Logical statement applicable to entire matrix

Hi all,

I'm running an iterative code which solves for a value Cd from an initial guess, Cd_org. Currently, I have the code running, but it stops after one iteration because a few of my 27832 data points are within the required tolerance. Is there some way to ensure that every value of my delta_Cd matrix falls within the tolerance?

 Cd_org = ones(length(t),1); 
 tolerance = 0.00001;
 delta_Cd = 0;
 while delta_Cd < tolerance
     for n = 1:length(t)
.
.
.
% Establishes variable Cd, calculates new Cd using Cd_org.
.
.
.
     end
 delta_Cd = Cd - Cd_org;
 Cd_org = Cd;
 end

Thanks, Aisha

Answer 1

[Honglei Chen]

You can try

while all(delta_Cd(:) < tolerance)

HTH

[Aisha McKee]

No luck. Still only runs the one time....

[Honglei Chen]

Looks like the original condition is problematic, as it says do it while the error is within the tolerance. Probably should be

while any(delta_Cd(:) > tolerance)

as Walter mentioned below

HTH

Answer 2

[Walter Roberson]

   Cd_org = ones(length(t),1); 
   tolerance = 0.00001;
   delta_Cd = inf;
   while any(delta_Cd > tolerance)
       for n = 1:length(t)
  .
  .
  .
  % Establishes variable Cd, calculates new Cd using Cd_org.
  .
  .
  .
       end
   delta_Cd = Cd - Cd_org;
   Cd_org = Cd;
   end

[Aisha McKee]

This ran for infinity iterations and eventually all my points went to infinity with this error message.

>>Warning: Rank deficient, rank = 0, tol = 1.#INF00e+00.

Thoughts?

[Walter Roberson]

Do you use delta_Cd in your calculations? If so then initialize it to something like tolerance*2 instead of inf.

If you do not... remember that there are calculations where it is not possible to have all the outputs simultaneously less than the tolerance.

Also I would recommend modifying to

     while any(abs(delta_Cd) > tolerance)