MATLAB: Problem in using fsolve

fsolvenon-linear

I am trying to solve the following set of equations but the solver is prematurely stopped and If I alter the MaxIter and MaxEvals then also its showing stopped prematurelly and then I tried to alter the tolerance then its shows no solution found. Can anyone guide how can i obtain the solution for the following set of equations.

function F = compare(X)
rho=1000.0;
F=[
1.0*7.018e-4*4.2850022315+-1.0*1.25e-3*X(2)+-1.0*7.36e-5*X(3)+-1.0*4.47e-5*(4);

-1.0*1.25e-3*X(1)+1.0*1.66e-5*X(4)+1.0*4.47e-5*X(5)+1.0*1.07e-4*X(6);

+1.0*7.36e-5*X(2)-1.0*1.07e-4*X(6)+1.0*1.07e-4*X(7);

1.0*4.47e-5*X(3)+-1.0*4.47e-5*X(5)+-1.0*1.07e-4*X(7);

1.0*0.4*rho*4.2850022315979^2+1.0*807362.4375+-1.0*0.4*rho*X(2)^2 + -1.0*X(9);

1.0*0.4*rho*4.2850022315979^2 + 1.0*807362.4375+-1.0*0.4*rho*X(3)^2+-1.0*X(10);

1.0*0.4*rho*4.2850022315979^2 + 1.0*807362.4375+-1.0*0.4*rho*X(4)^2+-1.0*X(11);

-1.0*0.4*rho*X(1)^2 + -1.0*X(8)+ 1.0*0.4*rho*X(4)^2 + 1.0*X(11);

-1.0*0.4*rho*X(1)^2 + -1.0*X(8) + 1.0*0.4*rho*X(5)^2 + 1.0*X(12);

-1.0*0.4*rho*X(1)^2 + -1*X(8) + 1.0*0.4*rho*X(6)^2 + 1.0*X(13);

-1.0*0.4*rho*X(6)^2 + -1.0*X(13) + 1.0*0.4*rho*X(2)^2 + 1.0*X(9);

-1.0*0.4*rho*X(6)^2 + -1.0*X(13) + 1.0*0.4*rho*X(7)^2 + 1.0*X(14);

 1.0*0.4*rho*X(3)^2 + 1.0*X(10) + -1.0*0.4*rho*X(5)^2 + -1.0*X(12);
 1.0*0.4*rho*X(3)^2 + 1.0*X(10)   +  -1.0*0.4*rho*X(7)^2 + -1.0*X(14)
 ];

end

options = optimset('Display','iter','TolX',1e-7,'MaxFunEvals',18000,'MaxIter',2200);

[Sol,fval,exitflag]=fsolve(@compare,X,options);

following error comes

No solution found.
fsolve stopped because the last step was ineffective. However, the vector of function
values is not near zero, as measured by the default value of the function tolerance. 
<stopping criteria details>

where X's are the variables

Please Guide!!!

Regards

Best Answer

You have to write your function so that all of your unknowns are in the same parameter vector.

So calling your parameter vector ‘B’:

B = [V(1) ... V(7) P(1) ... P(7)];

then replace V(1) with B(1), P(1) with B(8), and so forth.

If you intend ‘F’ to be an anonymous function, you have to write it as:

F = @(B,rho) [1.0*10.0*25+-1.0*17.0*B(2)+ ... +  -1.0*0.4*rho*B(7)^2 + -1.0*B(14)];

(I abbreviated the function here.) Use the ‘Search’ and ‘Replace’ utilities in the MATLAB Editor to do this more easily.

Your call to fsolve is then:

[x,fval] = fsolve(@(B) F(B,rho),x0)

Your F function will get ‘rho’ from your workspace.

EDIT — Try using different starting values for ‘X’ that are reasonably close to what you expect them to be. Otherwise experiment with them. If it doesn’t converge on a solution after several attempts, it may not have a solution. If you believe it should, go back to your original equations and be sure you wrote ‘F’ correctly.

Related Solutions

MATLAB: Trying to minimise a function subject to nonlinear constraints

Simple mistake(s) here. You have TWO equality constraints. But your constraint function is written to have ONE equality constraint, and ONE inequality constraint. And you also wrote the x*z=3 constraint wrong. You wrote it as y*z=3.

Instead, try this:

function [c,ceq] = confuneq(x)
    c = [];
    ceq = [x(2)^2+x(3)^2-1; x(1)*x(3)-3];
end

That should now be a correct way to write the constraints.

Of course, don't expect fmincon to provide both solutions in one call. You need to start the optimizer out in the correct basin of attraction to get the solution you might want to see.

x = fmincon(fun,[1 1 2],A,b,Aeq,beq,lb,ub,nonlcon,options)
 Iter  Func-count            Fval   Feasibility   Step Length       Norm of   First-order  
                                                                       step    optimality
    0           4    3.000000e+00     4.000e+00     1.000e+00     0.000e+00     2.000e+00  
    1           9    2.928889e+00     3.352e+00     7.000e-01     1.504e+00     9.644e-01  
    2          13    2.571265e+00     7.559e-01     1.000e+00     1.060e+00     3.470e-01  
    3          17    2.789068e+00     2.110e-01     1.000e+00     7.947e-01     3.581e-01  
    4          21    2.988095e+00     1.191e-02     1.000e+00     3.066e-01     4.893e-02  
    5          25    2.999988e+00     1.161e-05     1.000e+00     1.437e-02     9.121e-04  
    6          29    3.000000e+00     5.045e-09     1.000e+00     7.238e-05     1.451e-04  
    7          33    3.000000e+00     5.265e-09     1.000e+00     7.255e-05     1.646e-07  
Local minimum found that satisfies the constraints.
Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the value of the optimality tolerance,
and constraints are satisfied to within the value of the constraint tolerance.
<stopping criteria details>
x =
          2.99999999210178     -1.73125747607761e-08          1.00000000263274
          
>> x = fmincon(fun,[-1 1 0],A,b,Aeq,beq,lb,ub,nonlcon,options)
 Iter  Func-count            Fval   Feasibility   Step Length       Norm of   First-order  
                                                                       step    optimality
    0           4    1.000000e+00     3.000e+00     1.000e+00     0.000e+00     2.000e+00  
    1          12    1.720300e+00     2.280e+00     2.401e-01     7.203e-01     1.520e+00  
    2          19    1.688289e+00     1.455e+00     3.430e-01     9.684e-01     6.358e-01  
    3          23    2.587102e+00     4.133e-01     1.000e+00     1.342e+00     6.046e-01  
    4          27    2.929079e+00     7.341e-02     1.000e+00     6.240e-01     2.038e-01  
    5          31    3.000225e+00     6.050e-04     1.000e+00     9.457e-02     5.761e-02  
    6          35    3.000027e+00     5.576e-04     1.000e+00     2.362e-02     1.620e-02  
    7          39    3.000000e+00     2.761e-05     1.000e+00     5.308e-03     4.556e-04  
    8          43    3.000000e+00     2.435e-09     1.000e+00     6.453e-05     2.124e-05  
    9          47    3.000000e+00     0.000e+00     1.000e+00     4.398e-09     1.192e-06  
Local minimum found that satisfies the constraints.
Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the value of the optimality tolerance,
and constraints are satisfied to within the value of the constraint tolerance.
<stopping criteria details>
x =
         -3.00000000000053      5.92273862321821e-07        -0.999999999999825

MATLAB: Nonlinear constraints violated by fmincon in output with sqp and interior point algorithms

You misunderstand the documentation: the 'sqp' and 'interior-point' algorithms satisfy BOUNDS at all iterations, not nonlinear constraints.

For your problem, it is clear that the merit function that the solvers use internally is not working well. This merit function balances the two goals of lowering infeasibility and lowering the objective function. Therefore, I suggest that you artificially make your constraint function much larger, to have fmincon pay more attention to it. Multiply mycon by 1e4 or so (that is, have mycon return a value that is a large factor multiplied by the true constraint value)and see whether that helps.

Alan Weiss

MATLAB mathematical toolbox documentation

Best Answer

Related Solutions

MATLAB: Trying to minimise a function subject to nonlinear constraints

MATLAB: Nonlinear constraints violated by fmincon in output with sqp and interior point algorithms

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