Piecewise static policies for two-stage adjustable robust linear optimization

Abstract

In this paper, we consider two-stage adjustable robust linear optimization problems under uncertain constraints and study the performance of piecewise static policies. These are a generalization of static policies where we divide the uncertainty set into several pieces and specify a static solution for each piece. We show that in the worst-case, there is no piecewise static policy with a polynomial number of pieces that has a significantly better performance than an optimal static policy. This is quite surprising as piecewise static policies are significantly more general than static policies. The proof is based on a combinatorial argument and the structure of piecewise static policies.

Appendix: Proof of Lemma 5

First, note that for $h >0$,

$$\begin{aligned} T(\mathcal{U},\varvec{h})=\left\{ \left( \frac{v_1}{h_1},\frac{v_2}{h_2},\ldots ,\frac{v_m}{h_m}\right) \; \Bigg \vert \; (v_1,v_2,\ldots ,v_m) \in T(\mathcal{U},\varvec{e}) \right\} . \end{aligned}$$

Then we can easily prove that $\kappa (T(\mathcal{U},\varvec{h})) = \kappa (T(\mathcal{U},\varvec{e})).$ In fact, let $\varvec{v} \in \mathsf{conv}(T(\mathcal{U},\varvec{h}))$. Then,

$\sum _{i=1}^m v_ih_i \varvec{e}_i \in \mathsf{conv} (T(\mathcal{U},\varvec{e}))$. Therefore,

$$\begin{aligned} \frac{1}{\kappa (T(\mathcal{U},\varvec{e}))}\cdot \left( \sum _{i=1}^m v_i h_i \varvec{e}_i \right) \in T(\mathcal{U},\varvec{e}). \end{aligned}$$

Then,

$$\begin{aligned} \frac{1}{\kappa T((\mathcal{U},\varvec{e}))} \cdot \varvec{v} \in T(\mathcal{U},\varvec{h}), \end{aligned}$$

which implies,

$$\begin{aligned} \mathsf{conv} (T(\mathcal{U},\varvec{h}) )\subseteq \kappa (T(\mathcal{U},\varvec{e}))\cdot T(\mathcal{U},\varvec{h}), \end{aligned}$$

and finally $\kappa (T(\mathcal{U},\varvec{h})) \le \kappa ( T(\mathcal{U},\varvec{e}))$. Similarly, we also have $\kappa (T(\mathcal{U},\varvec{h})) \ge \kappa (T(\mathcal{U},\varvec{e})) .$ Now, it’s sufficient to show that $\kappa (T(\mathcal{U},\varvec{e})) = \sum _{i=1}^m \tau _i .$ Let first show that

$$\begin{aligned} \mathsf{conv} (T(\mathcal{U},\varvec{e})) = \left\{ (v_1,v_2,\ldots ,v_m) \in [0,1]^m \; \Bigg \vert \; \sum _{i=1}^m \frac{v_i}{\tau _i}\le 1 \right\} . \end{aligned}$$

(5.1)

Let $\varvec{v} \in \mathsf{conv}(T(\mathcal{U},\varvec{e}))$. From Lemma 1, we have $\varvec{v} = \sum _{i=1}^m \lambda _i a_i \varvec{e}_i ,$ where $\sum _{i=1}^m \lambda _i =1$, ${\lambda }_i \in [0,1]$ and $0 \le a_i \le {\tau }_i, \quad \forall i \in [m]$. We have,

$$\begin{aligned} \sum _{i=1}^m \frac{v_i}{{\tau }_i} = \sum _{i=1}^m {\lambda }_i \cdot \frac{a_i}{{\tau }_i} \le \sum _{i=1}^m {\lambda }_i = 1. \end{aligned}$$

Conversely, let $\varvec{v} \in {\mathbb {R}}^m_+$ such that,

$$\begin{aligned} \sum _{i=1}^m \frac{v_i}{\tau _i}\le 1. \end{aligned}$$

We have

$$\begin{aligned} \varvec{v}= \sum _{j=1}^m \lambda _j a_j \varvec{e}_j, \end{aligned}$$

where for all $j \in [m]$,

$$\begin{aligned} \lambda _j = \frac{\frac{v_j}{\tau _j}}{\sum _{i=1}^m \frac{v_i}{\tau _i}} \qquad \text {and} \qquad a_{j}=\tau _j \sum _{i=1}^m \frac{v_i}{\tau _i}. \end{aligned}$$

We have $\sum _{j=1}^m \lambda _j =1$ and $a_j \le {\tau }_j \; \forall j \in [m]$. Then, $\varvec{v} \in \mathsf{conv}(T(\mathcal{U},\varvec{e}))$.

Now, we would like to find a lower bound for $\kappa (T(\mathcal{U},\varvec{e}))$. Let $\alpha \ge 1$ such that $\mathsf{conv}(T(\mathcal{U},\varvec{e})) \subseteq \alpha \cdot T(\mathcal{U},\varvec{e}).$ From (5.1), we have

$$\begin{aligned} \left( \frac{\tau _1^2}{\sum _{i=1}^m \tau _i}, \frac{\tau _2^2}{\sum _{i=1}^m \tau _i},\ldots ,\frac{\tau _m^2}{\sum _{i=1}^m \tau _i} \right) \in \mathsf{conv}(T(\mathcal{U},\varvec{e})) \end{aligned}$$

Then, there exists $\mathsf{diag}(\varvec{v} )\in \mathcal{U}$ and $\varvec{\mu } \in {\mathbb {R}}_+^m$, $\sum _{i=1}^m \mu _i =1$, such that

$$\begin{aligned} \left( \frac{\tau _1^2}{\sum _{i=1}^m \tau _i}, \frac{\tau _2^2}{\sum _{i=1}^m \tau _i},\ldots ,\frac{\tau _m^2}{\sum _{i=1}^m \tau _i} \right) = \alpha \cdot \mathsf{diag}(\varvec{v})^T \varvec{\mu } , \end{aligned}$$

i.e. $\forall 1 \le i \le m$,

$$\begin{aligned} \frac{\tau _i^2}{\sum _{j=1}^m \tau _j} = \alpha \mu _i v_i \end{aligned}$$

From Cauchy–Shwartz inequality we have,

$$\begin{aligned} \sum _{i=1}^m \frac{ \tau _i^2 }{\mu _i} \ge \left( \sum _{i=1}^m \tau _i \right) ^2, \end{aligned}$$

Then,

$$\begin{aligned} \alpha \left( \sum _{i=1}^m \tau _i\right) \left( \sum _{i=1}^m v_i\right) \ge \left( \sum _{i=1}^m \tau _i \right) ^2, \end{aligned}$$

i.e.

$$\begin{aligned} \alpha \left( \sum _{i=1}^m v_i\right) \ge \left( \sum _{i=1}^m \tau _i \right) , \end{aligned}$$

therefore,

$$\begin{aligned} \alpha \ge \sum _{i=1}^m \tau _i, \end{aligned}$$

where the last inequality follows from $\sum _{i=1}^m v_i \le 1.$ To finish our proof we show that,

$$\begin{aligned} \mathsf{conv}(T({ \mathcal U } ,\varvec{e})) \subseteq \left( \sum _{i=1}^m \tau _i\right) \cdot T({ \mathcal U } ,\varvec{e}). \end{aligned}$$

Let $\varvec{v} \in \mathsf{conv}(T({ \mathcal U } ,\varvec{e}))$, we have from (5.1),

$$\begin{aligned} \sum _{i=1}^m \frac{v_i}{\tau _i} \le 1 . \end{aligned}$$

For all $1 \le j \le m$, let define,

$$\begin{aligned} \mu _j = \frac{\frac{v_j}{\tau _j}}{\sum _{i=1}^m \frac{v_i}{\tau _i}} \qquad \text {and} \qquad b_j= \tau _j \frac{\sum _{i=1}^m \frac{v_i}{\tau _i}}{\sum _{i=1}^m \tau _i}. \end{aligned}$$

Then

$$\begin{aligned} \varvec{v} = \left( \sum _{i=1}^m \tau _i\right) \cdot \mathsf{diag}(\varvec{b})^T \varvec{\mu } \end{aligned}$$

We have $\forall j \in [m]$,

$$\begin{aligned} b_j \le \frac{\tau _j}{\sum _{i=1}^m \tau _i} \le \tau _j \end{aligned}$$

where the second inequality holds because $\sum _{i=1}^m \tau _i \ge 1$. Furthermore,

$$\begin{aligned} \sum _{j=1}^m b_j = \sum _{i=1}^m \frac{v_i}{\tau _i} \le 1. \end{aligned}$$

Therefore, $\mathsf{diag}(\varvec{b}) \in \mathcal{U}$. Since $\sum _{j=1}^n \mu _j =1$, $\mathsf{diag}(\varvec{b})^T \varvec{\mu }\in T({ \mathcal U } ,\varvec{e}).$ We conclude that

$$\begin{aligned} \varvec{v} \in \left( \sum _{i=1}^m \tau _i\right) \cdot T({ \mathcal U } ,\varvec{e}). \end{aligned}$$

$\square$