq {{\mathrm{conv}}}(S_1^+\cup S_2^+)$. Furthermore, ${{\mathrm{conv}}}(S_1^+\cup S_2^+)$ is closed by Corollary 9.8.1 in [31] since $S_1^+$ and $S_2^+$ have the same recession cone. Hence, ${{\mathrm{\overline{conv}}}}(S_1\cup S_2)\subseteq {{\mathrm{conv}}}(S_1^+\cup S_2^+)$. We claim ${{\mathrm{\overline{conv}}}}(S_1\cup S_2)={{\mathrm{conv}}}(S_1^+\cup S_2^+)$. Let $x^+\in {{\mathrm{conv}}}(S_1^+\cup S_2^+)$. Then there exist $u_1\in S_1$, $v_2\in {{\mathrm{rec}}}S_2$, $u_2\in S_2$, and $v_1\in {{\mathrm{rec}}}S_1$ such that $x^+\in {{\mathrm{conv}}}\{u_1+v_2,u_2+v_1\}$. To prove the claim, it is enough to show that $u_1+v_2,u_2+v_1\in {{\mathrm{\overline{conv}}}}(S_1\cup S_2)$. Consider the point $u_1+v_2$ and the sequence

$$\begin{aligned} \left\{ \left( 1-\frac{1}{k}\right) u_1+\frac{1}{k}\left( u_2+k v_2\right) \right\} _{k\in \mathbb {N}}. \end{aligned}$$

For any $k\in \mathbb {N}$, we have $u_1\in S_1$ and $u_2+k v_2\in S_2$. Therefore, this sequence is in ${{\mathrm{conv}}}(S_1\cup S_2)$. Furthermore, it converges to $u_1+v_2$ as $k\rightarrow \infty$ which implies $u_1+v_2\in {{\mathrm{\overline{conv}}}}(S_1\cup S_2)$. A similar argument shows $u_2+v_1\in {{\mathrm{\overline{conv}}}}(S_1\cup S_2)$ and proves the claim. $\square$

Proof of Proposition 6

Let $S_1^+:=S_1+{{\mathrm{rec}}}S_2$ and $S_2^+:=S_2+{{\mathrm{rec}}}S_1$. We have ${{\mathrm{conv}}}(S_1\cup S_2)\subseteq {{\mathrm{\overline{conv}}}}(S_1\cup S_2)={{\mathrm{conv}}}(S_1^+\cup S_2^+)$ by Lemma 1. We are going to show ${{\mathrm{conv}}}(S_1^+\cup S_2^+)\subseteq {{\mathrm{conv}}}(S_1\cup S_2)$ to prove that ${{\mathrm{conv}}}(S_1\cup S_2)$ is closed when (19) is satisfied. Let $x^+\in S_1^+$. Then there exist $u_1\in S_1$ and $v_2\in {{\mathrm{rec}}}(S_2)$ such that $x^+=u_1+v_2$. If $c_2^\top v_2>0$, then there exists $\epsilon \ge 1$ such that $x^+ +\epsilon v_2\in S_2$ and we have $x^+\in {{\mathrm{conv}}}(S_1\cup S_2)$. Otherwise, $c_2^\top v_2=0$, and by the hypothesis, $c_1^\top v_2\ge 0$. This implies $x^+\in S_1$, and thus $S_1^+\subseteq {{\mathrm{conv}}}(S_1\cup S_2)$. Through a similar argument, one can show $S_2^+\subseteq {{\mathrm{conv}}}(S_1\cup S_2)$. Hence, $S_1^+\cup S_2^+\subseteq {{\mathrm{conv}}}(S_1\cup S_2)$. Taking the convex hull of both sides yields ${{\mathrm{conv}}}(S_1^+\cup S_2^+)\subseteq {{\mathrm{conv}}}(S_1\cup S_2)$.

For the converse, suppose Condition (1) holds, and let $x^*\in S_1$ be such that $c_2^\top x^*\le c_2^\top x$ for all $x\in S_1$. Note that $c_2^\top x^*<c_{2,0}$ since otherwise, $S_1\subseteq S_2$. Pick $\delta >0$ such that $x':=x^*+\delta r^*\notin S_1$. Then $x'\notin S_2$ too because $c_2^\top x'=c_2^\top x^*<c_{2,0}$. For any $0<\lambda <1$, $x_1\in S_1$, and $x_2\in S_2$, we can write $c_2^\top (\lambda x_1+(1-\lambda )x_2)\ge \lambda c_2^\top x^*+(1-\lambda )c_{2,0}>c_2^\top x'$. Hence, $x'\notin {{\mathrm{conv}}}(S_1\cup S_2)$. On the other hand, $x'\in S_1^+\subseteq {{\mathrm{conv}}}(S_1^+\cup S_2^+)={{\mathrm{\overline{conv}}}}(S_1\cup S_2)$ where the last equality follows from Lemma 1. $\square$

Proof of Corollary 2

Suppose there exist $\beta _1,\beta _2\in \mathbb {R}$ such that $c_1-\beta _2c_2\in \mathbb {K}^*$ and $c_2-\beta _1c_1\in \mathbb {K}^*$. Consider the following minimization problem

$$\begin{aligned} \inf _u\{c_1^\top u:\;c_2^\top u=0,u\in \mathbb {K}\} \end{aligned}$$

and its dual

$$\begin{aligned} \sup _\delta \{0:\;c_1-\delta c_2\in \mathbb {K}^*\}. \end{aligned}$$

Because $\beta _2$ is a feasible solution to the dual problem, we have $c_1^\top u\ge 0$ for all $u\in \mathbb {K}$ such that $c_2^\top u=0$. Similarly, one can use the existence of $\beta _1$ to show that the second part of (19) holds too. Then by Proposition 6, ${{\mathrm{conv}}}(C_1\cup C_2)$ is closed. $\square$

Proof of Proposition 7

Every undominated valid inequality has to be tight on either $C_1$ or $C_2$; otherwise, we can just increase the right-hand side to obtain a dominating valid inequality. By Proposition 2, undominated valid inequalities are of the form $\mu ^\top x\ge \mu _0$ where $(\mu ,\mu _0,\alpha _1,\alpha _2,\beta _1,\beta _2)$ satisfies the first system in (8). In particular, we have $\beta _1>0$, $\beta _1c_{1,0}\ge c_{2,0}$, and $\mu _0=c_{2,0}$. Now consider the following minimization problem

$$\begin{aligned} \inf _u\{\mu ^\top u:\;u\in C_1\} \end{aligned}$$

and its dual

$$\begin{aligned} \sup _\delta \{\delta c_{1,0}:\;\mu -\delta c_1\in \mathbb {K}^*,\;\delta \ge 0\}. \end{aligned}$$

Note that $\beta _1$ is a feasible solution to the dual problem. The set $C_1$ is strictly feasible by Condition 2, so strong duality applies to this pair of conic optimization problems. The dual problem admits an optimal solution $\delta ^*$ which satisfies $\delta ^*\ge \beta _1>0$ because $c_{1,0}\ge 0$. Then

$$\begin{aligned} {{\mathrm{sign}}}\{\delta ^* c_{1,0}\}={{\mathrm{sign}}}\{c_{1,0}\}=c_{1,0}>c_{2,0}=\mu _0. \end{aligned}$$

Hence, the inequality $\mu ^\top x\ge \mu _0$ cannot be tight on $C_1$. $\square$