A queue is a chain on 0, 1, 2, … whose only moves are up one and down one. Draw a line between n and n+1 and the chain crosses it upward exactly as often as it crosses back down in the long run — it cannot do otherwise, because to cross up twice running it would have to be on both sides at once. That single equation πₙλₙ = πₙ₊₁μₙ₊₁ gives every ratio, and chaining them from state 0 and dividing by the total gives π with no matrix anywhere in it. Because cutting is the step you are being asked to believe, the page also builds the full balance system and solves it by elimination, and compares the two as fractions rather than to a tolerance.
Core ideas
Three ideas. The first is the equation, the second is why it is legitimate, and the third is why this course meets it before any named queue.
Between two consecutive upward crossings of the line between n and n+1 there must be a downward crossing, since the chain moves one state at a time and cannot be on both sides at once. So in the long run the two rates are equal: the rate up is πₙλₙ and the rate down is πₙ₊₁μₙ₊₁, and those are the same number. Nothing about the sizes of the rates entered that argument.
The full balance equations — flow into each state equals flow out, with one dependent row replaced by Σπ = 1 — are built and solved by exact Gauss-Jordan on the same page. The comparison is Requ on fractions, not a tolerance, so a cut argument that were merely a convenient shortcut would show up as a disagreement rather than as two decimals that look alike.
That is the reason to meet the cut before any named model. M/M/1 is the case where every λₙ is the same number; s servers is μₙ = min(n, s)μ; a finite room is the same chain stopped; a customer who balks is λₙ falling with n. The moment a rate depends on the state, the closed forms stop applying and this equation does not.
The material
The chain, the crossing argument, the product that comes out of it, the independent check, and Little's Law on the rate that actually gets in.
States 0, 1, 2, …, N. From state n the chain moves to n + 1 at rate λₙ (an arrival) and to n − 1 at rate μₙ (a service completion), and nowhere else. There is no arrival out of the top state and no service out of state 0.
The lab takes two lists: one arrival rate per state starting at 0, and one service rate per state starting at 1. A chain with N arrival rates has states 0 to N.
In the long run, for every n, πₙ λₙ = πₙ₊₁ μₙ₊₁. Hence πₙ₊₁ = πₙ · λₙ/μₙ₊₁, and chaining from π₀ gives πₙ as a product of ratios, fixed by Σπ = 1.
Draw the line between n and n + 1. The chain moves one state at a time, so every upward crossing of that line is followed by a downward crossing before the next upward one: it cannot cross up twice in a row without being on both sides at once.
So over a long run the numbers of crossings in the two directions differ by at most one, and the long-run rates are equal. The rate of upward crossings is the long-run fraction of time at n times the arrival rate there, which is πₙλₙ; the rate downward is πₙ₊₁μₙ₊₁. Setting them equal and solving for πₙ₊₁ gives the ratio.
□
The argument used the one-step-at-a-time structure and nothing else. It did not assume the rates were constant, that the chain was infinite, that arrivals were Poisson, or that anything was stable. That is why one function serves every model on this material: there is one chain and several readings of it rather than several models.
The two routes share no arithmetic. The left-hand one multiplies ratios and divides by a total; the right-hand one builds the equation πₙ(λₙ + μₙ) = πₙ₋₁λₙ₋₁ + πₙ₊₁μₙ₊₁ for every state, drops one by name because the equations are dependent for exactly the reason they were dependent on a general chain, puts the normalisation in its place, and eliminates. They agree entry for entry, and the page says so with an exact equality rather than a comparison of decimals.
Dropping a different equation changes nothing, as it did not on a general chain: all four choices give (4/19, 6/19, 6/19, 3/19). The dependency is the same one — sum the balance equations and every term cancels — and so is the repair.
L = Σ n πₙ is the mean number in the system, and it comes straight out of π. W is L divided by a rate, and the rate is not always λ: on a chain whose arrival rate depends on the state, the long-run rate of admissions is λ_eff = Σ πₙ λₙ. On the repairer example that is (4/19)(3) + (6/19)(2) + (6/19)(1) = 30/19, so W = (27/19)/(30/19) = 9/10. The identity is the one “Little's Law from a Trace” proves by counting; what this page contributes is the right rate to divide by.
Set λ = 4, 4, 4, 4, 4 and μ = 3, 6, 6, 6, 6. The service rate out of state 1 is 3 and out of every higher state is 6: a second server that comes into use once there are two customers and then has nobody else to add.
The ratios are 4/3 for the first cut and 2/3 thereafter, giving π = (243, 324, 216, 144, 96, 64)/1087. The mode is state 1 rather than state 0, which is what the single early ratio above 1 does, and no formula was needed to see it.
Set λ = 5, 5, 5, 5 and μ = 3, 3, 3, 3. Every ratio is 5/3, so π rises geometrically to π = (81, 135, 225, 375, 625)/1441 with its mass at the top.
The page flags that at the top of the chain arrivals are at least as fast as service, so the untruncated chain would have no distribution at all. What you are looking at is a finite chain and its answers are answers about that chain, not about an unbounded queue. Which model you meant is a modelling question and the page will not decide it for you.
One restriction: every service rate has to be strictly positive. A zero μ closes the chain off — nothing can come back down from that state — so the cut above it has no solution with probability on both sides, and the lab refuses rather than dividing by zero. This mode also stops at eight cuts, because it prints every cut equation and the whole balance system; the constant-rate readings take chains as long as you like.
Interactive
cut the chain, balance the crossings, and check it against the full linear system
Method
Four steps and no matrix. The last one is the one that decides whether your W means anything.
One arrival rate out of each state from 0 upward, one service rate out of each state from 1 upward, the same count in each. Getting the offset wrong by one is the standard error here and it produces a plausible distribution for a different chain.
πₙ₊₁/πₙ = λₙ/μₙ₊₁. Multiply them up from π₀ = 1 to get the unnormalised weights. Keep them as fractions: the product of ratios is where a decimal starts losing digits, and the whole point of the second route is a comparison that has to be exact.
Add the weights and divide. This is the only place normalisation enters, and it is the same replacement for the same dependency that a general chain needed: the balance equations fix π up to a scale and nothing more.
L = Σ n πₙ. For W, divide by λ_eff = Σ πₙ λₙ, not by λ. On a chain with constant rates and no losses those coincide; on every other chain on this material they do not, and the difference is not small.
Three machines, each failing at rate 1, and one repairer working at rate 2. With n already broken there are 3 − n left to fail, so the arrival rate falls: λ = 3, 2, 1. The repairer does not speed up: μ = 2, 2, 2.
Two things are being claimed at once here and it is worth separating them. The first is that the cut equations are correct, and the check for that is the balance system: two routes, no shared arithmetic, the same four fractions. The second is that they are easy, and the check for that is the line count: three multiplications and one division against a four-by-five elimination.
The λ_eff line is the one that generalises. It is 30/19 ≈ 1.579, not 3, because the arrival rate is 3 only while nothing is broken. Dividing L by 3 would give 0.474 and would be a statement about a queue in which all three machines can fail while three are already broken.
For a rehearsal, change the repairer's rate from 2 to 3 and predict the direction of every figure before recomputing. The supplied first move is that each ratio is multiplied by 2/3, so all the weight shifts towards state 0. Say what that does to L, to λ_eff and to W — and note that two of those move in opposite directions, which is why W is worth computing rather than guessing.
You should be able to state the crossing argument, write one cut equation per boundary and chain them into an unnormalised product, normalise, build the full balance system and recognise its dependency as the same one a general chain has, compute L and λ_eff from π, and divide by the right one of them.
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