## 3 Optimal Frame Length and Maximum Channel Utilization  
In the FSA-based networks, we assume that the current frame length is \(L\), which is divided into slots with equal duration. Also, we consider that \(N\) users want to transmit packets and each randomly choosing one slot for transmission in the current frame. For a given slot, there are three probable outcomes: no packet transmitted (an idle slot), only one packet transmitted (a successful slot) and at least two packets transmitted (a collided slot). The procedure that \(N\) packets select slots in frame length \(L\) can be seen as \(N\)-fold Bernoulli trials with probability \(\frac{1}{L}\). Therefore, the probability of \(r\) packets transmitted in the same slot is given by  
\[p(r)=\left( \begin{array}{c}N\\ r\end{array} \right)\left(\frac{1}{L}\right)^{r}\left(1-\frac{1}{L}\right)^{N-r} \tag{31}\]  
where\(\left( \begin{array}{c}N\\ r\end{array} \right)=\frac{N!}{r!(N-r)!}\). Then, the probabilities that a slot is an idle slot (\(r=0\)), a successful slot (\(r=1\)) or a collided slot (\(r\geq 2\)) are \(p_{idl}\), \(p_{suc}\) and \(p_{col}\), respectively, which can be given by  
\[p_{idl} = \left(1-\frac{1}{L}\right)^{N} \tag{32}\] \[p_{suc} = \frac{N}{L}\left(1-\frac{1}{L}\right)^{N-1}\] (33) \[p_{col} = 1-\left(1-\frac{1}{L}\right)^{N}-\frac{N}{L}\left(1-\frac{1}{L} \right)^{N-1} \tag{34}\]  
Therefore, when the capture effect occurs (A collided slot possibly becomes a successful slot), the expected number of idle, successful and collided slots are respectively given by \[c_{iall} = p_{iall}L \tag{35}\] \[c_{suc} = p_{suc}L+L\sum_{i=2}^{i_{max}}p_{cap}(n=i)p(r=i)\] (36) \[c_{col} = L\sum_{i=2}^{i_{max}}\left[1-p_{cap}(n=i)\right]p(r=i) \tag{37}\]  
where \(p_{cap}(\cdot)\) is the capture probability, which can be obtained from (1) on the basis of different values of \(n\). In addition, \(i_{max}\) is the maximum number of packets colliding in the same slot. Then, the channel utilization can be defined by  
\[\eta=\frac{c_{suc}}{L}=\frac{N}{L}\left(1-\frac{1}{L}\right)^{N-1}+\sum_{i=2} ^{i_{max}}\frac{N!}{i!(N-i)!}\left(\frac{1}{L}\right)^{i}\left(1-\frac{1}{L} \right)^{N-i}p_{cap}(n=i) \tag{38}\]  
Here, we take a linear model into account, i.e., \(L=\lambda N\) with \(0<\lambda\leq 1\). Then we can obtain  
\[\lim_{N\rightarrow+\infty}\frac{N-1}{N} = 1 \tag{39}\] \[\lim_{N\rightarrow+\infty}\left(1-\frac{1}{\lambda N}\right)^{N} = e^{-\frac{1}{\lambda}} \tag{40}\]  
Substituting (39) and (40) into (38), the channel utilization is given by  
\[\eta\approx\frac{1}{\lambda}e^{-\frac{1}{\lambda}}+\sum_{i=2}^{i_{max}}\frac{ \lambda^{-i}}{i!}e^{-\frac{1}{\lambda}}p_{cap}(n=i) \tag{41}\]  
Let (41) be the maximum, we can obtain  
\[p_{suc} = \frac{N}{L}\left(1-\frac{1}{L}\right)^{N-1}\]0  
Then, the optimal frame length can be calculated by  
\[p_{suc} = \frac{N}{L}\left(1-\frac{1}{L}\right)^{N-1}\]1  
where \(\lceil\cdot\rceil\) is a ceiling integer function. After substituting (43) into (38) or (41), the maximum channel utilization can be obtained. For an intuitive explanation, we give one special case to show the capture effect impacts on the optimal frame length and the channel utilization. We assume the capture probability \(p_{cap}\) keeps unchanged in every collided slot of frame \(L\) like literature [11] (note the reality is not like this).