On convex envelopes for bivariate functions over polytopes

Abstract

In this paper we discuss convex envelopes for bivariate functions, satisfying suitable assumptions, over polytopes. We first propose a technique to compute the value and a supporting hyperplane of the convex envelope over a general two-dimensional polytope through the solution of a three-dimensional convex subproblem with continuously differentiable constraint functions. Then, for quadratic functions as well as for some polynomial and rational ones, again satisfying suitable assumptions, we show how the same computations can be carried out through the solution of a single semidefinite problem.

Appendix: Example 2.1

Since the computations are based on elementary (though tedious) computations, we only give a brief sketch about how to solve subproblem (9) without deriving the solutions for subproblems (10) and (11).

The general structure of the subproblems to be solved is the following

$$\begin{aligned} \begin{array}{lll} \max&c&\\&c\le \eta _j(a,b)+a x_0+ b y_0&\quad j=1,\ldots ,r \\&(a,b)\in S,&\end{array} \end{aligned}$$

(23)

where $\eta _j,\,j=1,\ldots ,r$, are either affine or concave functions, and $S$ is some polyhedral region defined by the derivative conditions at the extremes of the convex edges. Then, in order to solve the subproblem, we can first split its feasible region into the following $r$ subregions

$$\begin{aligned} S_i=\{(a,b)\in S\ :\ \eta _i(a,b)\le \eta _j(a,b),\ j=1,\ldots ,r\}\ \ \ i=1,\ldots ,r; \end{aligned}$$

(24)

next, we solve the problem over each subregion $S_i$, which amounts at solving the following problem

$$\begin{aligned} \begin{array}{lll} \max&\eta _i(a,b)+ax_0+ by_0&\\&(a,b)\in S_i.&\end{array} \end{aligned}$$

Finally, we take the maximum of all the $r$ optimal values. The procedure will now be illustrated with subproblem (9).

First, we make the change of variable

$$\begin{aligned} z=-\left(a+\frac{1}{2}b\right), \end{aligned}$$

so that the subproblem can be rewritten as

$$\begin{aligned} \begin{array}{ll} \max&c \\&c\le -a+a x_0-2a y_0-2z y_0 \\&c\le -2a+a x_0-2a y_0-2z y_0 \\&c\le \frac{1}{2}+2\sqrt{z}+2a + 2z+a x_0-2a y_0 -2z y_0 \\&\frac{1}{4}\le z\le 1. \end{array} \end{aligned}$$

(25)

Note that with respect to the general structure (23) we have

$$\begin{aligned} S=\left\{ z\ :\ \frac{1}{4}\le z\le 1\right\} , \end{aligned}$$

and

$$\begin{aligned} \begin{array}{lll} \eta _1(a,z)&= -a \\ \eta _2(a,z)&= -2a\\ \eta _3(a,z)&= \frac{1}{2}+2\sqrt{z}+2a + 2z. \end{array} \end{aligned}$$

We will also use the following notation

$$\begin{aligned} l(a,z;x_0,y_0)=a x_0-2a y_0 -2z y_0 . \end{aligned}$$

The feasible region can now be split into the three sets $S_i,\,i=1,\ldots ,3$, defined in (24).

Subregion $S_1=\{\eta _2(a,z)\ge \eta _1(a,z),\ \eta _3(a,z)\ge \eta _1(a,z),\ z\in [1/4,1]\}$.

Note that

$$\begin{aligned} -a\le \frac{1}{2}+2\sqrt{z}+2a + 2z \Rightarrow a\ge -\frac{1}{3}\left(\frac{1}{2}+2\sqrt{z}+2z\right). \end{aligned}$$

The optimal value over $S_1$ is equal to

$$\begin{aligned} \max _{(a,z)\in S_1}\ \eta _1(a,z)+l(a,z;x_0,y_0)=a\left(x_0-2y_0-1\right)-2z y_0. \end{aligned}$$

If $x_0-2y_0-1\ge 0$ (i.e., over $T^{\prime }_1$), the optimal solution is $a^*=0, z^*=\frac{1}{4}$ with optimal value $-\frac{1}{2}y_0$. If $x_0-2y_0-1\le 0$, then $a^*=-\frac{1}{3}\left(\frac{1}{2}+2\sqrt{z}+2z\right)$, so that we are left with the following problem

$$\begin{aligned} \max _{z\in [1/4,1]} \left(-\frac{1}{6}-\frac{2}{3}\sqrt{z}-\frac{2}{3}z\right)(x_0-2y_0-1) - 2 z y_0. \end{aligned}$$

(26)

The optimal solution is

$$\begin{aligned} z^*= \left\{ \begin{array}{ll} \frac{1}{4}&(x_0,y_0)\in T^{\prime \prime }_1 \\ [2mm] \frac{1}{4}\left[\frac{1-x_0+2 y_0}{x_0+y_0-1}\right]^2&(x_0,y_0)\in T_2, \end{array} \right. \end{aligned}$$

with the optimal value

$$\begin{aligned} \left\{ \begin{array}{ll} \frac{2}{3}-\frac{2}{3}x_0+\frac{5}{6} y_0&(x_0,y_0)\in T^{\prime \prime }_1 \\ [2mm] \frac{y_0(1-x_0+2 y_0)}{2(x_0+y_0-1)}&(x_0,y_0)\in T_2. \end{array} \right. \end{aligned}$$

Subregion $S_2=\{\eta _2(a,z)\le \eta _1(a,z),\ \eta _2(a,z)\le \eta _3(a,z),\ z\in [1/4,1]\}$.

It can be easily checked that over this subregion $\eta _2(a,z)\ge \eta _1(a,z)$ implies $a\ge 0$. In this case the optimal value is equal to

$$\begin{aligned} \max _{(a,z)\in S_2}\ a\underbrace{(x_0-2y_0-2)}_{\le 0\ \text{ over}\ \ P }-2z y_0. \end{aligned}$$

Therefore, the optimal solution is $a^*=0, z^*=\frac{1}{4}$, with optimal value $-\frac{1}{2} y_0$.

Subregion $S_3=\{\eta _3(a,z)\le \eta _1(a,z),\ \eta _3(a,z)\le \eta _2(a,z)\ z\in [1/4,1]\}$.

We have

$$\begin{aligned} a\le -\frac{1}{3}\left(\frac{1}{2}+2 \sqrt{z}+2z\right). \end{aligned}$$

and we need to solve the problem

$$\begin{aligned} \max _{(a,z)\in S_3}\ a\underbrace{(2+x_0-2y_0)}_{\ge 0\ \text{ over}\ \ P }+\frac{1}{2}+2\sqrt{z}+2z(1-y_0). \end{aligned}$$

Therefore, $a^*=-\frac{1}{3}\left(\frac{1}{2}+2 \sqrt{z}+2z\right)$ and we are left with the problem (26), whose optimal solution has been already obtained (to be more precise, we notice that the optimal solution is $z^{*}=\frac{1}{4}$ also over $T^{\prime }_1$).

By combining the different optimal values, we can conclude that the optimal value for subproblem (9) is given by (12).