Gaining or losing perspective for convex multivariate functions on box domains

Abstract

Mixed-integer nonlinear optimization formulations of the disjunction between the origin and a polytope via a binary indicator variable is broadly used in nonlinear combinatorial optimization for modeling a fixed cost associated with carrying out a group of activities and a convex cost function associated with the levels of the activities. The perspective relaxation of such models is often used to solve to global optimality in a branch-and-bound context, but it typically requires suitable conic solvers and is not compatible with general-purpose NLP software in the presence of other classes of constraints. This motivates the investigation of when simpler but weaker relaxations may be adequate. Comparing the volume (i.e., Lebesgue measure) of the relaxations as a measure of tightness, we lift some of the results related to the simplex case to the box case. In order to compare the volumes of different relaxations in the box case, it is necessary to find an appropriate concave upper bound that preserves the convexity and is minimal, which is more difficult than in the simplex case. To address the challenge beyond the simplex case, the triangulation approach is used.

Appendix: Triangulation method for the integral of an exponential function on a hypercube

Let $\tilde{\textbf{c}}^\top :=\textbf{c}^\top \textbf{A}$. In this “Appendix”, we verify that the result obtained from the triangulation method to compute $\int _{[0,1]^n} e^{\tilde{\textbf{c}}^\top \textbf{y}} d\textbf{y}$ is the same as $\prod _{j=1}^n \frac{e^{\tilde{c}_j}-1}{\tilde{c}_j}$ in Proposition 3.8 under a simple genericity assumption: for any nonempty subset S of [n], $\sum _{j\in S}\tilde{c}_j\ne 0$. This assumption ensures that we can use the short formulae of Brion to compute over each simplex. With Kuhn’s triangulation, we obtain

$$\begin{aligned} \int _{[0,1]^n} e^{\tilde{\textbf{c}}^\top \textbf{y}} d\textbf{y}&= \sum _{(i_1,\dots ,i_n) \in \Omega }\int _{\Delta _{i_1,\dots ,i_n}} e^{\tilde{\textbf{c}}^\top \textbf{y}} d\textbf{y}\\&=\sum _{(i_1,\dots ,i_n)\in \Omega }\sum _{j=0}^n \frac{e^{\tilde{\textbf{c}}_{(i_1,\dots ,i_n)}^\top \textbf{w}_j}}{\prod _{k\ne j} \left( \tilde{\textbf{c}}_{(i_1,\dots ,i_n)}^\top (\textbf{w}_j-\textbf{w}_k)\right) }, \end{aligned}$$

where $\Delta _{i_1,\dots ,i_n}:=\{\textbf{x}: 0\le x_{i_1}\le \dots \le x_{i_n}\le 1\}$, $\tilde{\textbf{c}}_{(i_1,\dots ,i_n)}:=(\tilde{c}_{i_1},\dots ,\tilde{c}_{i_n})$, $\textbf{w}_0:=\textbf{0}$, $\textbf{w}_j:=\sum _{\ell =n+1-j}^n \textbf{e}_\ell$, and $\Omega$ is the set of all permutations of [n].

We are going to verify that

$$\begin{aligned} \sum _{(i_1,\dots ,i_n)\in \Omega }\sum _{j=0}^n \frac{e^{\tilde{\textbf{c}}_{(i_1,\dots ,i_n)}^\top \textbf{w}_j}}{\prod _{k\ne j}\left( \tilde{\textbf{c}}_{(i_1,\dots ,i_n)}^\top (\textbf{w}_j-\textbf{w}_k)\right) } = \prod _{j=1}^n \frac{e^{\tilde{c}_j}-1}{\tilde{c}_j}, \end{aligned}$$

by comparing the coefficient of the term $e^{\tilde{c}_{j_1}+\tilde{c}_{j_2}+\dots +\tilde{c}_{j_k}}$ on both sides. Because

$$\begin{aligned} \prod _{j=1}^n \frac{e^{\tilde{c}_j}-1}{\tilde{c}_j} = \frac{1}{\prod _{j=1}^n\tilde{c}_j}\sum _{k=0}^n (-1)^{n-k}\sum _{1\le j_1<j_2<\dots <j_k\le n}e^{\tilde{c}_{j_1}+\tilde{c}_{j_2}+\dots +\tilde{c}_{j_k}}, \end{aligned}$$

we know the coefficient on the right-hand side is $\frac{(-1)^{n-k}}{\prod _{j=1}^n\tilde{c}_j}$.

Before computing the coefficient on the left-hand side, we first prove that for $\ell := |T|$,

$$\begin{aligned} \sum _{(i_1,\dots ,i_{\ell })\in \Omega (T)} \frac{1}{\tilde{c}_{i_{1}}(\tilde{c}_{i_{1}}+ \tilde{c}_{i_{2}})\dots (\tilde{c}_{i_{1}}+\dots + \tilde{c}_{i_{\ell }})}=\frac{1}{\prod _{j\in T}\tilde{c}_j}, \end{aligned}$$

(3)

by induction on $\ell$. The result is trivial when $\ell =1$. Suppose the result holds for $\ell -1\ge 1$, then we can compute by the inductive hypothesis for each fixed $i_{\ell }$,

$$\begin{aligned}&\sum _{(i_1,\dots ,i_{\ell })\in \Omega (T)}\frac{1}{\tilde{c}_{i_{1}}(\tilde{c}_{i_{1}}+\tilde{c}_{i_{2}})\dots (\tilde{c}_{i_{1}}+\dots +\tilde{c}_{i_{\ell }})}\\&\quad = \sum _{j\in T} \frac{1}{\sum _{s\in T}\tilde{c}_s}\sum _{(i_1,\dots ,i_{\ell })\in \Omega (T),i_{\ell }=j}\frac{1}{\tilde{c}_{i_{1}}(\tilde{c}_{i_{1}}+\tilde{c}_{i_{2}})\dots (\tilde{c}_{i_{1}}+\dots +\tilde{c}_{i_{\ell -1}})}\\&\quad = \sum _{j\in T} \frac{1}{\sum _{s\in T}\tilde{c}_s} \frac{1}{\prod _{s\in T\setminus \{j\}}\tilde{c}_s}~\quad (\text {by the inductive hypothesis})\\&\quad = \frac{1}{\prod _{s\in T}\tilde{c}_s}. \end{aligned}$$

Thus, (3) holds.

Let $S:=\{j_1,j_2,\dots ,j_k\}$, $S^c:=[n]{\setminus } S$. The coefficient of the term $e^{\tilde{c}_{j_1}+\tilde{c}_{j_2}+\dots +\tilde{c}_{j_k}}$ on the left-hand side is

$$\begin{aligned}&\sum _{(i_1,\dots ,i_n)\in \Omega , (i_1,\dots ,i_{n-k})\in S^c}\frac{1}{\prod _{k\ne j}(\tilde{\textbf{c}}_{(i_1,\dots ,i_n)}^\top (\textbf{w}_j-\textbf{w}_k))}\\&\quad =\sum _{(i_1,\dots ,i_{n-k})\in \Omega (S_c)}\frac{1}{-\tilde{c}_{i_{n-k}}(-\tilde{c}_{i_{n-k-1}}-\tilde{c}_{i_{n-k}})\dots (-\tilde{c}_{i_{1}}-\dots -\tilde{c}_{i_{n-k}})}\\&\qquad \cdot \sum _{(i_{n-k+1,\dots ,i_{n}})\in \Omega (S)}\frac{1}{\tilde{c}_{i_{n-k+1}}(\tilde{c}_{i_{n-k+1}}+\tilde{c}_{i_{n-k+2}})\dots (\tilde{c}_{i_{n-k+1}}+\dots +\tilde{c}_{i_{n}})}\\&\quad =\frac{1}{(-1)^{n-k}\prod _{j\in S_c}\tilde{c}_j} \cdot \frac{1}{\prod _{j\in S}\tilde{c}_j}\\&\quad =\frac{(-1)^{n-k}}{\prod _{j\in [n]}\tilde{c}_j}, \end{aligned}$$

where $\Omega (T)$ is the set of all permutations of T, and the penultimate equation follows from (3). Therefore, we have demonstrated that the results from both methods are the same.