Ratioing Errors

Date: 01/07/2002 at 09:51:25
From: Rich Thomas
Subject: Ratioing errors

Hello.

I'm sure this is a daft question, but my maths is rusty and I cannot 
be confident of my workings. The problem involves errors and ratioing.  
I'm dealing with two 'elements', A and B.  I have their values, and 
their respective errors. I wish to plot A versus A/B, and on the plot 
include error bars. My problem is that I cannot recall (or find 
elsewhere) how to calculate the error bars for the ratios.  

For example, if A = 10ppm +/- 2%, and B = 20ppm +/- 3%; if I plot 
A vs B, the errors will be +/- 0.2ppm and 0.6ppm respectively, won't 
they?  And A/B will equal 0.5, but what is the error (i.e. maximum 
error)?

Many thanks,
Rich Thomas

Date: 01/08/2002 at 09:13:46
From: Doctor Peterson
Subject: Re: Ratioing errors

Hi, Rich.

The percentage error in a ratio can be approximated by adding the 
percentage errors in the two values. So A/B in your example would be 
0.5 +/- 5%.

You could get the exact error by working out the actual ratio when A 
and B are at opposite extremes; for instance, when

    A = 10 + 2% = 10 + 0.2 = 10.2

and

    B = 20 - 3% = 20 - 0.6 = 19.4

then

    A/B = 10.2/19.4 = 0.526

which is about 0.5 + 5%, as I predicted.

This rule can be derived using calculus. The differential of x/y is

    d(x/y) = (y dx - x dy)/y^2
           = 1/y dx - x/y^2 dy
           = x/y dx/x - x/y dy/y
           = x/y (dx/x - dy/y)

so

    d(x/y)   dx   dy
    ------ = -- - --
      x/y     x    y

Therefore, for small errors, the maximum proportional change of x/y is 
the sum of the absolute values of the proportional changes in x and y 
(since it will be greatest when the signs of dx and dy are opposite).

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   

Date: 01/08/2002 at 09:32:24
From: Rich Thomas
Subject: Ratioing errors

Very many thanks to Dr Peterson for his great help. The (very quick!) 
response was detailed yet very straightforward to follow. You've 
helped greatly. Once again, many thanks.

Rich Thomas