Budgeted Adversarial Network Resource Utilization Games

Abstract

This paper studies budgeted adversarial resource utilization game, where one of the player’s (designer) strategy is the utilization of resources while the other player’s (adversary) role is to police the resources for misuse. In this context, we consider routing games where a designer plans routes on a computer network and the adversary intercepts the routes on the network. Another example is in determining adversarial strategies to block access to travel or resources that may be considered to pose a risk to society, e.g. during a pandemic where the population (designer) goal may not be coincide with the societal goal of minimizing accessing a banned resource. We model this as a zero-sum game with constraints on the adversary or designer budgets. While zero-sum games can be solved using linear programs, we illustrate faster combinatorial methods to solve the problem. We first consider the resource access problem game on a bipartite graph where both the designer and the adversary have independent budget constraints and distinct costs and show a fast algorithm to determine a Nash equilibrium. We also consider the situation where the designer would strategize on paths in a general graph. In this application of determining network paths, where the adversary would attack edges in order to block the paths, we also discuss the case of multiple designers and, in particular show faster algorithms when there are 2 designers. These results utilize properties of minimum cuts in 2-commodity flow routing.

A Appendix

1.1 A.1 Proofs

Proof of Lemma 1

Proof

$$\begin{aligned} \begin{aligned} B_D[k]-B_D[k-1]=&\frac{CD(k)}{C(k)}-\frac{CD(k-1)}{C(k-1)} =\frac{C(k-1)\cdot CD(k) -C(k)\cdot CD(k-1)}{C(k)\cdot C(k-1)}\\ =\,&\frac{c[k]}{C(k)\cdot C(k-1)}\cdot (d[k]\cdot C(k-1)-CD(k-1))\\ \end{aligned} \end{aligned}$$

$$\begin{aligned} \begin{aligned} d[k]>d[i],\forall i<k\implies (d[k]\cdot C(k-1)-CD(k-1))>0\implies B_D[k]-B_D[k-1]>0 \end{aligned} \end{aligned}$$

Proof of Lemma 2

Proof

Denote the attacks used in calculating $B_A[t-1]$ and $B_A[t]$ by $X'$ and $X''$ respectively.

$$ \left\{ \begin{array}{l} \begin{aligned} X'[i]&{}=\frac{d[t-1]-d[i]}{d[t-1]-d[1]},&{}\forall &{}i=1\dots t-1\\ X''[i]&{}=\frac{d[t]-d[i]}{d[t]-d[1]},&{}\forall &{}i=1\dots t \end{aligned} \end{array}\right. \implies $$

$$\begin{aligned} \begin{aligned} X''[i]-X'[i]=&\frac{d[t]d[i]+d[t-1]d[1]-d[t-1]d[i]-d[t]d[1]}{(d[t]-d[1])(d[t-1]-d[1])}\\ =&\frac{(d[i]-d[1])(d[t]-d[t-1])}{(d[t]-d[1])(d[t-1]-d[1])}>0,\forall i=2\dots t-1 \end{aligned} \end{aligned}$$

Therefore

$$ \left\{ \begin{array}{l} \begin{aligned} &{}X'[1]=X''[1]=1&{}\\ &{}X'[t]=X''[t]=0&{}\\ &{}X'[i]<X''[i],\\ &{}\forall i=2\dots t-1 \end{aligned} \end{array}\right. \implies B_A[t-1]=\sum _{i=1}^m c[i]X'[i]<\sum _{i=1}^m c[i]X''[i]=B_A[t]$$

Proof of Lemma 3

Proof

• 1. $\gamma \ge 0$ The numerator: $B_D[\hat{k}]=\frac{CD(\hat{k})}{C(\hat{k})}\le \frac{d[\hat{k}]C(\hat{k})}{C(\hat{k})}=d[\hat{k}]$, $d[\hat{k}]\ge B_D[\hat{k}]\ge B_D \implies d[\hat{k}] - B_D \ge 0$ The denominator: $d[\hat{k}]\ge d[i],\forall i<\hat{k}\implies d[\hat{k}]C(k-1)-CD(\hat{k}-1)> d[\hat{k}]C(k-1)-d[\hat{k}]C(k-1)=0$

• 2. $0\le f^*[i]\le 1,\forall i$ $\forall i=1\dots \hat{k}-1,f^*[i]\ge 0$, since $f^*[i]=c[i]\gamma$. For $f^*[\hat{k}]$, the numerator: $B_D>B_D[\hat{k}-1]\implies B_D \cdot C(\hat{k}-1)> \frac{CD(\hat{k}-1)}{C(\hat{k}-1)}C(\hat{k}-1)\implies B_D \cdot C(\hat{k}-1)-CD(\hat{k}-1)>0$. The denominator $>0$ is proven above. Since $f^*[i]\ge 0$, $\sum _{i=1}^{\hat{k}}f^*[i]=1$, $f^*[i]\le 1,\forall i$. Therefore $0\le f^*[i]\le 1,\forall i$.

• 3. $f^*[i]\le c[i]\gamma ,\forall i$ $\forall i=1\dots \hat{k}-1$, $f^*[i]=c[i]\gamma$. $c[\hat{k}]\gamma -f^*[\hat{k}]= \frac{c[\hat{k}](d[\hat{k}]-B_D)-B_D\cdot C(\hat{k}-1)+CD(\hat{k}-1)}{c[\hat{k}]S(\hat{k}-1)-SQ(\hat{k}-1)}$ The numerator: $c[\hat{k}](d[\hat{k}]-B_D)-B_D\cdot C(\hat{k}-1)+CD(\hat{k}-1)= CD(\hat{k}) - B_D\cdot C(\hat{k}) \ge CD(\hat{k}) - B_D[\hat{k}]\cdot C(\hat{k}) =CD(\hat{k}) - \frac{CD(\hat{k})}{C(\hat{k})}C(\hat{k}) =0$. The denominator $>0$ is proven above. Thus $\forall i=1\dots \hat{k}$, $f^*[i]\le c[i]\gamma$.

Therefore $f^*,\gamma$ are feasible.

Proof of Lemma 4

Proof

• 1. $X_D,\varOmega \ge 0$ The numerators $>0$ obviously. The denominator: $d[\hat{k}]>d[i],\forall i<k\implies d[\hat{k}]C(\hat{k})-CD(\hat{k})>d[\hat{k}]C(\hat{k}) - d[\hat{k}]C(\hat{k})=0$

• 2. $0\le X^*[i]\le 1$ The numerator: $d[\hat{k}]>d[i],\forall i<k\implies (d[\hat{k}]-d[i])B_A>0$ The denominator $>0$ is proven above. Therefore $X^*[i]\ge 0$. Since $X^*[1]\ge X^*[i],\forall i=1\dots \hat{k}$, we need only to show $X^*[1]\le 1$.

  $$\begin{aligned} \begin{aligned} X^*[1]=&\frac{d[\hat{k}]-d[1]}{d[\hat{k}]C(\hat{k}) - CD(\hat{k})}B_A \le \frac{d[\hat{k}]-d[1]}{d[\hat{k}]C(\hat{k}) - CD(\hat{k})}B_A[\hat{k}]\\ =&\frac{d[\hat{k}]-d[1]}{d[\hat{k}]C(\hat{k}) - CD(\hat{k})}\cdot \frac{d[\hat{k}]C(\hat{k}) - CD(\hat{k})}{d[\hat{k}]-dc[1]} =1 \end{aligned} \end{aligned}$$

  Thus $0\le X^*[i]\le 1,\forall i=1\dots \hat{k}$.

Therefore $X^*,X_D,\varOmega$ are feasible.

Proof of Lemma 5

Proof

The complementary slackness conditions (ii)(iii)(v) are obviously satisfied due to the equations.

$\forall i=1\dots \hat{k}-1$, $c[i]\gamma =f[i]$, $\forall i=\hat{k}\dots m$, $X^*[i]=0$. Therefore condition (i) is satisfied.

$\forall i=1\dots \hat{k}$, $X_D-X^*[i]=d[i]\varOmega$, $\forall i=\hat{k}+1\dots m$, $f^*[i]=0$. Therefore condition (iv) is satisfied.

$\forall i=1\dots m$, $\delta [i]=0$. Therefore condition (vi) is satisfied.

All the complementary slackness conditions are met, the proposed solution is the optimal.

Proof of Lemma 6

Proof

• 1. $\gamma \ge 0$ The numerator: $B_D - d[1]\ge 0$ by assumption. The denominator: $CD(t)-d[1]C(t)=\sum _{i=1}^t c[i](d[i]-d[1])>0$.

• 2. $\delta [1]\ge 0$ The denominator$>0$ is proven above. Recall that $\hat{k}$ is the largest integer with $B_D(\hat{k}-1)<B_D$. Since $t\ge \hat{k}$, the numerator $CD(t)-B_D C(t)\ge CD(t)-B_D[t]C(t) =CD(t)-\frac{CD(t)}{C(t)}C(t)=0$.

• 3. $0\le f^*\le 1$ Since $\gamma ,\delta [1]\ge 0$, $f^*\ge 0$. Since $\sum _{i=1}^t f^*[i]=1$, $f^*\le 1$.

Therefore $f^*,\gamma ,\delta [1]$ are feasible.

Proof of Lemma 7

Proof

• 1. $\varOmega \ge 0$ The denominator: $c[i]>c[1],\forall i>1\implies CD(t)-d[1]C(t)>0$. The numerator: $C(t)-B_A>C(t)-B_A[t+1] = C(t)-\sum _{i=1}^{t}c[i]X[i]+0\cdot X[t+1]> C(t) - \sum _{i=1}^t c[i]\cdot 1= C(t) - C(t)=0$.

• 2. $X_D\ge 0$ $X_D-X^*[1]=d[1]\varOmega \implies X_D=1+d[1]\varOmega \ge 0$.

• 3. $0\le X^*[i]\le 1,\forall i=1\dots t$ Since $X_D-X^*[i]=d[i]\varOmega$, $X^*[1]>X^*[2]>\dots >X^*[t]$. $X^*[1]=1\implies X^*[i]\le 1$. To show $X^*[i]\ge 0$, it suffices to show $X^*[t]\ge 0$. The denominator $>0$ is proven above. The numerator: ${CD(t)+(d[t]-d[1])B_A-d[t]C(t)}\ge CD(t)+(d[t]-d[1])B_A[t]-d[t]C(t) =CD(t)+(d[t]-d[1])\frac{d[t]C(t)-CD(t)}{d[t]-d[1]}-d[t]C(t)=0$. Thus $0\le X^*[i]\le 1,\forall i=1\dots t$.

Therefore $X_D,\varOmega ,X^*$ are feasible.

Proof of Lemma 8

Proof

The complementary slackness conditions (ii)(iii)(v) are obviously satisfied due to the equations.

$\forall i=1\dots t$, $c[i]\gamma +\delta [i]=f^*[i]$, $\forall i=t+1\dots m$, $X^*[i]=0$. Therefore condition (i) is satisfied.

$\forall i=1\dots t$, $X_D-X^*[i]=d[i]\varOmega$, $\forall i=t+1\dots m$, $f^*[i]=0$. Therefore condition (iv) is satisfied.

$X^*[1]=1$, $\forall i=2\dots m$, $\delta [i]=0$. Therefore condition (vi) is satisfied.

All the complementary slackness conditions are met, the proposed solution is the optimal.

Proof of Lemma 9

Proof

• 1. $\gamma \ge 0$ The numerator: $B_D - d[1]\ge 0$ by assumption. The denominator: $CD(m)-d[1]C(m)=\sum _{i=1}^m c[i](d[i]-d[1])>0$.

• 2. $\delta [1]\ge 0$ The denominator $>0$ is proven above. The numerator: Recall that $\hat{k}$ is the largest integer with $B_D(\hat{k}-1)<B_D$. Since $m\ge \hat{k}$, the numerator $CD(m)-B_D\cdot C(m)\ge CD(m)-B_D[m]C(m) =CD(m)-\frac{CD(m)}{C(m)}C(m)=0$.

• 3. $0\le f^*\le 1$ Since $\gamma ,\delta [1]\ge 0$, $f^*\ge 0$. Since $\sum _{i=1}^t f^*[i]=1$, $f^*\le 1$.

Therefore $f^*,\gamma ,\delta [1]$ are feasible.

Proof of Lemma 10

Proof

• 1. $\varOmega \ge 0$ The numerator $S(m)-B_A\ge 0$ by assumption. The denominator $CD(m)-d[1]C(m)=\sum _{i=1}^m c[i](d[i]-d[1])>0$.

• 2. $X_D\ge 0$ $X_D-X^*[1]=c[1]\varOmega \implies X_D=1+c[1]\varOmega \ge 0$.

• 3. $0\le X^*[i]\le 1,\forall i=1\dots m$ Since $X_D-X^*[i]=d[i]\varOmega$, $X^*[1]>X^*[2]>\dots >X^*[m]$. $X^*[1]=1\implies X^*[i]\le 1$. To show $X^*[i]\ge 0$, it suffices to show $X^*[m]\ge 0$. The denominator $>0$ is proven above. The numerator:

  

  Thus $0\le X^*[i]\le 1,\forall i=1\dots m$.

Therefore $X_D,\varOmega ,X^*$ are feasible.

Proof of Lemma 11

Proof

The complementary slackness conditions (ii)(iii)(v) are obviously satisfied due to the equations.

$\forall i=1\dots m$, $c[i]\gamma +\delta [i]=f^*[i]$. Therefore condition (i) is satisfied.

$\forall i=1\dots m$, $X_D-X^*[i]=d[i]\varOmega$. Therefore condition (iv) is satisfied.

$X^*[1]=1$, $\forall i=2\dots m$, $\delta [i]=0$. Therefore condition (vi) is satisfied.

All the complementary slackness conditions are met, the proposed solution is the optimal.

Proof of Theorem 2

Proof

Since every possible path in the network has to go through $C_{min}$ at least once, $X_\mathcal {P}[j]\ge B/c^*,\ \forall j$. The attack above satisfies Property A1 with $\chi =B/c^*$, therefore the adversary’s gain is $B/c^*$, and the designer doesn’t want to change the flow design and use any path $p_j \notin \mathcal {P'}$.

Since neither the designer nor the adversary wants to move from the current strategy, this is a Nash equilibrium.

Proof of Lemma 15

Proof

Use the costs as capacities of edges. Since $\frac{c_{12}}{2}\le c_1\le c_2$, by Theorem 3 it is possible to compute a feasible 2-commodity flow $f^*$ with requirements $(\frac{c_{12}}{2},\frac{c_{12}}{2})$ in $O(n^3)$ steps [10]. This flow satisfies the following:

$$ \left\{ \begin{array}{l} \begin{aligned} &{}\forall e_i\in E,&{}f^*_e[i]\le c[i]\\ &{}\forall e_i\in C_{12},&{}f^*_e[i]=c[i] \end{aligned} \end{array}\right. $$

Scale $f^*$ down by $\frac{2}{c_{12}}$, and it satisfies PROPERTY A3 and each designer’s flow sums up to be 1.

Proof of Lemma 16

Proof

We construct the following dual variables feasible for LP12, satisfying complementary slackness conditions.

$$ \left\{ \begin{array}{l} \begin{aligned} \delta [i]&{}=0,\forall e_i\\ \alpha [j]&{}=0,\forall p_j\\ \beta [j]&{}=f^*[j],\forall p_j\\ \gamma &{}=2/c_{12} \end{aligned} \end{array}\right. $$

Since the adversary’s program does not distinguish the 2 designers, by substituting $C^*$ in the proof by $C_{12}$, the proof becomes identical to that of Lemma 14.

Proof of Lemma 17

Proof

Every possible path of each designer goes through $C_{12}$ at least once, therefore this attack satisfies PROPERTY A1 with $\chi =\frac{B}{c_{12}}$. By Lemma 12 the designers don’t want to change the flow designs. The adversary’s gain is $\frac{2B}{c_{12}}$.

Since neither the designers nor the adversary wants to move from the current strategy, this is a Nash equilibrium.

Proof of Lemma 18

Proof

Use the costs as capacities of edges. Since $c_1\le \frac{c_{12}}{2}$, by Theorem 3 it is possible to compute a feasible 2-commodity flow $f^*$ with requirements $(c_1,c_1)$ in $O(n^3)$ steps. This flow satisfies the following:

$$ \left\{ \begin{array}{l} \begin{aligned} &{}\forall e_i\in E,&{}f^*_e[i]\le c[i]\\ &{}\forall e_i\in C_1,&{}f^*_e[i]=c[i] \end{aligned} \end{array}\right. $$

Scale $f^*$ down by $\frac{1}{c_{1}}$, and it satisfies PROPERTY A4 and each designer’s flow sums up to be 1.

Proof of Lemma 19

Proof

We construct the following dual variables of LP12

$$ \left\{ \begin{array}{l} \begin{aligned} \delta [i]&{}=0,\forall e_i\\ \alpha [j]&{}=0,\forall p_j\\ \beta [j]&{}=f^*[j],\forall p_j\\ \gamma &{}=1/c_1 \end{aligned} \end{array}\right. $$

We proceed to show that all complementary slackness conditions of LP11, LP12 are satisfied.

Condition (i)(iv)(v) are clearly satisfied.

Condition (ii) is satisfied since

• 1. $\forall p_j\in \mathcal {P'}_1$, since every path goes through $C_1$ only once, $\sum _{i|e_i\in p_j} X[i] =B/c_1=X_\mathcal {P}[j]$.

• 2. $\forall p_j\in \mathcal {P'}_2$, since no edge or path of Designer 2 is attacked, $\sum _{i|e_i\in p_j} X[i]=X_\mathcal {P}[j]=0$.

Condition (iii) is satisfied since

$$ \begin{aligned}&\sum _{i=1}^m c[i]X[i]=\sum _{i|e_i\in C_1}c[i]X[i] =B/c_1\cdot \sum _{i|e_i\in C_1}c[i]=B/c_1\cdot c_1=B \end{aligned} $$

Condition (vi) is satisfied since

• 1. $\forall e_i\notin C_1$, $X[i]=0$.

• 2. $\forall e_i\in C_1$,

  $$ \begin{aligned}&\delta [i]+c[i]\gamma -\sum _{j|p_j\ni e_i}\beta [j] =c[i]\gamma -\sum _{j|p_j\ni e_i}\beta [j]=c[i]\gamma -f^*_e =c[i]\gamma -\frac{c[i]}{c_1} =0 \end{aligned} $$

Since all the conditions are satisfied, LP11, LP12 are at optimum with the given variables. The adversary is not incentivized to deviate.

Proof of Lemma 20

Proof

No path in $\mathcal {P}_2$ is attacked, so Designer 2 doesn’t want to deviate from the current flow design. Every possible path of Designer 1 goes through $C_1$ at least once, therefore the attack satisfies PROPERTY A1 with $\chi =\frac{B}{c_1}$ for Designer 1. The adversary’s gain is $\frac{B}{c_1}$, and the designers don’t want to change the flow designs.

Since neither the designers nor the adversary wants to change from the current strategy, this is a Nash equilibrium.