M/M/1 Average Time in System

The average time a job spends in a single-server queue, from arriving to leaving, is $W = 1/(\mu - \lambda)$: one over the spare capacity. A 40-an-hour server handling 30 an hour has 10 an hour of slack, so $W = 1/10$ hour, six minutes. The actual service takes only 1.5 minutes of that, so three quarters of the time is spent waiting behind other people.

The reciprocal form makes the sensitivity vivid. Ten jobs an hour of headroom gives a six-minute wait. Five gives twelve minutes. One gives an hour. Nothing about the server changed except how close the load crept to capacity. When someone reports that response times tripled after a modest traffic increase, this is almost always the mechanism, and adding a small amount of capacity back usually fixes it out of proportion to its cost.

Worth knowing: this quantity and the average number in system are linked by Little's law, $L = \lambda W$, which holds for essentially any queue regardless of the arrival or service distribution. Check the numbers above against it and they agree, $3 = 30 \ imes 0.1$, which is a good sanity test on any queueing calculation. Note also that $W$ here counts the whole stay including service. The waiting-only figure is $W - 1/\mu$, and mixing the two is the most common error in queueing homework.