L = λW is not a model and not an approximation: it is one area counted two ways. Slice the graph of the number in system vertically and you get the time-average L; slice it horizontally and you get one bar per customer, whose total is n·W. The library proves it here, on a trace of five customers.
Core ideas
The proof is a picture, and once the picture is seen the formula stops needing to be remembered.
Each customer adds exactly 1 to N(t) for exactly as long as it is in the system, and nothing to it otherwise. So the region under the staircase can be added in vertical strips — one per unit of time, height N(t) — or in horizontal bars, one per customer, length dᵢ − aᵢ. The two totals are the same number because they are the same region.
There is no Poisson here, no memorylessness, no steady state, no service discipline and no independence. L = λW holds for one particular trace of five customers, for a queue served in reverse order, for a system that is not a queue at all. It is arithmetic about a window.
The identity compares an area that has accrued with stays that have finished. If someone is still inside at time T, the area has counted time that no completed stay has counted yet, and the two sides differ — not because the law failed but because the measurement stopped early.
The material
The definitions, the proof, the arithmetic on a five-customer trace, and the one way to get it wrong.
Over a window [0, T] in which n customers arrive at times aᵢ and leave at times dᵢ: N(t) is the number present at time t; L is the time-average of N(t), the area under it divided by T; λ is n/T; and W is the mean of the sojourn times dᵢ − aᵢ.
Take five customers arriving at 0, 1, 2, 5, 6 and leaving at 3, 4, 6, 8, 10. Nothing is assumed about why they arrived when they did or how long they were served. This is a record, not a model.
The two seventeens are not a coincidence and they are not a check that might have failed. They are the same area, added in two directions.
If every customer who arrives in [0, T] has also departed by T, then L = λW exactly, where L, λ and W are as defined above.
Customer i contributes 1 to N(t) for t in [aᵢ, dᵢ) and 0 elsewhere, so the total area under N(t) is Σᵢ (dᵢ − aᵢ): adding the strips vertically and adding the bars horizontally are two orders of summation over the same finite set of unit cells.
Divide that equality by T. The left side is area/T = L. The right side is Σᵢ(dᵢ − aᵢ)/T, and multiplying and dividing by n makes it (n/T)·(Σᵢ(dᵢ − aᵢ)/n) = λW. No property of the arrival times or the service times was used, only that every stay lies inside the window.
□
On the trace above: L = 17/10, λ = 5/10 = 1/2, W = 17/5, and λW = (1/2)(17/5) = 17/10, which is L to the digit. The lab prints all three as exact fractions for whatever trace you type into it.
Cut the same trace at T = 8, while the last customer is still inside. Now L = 15/8 = 1.875 and λW = (5/8)(13/4) = 65/32 = 2.03125. The two sides part company because λ counted five arrivals while W averaged the four stays that had finished. Extend the window to 10 and they agree again. Every report of “Little's Law not holding” on a real system is this: a measurement window with work still in flight, or three quantities measured at three different boundaries.
Arrivals at 0, 2, 4, 6, 8 and departures at 2, 4, 6, 8, 10: exactly one customer is present at every instant, so L = 1. With λ = 1/2 and W = 2, λW = 1. Here L is the fraction of time the system is occupied, which is the ρ of the previous lesson seen from this side.
What the identity does not give you is any information about the distribution of the waiting. W is a mean, and a mean says nothing about the p99 that Latency and the Tail measured. Two systems with identical L, λ and W can have entirely different tails.
The same counting argument works across any boundary you can draw, with λ the rate that crosses into it: a component, a thread pool, a whole datacentre. That generalisation is Operations Research, markov-chains-decisions-and-queues/littles-law, which cites this proof and carries the boundary version and the rule that the rate must be the one actually admitted — which is exactly what a lossy system later in this course will need.
Interactive
L = λW with no distribution in it anywhere
Method
Do this once by hand. It takes a minute and it is the difference between a formula and a fact.
Step up at each arrival, down at each departure. On a slotted trace it is enough to write the count in each slot in a row.
Sum the heights. On a unit-slot trace that is a sum of whole numbers, and it is the left-hand side of the identity.
dᵢ − aᵢ for each, then total them. This is the same number, and getting a different one means a time was mis-transcribed.
L = area/T, λ = n/T, W = total/n. Multiply λ by W and compare with L; as exact fractions they are equal, not close.
L = λW is a statement about a window, so the window is part of the answer. If work was still in flight at the end, say so — that is the one condition the proof needed.
The first column is the identity. The second is the only way to break it, done deliberately so that the break is recognisable later.
The second column did not break the law. It measured λ over five arrivals and W over four completions, which are two different populations, and the gap between 15/8 and 65/32 is exactly the stay that had not finished.
For a faded rehearsal, take arrivals 0, 1, 4 with departures 2, 5, 6. The supplied first move is that the horizon is T = 6 and every customer has left by it. Build the staircase, get the area both ways, and predict L before dividing anything.
You should be able to build N(t) from arrival and departure times, get the area by counting in either direction, produce L, λ and W as exact fractions, and diagnose a failed check as a window or a boundary problem rather than a broken law.
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