Fmincon stopped because the size of the current step is less than the default value of the step size tolerance...How can I solve this?
2 views (last 30 days)
Show older comments
This is my coding
U=[0.04854 0 0 0; 0.03483 0.03125 0 0; 0.02477 0.02223 0.04000 0; 0.01738 0.01550 0.02807 0.05263];
m=[1.6378; 1.1753; 0.8359; 0.5867];
n=4;
fun=inline('200*0.5.^0.8+1000*0.5.^0.8*x(1).^2+200*0.3.^0.8+1000*0.3.^0.8*x(2).^2+200*0.4.^0.8+1000*0.4.^0.8*x(3).^2+200*0.5.^0.8+1000*0.5.^0.8*x(4).^2');
x0=[1 1 1 1];
a=[5;5;5;5];
b=[200;200;200;200];
A=[-200 0 0 0; 0 -200 0 0; 0 0 -200 0; 0 0 0 -200];
B=inv(U)*(a-m);
LB=[0 0 0 0];
UB=[1 1 1 1];
[x,fval]=fmincon(fun,x0,A,B,[],[],LB,UB,[]);
x;
fval;
L=ones(4,1);
for i=1:n
L(i)=200*(1-x(i));
end;
L;
L2=U*L+m;
But after running it, it shows
Local minimum possible. Constraints satisfied.
fmincon stopped because the size of the current step is less than
the default value of the step size tolerance and constraints are
satisfied to within the default value of the constraint tolerance.
<stopping criteria details>
May I ask, how can I solve this?
0 Comments
Accepted Answer
More Answers (0)
See Also
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!