# Call Blocking Probabilities under a Probabilistic Bandwidth Reservation Policy in Mobile Hotspots
## Appendix A  
Consider the capacity of the AP's WLAN to be \(C=4\) b.u. During the stop phase, the vehicle remains for an exponentially distributed time in a location with mean \(\theta_{s}^{-1}=20\). To facilitate handover users, the \"strict\" BR policy is considered whereby \(t_{new}=2\) b.u. are reserved for them and \(p_{s,new}=0\) when \(j=C-t_{new}\),..., \(C-1\). The duration of the moving phase is also exponentially distributed with mean \(\theta_{m}^{-1}=100\). In addition, let \(\lambda_{new}=\lambda_{h}=\mu_{s}^{-1}=\mu_{m}^{-1}=1.0\). The corresponding 2-D Markov chain of this tutorial example consists of 10 states with a steady-state probability \(P(i,n)\), where \(i=0\) refers to the stop phase, \(i=1\) refers to the moving phase and there exist \(n\) users in-service with \(0\leq n\leq 4\). The computation of the various performance measures (e.g., CBP and link utilization) can be executed via the solution of the following system of 10 global balance equations:  
The solution of the previous system of equations is:  
\(P(0,0)=0.029488;P(1,0)=0.036133;P(0,1)=0.057389;P(1,1)=0.30772;P(0,2)=0.056492;P(1,2)=0.154757;P(0,3)=0.018658;P(1,3)=0.051758;P(0,4)=0.004639;P(1,4)=0.012965.\)  
In what follows, we consider the proposed algorithm of Section 2.2 and present the calculations of the coefficients \(S_{i,n}^{0}\) and \(S_{i,n}^{1}\). According to (6) and since \(S_{0,0}^{0}=1\), \(S_{1,0}^{1}=0\) and \(S_{0,y}^{0}=0\) for \(y<0\), we have the following values of \(S_{i,n+1}^{0}\):  
state (0,0): \(S_{1,1}^{0}=1.015_{1,0}^{0}-0.05_{0,0}^{0}\Rightarrow S_{1,1}^{0}=2.05\)  
state (0,1): \(S_{0,2}^{0}=1.5255_{0,1}^{0}-5_{0,0}^{0}-0.0055_{1,1}^{0}\Rightarrow S_{0,2}^ {0}=2.1265\)  
state (1,1): \(S_{1,2}^{0}=1.0055_{1,1}^{0}-0.0255_{0,1}^{0}\Rightarrow S_{1,2}^{0}=-0.1015\)  
Following the same procedure, we obtain the values:state (0,2): \(S^{0}_{0,3}=0.7956133\) state (1,2): \(S^{0}_{1,3}=-\) 0.1206133 state (0,3): \(S^{0}_{0,4}=0.27423503\) state (1,3): \(S^{0}_{1,4}=-\) 0.10548503  
According to (7) and since \(S^{1}_{0,0}=0\), \(S^{1}_{1,0}=1\) and \(S^{1}_{0,y}=0\) for \(y<0\), we have the following values of \(S^{1}_{i,n+1}\):  
state (0,0): \(S^{1}_{0,1}=-0.01\) state (1,0): \(S^{1}_{1,1}=1.01\) state (0,1): \(S^{1}_{0,2}=-0.0203\) state (1,1): \(S^{1}_{1,2}=0.5153\) state (0,2): \(S^{0}_{0,3}=-0.0156893\) state (1,2): \(S^{1}_{1,3}=\) 0.1806893 state (0,3): \(S^{1}_{0,4}=\) 0.01126217 state (1,3): \(S^{1}_{1,A}=\) 0.05251217  
Having computed the coefficients \(S^{0}_{i,n}\) and \(S^{1}_{i,n}\), we consider (9) which takes the form:  
\[0.3160934P(0,0)-0.03044759P(1,0)=0\] (A1)  
Similarly, based on (11), we have the second equation between \(P(0,0)\) and \(P(1,0)\):  
\[5.86875P(0,0)+2.70125P(1,0)=1\] (A2)  
The system of (A1) and (A2) results in: \(P(0,0)=0.029488\), \(P(1,0)=0.306133\). Having determined the values of \(P(0,0)\) and \(P(1,0)\), together with the values of \(S^{0}_{i,n}\) and \(S^{1}_{i,n}\), we can now calculate the values of \(P(i,n)\) via (4) and obtain the following results: \(P(0,1)=0.057389\); \(P(1,1)=0.30772\); \(P(0,2)=0.056492\); \(P(1,2)=0.154757\); \(P(0,3)=0.018658\); \(P(1,3)=0.051758\); \(P(0,4)=0.004639\); \(P(1,4)=0.012965\). Having determined the values of \(P(0,0)\) and \(P(1,0)\), (via (9) and (11), respectively) and the values of \(S^{0}_{i,n}\) and \(S^{1}_{i,n}\)(via (6) and (7), respectively), we can calculate the values of \(P(i,n)\) via (4) and, consequently the CBP of new and handovers calls together with the link utilization. For reference, we present the values of the various performance measures.