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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
up vote 2 down vote accepted

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 $n \mu$.

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Thanks for your reply, but I am still a bit unsure. Why do I need the minimum and also I thought there are infinite servers so why are we considering the case for n servers? Thanks – Rosie Mar 5 '13 at 18:31
@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
thanks again, it's just I don't know what you mean by packets. The way we do it in lectures is that this is a queue of people, with infinite servers, where do the packets come in? Thanks again – Rosie Mar 5 '13 at 21:24
@Rosie: Sorry, pls. replace packets with people :) I study this from a communication networks perspective... – Bravo Mar 5 '13 at 22:07
thanks a lot makes much more sense – Rosie Mar 5 '13 at 22:44

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