## Appendix A  
Solution for the balance equations shown in Section 3 are provided here. Equation (A1) represents the steady state probability for \(P_{1}\).  
\[\bigg{(}\lambda+2\lambda+3\lambda+\ \cdots+(n-1)\lambda+\lambda_{F}+\frac{\mu}{n- 0}\bigg{)}P_{1}=\lambda P_{0}\] (A1)  
Equations (A2)-(A5) process the simplification.  
\[\bigg{(}\frac{\lambda(n-1)(n-2)}{2}+\lambda_{F}+\frac{\mu}{n-0}\bigg{)}P_{1} =\lambda P_{0}\] (A2)  
\[P_{1}=\lambda\bigg{(}\frac{\lambda(n-1)(n-2)}{2}+\lambda_{F}+\frac{\mu}{n-0} \bigg{)}P_{1}=\lambda P_{0}\] (A3)  
\[P_{1}=\lambda\bigg{(}\frac{\lambda(n-1)(n-2)}{2}+\lambda_{F}+\frac{\mu}{n-0} \bigg{)}^{-1}P_{0}\] (A4)  
\[P_{1}=\lambda\eta_{1}P_{0}\] (A5)  
Similarly, \(P_{2}\) can be expressed as follows:  
\[\bigg{(}\lambda+2\lambda+3\lambda+\ \cdots+(n-2)\lambda+\lambda_{F}+\frac{\mu}{n- 1}\bigg{)}P_{2}=\lambda P_{1}+2\lambda P_{0}\] (A6)  
\[\bigg{(}\frac{\lambda(n-2)(n-3)}{2}+\lambda_{F}+\frac{\mu}{n-1}\bigg{)}P_{2}= \lambda(P_{1}+2P_{0})\] (A7)  
\[P_{2}=\lambda\bigg{(}\frac{\lambda(n-2)(n-3)}{2}+\lambda_{F}+\frac{\mu}{n-1} \bigg{)}^{-1}(Q_{1}P_{0}+2P_{0})\] (A8)\[P_{2}=\lambda q_{2}(Q_{1}P_{0}+2P_{0})\] (A9)  
\[P_{2}=\lambda q_{2}(Q_{1}+2)P_{0}\] (A10)  
\[P_{2}=Q_{2}P_{0}\] (A11)  
\(P_{3}\) can be expressed as follows, and simplified in Equations (A12) and (A13).  
\[\bigg{(}\lambda+2\lambda+3\lambda+\ \cdots+(n-3)\lambda+\lambda_{F}+\frac{ \mu}{n-2}\bigg{)}P_{3}=\lambda P_{2}+2\lambda P_{1}+3\lambda P_{0}\] (A12)  
\[\bigg{(}\frac{\lambda(n-3)(n-4)}{2}+\lambda_{F}+\frac{\mu}{n-2}\bigg{)}P_{3}= \lambda(P_{2}+2P_{1}+3P_{0})\] (A13)  
Similarly, \(P_{i}\), \(0<i<n\) can be expressed as follows:  
\[\bigg{(}\lambda+2\lambda+3\lambda+\ \cdots+(n-i)\lambda+\lambda_{F}+\frac{ \mu}{n-i+1}\bigg{)}P_{i}=\lambda P_{i-1}+2\lambda P_{i-2}+\ldots+i\lambda P_{0}\] (A14)  
\[\bigg{(}\frac{\lambda(n-i)(n-i+1)}{2}+\lambda_{F}+\frac{\mu}{n-i+1}\bigg{)}P_ {i}=\lambda(P_{i-1}+2P_{i-2}+\ldots+iP_{0})\] (A15)  
Lastly, \(P_{n}\) can be expressed as follows:  
\[(\lambda_{F}+\mu)P_{n}=\lambda P_{n-1}+2\lambda P_{n-2}+\ldots+n\lambda P_{0}\] (A16)  
\[(\lambda_{F}+\mu)P_{n}=\lambda(P_{n-1}+2P_{n-2}+\ldots+nP_{0})\] (A17)  
\(P_{0}\) can be solved by substituting Equation (4) into Equation (3):  
\[P_{0}+P_{1}+P_{2}+\ldots+P_{n}=1\] (A18)  
\[P_{0}+q_{1}^{-1}P_{0}\Bigg{[}\sum_{j=1}^{1}j\Bigg{[}\sum_{k=1}^{0}\Bigg{[} \prod_{l=1}^{k-1}q_{l}^{-1}\Bigg{]}\Bigg{]}\Bigg{[}k\Bigg{]}+q_{2}^{-1}P_{0} \Bigg{[}\sum_{j=1}^{2}j\Bigg{[}\sum_{k=1}^{1}\Bigg{[}\prod_{l=1}^{k-1}q_{l}^{- 1}\Bigg{]}\Bigg{]}\Bigg{[}k\Bigg{]}+\ldots+P_{n}=1\] (A19)  
As, \(P_{0}\) is a term that is common in all the terms in Equation (A19):  
\[P_{0}\Bigg{[}1+q_{1}^{-1}\Bigg{[}\sum_{j=1}^{1}j\Bigg{[}\sum_{k=1}^{0}\Bigg{[} \prod_{l=1}^{k-1}q_{l}^{-1}\Bigg{]}\Bigg{]}\Bigg{]}+q_{2}^{-1}\Bigg{[}\sum_{ j=1}^{2}j\Bigg{[}\sum_{k=1}^{1}\Bigg{[}\prod_{l=1}^{k-1}q_{l}^{-1}\Bigg{]} \Bigg{]}\Bigg{[}k\Bigg{]}+\ldots\Bigg{]}=1\] (A20)  
Once the value of \(P_{0}\) is known, the other steady state probabilities of the Markovian model can be obtained by using Equation (4) consecutively. Given that all the steady state probabilities are known, the values of \(L\), \(L_{Q}\), \(W\) and \(W_{Q}\) are calculated.