Reformulate a Constrained Linear Least Square Problem
2 次查看(过去 30 天)
显示 更早的评论
I have a problem of the form
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/190884/image.png)
Normally you solve it like this:
u = (0:0.1:10)';
v = sin(2*pi*1/30*(u-0.5));
plot(u,v);
xlabel('u');
ylabel('v');
C = u.^[0 1 2 3 4 5 6 7];
d = v;
u2 = (0:0.01:11)';
A = -u2.^[0 1 2 3 4 5 6 7];
b = zeros(size(A,1),1);
% normally just use lsqlin to solve this
x = lsqlin(C,d,A,b);
hold all
plot(u,C*x)
However, my problem is actually quite large and ill-conditioned because of the high degree of polynomial that I am fitting. The interior-point solver uses C'*C to solve this problem. That is a problem because it squares the condition number of C. The interior-point algorithm struggles to converge while the older, now deprecated Active-set algorithm works well. The Active-set algorithm wasn't large scale. It was medium scale. It worked great. However, I don't have that available. What are some ways I can stabilize and solve this problem?
Some ideas:
- Maybe I can use the null space of A?
- Maybe I can reformulate the problem to fmincon with proper settings.
- I know about orthogonal transformation for a single dimension polynomial. However, I work with multidimensional polynomials and even though orthogonal decomposition and formulations exist, I don't know them or how to use them. If you propose this, please refer to how I can use these and where I can learn about them and understand them.
0 个评论
采纳的回答
Steve Grikschat
2018-5-29
You can try a trick from Lawson & Hanson and re-formulate to a minimal distance problem and solve via lsqnonneg. lsqnonneg is an active-set method, so if those work for you, then it might as well.
The minimum distance problem looks like:
minimize ||x||^2
x
s.t. Abar*x <= bbar
Code:
% Transform into minimal distance
[Q,R] = qr(C,0);
dbar = Q'*d;
Abar = A/R;
bbar = b - ARinv*dbar;
% Get min-distance into lsqnonneg form
n = size(Abar,2);
E = [Abar';
bbar'];
f = [zeros(n,1); -1];
[u,~,residual] = lsqnonneg(E,f);
xbar = -residual(1:n)/residual(end);
% Map back
x = R\(xbar+dbar);
6 个评论
Steve Grikschat
2018-6-4
Glad you found my typo and that this worked for you.
fmincon is going to be slower on structured problems like this because it doesn't assume any structure.
更多回答(1 个)
Nikhil Negi
2018-5-29
Hello Jason,
let me see if i understand your problem coorectly you want to minimize the function J which is constrained by the inequality A*x < 0, here im assuming you want A*x < b where b is a mineq x 1 zero vector. please correct me if i'm wrong.
i think you can use fmincon function of matlab for this problem very efficiently.
%define J as
J=@(x)0.5*((rssq(C*x-d))^2);
x0=rand(8,1);
b=zeros(mineq,1);
x=finmincon(J,x0,A,b);
use different random values of x0 because it might give local minima (fmincon is generally used for convex functions because we can not be sure if the minima given is local or global) and compare J(x) for all these x obtained and compare J(x) for these x's and the minimum of these will give you your answer. now its not 100% certain because there are chances that it will always get stuck on the local minima and never reach global minima but i think if you do it for 100 or 1000 random x0 it should give u the correct anwer.
2 个评论
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Linear Least Squares 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!