# Analyzing Zone-Based Registration under 2-Step Paging in Mobile Communication Network
## 3 Analysis of ZBR under 2-Step Paging
### Analysis of Time Spent in a Zone in 1 ZR  
To compare the performance with 2 ZR, which only considers movement between the zones, let us also find the time spent in the zone for 1 ZR. First, let us determine the average number of cells until a UE goes through a zone in 1 ZR with a zone composed of several cells and then find the time spent in the zone based on this number. Let us then find the average number of cells until a UE that has just entered the current zone passes through the zone. For example, consider the case of \(n=9\). The nine cells on the right are the ones in the current zone and the three cells on the left are the ones in the zone visited immediately before. If the case of leaving the current zone and performing location registration is defined as the absorbing state of state 0, then the number of cells that a UE that has just entered the current zone (states 1 or 2) enters until it leaves this zone can be obtained using the following method:  
The state transition diagram and state transition probability matrix for the state 0 (absorbing state) that enters the new zone and performs location registration and six states of nine cells in the current zone are as follows:  
In Figure 9, \(R\) represents the transition probability matrix from the transit states to the absorbing state, \(T\) represents the transition probability matrix between the transient states, and \(I\) represents the identity matrix. Starting from the current state \(i\), let \(E_{ij}\) be the average number of visits to state \(j\) before entering the absorbing state. Then, \(E\)--a matrix of \(E_{ij}\)--can be obtained as follows ([20], p. 231):  
\[E=(I-T)^{-1} \tag{3}\]  
Defining \(N_{i}\) as the number of transitions (number of cells visited) from the transit state \(i\) to the absorbing state 0, when \(n=9\), the average number of cells passing through in a zone, \(E(N)\), can be expressed as follows:  
\[E(N)=p_{1}\times E(N_{1})+p_{2}\times E(N_{2}) \tag{4}\]  
Figure 9: State transition diagram and state transition probability matrix when absorbing state 0 is introduced (\(n=9\)). where \(p_{i}\) (\(i=1,2\)) is the probability that, among the UEs that have just entered the current zone, the state is \(i\). \(E(N_{i})\) can be obtained by adding all the values of row \(i\) of \(E=(I-T)^{-1}\). \[E(N_{i})=\sum_{j=1}^{6}E_{ij}\text{ where }E=(I-T)^{-1} \tag{5}\]  
Suppose \(A_{i}\) (\(i=1,2\)) is the event that a UE enters the current zone from the zone visited immediately before and becomes the state \(i\) and that \(B_{i}\) (\(i=1,2,\ldots,6\)) is the event that a UE is in the cell corresponding to the state \(i\) (in the zone visited just before). Then, considering states in Figure 10, we can obtain the following equation:  
\[ \begin{split} P[A_{1}]&=P[B_{1}]\times P[\text{ location registration occurs with one cell entrance }\mid B_{1}]\\ &+P[B_{5}]\times P[\text{ location registration occurs with one cell entrance }\mid B_{5}]\\ &=\pi_{1}\times(q+r)+\pi_{5}\times 2r\end{split}  \tag{6}\]  
\[ \begin{split} P[A_{2}]&=P[B_{2}]\times P[\text{ location registration occurs with one cell entrance }\mid B_{2}]\\ &+P[B_{3}]\times P[\text{ location registration occurs with one cell entrance }\mid B_{3}]\\ &+P[B_{6}]\times P[\text{ location registration occurs with one cell entrance }\mid B_{6}]\\ &=\pi_{2}\times q+\pi_{3}\times r+\pi_{6}\times r\end{split}