# Stochastic Game Analysis of Cooperation and Selfishness in a Random Access Mechanism
## 4 Problem Formulation
### Performance Evaluation
Thus, since the state of the Markov chain corresponds to the number of backlogged packets, then the average number of backlogged cooperative users can be derived by considering the current number of backlogged users \(m\) and taking the sum over all the possible states of the non-cooperative users which is given by  
\[m\cdot\left[\sum_{n=0}^{N}\Pi_{m,n}\right].\]  
Then, we can sum all the possible states of \(m\). Using a similar approach, we can derive the average number of backlogged nodes among non-cooperative users. **Corollary 1**. : _Combining Equation (16) with (26), and Equation (17) with (27), we can derive an expression for the average throughput as follows_  
\[TH_{c} =\frac{p_{c}}{T_{s}}(M-S_{c}), \tag{28}\] \[TH_{nc} =\frac{p_{nc}}{T_{s}}(N-S_{nc}). \tag{29}\]  
Proof. : We can obtain the two results of the corollary by developing Equations (16) and (17) as follows:  
\[TH_{c} =\frac{p_{c}}{T_{s}}\sum_{m=0}^{M}\left[M\sum_{n=0}^{N}\Pi_{m,n}- m\sum_{n=0}^{N}\Pi_{m,n}\right], \tag{30}\] \[=\frac{p_{c}}{T_{s}}\bigg{[}M\sum_{m=0}^{M}\sum_{n=0}^{N}\Pi_{m,n }-\sum_{m=0}^{M}m\sum_{n=0}^{N}\Pi_{m,n}\bigg{]},\] (31) \[=\frac{p_{c}}{T_{s}}(M-S_{C}), \tag{32}\]  
and  
\[TH_{nc} =\frac{p_{nc}}{T_{s}}\sum_{n=0}^{N}\left[N\sum_{m=0}^{M}\Pi_{m,n} -n\sum_{m=0}^{M}\Pi_{m,n}\right], \tag{33}\] \[=\frac{p_{nc}}{T_{s}}\left[N\sum_{n=0}^{N}\sum_{m=0}^{M}\Pi_{m,n} -\sum_{n=0}^{N}n\sum_{m=0}^{M}\Pi_{m,n}\right],\] (34) \[S_{nc} =\sum_{n=0}^{N}n\cdot\left[\sum_{m=0}^{M}\Pi_{m,n}\right]. \tag{27}\]0  
**Proposition 3**. : _The access delays of the transmitted packets of cooperative and non-cooperative users are given by_  
\[S_{nc} =\sum_{n=0}^{N}n\cdot\left[\sum_{m=0}^{M}\Pi_{m,n}\right]. \tag{27}\]1 \[S_{nc} =\sum_{n=0}^{N}n\cdot\left[\sum_{m=0}^{M}\Pi_{m,n}\right]. \tag{27}\]2  
Proof. : According to [PERSON]'s result [30], the average number of packets in a stationary system is equal to the average effective throughput multiplied by the average time that a packet spends in the system. Note that the actual number of packets in the system includes the backlogged packets and transmitted packets. Thus, the packet delays for both users are:  
\[S_{nc} =\sum_{n=0}^{N}n\cdot\left[\sum_{m=0}^{M}\Pi_{m,n}\right]. \tag{27}\]3 \[S_{nc} =\sum_{n=0}^{N}n\cdot\left[\sum_{m=0}^{M}\Pi_{m,n}\right]. \tag{27}\]4  
In order to accurately evaluate the system implementation, we should explore the performance of the backlogged packets. The next proposition gives the average throughput of backlogged users. **Proposition 4**. : _Consider the transition matrix \(P\), where \(P_{(ij)(i^{\prime},j^{\prime})}\) is the transition probability from state \((i,j)\) to state \((i^{\prime},j^{\prime})\). The average throughput of the backlogged packets transmitted by cooperative and non-cooperative users is given, respectively, as follows:_  
\[S_{nc} =\sum_{n=0}^{N}n\cdot\left[\sum_{m=0}^{M}\Pi_{m,n}\right]. \tag{27}\]5  
\[S_{nc} =\sum_{n=0}^{N}n\cdot\left[\sum_{m=0}^{M}\Pi_{m,n}\right]. \tag{27}\]6  
Proof. : In the stationary regime, the average number of backlogged packets that enter the system in a slot time \(T_{s}\) corresponds to the average number of packets that leave the system in the same slot time, and thus to the backlogged throughput. Furthermore, for a given system state \((m,n_{1})\), the probability that \(k\) backlogged packets corresponding to cooperative users enter the system, or equivalently, the probability that cooperative-users packets become backlogged is given by \(P_{(m,n_{1})(m+k,n_{2})}\). Thus, the number of backlogged packets that enter the system at a given state \((m,n_{1})\) is \(k\cdot P_{(m,n_{1})(m+k,n_{2})}\), where \(k=1,\ldots,M-m\). Thus, by considering all possible states \(n_{1}\in\{0,\ldots,N\}\), \(n_{2}\in\{0,\ldots,N\}\), and \(m\in\{0,\ldots,M\}\), we can obtain the expression (40). A similar approach can be used to derive the second equation of Proposition 4. **Corollary 2**. : _Equations (40) and (41) can be expressed in terms of the average throughput leaving the system at the stationary regime.