## 4 Model Solution
### Solving the Semi-Open-Loop Queuing Network Model under a Serial Strategy  
#### 4.2.1 Solving the Transfer Rate Matrix Q  
Define the system state as \((n_{1},n_{2},\ i,j,k)\), where \(i\geq 0\) is the number of external queue pickup orders, \(0\leq j\leq M\) is the number of hoist node pickup orders and \(0\leq k\leq M\) is the number of shuttle node pickup orders. Set \(n_{1}=i+j\) as the sum of the number of the external queues and first-node pickup orders for the sake of the solution convenience of solving and combine it with \(n_{2}=k\) to form the birth and death process of the two states. If the sum of the external queue, the first node and the second node pickup order is greater than M, then there must be idle four-way shuttles in the system, and the external queue waiting for pickup orders will be 0. If the sum of the external queue, the first node and the second node pickup order number is more than M, then there are no idle shuttles in the system, and the whole system is busy, so the external queue is defined as [39]:  
\[\mathrm{i}= \begin{cases}0&n_{1}+n_{2}\leq M\\ n_{1}+n_{2}-M&n_{1}+n_{2}>M\end{cases}  \tag{44}\]  
When the performance metrics are calculated (e.g., average queue length), only \(n_{1}\) and \(n_{2}\) are relevant. Therefore, for the sake of brevity, this semi-open-loop intraqueueing network state variable is modified to \((n_{1},n_{2})\). Defining the steady-state probability vector as \(\pi_{0}=\left(\pi_{(0,0)},\pi_{(0,1)},\pi_{(0,2)}\ldots\right)\),\(\pi_{0}=\left(\pi_{(i,0)},\pi_{(i,1)},\pi_{(i,2)}\ldots\right)\pi_{i}= \left(\pi_{(i,0)},\pi_{(i,1)},\pi_{(i,2)}\ldots\right)\), on the basis of the Coxian distribution, where \((\eta,\ H)\) denotes the arrival interval time distribution, \((\omega,\ W)\) denotes the service time of node 1 and \((\gamma,\ U)\) denotes the service time of node 2, the transfer rate matrix can be expressed as:  
\[\mathcal{Q}= \begin{bmatrix}B_{0}&C_{0}\\ A_{1}&B&C\\ &A&B&C\\ &\ddots&\ddots&\ddots\end{bmatrix}  \tag{45}\]where \(B_{0}\) is the rate matrix for the transfer from the \(\pi_{0}=\left(\pi_{(0,0)},\pi_{(0,1)},\pi_{(0,2)}\ldots\right)\) state to the \(\pi_{0}=\left(\pi_{(0,0)},\pi_{(0,1)},\pi_{(0,2)}\ldots\right)\) state, which is an \((M+1)\times(M+1)\)-squared matrix, M is the number of shuttles served by the hoist, \(I^{H}\) is a unit matrix of the same order as H, \(U^{\circ}\) is the absorption probability matrix of U, \(\otimes\) is the Kronecker product and \(\oplus\) is the Kronecker sum. A transfer diagram of the generation and extinction process is shown in Figure 10.  
\[B_{0}= \begin{bmatrix}(n_{1},n_{2})&(0,0)&(0,1)&(0,2)&\ldots&(0,M)\\ (0,0)&H&&&&\\ (0,1)&I^{H}\otimes U^{\circ}&H\oplus U&&&&\\ (0,2)&&I^{H}\otimes U^{\circ}\gamma&H\oplus U&&\\ \vdots&&\ddots&&\ddots&\\ (0,M)&&&&I^{H}\otimes U^{\circ}\gamma&H\oplus U\end{bmatrix}  \tag{46}\]  
\(C_{0}\) is the rate matrix of the transfer from the \(\pi_{0}=\left(\pi_{(0,0)},\pi_{(0,1)},\pi_{(0,2)}\ldots\right)\) state tothe \(\pi_{1}=\left(\pi_{(1,0)},\pi_{(1,1)},\pi_{(1,2)}\ldots\right)\) state, \(I^{U}\) is the unit matrix of the same order as U, and \(H^{\circ}\) is the absorption probability matrix of H.  
\[C_{0}= \begin{bmatrix}(n_{1},n_{2})&(1,0)&(1,1)&(1,2)&\ldots&(1,M)\\ (0,0)&H^{\circ}\eta\otimes\omega&\\ (0,1)&&H^{\circ}\eta\otimes\omega\otimes I^{U}&&\\ (0,2)&&&&H^{\circ}\eta\otimes\omega\otimes I^{U}&&\\ \vdots&&&&\ddots&\\ (0,M)&&&&H^{\circ}\eta\otimes I^{U}\end{bmatrix}  \tag{47}\]  
\(A_{1}\) is the rate matrix of the transfer from that state to the \(\pi_{0}=\left(\pi_{(0,0)},\pi_{(0,1)},\pi_{(0,2)}\ldots\right)\) state and \(W^{\circ}\) is the absorption probability matrix of the same order as \(W\). As the upper limit of the capacity of node 2 is \(M\), the transfer rate is 0 in the transfer from the state \((1,M)\) to \((0,M)\). \[A_{1}=