Question about accuracy of two different methods

2 次查看（过去 30 天）
Devin
Devin 2018-11-14
评论： Devin 2018-11-15
Greetings,
I have a question about accuracy. I have mathmatically equivalent method 1 and method 2. Practically, the final diffference beteween vol and vol1 is 8.8818e-15. In my program, this tiny difference can be amplified. But I still need to use method 1 form to get results as method 2.
%Method 1:
vol = 6
for i = 1:5000
vol = vol + 18.3 * 1e-3 * 0.1;
end
%Method 2:
vol1 = 6 + 18.3 * 1e-3 * 0.1 * 5000;
Does anyone can help me to explain why these two methods are different? And how can I get accurate results to avoid this tiny difference to be amplified?

请先登录，再回答此问题。

采纳的回答

Stephen
Stephen 2018-11-14
Method 1 has 5000 operations, each of which is bound to generate some error. Your machine epsilon may vary from mine, but mine is 2.2204 e-16 on a 64 bit installation of Matlab. (Type "eps" on your command line) Every Matlab operation is rounded to the nearest eps, so if you have 5000 operations, that is 5000 opportunities to accumulate error.
If you are extraordinarily unlucky, and each of those rounding errors went in the same direction, you might see as much as 5.5511e-13 in error accumulation. (max error = 5000*eps/2). So in reality, your accumulation of 8.88e-15 in error over 5000 operations is really not all that bad, as the errors seem to be distributed near the desired mean of zero. If your system is sensitive to that level of error, I would suggest you reconsider the reasons why you "must" use method one.
  3 个评论
显示 1更早的评论
Stephen
Stephen 2018-11-14
Since you're using 64 bit, double precision variables in your code, there isn't much that you can do at this time. Some 64 bit applications use a "long double" variable type which would reduce machine epsilon to 2^-63 or 1e-19, but to my knowledge that is not offered by Matlab.
Devin
Devin 2018-11-15
Thank you very much, I will discuss with my prof.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Get Started with MATLAB 的更多信息

标签

产品


版本

R2018a

Community Treasure Hunt

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

Start Hunting!

Translated by