lsqnonlin question
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Consider the following code:
X(1:10304)=xlsread('qs_comp_2.xls',2,'G3822:G14125');
Y(1:10304)=xlsread('qs_comp_2.xls',2,'H3822:H14125');
X(10305:12631)=xlsread('comp.xls',3,'T4259:T6585');
Y(10305:12631)=xlsread('comp.xls',3,'U4259:U6585');
X0=[1012 1400 0.17 -0.0001];
lb = [1012;900;0.17;-0.0001];
ub=[1300;1600;0.3119;10];
StartAt = [1012;900;0.17;-0.0001];
options = optimset('MaxFunEvals',10000);
x=lsqnonlin(@(X0)fit_simp(X0,X,Y),StartAt,lb,ub,options);
[total_readings,epsilon_dot_QS,epsilon_dot_MR] = GetMRDetails;
for i=1:total_readings
Y_new(i)=(X0(1)+X0(2)*(X(i)^X0(3))+X0(4)*log(epsilon_dot_QS/epsilon_dot_MR));
end
Please can anyone help me here:
1) Is the final optimised vector X0 or is it x? That is: which id the final answer X0 or X 2)Is there anything I'm missing in this code?
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Respuestas (2)
Sean de Wolski
el 18 de Abr. de 2012
It is x, the output from lsqnonlin. It doesn't look like the above code ever uses the actual optimized value.
Shalini
el 18 de Abr. de 2012
2 comentarios
Sean de Wolski
el 18 de Abr. de 2012
That looks fine, it's that your call to LSQNONLIN generates a vector called 'x'. This vector is not used in the remainder of the code you showed us.
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