# m/m/1/k=1 variance

Is there a nice formula for the finite time variance of the m/m/1/k=1 queue in steady state? (Exponential arrivals $$\lambda$$, exponential service time $$\mu$$, 1 server and no queue, so if a job arrives while one is in service it is lost.) In particular I'm interested in k=1 where this is a simply continuous time markov chain with 2 states, busy and idle.

Formally, let f be the fraction of time there is a job in service, ie how busy is the server. (I don't care about the lost jobs.) Then for any length of time $$E[f]=\mu/(\lambda+\mu)$$ in steady state, but what is the variance of $$f$$ over a period of time $$T$$.

For the special case where $$\lambda=\mu$$ then over an interval of length $$T$$ the number of jobs serviced is Poisson and conditional on this number, the times at which the jobs begin and end are uniformly distributed over the interval, so you can compute all the probabilities as (ugly) sums. Is there a similar result when $$\lambda\not=\mu$$?

• Ah, queueing model without a queue! 😆 Dec 20, 2023 at 3:39
• I would think that something like $$\Delta \ell_t := \max \left(0, \min \left( k, \ell_t + A_t \right) - D_t \right)$$ where $\ell$ is the number of jobs being served, $A_t$ is the number of arrival into the service center (not the queue) at time $t$, and $D_t$ are the number departures from the service center, would be a starting point. Dec 20, 2023 at 3:46