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.
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Notes
This can be checked in polynomial time (though cubic) with respect to the dimension of the matrix A.
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Appendix: Example 2.1
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
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
next, we solve the problem over each subregion \(S_i\), which amounts at solving the following problem
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
so that the subproblem can be rewritten as
Note that with respect to the general structure (23) we have
and
We will also use the following notation
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
The optimal value over \(S_1\) is equal to
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
The optimal solution is
with the optimal value
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
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
and we need to solve the problem
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).
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Locatelli, M., Schoen, F. On convex envelopes for bivariate functions over polytopes. Math. Program. 144, 65–91 (2014). https://doi.org/10.1007/s10107-012-0616-x
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DOI: https://doi.org/10.1007/s10107-012-0616-x