# Continuous-Service M/M/1 Queuing Systems
## 2 Derivation of the Steady-State Distribution
\[(\text{Between States }-2\text{ and }-3): \lambda\pi_{-2}=\mu(\pi_{-3}+\pi_{-4}+\pi_{-5}+\cdots)\] \[\vdots\] \[(\text{Between States }-j\text{ and }-j-1): \lambda\pi_{-j}=\mu\Big{(}\pi_{-j-1}+\pi_{-j-2}+\pi_{-j-3}+\cdots \Big{)} \tag{1}\]  
For the cuts on the right side of State 0, we have  
\[(\text{Between States }0\text{ and }1): \mu\pi_{1}=\mu(\pi_{-1}+\pi_{-2}+\pi_{-3}+\cdots)\] \[(\text{Between States }1\text{ and }2): \mu\pi_{2}=\lambda\pi_{1}+\mu(\pi_{-2}+\pi_{-3}+\pi_{-4}+\cdots)\] \[(\text{Between States }2\text{ and }3): \mu\pi_{3}=\lambda\pi_{2}+\mu(\pi_{-3}+\pi_{-4}+\pi_{-5}+\cdots)\] Figure 2.0 Figure 2.1  
Combining Equations (1) and (2), we may see that  
Figure 2.2 Figure 2.3 Figure 2.4 Figure 2.5 Figure 2.6  
Adding up Equations in (3) yields  
Figure 2.7  
Dividing through by \(\mu\) gives us  
Figure 2.8  
recall that \(\rho=\frac{\lambda}{\mu}\) and that total probabilities sum to 1. Thus,  
Figure 2.9  
\[(\text{Between States }-1\text{ and }-2):\;\lambda\pi_{-1}=\mu(\pi_{-2}+\pi_{-3}+ \pi_{-4}+\cdots)\]0 A rate diagram for the continuous-service M/M/1 queueing system with a cut. We now seek to express \(\pi_{-j}\) in terms of \(\pi_{0}\). To this end, we divide Equations (1) through by \(\mu\) to produce  
\[ \begin{array}{rl}\pi_{0}&=\frac{1}{\rho}(\pi_{-1}+\pi_{-2}+\pi_{-3}+\cdots)\\ &=\frac{1}{\rho}\pi_{-1}+\frac{1}{\rho}(\pi_{-2}+\pi_{-3}+\pi_{-4}+\cdots).\end{array}  \tag{5}\]  
We may also obtain \(\pi_{-1}=\frac{1}{\rho}(\pi_{-2}+\pi_{-3}+\pi_{-4}+\cdots)\). Thus,  
\[ \begin{array}{rl}\pi_{0}&=\frac{1}{\rho}\pi_{-1}+\pi_{-1}\\ &=\Big{(}1+\frac{1}{\rho}\Big{)}\pi_{-1},\end{array} \]  
and hence  
\[\pi_{-1}=\Bigg{(}\frac{\rho}{1+\rho}\Bigg{)}\pi_{0}. \tag{6}\]  
Similarly,  
\[ \begin{array}{rl}\pi_{-1}&=\frac{1}{\rho}(\pi_{-2}+\pi_{-3}+\pi_{-4}+\cdots) \\ &=\frac{1}{\rho}\pi_{-2}+\frac{1}{\rho}(\pi_{-3}+\pi_{-4}+\pi_{-5}+\cdots)\\ &=\frac{1}{\rho}\pi_{-2}+\pi_{-2}\\ &=\Big{(}1+\frac{1}{\rho}\Big{)}\pi_{-2},\end{array} \]  
which gives us  
\[ \begin{array}{rl}\pi_{-2}&=\Big{(}\frac{\rho}{1+\rho}\Big{)}\pi_{-1}\\ &=\Big{(}\frac{\rho}{1+\rho}\Big{)}^{2}\pi_{0}.\end{array}  \tag{7}\]  
Likewise for \(\pi_{-2}\),  
\[ \begin{array}{rl}\pi_{-2}&=\frac{1}{\rho}(\pi_{-3}+\pi_{-4}+\pi_{-5}+\cdots) \\ &=\frac{1}{\rho}\pi_{-3}+\frac{1}{\rho}(\pi_{-4}+\pi_{-5}+\pi_{-6}+\cdots)\\ &=\frac{1}{\rho}\pi_{-3}+\pi_{-3}\\ &=\Big{(}1+\frac{1}{\rho}\Big{)}\pi_{-3},\end{array} \]  
which gives rise to  
\[ \begin{array}{rl}\pi_{-3}&=\Big{(}\frac{\rho}{1+\rho}\Big{)}\pi_{-2}\\ &=\Big{(}\frac{\rho}{1+\rho}\Big{)}^{3}\pi_{0}.\end{array}  \tag{8}\]  
As we continue in this manner to solve for each successive \(\pi_{-j}\), we may see a trend that gives rise to the following in terms of \(\pi_{0}\),  
\[\pi_{-j}=\Bigg{(}\frac{\rho}{1+\rho}\Bigg{)}^{j}\pi_{0}. \tag{9}\]  
We are now in a position to determine the value of \(\pi_{0}\) in the following. Since total probabilities sum to 1, Equation (4) results in  
\[\pi_{0}+\pi_{-1}+\pi_{-2}+\pi_{-3}+\cdots=1-\rho. \tag{10}\]  
Combining Equations (9) and (10) yields  
\[ \begin{array}{rl}\pi_{0}+\Big{(}\frac{\rho}{1+\rho}\Big{)}\pi_{0}+\Big{(} \frac{\rho}{1+\rho}\Big{)}^{2}\pi_{0}+\Big{(}\frac{\rho}{1+\rho}\Big{)}^{3} \pi_{0}+\cdots=1-\rho\\ &\pi_{0}\Big{[}1+\Big{(}\frac{\rho}{1+\rho}\Big{)}+\Big{(}\frac{\rho}{1+\rho} \Big{)}^{2}+\Big{(}\frac{\rho}{1+\rho}\Big{)}^{3}+\cdots\Big{]}=1-\rho.\end{array} \]  
The infinite series in the square brackets above is a geometric series, so we may write it as  
\[\pi_{0}\Bigg{(}\frac{1}{1-\frac{\rho}{1+\rho}}\Bigg{)}=1-\rho\pi_{0}(1+\rho)=1 -\rho.\]Thus,  
\[\pi_{0}=\frac{1-\rho}{1+\rho}. \tag{11}\]  
We proceed to next determine the value of \(\pi_{-j}\). Substituting Equation (11) into Equation (9) returns  
\[