trying to use fminbnd to minimize a function with resolution 0.01
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AJAY CHANDRA DORAGARI
le 5 Juin 2020
Commenté : AJAY CHANDRA DORAGARI
le 5 Juin 2020
function [e,t,lf]=f2(q,w)
e=fminbnd(@fun,q,w)
t=(1000/e)-(0.25*pi*e) (%these are final values please ignore them)
lf=(50*pi*r)+(((2*r)+(2*l))*40); (%these are final values please ignore them)
function [c]=fun(r)
r=q:0.01:w;
l=(1000./r)-(0.25*pi.*r);
c=(50*pi.*r)+(((2.*r)+(2.*l))*40);(%trying to minimize this function but i couldnt using fminbnd )
end
end
but im getti
ng these errors
Error using fminbnd (line 220)
User supplied objective function must return a scalar value.
Error in f2 (line 2)
e=fminbnd(@fun,q,w)
trying to learn matlab bascis on my own
please help me in this regard
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Matt J
le 5 Juin 2020
Modifié(e) : Matt J
le 5 Juin 2020
You cannot use fminbnd to guarantee a distance of 0.01 (or any other distance) to the global minimum, but typically you will do much better than that with the default tolerances. To eliminate your fminbnd error message remove the line, r=q:0.01:w,
e=fminbnd(@fun,q,w)
function [c]=fun(r)
%r=q:0.01:w;
l=(1000./r)-(0.25*pi.*r);
c=(50*pi.*r)+(((2.*r)+(2.*l))*40);
end
You can however guarantee a distance of 0.01 if you use exhaustive search, instead of fminbnd. That would look like,
r=q:0.01:w;
[~,minloc]=min(fun(r));
e=r(minloc)
function [c]=fun(r)
l=(1000./r)-(0.25*pi.*r);
c=(50*pi.*r)+(((2.*r)+(2.*l))*40);
end
5 commentaires
Matt J
le 5 Juin 2020
Modifié(e) : Matt J
le 5 Juin 2020
It is probably intended that you to set the TolX parameter to 0.01, as in the example below. But note that this is not the same as varying r in increements of 0.01.
q=5;w=10;
e=fminbnd(@fun,q,w,optimset('TolX',0.01))
function [c]=fun(r)
l=(1000./r)-(0.25*pi.*r);
c=(50*pi.*r)+(((2.*r)+(2.*l))*40);
end
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