Suppose we have a queue with parameters M/G/1 where M denotes a Markovian input describing the arrival times of customers into a system. Since M is Markovian, this implies that the inter-arrival times of customers into a system has a poisson process with rate $\lambda$. Now, the queue follows a general input describing the service times for any ith customers. Denote the service time for any ith customer to be $s_{i}$. The service time is a poisson process with rate $\mu$.

Now, for the case where $\frac{\lambda}{\mu}$:

For any unit time, the server clears the queue at a rate faster than the arrival of customers into the system. From this, it is then reasonable to argue that there exists a duration of time where the servers are idle. This suggests that the Markovian process alternate between an idle and busy state.

From here, I am told that the idle duration follows a poisson process with rate $\lambda$.

I am unable to understand why the occurence of the idle duration $I_{n}$ has a rate $\lambda$.

Any help is appreciated.


1 Answer 1


The inter-arrival times are exponentially distributed, which means they're memoryless. This means that at the precise moment when the queue goes idle, the time until the next arrival (and therefore the idle time) is exponential with rate $\lambda$, regardless of how long it's been since the previous arrival.


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