## 5 Performance Analysis of the CL-APCC Protocol
### Energy Complexity Analysis of CL-APCC Protocol  
**Theorem 3.**\(O(N^{2}/K)\) is the complexity of energy in any grid, it is further known that \(O(N^{2})\) is the energy overhead of the whole network.  
_Proof._ It is mainly divided into four parts for the whole network's energy overhead which are data sending, data receiving, node monitoring and node calculation. Owing to the fact that the energy overhead of node monitoring and calculation is much smaller than that of data sending and receiving, we only take into account the energy overhead of data sending and receiving.  
From figure 3, the outputting rate of \(i-1\) is equal to the inputting rate of \(i\). If it gets the data rate generated by data sources, in accordance with the relationship between the average rates of inputting and outputting in section 4.1.2, we can analyze the average energy overhead of the grid in the network. Using the energy overhead model of the literature [20], the energy overhead is \(E_{TX}=E_{elec}\times L+\varepsilon_{f_{S}}\times L\times d^{2}\) which is transmitted by node from data packets with the length of \(L\) to the distance of \(d\), and energy overhead of data receiving is \(E_{RX}=E_{elec}\times L\). Thus, for one grid, data transmission of energy overhead within period \(t\) is shown as below:  
\[\begin{array}{c}E_{TX}^{k}\left(t\right)=n\times t\times E_{TX}\times\mu_{ avg}^{k}\left(t\right)=n\times t\times\left(E_{elec}\times L+\varepsilon_{f_{S}} \times L\times d^{2}\right)\times\mu_{avg}^{k}\left(t\right)\\ \\ E_{RX}^{k}\left(t\right)=n\times t\times E_{RX}\times\lambda_{avg}^{k}\left(t \right)=n\times t\times E_{elec}\times L\times\lambda_{avg}^{k}\left(t\right) \end{array} \tag{28}\]  
The total energy overhead of the grid within period \(t\) is as below:  
\[E_{total}^{k}\left(t\right)=E_{TX}^{k}\left(t\right)+E_{RX}^{k}\left(t\right) =n\times t\times\left(E_{elec}\times L+\varepsilon_{f_{S}}\times L\times d^{2 }\right)\times\mu_{avg}^{k}\left(t\right)+n\times t\times E_{elec}\times L \times\lambda_{avg}^{k}\left(t\right) \tag{29}\]  
From the Equation (29) and \(n<N/K\), we know that:  
\[Max\left(E_{total}^{k}\left(t\right)\right)=\frac{N^{2}\times\pi\times r^{2} \times L_{Data}^{q}}{K\times Min\left(\mu_{con}^{i}\left(t\right)\right)}\times \left(\left(E_{elec}\times L+\varepsilon_{f_{S}}\times L\times d^{2}\right) \times\mu_{avg}^{k}\left(t\right)+E_{elec}\times L\times\lambda_{avg}^{k}\left( t\right)\right) \tag{30}\]  
Owing to \(d\leq r\), \(O(N^{2}/K)\) is the complexity of energy overhead in the grid, it is further known that \(O(N^{2})\) is the complexity of energy overhead in the whole network. Therefore, theorem 3 is proved to be correct.  
(3) Within one period, the storage overhead of the whole network equals to the product. This means the storage overhead of each grid multiplies the number of grids, \(K\). Therefore, the overall storage overhead of network \(M_{total}(t)\) is shown as follows.  
\[M_{total}\left(t\right)=K\times E_{real}\left(t\right)=K\times\frac{n\times t \times\left(\lambda_{avg}^{k}\left(t-1\right)-\mu_{avg}^{k}\left(t-1\right) \right)}{e^{\left(\lambda_{avg}^{k}\left(t-1\right)-\mu_{avg}^{k}\left(t-1 \right)\right)\lambda t}}{e^{\left(\lambda_{avg}^{k}\left(t-1\right)-\mu_{avg}^ {k}\left(t-1\right)\right)\lambda t}}\times e^{\frac{\left(\lambda_{avg}^{k} \left(t-1\right)-\mu_{avg}^{k}\left(t-1\right)\right)\lambda t}{\left(\lambda_ {avg}^{k}\left(t-1\right)-\mu_{avg}^{k}\left(t-1\right)\right)\lambda t}} \tag{36}\]  
Therefore, the theorem 4 is proved to be correct.