I'm trying to solve the following exercise (not homework):
Consider a $M/M/1$ queue with an arrival rate of 60 customers per hour and a mean service time of 45 seconds. A period during which there are 5 or more customers in the system is called crowded, when there are less than 5 customers it is quiet. What is $(i)$ the mean number of crowded periods per day (8 hours) and $(ii)$ how long do they last on average?
So the arrival and service processes occur according to Poisson processes.
Converting everything to minutes, we see that we have an arrival rate of $\lambda = 1$ customer per minute and a service rate of $\mu = \frac{4}{3}$. So we have a workload of $\rho = \frac{3}{4}$.
Now, there is a section in the reader I'm using that discusses the distribution and mean of a 'busy period', ie. a period where the queue is not empty (as opposed to 'idle periods', where there are no customers in the queue). I'm trying to reproduce a similar argument for this case, in that I'm naming a period where there are 5 or more customers in the queue a crowded period and a period where there are less than 5 customers a 'quiet period'. Unfortunately I have made no progress at this point. Any advice and hints are welcome.
EDIT
I'll provide the results the reader gives for the busy period here (where, again, a busy period is defined as the period that the system (queue + server) is not empty).
It is given on pages 37 through 39 of the following pdf: http://www.win.tue.nl/~iadan/queueing.pdf
So I need to adapt the argument given in section 4.6.2.
Best Answer
(i) A new crowded period starts every time the system transitions from having 4 customers to having 5 customers. Thus the mean number of crowded periods per day is the mean number of times per day that the system transitions from state 4 to state 5. This is $8(60)\lambda \pi_4$, where $\lambda$ is the per-minute arrival rate and $\pi_4$ is the long-term probability that there are 4 customers in the system. (See, for example, the second-to-last paragraph on page 30 in the notes you link to.) Since $\pi_4 = (1 - \rho) \rho^4$ (see eq. 4.6 in the linked notes), there are about 38 crowded periods per day on average.
(ii) The average length of time of a crowded period is the average amount of time between a state 4 to state 5 transition and the next state 5 to state 4 transition. Thanks to the fact that you have an $M/M/1$ queue (in particular, one server, an infinite-length queue, and the memoryless property that comes from having Poisson arrivals and exponential service times), if you draw out the state transition diagram you will see that this is exactly equivalent to the average amount of time between a state 0 to state 1 transition and the next state 1 to state 0 transition. So the average length of time of a crowded period is the same as that for a "busy period" in the example given in the notes; i.e., $$\frac{1/\mu}{1 - \rho} = \frac{3/4}{1 - 3/4} = 3 \text{ minutes}.$$