A Conic Representation of the Convex Hull of Disjunctive Sets and Conic Cuts for Integer Second Order Cone Optimization

Abstract

We study the convex hull of the intersection of a convex set E and a disjunctive set. This intersection is at the core of solution techniques for Mixed Integer Convex Optimization. We prove that if there exists a cone K (resp., a cylinder C) that has the same intersection with the boundary of the disjunction as E, then the convex hull is the intersection of E with K (resp., C).The existence of such a cone (resp., a cylinder) is difficult to prove for general conic optimization. We prove existence and unicity of a second order cone (resp., a cylinder), when E is the intersection of an affine space and a second order cone (resp., a cylinder). We also provide a method for finding that cone, and hence the convex hull, for the continuous relaxation of the feasible set of a Mixed Integer Second Order Cone Optimization (MISOCO) problem, assumed to be the intersection of an ellipsoid with a general linear disjunction. This cone provides a new conic cut for MISOCO that can be used in branch-and-cut algorithms for MISOCO problems.

Appendices

Appendix 1: The Proofs of Lemmas 9 and 10

For the sake of simplifying the algebra of these proofs we use the following observation. If Q ≻ 0 and the quadric $\mathcal{Q}$ is not single point, $\mathcal{Q}$ can be transformed to a unit hypersphere $\{y \in \mathbb{R}^{n}\mid \left \Vert y\right \Vert ^{2} \leq 1\}$ using the affine transformation

$$\displaystyle{ y = \frac{Q^{1/2}(x + Q^{-1}q)} {\sqrt{\left \Vert u_{q } \right \Vert ^{2 } -\rho }}. }$$

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Observe that this transformation preserves the inertia of Q, hence the classification of the quadric is not changed. Additionally, observe that if we apply the same transformation to two parallel hyperplanes, the resulting hyperplanes are still parallel. Hence, throughout this proof, if Q ≻ 0, we assume w.l.o.g. that the quadric $\mathcal{Q}$ is a unit hypersphere centered at the origin. In this case, we have that the positive definite matrix Q of Section 4 is the identity matrix, the vector q is the zero vector, $\rho = -1$. Additionally, given Assumption 2 and the assumption that $\left \Vert a\right \Vert = \left \Vert b\right \Vert = 1$, we have that $\left \vert \alpha \right \vert \leq 1$, and $\left \vert \beta \right \vert \leq 1$. Finally, recall Assumption 1, then we may assume w.l.o.g. that α > β.

1.1 Proof of Lemma 9

From Section 4.1 we have that $\hat{\tau }= -1$, and the numerator of the right-hand side of (9) reduces to

$$\displaystyle\begin{array}{rcl} f(\tau ) =\tau ^{2}\frac{(\alpha -\beta )^{2}} {4} +\tau (1-\alpha \beta ) + 1.& & {}\\ \end{array}$$

Recall from Section 4.1.1 that the quadrics $\mathcal{Q}(\bar{\tau }_{1})$ and $\mathcal{Q}(\bar{\tau }_{2})$ in the family $\{\mathcal{Q}(\tau )\mid \tau \in \mathbb{R}\}$, are computed using the roots $\bar{\tau }_{1}$ and $\bar{\tau }_{2}$ of the function f(τ). Particularly, we have that

$$\displaystyle\begin{array}{rcl} \bar{\tau }_{1}& =& 2\left (\frac{\alpha \beta -1 -\sqrt{(1 -\alpha ^{2 } )(1 -\beta ^{2 } )}} {(\alpha -\beta )^{2}} \right ), {}\\ \bar{\tau }_{2}& =& 2\left (\frac{\alpha \beta -1 + \sqrt{(1 -\alpha ^{2 } )(1 -\beta ^{2 } )}} {(\alpha -\beta )^{2}} \right ), {}\\ \end{array}$$

where $\bar{\tau }_{1} \leq \bar{\tau }_{2}$. Note that if α = β, then f(τ) is a linear function. In this case we would have that $\mathcal{A}\cup \mathcal{B} = \mathbb{R}^{n}$ and it is easy to verify that $\mbox{ $\text{conv}\left (\mathcal{Q} \cap (\mathcal{A}\cup \mathcal{B})\right )$} = \mathcal{Q}$. However, recall that our assumption is β ≠ α. Hence, for the rest of this proof we assume w.l.o.g. that α > β, which results from Assumption 1.

The vertices of the cones $\mathcal{Q}(\bar{\tau }_{1})$ and $\mathcal{Q}(\bar{\tau }_{2})$ are $x(\bar{\tau }_{i}) = -Q(\bar{\tau }_{i})^{-1}q(\bar{\tau }_{i})$, i = 1, 2. We can express $x(\bar{\tau }_{i})$ in terms of a, α, and β as follows:

$$\displaystyle\begin{array}{rcl} x(\bar{\tau }_{i}) = -Q(\bar{\tau }_{i})^{-1}q(\bar{\tau }_{ i})& =& -\left (I - \frac{\bar{\tau }_{i}} {(1 +\bar{\tau } _{i})}aa^{\top }\right )\left (-\bar{\tau }_{ i}\frac{(\alpha +\beta )} {2} a\right ) {}\\ & =& \bar{\tau }_{i}\frac{(\alpha +\beta )} {2} \left (1 - \frac{\bar{\tau }_{i}} {(1 +\bar{\tau } _{i})}\right )a {}\\ & =& \bar{\tau }_{i} \frac{(\alpha +\beta )} {2(1 +\bar{\tau } _{i})}a. {}\\ \end{array}$$

Consider the inner product

$$\displaystyle{ a^{\top }x(\bar{\tau }_{ i}) = -a^{\top }Q(\bar{\tau }_{ i})^{-1}q(\bar{\tau }_{ i}) =\bar{\tau } _{i} \frac{(\alpha +\beta )} {2(1 +\bar{\tau } _{i})}a^{\top }a =\bar{\tau } _{ i} \frac{(\alpha +\beta )} {2(1 +\bar{\tau } _{i})}. }$$

Note that if $\alpha = -\beta$ then $a^{\top }x(\bar{\tau }_{i}) = 0$. Recall from Theorem 2 that in that case $Q(\bar{\tau }_{1})$ is a cylinder. For that reason, we assume that α ≠ −β for the rest of this proof.

Next, note that since $\mathcal{A}^{=}$ and $\mathcal{B}^{=}$ are parallel, then $\mathcal{A} \cap \mathcal{B} =\emptyset$. Then, we need to show that in the first and third cases of Theorem 2 the vertex $x(\bar{\tau }_{2})$ cannot be in the set $\overline{\mathcal{A}} \cap \overline{\mathcal{B}}$. Assume to the contrary that $x(\bar{\tau }_{2}) \in \overline{\mathcal{A}} \cap \overline{\mathcal{B}}$. Now, since we are analyzing the first and third cases of Theorem 2 we know that $\bar{\tau }_{2} <-1$. Thus, if $a^{\top }x(\bar{\tau }_{2}) <\alpha$ and $a^{\top }x(\bar{\tau }_{2})>\beta$, then

$$\displaystyle{ \bar{\tau }_{2}(\beta -\alpha )> 2\alpha \quad \text{ and }\quad \bar{\tau }_{2}(\alpha -\beta ) <2\beta. }$$

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Substituting $\bar{\tau }_{2}$ in (23) we obtain that $\sqrt{\frac{(1-\alpha ^{2 } )} {(1-\beta ^{2})}} = 1$. The last inequality is possible only if either $\alpha = -\beta$ or α = β. Hence, in the first and third cases of Theorem 2 the vertex $x(\bar{\tau }_{2})$ cannot be in the set $\overline{\mathcal{A}} \cap \overline{\mathcal{B}}$.

Thus, since the intersections $\mathcal{Q}(\bar{\tau }_{2}) \cap \mathcal{A}^{=}$ and $\mathcal{Q}(\bar{\tau }_{2}) \cap \mathcal{B}^{=}$ are bounded, then one of the following two cases holds:

• Case 1: $\mathcal{Q}^{+}(\bar{\tau }_{2}) \cap \mathcal{A}^{=} = \mathcal{E} \cap \mathcal{A}^{=}$ and $\mathcal{Q}^{+}(\bar{\tau }_{2}) \cap \mathcal{B}^{=} = \mathcal{E} \cap \mathcal{B}^{=}$;

• Case 2: $\mathcal{Q}^{-}(\bar{\tau }_{2}) \cap \mathcal{A}^{=} = \mathcal{E} \cap \mathcal{A}^{=}$ and $\mathcal{Q}^{-}(\bar{\tau }_{2}) \cap \mathcal{B}^{=} = \mathcal{E} \cap \mathcal{B}^{=}$.

Consequently, we have that one of the cones $\mathcal{Q}^{+}(\bar{\tau }_{2})$ and $\mathcal{Q}^{-}(\bar{\tau }_{2})$ found at the root $\bar{\tau }_{2}$ satisfy Proposition 1.

 □ 

1.2 Proof of Lemma 10

Recall from Section 4.2.1 that the quadrics $\mathcal{Q}(\bar{\tau }_{1})$ and $\mathcal{Q}(\bar{\tau }_{2})$ in the family $\{\mathcal{Q}(\tau )\mid \tau \in \mathbb{R}\}$ of Theorem 4 are computed using the roots of the function (15), which in this case simplifies to

$$\displaystyle\begin{array}{rcl} f(\tau ) = \left ((\alpha \beta -a^{\top }b)^{2} - (1 -\alpha ^{2})(1 -\beta ^{2})\right )\tau ^{2} + 4(a^{\top }b-\alpha \beta )\tau + 4.& & {}\\ \end{array}$$

The roots of f(τ) are

$$\displaystyle\begin{array}{rcl} \bar{\tau }_{1}& =& 2\left ( \frac{\alpha \beta -a^{\top }b -\sqrt{(1 -\alpha ^{2 } )(1 -\beta ^{2 } )}} {(\alpha \beta -a^{\top }b)^{2} - (1 -\alpha ^{2})(1 -\beta ^{2})}\right ) = \frac{2} {\alpha \beta -a^{\top }b + \sqrt{(1 -\alpha ^{2 } )(1 -\beta ^{2 } )}}, {}\\ \bar{\tau }_{2}& =& 2\left ( \frac{\alpha \beta -a^{\top }b + \sqrt{(1 -\alpha ^{2 } )(1 -\beta ^{2 } )}} {(\alpha \beta -a^{\top }b)^{2} - (1 -\alpha ^{2})(1 -\beta ^{2})}\right ) = \frac{2} {\alpha \beta -a^{\top }b -\sqrt{(1 -\alpha ^{2 } )(1 -\beta ^{2 } )}}, {}\\ \end{array}$$

where $\bar{\tau }_{1} \leq \bar{\tau }_{2}$.

Also, recall that the classification of the quadrics $\mathcal{Q}(\bar{\tau }_{1})$ and $\mathcal{Q}(\bar{\tau }_{2})$ is done based on the ratio f(τ)∕g(τ), where g(τ) simplifies in this case to

$$\displaystyle{ g(\tau ) = ((a^{\top }b)^{2} - 1)\tau ^{2} + 4a^{\top }b\tau + 4. }$$

Note that if $(a^{\top }b)^{2} - 1 = 0$, then we obtain that g is an affine function with a zero at − 1. However, since $\left \Vert a\right \Vert = \left \Vert b\right \Vert = 1$, in this case we either obtain that a ^⊤ b = cos(0), which implies that a = b, or we obtain that $a^{\top }b = -\cos (0)$, which implies that $a = -b$. This is the case when we have parallel hyperplanes, which was already analyzed in section “Proof of Lemma 9” in Appendix 1 and will not be considered in the rest of this proof. Now, the roots of g(τ) are

$$\displaystyle{ \hat{\tau }_{1} = - \frac{2} {a^{\top }b + 1} <0\qquad \text{and}\qquad \hat{\tau }_{2} = - \frac{2} {a^{\top }b - 1}> 0. }$$

The vertex of the cone $\mathcal{Q}(\bar{\tau }_{2})$ is $x(\bar{\tau }_{2}) = -Q(\bar{\tau }_{2})^{-1}q(\bar{\tau }_{2})$. We can express $x(\bar{\tau }_{2})$ in terms of a, b, α, and β as follows:

$$\displaystyle\begin{array}{rcl} x(\bar{\tau }_{2})& =& -Q(\bar{\tau }_{2})^{-1}q(\bar{\tau }_{ 2}) {}\\ & =& -\left (I -\frac{(aa^{\top } + bb^{\top })\bar{\tau }_{2}^{2} - (a^{\top }b\bar{\tau }_{2}^{2} + 2\bar{\tau }_{2})(ba^{\top } + ab^{\top })} {(1 - (a^{\top }b)^{2})\bar{\tau }_{2}^{2} - 4a^{\top }b\bar{\tau }_{2} - 4} \right )\left (-\bar{\tau }_{2}\frac{\beta a +\alpha b} {2} \right ) {}\\ & =& \frac{\bar{\tau }_{2}\left (\left ((\alpha -a^{\top }b\beta )\bar{\tau }_{2} - 2\beta \right )a + \left ((\beta -a^{\top }b\alpha )\bar{\tau }_{2} - 2\alpha \right )b\right )} {(1 - (a^{\top }b)^{2})\bar{\tau }_{2}^{2} - 4a^{\top }b\bar{\tau }_{2} - 4}. {}\\ \end{array}$$

Consider the inner products

$$\displaystyle\begin{array}{rcl} a^{\top }x(\bar{\tau }_{ 2}) = \frac{\bar{\tau }_{2}\left ((1 - (a^{\top }b)^{2})\alpha \bar{\tau }_{2} - 2(a^{\top }b\alpha +\beta )\right )} {(1 - (a^{\top }b)^{2})\bar{\tau }_{2}^{2} - 4a^{\top }b\bar{\tau }_{2} - 4} & & {}\\ \end{array}$$

and

$$\displaystyle\begin{array}{rcl} b^{\top }x(\bar{\tau }_{ 2}) = \frac{\bar{\tau }_{2}\left ((1 - (a^{\top }b)^{2})\beta \bar{\tau }_{2} - 2(a^{\top }b\beta +\alpha )\right )} {(1 - (a^{\top }b)^{2})\bar{\tau }_{2}^{2} - 4a^{\top }b\bar{\tau }_{2} - 4}.& & {}\\ \end{array}$$

Next, we show that in the first and fourth cases of Theorem 4 the vertex $x(\bar{\tau }_{2})$ cannot be in the set $\overline{\mathcal{A}} \cap \overline{\mathcal{B}}$. Assume to the contrary that $x(\bar{\tau }_{2}) \in \overline{\mathcal{A}} \cap \overline{\mathcal{B}}$. Note that $\hat{\tau }_{1}$ and $\hat{\tau }_{2}$ are the roots of $(1 - (a^{\top }b)^{2})\tau ^{2} - 4a^{\top }b\tau - 4 = -g(\tau )$. Now, since we are analyzing the first and fourth cases of Theorem 4 we know that $\hat{\tau }_{2} <\bar{\tau } _{1}$, or $\bar{\tau }_{2} <\hat{\tau } _{1}$, or $\bar{\tau }_{1} <\hat{\tau } _{1} <\hat{\tau } _{2} <\bar{\tau } _{2}$. Even more, since 1 − (a ^⊤ b)² ≥ 0 we have that $(1 - (a^{\top }b)^{2})\bar{\tau }_{2}^{2} - 4a^{\top }b\bar{\tau }_{2} - 4 \geq 0$. Thus, if $a^{\top }x(\bar{\tau }_{2}) <\alpha$ and $b^{\top }x(\bar{\tau }_{2})>\beta$, then

$$\displaystyle{ (a^{\top }b\alpha -\beta )\bar{\tau }_{ 2} <-2\alpha \quad \text{ and }\quad (a^{\top }b\beta -\alpha )\bar{\tau }_{ 2}> -2\beta. }$$

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Substituting $\bar{\tau }_{2}$ in (24) we obtain that $\frac{\alpha }{\sqrt{1-\alpha ^{2 }}} = - \frac{\beta }{\sqrt{1-\beta ^{2 }}}$, which implies that $\alpha = -\beta$. This is possible if $\bar{\tau }_{2} =\hat{\tau } _{1}$, which is not in the cases being considered. Hence, in the first and fourth cases of Theorem 4 $x(\bar{\tau }_{2})$ cannot be in the set $\overline{\mathcal{A}} \cap \overline{\mathcal{B}}$.

Similarly, we can show that in the first and fourth cases of Theorem 4 the vertex $x(\bar{\tau }_{2})$ cannot be in the set $\mathcal{A} \cap \mathcal{B}$. In particular, if $a^{\top }x(\bar{\tau }_{2})>\alpha$ and b ^⊤ $x(\bar{\tau }_{2}) <\beta$, then

$$\displaystyle{ (a^{\top }b\alpha -\beta )\bar{\tau }_{ 2}> -2\alpha \quad \text{ and }\quad (a^{\top }b\beta -\alpha )\bar{\tau }_{ 2} <-2\beta. }$$

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Substituting $\bar{\tau }_{2}$ in (25) we obtain that $\frac{\alpha }{\sqrt{1-\alpha ^{2 }}} = - \frac{\beta }{\sqrt{1-\beta ^{2 }}}$. This implies that $\bar{\tau }_{2} =\hat{\tau } _{1}$, which is not in the cases being considered. Hence, the vertex $x(\bar{\tau }_{2})$ cannot be in the set $\mathcal{A} \cap \mathcal{B}$.

Thus, since the intersections $\mathcal{Q}(\bar{\tau }_{2}) \cap \mathcal{A}^{=}$ and $\mathcal{Q}(\bar{\tau }_{2}) \cap \mathcal{B}^{=}$ are bounded, then one of the following two cases is true:

• Case 1: $\mathcal{Q}^{+}(\bar{\tau }_{2}) \cap \mathcal{A}^{=} = \mathcal{E} \cap \mathcal{A}^{=}$ and $\mathcal{Q}^{+}(\bar{\tau }_{2}) \cap \mathcal{B}^{=} = \mathcal{E} \cap \mathcal{B}^{=}$.

• Case 2: $\mathcal{Q}^{-}(\bar{\tau }_{2}) \cap \mathcal{A}^{=} = \mathcal{E} \cap \mathcal{A}^{=}$ and $\mathcal{Q}^{-}(\bar{\tau }_{2}) \cap \mathcal{B}^{=} = \mathcal{E} \cap \mathcal{B}^{=}$.

Consequently, we have that one of the cones $\mathcal{Q}^{+}(\bar{\tau }_{2})$, $\mathcal{Q}^{-}(\bar{\tau }_{2})$ found at the root $\bar{\tau }_{2}$ satisfies Proposition 1. □ 

Appendix 2: Additional Lemma

Lemma 11.

Let $\mathcal{C}\subset \mathbb{R}^{n}$ be a cylinder with a compact base. Then $\mathcal{C}$ is closed.

Proof.

Let $\mathcal{D}$ be a compact base for $\mathcal{C} =\{ x \in \mathbb{R}^{n}\vert x = d +\sigma d_{0},d \in \mathcal{D},\sigma \in \mathbb{R}\}$ and let $u \in \mathbb{R}^{n}$ be a vector such that $u\notin \mathcal{C}$. Our goal is to show that there is a neighborhood $\mathcal{U}$ of u such that $\mathcal{U} \cap \mathcal{C} =\emptyset$.

Let $\delta =\max \{ \left \Vert u - x\right \Vert \mid x \in \mathcal{D}\}> 0$ be the maximum distance from a point $x \in \mathcal{D}$ to u. Let us choose $\sigma _{o} = (\delta +1)/\left \Vert d_{0}\right \Vert$ and let $\mathcal{B}$ be the open ball of radius 1 centered at u. Define the set $\mathcal{C}_{1} =\{ x \in \mathbb{R}^{n}\mid x = d +\sigma d_{o},d \in \mathcal{D},\sigma \leq -\sigma _{o}\} \cup \{ x \in \mathbb{R}^{n}\mid x = d +\sigma d_{0},d \in \mathcal{D},\sigma \geq \sigma _{o}\}$. Then, we have that $\mathcal{B} \cap \mathcal{C}_{1} =\emptyset$.

Let $\mathcal{X} = \mathcal{D}\times [-\sigma _{o},\sigma _{o}]$, and consider the map $h: \mathcal{X}\mapsto \mathbb{R}^{n}$, defined by $h(\sigma,x) = x +\sigma d_{o}$. Since $\mathcal{D}$ and [−σ _o , σ _o ] are compact we have that $\mathcal{X}$ is compact. Since h is continuous in $\mathcal{X}$ we have that the image $h(\mathcal{X})$ is a compact set as well, and hence closed in $\mathbb{R}^{n}$. Furthermore, note that $h(\mathcal{X}) \subset \mathcal{C}$, thus $u\notin h(\mathcal{X})$. Hence, there is a neighborhood $\mathcal{N}$ of u such that $\mathcal{N} \cap h(\mathcal{X}) =\emptyset$. Let $\mathcal{U} = \mathcal{B} \cap \mathcal{N}$, then for any $\sigma \in \mathbb{R}$ we have that $\mathcal{U} \cap (\sigma d_{0} + \mathcal{D}) =\emptyset$. This proves that the complement of $\mathcal{C}$ is open, thus $\mathcal{C}$ is closed. □