Rule of thumb · MathematicsNº 173 / 190

Five sevenths, then take off the holidays

How many working days is that?

Working days ≈ calendar days × 5/7, minus the public holidays in between. Over a whole number of weeks it is exact; otherwise it is never out by as much as a day and a half.

Why it works

Five weekdays in every seven days is not an approximation over a whole week — it is the definition, so any span of 7, 14 or 21 days contains exactly 5, 10 or 15 weekdays whatever day it starts on. The error only comes from the leftover days at the ends, and the extremes are the partial weeks at the ends: two days covering a Saturday and Sunday hold no weekdays at all, and five days covering Monday to Friday hold five. Work those two out and the bound falls out exactly — 10/7 of a day, about 1.43, in either direction and never more. That bound does not grow with distance: the 5/7 estimate is as good over five years as over five weeks, which is why it is worth trusting. What does grow is the holiday correction — eight a year in England and Wales, nine in Scotland, eleven US federal days — and that is the part people forget. Six weeks sounds like plenty until six weeks crosses Christmas.

When it fails

It says nothing about whether anyone is actually working. Annual leave, industry shutdowns, part-time patterns and the dead week between Christmas and New Year are all invisible to it — those days are working days by this definition and by nobody else’s. And the holiday count is jurisdictional: a one-off holiday declared for a coronation or a jubilee is in nobody’s formula.

Do it exactly

Estimate with the rule, then check it against the calculator that models it properly.

Open Countdown & Working Days →

More in Mathematics

← PreviousThe framing sets the depth of field, not the lens Next →The hour hand keeps moving

Browse all 190 rules of thumb →