Worked M/G/1 Calculations for Exponential and Deterministic Service

From Queueing Formulas to Exact Calculations

For an M/G/1 queue, the arrival process is Poisson and there is one server, while service times follow a general distribution. Two familiar special cases are M/M/1, with exponential service times, and M/D/1, with a fixed service time. In both cases, the calculation begins with the same stability check and the same mean-waiting-time formula. The service-time distribution enters through its first two moments, so the two queues can have the same arrival rate and mean service time but different waiting times.

We consider a single-server, infinite-capacity queue with first-come, first-served service. Arrivals form a Poisson process of rate \(\lambda>0\). The successive service times are independent and identically distributed, independent of the arrival process, with a finite first and second moment. Write \(S\) for a service time, \(\mathbb{E}[S]\) for its mean, and \(\mathbb{E}[S^2]\) for its second moment. The service rate is \(\mu=1/\mathbb{E}[S]\). The traffic intensity is \(\rho=\lambda\mathbb{E}[S]=\lambda/\mu\). We use \(W_q\) for a typical customer's mean waiting time before service, \(W\) for mean time in the system (waiting plus service), \(L_q\) for mean number waiting, and \(L\) for mean number in the system.

The Pollaczek–Khinchine formula gives the stationary mean waiting time for a stable M/G/1 queue; its derivation is part of the earlier material on the M/G/1 waiting-time formula. It requires \(\rho<1\) and \(\mathbb{E}[S^2]<\infty\). The M/M/1 and M/D/1 calculations below use it only after checking those conditions. Little's law relates a long-run mean number in a system to the arrival rate and the mean time spent there; here all arrivals are accepted and the stable system's long-run departure rate equals \(\lambda\). The distinction between \(W_q,L_q\), which exclude service, and \(W,L\), which include it, is essential.

The Two Service-Time Moment Calculations

Derivation. For an exponential random variable with rate \(\mu\), its first and second moments are \(1/\mu\) and \(2/\mu^2\). Substituting the second moment into the Pollaczek–Khinchine formula gives \(W_q=\lambda/(\mu^2(1-\rho))\). Since \(1-\rho=(\mu-\lambda)/\mu\), this becomes \(W_q=\lambda/(\mu(\mu-\lambda))=\rho/(\mu-\lambda)\). Adding the mean service time gives \(W=\rho/(\mu-\lambda)+1/\mu=1/(\mu-\lambda)\). Little's law then gives \(L_q=\lambda\rho/(\mu-\lambda)=\rho^2/(1-\rho)\) and \(L=\lambda/(\mu-\lambda)=\rho/(1-\rho)\). These expressions require \(\lambda<\mu\); at or above that threshold, the stable-queue formulas do not apply.

These formulas are direct substitutions into the same M/G/1 result. Exponential service has a larger second moment than deterministic service with the same mean: for mean \(d\), the exponential second moment is \(2d^2\), whereas the deterministic second moment is \(d^2\). Therefore, at a fixed arrival rate and fixed mean service time, the exponential model has twice the mean queueing delay of the deterministic model. This comparison concerns the mean waiting time in a stable single-server queue; it does not say that every customer's wait is twice as long.

Worked Calculations

When the Formulas Do Not Apply

The stability check is not optional. For example, if \(\lambda=1\) per hour and deterministic service takes \(d=1\) hour, then \(\rho=\lambda d=1\). Substitution into \(\lambda d^2/(2(1-\rho))\) would put zero in the denominator. This is not a finite waiting-time answer: the stable stationary mean formula is inapplicable because the queue is at the boundary \(\rho=1\). If \(\rho>1\), offered work arrives faster than the server can process it on average, so there is likewise no stable stationary queue with the finite means calculated here.

Common mistake: using the service rate where a second moment is required. In the first example, incorrectly replacing the exponential second moment \(2/25\) by \(1/25\) would yield \(W_q=2(1/25)/(2(3/5))=1/15\) hour. That is not the M/M/1 waiting time: the exponential second moment is \(2/\mu^2=2/25\), not \((1/\mu)^2=1/25\). Using the correct moment gives \(W_q=2/15\) hour. The mistake amounts to treating exponential service as deterministic service with the same mean.

For a reliable calculation, keep the sequence explicit: identify the service-time distribution, compute its first and second moments, check \(\rho<1\), find \(W_q\), add the mean service time to get \(W\), and use Little's law for \(L_q\) and \(L\). The final consistency check \(L=L_q+\rho\) catches common mix-ups between the queue and the whole system.