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I have a problem of the form

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.

Steve Grikschat
on 29 May 2018

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);

Steve Grikschat
on 4 Jun 2018

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.

Nikhil Negi
on 29 May 2018

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.

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