# Continuous-Service M/M/1 Queuing Systems
## 2 Derivation of the Steady-State Distribution
\begin{array}{rl}\pi_{-1}&=\Big{(}\frac{\rho}{1+\rho}\Big{)}\Big{(}\frac{1- \rho}{1+\rho}\Big{)}\\ &=\frac{\rho(1-\rho)}{(1+\rho)^{2}}\\ \pi_{-2}&=\Big{(}\frac{\rho}{1+\rho}\Big{)}^{2}\Big{(}\frac{1-\rho}{1+\rho} \Big{)}\\ &=\frac{\rho^{2}(1-\rho)}{(1+\rho)^{3}}\\ \pi_{-3}&=\Big{(}\frac{\rho}{1+\rho}\Big{)}^{3}\Big{(}\frac{1-\rho}{1+\rho} \Big{)}\\ &=\frac{\rho^{3}(1-\rho)}{(1+\rho)^{4}}\\ &\vdots\\ \pi_{-j}&=\Big{(}\frac{\rho}{1+\rho}\Big{)}^{j}\Big{(}\frac{1-\rho}{1+\rho} \Big{)}\\ &=\frac{\rho^{j}(1-\rho)}{(1+\rho)^{j+1}}.\end{array}  \tag{12}\]  
Finally, we are now ready to determine the value of \(\pi_{j}\). To do this, dividing through Equations (2) by \(\mu\), we obtain the following  
\[ \begin{array}{rl}\pi_{1}=\pi_{-1}+\pi_{-2}+\pi_{-3}+\cdots\\ \pi_{2}=\rho\pi_{1}+\pi_{-2}+\pi_{-3}+\pi_{-4}+\cdots\\ \pi_{3}=\rho\pi_{2}+\pi_{-3}+\pi_{-4}+\pi_{-5}+\cdots\\ \pi_{4}=\rho\pi_{3}+\pi_{-4}+\pi_{-5}+\pi_{-6}+\cdots\\ \vdots\\ \pi_{j}=\rho\pi_{j-1}+\pi_{-j}+\pi_{-j-1}+\pi_{-j-2}+\cdots\end{array}  \tag{13}\]  
To simplify \(\pi_{1}\) in Equations (13), we see that \(\pi_{1}=\rho\pi_{0}\) from Equation (5). Then, the value of \(\pi_{1}\) follows from Equation (11),  
\[\pi_{1}=\frac{\rho(1-\rho)}{1+\rho}. \tag{14}\]  
Similarly, we see that \(\pi_{2}=\rho\pi_{1}+\rho\pi_{-1}\) from Equations (7). Thus, the value of \(\pi_{2}\) follows from Equations (12) and (14),  
\[ \begin{array}{rl}\pi_{2}=\rho\cdot\frac{\rho(1-\rho)}{1+\rho}+\rho\cdot \frac{\rho(1-\rho)}{(1+\rho)^{2}}\\ =\rho^{2}(1-\rho)\Big{[}\frac{1}{1+\rho}+\frac{1}{(1+\rho)^{2}}\Big{]}\\ =\rho^{2}(1-\rho)\Big{[}\frac{(1+\rho)^{2}-1}{\rho(1+\rho)^{2}}\Big{]}\\ =\rho(1-\rho)\Big{(}1-\frac{1}{(1+\rho)^{2}}\Big{)}.\end{array} \]In a similar fashion, we may obtain the following results,  
\[ \begin{split}\pi_{3}=\rho^{3}&(1-\rho)\bigg{[}\frac{1}{ 1+\rho}+\frac{1}{\left(1+\rho\right)^{2}}+\frac{1}{\left(1+\rho\right)^{3}} \bigg{]}\\ &=\rho^{3}(1-\rho)\bigg{[}\frac{\left(1+\rho\right)^{3}-1}{\rho \left(1+\rho\right)^{3}}\bigg{]}\\ &=\rho^{2}(1-\rho)\bigg{(}1-\frac{1}{\left(1+\rho\right)^{3}} \bigg{)}\\ \pi_{4}=\rho^{4}(1-\rho)\bigg{[}\frac{1}{1+\rho}+\frac{1}{\left( 1+\rho\right)^{2}}+\frac{1}{\left(1+\rho\right)^{3}}+\frac{1}{\left(1+\rho \right)^{4}}\bigg{]}\\ &=\rho^{4}(1-\rho)\bigg{[}\frac{\left(1+\rho\right)^{4}-1}{\rho \left(1+\rho\right)^{4}}\bigg{]}\\ &=\rho^{3}(1-\rho)\bigg{(}1-\frac{1}{\left(1+\rho\right)^{4}} \bigg{)}\\ &\qquad\qquad\qquad\vdots\\ \pi_{j}=\rho^{j-1}(1-\rho)\bigg{(}1-\frac{1}{\left(1+\rho \right)^{j}}\bigg{)}.\end{split}  \tag{15}\]  
We have now found the steady-state distribution \(\pi=\{\ldots,\pi_{-2},\pi_{-1},\pi_{0},\pi_{1},\pi_{2},\ldots\}\) for our continuous-service M/M/1 systems. Piecing together Equations (11), (12), and (15) yields  
\[\pi_{j}=\left\{ \begin{array}{ll}\frac{\rho^{-j}(1-\rho)}{\left(1+\rho\right) ^{\left(1-j\right)}}&\text{for $j=0,-1,-2,\ldots$}\\ \rho^{j-1}(1-\rho)\bigg{(}1-\frac{1}{\left(1+\rho\right)^{j}}\bigg{)}&\text{ for $j=1,\;2,\ldots$}\end{array} \right. \tag{16}\]  
Recall that the steady-state distribution for the standard M/M/1 system is \(\pi=\{\pi_{0},\pi_{1},\pi_{2},\ldots\}\) where \(\pi_{j}=\rho^{j}(1-\rho)\) with \(j=0,\;1,\;2,\cdots\)[25]. Interestingly, our results are similar to that of the standard system but differ largely by some factors.