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Probability, Statistics and Queueing

There are two service windows of the same speed. In scheme A, each customer picks one of the windows with equal probability on arrival and joins that window's own queue. In scheme B, customers join a single queue and are served by whichever window becomes free first. Arrivals are random, and the total arrival rate and service speed are the same in both schemes. Which of the following correctly compares the average waiting time?

  1. AScheme A is shorter, because the queue is split in two and the arrival rate per queue is halved
  2. BScheme B is shorter, because in scheme A one window can be idle while the other has a queue
  3. CThey are the same, because with the same number and speed of windows the average waiting time does not depend on how the queue is formed
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Answer: B. Scheme B is shorter, because in scheme A one window can be idle while the other has a queue

Scheme A can be regarded as two M/M/1 queues side by side, each with half the arrival rate. Scheme B is an M/M/2 queue in which two windows share one queue, so a customer is served without waiting whenever either window is free. For example, with an average service time of 1 minute and a total arrival rate of 1 per minute (utilization 0.5), the average waiting time is 0.5/(1 − 0.5) × 1 = 1 minute in scheme A but about 0.33 minutes in scheme B. This is why a single queue shared by several windows is widely used, for example at the counters in banks and in server processing.

  • A. The arrival rate per queue is halved, but each queue has only one window, so the utilization per window is the same as in scheme B. Moreover, customers in one queue cannot use the other window when it is free.
  • B. Correct. With a shared queue, a window is never idle while customers are waiting.
  • C. Even with the same number and speed of windows, the way the queue is formed changes the waiting time.
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