How to make my code run faster?

I'm new to matlab and this is one of the very first functions I wrote. My function works very well except that it's very slow (~27 seconds). How can I take advantage of Matlab's syntax to better optimize it?

function [ amin,bmin,cmin ] = perm( x,y,xmin )

A=(1:0.1:9);

B=(4:0.15:16);

C=(0.003:0.0005875:0.05);

result = 0.0;

amin = 15;

bmin = 16;

cmin = 0.05;

rmin = 1000;

temp = 0;

for i=1:81

    for j=1:81

        for k=1:81

            result  = -A(k)* erf((x-xmin)*C(i))+B(j);
            Error = minus(y,result).^2;

            temp = norm(Error);

                if  temp < rmin;
                    rmin = temp;
                    amin = A(k);
                    bmin = B(j);
                    cmin = C(i);
                end

        end
    end
end

end

Answer 1

[Roger Stafford]

If you ignore the limits you have placed on A and B, and you want to do this minimization without using matlab's optimization functions, you can take advantage of the fact that your expression is linear in A and B and thereby cut down the number of iterations from 81^3 to only 81 as C varies. This is because, for any given value of C, the optimum values for A and B can be directly calculated as given in the code below. This the linear regression technique.

 function [amin,bmin,cmin] = perm(x,y,xmin)

 dmax = -inf;
 ym = mean(y);
 C=(0.003:0.0005875:0.05);
 for i = 1:length(C)
   ei = erf((x-xmin)*C(i));
   em = mean(ei);
   d = (sum((y-ym).*(ei-em)))^2/sum((ei-em).^2);
   if d > dmax
     dmax = d;
     cmin = C(i);
   end
 end
 ei = erf((x-xmin)*cmin);
 em = mean(ei);
 amin = -sum((y-ym).*(ei-em))/sum((ei-em).^2);
 bmin = ym+amin*em;

Instead of the above you can use the above -d as your objective function to be minimized and call on one of the optimization routines to find the best C value. Afterwards you can calculate A and B as above.

[Ali]

Thank you, this was very helpful

Answer 2

[Jan]

The general rule applies for all computer languages and the real life also:

Avoid repeated calculations!

rmin2 = rmin * rmin;
for i=1:81
  c1 = erf((x-xmin)*C(i));
  for j=1:81
    for k=1:81
       result = -A(k)* c1 +B(j);
       Error  = (y - result) .^ 2;    % Is y a vector?
       temp   = sum(Error .* Error);  % SQRT is more expensive then squaring
       if temp < rmin2
         rmin  = temp;
         rmin2 = rmin * rmin;
         amin  = A(k);
         bmin  = B(j);
         cmin  = C(i);
       end 
     end
   end
end

[Ali]

thank you

[Ali]

yes, y is a vector