CODEWITHSUNDEEP
matlaboptimization
user2042145
user2042145

Reputation: 141

Optimization of matrix on matlab using fmincon

I have a 30x30 matrix as a base matrix (OD_b1), I also have two base vectors (bg and Ag). My aim is to optimize a matrix (X) who's dimensions are 30X30 such that: 1) the squared difference between vector (bg) and vector of sum of all the columns is minimized. 2)the squared difference between vector (Ag) and vector of sum of all rows is minimized. 3)the squared difference between the elements of matrix (X) and matrix (OD_b1) is minimized.

The mathematical form of the equation is as follows:

enter image description here

I have tried this:

 fun=@(X)transpose(bg-sum(X,2))*(bg-sum(X,2))+ (Ag-sum(X,1))*transpose(Ag-sum(X,1))+sumsqr(X_b-X);
[val,X]=fmincon(fun,OD_b1,AA,BB,Aeq,beq,LB,UB)

I don't get errors but it seems like it's stuck.

Is it because I have too many variables or is there another reason?

Thanks in advance

Upvotes: 1

Views: 2576

Answers (2)

Erwin Kalvelagen
Erwin Kalvelagen

Reputation: 16797

I would probably solve this as a QP using quadprog using the following reformulation (keeping the objective as simple as possible to make the problem "less nonlinear"):

min sum(i,v(i)^2)+sum(i,w(i)^2)+sum((i,j),z(i,j)^2)
v = bg - sum(c,x)
w = ag - sum(r,x)
Z = xbase-x

The QP solver is more precise (no gradients using finite differences). This approach also allows you to add additional bounds and linear equality and inequality constraints.

The other suggestion to form the first order conditions explicitly is also a good one: it also has no issue with imprecise gradients (the first order conditions are linear). I usually prefer a quadratic model because of its flexibility.

Upvotes: 0

Matthew Gunn
Matthew Gunn

Reputation: 4529

This is a simple, unconstrained least squares problem and hence has a simple solution that can be expressed as the solution to a linear system.

I will show you (1) the precise and efficient way to solve this and (2) how to solve with fmincon.

The precise, efficient solution:

Problem setup

Just so we're on the same page, I initialize the variables as follows:

n = 30;
Ag  = randn(n, 1);    % observe the dimensions
X_b = randn(n, n);
bg  = randn(n, 1);

The code:

A1 = kron(ones(1,n), eye(n));
A2 = kron(eye(n), ones(1,n));
A = (A1'*A1 + A2'*A2 + eye(n^2));
b = A1'*bg + A2'*Ag + X_b(:);
x = A \ b;                      % solves A*x = b
Xstar = reshape(x, n, n);

Why it works:

I first reformulated your problem so the objective is a vector x, not a matrix X. Observe that z = bg - sum(X,2) is equivalent to:

x = X(:)                        % vectorize X
A1 = kron(ones(1,n), eye(n));   % creates a special matrix that sums up
                                % stuff appropriately
z = A1*x;

Similarly, A2 is setup so that A2*x is equivalent to Ag'-sum(X,1). Your problem is then equivalent to:

minimize (over x) (bg - A1*x)'*(bg - A1*x) + (Ag - A2*x)'*(Ag - A2*x) + (y - x)'*(y-x) where y = Xb(:). That is, y is a vectorized version of Xb.

This problem is convex and the first order condition is a necessary and sufficient condition for the optimum. Take the derivative with respect to x and that equation will define your solution! Sample example math for almost equivalent (but slightly simpler problem is below):

minimize(over x) (b - A*x)'*(b - A*x) + (y - x)' * (y - x)

rewriting the objective:

b'b- b'Ax - x'A'b + x'A'Ax +y'y - 2y'x+x'x

Is equivalent to:

minimize(over x) (-2 b'A - 2y'*I) x + x' ( A'A + I) * x

the first order condition is:

(A'A+I+(A'A+I)')x -2A'b-2I'y = 0
(A'A+I) x = A'b+I'y

Your problem is essentially the same. It has the first order condition:

(A1'*A1 + A2'*A2 + I)*x = A1'*bg + A2'*Ag + y

How to solve with fmincon

You can do the following:

f = @(X) transpose(bg-sum(X,2))*(bg-sum(X,2)) + (Ag'-sum(X,1))*transpose(Ag'-sum(X,1))+sum(sum((X_b-X).^2));
o = optimoptions('fmincon');%MaxFunEvals',30000);
o.MaxFunEvals = 30000;
Xstar2 = fmincon(f,zeros(n,n),[],[],[],[],[],[],[],o);

You can then check the answers are about the same with:

normdif = norm(Xstar - Xstar2)

And you can see that gap is small, but that the linear algebra based solution is somewhat more precise:

gap = f(Xstar2) - f(Xstar)

If the fmincon approach hangs, try it with a smaller n just to gain confidence that my linear algebra based solution is more precise, way way faster etc... n = 30 is solving a 30^2 = 900 variable optimization problem: not easy. With the linear algebra approach, you can go up to n = 100 (i.e. 10000 variable problem) or even larger.

Upvotes: 1

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