A copositive approach for two-stage adjustable robust optimization with uncertain right-hand sides

Abstract

We study two-stage adjustable robust linear programming in which the right-hand sides are uncertain and belong to a convex, compact uncertainty set. This problem is NP-hard, and the affine policy is a popular, tractable approximation. We prove that under standard and simple conditions, the two-stage problem can be reformulated as a copositive optimization problem, which in turn leads to a class of tractable, semidefinite-based approximations that are at least as strong as the affine policy. We investigate several examples from the literature demonstrating that our tractable approximations significantly improve the affine policy. In particular, our approach solves exactly in polynomial time a class of instances of increasing size for which the affine policy admits an arbitrarily large gap.

Appendix

1.1 Proof of Lemma 2

Proof

Any feasible $y(\xi )$ satisfies

$$\begin{aligned} y(\xi )_s&\ge \max \{\xi _s, 1 - \xi _s\} + y(\xi )_{s-1} \\&\ge \max \{\xi _s, 1 - \xi _s\} + \max \{\xi _{s-1}, 1 - \xi _{s-1}\} + y(\xi )_{s-2} \\&\ge \cdots \ge \sum _{i=1}^s \max \{\xi _i, 1 - \xi _i\} \end{aligned}$$

Hence, applying this inequality at an optimal $y(\cdot )$, it follows that

$$\begin{aligned} v_{{{\mathrm{RLP}}},2}^* \ge \max \limits _{\xi \in \Xi _2} \sum _{i=1}^s \max \{\xi _i, 1 - \xi _i\}. \end{aligned}$$

Under the change of variables $\mu := 2 \xi - \mathbbm {1}_s$, we have

$$\begin{aligned} v_{{{\mathrm{RLP}}},2}^*&\ge \max \limits _{\xi \in \Xi _2} \sum _{i=1}^s \max \{\xi _i, 1 - \xi _i\} = \max \limits _{\Vert \mu \Vert \le 1} \sum _{i=1}^s \tfrac{1}{2} \max \{1 + \mu _i, 1 - \mu _i\} \\&= \frac{1}{2} \max \limits _{\Vert \mu \Vert \le 1} \sum _{i=1}^s (1 + |\mu _i|) = \frac{1}{2} \left( s + \max \limits _{\Vert \mu \Vert \le 1} \Vert \mu \Vert _1 \right) = \tfrac{1}{2} (\sqrt{s} + s), \end{aligned}$$

where the last equality follows from the fact that the largest 1-norm over the Euclidean unit ball is $\sqrt{s}$. Moreover, one can check that the specific, sequentially defined mapping

$$\begin{aligned} y(\xi )_1&:= \max \{\xi _1, 1 - \xi _1\} \\ y(\xi )_2&:= \max \{\xi _2, 1 - \xi _2\} + y(\xi )_1 \\&\vdots \\ y(\xi )_s&:= \max \{\xi _s, 1 - \xi _s\} + y(\xi )_{s-1} \end{aligned}$$

is feasible with objective value $\tfrac{1}{2}(\sqrt{s} + s)$. So $v_{{{\mathrm{RLP}}}, 2}^* \le \tfrac{1}{2}(\sqrt{s} + s)$, and this completes the argument that $v_{{{\mathrm{RLP}}}, 2}^* = \tfrac{1}{2}(\sqrt{s} + s)$. $\square$

1.2 Proof of Proposition 6

The proof of Proposition 6 requires the following lemma.

Lemma 3

If a symmetric matrix V is positive semidefinite on the null space of the rectangular matrix E (that is, $z \in \mathrm {Null}(E) \Rightarrow z^T V z \ge 0$), then there exists $\mu > 0$ such that $\rho I + V + \mu E^TE \succ 0$.

Proof

We prove the contrapositive. Suppose $\rho I + V + \mu E^TE$ is not positive definite for all $\mu > 0$. In particular, there exists a sequence of vectors $\{ z_\ell \}$ such that

$$\begin{aligned} z_\ell ^T(\rho I + V + \ell E^TE) z_\ell \le 0, \ \ \Vert z_\ell \Vert = 1. \end{aligned}$$

Since $\{ z_\ell \}$ is bounded, there exists a limit point $\bar{z}$ such that

$$\begin{aligned} z_\ell ^T(\tfrac{1}{\ell } (\rho I + V) + E^TE)z_\ell \le 0 \ \ \Rightarrow \ \ \bar{z}^T E^T E \bar{z} = \Vert E z \Vert ^2 \le 0 \ \ \Leftrightarrow \ \ \bar{z} \in \mathrm {Null}(E). \end{aligned}$$

Furthermore,

$$\begin{aligned} z_\ell ^T(\rho I + V) z_\ell \le -\ell z_\ell ^TE^TEz_\ell = -\ell \Vert E z_\ell \Vert ^2 \le 0&\Rightarrow \ \ \bar{z}^T (\rho I + V) \bar{z} \le 0 \\&\Leftrightarrow \ \ \bar{z}^T V \bar{z} \le -\rho \Vert \bar{z}\Vert ^2 < 0. \end{aligned}$$

Thus, V is not positive semidefinite on $\mathrm {Null}(E)$. $\square$

Proof of Proposition 6

For fixed $\rho > 0$, let us construct the claimed feasible solution. Set

$$\begin{aligned} \lambda = v^*_{{{\mathrm{RLP}}},2} = \tfrac{1}{2}(\sqrt{s} + s), \quad \alpha = 0, \quad \tau = \tfrac{1}{4} \sqrt{s}, \quad S_{21} = 0, \end{aligned}$$

and

$$\begin{aligned} S_{22} = \frac{1}{2\sqrt{s}} \sum _{i=1}^{s} \left( f_{2i}f_{2i-1}^T + f_{2i-1}f_{2i}^T \right) \ge 0, \end{aligned}$$

where $f_{j}$ denotes the j-th standard basis vector in ${\mathbb {R}}^m = {\mathbb {R}}^{2s}$. Note that clearly $\alpha \in \widehat{\mathcal {U}}_2$ and $\mathrm{Rows}(S_{21}) \in \widehat{\mathcal {U}}_2$. Also forcing $v = \mu \mathbbm {1}_k$ for a single scalar variable $\mu$, where $\mathbbm {1}_k$ is the all-ones vector of size $k = s+1$, the feasibility constraints of (15) simplify further to

$$\begin{aligned} \rho I + \begin{pmatrix} \tfrac{1}{2} (s + \sqrt{s}) e_1e_1^T - \tfrac{1}{4} \sqrt{s} J &{}\quad -\tfrac{1}{2}F^T \\ -\tfrac{1}{2}F &{}\quad -S_{22} \end{pmatrix} + \mu E^TE \succeq 0, \end{aligned}$$

(19)

where $e_1 \in {\mathbb {R}}^k = {\mathbb {R}}^{s+1}$ is the first standard basis vector. For compactness, we write

$$\begin{aligned} V := \begin{pmatrix} \tfrac{1}{2} (s + \sqrt{s}) e_1e_1^T - \tfrac{1}{4} \sqrt{s} J &{}\quad -\tfrac{1}{2}F^T \\ -\tfrac{1}{2}F &{}\quad -S_{22} \end{pmatrix} \end{aligned}$$

(20)

so that (19) reads $\rho I + V + \mu E^TE \succeq 0$.

We next show that the matrix V is positive semidefinite on $\mathrm {Null}(E)$. Recall that $E \in {\mathbb {R}}^{n_2 \times (k+m)} = {\mathbb {R}}^{s \times (3s+1)}$. For notational convenience, we partition any $z \in {\mathbb {R}}^{k+m}$ into $z = {u \atopwithdelims ()w}$ with $u \in {\mathbb {R}}^k = {\mathbb {R}}^{s+1}$ and $w \in {\mathbb {R}}^m = {\mathbb {R}}^{2s}$. Then, from the definition of E, we have

$$\begin{aligned} z = {u \atopwithdelims ()w} \in \mathrm {Null}(E)&\Longleftrightarrow \ \ \left\{ \begin{array}{l} w_1 + w_2 = w_3 + w_4 \\ w_3 + w_4 = w_5 + w_6 \\ \vdots \\ w_{2s - 3} + w_{2s - 2} = w_{2s - 1} + w_{2s} \\ w_{2s - 1} + w_{2s} = u_1 \end{array} \right\} \\&\Longleftrightarrow \ \ w_{2i - 1} = u_1 - w_{2i} \ \ \forall \ i = 1, \ldots , s. \end{aligned}$$

So, taking into account the definition (20) of V,

$$\begin{aligned} 4 \, z^T Vz = 4 \, \textstyle {{u \atopwithdelims ()w}}^T V {u \atopwithdelims ()w} = u^T \left( 2(s + \sqrt{s}) e_1e_1^T - \sqrt{s} J \right) u - 4 \, w^T F u - 4 \, w^T S_{22} w, \end{aligned}$$

which breaks into the three summands, and we will simplify each one by one. First,

$$\begin{aligned} u^T \left( 2(s + \sqrt{s}) e_1e_1^T - \sqrt{s} J \right) u&= 2(s + \sqrt{s}) u_1^2 - \sqrt{s} u_1^2 + \sqrt{s} \sum _{j=2}^{s+1} u_j^2 \\&= 2 \, s \, u_1^2 + \sqrt{s} \, u^T u. \end{aligned}$$

Second,

$$\begin{aligned} -4w^T F u&= -4 \sum _{j = 1}^{2s} w_j [F u]_j = -4 \sum _{i = 1}^{s} \left( w_{2i - 1} [Fu]_{2i - 1} + w_{2i} [Fu]_{2i} \right) \\&= -2 \sum _{i = 1}^{s} \left( w_{2i - 1} (u_1 + u_{i+1}) + w_{2i} (u_1 - u_{i+1}) \right) \\&= -2 \sum _{i = 1}^{s} \left( (w_{2i - 1} + w_{2i}) u_1 + u_{i+1}(w_{2i-1} - w_{2i}) \right) \\&= -2 \sum _{i = 1}^{s} \left( u_1^2 + u_{i+1}(w_{2i-1} - w_{2i}) \right) \\&= -2 \, s \, u_1^2 - 2 \sum _{i = 1}^{s} u_{i+1}(w_{2i-1} - w_{2i}) \\&= -2 \, s \, u_1^2 + 2 \sum _{i = 1}^{s} u_{i+1}(w_{2i} - w_{2i-1})\\&= -2 \, s \, u_1^2 + 2 \sum _{i = 1}^{s} u_{i+1}(2 w_{2i} - u_1). \end{aligned}$$

Finally,

$$\begin{aligned} - 4 w^T S_{22} w&= -4 w^T \left( \frac{1}{2\sqrt{s}} \sum _{i=1}^{s} \left( f_{2i}f_{2i-1}^T + f_{2i-1}f_{2i}^T \right) \right) w \\&= -\frac{4}{\sqrt{s}} \sum _{i=1}^{s} w_{2i - 1} w_{2i} = -\frac{4}{\sqrt{s}} \sum _{i=1}^{s} (u_1 - w_{2i}) w_{2i}. \end{aligned}$$

Combining the three summands, we have as desired

$$\begin{aligned} 4 z^T Vz&= \left( 2s \, u_1^2 + \sqrt{s} \, u^T u \right) + \left( -2s \, u_1^2 + 2 \sum _{i = 1}^{s} u_{i+1}(2w_{2i} - u_1) \right) \\&\quad + \left( -\frac{4}{\sqrt{s}} \sum _{i=1}^{s} (u_1 - w_{2i}) w_{2i} \right) \\&= \sqrt{s} \, u^T u + 2 \sum _{i = 1}^{s} u_{i+1}(2 w_{2i} - u_1) -\frac{4}{\sqrt{s}} \sum _{i=1}^{s} (u_1 - w_{2i}) w_{2i} \\&= \sum _{i = 1}^{s} \left( \frac{1}{\sqrt{s}} \, u_1^2 + \sqrt{s} \, u_{i+1}^2 + 2 \, u_{i+1}(2 w_{2i} - u_1) -\frac{4}{\sqrt{s}} (u_1 - w_{2i}) w_{2i} \right) \\&= \sum _{i = 1}^{s} \left( \frac{1}{\sqrt{s}} \, u_1^2 - 2 \, u_1 \, u_{i+1} -\frac{4}{\sqrt{s}} \, u_1 \, w_{2i} + \sqrt{s} \, u_{i+1}^2 + 4 \, u_{i+1} \, w_{2i} + \frac{4}{\sqrt{s}} \, w_{2i}^2 \right) \\&= \sum _{i=1}^{s} \left( - (s)^{-1/4} \, u_1 + (s)^{1/4} \, u_{i+1} + 2(s)^{-1/4} \, w_{2i} \right) ^2 \\&\ge 0. \end{aligned}$$

Given that $\rho$, V, and E are defined as above. By Lemma 3, $\mu$ can be chosen so that (19) is indeed satisfied. $\square$

1.3 Proof of Proposition 7

Proof

By construction, Z is positive semidefinite, and one can argue in a straightforward manner that

$$\begin{aligned} Z_{11} = {{\mathrm{Diag}}}(1, \tfrac{1}{s}, \ldots , \tfrac{1}{s}), \quad Z_{22} = \frac{1}{4} \left( I + \mathbbm {1}_m\mathbbm {1}_m^T - \sum _{i=1}^{s}(f_{2i}f_{2i-1}^T + f_{2i-1}f_{2i}^T) \right) , \end{aligned}$$

and

$$\begin{aligned} Z_{21} = \frac{1}{2} \begin{pmatrix} 1 &{}\quad \tfrac{1}{\sqrt{s}} &{}\quad 0 &{}\quad \cdots &{}\quad 0 \\ 1 &{}\quad -\tfrac{1}{\sqrt{s}} &{}\quad 0 &{}\quad \cdots &{} \quad 0 \\ 1 &{} \quad 0 &{} \quad \tfrac{1}{\sqrt{s}} &{}\quad \cdots &{}\quad 0 \\ 1 &{} \quad 0 &{}\quad -\tfrac{1}{\sqrt{s}} &{}\quad \cdots &{} \quad 0 \\ \vdots &{} \quad \vdots &{} \quad \vdots &{}\quad \ddots &{} \quad \vdots \\ 1 &{} \quad 0 &{} \quad 0 &{}\quad \cdots &{} \quad \tfrac{1}{\sqrt{s}} \\ 1 &{} \quad 0 &{} \quad 0 &{}\quad \cdots &{}\quad -\tfrac{1}{\sqrt{s}} \end{pmatrix}. \end{aligned}$$

Then Z clearly satisfies $g_1g_1^T \bullet Z = 1$, $Z_{11}e_1 \in \widehat{\mathcal {U}}_2$, $J \bullet Z_{11} \ge 0$, $Z_{22} \ge 0$, and $\mathrm{Rows}(Z_{21}) \in \widehat{\mathcal {U}}_2$. Furthermore, the constraint $I \bullet Z \le r$ is easily satisfied for sufficiently large r. To check the constraint ${{\mathrm{diag}}}(E Z E^T) = 0$, it suffices to verify $EZ = 0$, which amounts to two equations. First,

$$\begin{aligned} 0 = E { 2e_1 \atopwithdelims ()\mathbbm {1}_m } = -2 \, d e_1^Te_1 + B^T \mathbbm {1}_m = -2d + 2d = 0, \end{aligned}$$

and second, for each $i=1, \ldots , s$,

$$\begin{aligned} 0= & {} E { \tfrac{2}{\sqrt{s}}e_{i+1} \atopwithdelims ()f_{2i-1} - f_{2i} } = - \frac{2}{\sqrt{s}} \, de_1^Te_{i+1} \\&\quad +\, B^T(f_{2i-1} - f_{2i}) = 0 + B^Tf_{2i-1} - B^Tf_{2i} = 0. \end{aligned}$$

So the proposed Z is feasible. Finally, it is clear that the corresponding objective value is $F \bullet Z_{21} = \tfrac{1}{2} (\sqrt{s} + s)$. So Z is indeed optimal. $\square$

1.4 Proof of Proposition 8

Proof

The inclusion $\Xi _1 \subseteq \Xi _2$ implies $\widehat{\mathcal {U}}_1 \subseteq \widehat{\mathcal {U}}_2$ and ${\mathrm {CPP}}(\widehat{\mathcal {U}}_1 \times {\mathbb {R}}_+^{2s}) \subseteq {\mathrm {CPP}}(\widehat{\mathcal {U}}_2 \times {\mathbb {R}}_+^{2s})$. Hence, ${\mathrm {COP}}(\widehat{\mathcal {U}}_1 \times {\mathbb {R}}_+^{2s}) \supseteq {\mathrm {COP}}(\widehat{\mathcal {U}}_2 \times {\mathbb {R}}_+^{2s})$. Moreover, it is not difficult to see that the construction of $\mathrm {IB}(\widehat{\mathcal {U}}_1 \times {\mathbb {R}}_+^{2s})$ introduced at the end of Sect. 4 for the polyhedral cone $\widehat{\mathcal {U}}_1$ satisfies $\mathrm {IB}(\widehat{\mathcal {U}}_1 \times {\mathbb {R}}_+^{2s}) \supseteq \mathrm {IB}(\widehat{\mathcal {U}}_2 \times {\mathbb {R}}_+^{2s})$. Thus, we conclude $v_{\mathrm {IB},1}^* \le v_{\mathrm {IB},2}^* = \tfrac{1}{2}(\sqrt{s} + s)$.

We finally show $v_{\mathrm {IB},1}^* \ge v_{\mathrm {IB},2}^*$. Based on the definition of $\widehat{\mathcal {U}}_1$ using the matrix P, similar to (16) the corresponding dual problem is

$$\begin{aligned} \begin{array}{lll} v_{\mathrm {IB},1}^* = &{} \max &{} F \bullet Z_{21} \\ &{} \text {s.t.}&{} {{\mathrm{diag}}}(E Z E^T) = 0, \ I \bullet Z \le r \\ &{} &{} PZ_{11}e_1 \ge 0, \ PZ_{11}P^T \ge 0, \ Z_{22} \ge 0, \ P Z_{21}^T \ge 0 \\ &{} &{} Z \succeq 0, \ g_1g_1^T \bullet Z = 1. \end{array} \end{aligned}$$

(21)

To complete the proof, we claim that the specific Z detailed in the previous subsection is also feasible for (21). It remains to show that $PZ_{11}e_1 \ge 0$, $PZ_{11}P^T \ge 0$, and $PZ_{21}^T \ge 0$.

Recall that $Z_{11} = {{\mathrm{Diag}}}(1, \tfrac{1}{s}, \ldots , \tfrac{1}{s})$ and every row of P has the form $(1, \pm 1, \ldots , \pm 1)$. Clearly, we have $PZ_{11}e_1 \ge 0$. Moreover, each entry of $PZ_{11}P^T$ can be expressed as ${1 \atopwithdelims ()\alpha }^T Z_{11} {1 \atopwithdelims ()\beta }$ for some $\alpha , \beta \in {\mathbb {R}}^s$ each of the form $(\pm 1, \ldots , \pm 1)$. We have

$$\begin{aligned} \textstyle {{1 \atopwithdelims ()\alpha }}^T Z_{11} {1 \atopwithdelims ()\beta } = 1 + \frac{1}{s} \cdot \alpha ^T \beta \ge 1 + \frac{1}{s} (-s) \ge 0. \end{aligned}$$

So indeed $PZ_{11}P^T \ge 0$. To check $PZ_{21}^T \ge 0$, recall also that every column of $Z_{21}^T$ has the form $\tfrac{1}{2} (e_1 \pm \tfrac{1}{\sqrt{s}} e_{i+1})$ for $i=1,\ldots ,s$, where $e_{\bullet }$ is a standard basis vector in ${\mathbb {R}}^k = {\mathbb {R}}^{s+1}$. Then each entry of $2 P Z_{21}^T$ can be expressed as

$$\begin{aligned} \textstyle {{1 \atopwithdelims ()\alpha }}^T e_1 \pm \tfrac{1}{\sqrt{s}} {1 \atopwithdelims ()\alpha }^T e_{i+1} \ge 1 - \tfrac{1}{\sqrt{s}} > 0. \end{aligned}$$

So $PZ_{21}^T \ge 0$, as desired. $\square$