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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.

An open migration process has finitely many colonies and state $n\in {\mathbb{N}}^J$. External arrivals to colony $j$ form independent Poisson processes with rates ${\nu }_j$. When $n_j=k>0$, individuals leave that colony at total rate $g_j(k)>0$, with $g_j(0)=0$. A departing individual moves to $k$ with probability $p_{jk}$ or exits with probability $p_{j0}=1-{\sum }_k{p}_{jk}$. The routing choices and all arrival and service randomness are independent. The routing matrix $P=(p_{jk})$ is substochastic and transient: every individual eventually exits, equivalently its spectral radius is less than one. Self-routing is allowed, but creates no state change.

The traffic equations of an open migration process, using row vectors, areThe inverse matrix exists because ${\sum }_{m\ge 0}{P}^m$ converges; its $(k,j)$ entry counts expected visits to $j$ by an individual entering at $k$. Thus ${\alpha }_j$ includes both external arrivals and internal migrations. Restrict first to the accessible colonies with ${\alpha }_j>0$.

SetIf each $Z_j<\infty$, the candidate product-form stationary distribution of an open migration process isHere is a global balance for a continuous-time Markov chain proof, which does not incorrectly assume that the routing is reversible. The weight ratio is $\pi (n+e_j)/\pi (n)={\alpha }_j/g_j(n_j+1)$. After dividing the incoming balance equation at $n$ by $\pi (n)$, external arrivals contribute ${\sum }_{j:n_j>0}{\nu }_j{g}_j(n_j)/{\alpha }_j$, exits contribute ${\sum }_j{\alpha }_j{p}_{j0}$, and migrations from $k$ to $j\ne k$ contribute ${\sum }_{j:n_j>0}{\sum }_{k\ne j}{g}_j(n_j){\alpha }_k{p}_{kj}/{\alpha }_j$. Combining the first and third contributions using the traffic equations of an open migration process yields ${\sum }_j{g}_j(n_j)(1-p_{jj})$. Summing those same traffic equations of an open migration process gives ${\sum }_j{\alpha }_j{p}_{j0}={\sum }_j{\nu }_j$. The incoming rate is therefore exactly the outgoing rateThis proves stationarity. With finite arrival rates, finite service rates in each state and transient routing, the process is nonexplosive: only finitely many individuals enter in a finite interval, and each makes finitely many routing transitions before exit, almost surely. On the accessible irreducible Markov chain, a finite normalizer gives the unique stationary distribution and positive recurrence. Colonies with zero effective arrivals are empty in this equilibrium. For a nonempty accessible colony, a divergent $Z_j$ prevents normalization of the equilibrium weights.

For the single-server specialization, $g_j(k)={\varphi }_j$ for $k>0$. Each colony has the stationary distribution of an M-M-1 queue:requiring ${\varphi }_j>{\alpha }_j$ on every colony with positive traffic. The service rates do not change the traffic equations of an open migration process.

For positive ${\alpha }_j$, write $B=F-{\sum }_j{\alpha }_j$ and $s_j={\varphi }_j-{\alpha }_j$. A stable allocation requires $B>0$ and $s_j>0$, with ${\sum }_j{s}_j=B$. The Cauchy-Schwarz inequality givesEquality holds precisely when $s_j$ is proportional to $\sqrt{{\alpha }_j}$. Hence the unique optimal service-capacity allocation in an open migration process and its mean population areThere is no stable allocation when $F\le {\sum }_j{\alpha }_j$ and some traffic is positive. If some ${\alpha }_j=0$, those empty colonies contribute no population; allocate the spare capacity to the positive-traffic colonies by the same rule. Allowing zero capacity at an unused colony attains this optimum. If strictly positive ${\varphi }_j$ are required even there, the displayed value is only an infimum, approached as their capacities tend to zero. If every ${\alpha }_j=0$, the equilibrium is empty and any admissible allocation minimizes its mean population at zero.

In Kendall notation, the first two letters $M$ mean Poisson arrivals and exponential service times. An M/M/1 queue has arrival rate $\lambda$, one server of rate $\mu$, and queue-length transitionsIt has a stationary distribution of an M-M-1 queue exactly whenand then

An M-M-infinity queue has infinitely many servers, so every customer begins service immediately. Its occupancy transitions areThe detailed balance for a birth-death process equations giveThus for every $\lambda ,\mu >0$ the stationary distribution of an M-M-infinity queue exists and isthe Poisson distribution of mean $\rho$. The Burke theorem for an M-M-infinity queue states that in this stationary regime the departure process is a Poisson process of rate $\lambda$; equivalently, time reversal turns departures into arrivals. The past departure process is also independent of the number in the system at the observation time.