## 4 Model Solution
### Solving the Separation-Aggregation Queueing Network Model Using the Parallel Strategy
#### 4.1.1 Constructing a Closed-Loop Queuing Network to Compute State-Based Service Rates
The equilibrium equation is shown in Table 3, and the birth and death process is shown in Figure 9.  
The approximate probability of finding node 1 is:  
\[\pi_{1}(0)=\frac{1}{1+\sum_{k_{1}}^{+\infty}\prod_{i=1}^{k_{1}}\frac{\lambda_{ r}}{\mu_{FES}(i)}} \tag{31a}\] \[\pi_{1}(k_{1})=\pi_{1}(0)\prod_{i=1}^{k_{1}}\frac{\lambda_{r}}{\mu_{FES}(i)} \tag{31b}\]  
Figure 9: Extinction process. \begin{table}
\begin{tabular}{c c c} \hline \hline
**Number of Customers (Status)** & **Equilibrium Equation** & **Simplified Equations** \\ \hline
0 & \(\pi_{1}(0)\lambda_{r}=\pi_{1}(1)\mu_{FES}(1)\) & \(\pi_{1}(1)=\pi_{1}(0)\frac{\lambda_{r}}{\mu_{FES}(1)}\) \\
1 & \(\pi_{1}(1)(\mu_{FES}(1)+\lambda_{r})=\pi_{1}(0)\lambda_{r}+\pi_{1}(2)\mu_{FES}(2)\) & \(\pi_{1}(2)=\pi_{1}(0)\frac{\lambda_{r}^{2}}{\mu_{FES}(1)+\mu_{FES}(2)}\) \\
2 & \(\pi_{1}(2)(\mu_{FES}(2)+\lambda_{r})=\pi_{1}(1)\lambda_{r}+\pi_{1}(3)\mu_{FES}(3)\) & \(\pi_{1}(3)=\pi_{1}(0)\frac{\lambda_{r}^{2}}{\mu_{FES}(1)+\mu_{FES}(2)}\).\(\mu_{FES}(1)\) \\... &... &... \\ \hline \hline \end{tabular}
\end{table} 
Table 3: Equilibrium equation transfer table. As the service rate of node 1 is equivalently represented by the throughput rate calculated in Section 4.1.1 and the throughput rate is related to the load of node 1, the average service rate of node 1 is:  
\[\mu_{1}=\sum\limits_{k_{1}}^{+\infty}\pi(k_{1})\cdot\mu_{FES}(k_{1}) \tag{32}\]  
The average number of customers served at node 1, \(\overline{K}_{1}\), can be calculated from the approximate probability:  
\[\overline{K}_{1}=\sum\limits_{k_{1}}^{+\infty}k_{1}\cdot\pi_{1}(k_{1}) \tag{33}\]  
(2) Solving node 2  
As node 1 obeys the negative exponential distribution and node 2 obeys the general distribution, the queueing network formed is a nonproduct solution and the queueing network formed is not easy to solve; therefore, in this paper, the MEM is applied to the solution of the open-loop queueing network, where the entropy is the measure of uncertainty that can be predicted for something. For node 2, the entropy function is introduced according to the maximum entropy principle in the form of a universal ME solution \(\pi_{2}(n)\) by maximising the entropy generalisation function \(H(p)\) under the following constraints [34; 35]:  
\[\mathrm{H}(\pi)=-\sum\limits_{n=1}^{k_{2}}\pi_{2}(n)\ln[\pi_{2}(n)],k_{2}\geq 1 \tag{34}\]  
The optimisation model of the maximum is:  
\[\mathrm{max}H(\pi)=-\sum\limits_{n=1}^{k_{2}}\pi_{2}(n)\ln[\pi_{2}(n)] \tag{35}\]  
\[\mathrm{s.t. }\left\{ \begin{array}{l}\sum\limits_{n=0}^{k_{2}}p(n)=1\\ \sum\limits_{n=1}^{k_{2}}h(n)\pi_{2}(n)=\rho\\ \sum\limits_{k=1}^{k_{2}}n\pi_{2}(n)=L\end{array} \right. \tag{36}\]  
The approximate probability of node 2 is:  
\[\pi_{2}(n)=\left\{\begin{array}{l}(1-\rho)\ \Setting the average number of customers at node 2 to \(\overline{K}_{2}\), the approximate probability can be derived from:  
\[\overline{K}_{2}=\sum_{k_{2}=1}^{+\infty}k_{2}\cdot\pi_{2}(k_{2}) \tag{39}\]  
Solving for system performance metrics  
1. The system response time, which is the time required to complete a pickup operation, including the waiting time and the service time, can be derived using [PERSON]'s theorem. \[E[T]=\frac{\overline{K}_{1}}{\lambda_{r}}+\frac{1}{\mu_{2}} \tag{40}\]  
2. The waiting time in the external queue for pickup is:  
\[W_{r}=\frac{\overline{K}_{1}}{\lambda_{r}}-\frac{1}{\overline{\mu}_{1}} \tag{41}\]  
3. The external queue waiting for pickup is:  
\[L_{eq}=W_{r}\cdot\lambda_{r} \tag{42}\]  
4. As this paper studies multiple shuttles in a cell, the arrival rate of each shuttle is \(\frac{\lambda_{r}}{M}\) and the utilisation rate of the shuttles can be calculated as:  
\[\rho_{s}=\frac{\lambda_{r}}{M\cdot\mu_{s}} \tag{43}\]