# Translate percantage increases / decreases into 0 - 100 range?

I am tracking some percentage increase and decrease numbers and I need to plot them on a line graph. However I can only plot them on a line graph with a range of 0 - 100.

What formula would I use translate these percentage values into a range of 0 - 100 if this is possible?

Here is an example dataset:

-15.79, -41.33, -86.59, -41.67, 17.65, 66.67, 191.67, 31.25, 24.39, 0.00

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What do you want the values between 0-100 to represent? Are they going to be completely arbitrary? Why do you have this restriction of only being able to plot values between 0 and 100 in the first place? – Chris Taylor May 30 '12 at 10:34
You might want to use the linear transformation $x\mapsto\frac{x+100}{2}$. However, I see no good reason to do this. – Rasmus May 30 '12 at 10:34
@Chris Taylor The data is going to be used in a program. The service used to generate the graph has this restriction of values between 0 - 100. The values should represent the percentage movements. I just don't know how to transform the numbers and scale them exactly. For example if I had a range of -50 to 50 I could transform the values to o - 100 and a value of 0 would be plotted at 50. How do I accurately do this without know the upper and lower limits being fixed? – lanks May 30 '12 at 10:45

The maximum is 191.67 and the minimum -86.59 (though presumably the minimum could in theory go down as far as -100). So one possible linear transformation might be to have the top of the graph representing +200 and the bottom -100 using $$f(x) = \frac{x+100}{3}$$ which would give points to plot at

28.07, 19.5567, 4.47, 19.4433, 39.2167, 55.5567, 97.2233, 43.75, 41.4633, 33.3333


You can round these if you are restricted to integer values. If your graph looks upside down, then subtract all these values from 100.

More generally, if $M$ is the maximum or some convenient number above it and $m$ the minimum or some convenient number below it then you can use $$f(x)=100 \times \frac{x-m}{M-m}$$ or if you want it the other way up $$f(x)=100 \times \frac{M-x}{M-m}.$$

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Thanks Henry. How would I adjust the transformation if the max and min changes? – lanks May 30 '12 at 10:54
@lanks: I have added more general formulae – Henry May 30 '12 at 11:08