# Improvement and Performance Evaluation When Implementing a Distance-Based Registration
## 5 Registration Cost  
Assuming the registration cost for one registration is \(U\), then registration cost in an hour, \(C_{U}\), is as follows:  
\[C_{U}=U\sum_{b\in B}\widetilde{\pi_{b}};p_{b0}\cdot\lambda_{m} \tag{4}\]  
\(B\): set of boundary cells;  
\(\widetilde{\pi_{b}}\): steady-state probability of a boundary cell \(b\);  
\(p_{b0}\): transition probability from boundary cell b to central cell 0. To obtain the registration cost using this equation, the key is to obtain the steady state probability of each cell. For example, if \(N=5\), the steady state probability of each cell in the LA can be obtained. If \(N=5\), three states can be defined as shown in Table 1 and Figure 6.  
Similarly, the UE in state 1 can enter a state 0 with a probability of \((1/6)P[T_{c}>T_{m}]\). It can move to state 2 with a probability of \((1/6)P[T_{c}>T_{m}]\), or move to a new LA (and be in state 0 again) with a probability of \((1/2)P[T_{c}>T_{m}]\). When a call is generated, the UE in state 1 can also transit to state 0\({}^{\prime}\) with a probability of \(P[T_{c}<T_{m}]\). The UE in state 0\({}^{\prime}\) can enter state 1 with a probability of \((1/3)P[T_{c}>R_{m}]\). It can move to state 2 with a probability of \((1/3)P[T_{c}>R_{m}]\), or move to a new LA (and be in state 0) with a probability of \((1/3)P[T_{c}>R_{m}]\). Finally, when a call is generated, the UE in state 0\({}^{\prime}\) can transit to state 0\({}^{\prime}\) again with a probability of \(P[T_{c}<R_{m}]\). For convenience, using the following characters:  
\[m=P[T_{c}>T_{m}],\;\;m^{\prime}=P[T_{c}>R_{m}] \tag{5}\]  
we can get the final state transition diagram as shown in Figure 7. The corresponding transition probability matrix is shown in Figure 8.  
Similarly, we can get the state transition diagram and the corresponding transition probability matrix for various \(N\) to calculate the steady-state probability of each state. Table 2 below summarizes the state of each cell in an LA and the corresponding transition probability matrix for various \(N\). Figure 8: Transition probability matrix. Figure 7: State transition diagram (\(N=5\)). \begin{table}
\begin{tabular}{c c c} \hline \hline \(N\) & **Definition of States** & **Transition Probability Matrix P** \\ \hline
2 & 0 & 0 \\ \hline
3 & 0 & 0 \\ \hline \(\begin{bmatrix}m&1-m\\ m^{\prime}&1-m^{\prime}\end{bmatrix}\) \\ \hline
4 & 1 & 0 \\ \hline \(\begin{bmatrix}\frac{2}{3}m&\frac{1}{3}m&1-m\\ m&0&1-m\\ \frac{2}{3}m^{\prime}&\frac{1}{3}m^{\prime}&1-m^{\prime}\end{bmatrix}\) \\ \hline
5 & 2 & 0 & 2 \\ \hline \(\begin{bmatrix}\frac{1}{3}m&\frac{1}{3}m&\frac{1}{3}m&1-m\\ \frac{2}{3}m&\frac{1}{6}m&\frac{1}{6}m&1-m\\ \frac{5}{6}m&\frac{1}{6}m&0&1-m\\ \frac{1}{3}m^{\prime}&\frac{1}{3}m^{\prime}&\frac{1}{3}m^{\prime}&1-m^{\prime} \end{bmatrix}\) \\ \hline
6 & 1 & 1 \\ \hline \(\begin{bmatrix}\frac{1}{3}m&\frac{1}{3}m&\frac{1}{3}m&1-m\\ \frac{2}{3}m&\frac{1}{6}m&\frac{1}{6}m&1-m\\ \frac{5}{6}m&\frac{1}{6}m&0&1-m\\ \frac{1}{3}m^{\prime}&\frac{1}{3}m^{\prime}&\frac{1}{3}m^{\prime}&1-m^{\prime} \end{bmatrix}\) \\ \hline \end{tabular}
\end{table} 
Table 2: State of each cell in location area and transition probability matrix for various \(N\). Now, let us obtain m and m\({}^{\prime}\) as well as the steady-state probability. The probability that a UE enters other cells before a call occurs can be calculated as below [2,3]:  
\[m=P[T_{c}>T_{m}]=\int_{0}^{\infty}\int_{t_{m}}^{\infty}\lambda_{c}e^{-\lambda_{ c}t_{c}}f_{m}(t_{m})dt_{c}dt_{m}=f_{m}^{*}(\lambda_{c}) \tag{6}\]  
To calculate \(m^{\prime}=P[T_{c}>R_{m}]\), let us consider the density function of \(R_{m}\) first.