Failure in initial objective function evaluation. FSOLVE cannot continue.

I have some problems with fsolve:

My function:

function F = simple(x)
global c1 c2

F(1) = x(1)-c1;
F(2) = x(2)-c2;
end

Here is my code that calls this function in my main file:

global c1 c2
c1 = 150
c2 = 130

options = optimoptions('fsolve','Display','iter');
[x,fval,exitflag,output] = fsolve(@simple,[0 0],options)

Matlab's error message is as follows. If this is caused by some other part of my code. What can be the cause? How do I locate and correct that part? Thank you very much.

Error message:

Unable to perform assignment because the indices on the left side are not

compatible with the size of the right side.

Error in simple (line 4)

F(1) = x(1)-c1;

Error in fsolve (line 264)

fuser = feval(funfcn{3},x,varargin{:});

Error in nlogit (line 478)

[x,fval,exitflag,output] = fsolve(@simple,[0 0],options)

Caused by:

Failure in initial objective function evaluation. FSOLVE cannot continue.

[Ludovico Cucciniello]

The problem might be the way you're defining the variables c1 and c2. Anyway, I tried this and seems to work.

c = [150, 300];
x0 = [0,0];

options = optimoptions('fsolve','Display','iter');
[x,fval,exitflag,output] = fsolve(@simple,x0,options,c);

function F = simple(x,c)
F(1) = x(1)-c(1);
F(2) = x(2)-c(2);
end

[Chang Li]

I figured out that I forget to declare global variables somewhere. Now it works. Thank you very much.

Answer 1

[Walter Roberson]

Works for me.

format long g

global c1 c2
c1 = 150

c1 = 
   150

c2 = 130

c2 = 
   130

options = optimoptions('fsolve','Display','iter');
[x,fval,exitflag,output] = fsolve(@simple,[0 0],options)

                                         Norm of      First-order   Trust-region
 Iteration  Func-count     f(x)          step         optimality    radius
     0          3           39400                           150               1
     1          6           39004              1            149               1
     2          9         38022.8            2.5            147             2.5
     3         12         35624.4           6.25            143            6.25
     4         15         29970.3         15.625            131            15.6
     5         18         17971.2        39.0625            101            39.1
     6         21            1325        97.6563           27.5            97.7
     7         24     1.23776e-14        36.4006        9.4e-08             244

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

x = 1×2

150.000000094 129.999999940489

fval = 1×2

1.0e+00 * 9.40000859372958e-08 -5.95112510382023e-08

exitflag = 
     1

output = struct with fields:
       iterations: 7
        funcCount: 24
        algorithm: 'trust-region-dogleg'
    firstorderopt: 9.4000085878389e-08
          message: '↵Equation solved.↵↵fsolve completed because the vector of function values is near zero↵as measured by the value of the function tolerance, and↵the problem appears regular as measured by the gradient.↵↵<stopping criteria details>↵↵Equation solved. The sum of squared function values, r = 1.237761e-14, is less than↵sqrt(options.FunctionTolerance) = 1.000000e-03. The relative norm of the gradient of r,↵9.400009e-08, is less than options.OptimalityTolerance = 1.000000e-06.↵↵'

function F = simple(x)
global c1 c2

F(1) = x(1)-c1;
F(2) = x(2)-c2;
end