A Game Theoretic Framework for Distributed Mission Slice Allocation and Management for Tactical Networks

Abstract

In this paper, we address the problem of implementing distributed slice allocation algorithms for tactical networks deployed in regions lacking a communication infrastructure. We propose a multi-agent game implementation in which Application Slice Agents (ASAs) compete for shared computational and bandwidth resources in the network. We prove that the game can be implemented as a two stage potential game, with the first stage ensuring the selection of computational servers with load balancing, and the second stage improving connectivity and network throughput by promoting inter-slice cooperation. Our proposed analysis framework also incorporates a theoretical model to capture the impact of mitigation methods for timing channel leaks for the shared resources for different slices, and proposes a theoretical framework that allows for a security-delay tradeoff analysis.

Appendices

Appendix A: Proof of Theorem 1

We construct the potential function as:

$$\begin{aligned} \Phi (a) = -\sum _{i =1}^NC^I_{ST_i}(a_i) - \eta * \sum _{i =1}^N \sum _{s \epsilon a_i} \sum _{t \epsilon c_{k_i}} D(s, c_{t}^{clt})- \omega *\sum _{i =1}^N\sum _{j \epsilon a_i}l_j(n_j^i(a)) \end{aligned}$$

(A.1)

We show that this is an exact potential function for the game.

Let a single slice agent i change its strategy from $a_i$ to $b_i$, the change in potential function $\bigtriangleup \Phi$ should exactly be equal to the change in the players’ utility $\bigtriangleup u_i$: i.e., $\bigtriangleup \Phi = \bigtriangleup U$.

$$\begin{aligned}&\bigtriangleup U^I_i = U^I_i(b_i, a_{-i}) -U^I_i(a_i, a_{-i}) \nonumber \\&\quad = (C^I_{ST_i}(a_i) - C^I_{ST_i}(b_i)) - \eta * \left(\sum _{s \epsilon b_i} \sum _{t \epsilon c_{k_i}} D(s, c_{t}^{clt}) - \sum _{s \epsilon a_i} \sum _{t \epsilon c_{k_i}} D(s, c_{t}^{clt})\right) \nonumber \\&\qquad -\omega *\left(\sum _{j\epsilon {b_i|a_{-i}}} l_j(n_j(a)) - \sum _{j \epsilon a_i |a_{-i}} l_j(n_j(a))\right) \end{aligned}$$

(A.2)

Let,

$$\begin{aligned} U_i^1&= (C^I_{ST_i}(a_i) - C^I_{ST_i}(b_i)) \nonumber \\&\quad + \eta * \left(\sum _{s \epsilon a_i} \sum _{t \epsilon c_{k_i}} D(s, c_{t}^{clt}) - \sum _{s \epsilon b_i} \sum _{t \epsilon c_{k_i}} D(s, c_{t}^{clt})\right) \end{aligned}$$

(A.3)

$$\begin{aligned} U_i^2&= \omega *\left(\sum _{j \epsilon a_i |a_{-i}}l_j(n_j(a)) - \sum _{j \epsilon b_i|a_{-i}}l_j(n_j(a))\right) \end{aligned}$$

(A.4)

Therefore, (A.2) can be written as,

$$\begin{aligned} \bigtriangleup U^I_i = U_i^1 + U_i^1 \end{aligned}$$

Then,

$$\begin{aligned}\Phi (a_i, a_{-i}) &= -\left(\sum _{v = 1, v \ne i}^NC^I_{ST_i}(a_v) + C^I_{ST_i}(a_i)\right) \nonumber \\&\quad - \eta *\left(\sum _{v = 1, v \ne i}^N \sum _{s \epsilon a_v} \sum _{t \epsilon c_{k_v}} D(s, c_{t}^{clt})\right) + \sum _{s \epsilon a_i} \sum _{t \epsilon c_{k_i}} D(s, c_{t}^{clt})) \nonumber \\&\quad - \omega *\sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(a_i, a_{-i}) \end{aligned}$$

(A.5)

$$\begin{aligned}\Phi (b_i, a_{-i})& = -\left(\sum _{v = 1, v \ne i}^NC^I_{ST_i}(a_v) + C^I_{ST_i}(b_i)\right) \nonumber \\&\quad - \eta *\left(\sum _{v = 1, v \ne i}^N \sum _{s \epsilon a_v} \sum _{t \epsilon c_{k_v}} D(s, c_{t}^{clt})\right) + \sum _{s \epsilon b_i} \sum _{t \epsilon c_{k_i}} D(s, c_{t}^{clt})) \nonumber \\&\quad - \omega *\sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(b_i, a_{-i}) \end{aligned}$$

(A.6)

Let,

$$\begin{aligned} Q(a_{-i}) = -\sum _{v = 1, v \ne i}^NC^I_{ST_i}(a_v) - \eta *\sum _{v = 1, v \ne i}^N \sum _{s \epsilon a_v} \sum _{t \epsilon c_{k_v}} D(s, c_{t}^{clt})) \end{aligned}$$

(A.7)

Then, (A.5) and (A.6) can be written as,

$$\begin{aligned} \Phi (a_i, a_{-i})&= Q(a_{-i}) -C^I_{ST_i}(a_i) - \eta *\sum _{s \epsilon a_i} \sum _{t \epsilon c_{k_i}} D(s, c_{t}^{clt})) \nonumber \\&\quad - \omega *\sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(a_i, a_{-i})) \end{aligned}$$

(A.8)

$$\begin{aligned} \Phi (b_i, a_{-i})&= Q(a_{-i})-C^I_{ST_i}(b_i) - \eta *\sum _{s \epsilon b_i} \sum _{t \epsilon c_{k_i}} D(s, c_{t}^{clt})) \nonumber \\&\quad - \omega *\sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(b_i, a_{-i})) \end{aligned}$$

(A.9)

Because $Q(a_{-i})$ is not influenced by the strategy changing of player i from $a_i$ to $b_i$, the equation of $\bigtriangleup \Phi$ can be written as,

$$\begin{aligned} \bigtriangleup \Phi&= \Phi (b_i, a_{-i}) - \Phi (a_i, a_{-i}) \nonumber \\&= Q(a_{-i}) - Q(a_{-i}) + C^I_{ST_i}(a_i) - C^I_{ST_i}(b_i) \nonumber \\&\quad + \eta *\sum _{s \epsilon a_i} \sum _{t \epsilon c_{k_i}} D(s, c_{t}^{clt}) -\eta *\sum _{s \epsilon b_i} \sum _{t \epsilon c_{k_i}} D(s, c_{t}^{clt}) \nonumber \\&\quad + \omega *\sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(a_i, a_{-i}))- \omega *\sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(b_i, a_{-i})) \nonumber \\&= C^I_{ST_i}(a_i) - C^I_{ST_i}(b_i) \nonumber \\&\quad + \eta *\left(\sum _{s \epsilon a_i} \sum _{t \epsilon c_{k_i}} D(s, c_{t}^{clt}\right) - \sum _{s \epsilon b_i} \sum _{t \epsilon c_{k_i}} D(s, c_{t}^{clt})) \nonumber \\&\quad + \omega *\left(\sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(a_i, a_{-i})) - \sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(b_i, a_{-i}))\right) \end{aligned}$$

(A.10)

Let,

$$\begin{aligned} T_1= & {} C^I_{ST_i}(a_i) - C^I_{ST_i}(b_i) + \eta *\left(\sum _{s \epsilon a_i} \sum _{t \epsilon c_{k_i}} D(s, c_{t}^{clt}) - \sum _{s \epsilon b_i} \sum _{t \epsilon c_{k_i}} D(s, c_{t}^{clt})\right) \end{aligned}$$

(A.11)

$$\begin{aligned} T_2= & {} + \omega *\left(\sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(a_i, a_{-i})) - \sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(b_i, a_{-i}))\right) \end{aligned}$$

(A.12)

Therefore, (A.10) can be written as,

$$\begin{aligned} \bigtriangleup \Phi = T_1 + T_2 \end{aligned}$$

(A.13)

According to (A.3) and (A.12), it is obvious that $U_i^1 = T_1$. Therefore, (7) holds if $U_i^2 = T_2$.

Considering,

$$\begin{aligned} T_2&= \omega *\left(\sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(a_i, a_{-i})) - \sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(b_i, a_{-i}))\right) \nonumber \\&= \omega *\left(\sum _{j \epsilon F}\sum _{k = 1}^{n_j(a_i, a_{-i})}l_j(k) - \sum _{j \epsilon F}\sum _{k = 1}^{n_j(b_i, a_{-i})}l_j(k)\right) \end{aligned}$$

(A.14)

where the second equality comes from exchanging two sums.

Let,

$$\begin{aligned} E(a_i, b_i) = \sum _{k = 1}^{n_j(a_i, a_{-i})}l_j(k) - \sum _{k = 1}^{n_j(b_i, a_{-i})}l_j(k) \end{aligned}$$

(A.15)

We can write (A.14) as,

$$\begin{aligned} T_2&= \omega *\left(\sum _{j \epsilon F}\sum _{k = 1}^{n_j(a_i, a_{-i})}l_j(k) - \sum _{j \epsilon F}\sum _{k = 1}^{n_j(b_i, a_{-i})}l_j(k)\right) \nonumber \\&= \omega * \sum _{j \epsilon F}\left(\sum _{k = 1}^{n_j(a_i, a_{-i})}l_j(k) - \sum _{k = 1}^{n_j(b_i, a_{-i})}l_j(k)\right) \end{aligned}$$

(A.16)

We enumerate and discuss all the cases to check if each $j \epsilon F$ in $a_i$ and $b_i$, then we can get:

• Case 1: If j is in both $a_i$ and $b_i$, then $n_j(b_i, a_{-i}) = n_j(a_i, a_{-i})$. We can get:

  $$\begin{aligned} E(a_i, b_i) = 0 \end{aligned}$$

  (A.17)

• Case 2: If j is in $a_i$ but not in $b_i$. then $n_j(a_i, a_{-i}) = n_j(b_i, a_{-i}) + 1$. We can get:

  $$\begin{aligned} E(a_i, b_i) = n_j(a_i, a_{-i}) \end{aligned}$$

  (A.18)

• Case 3: If j is in $b_i$ but not in $a_i$. then $n_j(b_i, a_{-i}) = n_j(a_i, a_{-i}) + 1$. We can get:

  $$\begin{aligned} E(a_i, b_i) = -n_j(b_i, a_{-i}) \end{aligned}$$

  (A.19)

Therefore, We can write (A.16) as,

$$\begin{aligned} T_2 = \omega *\left(\sum _{j \epsilon a_i |a_{-i}}l_j(n_j(a)) - \sum _{j \epsilon b_i |a_{-i}}l_j(n_j(a))\right) \end{aligned}$$

(A.20)

According to (A.4) and (A.20), we can see that $U_i^2 = T_2$. This proves that (7) holds for the game $\langle P, F, A, (u_i)_{i \epsilon P}^I \rangle$, and the game is an exact potential game.

Appendix B: Proof of Theorem 2

We construct the potential function as:

$$\begin{aligned} \Phi (a) = -\sum _{i \epsilon P}C^{II}_{ST_i}(a_i) - \delta * \sum _{i \epsilon P} \sum _{s \epsilon a_i} (1 / P_s) - \alpha *\sum _{i =1}^N\sum _{j \epsilon a_i}l_j(n_j^i(a)) \end{aligned}$$

(B.1)

We show that this is an exact potential function for the game.

Let a single slice agent i change its strategy from $a_i$ to $b_i$, the change in potential function $\bigtriangleup \Phi$ should exactly be equal to the change in the players’ utility $\bigtriangleup U_i$: i.e., $\bigtriangleup \Phi = \bigtriangleup U$.

$$\begin{aligned}&\bigtriangleup U^{II}_i = U^{II}_i(bi, a_{-i}) - U^{II}_i(ai, a_{-i}) \nonumber \\&\quad = (C^{II}_{ST_i}(a_i) - C^{II}_{ST_i}(b_i)) - \delta * \left(\sum _{s \epsilon b_i} (1 / P_s) - \sum _{s \epsilon a_i} (1 / P_s)\right) \nonumber \\&\qquad -\alpha *\left(\sum _{j \epsilon b_i |a_{-i}}l_j(n_j(a)) - \sum _{j \epsilon a_i |a_{-i}}l_j(n_j(a))\right) \end{aligned}$$

(B.2)

Let,

$$\begin{aligned} U_i^3= & {} (C^{II}_{ST_i}(a_i) - C^{II}_{ST_i}(b_i)) + \delta * \left(\sum _{s \epsilon a_i} (1 / P_s) - \sum _{s \epsilon b_i} (1 / P_s)\right) \end{aligned}$$

(B.3)

$$\begin{aligned} U_i^4= & {} \alpha *\left(\sum _{j \epsilon a_i |a_{-i}}l_j(n_j(a)) - \sum _{j \epsilon b_i |a_{-i}}l_j(n_j(a))\right) \end{aligned}$$

(B.4)

Therefore, (B.2) can be written as,

$$\begin{aligned} \bigtriangleup U^{II}_i = U_i^3 + U_i^4 \end{aligned}$$

(B.5)

Then,

$$\begin{aligned} \Phi (a_i, a_{-i})&= -\left(\sum _{v = 1, v \ne i}^NC^{II}_{ST_i}(a_v) + C^{II}_{ST_i}(a_i))- \delta *(\sum _{v = 1, v \ne i}^N \sum _{s \epsilon a_v} (1 / P_s)\right) \nonumber \\&\quad + \sum _{s \epsilon a_i} (1 / P_s))- \alpha *\sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(a_i, a_{-i})) \end{aligned}$$

(B.6)

$$\begin{aligned} \Phi (b_i, a_{-i})&= -\left(\sum _{v = 1, v \ne i}^NC^{II}_{ST_i}(a_v) + C^{II}_{ST_i}(b_i))- \delta *(\sum _{v = 1, v \ne i}^N \sum _{s \epsilon a_v} (1 / P_s)\right) \nonumber \\&\quad + \sum _{s \epsilon b_i} (1 / P_s)) - \alpha *\sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(b_i, a_{-i})) \end{aligned}$$

(B.7)

Let,

$$\begin{aligned} Q(a_{-i})^{II} = -\sum _{v = 1, v \ne i}^NC^{II}_{ST_i}(a_v)- \delta *\sum _{v = 1, v \ne i}^N \sum _{s \epsilon a_v} (1 / P_s) \end{aligned}$$

(B.8)

Then, (B.6) and (B.7) can be written as,

$$\begin{aligned} \Phi (a_i, a_{-i})&= Q(a_{-i})^{II} -C^{II}_{ST_i}(a_i) - \delta *\sum _{s \epsilon a_i} (1 / P_s) \nonumber \\&\quad - \alpha *\sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(a_i, a_{-i})) \end{aligned}$$

(B.9)

$$\begin{aligned} \Phi (b_i, a_{-i})&= Q(a_{-i})^{II} -C^{II}_{ST_i}(b_i) - \delta *\sum _{s \epsilon b_i} (1 / P_s) \nonumber \\&- \alpha *\sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(b_i, a_{-i})) \end{aligned}$$

(B.10)

Because $Q(a_{-i})^{II}$ is not influenced by the strategy changing of player i from $a_i$ to $b_i$, the equation of $\bigtriangleup \Phi$ can be written as,

$$\begin{aligned} \bigtriangleup \Phi&= \Phi (b_i, a_{-i}) - \Phi (a_i, a_{-i}) \nonumber \\&= Q(a_{-i}) - Q(a_{-i}) + C^{II}_{ST_i}(a_i) - C^{II}_{ST_i}(b_i) \nonumber \\&\quad + \delta *\sum _{s \epsilon a_i} (1 / P_s) -\delta *\sum _{s \epsilon b_i} (1 / P_s) \nonumber \\&\quad + \alpha *\sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(a_i, a_{-i}))- \alpha *\sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(b_i, a_{-i})) \nonumber \\&= C^{II}_{ST_i}(a_i) - C^{II}_{ST_i}(b_i) \nonumber \\&\quad + \delta *\left(\sum _{s \epsilon a_i} (1 / P_s) - \sum _{s \epsilon b_i} (1 / P_s)\right) \nonumber \\&\quad + \alpha *\left(\sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(a_i, a_{-i}))- \sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(b_i, a_{-i}))\right) \end{aligned}$$

(B.11)

Let,

$$\begin{aligned} T_3= & {} C^{II}_{ST_i}(a_i) - C^{II}_{ST_i}(b_i)+ \delta *\left(\sum _{s \epsilon a_i} (1 / P_s) - \sum _{s \epsilon b_i} (1 / P_s)\right) \end{aligned}$$

(B.12)

$$\begin{aligned} T_4= & {} \alpha *\left(\sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(a_i, a_{-i})) - \sum _{i = 1}^N\sum _{j \epsilon a_i}l_j(n_j(b_i, a_{-i}))\right) \end{aligned}$$

(B.13)

Therefore, (B.11) can be written as,

$$\begin{aligned} \bigtriangleup \Phi = T_3 + T_4 \end{aligned}$$

(B.14)

According to (B.3) and (B.12), it is obvious that $U_i^3 = T_3$. Therefore, (14) holds if $U_i^4 = T_4$.

Referring to the proof steps in Theorem 1, we follow the same steps as (A.14)-(A.20) and prove that $U_i^4 = T_4$.

Consequently, (14) holds for the game $\langle P, F, A, (u_i)_{i \epsilon P}^{II} \rangle$.

Appendix C: Distributed Algorithms