Scenario reduction revisited: fundamental limits and guarantees

Abstract

The goal of scenario reduction is to approximate a given discrete distribution with another discrete distribution that has fewer atoms. We distinguish continuous scenario reduction, where the new atoms may be chosen freely, and discrete scenario reduction, where the new atoms must be chosen from among the existing ones. Using the Wasserstein distance as measure of proximity between distributions, we identify those n-point distributions on the unit ball that are least susceptible to scenario reduction, i.e., that have maximum Wasserstein distance to their closest m-point distributions for some prescribed $m<n$. We also provide sharp bounds on the added benefit of continuous over discrete scenario reduction. Finally, to our best knowledge, we propose the first polynomial-time constant-factor approximations for both discrete and continuous scenario reduction as well as the first exact exponential-time algorithms for continuous scenario reduction.

Appendix: Auxiliary results

The proof of Theorem 2 relies on the following two lemmas.

Lemma 1

The semidefinite program (6) admits an optimal solution $(\tau , \mathbf {S})$ with $\mathbf {S} = \alpha \mathbb {I} + \beta \varvec{11}^\top$ for some $\alpha , \beta \in \mathbb {R}$.

Proof

Let $(\tau , \mathbf {S}^\star )$ be any optimal solution to (6), which exists because (6) has a continuous objective function and a compact feasible set, and denote by $\mathfrak {S}$ the set of all permutations of I. For any $\sigma \in \mathfrak {S}$, the permuted solution $(\tau , \mathbf {S}^\sigma )$, with $s^\sigma _{ij} = s^\star _{\sigma (i)\sigma (j)}$ is also optimal in (6). Note first that $(\tau , \mathbf {S}^\sigma )$ is feasible in (6) because

$$\begin{aligned} \begin{aligned}&\tau \le \sum _{j \in J} \frac{1}{\vert I_j \vert ^2} \sum _{i \in I_j} \Bigg ( \vert I_j \vert ^2 s^\sigma _{ii} - 2\vert I_j \vert \sum _{k \in I_j} s^\sigma _{ik} + \sum _{k \in I_j} s^\sigma _{kk} + \sum _{\begin{array}{c} k,k^\prime \in I_j \\ k \ne k^\prime \end{array}} s^\sigma _{k k^\prime } \Bigg ) \\ \iff&\tau \le \sum _{j \in J} \frac{1}{\vert I^\sigma _j \vert ^2} \sum _{i \in I^\sigma _j} \Bigg ( \vert I^\sigma _j \vert ^2 s^\star _{ii} - 2\vert I^\sigma _j \vert \sum _{k \in I_j} s^\star _{ik} + \sum _{k \in I_j} s^\star _{kk} + \sum _{\begin{array}{c} k,k^\prime \in I_j \\ k \ne k^\prime \end{array}} s^\star _{k k^\prime } \Bigg ) , \end{aligned} \end{aligned}$$

where the index sets $I^\sigma _j=\{\sigma (i):~ i \in I_j \}$ for $j\in J$ form an m-set partition from within $\mathfrak {P}(I,m)$, and because $\mathbf {S}^\sigma \succeq \varvec{0}$ and $s^\sigma _{ii} = s^\star _{\sigma (i)\sigma (i)}\le 1$ for all $i\in I$ by construction. Moreover, it is clear that $(\tau , \mathbf {S}^\sigma )$ and $(\tau , \mathbf {S}^\star )$ share the same objective value in (6). Thus, $(\tau , \mathbf {S}^\sigma )$ is optimal in (6) for every $\sigma \in \mathfrak {S}$.

The convexity of problem (6) implies that $(\tau ,\mathbf {S})$ with $\mathbf {S}=\frac{1}{n!} \sum _{\sigma \in \mathfrak {S}} \mathbf {S}^\sigma$ is also optimal in (6). The claim follows by noting that $\mathbf {S}$ is invariant under permutations of the coordinates and thus representable as $\alpha \mathbb {I} + \beta \varvec{11}^\top$ for some $\alpha , \beta \in \mathbb {R}$. $\square$

Lemma 2

For $\alpha , \beta \in \mathbb {R}$ the eigenvalues of $\mathbf {S} = \alpha \mathbb {I} + \beta \varvec{11}^\top \in \mathbb {S}^n$ are given by $\alpha + n \beta$ (with multiplicity 1) and $\alpha$ (with multiplicity $n-1$).

Proof

Note that $\mathbf {S}$ is a circulant matrix, meaning that each of its rows coincides with the preceding row rotated by one element to the right. Thus, the eigenvalues of $\mathbf {S}$ are given by $\alpha + \beta (1 + \rho _j^1 + \ldots \rho _j^{n-1})$, $j=0,\ldots ,n-1$, where $\rho _j=e^{2\pi i j/n}$ and i denotes the imaginary unit; see e.g. Gray [12]. For $j=0$ we then obtain the eigenvalue $\alpha + n\beta$, and for $j= 1,\ldots , n-1$ we obtain the other $n-1$ eigenvalues, all of which equal $\alpha$ because $\sum _{k=0}^{n-1}e^{2\pi i jk/n}=(1-e^{2\pi i j})/(1-e^{2\pi i j/n})=0$. $\square$