Jason Nicholson
님이 질문을 제출함. 28 May 2018

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.

Answer by Steve Grikschat
on 29 May 2018

Accepted Answer

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);

Jason Nicholson
3 Jun 2018

Using lsqlin with the interior-point solver gives the following picture:

With "re-formulation to a minimal distance problem" and using the lsqnonneg I get the following:

As you can see, the re-formulation and use of the Active-set (lsqnonneg) solver works really well for this problem. Again, thank you.

Also, it should be noted that this is very fast. The code execute almost instantaneously. The solution with fmincon was a factor 10-100 slower.

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.

Jason Nicholson
on 13 Jun 2018

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Answer by 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.

Jason Nicholson
on 29 May 2018

Jason Nicholson
on 3 Jun 2018

This is a valid solution but it seems slow. I will keep it mind. Below is more information.

First of all, here is a better cost function that includes the gradient:

function [cost, gradient] = costFunction(x,C,d,CC,Cd)

residuals = C*x-d;

cost = 1/2*(residuals'*residuals);

if nargout > 1

gradient = CC*x-Cd;

end

end

Next, we set the options to fmincon with the active-set solver

options = optimoptions('fmincon', 'Algorithm', 'active-set', ...

'SpecifyObjectiveGradient', true, 'Display', 'iter-detailed');

or you can use the sqp solver

options = optimoptions('fmincon', 'Algorithm', 'active-set', ...

'SpecifyObjectiveGradient', true, 'Display', 'iter-detailed');

Next we make a find a reasonable guess:

x0 = C\d;

Then we call fmincon:

x = fmincon(@(x) costFunction(x,C,d,C'*C,C'*d),x0,A,b, ...

[],[],[],[],[],options);

I get the following graph (for both active-set and sqp) which shows the fit matches the test points.

The only downside is this is slow compared to the "re-formulate to a minimal distance problem" posted by Steve Grikschat.

It should be noted that the sqp and the active-set solver both gives the same results. This is good news. The sqp was faster for my test problem by a factor of 10. active-set took 127 seconds. sqp took 13 seconds. "reformulate to a minimal distance problem" took 0.063 seconds.

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