A Robust Optimization Approach to Designing Near-Optimal Strategies for Constant-Sum Monitoring Games

Abstract

We consider the problem of monitoring a set of targets, using scarce monitoring resources (e.g., sensors) that are subject to adversarial attacks. In particular, we propose a constant-sum Stackelberg game in which a defender (leader) chooses among possible monitoring locations, each covering a subset of targets, while taking into account the monitor failures induced by a resource-constrained attacker (follower). In contrast to the previous Stackelberg security models in which the defender uses mixed strategies, here, the defender must commit to pure strategies. This problem is highly intractable as both players’ strategy sets are exponentially large. Thus, we propose a solution methodology that automatically partitions the set of adversary’s strategies and maps each subset to a coverage policy. These policies are such that they do not overestimate the defender’s payoff. We show that the partitioning problem can be reformulated exactly as a mixed-integer linear program (MILP) of moderate size which can be solved with off-the-shelf solvers. We demonstrate the effectiveness of our proposed approach in various settings. In particular, we illustrate that even with few policies, we are able to closely approximate the optimal solution and outperform the heuristic solutions.

Appendices

A Exact Scenario-Based MILP

In Problem (5), the optimal pure strategy for the defender can be obtained from the solution of the following deterministic MILP problem which enumerates all the attacker pure strategies. This reformulation is exact, however, it requires a number of variables and constraints which is exponential in N. In this formulation $y_{\varvec{\xi }, n}$ is a binary variable and it is equal to 1 iff under attack scenario $\varvec{\xi }$, target n is covered.

(19)

Exact MILP Formulation of the K-Adaptability

The following reformulation is based on [11]. The objective function of the Problem (10) is identical to:

(20)

where $\varDelta _{K}(\varvec{l}) = \{ \varvec{\lambda }\in \mathbb {R}_{+}: \varvec{e}^\top \varvec{\lambda }= 1, \lambda _{k} = 0, \forall k\in \mathcal K : l_{k} \ne 0\}$. We define , and $\mathcal {L}_{+} := \{\varvec{l} \in \mathcal {L} > \varvec{0}\}$. Note that $\varDelta _{K}(\varvec{l}) = \emptyset$ if and only if $\varvec{l}>\varvec{0}$. If $\varXi _{\text {c}}(\varvec{x}, \varvec{y}, \varvec{l}) =\emptyset$ for all $\varvec{l} \in \mathcal {L}_{+}$, then the problem is equivalent to:

(21)

By applying the classical min-max theorem:

(22)

This problem is also equivalent to:

(23)

We note that if $\varXi _{\text {c}}(\varvec{x}, \varvec{y}, \varvec{l}) \ne \emptyset$, for some $\varvec{l} \in \mathcal {L}_{+}$ the objective of Problem (10) evaluates to $-\infty$. Using the epigraph form, Problem (10) is equivalent to:

$$\begin{aligned} \begin{array}{clll} \max \displaystyle &{}&{} \tau &{} \\ \text {s.t.} &{}&{} {\varvec{x}\in \mathcal {U}}, \varvec{y}^{k} \in \{0,1\}^{N}, k \in \mathcal {K} &{}\\ &{}&{}\tau \in \mathbb {R}, \; \varvec{\lambda }(\varvec{l}) \in \varDelta _{K}(\varvec{l}), \; \varvec{l} \in \partial \mathcal {L} &{} \\ &{}&{} \tau \le \displaystyle \sum _{k\in \mathcal {K}}\lambda _{k}(\varvec{l})\sum _{n\in \mathcal T}U_n y^{k}_{n} , &{}\; \forall \varvec{l} \in \partial \mathcal {L}, \; \varvec{\xi }\in \varXi _{\text {c}}(\varvec{x},\varvec{y},\varvec{l}) \\ &{}&{}\varXi _{\text {c}}(\varvec{x}, \varvec{y}, \varvec{l}) = \emptyset ,&{} \; \forall \varvec{l} \in \mathcal {L}_{+}. \end{array} \end{aligned}$$

(24)

The semi-infinite constraint associated with $\varvec{l} \in \partial \mathcal {L}$ is satisfied if and only if:

$$\begin{aligned} \begin{array}{cll} \min &{} \displaystyle \sum _{k\in \mathcal {K}}\lambda _{k}(\varvec{l})\sum _{n\in \mathcal T}U_n y^{k}_{n}\\ \text {s.t.} &{} {0 \le \xi _{n'} \le 1, \; \forall n'\in \mathcal N}\\ &{} \displaystyle \sum _{n'\in \mathcal N}\xi _{n'} \ge N - J \\ &{} \displaystyle y^{k}_{l_{k}} \ge \sum _{n' \in \delta (l_{k})}\xi _{n'}x_{n'} + 1, &{}\; \text {if } l_{k}>0 , \; \forall k \in \mathcal {K}\\ &{} \displaystyle y^{k}_{n} \le \sum _{n' \in \delta (n)} \xi _{n'}x_{n'}, \; \forall n\in \mathcal {T}, &{}\; \text {if } l_{k} = 0 , \; \forall k \in \mathcal {K} \\ \end{array} \end{aligned}$$

(25)

is greater than $\tau$.

In order to obtain the dual formulation, we introduce an auxiliary variable $\xi _{T+1} = 1$, and we rewrite the objective as: $(\sum _{k\in \mathcal {K}}\lambda _{k}(\varvec{l})\sum _{n\in \mathcal T}U_n y^{k}_{n})\;\xi _{T+1}$. Using strong linear programming duality:

$$\begin{aligned} \begin{array}{cll} \max &{} \displaystyle \sum _{n\in \mathcal {N}}-\alpha _{n}(\varvec{l}) + (N-J)\alpha _{N+1}(\varvec{l}) - \sum _{{\begin{matrix} k\in \mathcal {K}\\ l_{k} \ne 0 \end{matrix}}}(y^{k}_{l_{k}} -1)\nu _{k}(\varvec{l})+ \sum _{{\begin{matrix} k\in \mathcal {K} \\ l_{k} = 0 \end{matrix}} } \sum _{n\in \mathcal T}{y^{k}_{n}}\beta _{n}^{k}(\varvec{l}) + \alpha _{N+2}(\varvec{l}) &{}\\ \text {s.t.} &{} \alpha _{n}(\varvec{l}) \ge 0, n \in \{1,\ldots ,N+1\},\; \varvec{\beta }^{k}(\varvec{l}) \in \mathbb {R}^{N}_{+},\; \forall k \in \mathcal {K}, \; \varvec{\nu }(\varvec{l}) \in \mathbb {R}^{K}_{+} &{} \\ &{} - \alpha _{n}(\varvec{l}) + \alpha _{N+1}(\varvec{l}) - \displaystyle \sum _{{\begin{matrix} k\in \mathcal {K}\\ l_{k}\ne 0 \end{matrix}}}\sum _{n'\in \delta (l_{k})}{x_{n'}}\nu _{k}(\varvec{l}) + \displaystyle \sum _{{\begin{matrix} k\in \mathcal {K}\\ l_{k} = 0 \end{matrix}}}\sum _{n'\in \delta (n)}{x_{n'}}\beta _{n}^{k}(\varvec{l}) \le 0, \forall n \in \mathcal {T}, &{}\\ &{} \alpha _{N+2}(\varvec{l}) = \displaystyle \sum _{k\in \mathcal {K}}\lambda _{k}(\varvec{l})\sum _{n\in \mathcal T}U_n y^{k}_{n}. \end{array} \end{aligned}$$

(26)

Also, the last constraint in formulation (24) is satisfied if the following linear program is infeasible:

$$\begin{aligned} \begin{array}{clll} \min &{} \displaystyle 0 \\ \text {s.t.} &{} 0 \le \xi _{n} \le 1, &{} \;\forall n\in \mathcal N\\ &{} \displaystyle \sum _{n\in \mathcal N}\xi _{n} \ge N - J \\ &{} \displaystyle y^{k}_{l_{k}} \ge \sum _{n' \in \delta (l_{k})}\xi _{n'}x_{n'} + 1, \; &{} \forall k \in \mathcal {K},\; l_{k} \ne 0. \end{array} \end{aligned}$$

(27)

Using strong duality, this occurs if the dual problem is unbounded. Since the feasible region of the dual problem constitutes a cone, the dual problem is unbounded if and only if there is a feasible solution with an objective value of 1 or more. The dual problem is as below:

$$\begin{aligned} \begin{array}{cll} \max &{} \displaystyle \sum _{n\in \mathcal {N}}-\alpha _{n}(\varvec{l})+(N-J)\alpha _{N+1}(\varvec{l}) - \sum _{{\begin{matrix} k\in \mathcal {K}\\ l_{k} \ne 0 \end{matrix}}}(y^{k}_{l_{k}} - 1)\nu _{k}(\varvec{l})\\ \text {s.t.} &{} \varvec{\alpha }(\varvec{l}) \in \mathbb {R}_{+}^{N+1}, \; \varvec{\nu }(\varvec{l}) \in \mathbb {R}_{+}^{K}\\ &{} -\alpha _{n}(\varvec{l}) + \alpha _{N+1}(\varvec{l}) - \displaystyle \sum _{{\begin{matrix} k\in \mathcal {K}\\ l_{k} \ne 0 \end{matrix}}}\sum _{n'\in \delta (l_{k})}{x_{n'}}\nu _{k}(\varvec{l}) = 0 , \; \forall n \in \mathcal {N} \end{array} \end{aligned}$$

(28)