Last updated: 2026-07-01

M/M/c Queue

The M/M/c queue extends the single-server model to several identical servers sharing one waiting line. It is the workhorse of staffing and capacity planning: from the arrival rate, the per-server service rate, and the number of servers, it predicts how often requests have to wait, how long they wait, and how long the queue grows. The calculator returns all of these along with the per-server utilization.

What the name means

The notation follows Kendall's classification. The first M says arrivals are Markovian — they follow a Poisson process, so each arrival is independent. The second M says service times are exponentially distributed. The c is the number of servers. So M/M/c is "random arrivals, random service, c servers, one shared unlimited queue, served in order." With $c = 1$ it collapses to the ordinary M/M/1 queue.

Utilization and offered load

Two quantities drive everything. The offered load (in Erlangs) is the work arriving per unit of service capacity:

$a = \dfrac{\lambda}{\mu}$

and the utilization is that load spread across the servers:

$\rho = \dfrac{a}{c} = \dfrac{\lambda}{c\,\mu}$

where $\lambda$ is the arrival rate, $\mu$ the per-server service rate, and $c$ the number of servers. The system is stable only when $\rho < 1$; otherwise requests arrive faster than the servers can clear them and the backlog grows without bound.

The Erlang C formula

The key output is the probability that an arriving request finds every server busy and must wait. This is the Erlang C formula:

$P_{\text{wait}} = \dfrac{\dfrac{a^{c}}{c!\,(1-\rho)}}{\displaystyle\sum_{n=0}^{c-1}\dfrac{a^{n}}{n!} + \dfrac{a^{c}}{c!\,(1-\rho)}}$

Once the wait probability is known, the steady-state averages follow:

$L_q = \dfrac{P_{\text{wait}}\,\rho}{1 - \rho}, \qquad W_q = \dfrac{L_q}{\lambda}$

$L = L_q + a, \qquad W = W_q + \dfrac{1}{\mu}$

Here $L_q$ is the average number waiting, $W_q$ the average wait before service, $L$ the average number in the system, and $W$ the average response time.

Worked example

A service desk receives 5 tickets per minute ($\lambda = 5$) and each of its 3 agents resolves 2 per minute ($\mu = 2$, $c = 3$). The offered load is $a = 5/2 = 2.5$ Erlangs and the utilization is $\rho = 2.5/3 \approx 0.833$. The Erlang C formula gives:

So about 70% of arriving tickets wait, the queue holds roughly 3.5 tickets on average, and the typical wait is about 0.7 minutes before an agent is free. Adding a fourth agent at the same load drops the wait probability and queue sharply, which is the kind of trade-off this model is built to expose.

Pooling beats splitting

A central result of multi-server queueing is that one shared line in front of $c$ servers outperforms $c$ separate single-server queues with the same total capacity. Pooling smooths out the moments when one server happens to be idle while another is swamped, so the average wait falls. This is why a single ticket queue feeding many agents beats assigning each agent a private inbox. The same averaging idea underlies Little's Law Calculator, which ties together concurrency, throughput, and latency without assuming any particular distribution, and the single-server case is covered by M/M/1 Queue Calculator.

The M/M/c model assumes exponential service times and unlimited queue space, which real systems rarely match exactly. Treat its predictions as a planning guide: when service times are more regular than exponential, actual waits are usually shorter than the model suggests.