Which is the machine precision of Matlab?

169 views (last 30 days)
MRC
MRC on 17 Feb 2014
Edited: John D'Errico on 18 Feb 2014
Hi all, I have the following problem: I want to find the argmax (not necessarily unique) of a function f(alpha,beta). Under some assumptions on f it is possible to obtain the argmax set following this procedure:
1) compute F=max f(alpha,beta) wrto alpha and beta
2) for each alpha compute H(alpha)=max f(alpha,beta) with respect to beta
3) pick alpha s.t. H(alpha)=F;
4) pick corresponding beta from step 2)
I'm trying to perform this procedure in Matlab; my question is: assuming that the maximization problems have been correctly solved, how much the result given by Matlab is precise? i.e. in step 3) I should pick alpha s.t. H(alpha)~= F+...? or I should round the numbers to which digit?
I have attached F and H(alpha).
Thank you very much!
  1 Comment
John D'Errico
John D'Errico on 18 Feb 2014
I'm intrigued here. At one point you state you know nothing of numerical analysis. At another point, you state that "under some assumptions on f, it is possible to obtain the argmax".
So which is it? You know nothing, or a fair amount?
By the way, as optimization schemes go, I'm not terribly impressed with this one.
This almost feels like your homework assignment, given the statement above, as if you were fed that line.

Sign in to comment.

Accepted Answer

Iain
Iain on 18 Feb 2014
Matlab stores and operates on doubles. Doubles have 15-ish significant figures of precision.
If you keep your numbers all roughly the same, you'll manage to get 15ish accurate digits on your answer. If you don't, then, to demonstrate try:
zero = 10^17 + 1 - 10^17
one = 10^17 - 10^17 + 1
You and I both know that the answer to both equations is 1. Double format numbers get that wrong. If you do that sum in unsigned 64 bit integers, it gets it right.
zero = uint64(10^17) + uint64(1) - uint64(10^17)
one = uint64(10^17) - uint64(10^17) + uint64(1)
unsigned 64 bit integers have a precision of "1", and a maximum value of 2^64 - 1 (approx 4*10^18), to which you may apply an offset and scaling factor
  2 Comments
Iain
Iain on 18 Feb 2014
1) No. I mean that matlab rounds every double format number to the nearest number it can represent. This allows 15ish significant figures, significant figures are NOT decimal places. 156965468456231 has 15 significant figures, just like 0.000156965468456231 has 15 significant figures.
2) Matlab does not do that automatically. You need to understand your scenario well to establish where your numerical issues will occur.

Sign in to comment.

More Answers (1)

dpb
dpb on 18 Feb 2014
Edited: dpb on 18 Feb 2014
Default in Matlab is double which is stored with 53 bits of precision. This converts to roughly 15-16 decimal digits of precision (log10(2^53)-->15.95).
  5 Comments
dpb
dpb on 18 Feb 2014
Ooops...typo indeed, edited/corrected. Thanks for pointing it out...

Sign in to comment.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by