Little's law: throughput survives a boundary move, cycle time does not

Scale

The four fixed panels share one scale, 0 to 15 working days across and 0 to 20 cases a day up. The live chart above rescales with what you type.

Assumptions

Long run averages; the flow did not drift; one clock per boundary.

Model version

1.0

The formula, in words

• Three averages, and any two of them settle the third. How much work is in progress, how fast work is finishing, and how long one piece of work takes are not three independent facts. Over a long enough stretch, work in progress equals throughput multiplied by cycle time, and it holds for a whole route, one stage, a queue, or a shipment in transit. It needs no particular pattern of arrivals and no absence of variation. It is an accounting identity over three long run averages, which is exactly why it is useful for finding a number that cannot be true.
• The units are the check, and they are the reason one control sets both numbers. Cases a working day multiplied by working days leaves cases. Cases a working day multiplied by calendar days leaves nothing at all, because the two clocks count different things and the day in one is not the day in the other. On this page one control fixes both of a boundary's numbers, so that multiplication cannot be built. Across two boundaries the page will not convert, because turning working days into calendar days needs the length of your working week and that is a fact about you, not about arithmetic. Weeks and calendar days do convert, at seven days to the week, which is a definition rather than an assumption.
• All three numbers belong to a boundary, and only one of them survives moving it. Count the work in progress inside a narrower stretch and there is less of it. Measure the cycle time across a narrower stretch and it is shorter. Measure the throughput across a narrower stretch and, over a period where nothing is accumulating between the two, it is the same, because everything that finishes inside the wider boundary passed through the narrower one. That is why two people can quote two cycle times for the same work and both be right, and why neither number means anything until somebody says where it starts and stops.
• The gap between two nested counts is a multiplication, not a mystery. When both boundaries carry the same throughput, the difference between the work in progress they imply is that throughput times the time the narrower boundary does not count. On the shipped numbers that is 12 cases a working day across 13 working days, and 156 cases. Those cases are real, they are somewhere, and no report on the page mentions them.
• A zero queue is not zero work in progress. Where enough capacity runs in parallel that nothing ever waits, the work in progress is still the arrival rate times the time each piece is worked on, and it is a real count of things that have started and not finished. Twenty pieces of work arriving an hour, each taking half an hour, leaves ten in progress at any moment with nothing waiting behind anything. A boundary drawn around active work only will always return a smaller number than a boundary drawn around the whole route, and neither of them is the wait.

A worked example

The chain as shipped

Five stages between six edges: waiting to start, being worked, waiting to check, being checked, waiting to leave. The chain carries no durations and no counts. It is an order, not a measurement.

Boundary A, the whole chain

Called from the day it arrives to the day it leaves. Two numbers typed: 12 cases finish a working day, and one case takes 15 working days from end to end.

$12$ cases a working day, $15$ working days, $180$ cases. The working days cancel and cases remain, so the answer is a count of cases and not a rate. Boundary A implies 180 cases in progress.

Boundary B, one stage

Called the time my team has it open, running from work starts to work finishes. Two numbers typed: the same 12 cases a working day, and 2 working days. $12 \times 2 = 24$ cases. Boundary B implies 24 cases in progress.

The gap, two ways, and they agree

Directly: $180 - 24 = 156$ cases. Through the identity: $12$ cases a working day, $(15-2)$ working days, $156$ cases. The two routes agree because the throughput is the same on both boundaries, and that agreement is the check.

Where the 156 cases are

They are in the four stages boundary A covers and boundary B does not: waiting to start, waiting to check, being checked and waiting to leave. How they divide among those four is not on this page, because nothing was typed that could decide it. What is on the page is that there are 156 of them.

The share of the clock

Boundary B covers $\frac{2}{15} = 13.3\%$ of the elapsed time. The other 86.7% of the elapsed time is outside the boundary the shorter report is measured on, and the throughput, being 12 cases a working day on both, is the only one of the three numbers that does not notice.

The rectangles, and why the picture is the arithmetic

Draw a rectangle 15 working days wide and 12 cases a working day tall. Its area is 180, in cases, because the two units multiply that way and for no other reason. Draw boundary B's rectangle from the same corner, 2 wide and 12 tall, area 24. Because the heights are equal the second rectangle sits inside the first sharing three of its four edges, and what is left is a strip 13 wide and 12 tall. 156 cases, drawn.

What a second throughput does to the picture

Take the same boundaries and let boundary B's report count 15 cases a working day rather than 12, which is chip 1. Boundary B now implies $15 \times 2 = 30$ cases. The two rectangles no longer share a top edge, the strip between them is not a rectangle, and the subtraction $180 - 30 = 150$ is arithmetic without a meaning: it is the difference between two counts taken on two different flows. On these numbers the page draws no shaded difference at all, and that absence is the finding.

What two calendars do to it

Chip 2 leaves boundary A at 12 a day and 21 days but moves it to the calendar, so $12 \times 21 = 252$ cases, while boundary B stays at 12 a working day and 2 working days for 24 cases. Both multiplications are sound. Neither answer is wrong. The subtraction $252 - 24$ is still refused, because saying whether 12 a calendar day is more or less work than 12 a working day needs the number of working days in the week, and nobody typed one.

The fixed panels, one parameter, four values

Boundary A holds at 15 by 12 in all four. Boundary B holds at 2 working days and takes 8, 12, 16 and 20 cases a working day, implying 16, 24, 32 and 40 cases. Only the second panel has two rectangles that share a top edge, and it is the only one of the four in which subtracting the two counts describes a stretch of chain.

What the identity will not do

It will not say whether 180 cases in progress is a lot. It will not say which of the five stages holds most of them. It will not say whether 15 working days was a promise or an outcome. It says that 12 and 15 imply 180, and that if the floor holds 60, then one of the three numbers is measured on something other than the boundary its owner thinks it is. That is the whole of the claim, and it is enough to end most arguments about whose figure is right.

What this does not model

Whether the flow drifted during the period the three numbers were taken from, which the arithmetic assumes and cannot test. Whether the period opened or closed on an unusual backlog, which pushes the observed cycle time away from the long run figure in a direction the page cannot sign. Whether cases return to a stage they have already passed, which makes one case consume several passes through the route, so a count of arrivals and a count of visits are two different populations and only one of them belongs in the throughput. Whether a case in one week is the same kind of thing as a case in another, since a mix that moves turns a common unit into an average of unlike items. Whether the boundary is crossed once, since anything that leaves and comes back is counted twice by one of the three numbers and once by another. How the work divides among the stages, which this page refuses to guess. What the cycle time's spread is, which the average conceals entirely and which decides how often the work is late even when the average is comfortable. And whether the cycle time was quoted or observed, since a promise and an outcome are two different quantities and a measured service figure rises when the quote is loosened, without any work moving faster.

Check it yourself

Boundary A takes 180 cases in progress and 12 cases a working day and implies a cycle time of 15 working days; boundary B takes 24 cases and 2 working days and implies a throughput of 12 cases a working day. Every figure on the page is identical to the shipped state, reached from four different typed numbers, because the identity does not care which two you have. Only the question they answer did.