Results inside for are different from outside

Hi,

I have implemented an optimization algorithm. Now I am trying to understand which are the value of some parameters that make the error the smallest possible. Therefore I created a function that receives three files and the value of the parameter I am varying. All the other parameters are mantained fixed.

So whenever I do this:

J=GradMomChemo_TESTE('Parametros_L2_JN','Controlo_5_tudo','Cond_Iniciais_1',1.2);

the result is:

J=1.2708*10^7

However, if I calculate this inside a for the result is different:

up=1.1:0.05:2;
mini=zeros(2,length(up));
for i=1:length(up)
    mini(1,i)=up(i);
    mini(2,i)=GradMomChemo_TESTE('Parametros_L2_JN','Controlo_5_tudo','Cond_Iniciais_1',up(i));
end

In this case the value of J correpondent to up=1.2 is:

    J=2.712*10^7

Why does this happen? What can I do to change this situation?

Answer 1

[Cam Salzberger]

Hello Elisa,

There are a couple of possible reasons, but without the CradMomChemo_TESTE function, it's difficult to tell which might be the root cause. One possible cause, though, is precision difficulties. If your function is sensitive to very small perturbations, then even simple precision differences can cause drastically different results.

Technically, neither of your pieces of code are calling that function with 1.2 exactly. That is because 1.2 cannot be represented exactly in double precision. In the first snippet, this is the value you call with:

fprintf('%.20f\n',1.2)
1.19999999999999995559

In the next snippet, this is the value:

x = 1.1:0.05:2;
fprintf('%.20f\n',x(3))
1.20000000000000017764

Here is some information on floating point accuracy. There have also been several questions here on MATLAB Answers that address this.

If you run your function outside the loop with the exact same value x(3), and it is still giving different results, then it is likely that something different is happening with the function itself. It's possible that there are two versions of the function on your path, and the code calls one while your test from the Command Window calls the other. Check for duplicate copies with which -all.

It's also possible that your function has some time, parent, or otherwise random properties to it. Put a breakpoint in at the entrance to the function and step through it, to see where differences start to happen.

Hope this helps!

-Cam

[Elisa Pacheco]

This is exactly the problem! I tried the vector

up=[1 1.05 1.1 1.15 1.2 (...) 2]

and everything went right! thank you soo much!

[Cam Salzberger]

Sure thing. I would have a look at the function that is using these numbers though. If it's varying by as much as 50% from an input change of about 0.0000000000001%, then your function is either extremely sensitive to input changes, or your function is unstable about the input point of 1.2.

Does that function use any conditionals of the form:

if inputVal < 1.2
    ...

or anything like that? Or does it compare equality with 1.2 exactly at some point? If so, you may either want to ensure that the input at that point is exactly what you expect it to be, or you might consider adding some tolerance terms. For example:

if abs(inputVal-1.2) < 1e-7
    ...

Or

if inputVal >= 1.2-1e-7
    ...

Answer 2

[Henrik Jacobsen]

I have two guesses. It's hard to know without having the functions.

1) You are not actually looking at up=1.2 in your for loop.

2) You have some global variable that changes when running the for loop. To test both cases, try running

up=[1.2 1.1 1.2];
mini=zeros(2,length(up));
for i=1:length(up)
  mini(1,i)=up(i);
  mini(2,i)=GradMomChemo_TESTE('Parametros_L2_JN','Controlo_5_tudo','Cond_Iniciais_1',up(i));
end

if mini(2,1) is different from mini(2,3), then there is a variable that changes when the function is running. If they are the same and equal to 1.2708*10^7, then you were not looking at the value of up you thought you were.

[Elisa Pacheco]

Thank you Henrik! Your first guess was right. In fact, when I use

up=[1.2 1.1 1.2]

I obtain 1.2708*10^7! Thank you for your help!