## 4 Proposed Method for Estimating the Tail Probability of the MMPP/D/1 Queue  
### Optimal Parameter for Estimation  
Now, we are identifying the optimal parameter for estimating the Importance Sampling (IS) of the MMPP/D/1 queue. The queue length and the state of the Markov-Modulated Poisson Process (MMPP) at time slot \(t\) are denoted by \(Q_{t}\) and \(S_{t}\), respectively. The process \(X_{t}=(Q_{t},S_{t})\) denotes a series of states at various time intervals, with \(t\) having a range of values of \(0,1,2\), and so forth. The probability of transitioning from state \(X_{t}\) to state \(X_{t+1}\) is represented by \(P(X_{t},X_{t+1})\). Due to the deterministic service time of the MMPP/D/1 queue system, the packet length remains constant. The disparity in the length of the queue between the \(t\)th and \((t+1)\)st time slots is represented as \(y\) and can be calculated as the difference between \(Q_{t+1}\) and \(Q_{t}\).  
We will provide a comprehensive elucidation on the process of determining the different queue lengths, represented as \(y\), in the MMPP/D/1 queue. At time slot \((t+1)\), the state of the Markov-Modulated Poisson Process (MMPP) is represented as \(S_{t+1}=j\), indicating that \(A_{j}\) packets have been received. In this context, \(A_{j}\) denotes a Poisson random variable that corresponds to state \(j\) and has an arrival rate of \(\lambda_{j}\). If the value of \(Q_{t}\) is greater than zero, then the value of \(y\) is equal to \(A_{j}\) minus one. The equation \(y=A_{j}-1\) representsthe relationship between the number of packets served, denoted by \(y\), and the number of consecutive slots, denoted by \(A_{j}\). This equation accounts for the fact that there is exactly one packet served between two consecutive slots. If \(Q_{t}=0\), then \(y\) is equal to \(A_{j}\).  
Denoting by \(h=j-i,z=(y,h)\), we have the state transition probability \(P(X_{t},X_{t}+z)\) as follows:  
\[P(X_{t},X_{t}+z)=\begin{cases}\dfrac{(\lambda_{j}/\mu)^{y+1}}{(y+1)!}e^{- \lambda_{j}/\mu}r_{ij},&\text{if }Q_{t}>0,\\ \dfrac{(\lambda_{j}/\mu)^{y}}{y!}e^{-\lambda_{j}/\mu}r_{ij},&\text{if }Q_{t}=0 \end{cases} \tag{8}\]  
To obtain the optimal distribution simulation, we solve the equation  
\[M_{X_{t}}(\theta_{0},\theta_{1})=1 \tag{9}\]  
According [4], we have \(\theta_{0}=\theta_{0}^{*}\), and \(\theta_{1}=\theta_{1}^{*}\) are the solutions of Equation (9). Hence, the optimal distribution simulation \(P^{*}(X_{t},X_{t}+z)\) is given by  
\[P^{*}(X_{t},X_{t}+z)=e^{\theta_{0}^{*}y}e^{\theta_{1}^{*}(j-i)}P(X_{t},X_{t}+z) \tag{10}\]  
In case of \(Q>0\), from (10), we have  
\[\dfrac{(\lambda_{j}^{*}/\mu^{*})^{y+1}}{(y+1)!}e^{-\lambda_{j}^{*}/\mu^{*}}r_ {ij}^{*}=\dfrac{(\lambda_{j}/\mu)^{y+1}}{(y+1)!}e^{-\lambda_{j}/\mu}r_{ij}\]  
\[\Longleftrightarrow(\lambda_{j}^{*}/\mu^{*})^{y+1}e^{-\lambda_{j}^{*}/\mu^{*} }r_{ij}^{*}=(\lambda_{j}/\mu)^{y+1}e^{-\lambda_{j}/\mu}r_{ij} \tag{11}\]  
To compare both sides of (11), we have the optimal parameter \(\lambda_{i}^{*}/\mu^{*}\) and \(r_{ij}^{*}\) as follows:  
\[\begin{cases}\lambda^{*}/\mu^{*}=e^{\theta_{0}}\lambda/\mu\\ r_{ij}^{*}=e^{-\theta_{0}-\lambda_{j}/\mu-\lambda_{j}^{*}/\mu^{*}+\theta_{1}(j- i)}r_{ij}.\end{cases} \tag{12}\]  
Similarly, we have the optimal parameter in case of \(Q=0\) as follows:  
\[\begin{cases}\lambda^{*}/\mu^{*}=\lambda/\mu\\ r_{ij}^{*}=r_{ij}.\end{cases} \tag{13}\]  
In (12), we can change the optimal service rate \(\lambda^{*}\) and optimal service rate \(\mu^{*}\) to obtain the optimal distribution simulation. In this case, the value of the arrival rate is kept unchanged; Equation (12) becomes  
\[\begin{cases}\lambda_{i}^{*}=\lambda_{i}\\ \mu^{*}=e^{-\theta_{0}^{*}}\mu\end{cases} \tag{14}\]  
Thus, the optimal processing rate \(\mu^{*}=e^{-\theta_{0}^{*}}\mu\) was determined for the IS simulation.