## 3 System Throughput and Packet Loss Rate
### Packet Loss Rate  
A packet is lost if none of the \(K\) satellites are able to correctly receive it. In order to estimate the packet loss rate, we assume that a target user always transmits, and then we determine the probability that its packets cannot be successfully decoded.  
Assuming that \(u\) packets were simultaneously transmitted by other users in the same given time slot, the event of the successful reception of the target packet (from the target user) at the \(k\)th satellite occurs with probability \(p_{k}=(1-\varepsilon_{k})\varepsilon_{k}^{u}\). Then, the packet loss rate, i.e., the packet loss probability from the target user, at the \(k\)th satellite, is:  
\[\mathcal{P}_{k} =1-\sum_{u=0}^{\infty}p_{k}\,\mathbb{P}[U=u]\] \[=1-\sum_{u=0}^{\infty}\left[(1-\varepsilon_{k})\varepsilon_{k}^{ u}\frac{G^{u}e^{-G}}{u!}\right]\] \[=1-(1-\varepsilon_{k})e^{-G(1-\varepsilon_{k})}. \tag{7}\]  
From (3) and (7), we can verify that \(\mathcal{P}_{k}(G)=1-\frac{\mathcal{T}_{2}(G)}{G}\), which is consistent with the fact that the maximum throughput achievable by a user (target user) is 1 and the maximum traffic load generated by a single user is one packet per time slot.  
Then, the event that specifically the target packet is successfully received by \(|\mathbb{J}|\) satellites, of the subset \(\mathbb{J}\subseteq\{1,\ldots,K\}\), occurs with probability  
\[p_{\mathbb{J}}=\prod_{k\in\mathbb{J}}(1-\varepsilon_{k})\varepsilon_{k}^{u}. \tag{8}\]  
After the realization of many time-slots \(N\), which implies the transmission of \(N\) packets from the target user, the average number of target packets jointly received by \(|\mathbb{J}|\)satellites, for all \(u\), is \(\left|\bigcap_{j\in\mathbb{J}}D_{j}\right|=N\sum_{u=0}^{\infty}p_{j}\mathbb{P}[U=u]\). Thus, the system packet loss rate \(\mathcal{P}(G)\) is  
\[\mathcal{P}(G) =\lim_{N\to\infty}\frac{1}{N}\Bigg{[}N-\sum_{\exists j\
eq\emptyset,\;\mathbb{J}\subseteq\{1,\ldots,K\}}(-1)^{|\mathbb{J}|+1}\Bigg{|}\bigcap_{j \in\mathbb{J}}D_{j}\Bigg{|}\Bigg{]}\] \[=1-\sum_{\exists j\
eq\emptyset,\;\mathbb{J}\subseteq\{1,\ldots,K \}}(-1)^{|\mathbb{J}|+1}\sum_{u=0}^{\infty}\Bigg{[}\frac{G^{u}e^{-G}}{u!} \prod_{k\in\mathbb{J}}(1-\varepsilon_{k})\varepsilon_{k}^{u}\Bigg{]}\] \[=1-\sum_{\exists j\
eq\emptyset,\;\mathbb{J}\subseteq\{1,\ldots,K \}}(-1)^{|\mathbb{J}|+1}e^{-G}\prod_{k\in\mathbb{J}}(1-\varepsilon_{k})\sum_ {u=0}^{\infty}\Bigg{[}\frac{G^{u}}{u!}\Bigg{(}\prod_{k\in\mathbb{J}} \varepsilon_{k}\Bigg{)}^{u}\Bigg{]}\] \[=1-\sum_{\exists j\
eq\emptyset,\;\mathbb{J}\subseteq\{1,\ldots,K \}}(-1)^{|\mathbb{J}|+1}\;\prod_{k\in\mathbb{J}}(1-\varepsilon_{k})\;e^{-G} \big{(}1-\prod_{k\in\mathbb{J}}\varepsilon_{k}\big{)}. \tag{9}\]  
Now, from (6) and (9) we can verify that \(\mathcal{P}(G)=1-\frac{\mathcal{T}(G)}{G}\). Again, we give an example for the case of \(K=2\),  
\[\mathcal{P}_{2}(G) =\sum_{u=0}^{\infty}\Big{[}\frac{G^{u}e^{-G}}{u!}\Big{]}\Big{[}1 -(1-\varepsilon_{1})\varepsilon_{1}^{u}\Big{]}[1-(1-\varepsilon_{2}) \varepsilon_{2}^{u}]\] \[=1-(1-\varepsilon_{1})e^{-G(1-\varepsilon_{1})}-(1-\varepsilon _{2})e^{-G(1-\varepsilon_{2})}+(1-\varepsilon_{1})(1-\varepsilon_{2})e^{-G(1 -\varepsilon_{1}\varepsilon_{2})}. \tag{10}\]  
Notice that, similarly to the previous subsection when \(\varepsilon_{1}=\varepsilon_{2}=\varepsilon\), the system packet loss rate becomes in \(\mathcal{P}_{2}(G)=1-2(1-\varepsilon)e^{-G(1-\varepsilon)}+(1-\varepsilon)^{ 2}e^{-G\big{(}1-\varepsilon^{2}\big{)}}\), exactly the result of ([36], Equation (7)) for \(K=2\), showing that the model of equal erasure probabilities is a particular case of the formulation presented in this subsection too.