## 2 Modeling and Analysis of State Transitions for Energy-Efficient Femto BS
### Performance Analysis  
For the performance analysis, we have made the following assumptions regarding the density functions of random variables in each state:  
* Session arrivals at an MS occur according to a Poisson process, and session arrivals at a femto BS from all of the MSs in the coverage of the femto BS occur according to a Poisson process with parameters \(\lambda_{s}\);
* The time duration that a femto BS remains in the active (data) state follows a busy period of \(M/M/\infty\) queuing model;
* Location registration arrival at an MS occurs according to a Poisson process, and location registration arrivals at a femto BS from all of the MSs in the coverage of the femto BS occur according to a Poisson process with parameters \(\lambda_{r}\);
* The time duration that a femto BS remains in the active (signaling) state follows an exponential distribution of \(\mu_{r}\), since it is assumed that the service time of a location registration signaling is very short compared to the inter-arrival time of two consecutive location registrations, and thus, more than one location registrations do not occur during the service time of the active (signaling) state;
* The value of inactivity timer is assumed as constant and is denoted by \(T_{I}\). * The time durations that a femto BS remains in the sleep entering and waking up states are assumed to be a fixed, and they are denoted by \(T_{E}\) and \(T_{W}\), respectively [24]. We denote the active (data), idle, active (signaling), sleep entering, sleep and waking up states as States \(1,2,3,4,5\) and \(6\), respectively, for notational convenience. Since the residence times of all states of a femto BS do not follow an exponential distribution, the femto BS state transitions behavior is analyzed using a semi-Markov process approach [25]. The steady-state probability of each femto BS state can be obtained as:  
\[P_{k}=\frac{\pi_{k}\overline{t_{k}}}{\sum_{i=1}^{6}\pi_{i}\overline{t_{i}}}, \qquad\quad k=1,2,3,4,5,\mathrm{and}\;6 \tag{1}\]  
where \(\pi_{k}\) denotes the stationary probability of state \(k\) and \(\overline{t_{k}}\) is the mean residence time of the femto BS in state \(k\). The stationary probability is obtained by solving the following balancing equations [25]:  
\[\pi_{j} = \sum_{k=1}^{6}\pi_{k}P_{kj},\;\;\;\;\;\;j=1,2,3,4,5,\mathrm{and}\;6 \tag{2}\] \[1 = \sum_{k=1}^{6}\pi_{k} \tag{3}\]  
where \(P_{kj}\) represents the state transition probability from state \(k\) to state \(j\). The state transition probability matrix \(P=[p_{kj}]\) of the state transition model is given by Figure 1:  
\[P=\left( \begin{array}{cccccc}0&P_{12}&0&0&0&0\\ P_{21}&0&P_{23}&P_{24}&0&0\\ 0&P_{32}&0&0&0&0\\ 0&0&0&0&P_{45}&0\\ 0&0&0&0&0&P_{56}\\ P_{61}&0&0&0&0&0\end{array} \right)\]  
Then, stationary probabilities can be solved as:  
\[\pi_{1} = \frac{P_{21}+P_{24}P_{45}P_{56}P_{61}}{Den} \tag{4}\] \[\pi_{2} = \frac{1}{Den}\] (5) \[\pi_{3} = \frac{P_{23}}{Den}\] (6) \[\pi_{4} = \frac{P_{24}}{Den}\] (7) \[\pi_{5} = \frac{P_{24}P_{45}}{Den}\] (8) \[\pi_{6} = \frac{P_{24}P_{45}P_{56}}{Den} \tag{9}\]  
where \(Den\) is obtained by:  
\[Den=1+P_{21}+P_{23}+P_{24}(1+P_{45}(1+P_{56}(1+P_{61}))) \tag{10}\]  
State transition probability \(P_{kj}\) can be derived based on the distribution of time from states \(k\) to \(j\), \(T_{kj}\). In the active (data), active (signaling), sleep entering, sleep and waking up states, since exits from these states occur due to only one event, \(P_{12}=P_{32}=P_{45}=P_{56}=P_{61}=1\).