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It is common knowledge that an M/M/1 queueing system is stable if $\lambda<\mu$, where $\lambda$ is the arrival rate and $\mu$ is the service rate.

When $\lambda>\mu$, it is quite obvious that the system is unstable. Since the throughput is $TH=\mu$, so on average, an additional $\lambda-TH>0$ unserved customers will be added to the queue per time unit, so in long run the queue length reaches $+\infty$.

However, I am confused about the case $\lambda=\mu$. In this case, the throughput is $TH=\lambda=\mu$, so the average number of customers departing from and arriving to the system are equal, $\lambda-TH=0$, the system should be stable.

What is wrong with the above arguments?

Math. H
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1 Answers1

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If μ = λ, then the system is unstable. Indeed, in this case, with the first arrival, two timers start counting simultaneously: the timer of the next arrival and the timer of the first service. For the system to return to the initial state (X(t) = 0) the service must be done faster than the next arrival, but this only happens with a 50% probability, while also with a 50% probability the next arrival will happen before the previous departure. (formally you may show this by considering the difference of the two identical exponential time distributions (see for example here).

Thus, after a while it is certain that there will be one unserved customer to the queue and one that will be served. After the second arrival, by taking into account the "memoryless" property of the involved distributions, there is again 50% probability for another customer to arrive before the previous customer leave.

It turns out, that the total number of the customers on the system can be arbitrarily large with a corresponding probability large enough to ensure that the expected value converge to infinity with time. A formally detailed calculation is still needed, but I hope the above enlighten the situation describe.

Epameinondas
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