Modeling and mitigating the effects of supply chain disruption in a defender–attacker game

Abstract

The outcomes of a defender–attacker game depend on the defender’s resources delivered through military supply chains. These are subject to disruptions from various sources, such as natural disasters, social disasters, and terrorism. The attacker and defender are at war; the defender needs resources to defeat the attacker, but those resources may not be available due to a supply chain disruption that occurs exogenously to the game. In this paper, we integrate a defender–attacker game with supply chain risk management, and study the defender’s optimal preparation strategy. We provide analytical solutions, conduct numerical analysis, and compare the combined strategy with other protection strategies. Our results indicate that: (a) the defender benefits in a defender–attacker game by utilizing supply chain risk management tools; and (b) the attacker’s best response resource allocation would not be deterred by capacity backup protection and/or inventory protection. The feature of this paper is that the defender, being the downstream user of the supply chain, is involved in a strategic contest against the attacker. This model is different than game theory applied to private-sector supply chains because most game theoretic models of private sector supply chains usually explore relationships between suppliers and firms in the same supply chain or between multiple firms competing in the marketplace for customers. Therefore, supply chain risk management for such a military application imposes effects that have not been studied before.

Appendices

Appendices

In this section, we provide proofs for different protection strategies, including capacity backup protection, inventory protection, and combined protection, as well as dominated cases (Proposition 1).

1.1 Appendix 1: Proof of best response function for different protections

From the attacker’s utility function (Eq. 11), since the attacker’s resource allocation $R$ is a continuous variable, we could calculate the optimum of $R$ as a function of the defender’s resource allocation, $r$, decision on using capacity backup, $k$, and on using inventory, $I$. In particular, if $R\ge 0$, the first derivative should be zero:

$$\begin{aligned} \begin{aligned} \frac{dE[u_T]}{dR}&= \left( \frac{V(r+R)-RV}{(r+R)^2}-\gamma _T\right) \left[ E[X]+\int _l\int _d \min \left\{ \frac{I}{r},(1-k)l+kd\right\} f_D(d)f_L(l)\hbox {d}d\hbox {d}l\right. \\&\left. \quad +\,k(E[L]-E[D])\right] -\gamma _T\int _l\int _d \left[ (1-k)l+kd-\frac{I}{r}\right] ^{+}f_D(d)f_L(l)\hbox {d}d\hbox {d}l = 0\\ \end{aligned} \end{aligned}$$

We solve this equation and get:

$$\begin{aligned}&\hat{R}(r, k, I)\nonumber \\&\quad = {\left\{ \begin{array}{ll} G_{k,I}, &{}\hbox {if}\, r<\frac{V\cdot \left[ E[X]+k(E[L]-E[d])+\int _l\int _d\min \left\{ \frac{I}{r},(1-k)l+kd\right\} f_D(d)f_L(l)\hbox {d}d\hbox {d}l\right] }{\gamma _T\left[ E[X]+k(E[L]-E[d])+\int _l\int _d\min \left\{ \frac{I}{r},(1-k)l+kd\right\} f_D(d)f_L(l)\hbox {d}d\hbox {d}l+\int _l\int _d\left[ kd+(1-k)l-\frac{I}{r}\right] ^{+}f_D(d)f_L(l)\hbox {d}d\hbox {d}l\right] }\\ 0, &{}\hbox {if}\, r\ge \frac{V\cdot \left[ E[X]+k(E[L]-E[d])+\int _l\int _d\min \left\{ \frac{I}{r},(1-k)l+kd\right\} f_D(d)f_L(l)\hbox {d}d\hbox {d}l\right] }{\gamma _T\left[ E[X]+k(E[L]-E[d])+\int _l\int _d\min \left\{ \frac{I}{r},(1-k)l+kd\right\} f_D(d)f_L(l)\hbox {d}d\hbox {d}l+\int _l\int _d\left[ kd+(1-k)l-\frac{I}{r}\right] ^{+}f_D(d)f_L(l)\hbox {d}d\hbox {d}l\right] } \end{array}\right. } \end{aligned}$$

where $G_{k,I}=\sqrt{\frac{V\cdot r\cdot \left[ E[X]+k(E[L]-E[d])+\int _l\int _d\min \left\{ \frac{I}{r},(1-k)l+kd\right\} f_D(d)f_L(l)\hbox {d}d\hbox {d}l\right] }{\gamma _T\left[ E[X]+k(E[L]-E[d])+\int _l\int _d\min \left\{ \frac{I}{r},(1-k)l+kd\right\} f_D(d)f_L(l)\hbox {d}d\hbox {d}l+\int _l\int _d\left[ kd+(1-k)l-\frac{I}{r}\right] ^{+}f_D(d)f_L(l)\hbox {d}d\hbox {d}l\right] }}-r.$

The second order condition is satisfied, since

$$\begin{aligned}&\frac{d^2E[u_T]}{d^2R}\\&\quad =-\frac{2Vr\left[ E[X]+k(E[L]-E[d])+\int _l\int _d \min \left\{ \frac{I}{r},(1-k)l+kd\right\} f_D(d)f_L(l)\hbox {d}d\hbox {d}l\right] }{(r+R)^3}\le 0. \end{aligned}$$

Best response functions for no protection in Eq. (3), capacity backup protection in Eq. (6), and inventory protection in Eq. (9) are special cases of best response functions for combined protection in Eq. (12) when $I=k=0$, $I=0$, and $k=0$, respectively. Please see above for the proof.

1.2 Appendix 2: Proof of dominated cases as shown in Table 3

There are four cases as shown in the following:

• 1. When $k=1$ and $r=0$, from Eq. (10), the payoff function of the defender becomes:

  $$\begin{aligned} \max _{k=1, r=0, I\ge 0} E[u_G(k,r,R,I)] =-\left[ h\cdot I\cdot \int _{x} x\cdot f_X(x)\hbox {d}x+h\cdot \int _{l} \left[ I-rl\right] ^{+}f_L(l)\hbox {d}l\right] \end{aligned}$$

  When $I=0$, $E[u_G(k,r,R,I)]$ reaches a maximum at $0$ due to the first derivative $\frac{dE[u_G]}{dI}\le 0$. So, for the defender, case #4 is dominated by cases #6 and #8. Similarly, for the defender, case #7 is dominated by cases #6 and #8.

• 2. When $k=1$ and $r=0$, from Eq. (11), the payoff function of the attacker becomes:

  $$\begin{aligned} \max _{R\ge 0} E[u_T(k,r,R,I)]&= \left( V-\gamma _{T} R\right) \left( \int _x\int _l\int _d\left( x+min\left\{ \frac{I}{r},(1-k)l+kd\right\} +k(l-d)\right) \right. \\&\left. \times \,f_X(x)f_L(l)f_D(d)\hbox {d}d\hbox {d}l\hbox {d}x\right. \nonumber \\&\left. + \int _l\int _d \left[ kd+(1-k)l-\frac{I}{r}\right] ^{+} f_L(l)f_D(d)\hbox {d}d\hbox {d}l\right) \end{aligned}$$

  When $R=0$, $E[u_T(k,r,R,I)]$ reaches a maximum at $V\cdot \left( \int _{x}x\cdot f_{X}(x)\hbox {d}x +\int _l l\cdot f_L(l)\hbox {d}l\right)$ due to the first derivative $\frac{dE[u_T]}{dR}\le 0$. Therefore, for the attacker, case #6 is dominated by cases #7 and #8.

• 3. When $k=0$ and $r=0$, Eq. (10) implies:

  $$\begin{aligned} \max _{k=0, r=0, I\ge 0} E[u_G(k,r,R,I)] =-\left[ h\cdot I\cdot \int _{x} x\cdot f_X(x)\hbox {d}x+h\cdot \int _{l}\int _{z=0}^{l} \left[ I-rz\right] ^{+}f_L(l)\mathrm{d}z\hbox {d}l\right] \end{aligned}$$

  When $I=0$, $E[u_G(k,r,R,I)]$ reaches a maximum at $0$ due to the first derivative $\frac{dE[u_G]}{dI}\le 0$. So, for the defender, case #12 is dominated by cases #15 and #16. Similarly, case #15 is dominated by cases #14 and #16.

• 4. When $k=0$ and $r=0$, from Eq. (11), the payoff function of the attacker becomes:

  $$\begin{aligned}&\max _{R\ge 0} E[u_T(r,R,I)]\\&\quad = \left( V-\gamma _{T}\cdot R\right) \cdot \left[ \int _x\int _l\left( x+min\left\{ \frac{I}{r},l\right\} \right) f_X(x)f_L(l)\hbox {d}l\hbox {d}x + \int _l \left[ l-\frac{I}{r}\right] ^{+}f_L(l)\hbox {d}l \right] \end{aligned}$$

  When $R=0,\, E[u_T(r,R,I)]$ reaches a maximum at $\left[ \int _x\int _l\left( x+min\left\{ \frac{I}{r},l\right\} \right) f_X(x)f_L(l)\right.$ $\left. \hbox {d}l\hbox {d}x\right. \left. + \int _l \left[ l-\frac{I}{r}\right] ^{+}f_L(l)\hbox {d}l \right] \cdot V$ due to the first derivative $\frac{dE[u_T]}{dR}\le 0$. Therefore, for the attacker, case #14 is dominated by cases #15 and #16.

  In summary, there are nine possible cases: #1, #2, #3, #5, #9, #10, #11, #13 and #16, as shown in Table 3.