## 3 Energy Consumption Model  
In this paper, we use a radio model proposed in [4]. This radio energy model is utilized to measure energy consumption for the proposed CCBE algorithm. In [4], the radio model is a combination of three main models: the transmitter, the receiver and the power amplifier. The energy consumed by the transmitter consists of transmitter circuitry and the power amplifier, and the energy consumed in the receiver for receiving data consists of the receiver circuitry [4]. When a packet is transmitted from a transmitter to a receiver, where the distance between them is \(d\), the received signal power at the receiver is [26]:  
\[p_{r}(d)=\frac{p_{t}G_{t}G_{r}\lambda^{2}}{(4\pi)^{2}d^{\upbeta}Loss} \tag{1}\]  
where \(G_{r}\) and \(G_{t}\) are respectively receiver and transmitter gains. Furthermore, \(Loss\) represents any additional losses in the packet transmission, and \(\lambda\) represents carrier wavelength. The propagation loss factor \(\upbeta\) typically varies between two and four. Hence, considering \(G_{t}=G_{r}=1\), and \(Loss=1\), the received signal power at the receiver is:  
\[p_{r}(d)=\frac{p_{t}\lambda^{2}}{(4\pi)^{2}d^{\upbeta}} \tag{2}\]  
To receive the data packets successfully, the received signal power at the receiver must be above a minimum threshold power \((p_{thr})\). Hence, the transmitter signal power at the transmitter must be above this threshold \(\frac{p_{thr}(4\pi)^{2}d^{\upbeta}}{\lambda^{2}}\). The energy absorbed by the transmitter is:  
\[E_{t}=(e_{e}+\frac{p_{thr}(4\pi)^{2}R^{\upbeta}}{\lambda^{2}d_{r}})\times Packet =(e_{e}+e_{a}R^{\upbeta})\times Packet \tag{3}\]  
where \(e_{a}=\frac{p_{thr}(4\pi)^{2}}{\lambda^{2}d_{R}}\), which is considered as the \(energy/bit\) absorbed in the transmitter RF amplifier, \(d_{R}\) is the transmit or receive data rate \((bit/second)\) of each network node, \(Packet\) is the number of bitsin the packet and \(e_{e}\) is the \(energy/bit\) consumed in transmitter electronics. Hence, based on [27], the energy absorbed per second by a sensor node in three states can be calculated as follows:  
\[\begin{split} E_{t}&=(e_{e}+e_{a}R^{\text{\text{ \textregistered}}})N_{t}\\ E_{r}&=(e_{e})N_{r}\\ E_{l}&=e_{l}T_{l}=e_{l}(1-T_{t}-T_{r})=e_{e}d_{R}(1- T_{t}-T_{r})\end{split} \tag{4}\]  
The time for receiving and transmitting the data traffic between the cluster is denoted by \(T_{r}\) and \(T_{t}\), respectively, where \(N_{t}\) and \(N_{r}\) are the traffic data bits transmitted and received, respectively. Equation (5) represents the value of \(T_{r}\) and \(T_{t}\).  
\[\begin{split} T_{t}&=\frac{N_{t}}{d_{R}}\\ T_{r}&=\frac{N_{r}}{d_{R}}\end{split} \tag{5}\]  
The amount of time spent in one second for listening to the radio channel is represented as \(T_{l}\) : \(T_{l}=1-T_{t}-T_{r}\), \((0\leq T_{l}\leq 1)\); thus, \(0\leq(1-\frac{N_{t}}{d_{R}}-\frac{N_{r}}{d_{R}}\leq 1)\). Considering the static data traffic environment, where \(N_{t}=N_{r}=N\), the value of \(N\) is represented in Equation (6) as:  
\[0\leq N\leq\frac{1}{2}\times d_{R}\times 1 second \tag{6}\]  
As a result, \(\frac{1}{2}\times d_{R}\) bits represents the maximum amount of data that can be transmitted in each cluster per second, when nodes in the cluster do not listen to the radio environment and spend half a second for receiving the packets and another half for transmitting the packets. In simulations, we consider \(d_{R}=2.5\times 10^{5}\) bps [28]; the maximum data that can be relayed in WSNs based on Equation (6) is \(1.25\times 10^{5}\) bits per second, where the energy consumed for listening to the radio environment per second is represented as \(e_{l}\). \(e_{e}\), \(e_{a}\) and \(e_{l}\) are obtained from the design characteristics of the transceivers. From [4,28], the specific values of \(e_{e}\) is: \(e_{e}=3.32\times 10^{-7}\) J/bit. When \(p_{thr}=2\times 10^{-9}w\), \(d_{R}=2.5\times 10^{5}\) bps, \(f=2.4\times 10^{9}\) Hz, and \(e_{a}\approx 8\times 10^{-11}\) J/bit/m\({}^{2}\).