Comparing Arrival Reduction With Faster Service

Two Ways to Reduce Congestion

Congestion in an M/M/1 queue can be reduced by changing the demand placed on the server or by increasing the server’s service rate. These interventions need not have the same effects on every performance measure. In particular, utilization alone does not determine mean sojourn time: the difference between the arrival and service rates matters too. A deadline comparison also needs a clearly specified deadline. Here, the deadline is for a customer’s sojourn time, the total time in the system, including both queueing and service.

Consider an M/M/1 queue on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Arrivals form a Poisson process of rate \(\lambda>0\), and successive service times are independent exponential random variables of rate \(\mu>0\), independent of arrivals. There is one server, customers are served in arrival order, and the queue has unlimited capacity. Let \((N_t)_{t\ge0}\) be the continuous-time process giving the number of customers in the system, with state space \(S=\{0,1,2,\ldots\}\), and let \((\mathcal{F}_t)_{t\ge0}\) be its natural history. The stable case is \(\lambda<\mu\). All steady-state measures below assume this condition and refer to a customer arriving in steady state.

Measures Used in the Comparison

In a stable M/M/1 queue, the stationary number in the system has probabilities \(\mathbb{P}(N_t=n)=(1-\rho)\rho^n\), for \(n\ge0\). These probabilities sum to one because \(\sum_{n\ge0}(1-\rho)\rho^n=1\). The steady-state arrival-observation property for Poisson arrivals (PASTA) means an arriving customer sees this same distribution. If the customer sees \(n\) customers, their sojourn time is the sum of \(n+1\) independent exponential service times of rate \(\mu\): the \(n\) customers ahead and that customer’s own service.

The following calculation gives the sojourn-time distribution. For \(s\ge0\), an exponential service time has Laplace transform \(\mu/(\mu+s)\); independence makes the transform of \(n+1\) such times \((\mu/(\mu+s))^{n+1}\). Averaging over the number seen on arrival gives

The final expression is the Laplace transform of an exponential random variable of rate \(\mu-\lambda\); uniqueness of Laplace transforms identifies the distribution. Consequently, for \(t\ge0\), the mean sojourn time and deadline failure probability are

The mean service time is \(1/\mu\), so the mean queue waiting time is the mean sojourn time less the mean service time: \(\mathbb{E}[T]-1/\mu=\lambda/[\mu(\mu-\lambda)]=\rho/(\mu-\lambda)\). These formulas distinguish two effects: \(\rho\) measures the fraction of capacity used, while \(\mu-\lambda\) sets the rate in the sojourn-time distribution.

Proof. Write \(g=\mu-\lambda>0\) for the original rate difference. Under the arrival reduction, the new difference between service and arrival rates is \(\mu-(\lambda-\delta)=g+\delta\). Under the service increase it is \((\mu+\delta)-\lambda=g+\delta\). The formulas above therefore give mean sojourn time \(1/(g+\delta)\) and failure probability \(e^{-(g+\delta)d}\) in both cases, including \(d=0\).

The new utilizations are \(\rho_A=(\lambda-\delta)/\mu\) and \(\rho_S=\lambda/(\mu+\delta)\). Both denominators are positive, and \[ \rho_S-\rho_A =\frac{\lambda\mu-(\lambda-\delta)(\mu+\delta)} {\mu(\mu+\delta)} =\frac{\delta(\mu-\lambda+\delta)}{\mu(\mu+\delta)}>0. \] Thus \(\rho_A<\rho_S\). Mean queue waiting time is utilization divided by the rate difference. Since that difference is the same positive value \(g+\delta\) under both interventions, the lower utilization under arrival reduction also gives strictly lower mean queue waiting time. This proves all the claims. \(\square\)

The theorem is an exact comparison, not a general claim that either intervention is preferable. It compares equal absolute rate changes. A different change in rates can produce a different rate difference, and then mean sojourn time and deadline performance need not match.

Choosing the Right Comparison

To compare interventions for a particular queue, first state what is held fixed and how large each rate change is. Then check stability, calculate the new utilization, and calculate the rate difference separately. For a sojourn-time deadline \(d\), use \(e^{-(\mu-\lambda)d}\) with the rates after the intervention. A deadline for queue waiting alone is a different measure and must not be evaluated using the sojourn-time formula.

Common mistake: using \(1/\mu\) as mean sojourn time. In the first worked example, that calculation would give \(1/10\) hour at baseline and, after the service increase, \(1/11\) hour. Those are mean service times, not mean times in the system. The correct post-intervention mean sojourn time is \(1/(11-9)=1/2\) hour. The gap \(11-9\) accounts for queueing as well as service. Another tempting error is to conclude from equal utilization that two systems have equal delays; the third example disproves that by direct calculation.

Reducing arrivals can represent demand management, appointment smoothing, or redirecting some demand elsewhere; increasing service can represent faster processing or added capacity. The calculations quantify queue performance under the M/M/1 assumptions, but they do not assign costs to either intervention or account for changes in service quality or demand. Those considerations are needed to choose an intervention in practice.