Common Utilization and Delay Errors in M/M/c

Three Quantities That Are Easy to Confuse

In an M/M/c queue, several important quantities are related but not interchangeable. Offered load measures demand in units of server capacity; utilization measures the fraction of the total service capacity demanded. Erlang C measures the chance an arrival must wait in steady state. It does not, by itself, give a waiting time. Confusing these quantities can produce a plausible-looking calculation that is dimensionally or probabilistically impossible.

Assume throughout that arrivals form a Poisson process of rate $\lambda>0$, there are $c\ge1$ identical servers, service times are independent exponential random variables with rate $\mu>0$ per server, customers are served first-come, first-served, and the waiting room has unlimited capacity. Write $X_t$ for the number of customers in the system at time $t$, so the state space is $S=\{0,1,2,\ldots\}$. The process is considered on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. The steady-state quantities below apply when the queue has a stationary distribution.

Definition: The offered load is $a=\lambda/\mu$, measured in Erlangs. The utilization per server is $\rho=\lambda/(c\mu)=a/c$. The total service capacity is $c\mu$, so the stability condition is $\lambda<c\mu$, equivalently $\rho<1$.

Offered load is not generally a fraction: it can exceed $1$ when the arrival rate exceeds the rate of one server. Utilization per server is the offered load divided by the number of servers. Under stability, $0<\rho<1$. These are average-load quantities, not guarantees that every server is busy for exactly a fraction $\rho$ of every finite observation period.

Check Stability Before Using Steady-State Formulas

For this model, when the system contains $n$ customers, arrivals occur at rate $\lambda$, while service completions occur at rate $\min(n,c)\mu$. In particular, when $n\ge c$, all servers are busy and the total completion rate is $c\mu$. This explains why the comparison for stability is with the combined rate of all $c$ servers—not with the rate $\mu$ of one server.

Result (stability criterion): For an M/M/c queue with $\lambda>0$, $\mu>0$, and finite $c\ge1$, a stationary probability distribution exists exactly when $\lambda<c\mu$.

Proof. The birth-death balance equations in stationarity give $p_{n+1}=p_n\lambda/((n+1)\mu)$ for $0\le n<c$, and $p_{n+1}=p_n\lambda/(c\mu)=p_n\rho$ for $n\ge c$, where $p_n=\mathbb{P}(X_t=n)$ in stationarity. If $\rho\ge1$, the tail probabilities satisfy $p_{c+j}=p_c\rho^j$ for every integer $j\ge0$. The chain can reach state $c$ from state $0$, and the positive arrival and service rates make it irreducible, so a stationary distribution cannot assign zero probability to $c$. Thus $p_c>0$, and $\sum_{j\ge0}p_c\rho^j$ diverges when $\rho\ge1$. No stationary probability distribution can then be normalized. If $\rho<1$, the weights defined by the same balance equations have a finite geometric tail. Normalizing these weights gives a stationary probability distribution. Therefore stationarity exists exactly when $\rho<1$, which is equivalent to $\lambda<c\mu$. $\square$

The criterion is a condition for steady state, not merely a technical restriction on one formula. If $\lambda\ge c\mu$, steady-state probabilities and steady-state performance measures for the unlimited M/M/c queue cannot be obtained by substituting into stable-queue formulas.

Worked Example: Offered Load Is Not Utilization

Take $c=3$, $\lambda=4$ customers per unit time, and $\mu=2$ services per unit time per server. Find the offered load and utilization, and check stability.

The offered load is $a=\lambda/\mu=4/2=2$ Erlangs. The total service capacity is $c\mu=3\cdot2=6$ customers per unit time, so the utilization per server is

\rho=\frac{\lambda}{c\mu}=\frac{4}{3\cdot2}=\frac{4}{6}=\frac{2}{3}.

The equivalent calculation $a/c=2/3$ agrees. Since $\lambda=4<6=c\mu$, the queue is stable. In particular, the correct simplification is $4/6=2/3$, not $2/6$. The quantities have the expected interpretation: offered load is $2$ Erlangs, while utilization per server is $2/3$.

Check: $a/c=2/3$, and $\lambda/(c\mu)=4/6=2/3$. Both definitions give the same utilization.

Erlang C Is a Probability, Not a Delay

For a stable M/M/c queue, let $C$ denote the Erlang C probability that an arrival must wait for service. Under Poisson arrivals, PASTA (Poisson arrivals see time averages) identifies the stationary probability that an arrival finds all $c$ servers busy with the stationary probability that $X_t\ge c$. PASTA is a standard result; its proof is not needed for the calculations here.

Definition: For $a=\lambda/\mu$ and $\rho=a/c<1$, the Erlang C probability is $$ C=\frac{\dfrac{a^c}{c!(1-\rho)}}{\displaystyle\sum_{k=0}^{c-1}\frac{a^k}{k!}+\dfrac{a^c}{c!(1-\rho)}}. $$ It is dimensionless and lies between $0$ and $1$. It is the probability of a positive queueing delay for an arrival in steady state, not a duration of delay.

For the stable M/M/c model under the assumptions above, the waiting-time distribution gives a direct way to convert this probability into time-based measures. Write $W_q$ for the time from arrival until service begins. Its tail probability for $t\ge0$ is

\mathbb{P}(W_q>t)=C e^{-(c\mu-\lambda)t}.

Thus $\mathbb{P}(W_q>0)=C$, while $\mathbb{E}[W_q]=C/(c\mu-\lambda)$. The latter has units of time: $C$ is dimensionless and $c\mu-\lambda$ has units of inverse time. Conditional on an arrival having to wait, its mean wait is $1/(c\mu-\lambda)$. These are distinct summaries: the probability of any wait, the probability of exceeding a chosen time, and the expected wait need not be numerically similar.

Result (waiting-time tail): In a stable M/M/c queue, $\mathbb{P}(W_q>t)=C e^{-(c\mu-\lambda)t}$ for $t\ge0$. Consequently, $\mathbb{P}(W_q>0)=C$ and $\mathbb{E}[W_q]=C/(c\mu-\lambda)$.

Proof. Let $p_n$ be the stationary probability of $X_t=n$. The birth-death balance equations imply $p_{c+j}=p_c\rho^j$ for integers $j\ge0$. Therefore, conditional on finding at least $c$ customers in the system, the excess $j=X_t-c$ has probabilities $\mathbb{P}(X_t-c=j\mid X_t\ge c)=(1-\rho)\rho^j$. By PASTA, the same conditional distribution applies to an arrival that must wait. Given $j$ customers beyond the $c$ in service, the arrival waits for $j+1$ service completions. While it waits, all $c$ servers remain busy, so these completions occur at total rate $c\mu$. The waiting time, conditional on $j$, is therefore the time to $j+1$ events in a rate-$c\mu$ Poisson process; its tail at $t\ge0$ is $e^{-c\mu t}\sum_{k=0}^{j}(c\mu t)^k/k!$. Averaging over $j$, the conditional tail given that the arrival waits is $$ \begin{aligned} \sum_{j=0}^{\infty}(1-\rho)\rho^j e^{-c\mu t}\sum_{k=0}^{j}\frac{(c\mu t)^k}{k!} &=e^{-c\mu t}\sum_{k=0}^{\infty}\frac{(c\mu t)^k}{k!}\sum_{j=k}^{\infty}(1-\rho)\rho^j\\ &=e^{-c\mu t}\sum_{k=0}^{\infty}\frac{(c\mu t)^k}{k!}\rho^k\\ &=e^{-c\mu t}e^{\rho c\mu t} =e^{-(c\mu-\lambda)t}, \end{aligned} $$ where the geometric sum is $\sum_{j=k}^{\infty}(1-\rho)\rho^j=\rho^k$, and $\rho c\mu=\lambda$. The probability of having to wait is $C$, so multiplying this conditional tail by $C$ gives the stated unconditional tail. Setting $t=0$ gives $\mathbb{P}(W_q>0)=C$. Integrating the tail gives $\mathbb{E}[W_q]=\int_0^\infty C e^{-(c\mu-\lambda)t}\,dt=C/(c\mu-\lambda)$, since stability ensures $c\mu-\lambda>0$. $\square$

Worked Example: Compute the Probability and the Mean Delay Separately

Use the stable queue with $c=3$, $\lambda=4$, and $\mu=2$. Calculate Erlang C and the mean waiting time, checking each calculation.

From the earlier parameter calculation, $a=2$ and $\rho=2/3$. The terms in the Erlang C denominator are $\frac{a^0}{0!}=1$, $\frac{a^1}{1!}=2$, and $\frac{a^2}{2!}=2$. The remaining term is $\frac{a^3}{3!(1-\rho)}=\frac{8}{6(1/3)}=4$. Therefore

C=\frac{4}{1+2+2+4}=\frac{4}{9}.

The total service rate is $c\mu=3\cdot2=6$, so $c\mu-\lambda=6-4=2$. The mean queueing delay is not $C$; it is

\mathbb{E}[W_q]=\frac{C}{c\mu-\lambda}=\frac{4/9}{2}=\frac{2}{9}

time units. Check: the waiting-time tail gives $\mathbb{P}(W_q>0)=(4/9)e^0=4/9$, as required. Integrating that tail gives $(4/9)\int_0^\infty e^{-2t}\,dt=(4/9)(1/2)=2/9$, confirming the mean.

Worked Example: A Waiting Probability Is Not a Waiting-Time Measure

Let $c=2$, $\lambda=1$, and $\mu=1$. Find the probability an arrival waits, the probability its wait exceeds $\ln 2$, and the mean waiting time.

Here $a=1/1=1$, $\rho=1/2$, and $\lambda=1<2=c\mu$, so the queue is stable. The Erlang C numerator is $\frac{1^2}{2!(1-1/2)}=1$, and the denominator is $1+1+1=3$. Thus $C=1/3$. Since $c\mu-\lambda=2-1=1$, the tail formula gives

\mathbb{P}(W_q>\ln 2)=\frac{1}{3}e^{-\ln 2}=\frac{1}{6}, \qquad \mathbb{E}[W_q]=\frac{1/3}{1}=\frac{1}{3}.

The first quantity is a probability, the second is a probability for a specified threshold, and the third is a time. Check: at $t=0$, the tail formula gives $\mathbb{P}(W_q>0)=\frac13 e^0=\frac13=C$. Also, conditional on waiting, the mean is $1/(c\mu-\lambda)=1$; multiplying by the probability $1/3$ of waiting gives the unconditional mean $1/3$.

When the Formula Is Used Outside Its Conditions

Worked Example: An Unstable Queue Produces an Impossible Erlang C

Take $c=3$, $\lambda=7$, and $\mu=2$. The offered load is $a=7/2$, the total capacity is $c\mu=6$, and $\rho=7/6>1$. Thus $\lambda>c\mu$, and the queue has no stationary distribution.

If the stable Erlang C formula is incorrectly applied anyway, the numerator becomes $\frac{a^3}{3!(1-\rho)}=\frac{343/8}{6(-1/6)}=-343/8$. The finite-sum part of the denominator is $1+\frac72+\frac{(7/2)^2}{2}=1+\frac72+\frac{49}{8}=\frac{85}{8}$. The full denominator is therefore $\frac{85}{8}-\frac{343}{8}=-\frac{258}{8}$, and the formal ratio is

\frac{-343/8}{-258/8}=\frac{343}{258}>1.

This is not a probability. The calculation is not a new waiting probability; it is evidence that the stable formula has been used where its hypothesis fails. Check: $\rho=7/(3\cdot2)=7/6>1$, so the stability condition fails before any steady-state formula is evaluated.

Common mistake. Treating $C$ as a duration, or reporting $C$ as a probability that a customer waits longer than a particular threshold, loses essential information. For a threshold $t>0$, use $C e^{-(c\mu-\lambda)t}$; for the mean delay, use $C/(c\mu-\lambda)$. Another common error is to divide $\lambda$ by $\mu$ and call the result utilization per server. That ratio is offered load; the utilization is $\lambda/(c\mu)$.

A useful order of operations is to confirm the model assumptions, calculate $a$ and $\rho$ separately, check $\lambda<c\mu$, and only then use steady-state formulas. After calculating, check the units and interpretation: Erlang C must be a probability between $0$ and $1$, a mean waiting time must have units of time, and a threshold probability must specify the threshold.

Check Your Understanding

For each question, state the relevant condition or quantity before calculating.

An M/M/4 queue has $\lambda=6$ and $\mu=2$. Find the offered load and utilization per server, and check stability.
Why is an Erlang C calculation not valid for an unlimited M/M/c queue when $\lambda\ge c\mu$?
In a stable queue with $C=1/4$ and $c\mu-\lambda=2$ per unit time, find $\mathbb{P}(W_q>0)$ and $\mathbb{E}[W_q]$, including units for the mean.
For a stable M/M/c queue, write the formula for $\mathbb{P}(W_q>t)$ and explain what changes when $t=0$.
A calculation reports a steady-state probability of waiting equal to $6/5$. Give two checks that should be made before interpreting the result.