Chance-constrained set covering with Wasserstein ambiguity

Abstract

We study a generalized distributionally robust chance-constrained set covering problem (DRC) with a Wasserstein ambiguity set, where both decisions and uncertainty are binary-valued. We establish the NP-hardness of DRC and recast it as a two-stage stochastic program, which facilitates decomposition algorithms. Furthermore, we derive two families of valid inequalities. The first family targets the hypograph of a “shifted” submodular function, which is associated with each scenario of the two-stage reformulation. We show that the valid inequalities give a complete description of the convex hull of the hypograph. The second family mixes inequalities across multiple scenarios and gains further strength via lifting. Our numerical experiments demonstrate the out-of-sample performance of the DRC model and the effectiveness of our proposed reformulation and valid inequalities.

Appendix A: Proof of Proposition 3

Proof

Consider a graph $\mathcal {G} := (\mathcal {V}, \mathcal {E})$ with vertex set $\mathcal {V}$ and edge set $\mathcal {E}$, on which the classic NP-hard vertex cover problem has the following binary linear formulation:

$$\begin{aligned} \mathop {\text {min}}~~&\sum _{u \in \mathcal {V}} x_u, \\ \mathop {\text {s.t.}}~~&x^{\top }\xi _{u, v} \ge 1, \forall (u, v) \in \mathcal {E}, \\&x_u \in \{0, 1\}, \forall u \in \mathcal {V}, \end{aligned}$$

(VC)

where binary variables $x_u$ indicate whether node $u \in \mathcal {V}$ is part of the vertex cover and $\xi _{u,v}$ is a binary vector with two nonzero entries: $\xi _{u,v} = e_u + e_v$ and $x^{\top }\xi _{u,v} = x_u + x_v$ for all $x \in \{0, 1\}^{|\mathcal {V}|}$. In particular, a vertex cover $x$ can cover every edge twice if and only if all nodes are in the cover, i.e. , $x = \mathbf {1}$. We provide a polynomial reduction from (VC) to the following instance of (8) to finish the proof:

$$\begin{aligned} \mathop {\text {min}}_{x \in \{0, 1\}^{|\mathcal {V}'|}} ~~&L(x) := \frac{1}{|\mathcal {V}'|} \sum _{u \in \mathcal {V}} x_u + \frac{1}{|\mathcal {V}'|} x_{w'} + x_{w} - \mathop {\text {min}}_{(u, v) \in \mathcal {E}'} \left( x^{\top } \xi _{u,v} \right) ^{1/p}, \end{aligned}$$

(29)

where we add two new nodes $w$ and $w'$ and augment the graph $\mathcal {G}$ to obtain $\mathcal {G}' := (\mathcal {V}', \mathcal {E}')$, $\mathcal {V}' := \mathcal {V} \cup \Big \{w, w'\Big \}$, and $\mathcal {E}' := \mathcal {E} \cup \Big \{(w, w')\Big \}$. Since the optimal value of (VC) is bounded above by $|\mathcal {V}|$, we are only interested in whether there is a vertex cover of size less than or equal to $K \in \mathbb {Z}_+, 1 \le K \le |\mathcal {V}| - 1$. On the one hand, suppose that there exists a vertex cover $x$ with size less than or equal to $K$, then together with $x_{w'} := 1$ and $x_w := 0$, $x^{\prime } := (x, x_{w'}, x_w)$ form a feasible solution to (29) with objective value less than or equal to $(K + 1) / |\mathcal {V}'| - 1$ because

$$\begin{aligned} L(x^{\prime }) = \frac{1}{|\mathcal {V}'|} \sum _{u \in \mathcal {V}} x_u + \frac{1}{|\mathcal {V}'|} x_{w'} + x_{w} - \mathop {\text {min}}_{(u, v) \in \mathcal {E}'} \left( x^{\top } \xi _{u,v}\right) ^{1/p} \le \frac{1}{|\mathcal {V}'|}(K + 1) - 1. \end{aligned}$$

On the other hand, suppose that there is a $y := (y_0, y_{w'}, y_{w}) \in \{0, 1\}^{|\mathcal {V}'|}$ such that $L(y) \le (K + 1) / |\mathcal {V}'| - 1$. We discuss the following three cases, with regard to the coverage number $C(y) := \mathop {\text {min}}_{(u, v) \in \mathcal {E}'} \left( y^{\top } \xi _{u, v}\right) ^{1/p}$, to show that there exists a vertex cover with size at most K.

• 1. $C(y) = 0$ is impossible because

  $$\begin{aligned} C(y) = \mathop {\text {min}}_{(u, v) \in \mathcal {E}'} \left( y^{\top }\xi _{u, v} \right) ^{1/p}&\ge \frac{1}{|\mathcal {V}'|} \sum _{u \in \mathcal {V}} y_u + \frac{1}{|\mathcal {V}'|} y_{w'} + y_{w} - \frac{1}{|\mathcal {V}'|} (K + 1) + 1 \\&\ge -\frac{1}{|\mathcal {V}'|} (K + 1) + 1 \ge 1 - \frac{|\mathcal {V}| + 1}{|\mathcal {V}| + 2} > 0. \end{aligned}$$

• 2. $C(y) = 1$: suppose that $y_w = 1$ and $y_{w'}$ equals zero or one. Then an alternative solution $y' := (y_0, y'_{w'}, y'_w)$ with $y'_{w'} = 1$ and $y'_w = 0$ to formulation (29) satisfies:

  $$\begin{aligned} C(y')&= \mathop {\text {min}}\Big \{ \mathop {\text {min}}_{(u, v) \in \mathcal {E}} \left( y_0^{\top }\xi _{u, v} \right) ^{1/p}, (y'_{w'} + y'_{w})^{1/p} \Big \} = 1 \quad \text {and} \\ L(y')&\le L(y) \le \frac{1}{|\mathcal {V}'|} (K + 1) - 1. \end{aligned}$$

  It follows that $y_0$ is a vertex cover of $\mathcal {G}$, and we may assume that $y_{w'} = 1, y_w = 0$ in formulation (29) without loss of optimality. Hence,

  $$\begin{aligned}&L(y) = \frac{1}{|\mathcal {V}'|} \sum _{u \in \mathcal {V}} (y_0)_u + \frac{1}{|\mathcal {V}'|} y_{w'} + y_{w} - C(y) = \frac{1}{|\mathcal {V}'|} \left( \sum _{u \in \mathcal {V}} (y_0)_u + 1 \right) - 1\\&\quad \le \frac{1}{|\mathcal {V}'|} (K + 1) - 1, \end{aligned}$$

  which implies that $y_0$ is a vertex cover of size at most $K$.

• 3. $C(y) = 2^{1/p}$ is also impossible. Indeed, since $C(y) = 2^{1/p}$, every edge of $\mathcal {G}'$ is covered twice, implying that $y_{w'} = w_{w} = (y_0)_u = 1$ for all $u \in \mathcal {V}$. Then,

  $$\begin{aligned} L(y) = \frac{1}{|\mathcal {V}'|} ( |\mathcal {V}| + 1) + 1 - 2^{1/p} \le \frac{1}{|\mathcal {V}'|} (K + 1) - 1 \le \frac{1}{|\mathcal {V}'|} |\mathcal {V}| - 1. \end{aligned}$$

  Simplifying the two ends of the above inequalities gives us $2 + |\mathcal {V}'|^{-1} \le 2^{1/p}$, which is impossible for $p \ge 1$. Therefore, $C(y)$ cannot be $2^{1/p}$.

To sum up, if there is a feasible solution y to formulation (29) with $L(y) \le (K + 1) / |\mathcal {V}'| - 1$, then there is a vertex cover of $\mathcal {G}$ with size at most $K$. $\square$