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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].

Consider an M/M/1 queue with an exogenous Poisson process of arrivals of rate $\lambda$, one server, unlimited waiting space, first come first served service, and independent and identically distributed service times with an exponential distribution of rate $\mu$, independent of arrivals. There is no blocking or abandonment, so offered rate and admitted queue throughput coincide: ${\lambda }_{\mathrm{eff}}=\lambda$. The memoryless property makes the queue population a birth-death process. For $0<\lambda <\mu$, the stationary distribution of an M-M-1 queue isThis is a geometric distribution on the nonnegative integers. Its mean, together with Little's law, givesHere $L,W$ count the whole system, and $L_q,W_q$ count only waiting. These distinctions and formulas are derived in MIT 2.854, M/M/1 Queue notes, pages 1-2.

Take $\lambda =4$ customers per hour and $\mu =5$ customers per hour. The mean service time is $1/5$ hour, or 12 minutes, and the server utilization is $\rho =4/5=0.8$. Computing each quantity explicitly,Thus Little's law checks both boundaries: $4=4\cdot 1$ and $3.2=4\cdot 0.8$. The mean number in service is $L-L_q=0.8$, rather than one, because the server is idle 20 percent of the time. By PASTA, an arriving customer sees four customers already in the system on average. Under first come first served, the memoryless property gives mean residual service time $1/\mu$ for the customer in service; clearing those four customers takes $4/5=0.8$ hour on average, independently confirming the mean queueing delay.

Holding $\mu =5$ per hour fixed, increasing arrivals from $4$ to $4.5$ per hour raises server utilization from $0.8$ to $0.9$, but doubles mean customer sojourn time from one to two hours. At $\lambda =4.9$ per hour, server utilization is $0.98$, $L=49$, $W=10$ hours, $L_q=48.02$, and $W_q=9.8$ hours. Both mean population and mean duration diverge as $\lambda$ approaches $\mu$ from below. The formula describes an explosive congestion penalty near capacity even though all arrivals still eventually receive service at every fixed subcritical rate. At or above capacity there is no stationary distribution for this unlimited M/M/1 queue, so these finite stationary formulas no longer apply.

The inverse dependence of mean customer sojourn time on spare capacity, $W=1/(\mu -\lambda )$, makes server utilization a poor standalone measure of service quality. Adding capacity at a nearly saturated M/M/1 queue can sharply reduce queueing delay. Stanley B. Gershwin develops this capacity interpretation in Markov Processes and Queues, MIT 2.854, 2016, slides 71-72.

The capacity and quality targets alone do not determine flow counts: the per-flow traffic statistics must also be supplied. The following explicit model gives a computable sufficient region, with genuine upper bounds on both loss fractions.

Measure queue workload and buffer size in bits. Denote capacity by $C$, shared buffer size by $B$, voice delay limit by $D$, target voice loss by ${\ell }_v$, and target data loss by ${\ell }_d$, with $0<{\ell }_v,{\ell }_d<1$. Let there be $n_v$ voice flows and $n_d$ data flows. Model each class-$j$ flow, $j\in \{v,d\}$, by a stationary Compound Poisson process with packet rate ${\nu }_j$ and positive independent packet sizes $S_j$, and assume independence across flows and marks. Use a work-conserving preemptive-resume priority queue with a shared buffer, with voice priority and first come first served within each class. This preemptive or fluid approximation makes the predicted voice waiting time exactly $V/C$, where $V$ is the remaining voice queue workload just before arrival. Assume packet sizes are bounded by a known $s_{\max }<B$ and interpret the buffer gate as a packet-fit test; rejecting packets which do not fit is the packet version of the full-buffer rule.

For one flow the Compound Poisson process formula givesFor fixed-size packets of size $s_j$, substitute $e^{\theta s_j}$. Estimate the packet rate and size law from measurements or a declared traffic contract. Also require the strict stability inequalityDefine $B_0=B-s_{\max }>0$ and $K=CD$. Couple the switch with an unthinned infinite-buffer aggregate queue and an unthinned infinite-buffer voice-only queue, both driven by the same offered packets and both served at rate $C$. Under the preemptive service discipline, dropping arrivals can only decrease both the total queue workload and the voice queue workload. Thus total switch queue workload $Q$ and voice queue workload $V$ satisfy $Q\le W_T$ and $V\le W_V$ in the stationary coupling. A packet-fit loss implies $Q>B-S_j\ge B_0$, and a voice-delay rejection implies $V\ge K$.

Apply the continuous-time exponential workload bound to the two comparison queues. For any positive ${\theta }_T,{\theta }_V$ satisfyingthe stationary loss-event probabilities obeyThe second inequality is a union bound: voice can be lost either at the shared buffer or at its delay gate. Data congestion therefore contributes to voice loss despite its lower service priority. Poisson arrivals see time averages applies to each offered packet stream, whose future arrivals are independent of the prearrival state; the stationary event bounds are consequently bounds on the offered-packet loss fractions. Voice packets that are admitted have queueing delay below $D$, because later voice packets join behind them and data work cannot delay them in this model.

To turn the bounds into flow counts, allocate loss budgetsFor example, ${\eta }_B=\min ({\ell }_d,{\ell }_v/2)$ and ${\eta }_V={\ell }_v-{\eta }_B$ also work; equality with ${\ell }_d$ is harmless. The required sufficient effective-bandwidth admission region with a voice delay gate isIt ensures $p_d\le {\ell }_d$ and $p_v\le {\ell }_v$. A simple enumeration isdiscarding negative bounds and pairs failing strict mean-load stability. The formulas assume positive-rate nonzero-size flows. Optimize a chosen revenue or flow-count objective over these integer pairs. Searching over the budget split ${\eta }_B$ can enlarge the sufficient region. For a bit-fluid buffer gate with no packet-fit issue, $B_0$ can be replaced by $B$.

Real nonpreemptive packet service needs an additional residual-service allowance or an explicit priority queue workload analysis: a voice packet may have to wait for the data packet already in service. Burst-correlated voice sources also require richer descriptors, such as measured finite-time effective bandwidths ${\alpha }_j(\theta ,t)$ or a Markov-modulated source model. For such inputs, additivity still holds for independent flows, but the independent-increment supermartingale bound is not automatic; use a valid sample-path large deviation principle or a model-specific overflow bound, with prefactors or a safety margin when using asymptotics. If packet arrivals do not form a Poisson process, loss fractions must be assessed at arrival epochs rather than inferred solely from time-average queue workload probabilities. These qualifications explain both how to calculate an admission region from specified traffic statistics and why the five switch/QoS parameters by themselves cannot supply a numerical answer.

The queueing delay is the time between a customer's arrival and the start of its service, excluding its own service time. Under first come first served with service rate $C$, its queueing delay is the queue workload ahead of it divided by $C$. The customer sojourn time also includes the customer's own service. In a preemptive priority queue with a shared buffer, a voice customer's initial waiting time depends only on the voice queue workload already present.

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.

Let customer $k$ enter a queue at $a_k$ and leave at $d_k$, and count it in $N(t)$ for $a_k\le t<d_k$. On a finite observation interval $[0,T]$, the integral of the queue population is exactly the sum of the lengths of the overlaps of customer visits with that interval, as displayed above; $x_{+}=\max (x,0)$ is the positive part. This deterministic identity includes visits already in progress at time zero and visits still unfinished at time $T$. It needs no probability distribution for arrivals or service times. The area argument is also developed in Richard C. Larson and Amedeo R. Odoni, Urban Operations Research, section 4.4.

For a fully computed example, start empty and observe through $T=6$ minutes. Three customers arrive at $a=(0,1,4)$ minutes, begin service at $b=(0,2,4)$ minutes, and depart at $d=(2,3,5)$ minutes. These times describe a single first come first served server with service times $(2,1,1)$ minutes. Their customer sojourn times are $(2,2,1)$ minutes, and their queueing delays are $(0,1,0)$ minutes. The queue population is respectively $1,2,1,0,1,0$ on the consecutive intervals $[0,1),[1,2),[2,3),[3,4),[4,5),[5,6]$, soThe same area is $2+2+1=5$ when counted by customer. Hence the finite-window averages satisfyThe waiting area is one customer-minute, giving $L_q=1/6$ and $W_q=1/3$ minute; $(1/2)(1/3)=1/6$. The server works for four minutes, so its server utilization is $4/6=2/3$ and $L-L_q=5/6-1/6=2/3$. Repeating this six-minute pattern indefinitely makes all these values long-run averages, without a Poisson process of arrivals.

An observation window that cuts visits short requires the overlap terms, rather than full customer sojourn times. Starting empty, let $A(T)$ count admissions before $T$ and write $w_k=d_k-a_k$. ThenFor the example cut at $T=2.5$ minutes, the area is $1+2+0.5=3.5$ customer-minutes, whereas the full durations of the two admitted customers sum to $4$; the remaining half-minute of the second visit is $R(2.5)=0.5$. A precise sufficient sample path condition for the limiting Little's law is $A(T)/T\to {\lambda }_{\mathrm{eff}}\in (0,\infty )$, $A(T{)}^{-1}{\sum }_{a_k<T}{w}_k\to W<\infty$, and $R(T)/T\to 0$. Dividing the identity by $T$ then gives $L={\lambda }_{\mathrm{eff}}W$. Equating admission and departure rates additionally uses $N(T)/T\to 0$, by the balance $N(T)=A(T)-D(T)$ for an initially empty queue, where $D(T)$ counts all departures.