# Queueing theory M/M/k - probability of number of busy servers seen by next arrival process

Consider a $n$ server parallel queueing system, need to calculate the probability of $1$ busy server as seen by next arrival process.

$\lambda$$=$$arrival$ $rate$ $of$ $processes$ ; $\mu$$=$$service$ $rate$ $of$ $processes$

When there are $0$ servers busy then next arrival will find $0$ busy servers for sure. If there is currently $1$ busy server, the next arrival finds $1$ busy server if when the time to the next arrival is less than the remaining service time, this is given as $\lambda/(\lambda+\mu)$. Can any one please explain how it is $\lambda/(\lambda+\mu)$, please provide me the material so I can understand such basic concepts better.

## 1 Answer

If you arrive at time $t$, what is the probability the server is busy. It is the probability that the service time is at least $t$ which is $$P[busy|arrive\ at\ t]=\int_{t}^\infty ds\ \mu e^{-\mu s}=e^{-\mu t}.$$ Therefore, by the law of total probability, the probability that the server is busy is $$\begin{array}{rcl} P[busy]&=&\int_{0}^\infty dt\ P[busy|arrive\ at\ t] P(t)\\ &=&\int_{0}^\infty dt\ e^{-\mu t}\cdot\lambda e^{-\lambda t}\\ &=&\lambda\int_{0}^\infty dt\ e^{-(\lambda+\mu) t}\\ &=&\frac{\lambda}{\lambda+\mu} \end{array}$$