Part III · Chapter 5 · Evidence · lesson 5
How many of these should exist by now?
14 min
Builds on Designed, or running? Decide for each one
Part III · Chapter 5 · Evidence · lesson 5
14 min
Builds on Designed, or running? Decide for each one
A control that runs once a quarter. A period that opens on the first of January and closes on the thirtieth of September. Two hundred and seventy-three days.
Before anything else: how many completed instances of that control does the calendar say should exist by the end of it?
Predict
Two different questions, and a bundle can fail either one.
How many? Whole cadence periods inside the window. A quarterly control over nine months owes three; a monthly control over a year owes twelve, not twelve and a sixth. One missing instance is one promise the cadence made and the register did not keep.
How much of the window is spoken for? Each artefact speaks to one cadence period. Two quarterly artefacts speak to a hundred and eighty-two days of a two-hundred-and-seventy-three-day window, and say nothing at all about the rest.
The two can disagree. Four artefacts for three expected periods looks like a surplus and is usually the same quarter filed twice, which fills the arithmetic without filling the gap.
Below is the case: the nine-month window, a quarterly cadence, two artefacts that survive and are findable, and a hundred and eighty days between the two of them. Read the summary before touching anything.
Evidence coverage
Hypothetical model, not a forecast
How much of an observation window the recorded artefacts actually speak to, and how long the longest stretch is that nothing speaks to at all.
It assumes
It ignores
Leave the cadence at ninety-one days. Raise the artefacts recorded until the model reports artefacts beyond what the cadence expects, and no days of the window left unspoken for.
Days between scheduled operations of the control: 7 weekly, 30 monthly, 91 quarterly — the same figures the consistency engine uses, so a lesson's arithmetic and the referee's agree.
Operations with an artefact that survives and is findable.
| Window length, in days | Days an artefact speaks to | Days nothing speaks to |
|---|---|---|
| 7 | 7 | 0 |
| 14 | 14 | 0 |
| 30 | 30 | 0 |
| 60 | 60 | 0 |
| 90 | 90 | 0 |
| 120 | 120 | 0 |
| 180 | 180 | 0 |
| 270 | 182 | 88 |
| 273 | 182 | 91 |
Not there yet — keep moving the controls.
Three owed, two held, one missing. A hundred and eighty-two days of the window spoken for and ninety-one days — a third of it — spoken for by nothing at all.
Then the figure that is not about counting: the longest stretch between consecutive artefacts is a hundred and eighty days, and nearly two scheduled operations of the control fall inside it. That is the same shortfall described in a way that is much harder to live with, because one missing instance sounds like an administrative slip and six months with nothing in them does not.
Push the recorded count to four and the model reports one artefact beyond what the cadence expects, with nothing left unspoken for. Both figures look better and nothing about the quarter nobody did has changed.
This is why the model reports surplus separately instead of netting it against the missing instances. Four quarterly artefacts in a nine-month window almost never means the control ran four times. It means the same quarter is in the folder twice, or a monthly report got filed as a quarterly one, or somebody reconstructed March in October. The arithmetic cannot tell any of those apart from genuine diligence, so it refuses to average them away.
The division only works while there is one cadence.
Move the control from quarterly to monthly in May and the expected count is two arithmetics stitched together: whole quarters up to the change, whole months after it, and a stub in between belonging to neither. The model takes one cadence, so it will give you the right answer to a question that is no longer the one you have.
Two honest ways through. Run it twice, once over each part of the period with its own cadence, and add the two expected counts — which is correct and makes the stub a decision you have to state rather than one the arithmetic hides. Or run it at the slower cadence across the whole window, which understates what you owe and is the conservative direction for a count you are going to present.
The one thing worth avoiding is running it at the faster cadence across the whole window, which manufactures missing instances for months when nobody had promised them.
Check