Welcome to the blog of

Josh Deprez

Permalink link Published 23 April 2015

Nines of nines

In the operations business we like to talk about nines of things, especially regarding service levels.


  • “one nine of availability” = available 0.9 of the time,
  • “two nines of availability” = available 0.99 of the time,
  • and so on…

then generally,

  • “\(n\) nines of availability” = available \((1 - 10^{-n})\) of the time,


This works for any whole number n: e.g. 5 nines is $$\begin{align}1 - 10^{-5} &= 1 - 0.00001 \\ &= 0.99999.\end{align}$$

There’s a problem with this simple generalisation, and that is, when people say “three and a half nines” the number they actually mean doesn’t fit the pattern. “Three and a half nines” means 0.9995, but

  • \(1 - 10^{-3.5} \approx 0.9996838\), and going the other way,
  • \(0.9995 \approx 1 - 10^{-3.30103}\).

We could resolve this difficulty by saying “3.3ish nines” when we mean 0.9995, or by meaning ~0.9996838 when we say “three and a half nines.” But there’s at least one function that fits the half-nines points as well!

Let’s start with the function above: $$f(n) = 1 - 10^{-n}.$$ For every odd integer, it just has to be lower by a small, correspondingly decreasing amount. We can do this by increasing the exponent of 10 by $$\begin{align}k &= 0.5 + \log_{10}(0.5) \\ &\approx 0.19897.\end{align}$$

One function for introducing a perturbation for halfodd integers is $$p(n) = \sin^2(\pi n).$$ When n is a whole integer, \(p(n) = 0\), and when \(n\) is half an odd integer, \(p(n) = 1\). Multiply this function by some constant and you’re in business.

Thus, define a new function \(g(n)\) for all \(n\):

$$g(n) := 1 - 10^{-n + k p(n)}$$


$$g(n) = 1 - 10^{-n + (0.5 + \log_{10}(0.5))\sin^2(\pi n)}$$

which, when plotted, looks like this:

a negative exponential curve with a negative exponential wiggle. And it has the desired property that at every integer and half-integer it has a value with the traditional number of nines and trailing five (or not).