Applying the Nonlinear Least Squares Method to Minimize the Objective Function to Find the Parameters of the Equation

7 次查看（过去 30 天）
yu zhang
yu zhang 2022-7-4
编辑： yu zhang 2022-7-5
The liquid-liquid equilibrium data were fitted using the NRTL equation. The equation parameters of NRTL are derived from the experimental data.
below is my code:
clc,clear
options=optimset('MaxIter',4000,'MaxFunEvals',2000000,'algorithm','levenberg-marquardt');
b0=[0,0,0,0,0,0];
[b,gamma1,gamma2,gamma3,gamma4,gamma5,gamma6]=lsqnonlin('NRTL',b0,[],[],options);
Although the code gives the result, the result is not what I want. And the format of the output gamma is also wrong. It should be a matrix of the same order as x1, but it has become something I don't understand. Can someone please give a reason? It would be best to give some solutions. Thanks!

请先登录，再回答此问题。

采纳的回答

Matt J
Matt J 2022-7-4
编辑：Matt J 2022-7-4
Your code makes strange assumptions about the output syntax of lsqnonlin.
[x,resnorm,residual,exitflag,output,lambda,jacobian] = lsqnonlin(___)
Why do you think lsqnonlin will return values for the gamma variables in NRTL?
  6 个评论
显示 4更早的评论
Torsten
Torsten 2022-7-4
编辑：Torsten 2022-7-4
If you want
x(:,1)+x(:,2)+x(:,3)-1.0 = 0;
x(:,4)+x(:,5)+x66-1.0 = 0
to be satisfied exactly and if you want x >= 0, use
options=optimset('MaxIter',40000,'MaxFunEvals',2000000,'algorithm','levenberg-marquardt');
x0=0.1.*ones(5,5);
lb = zeros(5,5);
ub = ones(5,5);
x=lsqnonlin('NRTLyan',x0,lb,ub,options);
together with the x(:,:) squared in your equations in function NRTLyan.
If you want x>= 0 and if it suffices if the relations
x(:,1)+x(:,2)+x(:,3)-1.0 = 0;
x(:,4)+x(:,5)+x66-1.0 = 0
are only approximately satisfied, use
options=optimset('MaxIter',40000,'MaxFunEvals',2000000,'algorithm','levenberg-marquardt');
x0=0.1.*ones(5,5);
lb = zeros(5,5);
ub = ones(5,5);
x=lsqnonlin('NRTLyan',x0,lb,ub,options);
together with
f=[x(:,1).*gamma1-gamma4.*x(:,4);x(:,2).*gamma2-gamma5.*x(:,5);x(:,3).*gamma3-gamma6.*x66;x(:,1)+x(:,2)+x(:,3)-1.0;x(:,4)+x(:,5)+x66-1.0];
If you want the relations
x(:,1)+x(:,2)+x(:,3)-1.0 = 0;
x(:,4)+x(:,5)+x66-1.0 = 0
to be satisfied exactly, but x>= 0 does not matter, only solve for x(:,i) for i=1,2,4 and calculate x(:,j) for j=3,5 from the relation.
Matt J
Matt J 2022-7-4
Do you have any other suggestions?
@yu zhang I suggest you Acccept-click this answer, since your original question appears to have been resolved.
Since you have a new question and since it is related to code different from this question, I suggest you post that in a new thread.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Solver Outputs and Iterative Display 的更多信息

标签

产品


版本

R2022a

Community Treasure Hunt

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

Start Hunting!

Translated by