ga-fitness

Dear All,

I need to optimize parameters of a function using the ga module. I have a coefficient matrix say A(9:4):

A =

   -0.0996   -0.0323   -0.0640   -0.0365
    0.0859   -0.0636    0.0552    0.0315
    0.1805    0.0125    0.1159    0.0662
    0.0521    0.0513    0.0335    0.0191
   -0.3028    0.0647   -0.1946   -0.1111
   -0.2242   -0.0382   -0.1440   -0.0822
    0.0475   -0.0191    0.0305    0.0174
    0.2170   -0.0010    0.1394    0.0796
    0.0437    0.0257    0.0281    0.0160

and a vector B where i store the experimental data say B(9:1):

B =

   -4.8700
   -3.2100
    5.1000
    5.3400
   -1.6200
   -8.1700
   -0.4700
    4.8300
    3.0800

my chromosome will be a vector x(4:1)

I use this fitness function:

FitnessFcn = @(x)bsxfun(@minus,abs(bsxfun(@times,A,x)),B)

But when I am using

[x]=ga(FitnessFcn,4)

I get:

??? Subscripted assignment dimension mismatch.

Error in ==> fcnvectorizer at 14
            y(i,:) = feval(fun,(pop(i,:)));

Error in ==> makeState at 47
        Score = fcnvectorizer(state.Population(initScoreProvided+1:end,:),FitnessFcn,1,options.SerialUserFcn);

Error in ==> gaunc at 41
state = makeState(GenomeLength,FitnessFcn,Iterate,output.problemtype,options);

Error in ==> ga at 279
        [x,fval,exitFlag,output,population,scores] = gaunc(FitnessFcn,nvars, ...

Caused by:
    Failure in user-supplied fitness function evaluation. GA cannot continue.

Could you please help me resolve the problem.

Thanks

Answer 1

[Walter Roberson]

http://www.mathworks.com/help/toolbox/gads/ga.html

The objective function, fitnessfcn, accepts a vector x of size 1-by-nvars, and returns a scalar evaluated at x.

However your expression abs(bsxfun(@times,A,x)) is going to result in 9 x 4, and when you @minus the 9 x 1 B from that you are going to get 9 x 4. And that isn't a scalar.

[chris]

thanks for the reply. So the case is to take the result of matrix vector multiplication (AxB) as a 9 x 1 vector. How is it possible to do the matrix equation Ax=B

[Walter Roberson]

Andrei's answer looks plausible.

Note that getting a 9 x 1 out is still not going to be the scalar output you need for the fitness function. Perhaps you need something like the sum of the squares of the values?

Answer 2

[Andrei Bobrov]

Can so?

FitnessFcn = @(x)abs(A*x-B) ;