MATLAB: Do I get no solution found in Fsolve despite the results match the nonlinear equations

differential equationsfsolve

Hi All

I am solving a set of Non Linear equations. when I put the solution results in each of my equations, the result is zero or very close to zero ( 1e-15 as residual)

the options I use :

options = optimoptions(@fsolve,'Display','iter','MaxFunEvals',1e6,'MaxIter',1e6,'TolFun',1e-1,'DerivativeCheck','on','Diagnostics','on');

but on the message board I get the no solution found, that does not change with any tolerance value :

Here I put a copy of the last iterations and the message :

    23         72     1.44467e+07    0.000238419            1.4        0.000238
    24         73     1.44467e+07    0.000238419            1.4        0.000238
    25         77     1.44467e+07    5.96046e-05          0.325        5.96e-05
    26         78     1.44467e+07    5.96046e-05          0.325        5.96e-05
    27         82     1.44467e+07    1.49012e-05          0.147        1.49e-05
    28         86     1.44467e+07    1.49012e-05         0.0306        1.49e-05
No solution found.
fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
selected value of the function tolerance.
fsolve stopped because the sum of squared function values, r, has gradient with
relative norm 3.057666e-02; this is less than options.OptimalityTolerance = 1.000000e-01.
However, r = 1.444675e+07, exceeds sqrt(options.FunctionTolerance) = 3.162278e-01.
Optimization Metric                                          Options
norm(grad r) =   3.06e-02                        OptimalityTolerance =   1e-01 (selected)
r =   1.44e+07                               sqrt(FunctionTolerance) =  3.2e-01 (selected)

Best Answer

You're asking for guesses? Because you are not plugging the solution into the same equations that were given to fsolve. fsolve thinks you have a residual norm of sqrt(1.44e+07).

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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: How to plot with lines in a for loop and use legends for each line

There is no reason to use cell arrays for this code, and it actually causes the problem that you are experiencing. A very good data concept is to use the simplest array possible, unless a more complicated array is the only choice. In this case a simple numeric array holds all of that data quite nicely, so there is no need to use a cell array.

M = [0.0274 0.0268 0.0275;...
    0.0130 0.0128 0.0124;...
    0.0111 0.0108 0.0113;...
    0.0110 0.0107 0.0107;...
    0.0109 0.0107 0.0107;...
    0.0109 0.0107 0.0107;...
    0.0109 0.0107 0.0107;...
    0.0109 0.0107 0.0107]
plot(M,'-*')
legend({'first','second','third'})

creates this:

Summary: do not make data more complicate than it needs to be, it only makes the code more complicated!

PS: here is an easy way to convert that cell array into a simple numeric matrix:

M = vertcat(matrix{:});

Best Answer

Related Solutions

MATLAB: Trying to minimise a function subject to nonlinear constraints

MATLAB: How to plot with lines in a for loop and use legends for each line

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