MATLAB: Fzero or solve a file full of relations

fzerosolve

Hi all-

I was playing around with fzero here: http://www.mathworks.com/help/matlab/ref/fzero.html#description

Trying to get it to vary x below until y below is 0.

z=[-300.00 1082.00 7494.20];
x=1.750871192;
zprime=[z(1)*x/10000 z(2)*x/10000 z(3)*x/10000];
a=[1 zprime(3) zprime(2) zprime(1)]';
B=[1.637510856e-06 -2.252209129e-05 0.0001949906158 9.284333582e-06
-2.252209129e-05 0.217829072 0.03330113194 0.0449503853 
0.0001949906158 0.03330113194 0.5146272212 0.03759452473 
9.284333582e-06 0.0449503853 0.03759452473 0.7076169167];
Ba = B*a;
atBa = a'*Ba;
y=atBa-1;

I think it can be done with fzero which requires no license for solve. But I do have the license for solve if that's the way to go. I hate to fight over limited licenses 🙂

Hope this one isn't too tough. Need to call it many times, quickly.

Thanks a ton.

Dave

Best Answer

Try

function main
x0=1;
sol=fzero(@f,x0);
function y=f(x)
z=[-300.00 1082.00 7494.20];
B=[1.637510856e-06  -2.252209129e-05 0.0001949906158 9.284333582e-06
   -2.252209129e-05 0.217829072      0.03330113194   0.0449503853 
   0.0001949906158  0.03330113194    0.5146272212    0.03759452473 
   9.284333582e-06  0.0449503853     0.03759452473   0.7076169167];
for k=1:length(x)
  xk=x(k);
  a=[1 z(3)*xk/10000 z(2)*xk/10000 z(1)*xk/10000]';
  y(k)=a'*B*a-1;
end

Best wishes

Torsten.

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: How to implement the following

   x=rand(10,1);mu=randi(49,1,1)/100+0.0001% example
   for k=1:length(x)-1
       if  x(k)>=0 & x(k)<mu     
           x(k+1)=x(k)/mu 
       elseif x(k)>=mu & x(k)<0.5 
           x(k+1)=(x(k)-mu)/(0.5-mu)
       elseif x(k)>=0.5 & x(k)<1 
           x(k+1)=(1-x(k))/mu
       end
   end

Best Answer

Related Solutions

MATLAB: Trying to minimise a function subject to nonlinear constraints

MATLAB: How to implement the following

Related Question