Extended formulations for convex envelopes

Abstract

In this work we derive explicit descriptions for the convex envelope of nonlinear functions that are component-wise concave on a subset of the variables and convex on the other variables. These functions account for more than 30 % of all nonlinearities in common benchmark libraries. To overcome the combinatorial difficulties in deriving the convex envelope description given by the component-wise concave part of the functions, we consider an extended formulation of the convex envelope based on the Reformulation–Linearization-Technique introduced by Sherali and Adams (SIAM J Discret Math 3(3):411–430, 1990). Computational results are reported showing that the extended formulation strategy is a useful tool in global optimization.

Appendix

The following elementary lemma shows that the facets of $\mathcal S _{\scriptscriptstyle {[l,u]}}^{\scriptscriptstyle \;(n)}$ and $\mathcal L _{\scriptscriptstyle {[l,u]}}^{\scriptscriptstyle \;(n_x,n_y)}$ are also facets of $\mathcal U _{f}$ and $\mathcal E _{f}$, respectively.

Lemma 2

Let $h,g_i:D\subseteq \mathbf R ^n\rightarrow \mathbf R ,\,i=1,\ldots ,m$, be continuous functions over a convex, compact domain $D\subseteq \mathbf R ^n$, and let $g:D\subseteq \mathbf R ^n\rightarrow \mathbf R ^m$ be the vector-valued function given by $g(x):=\big (g_1(x),\ldots ,g_m(x)\big )^\top$. Furthermore, consider the two convex sets $\mathcal{L }:={{\mathrm{conv}}}(\{(x,\zeta )\in \mathbf R ^{n+m}\mid \zeta = g(x),\; x\in D\})$ and $\mathcal{U }:= {{\mathrm{conv}}}(\{(x,\zeta ,\mu )\in \mathbf R ^{n+m+1}\mid \zeta = g(x),\;\mu \ge h(x),\; x\in D \})$. Then, each facet-defining inequality of $\mathcal L$ also induces a facet for $\mathcal{U }$.

Proof

Let $a^\top x + b^\top \zeta \le \gamma$ be an arbitrary facet-defining inequality for $\mathcal{L }$ with $a\in \mathbf R ^n, b\in \mathbf R ^m,$ and $\gamma \in \mathbf R$. Then, $a^\top x + b^\top \zeta \le \gamma$ is valid for $\mathcal{U }$. As $a^\top x + b^\top \zeta \le \gamma$ is facet-defining for $\mathcal L$, there are $n+m$ points $x^r\in D$ such that the points $\big (x^r,g(x^r)\big ),\,r=1,\ldots ,n+m$, are affinely independent and each point satisfies $a^\top x+ b^\top \zeta \le \gamma$ with equality. Now, consider the set of points $(x^r,\zeta ^r,\mu ^r):=\big (x^r,g(x^r),h(x^r)\big )\in \mathcal{U },\,r=1,\ldots ,n+m$, and the point $(x^{n+m+1},\zeta ^{n+m+1},\mu ^{n+m+1}):= \big (x^1, g(x^1),h(x^1)+1\big )$. Then, $a^\top x^r +b^\top \zeta ^r=\gamma$ holds for all $r=1,\ldots ,n+m+1$. Furthermore, the points $(x^r,\zeta ^r,\mu ^r),\,r=1,\ldots ,n+m+1$ are affinely independent since the set $\{(x^r,\zeta ^r)-(x^1,\zeta ^1)\mid r=2,\ldots , n+m+1\}$ is linearly independent. $\square$

The next lemma gives the key argument to derive the extended formulations $\mathcal U _{f}$ and $\mathcal E _{f}$ for functions $f$ belonging to Classes 1 and 2. It deals with a slight modification of the equation systems underlying the works [2, 27, 28] by Sherali and Adams to derive equivalent extended linear formulations for certain polynomial mixed-discrete programs. We adapt their proofs to our setting. For $V_1=\emptyset$, the statement of Lemma 3 follows from the fact that the convex hull of $\{F^{n_x+1}(v,y) \mid (v,y)\in {{\mathrm{vert}}}([l^x,u^x]\times [l^y,u^y])\}$ equals $\mathcal S _{\scriptscriptstyle {[l^x,u^x]\times [l^y,u^y]}}^{\scriptscriptstyle \;(n_x+1)}$. The special case when $V_2=\emptyset$ is discussed in Adams and Sherali [2].

Lemma 3

Let $[l,u]:=[l^x,u^x]\times [l^y,u^y]\subseteq \mathbf R ^{n_x}\times \mathbf R$ be a full-dimensional box, $N_x:=\{1,\ldots ,n_x\},\,V_x:={{\mathrm{vert}}}([l^x,u^x])$ and $n:=n_x+1$. Moreover, let $V_1,V_2\subseteq V_x$ be a partition of $V_x$, i.e., $V_x=V_1\cup V_2,\,V_1\cap V_2=\emptyset$. For a given $z\in \mathbf R ^{2^{n}-1}$, consider the following nonlinear system in the variables $\lambda _v,\,y^v$ with $v\in V_1$, and $\lambda _{v,l},\,\lambda _{v,u}$ with $v\in V_2$:

$$\begin{aligned} z_J&= \sum _{v\in V_{1}}{\lambda _v F^{\scriptscriptstyle {(n_x)}}_J(v)}+\sum _{v\in V_{2}}{\left( \lambda _{v,l}F^{\scriptscriptstyle {(n_x)}}_J(v)+\lambda _{v,u} F^{\scriptscriptstyle {(n_x)}}_J(v)\right) },\end{aligned}$$

(16)

$$\begin{aligned} z_{J\cup \{n\}}&= \sum _{v\in V_{1}}{\lambda _v y^v F^{\scriptscriptstyle {(n_x)}}_J(v)} +\sum _{v\in V_{2}}{\left( \lambda _{v,l} l^yF^{\scriptscriptstyle {(n_x)}}_J(v)+\lambda _{v,u} u^y F^{\scriptscriptstyle {(n_x)}}_J(v)\right) }, \end{aligned}$$

(17)

for all $J\subseteq N_x$. Its solution is given by

$$\begin{aligned} \lambda _{v}=\frac{e_{\hat{v}}(z^x)}{\prod _{j=1}^{n_x} (u_j-l_j)}, \quad y^v=\frac{\sum _{J\subseteq N_x} (-1)^{|J|+\alpha (\hat{v})}\; F^{\scriptscriptstyle {(n_x)}}_{N_x\setminus J}(\hat{v}) z_{J\cup \{n\}}}{e_{\hat{v}}(z^x)}, \end{aligned}$$

(18)

for $v\in V_1$, where $\hat{v}$ denotes the vector opposite to $v$ in $[l^x,u^x],\,z^x$ denotes the subvector of $z$-variables with entries $z_J,\,\emptyset \ne J\subseteq N_x$, and $e_{\hat{v}}(z^x)$ according to Eq. (6). The solution of $\lambda _{v,l}$ and $\lambda _{v,u}$ with $v\in V_{2}$ reads

$$\begin{aligned} \lambda _{v,l}=\frac{e_{(\hat{v},u^y)}(z)}{\prod _{j=1}^{n} (u_j-l_j)} \quad {\mathrm{and}}\quad \lambda _{v,u}=\frac{e_{(\hat{v},l^y)}(z)}{\prod _{j=1}^{n} (u_j-l_j)}. \end{aligned}$$

(19)

Proof

We prove Lemma 3 in two steps. Initially, we consider subsystem (I) defined by Eq. (16) and subsystem (II) defined by Eq. (17) for all $J\subseteq N_x$ independently. Afterwards we combine the solutions of the two subsystems.

Let $T$ be the matrix whose columns are given by the vectors $(1,F^{\scriptscriptstyle {(n_x)}}(v)),\,v\in V$. We can then bring both subsystems into the form $\zeta =T\xi$. This system has the unique solution (see [2, 17, 27])

$$\begin{aligned} \xi _v\;=\; \frac{e_{\hat{v}}(\zeta )}{\prod _{j=1}^{n_x}(u_j-l_j)},\qquad v\in V. \end{aligned}$$

For subsystem (I) we replace $\left( \lambda _{v,l} F^{\scriptscriptstyle {(n_x)}}_J(v)+\lambda _{v,u} F^{\scriptscriptstyle {(n_x)}}_J(v)\right)$ by $(\lambda _vF^{\scriptscriptstyle {(n_x)}}_J(v))$ in Eq. (16). Hence, we obtain the system $(1,z^x)=T\lambda$ with unique solution

$$\begin{aligned} \lambda _v\;=\; \frac{e_{\hat{v}}(z^x)}{\prod _{j=1}^{n_x}(u_j-l_j)},\quad v\in V. \end{aligned}$$

For subsystem (II) we first substitute

$$\begin{aligned} \left( \lambda _{v,l} l_nF^{\scriptscriptstyle {(n_x)}}_J(v)+\lambda _{v,u} u_n F^{\scriptscriptstyle {(n_x)}}_J(v)\right) {\mathrm{\quad by\quad }} \left( \lambda _{v}y^vF^{\scriptscriptstyle {(n_x)}}_J(v)\right) \end{aligned}$$

in Eq. (17) and afterwards, $\lambda _{v}y^v$ by $r_v$. With $\zeta _J=z_{J\cup \{n\}}$ for all $J\subseteq N_x$, subsystem (II) is of the form $\zeta =Tr$ with unique solution $r_v=e_{\hat{v}}(\zeta )/\prod _{j=1}^{n_x}(u_j-l_j),\,v\in V$.

Finally, we consider the original system, where $r_v=\lambda _vy^v,\,v\in V_1$, and $r_v=\lambda _{v,l}l^y+\lambda _{v,u}u^y,\,\lambda _v=\lambda _{v,l}+\lambda _{v,u},\,v\in V_2$. To obtain $y^v,\,v\in V_1$, we can solve $r_v=\lambda _vy^v$ for $y^v$ if $\lambda _v\ne 0$. Then, $y^v=r_v/\lambda _v=e_{\hat{v}}(\zeta )/e_{\hat{v}}(z^x)$. If $\lambda _v=0,\,y^v$ can take any value as its corresponding summand cancels out.

To derive $\lambda _{v,l}$ and $\lambda _{v,u},\,v\in V_{2}$, we solve the linear system $\lambda _v y^v=\lambda _{v,l} l^y+\lambda _{v,u} u^y$ and $\lambda _v=\lambda _{v,l}+\lambda _{v,u}$. Then, $\lambda _{v,l}=\lambda _v(u^y-y^v)/(u^y-l^y)$ and $\lambda _{v,u}=\lambda _v(y^v-l^y)/(u^y-l^y)$.

We prove the formula for $\lambda _{v,l}$ in Eq. (19). An analogous argument holds for $\lambda _{v,u}$. We get $\lambda _{v,l}=\lambda _v(u^y-y^v)/(u^y-l^y)=e_{\hat{v}}(z^x)(u^y-y^v)/\prod _{j\in N}(u_i-l_i)$. To deduce Eq. (19), it is thus sufficient to show that $e_{(\hat{v},u^y)}(z)=e_{\hat{v}}(z^x)(u^y- y^v)$. This follows because $e_{(\hat{v},u^y)}(z)$ can be rewritten as

$$\begin{aligned}&\sum _{J\subseteq N}{(-1)^{|J|+\alpha (\hat{v})} F^{\scriptscriptstyle {(n)}}_{N\setminus J}(\hat{v},u^y)z_J}\\&\quad =\sum _{J\subseteq N_x}{(-1)^{|J|+\alpha (\hat{v})} F^{\scriptscriptstyle {(n)}}_{N\setminus J}(\hat{v},u^y)z_J}\\&\quad \quad + \sum _{J=T\cup \{n\}:T\subseteq N_x}{(-1)^{|J|+\alpha (\hat{v})} F^{\scriptscriptstyle {(n)}}_{N\setminus J}(\hat{v})z_{J}}\\&\quad =\sum _{J\subseteq N_x}{(-1)^{|J|+\alpha (\hat{v})} u^yF^{\scriptscriptstyle {(n_x)}}_{N_x\setminus J}(\hat{v})z_J}+ \sum _{J\subseteq N_x}{(-1)^{|J|+\alpha (\hat{v})+1} F^{\scriptscriptstyle {(n_x)}}_{N_x\setminus J}(\hat{v})z_{J\cup \{n\}}}\\&\quad =\; u^y e_{\hat{v}}(z^x)-y^v e_{\hat{v}}(z^x)=e_{\hat{v}}(z^x)(u^y- y^v). \end{aligned}$$

This concludes the proof. $\square$