# Optimal Base Station Density of Dense Network:
## 4 Throughput of MAC Layer in SmBS Tier
### Fraction of Time for a Productive Channel  
Described in the system model, the state transition diagram of a packet in the SmBS channel is depicted in Figure 3 (and more detailed state transition analysis can also be found in [29]). There are three states for each packet. Depending on the collision times \(i\), we use \(S_{i}\) (sensing), \(W_{i}\) (waiting) and \(T_{i}\) (transmission) to denote the states, where \(i=0,1,2,\ldots K\). We assume \(\varepsilon\) to be the probability that the channel is sensed idle and \(p\) is the successful probability of transmission. Considering a packet in state \(i\), it transfers from \(S_{i}\) to \(W_{i}\) if the channel is sensed busy. We assume the time of \(W_{i}\) is equal to packet transmission time. When the channel is detected as idle, the packet will transfer from \(S_{i}\) to \(T_{i}\) with the probability \(q^{\prime}\). If it is successfully sent, a new packet will begin again at the initial state \(S_{0}\). Otherwise,it goes to \(S_{i+1}\) and continues to sense the channel. The channel is sensed idle if there is no transmission. Thus, its probability is  
\[\varepsilon=\frac{a\pi_{\text{idle}}+a\pi_{\text{suc}}+a\pi_{\text{col}}}{a\pi_{ \text{idle}}+(1+a)\left(\pi_{\text{suc}}+\pi_{\text{col}}\right)}=\frac{a}{1+a-e ^{-aG}}. \tag{10}\]  
The packet transmission is successful only if all other devices do not send data in the duration of \(a\) at the beginning of each busy period of the channel. Thus, it has \(p=e^{-aG}\).  
We denote \(s_{i}\), \(w_{i}\) and \(t_{i}\) as the limiting probability of \(S_{i}\), \(W_{i}\) and \(T_{i}\). With the Markov chain in Figure 3, we can derive the following results:  
\[\begin{array}{l}s_{0}=w_{0}+pt_{0}+pt_{1}+\dots+pt_{K}\\ w_{0}=(1-\varepsilon)\,s_{0},t_{0}=\varepsilon s_{0};\\ \text{when}\,i=1,2,\dots,K-1\\ \left\{\begin{array}{l}s_{i}=\varepsilon\left(1-q^{i}\right)s_{i}+w_{i}+(1 -p)\,t_{i-1}\\ w_{i}=(1-\varepsilon)\,s_{i},t_{i}=eq^{q}s_{i};\\ s_{K}=\varepsilon\left(1-q^{K}\right)s_{K}+w_{K}+(1-p)\,t_{K-1}+(1-p)\,t_{K}\\ w_{K}=(1-\varepsilon)\,s_{K},t_{K}=eq^{K}s_{K}.\end{array}\right.\end{array} \tag{11}\]  
For network stability, let \(p+q>1\), so all states in Markov Chain are positive recurrent. Combining (11) and the duration of the three states \(t_{T_{i}}=1\), \(t_{W_{i}}=1\) and \(t_{S_{i}}=a\), the time-average probability of the states is expressed by  
\[\begin{array}{l}\text{when}\,i=0,1,\dots,K-1\\ \left\{\begin{array}{l}\tilde{s}_{i}=ap\,(p+q-1)\,(1-p)^{i}\,/\,\left(q^{i}D \right)\\ \tilde{t}_{i}=ep\,(p+q-1)\,(1-p)^{i}\,/\,D\\ \tilde{w}_{i}=(1-\varepsilon)\,p\,(p+q-1)\,(1-p)^{i}\,/\,\left(q^{i}D\right) \\ \tilde{s}_{K}=a\,(p+q-1)\,(1-p)^{K}\,/\,\left(q^{K}D\right)\\ \tilde{t}_{K}=\varepsilon\,(p+q-1)\,(1-p)^{K}\,/\,D\\ \tilde{w}_{K}=(1-\varepsilon)\,(p+q-1)\,(1-p)^{K}\,/\,\left(q^{K}D\right), \end{array}\right.\end{array} \tag{12}\]  
where \(D=(1+a-\varepsilon)\left[pq-(1-q)\,(1-p)^{K+1}\,/\,q^{K}\right]+\varepsilon\,( p+q-1)\).  
Lemma 4: _The fraction of time that there is a successful transmission in the channel is_  
\[f_{\text{m}}=\frac{aGe^{-aG}}{1+a-e^{-aG}}. \tag{13}\]  
Proof of Lemma 4: The result can be proofed using the method of Theorem 1 in [29].  
It is observed that the successful-transmission time fraction \(f_{\text{m}}\) depends on the channel load \(G\). It can be understood intuitively since the channel load (also the load of an SmBS) has a direct impact on channel collision in LBT mode. Excessive load will lead to frequent collisions and backoff, and degrade the utilization of the channel. Thus, we need to analyze the channel load \(G\).  
Figure 3: State transition diagram of a packet in the SmBS channel.