Difference between value of sum obtained from program as compared to and the one obtained from command window

1 次查看（过去 30 天）
Chinmay Gadre
Chinmay Gadre 2024-1-21
评论： Chinmay Gadre 2024-1-21
I am adding four very small numbers in my program. The issue I am facing is the sum obtained from the program "obj" is coming out to be different than when I add the same four numbers in the command window. Why is the sum getting calculated wrongly when I run the program? How to fix it?
The green box displays the expected value of the sum while the red box has the incorrect result of adding I1t, I2t, I3t and I4t.
  4 个评论
显示 2更早的评论
Stephen23
Stephen23 2024-1-21
Ran in:
"The issue I am facing is the sum obtained from the program "obj" is coming out to be different than when I add the same four numbers in the command window."
The only issue here is that you are adding different numbers.
"Why is the sum getting calculated wrongly when I run the program?"
It is not wrong: when you add different numbers you will get a different result:
sum([-0.001335458063828,0.000057425539310,0.000971336765721,0.000306708780422])
ans = 1.3022e-08
sum([-0.0013,5.7426e-5,9.7134e-4,3.0671e-4])
ans = 3.5476e-05
So far absolutely no surprises.
"How to fix it?"
What is broken?
If you want to add exactly the same numbers then you will need to add exactly the same numbers. Adding exactly the same numbers does NOT mean adding some other numbers with a much lower precision.
Chinmay Gadre
Chinmay Gadre 2024-1-21
Hi Stephen,
Thank you for the explanation, was kind of a brainfade moment for me to be checking the sum of the displayed variables instead of the "actual" values.

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Sai Teja G
Sai Teja G 2024-1-21
Ran in:
Hi Chinmay,
I understand that you are attempting to add four numbers using two distinct methods. Let's execute both approaches now to comprehend the differences between them.
Case 1:
I1t = -0.001335458063828;
I2t = 0.000057425539310;
I3t = 0.000971336765721;
I4t = 0.000306708780422;
%obj = 0.000000013021625
obj=I1t+I2t+I3t+I4t
obj = 1.3022e-08
Case 2:
I1t = -0.0013;
I2t = 0.000057426;
I3t = 0.00097134;
I4t = 0.00030671;
%obj = 3.5476e-05
obj=I1t+I2t+I3t+I4t
obj = 3.5476e-05
Now you can clearly discern the difference between the two additions due to the variance in precision. This is the reason why you received different outputs in both cases. I hope this clarifies your doubt!
  1 个评论
Chinmay Gadre
Chinmay Gadre 2024-1-21
Hello Sai,
Thank you for the clarification. Was wrong on my part to be comparing the sum of the displayed variables in the command window with that of the actual values.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Logical 的更多信息

标签

产品


版本

R2023b

Community Treasure Hunt

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

Start Hunting!

Translated by