Duality for mixed-integer convex minimization

Abstract

We extend in two ways the standard Karush–Kuhn–Tucker optimality conditions to problems with a convex objective, convex functional constraints, and the extra requirement that some of the variables must be integral. While the standard Karush–Kuhn–Tucker conditions involve separating hyperplanes, our extension is based on mixed-integer-free polyhedra. Our optimality conditions allow us to define an exact dual of our original mixed-integer convex problem.

Appendix

We display in this appendix some elementary technical observations needed in the proof of Theorems 5 and 8. The notation B(x, R) designate the open Euclidean ball centered in x of radius $R>0$ and B[x, R] its closure. The relative interrior of a set S is written ${\text {ri}}(S)$.

Observation 13

We know by Doignon’s Theorem [4] that if a possibly infinite intersection of convex sets is integer-free and if one of these sets is bounded, then a sub-selection of at most $2^n$ of them is integer-free.

As in the proof of Theorem 5, we consider the sets $L_z$ defined in (7) and their intersection L. Since the d last components of $h_z$ are null, we can restrict our attention to the projection $\hat{L}_z$ of $L_z$ on its n first component and the intersection $\hat{L}$.

First, we show that $\hat{L}$ is bounded. Suppose otherwise and consider one of its rays $\ell :=\{z_0+td:t\ge 0\}$. Pick a few points $z_1,\ldots ,z_N\in \mathbb {Z}^n$ so that the interior of their convex hull contains $B[z_0,\sqrt{n}]$. Since the level set corresponding to $\max _{1\le i\le N}f(z_i)$ is bounded, there exists a ball $B[z_0,M]$ containing it. Further, there exists an integral point $\hat{z}\notin B[z_0,M]$ at a distance smaller than $\frac{\sqrt{n}}{2}$ from $\ell$. The half-space $\{y\in \mathbb {R}^n:h^\mathsf {T}y < h^\mathsf {T}\hat{z}\}$, with $h\in \partial f(\hat{z})$, contains L. Now, $\ell$ cannot be entirely in this half-space. This contradiction shows that $\hat{L}$ is bounded.

Let us take a closed ball B[0, R] containing $\hat{L}$. Consider the compact set $C:=B[0,2R]\backslash B(0,R)$. The sets $B(0,3R)\backslash {\text {cl}}(\hat{L}_z)$ are open and cover C. Hence, a finite number of them are enough to cover C, say $C\subseteq \cup _{i=1}^NB(0,3R)\backslash {\text {cl}}(\hat{L}_{z_i})$. Thus $\cap _{i=1}^N\hat{L}_{z_i}\subseteq B[0,R]$ by connectivity of this intersection.

Observe that the previous observation holds also when we replace some $L_z$ by their closure. The second observation follows closely the proof of the main result in [1].

Observation 14

Let $C\subseteq \mathbb {R}^{n+d}$ be a mixed-integer-free convex set. Assume that $C = {\text {int}}(L)\cap S$, where L and S are the bounded level sets of two convex functions f and g. (In the context of Theorem 8, g plays the role of $g_{\max }$) Suppose that:

$$\begin{aligned}&{\text {int}}(L) := \bigcap _{\begin{array}{c} z\in {\text {bd}}(L),\\ u \in \partial f(z) \end{array}}\{y\in \mathbb {R}^{n+d}:u^\mathsf {T}(y-z)<0\}\ne \emptyset \quad \text {and} \\&\quad S := \bigcap _{\begin{array}{c} z\in {\text {bd}}(S),\\ v \in \partial g(z) \end{array}}\{y\in \mathbb {R}^{n+d}:v^\mathsf {T}(y-z)\le 0\}. \end{aligned}$$

Denote by $\Pi$ the projection operator on the first n components and $\hat{C}:=\Pi (C)$. By continuity of $\Pi$, we know that $\Pi ({\text {cl}}(C)) = {\text {cl}}(\hat{C})$. Since ${\text {int}}(C)$ is nonempty, it is easy to verify that $\Pi ({\text {int}}(C)) = {\text {ri}}(\hat{C})$. Define for every $x\in {\text {bd}}(\hat{C}):={\text {cl}}(\hat{C})\backslash {\text {ri}}(\hat{C})$ the sets $C_x:=(\{x\}\times \mathbb {R}^d)\cap C$ and $\bar{C}_x:=(\{x\}\times \mathbb {R}^d)\cap {\text {cl}}(C)$ (Note that $\bar{C}_x$ is not necessarily the closure of $C_x$). For every $x\in {\text {bd}}(\hat{C})$, we know $C_x\subseteq {\text {bd}}(C)$. If $C_x$ is empty (and then $x\notin \hat{C}$), there exists $y\in S\cap {\text {bd}}(L)$ with $\Pi (y)=x$ (for otherwise $x\notin {\text {cl}}(\hat{C})$). If $C_x$ is not empty, then there exists $y\in C$ with $\Pi (y) = x$, $y\in {\text {int}}(L)$, and $y\in {\text {bd}}(S)$ (for otherwise $x\in {\text {ri}}(\hat{C})$).

Let us apply our previous observation on Doignon’s Theorem. We obtain $N\le 2^n$ inequalities, say the k first from L and the $N-k$ last from S whose intersection is integer-free, namely:

$$\begin{aligned} \hat{P} = \bigcap _{i=1}^k\{y\in \mathbb {R}^n:u_i^\mathsf {T}(y-x_i)< 0\} \cap \bigcap _{j=k+1}^N\{y\in \mathbb {R}^n:u_j^\mathsf {T}(y-x_j)\le 0\}. \end{aligned}$$

Note that some $x_i$ with $1\le i\le k$ might coincide with some $x_j$ with $k<j\le N$. Now, let $x\in \{x_1,\ldots ,x_N\}$. By what precedes, $x\notin \hat{P}$ iff $C_x = \emptyset$, in which case $x\in {\text {bd}}(L)$ and $x = x_i$ for some $1\le i\le k$. As in [1], for every $x\in {\text {bd}}(\hat{C})$ we define a point $y\in {\text {ri}}(\bar{C}_x)$ and the closed convex cone $N_x$ of affine functions from $\mathbb {R}^{n+d}$ to $\mathbb {R}$ vanishing on y and nonnegative on C. The extreme rays of $N_x$ are in $\partial f(y)$ if $y\in {\text {bd}}(L)$ and in $\partial g(y)$ if $y\in {\text {bd}}(S)$. Since $\hat{C}$ is of dimension n, the dimension of $N_x$ is not larger than $d+1$.

Now, $(u_i,0)^\mathsf {T}\in N_x$, so we can represent this vector as a combination of at most $d+1$ extreme vectors in $N_x$, say $b^{(1)},\ldots ,b^{(d+1)}$, with possibly repeated vectors. If $C_x$ is not empty, then $y\in {\text {int}}(L)$ because $y\in {\text {ri}}(C_x)$, so all these vectors belong to $\partial g(y)$, yielding inequalities of the type $\{z\in \mathbb {R}^{n+d}:(b^{(j)})^\mathsf {T}(z-y)\le 0\}$. If $C_x$ is empty, the corresponding inequalities are of the form $\{z\in \mathbb {R}^n:(b^{(j)})^\mathsf {T}(z-y)< 0\}$ or $\{z\in \mathbb {R}^n:(b^{(j)})^\mathsf {T}(z-y)\le 0\}$, with at least one of the first type, depending on $b^{(j)}\in \partial f(y)$ or $b^{(j)}\in \partial g(y)$. Note that y is not in the intersection of these half-spaces, and in fact, no point of $\{x\}\times \mathbb {R}^d$.

The mixed-integer-freeness of the resulting intersection is now clear.