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For an M/M/1/4 queue with $\lambda=7,\mu=8,K=4$ and $\rho=7/8$, $$L=\frac{\rho}{1-\rho}-\frac{(K+1)\rho^{K+1}}{1-\rho^{K+1}}$$

I don't understand how the expression for $L$ is derived here. I know the first term in there $(\frac{\rho}{1-\rho})$ refers to $L$ in an M/M/1 queue with infinite capacity but the second term $(\frac{(K+1)\rho^{K+1}}{1-\rho^{K+1}})$ is throwing me off. I think it's something to do with the fact that $P_{(K+1)}=0$ but I'm not sure how it's derived.

pihyvev
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1 Answers1

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We have the detailed balance equations $$ \lambda\pi_{i-1} = \mu\pi_i,\ 1\leqslant i\leqslant K $$ which yield the recurrence $\pi_i = \rho^i\pi_0$, where $\rho=\lambda/\mu$. From $\sum_{i=0}^K \pi_i=1$ we have $$ \pi_i = \frac{\rho^i(1-\rho)}{1-\rho^{K+1}}, $$ and hence \begin{align} L &= \sum_{i=1}^K i\pi_i\\ &=\sum_{i=1}^K i\frac{\rho^i(1-\rho)}{1-\rho^{K+1}}\\ &= \frac{\rho(1-(K+1)\rho^K+K\rho^{K+1})}{(1-\rho)(1-\rho^{K+1})}. \end{align} From here a partial fraction decomposition yields $$ L = \frac\rho{1-\rho}-\frac{(K+1)\rho^{K+1}}{1-\rho^{K+1}}, $$ as desired.

Math1000
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    Thanks for the thorough derivation. Is there an intuitive way to understanding $(\frac{(K+1)\rho^{K+1}}{1-\rho^{K+1}})$ ? – pihyvev Oct 06 '19 at 19:47
  • Not off the top of my head. You could think of it as a factor to compensate for the fact that there can only be $K$ customers in the system at most. – Math1000 Oct 06 '19 at 20:00