MATLAB: Simple miscalculation: (1-0.95)*4000 = 2.000000000000002e+02

floating pointimprecisionprecisionroundingrounding error

Dear all,

I am using Matlab 2013a and I found this imprecision in a simple operation: (1-0.95)*4000. Matlab returned me 2.000000000000002e+02 instead of 200. Thus, I cannot access a vector position since it is not integer and when I use ceil Matlab returns 201.

Anyone has noticed this mistake before? Is it already corrected in new versions of Matlab?

Thanks, Bruno Fanzeres

Best Answer

This is not a bug, nor is it a mistake. This is the normal behavior of floating point numbers, and although it is not at all unexpected it often confuses beginners.

MATLAB uses floating point numbers which are stored with a finite number of binary digits. Because of the finite precision of these numbers, calculations often incur some rounding error, and it is your job as the programmer to account for this in the design of your algorithm.

If you really absolutely must avoid the precision limits of the standard number classes, then you basically have two choices:

use a symbolic mathematics tool, such as the Symbolic Toolbox.
use a higher precision toolbox, such as John D'Errico's HPF. Note that infinite precision is impossible.

For more information read these:

http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F

http://de.mathworks.com/matlabcentral/answers/57444-faq-why-is-0-3-0-2-0-1-not-equal-to-zero

You might like to try James Tursa's FEX submission too:

http://de.mathworks.com/matlabcentral/answers/41536-why-is-2-24-10-22-4-not-equal-to-0

Related Solutions

MATLAB: VERY odd error with simple multiplication … OK… not so odd.

"Does anyone have any idea why this is happening?"

Yes, the behaviors of floating point numbers are quite well documented, and often discussed on this forum.

"It's pretty disturbing to think about the ramifications of this kind of error."

Only if you write your code believing that floating point error does not exist. Experienced programmers understand that calculations on floating point numbers can accumulate floating point errors (like your example does), and do not expect floating point mathematics to be equivalent to the symbolic maths that they learned in high school (i.e. arithmetic operations on floating point values are not associative), so they write their code accordingly to allow for this.

You need to learn about the limits of floating point numbers. Start by reading these:

https://www.mathworks.com/matlabcentral/answers/57444-faq-why-is-0-3-0-2-0-1-not-equal-to-zero

https://www.mathworks.com/matlabcentral/answers/316889-why-two-equal-numbers-are-not-equal

http://www.mathworks.com/matlabcentral/answers/102419-how-do-i-determine-if-the-error-in-my-answer-is-the-result-of-round-off-error-or-a-bug

http://www.mathworks.com/matlabcentral/answers/140656-about-numerical-difficulty-of-equality-comparison-operation-between-two-double-real-numbers-in-matlab

This is worth reading as well:

https://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html

http://faculty.tarleton.edu/agapie/documents/cs_343_arch/papers/1991_Goldberg_FloatingPoint.pdf

MATLAB: Colon operator rounding problem

Try it like this.

a=(0:1200)/10;
 disp(a(20));
 disp(a(20)-1.9);
 isequal(a(20),1.9)

Your previous code was not returning 1 because of rounding errors when doing finite precision arithmetic.

Best Answer

Related Solutions

MATLAB: VERY odd error with simple multiplication … OK… not so odd.

MATLAB: Colon operator rounding problem

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