Is it a good practice to manually set numbers with large negative exponential like 1e-300 to zero to avoid underflow in Matlab?
If not how can we avoid harm of underflow when implementing functions like log(1+exp(x))?
Typically, you get trouble when adding very large and very small values, because this can lead to a high relative error. Get rid of this summation (1+exp(x)), it quickly runs out of the range of double values when x is large.
log(1+exp(x))
log(1+1/exp(x))*exp(x))
log(1+1/exp(x))+log(exp(x))
log(1+1/exp(x))+x
An alternative is the use of vpa:
log(1+exp(vpa(10^6)))
very slow, but you get a result with the configured precision.
I never saw a case where manually setting small values to zero was a good solution, typically comparing with a tolerance is better.
Related
In the course of testing an algorithm I computed option prices for random input values using the standard pricing function blsprice implemented in MATLAB's Financial Toolbox.
Surprisingly ( at least for me ) ,
the function seems to return negative option prices for certain combinations of input values.
As an example take the following:
> [Call,Put]=blsprice(67.6201,170.3190,0.0129,0.80,0.1277)
Call =-7.2942e-15
Put = 100.9502
If I change time to expiration to 0.79 or 0.81, the value becomes non-negative as I would expect.
Did anyone of you ever experience something similar and can come up with a short explanation why that happens?
I don't know which version of the Financial Toolbox you are using but for me (TB 2007b) it works fine.
When running:
[Call,Put]=blsprice(67.6201,170.3190,0.0129,0.80,0.1277)
I get the following:
Call = 9.3930e-016
Put = 100.9502
Which is indeed positive
Bit late but I have come across things like this before. The small negative value can be attributed to numerical rounding error and / or truncation error within the routine used to compute the cumulative normal distribution.
As you know computers are not perfect and small numerical error always persists in all calculations, in my view therefore the question one should must ask instead is - what is the accuracy of the input parameters being used and therefore what is the error tolerance for outputs.
The way I thought about it when I encountered it before was that, in finance, typical annual stock price return variance are of the order of 30% which means the mean returns are typically sampled with standard error of roughly 30% / sqrt(N) which is roughly of the order of +/- 1% assuming 2 years worth of data (so N = 260 x 2 = 520, any more data you have the other problem of stationarity assumption). Therefore on that basis the answer you got above could have been interpreted as zero given the error tolerance.
Also we typically work to penny / cent accuracy and again on that basis the answer you had could be interpreted as zero.
Just thought I'd give my 2c hope this is helpful in some ways if you are still checking for answers!
I am trying to turn off denormal number support in matlab, so that basically any two computations that would result in a denormal number would instead just result in zero (DAZ, FTZ)
I've researched several sites include the one below, but I haven't found anything about doing this.
http://blogs.mathworks.com/cleve/2014/07/21/floating-point-denormals-insignificant-but-controversial-2/
I've never heard of such an option in Matlab. It would likely require deep manipulation of a lot of the floating-point math, effectively requiring a new datatype to be supported if this were to be an easily toggle-able option in Matlab. You could write your own mex C code to do this (more here and here) for an individual function.
And of course you can get something like this with one line of Matlab – here's an example:
a = [1e-300 1e-310 1e-310];
b = [1e-301 1e-311 1e-310];
x = a-b;
x(abs(x(:)) < realmin(class(x))) = 0;
where realmin is the smallest normalized floating-point number. However, the floating point math is still performed using the extended denormal/subnormal values in a. It's just the output that's clipped to zero.
Unless you're doing this for fun an experimentation, or possibly running code on an embedded platform, I'd really recommend against disabling denormals as a form of optimization. Instead, focus on why your values are so small and how you might rescale your problem to avoid the issue entirely.
I have a question that I somehow surprising cannot find an answer to. The issue is about floating point errors. It is not a "why does a==b give 0" question, but rather if there are a way to fix floating point errors. The part of the code where I am right now is sensitive and I need to find a way to solve floating point errors. I have tried with
round(100000*myDouble)/100000;
and with
double(int64(100000*myDouble))/100000
but the output does still have floating point errors for some numbers (not all of them, but a few is enough to mess up my code). The problem is that the function I use in matlab is a polygon clipper which I use to calculate the union of many polygons. The function looks for common points and if there are even a small difference, this will mess everything up. The problem should really not be a problem anyway since it is a union and partly overlapping polygons should not cause trouble. However due to some issues with the function I need to make sure to not have these overlaps. The function works really fine in most cases but to speed up I have added a vector of nan separated polygons with holes and since the function is not expected to handle cases like this there are sometimes problems.
There are no point to use int64 for the calculation instead since the function end up with a mex file and is thus not possible to make it work for int64.
There is not a single answer to solve all floating point precision issues. Some strategies are:
Usage of vpa (or symbolic variables).
Make the errors deterministic. If the same formula is used to calculate the same number twice, the result should be equal.
Don't round using decimal factors. Use 2^x instead. In this case, try 2^16 instead of 100000. This way you round the fraction and keep the exponent.
I'm trying to write a basic digit counter (an integer is inputted and the number of digits of that integer is outputted) for positive integers. This is my general formula:
dig(x) := Math.floor(Math.log(x,10))
I tried implementing the equivalent of dig(x) in Ruby, and found that when I was computing dig(1000) I was getting 2 instead of 3 because Math.log was returning 2.9999999999999996 which would then be truncated down to 2. What is the proper way to handle this problem? (I'm assuming this problem can occur regardless of the language used to implement this approach, but if that's not the case then please explain that in your answer).
To get an exact count of the number of digits in an integer, you can do the usual thing: (in C/C++, assuming n is non-negative)
int digits = 0;
while (n > 0) {
n = n / 10; // integer division, just drops the ones digit and shifts right
digits = digits + 1;
}
I'm not certain but I suspect running a built-in logarithm function won't be faster than this, and this will give you an exact answer.
I thought about it for a minute and couldn't come up with a way to make the logarithm-based approach work with any guarantees, and almost convinced myself that it is probably a doomed pursuit in the first place because of floating point rounding errors, etc.
From The Art of Computer Programming volume 2, we will eliminate one bit of error before the floor function is applied by adding that one bit back in.
Let x be the result of log and then do x += x / 0x10000000 for a single precision floating point number (C's float). Then pass the value into floor.
This is guaranteed to be the fastest (assuming you have the answer in numerical form) because it uses only a few floating point instructions.
Floating point is always subject to roundoff error; that's one of the hazards you need to be aware of, and actively manage, when working with it. The proper way to handle it, if you must use floats is to figure out what the expected amount of accumulated error is and allow for that in comparisons and printouts -- round off appropriately, compare for whether the difference is within that range rather than comparing for equality, etcetera.
There is no exact binary-floating-point representation of simple things like 1/10th, for example.
(As others have noted, you could rewrite the problem to avoid using the floating-point-based solution entirely, but since you asked specifically about working log() I wanted to address that question; apologies if I'm off target. Some of the other answers provide specific suggestions for how you might round off the result. That would "solve" this particular case, but as your floating operations get more complicated you'll have to continue to allow for roundoff accumulating at each step and either deal with the error at each step or deal with the cumulative error -- the latter being the more complicated but more accurate solution.)
If this is a serious problem for an application, folks sometimes use scaled fixed point instead (running financial computations in terms of pennies rather than dollars, for example). Or they use one of the "big number" packages which computes in decimal rather than in binary; those have their own round-off problems, but they round off more the way humans expect them to.
I compute this simple sum on Matlab:
2*0.04-0.5*0.4^2 = -1.387778780781446e-017
but the result is not zero. What can I do?
Aabaz and Jim Clay have good explanations of what's going on.
It's often the case that, rather than exactly calculating the value of 2*0.04 - 0.5*0.4^2, what you really want is to check whether 2*0.04 and 0.5*0.4^2 differ by an amount that is small enough to be within the relevant numerical precision. If that's the case, than rather than checking whether 2*0.04 - 0.5*0.4^2 == 0, you can check whether abs(2*0.04 - 0.5*0.4^2) < thresh. Here thresh can either be some arbitrary smallish number, or an expression involving eps, which gives the precision of the numerical type you're working with.
EDIT:
Thanks to Jim and Tal for suggested improvement. Altered to compare the absolute value of the difference to a threshold, rather than the difference.
Matlab uses double-precision floating-point numbers to store real numbers. These are numbers of the form m*2^e where m is an integer between 2^52 and 2^53 (the mantissa) and e is the exponent. Let's call a number a floating-point number if it is of this form.
All numbers used in calculations must be floating-point numbers. Often, this can be done exactly, as with 2 and 0.5 in your expression. But for other numbers, most notably most numbers with digits after the decimal point, this is not possible, and an approximation has to be used. What happens in this case is that the number is rounded to the nearest floating-point number.
So, whenever you write something like 0.04 in Matlab, you're really saying "Get me the floating-point number that is closest to 0.04. In your expression, there are 2 numbers that need to be approximated: 0.04 and 0.4.
In addition, the exact result of operations like addition and multiplication on floating-point numbers may not be a floating-point number. Although it is always of the form m*2^e the mantissa may be too large. So you get an additional error from rounding the results of operations.
At the end of the day, a simple expression like yours will be off by about 2^-52 times the size of the operands, or about 10^-17.
In summary: the reason your expression does not evaluate to zero is two-fold:
Some of the numbers you start out with are different (approximations) to the exact numbers you provided.
The intermediate results may also be approximations of the exact results.
What you are seeing is quantization error. Matlab uses doubles to represent numbers, and while they are capable of a lot of precision, they still cannot represent all real numbers because there are an infinite number of real numbers. I'm not sure about Aabaz's trick, but in general I would say there isn't anything you can do, other than perhaps massaging your inputs to be double-friendly numbers.
I do not know if it is applicable to your problem but often the simplest solution is to scale your data.
For example:
a=0.04;
b=0.2;
a-0.2*b
ans=-6.9389e-018
c=a/min(abs([a b]));
d=b/min(abs([a b]));
c-0.2*d
ans=0
EDIT: of course I did not mean to give a universal solution to these kind of problems but it is still a good practice that can make you avoid a few problems in numerical computation (curve fitting, etc ...). See Jim Clay's answer for the reason why you are experiencing these problems.
I'm pretty sure this is a case of ye olde floating point accuracy issues.
Do you need 1e-17 accuracy? Is this merely a case of wanting 'pretty' output?
In that case, you can just use a formatted sprintf to display the number of significant digits you want.
Realize that this is not a matlab problem, but a fundamental limitation of how numbers are represented in binary.
For fun, work out what .1 is in binary...
Some references:
http://en.wikipedia.org/wiki/Floating_point#Accuracy_problems
http://www.mathworks.com/support/tech-notes/1100/1108.html