# Probabilistic Retry and Threshold Multirate Loss Models for Impatient Calls
## 3 The Proposed Probabilistic Threshold Loss Model
Based on (9)-(12) and since \(\alpha_{k}=\lambda_{k}\mu_{k}^{-1}\) for \(k=1\), \(2\), \(3\) and \(\alpha_{kc}=\lambda_{k}\mu_{kc}^{-1}\) for \(k=3\) are the corresponding offered traffic-loads, we can write the following equations:  
\[\alpha_{1}b_{1}q(j-b_{1})+\alpha_{2}b_{2}q(j-b_{2})+\alpha_{3}b_{3}q(j-b_{3}) =(y_{1}(j)b_{1}+y_{2}(j)b_{2}+y_{3}(j)b_{3})q(j),\text{for }j\leq J_{0}+b_{3c} \tag{13}\]  
\[\alpha_{1}b_{1}q(j-b_{1})+\alpha_{2}b_{2}q(j-b_{2})+\alpha_{3}b_{3}q(j-b_{3}) +\alpha_{3c}p_{3c}b_{3c}q(j-b_{3c})=jq(j),\text{for }J_{0}+b_{3c}<j\leq J_{0}+b_{3} \tag{14}\]  
\[\alpha_{1}b_{1}q(j-b_{1})+\alpha_{2}b_{2}q(j-b_{2})+\alpha_{3c}p_{3c}b_{3c}q( j-b_{3c})=(y_{1}(j)b_{1}+y_{2}(j)b_{2}+y_{3c}(j)b_{3c})q(j),\text{for }J_{0}+b_{3}<j\leq C. \tag{15}\]  
In order to combine (13)-(15) into a single recursive formula for the computation of \(q(j)\)'s, two necessary approximations are adopted: (1) in (13), the value of \(y_{3c}(j)=0\) when \(j\leq J_{0}+b_{3c}\) and (2) in (15), the value of \(y_{3}(j)=0\) when \(J_{0}+b_{3}<j\leq C\). Based on  
Figure 2: The call admission mechanism for a new call in the probabilistic threshold loss model. these approximations we can write the following recursive formula, assuming that \(z_{k}(j)\) expresses the second approximation and \(z_{3c}(j)\) expresses the first approximation:  
\[q(j)=\left\{ \begin{array}{l}1,\text{ if }j=0\\ \frac{1}{j}\left(\sum\limits_{k=1}^{K}\alpha_{k}b_{k}z_{k}(j)q(j-b_{k})+\alpha _{3c}p_{3c}p_{3c}z_{3c}(j)q(j-b_{3c})\right)\\ j=1,\ldots,C\\ 0,\text{ otherwise}\end{array} \right., \tag{16}\]  
where \(z_{k}(j)=1\), if \(b_{kc}>0\) and \(j\leq J_{0}+b_{k}\), or if \(b_{kc}=0\) and \(j\leq C\) while \(z_{3c}(j)=1\) when \(J_{0}+b_{3c}<j\). Assuming the general case where a link accommodates calls of \(K\) service-classes, Equation (16) takes the following form:  
\[q(j)=\left\{ \begin{array}{l}1,\text{ if }j=0\\ \frac{1}{j}\left(\sum\limits_{k=1}^{K}\alpha_{k}b_{k}z_{k}(j)q(j-b_{k})+\sum \limits_{k=1}^{K}\alpha_{kc}p_{kc}b_{kc}z_{3c}(j)q(j-b_{kc})\right)\\ j=1,\ldots,C\\ 0,\text{ otherwise}\end{array} \right., \tag{17}\]  
where \(z_{k}(j)=1\), if \(b_{kc}>0\) and \(j\leq J_{0}+b_{k}\), or if \(b_{kc}=0\) and \(j\leq C\) while \(z_{kc}(j)=1\) when \(J_{0}+b_{kc}<j\). Having computed, via (17), the unnormalized values of \(q(j)\)'s in the probabilistic threshold loss model, we can now calculate the CBP of service-class \(k\) calls with their reduced bandwidth requirement of \(b_{kc}\) b.u., \(B_{kc}\), as follows:  
\[B_{kc}=(1-p_{kc})\sum\limits_{j=J_{0}+1}^{C-\underline{b}_{kc}}G^{-1}q(j)+ \sum\limits_{j=C-b_{kc}+1}^{C}G^{-1}q(j), \tag{18}\]  
where \(G=\sum\limits_{j=0}^{C}q(j)\) is the normalization constant. Similarly, we may also obtain the conditional CBP denoted as \(B_{kc}^{*}=\text{Prob}\left\{B_{kc}|j>J_{0}\right\}\) via:  
\[B_{kc}^{*}=\text{Prob}\left\{B_{kc}|j>J_{0}\right\}=\frac{B_{kc}}{\sum\limits _{j=J_{0}+1}^{C}G^{-1}q(j)} \tag{19}\]  
The probabilistic threshold loss model provides the same CBP with those obtained via the probabilistic retry loss model if we set \(J_{0}=C-b_{k}\) and \(p_{kc}=p_{kc}\). Apart from the CBP, we can also determine the mean number of in-service calls of service-class \(k\), \(\overline{n}_{k}\), accepted in the link with \(b_{k}\) b.u. via (5) where \(y_{k}(j)\) can be computed via:  
\[y_{k}(j)=\frac{\alpha_{k}z_{k}(j)q(j-b_{k})}{q(j)}. \tag{20}\]  
Similarly, we can compute the mean number of in-service calls of service-class \(k\), \(\overline{n}_{kc}\), accepted in the link with \(b_{kc}\) b.u. via the formula:  
\[\overline{n}_{kc}=\sum\limits_{j=1}^{C}y_{kc}(j)\frac{q(j)}{G}, \tag{21}\]  
where \(y_{kc}(j)\) refers to the mean number of service-class \(k\) calls with \(b_{kc}\) b.u. in state \(j\). The latter can be determined via the formula:  
\[y_{kc}(j)=\frac{\alpha_{kc}p_{kc}z_{k}(j)q(j-b_{kc})}{q(j)}. \tag{22}\]The percentage of calls accepted in the system with \(b_{kc}\) b.u. can be obtained as \((\overline{\pi}_{kc}/(\overline{\pi}_{k}+\overline{\pi}_{kc}))\cdot 100\%\).