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?
Risposte (2)
Sean de Wolski
il 18 Apr 2012
0 voti
It is x, the output from lsqnonlin. It doesn't look like the above code ever uses the actual optimized value.
1 Commento
Shalini
il 18 Apr 2012
Shalini
il 18 Apr 2012
2 Commenti
Sean de Wolski
il 18 Apr 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.
Shalini
il 18 Apr 2012
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