## 3 System Model and Performance Analysis
### Throughput of SU  
If \(a\leq 1\), the SU randomly selects a subchannel to transmit the packet, and the transmission of the SU packet does not affect the next round of BSR frames. If \(a>1\), then the transmission of the SU packet will affect the next round of BSR frames, so a packet is divided into equal \(n_{s}\) copies. Thus, the number of SU is \(\overline{n_{2}}\), as follows:  
\[\overline{n_{2}}=\begin{cases}n_{2}\theta p_{1},&a\leq 1,\\ \left(Int\left(\frac{K_{f}}{R_{1}t_{s}}\right)+1\right)n_{2}\theta p_{1},&a>1. \end{cases} \tag{30}\]  
The probability of collision for SU is as follows:  
\[p_{3}=\begin{cases}1-\left(1-\frac{1}{r}\right)^{n_{2}\theta p_{1}-1},&a\leq 1,\\ 1-\left(1-\frac{1}{r}\right)^{(Int\left(\frac{K_{f}}{R_{1}(s)}+1\right)n_{2} \theta p_{1}-1},&a>1.\end{cases} \tag{31}\]  
The probability of a subchannel being selected by at least one SU is \(\overline{p_{tr}}\).  
\[\overline{p_{tr}}=\begin{cases}1-\left(1-\frac{1}{r}\right)^{n_{2}\theta p_{ 1}},&a\leq 1,\\ 1-\left(1-\frac{1}{r}\right)^{(Int\left(\frac{K_{f}}{R_{1}(s)}+1\right)n_{2} \theta p_{1}},&a>1.\end{cases} \tag{32}\]The probability of a successful transmission on the subchannel for SU is given as:  
\[\overline{p_{s}}=\begin{cases}\dfrac{n_{2}\theta p_{1}\dfrac{1}{r} \bigg{(}1-\dfrac{1}{r}\bigg{)}^{n_{2}\theta p_{1}-1}}{\overline{p_{tr}}},&a \leq 1,\\ \dfrac{\bigg{(}Int\bigg{(}\dfrac{K_{f}}{R_{1}t_{s}}\bigg{)}+1 \bigg{)}n_{2}\theta p_{1}\dfrac{1}{r}\bigg{(}1-\dfrac{1}{r}\bigg{)}^{(Int( \dfrac{K_{f}}{R_{1}t_{s}})+1)n_{2}\theta p_{1}-1}}{\overline{p_{tr}}},&a>1. \end{cases} \tag{33}\]  
When \(a\leq 1\), the transmission can be completed within \(t_{s}\) without changing the packet size. When \(a>1\), the size of the SU packet becomes \(1/n_{s}\) times; thus, the expression of packet size is obtained.  
\[K_{f}=\begin{cases}K_{f},&a\leq 1,\\ \dfrac{K_{f}}{Int\bigg{(}\dfrac{K_{f}}{R_{1}t_{s}}\bigg{)}+1},&a>1\end{cases}. \tag{34}\]  
Then, SU throughput is expressed as follows:  
\[T_{S}=\begin{cases}\dfrac{\min(r,n_{1})K_{f}n\dfrac{\theta}{r} \bigg{(}1-\dfrac{\theta}{r}\bigg{)}^{n-1}}{t_{succ}}+\dfrac{\min(r,n_{2} \theta p_{1})K_{f}n_{2}\theta p_{1}\dfrac{1}{r}\bigg{(}1-\dfrac{1}{r}\bigg{)} ^{n_{2}\theta p_{1}-1}}{t_{succ}},&a\leq 1,\\ \dfrac{\min\bigg{(}r,\bigg{(}Int\bigg{(}\dfrac{K_{f}}{R_{1}t_{s}} \bigg{)}+1\bigg{)}n_{2}\theta p_{1}\bigg{)}K_{f}n_{2}\theta p_{1}\dfrac{1}{r} \bigg{(}1-\dfrac{1}{r}\bigg{)}^{(Int(\dfrac{K_{f}}{R_{1}t_{s}})+1)n_{2}\theta p _{1}-1}}{t_{succ}},&a>1.\end{cases} \tag{35}\]  
Thus, the total system throughput is \(T=T_{P}+T_{S}\). We have:  
If \(a\leq 1\),  
\[T =\dfrac{\min(r,n)K_{f}n\theta(1-\theta)^{n-1}}{t_{succ}}+\dfrac{ \min(r,n_{2}\theta p_{1})K_{f}n_{2}\theta p_{1}\dfrac{1}{r}\bigg{(}1-\dfrac{1 }{r}\bigg{)}^{n_{2}\theta p_{1}-1}}{t_{succ}} \tag{36}\] \[+\dfrac{\min(r-n\theta(1-p_{1}),n_{1}\theta p_{1})K_{f}n_{1} \theta p_{1}\dfrac{1}{r-n\theta(1-p_{1})}\bigg{(}1-\dfrac{1}{r-n\theta(1-p_{ 1})}\bigg{)}^{n_{1}\theta p_{1}-1}}{t_{succ}}.\]  
If \(a>1\),  
\[T =\dfrac{\min(r,n)K_{f}n\theta(1-\theta)^{n-1}}{t_{succ}} \tag{37}\] \[+\dfrac{\min(r-n\theta(1-p_{1}),n_{1}\theta p_{1})K_{f}n_{1} \theta p_{1}\dfrac{1}{r-n\theta(1-p_{1})}\bigg{(}1-\dfrac{1}{r-n\theta(1-p_{ 1})}\bigg{)}^{n_{1}\theta p_{1}-1}}{t_{succ}}\] \[+\dfrac{\min\bigg{(}r,\bigg{(}Int\bigg{(}\dfrac{K_{f}}{R_{1}t_{s} }\bigg{)}+1\bigg{)}n_{2}\theta p_{1}\bigg{)}K_{f}n_{2}\theta p_{1}\dfrac{1}{r} \bigg{(}1-\dfrac{1}{r}\bigg{)}^{(Int(\dfrac{K_{f}}{R_{1}t_{s}})+1)n_{2}\theta p _{1}-1}}{t_{succ}}.\]