# How to get transition rates in a $M/M/\infty$ queue

I am told for an $M/M/\infty$ queue the transition rates $q$ are as follows.

• $q(n,n+1) = \lambda$
• $q(n,n-1) =n\mu$

Can anybody explain the intuition behind $q(n,n-1)$?

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what does $q(n,n+1)$ denote? –  oks Mar 5 '13 at 18:47

It is the property of exponential random variables. You have $n$ active servers each with service rate of $\mu$. Then there is an arrival process of rate $\lambda$.
By exponential random variable property, if $X_1,X_2,\ldots$ are respectively exponential random variables with means $\lambda_1,\lambda_2,\ldots$, then $\min(X_1,X_2,X_3,\ldots, X_n)$ is a random variable with mean being the sum of individual means ($\lambda_1+\lambda_2+\ldots+\lambda_n$).
Thus in M/M/$\infty$ case, the rate at which a packet is removed from the queue is the minimum of $n$ exponential random variables, each with service rate $\mu$ and hence is equal to $n \mu$.
@Rosie: It is not about the number of servers, but the number of packets, which is $n$. To understand about the minimum, say you are in a state with $n$ packets and ask "When will I change from this state?" The answer for this is the minimum of the interarrival time and the $n$ service times for each of the packets. –  Bravo Mar 5 '13 at 19:51