Avoiding negative values in fminsearch

Hello, I am doing optimisation of a guassian function to calculate the midpoint and spread when the function below goes to zero. But when using fminsearch i get negative values.What i am supposed to get is 24 and 4 respectively please help!!

AA=1.580740308;
BB=0.854284221;
format long g

   invalues=[1,10];

   fun1=@(x)((AA-sum(A.*real(exp(-((E-x(1)).^2/x(2).^2)))))^2+...
             (BB-sum(B.*real(exp(-((E-x(1)).^2/x(2).^2)))))^2);

   [x,fval]=fminsearch(fun1,invalues)

[John D'Errico]

What is E?

[Matt J]

Your attachment didn't make it. Make sure you confirm the upload.

Also, it would help if you show the output that you did get. Realize that x(2)=+4 and x(2)=-4 produce the same value of fun1, so they are equivalent.

[Rena Berman]

(Answers dev) Restored question.

Answer 1

[Walter Roberson]

All of your terms involving x are squared, so there is no way to tell a negative value of the term from a positive value of the term. Negative x are valid solutions.

It is not possible to constrain fminsearch to prevent it from using negative x. The closest you could get would be to add a penalty, such as

fun1=@(x)((AA-sum(A.*real(exp(-((E-x(1)).^2/x(2).^2)))))^2 + ...
           (BB-sum(B.*real(exp(-((E-x(1)).^2/x(2).^2)))))^2) + ...
          10^100 * any(x < 0);

[Walter Roberson]

[24, 4] is not the minima. There is a better minima near [23.9999999995143, 4.00000000010837]. You can see the better minima by using

[x,fval] = fminsearch(fun1, [24,4], struct('TolX', 1e-12))

fminsearch is not a global optimizer. How well it does depends upon the initial values you give it.

In my tests, if the guess for x(1) is not at least 6.94 then fminsearch will get caught near 0 like you see. If the guess for x(1) is more than 29.056 then fminsearch will get caught either near 37 (for x(1) up to 29.53), or caught at 1E+16 or 1E+17 for larger x(1)

[Walter Roberson]

You need a global minimizer if you are getting caught in local minima. fmincon and fminunc are not able to escape sufficiently deep local minima either.

The approach I would use would be something like using the fun1 with penalty, and selecting a range of values to be probed, and

x1min = 0; x1max = 50;
x2min = 0; x2max = 50;
[X1, X2] = ndgrid( linspace(x1min, x1max, 20), linspace(x2min, x2max, 20));
[xvals, fvals] = arrayfun(@(x1,x2) fminsearch(fun1,[x1,x2], struct('TolX', 1e-12, 'MaxFunEvals', 2000, 'MaxIter', 2000)), X1, X2, 'Uniform', 0);
[bestfval, minidx] = min(cell2mat(fvals(:)));
bestx1x2 = xvals{minidx};

Now the minima is at x1 = bestx1x2(1), x2 = bestx1x2(2), with value bestfval

If the value were still too high (which it isn't) then you would increase the "20" to a higher number to get a more refined starting grid.

If you try this using the f1 without penalty you end up with an even better minima at -120.423695515953 -96.8967338642581

[Walter Roberson]

If this solves the problem, please Accept the answer.

Answer 2

[Matt J]

Applying abs() to x(1) in the objective function will effectively constrain it to be positive,

     fun1=@(x)((AA-sum(A.*real(exp(-((E-abs(x(1))).^2/x(2).^2)))))^2+...
               (BB-sum(B.*real(exp(-((E-abs(x(1))).^2/x(2).^2)))))^2);

With the objective function written this way, even if fminsearch returns a negative solution x, a positive solution can be immediately derived from it as abs(x).