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 $q(n,n-1)$?
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 $n \mu$.