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A busy period is the interval from an arrival that finds the system empty until the next departure leaving it empty. It excludes the preceding idle waiting time. In a stable M/G/1 queue with arrival rate $\lambda$ and mean service time $m$, its mean is $m/(1-\lambda m)$.

In an M/G/1 queue, arrivals during the first service generate independent subtrees of work. Conditional on service time $S=s$, the number of descendants is Poisson with mean $\lambda s$. Thus $B\stackrel{d}{=}S+{\sum }_{j=1}^{N(S)}{B}_j$, with independent descendant busy periods. Conditioning and the Poisson generating function give the displayed transform equation wherever its moment-generating functions are finite. Taking expectations yields $\mathbb{E}B=\mathbb{E}S+\lambda \mathbb{E}S\mathbb{E}B$ in the stable regime.

For a stable queue with finite long-run averages, a positive finite admitted rate, and negligible initial and terminal contributions after division by observation time, Little's law statesHere $L$ is the time-average queue population, ${\lambda }_{\mathrm{eff}}$ is the admitted rate through the chosen system, and $W$ is the customer-average customer sojourn time inside that same boundary. Without abandonment, admitted rate equals completed-service queue throughput in this stable regime. It may differ from the offered arrival rate because of blocking. When customers abandon, count their admissions and exits too, and average their actual time inside; using only service completions would mismatch the population being measured.

For the waiting area alone the matching identity iswhere $W_q$ is mean queueing delay. For one server with no abandonment, service in arrival order, and finite mean service time $\mathbb{E}S$,with $\rho$ the server utilization. Mixing total population $L$ with waiting-only duration $W_q$ would count different portions of each customer's visit. Customers who start service immediately contribute zero to $W_q$, and must still be included in that customer average.

The strength of Little's law is that it does not require a Poisson process, an exponential distribution of service times, or service in arrival order. The sample-path area identity for Little's law explains why: counting customer-time across time slices or across customer visits gives the same area. A sufficient probabilistic setting is repeated independent and identically distributed empty-to-empty cycles with finite positive mean cycle duration, finite positive mean customer count, and finite mean population area; the renewal-reward theorem then supplies the long-run averages. Stability alone does not guarantee finite mean queueing delay: the Pollaczek-Khinchine mean formula shows why a finite second moment of service time matters for that mean in an M/G/1 queue.

For example, suppose a service desk receives six requests per hour but admits only four; measured mean customer sojourn time of admitted requests is half an hour. Its mean queue population is $4\cdot (1/2)=2$ customers. Multiplying the offered rate by that duration would incorrectly give three. Conversely, counting two customers on average and measuring admitted queue throughput of four per hour gives $W=2/4=1/2$ hour. This lets one estimate a hard-to-track duration from occupancy and completed-job counts.

John D. C. Little published a rigorous proof in 1961 under finite-mean, strict stationarity, and ergodicity assumptions on the arrival process. Those were assumptions of that proof, rather than a requirement that every useful form of the identity have random arrivals. ^[ref]. ^[ref].

An M/G/1 queue has Poisson process arrivals of rate $\lambda$, independent identically distributed service times of law $G$, one server, and here infinite waiting capacity. Use first-come-first-served service. Let $X_n$ be the number left just after the $n$th departure. During the next service, let $A_n$ be the number of arrivals. ThenConditional on the next service time $S_n=s$, $A_n$ is Poisson with mean $\lambda s$. If $X_{n-1}=0$, the server waits for one arrival to start service; that arrival is the customer being served, and is not counted in $A_n$. This constructs the embedded Markov chain.

An M/G/1 queue has Poisson arrivals of rate $\lambda$, independent identically distributed service times with a general law of mean $m$, and one server. During the first service time $S$, a mean $\lambda \mathbb{E}S=\lambda m$ customers arrive, and each initiates an independent descendant busy-period contribution. Thusso for traffic intensity $\rho =\lambda m<1$,For $\rho \ge 1$ the expected busy period is infinite.

For a stable first-come first-served M/G/1 queue, write $\rho =\lambda \mathbb{E}S<1$ and $\tilde{S}(s)=\mathbb{E}{e}^{-sS}$. The stationary waiting-time Laplace transform isWith finite second service moment, differentiating at zero gives the Pollaczek-Khinchine mean formula. These are the transform and mean forms of the same queueing identity.

Queueing theory studies how jobs accumulate while waiting for limited service: customers at a desk, requests at a computer, or calls competing for a connection. A queue model specifies how jobs arrive, their service times, the number of servers, the buffer size of a queue, and the service discipline. Its practical questions include the queue population, queueing delay, customer sojourn time, queue throughput, and server utilization.

Average spare capacity does not guarantee short waits. Even when the service rate of a queue exceeds the arrival rate, bursts of arrivals and unusually long service times can leave work waiting. The M/M/1 utilization and delay example shows that an 80 percent server utilization can turn a 12-minute mean service time into a 60-minute mean customer sojourn time. As capacity is approached, congestion grows sharply; reserving spare capacity can be much more valuable than its idle fraction suggests.

Two complementary ideas make queueing theory especially useful. Little's law converts a measured queue population and queue throughput into a mean customer sojourn time, using a counting identity rather than a particular arrival law. Poisson arrivals see time averages answers a different question: when do arriving customers see the same state frequencies as an observer sampling over time? That requires an exogenous Poisson process and a system that cannot anticipate its future arrivals. Neither result by itself determines all queueing delays; detailed models such as the M/M/1 queue and M/G/1 queue supply the remaining information.