One server, memoryless arrivals at λ, memoryless service at μ, room for everyone. Balancing the flow across the cut between n and n + 1 gives λπₙ = μπₙ₊₁, so the states decay geometrically and every quantity the course needs — L, L_q, W, W_q and P(N > k) — follows from a geometric series.
Core ideas
The derivation is three lines and a geometric series. What takes care is knowing which system it describes and which it does not.
Draw a line between the states with n in the system and those with n + 1. In the long run the system crosses that line upward exactly as often as downward, so λπₙ = μπₙ₊₁. That single conservation statement, applied at every n, determines the whole distribution.
πₙ = ρⁿπ₀ follows immediately, and the probabilities must sum to one, so π₀ = 1 − ρ by Algebra's infinite-geometric-series — which also needs ρ < 1, and that is the same condition as the queue settling at all. The mean L = Σ nπₙ is the same series differentiated.
L is the number in the system and L_q the number waiting, and the difference is not a rounding detail: L − L_q = ρ exactly, the probability that the server is busy. At ρ = 4/5 the system holds 4 on average and 16/5 of them are queueing.
The material
The derivation, the worked arithmetic at ρ = 4/5, and a careful statement of what this formula is and is not checked against.
Arrivals are memoryless at rate λ; service times are memoryless with rate μ, so the mean service time is S = 1/μ; there is one server, unlimited room, and customers are taken in order. N is the number in the system, including the one in service, and πₙ = P(N = n) in the long run. The model settles precisely when ρ = λ/μ < 1.
The notation S = 1/μ is fixed here and used by the rest of this path: it is the mean service time, the thing a trace measures directly, and writing the results in terms of it is what makes the knee legible in the next lesson.
For ρ = λ/μ < 1, πₙ = (1 − ρ)ρⁿ for every n ≥ 0.
Consider the boundary between {0, 1, …, n} and {n + 1, n + 2, …}. Every crossing upward is an arrival that happens while the system holds n, at rate λπₙ; every crossing downward is a completion while it holds n + 1, at rate μπₙ₊₁. In the long run the two counts differ by at most one, so the rates are equal: λπₙ = μπₙ₊₁.
Therefore πₙ₊₁ = ρπₙ, and by induction πₙ = ρⁿπ₀. The probabilities sum to one, so π₀ Σ ρⁿ = 1; the series converges exactly when ρ < 1, to 1/(1 − ρ), giving π₀ = 1 − ρ and πₙ = (1 − ρ)ρⁿ.
The mean is L = Σ nπₙ = (1 − ρ) Σ nρⁿ = (1 − ρ)·ρ/(1 − ρ)² = ρ/(1 − ρ), using the same geometric sum in its Σ nrⁿ = r/(1 − r)² form.
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The tail falls out of the same geometric shape: P(N > k) = ρ^(k+1), because being above k means the chain climbed k + 1 steps each costing a factor ρ. L_q is L minus the one in service, L − ρ, which rearranges to ρ²/(1 − ρ). And W comes from Little's Law rather than from any new argument: W = L/λ, which simplifies to 1/(μ − λ) and hence to S/(1 − ρ).
Every one of those is an exact fraction, and the two checks in the middle are free: L − L_q must be ρ, and W − W_q must be S. If either fails, an arithmetic slip has happened rather than a modelling one.
It is tempting to say the slotted simulation of “Why Queues Form at ρ < 1” confirms these closed forms. It does not, and the claim would be false. That simulation is a discrete-time chain: Bernoulli arrivals with probability p per slot, geometric service with parameter q, late arrivals. At p = 2/5 and q = 1/2 its server utilisation is p/q = 4/5, exactly as it should be, and its exact mean number in system is 12/5. ρ/(1 − ρ) at ρ = 4/5 is 4. The continuous formula overstates that system by a factor of 5/3.
The relation between the two is a limit, not an agreement. Hold λ and μ fixed and let the slot Δ shrink with p = λΔ and q = μΔ: the discrete chain's mean climbs toward ρ/(1 − ρ), and the gap closes in proportion to Δ.
The lab prints that sequence beside the closed form rather than claiming a single slot size agrees with it. The exact finite-slot answer at each step is Operations Research's, markov-chains-decisions-and-queues/queues-in-discrete-time; what is shown here is that it converges to this lesson's formula and how fast.
The same reconciliation runs the other way. The balance argument above is the steady state of a birth–death chain, and Operations Research builds that chain properly in markov-chains-decisions-and-queues/poisson-arrivals-and-the-mm1-queue, recovering this distribution as its continuous-time limit. System Design owns the closed forms — this lesson is the library's reference for L, L_q, W, W_q and P(N > k) — and nothing in the two treatments contradicts the other.
Four assumptions are doing the work, and the rest of this course takes each of them away in turn: memorylessness (“Variability: the Kingman Approximation” prices getting it wrong), the single server (“Many Servers: Pooling, and Erlang C”), the unlimited room (“Bounded Queues and Loss”), and the steady state itself (“Transient Overload and Draining the Backlog”).
Interactive
Flow balance across a cut, and the shrinking slot that reaches it
Method
Get ρ first. Every other quantity is a function of it alone.
Above one there is no stationary distribution to compute, and the formulas return negative numbers rather than refusing. S = 1/μ is worth writing down at the same time.
π₀ = 1 − ρ and πₙ = (1 − ρ)ρⁿ. The tail P(N > k) = ρ^(k+1) needs no sum: it is one power.
L = ρ/(1 − ρ), and L_q = L − ρ. Doing it in that order makes the relationship between the two visible and gives a free check.
W = L/λ and W_q = L_q/λ. Check that W − W_q = S; if it does not, the arithmetic went wrong rather than the model.
Poisson arrivals, exponential service, one server, unlimited room. Naming the weakest one is what turns a computed number into an estimate with a direction of error.
A service taking one second on average, offered work four fifths as fast as it can do it. Every figure below is exact.
Read the last two lines together. The server takes one second over a request, and a request spends five seconds in the system: four of those five seconds are spent waiting behind other people, at a utilisation most capacity plans would call comfortable.
And there is a 41% chance of finding four or more in the system. The mean of 4 is not a typical value in any useful sense — the distribution is geometric, so it has no characteristic scale below its tail.
For a faded rehearsal, take λ = 9/10 with μ = 1. The supplied first move is ρ = 9/10, so π₀ = 1/10. Predict L, W and P(N > 3) before computing them, and notice how much moved for one tenth of extra load.
You should be able to derive πₙ = (1 − ρ)ρⁿ from the cut, compute L, L_q, W, W_q and P(N > k) as exact fractions at a rational ρ, and state precisely why the slotted chain of “Why Queues Form at ρ < 1” does not produce ρ/(1 − ρ) at a finite slot size.
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