Minimise the sum of squared errors, with non linear constraints

1 次查看（过去 30 天）
Lewis Marshall
Lewis Marshall 2021-3-26
编辑： Matt J 2021-4-5
hello i am trying to find the coefficient vlaues that minimises the sum of the squared erorrs between the eqaution ->
sigma_therory=((-(x(1) + x(2)./lamda(i)).*(2./lamda(i) - 2*lamda(i).^2))
and experimental data "sigma_c"
x(1) and x(2) are coefficient values in which im trying to calculate and "lamda" is the input variable to the experiment. x(1) >0 and x(1)+x(2)>0
I have generated this code which i think does what i require, however i know what i have wirtten in the for loop is poor code and ineffifcient as it takes very long to run, any help on how to improve this would be greatly appreicated.
y=0
for i=1:length(lamda)
fun{i}=@(x) y+((-(x(1) + x(2)./lamda(i)).*(2./lamda(i) - 2*lamda(i).^2))-sigma_c(i))^2;
y=fun; %save the output of the function to be added to the next itteration
end
x0=[1 -1];
A=[-1,0];
b=[0];
[xf,fval]=fmincon(fun,x0,A,b,[],[],[],[],@nolin);
function [c,ceq]=nolin(x) %to implement the coefficient x(1)+x(2)>0
c(1)=-x(1)-x(2)
ceq=[]
end

请先登录，再回答此问题。

回答（2 个）

Shravan Kumar Vankaramoni
Matt J
Matt J 2021-4-4
编辑：Matt J 2021-4-5
The unnecessary use of nonlinear constraints can slow things down. Your constraints are in fact linear, and should be expressed like this,
x0=[1 -1];
A=[-1,-1];
b=[0];
lb=[0,-inf];
[xf,fval]=fmincon(fun,x0,A,b,[],[],lb,[]);
Also, the way you have expressed your objective function should be giving you errors. You almost surely want this, instead of what you have above:
lamda=lamda(:); sigma_c=sigma_c(:);
fun=@(x) norm( (-(x(1) + x(2)./lamda).*(2./lamda - 2*lamda.^2))-sigma_c).^2 ;

类别

在 Help Center 和 File Exchange 中查找有关 Least Squares 的更多信息

标签

产品


版本

R2020b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by