Why Performance Changes Sharply Near Saturation

Why a Small Change in Utilization Can Matter

In a stable queue, increasing utilization means that the server is busy for a larger fraction of time. But the effect on congestion is not proportional to that increase: when utilization is already high, a small further increase can produce a large change in the mean queue length and delay. The reason is that random variation in arrivals and service has less spare capacity to be absorbed.

Consider an M/M/1 queue on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with continuous time \(t\ge0\). Let \(X_t\in S=\{0,1,2,\ldots\}\) be the number of customers in the system at time \(t\). Arrivals occur at rate \(\lambda>0\); service times are exponential with rate \(\mu>0\), and there is one server. Write \(\rho=\lambda/\mu\) for utilization. In this tutorial, performance means stationary averages, so we require the stability condition \(0<\rho<1\).

Write \(L=\mathbb{E}[X_t]\) for the stationary mean number in the system, \(L_q\) for the stationary mean number waiting in queue, \(W\) for a customer's mean time in the system, and \(W_q\) for the mean time waiting before service. The standard M/M/1 formulas established in the earlier material are

These are stationary means, not guarantees for individual customers. In particular, a finite mean at any fixed \(\rho<1\) does not mean that every customer experiences a short delay. The formulas quantify how the averages behave as the stability boundary is approached.

The Divergence Near Saturation

Proof. Substitute the formulas above and simplify for each \(0<\rho<1\): \((1-\rho)L=(1-\rho)\rho/(1-\rho)=\rho\), which tends to \(1\) as \(\rho\uparrow1\). Likewise, \((1-\rho)L_q=\rho^2\), which tends to \(1\); \((1-\rho)W=1/\mu\), which is constant in \(\rho\); and \((1-\rho)W_q=\rho/\mu\), which tends to \(1/\mu\). Since \(1-\rho\) tends to zero through positive values and each product has a strictly positive limit, each corresponding mean tends to infinity. For example, if \((1-\rho)L\to1\), then for \(\rho\) sufficiently close to one, \(L>(1/2)/(1-\rho)\), which grows without bound. The same argument applies to the other three means. \(\square\)

The factor \(1/(1-\rho)\) is the key scale. It is not a finite value at \(\rho=1\): the theorem describes a limit taken through stable queues, each with \(\rho<1\). At fixed \(\mu\), the mean sojourn time \(W\) grows exactly as the reciprocal of the unused service fraction \(1-\rho\). The other three means have the same leading growth, although their exact numerators differ.

Proof. With \(\mu\) fixed, \(L(\rho)=\rho/(1-\rho)\). The quotient rule gives \(\frac{dL}{d\rho}=[(1-\rho)+\rho]/(1-\rho)^2=1/(1-\rho)^2\). Also \(W(\rho)=1/[\mu(1-\rho)]\), so differentiation gives \(\frac{dW}{d\rho}=1/[\mu(1-\rho)^2]\). For the ratios, divide the positive finite means at \(\rho_2\) by those at \(\rho_1\). For \(L\), this yields \([\rho_2/(1-\rho_2)]/[\rho_1/(1-\rho_1)] =\rho_2(1-\rho_1)/[\rho_1(1-\rho_2)]\). For \(W\), the common factor \(1/\mu\) cancels, leaving \([1/(1-\rho_2)]/[1/(1-\rho_1)]=(1-\rho_1)/(1-\rho_2)\). All denominators are positive under the stated assumptions. \(\square\)

The derivative of each mean becomes large as \(\rho\) approaches one. This is a precise sense in which performance changes sharply near saturation: an equal-sized increase in utilization has a larger effect when starting from a higher utilization. For \(L\), the exact ratio also includes the numerator \(\rho\); for delay \(W\), the ratio depends only on the remaining capacity fractions.

Comparing Stable Traffic Intensities

The comparison shows why the final increments are consequential. Increasing utilization from \(\tfrac9{10}\) to \(\tfrac{19}{20}\) is an increase of only \(\tfrac1{20}\), but mean time in the system rises from \(50\) to \(100\) minutes. The service rate stays fixed; the change comes from halving the remaining fraction \(1-\rho\).

The denominator explains the difference between “a small utilization increase” and “a small performance change.” Utilization itself changes by the same amount, but its effect depends on the starting slack. The exact derivative \(dW/d\rho=1/[\mu(1-\rho)^2]\) is four times as large at \(\rho=\tfrac9{10}\) as at \(\rho=\tfrac45\), since \((\tfrac15)^2/(\tfrac1{10})^2=4\).

Stability Boundary and Common Mistake

The condition \(\rho<1\) is essential: the displayed formulas describe a stationary queue only on the stable side of the boundary. For example, if \(\lambda=\mu\), then \(\rho=1\). The birth-and-death balance equations for a stationary M/M/1 queue require \(\pi_n=\pi_0\rho^n\) for \(n\ge0\). At \(\rho=1\), these weights are all \(\pi_0\); if \(\pi_0>0\), their sum is infinite, and if \(\pi_0=0\), their sum is zero. Neither choice gives a probability distribution. Thus there is no stationary distribution to which the stationary mean formulas could apply at \(\rho=1\). The limit theorem concerns stable queues with \(\rho<1\); it does not assign a stationary mean at the boundary.

Common mistake: treating congestion as linear in utilization. A mistaken approximation is \(L=\rho\), which would give \(L=\tfrac9{10}\) at \(\rho=\tfrac9{10}\). The correct substitution is \(L=\rho/(1-\rho)=(\tfrac9{10})/(\tfrac1{10})=9\). The incorrect calculation drops the denominator that records spare capacity; near saturation, that denominator is precisely what drives the sharp increase.

When comparing queues, first confirm that each utilization is below one and then use the same performance measure and units. In particular, distinguish the number waiting, \(L_q\), from the number in the system, \(L\), and the queue wait, \(W_q\), from the total time in the system, \(W\). The formulas make each comparison exact, but only if the quantities being compared are defined consistently.