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Outer-product-free sets for polynomial optimization and oracle-based cuts

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Abstract

This paper introduces cutting planes that involve minimal structural assumptions, enabling the generation of strong polyhedral relaxations for a broad class of problems. We consider valid inequalities for the set \(S\cap P\), where S is a closed set, and P is a polyhedron. Given an oracle that provides the distance from a point to S, we construct a pure cutting plane algorithm which is shown to converge if the initial relaxation is a polyhedron. These cuts are generated from convex forbidden zones, or S-free sets, derived from the oracle. We also consider the special case of polynomial optimization. Accordingly we develop a theory of outer-product-free sets, where S is the set of real, symmetric matrices of the form \(xx^T\). All maximal outer-product-free sets of full dimension are shown to be convex cones and we identify several families of such sets. These families are used to generate strengthened intersection cuts that can separate any infeasible extreme point of a linear programming relaxation efficiently. Computational experiments demonstrate the promise of our approach.

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Notes

  1. This BoxQP relaxation only adds the “diagonal” McCormick estimates \(X_{ii} \le x_i\).

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Acknowledgements

The authors thank Eli Towle for pointing out an error in the presentation of the intersection cut strengthening procedure, Felipe Serrano for useful comments and suggestions that led to Lemma 6, and to the anonymous reviewers whose thorough feedback greatly improved the article. The authors would also like to thank the Institute for Data Valorization (IVADO) for their support through the IVADO Postdoctoral Fellowship program.

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Authors and Affiliations

  1. IEOR, Columbia University, New York, NY, USA

    Daniel Bienstock

  2. ISE, The Ohio State University, Columbus, OH, USA

    Chen Chen

  3. Institute of Engineering Sciences, Universidad de O’Higgins, Rancagua, Chile

    Gonzalo Muñoz

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  1. Daniel Bienstock
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Appendices

Appendix

A radius of the conic hull of a ball

Suppose we have a ball of radius r and with centre that is distance \(m>r\) from the origin. We wish to determine the radius of the conic hull of the ball at a specific point along its axis. Consider a 2-dimensional cross-section of the conic hull of the ball containing the axis; this is shown in Fig. 4 in rectangular (x, y) coordinates. A line passing through the origin and tangent to the boundary of the ball in the nonnegative orthant may be written in the form \(y=ax\) for some \(a>0\); let \((\bar{r}, \bar{m})\) be the point of intersection between line and ball. At \((\bar{r}, \bar{m})\) we have

$$\begin{aligned} (a\bar{r}-m)^2+\bar{r}^2=r^2 \iff (1+a^2)\bar{r}^2-2am\bar{r} + m^2-r^2 = 0. \end{aligned}$$
(21)

Now Eq. (21) should only have one unique solution with respect to \(\bar{r}\) since the line is tangent to the ball; thus the discriminant must be zero,

$$\begin{aligned} 4a^2m^2-4(1+a^2)(m^2-r^2)=0 \implies a=\frac{\sqrt{m^2-r^2}}{r}. \end{aligned}$$
(22)

Solving Eq. (21) for \(\bar{r}\) with Eq. (22),

$$\begin{aligned} \bar{r}= & {} \frac{2am}{2(1+a^2)}, \\= & {} \frac{r}{m}\sqrt{m^2-r^2},\\ \bar{m}= & {} a\bar{r}, \\= & {} \frac{m^2-r^2}{m}. \end{aligned}$$

Hence at distance d from the origin along the axis of the cone, the radius of the cone is \(\frac{\bar{r}}{\bar{m}}d\), or

$$\begin{aligned} \frac{r}{\sqrt{m^2-r^2}}d. \end{aligned}$$
(23)
Fig. 4
figure 4

In grey, a ball with radius r and distance \(m>r\) from the origin. In red, the boundary of its conic hull. In black, an intersection point between the boundary of the ball and its conic hull (color figure online)

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B Additional BoxQP experiments

See Table 5.

Table 5 Comparison of intersection cuts and wRLT+SDP on larger BoxQP instances
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Bienstock, D., Chen, C. & Muñoz, G. Outer-product-free sets for polynomial optimization and oracle-based cuts. Math. Program. 183, 105–148 (2020). https://doi.org/10.1007/s10107-020-01484-3

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