How to formalize an optimization problem in Matlab?

Hello.

How to issue a task in Matlab?

Given:

Vectors a1 and a2.

Find a vector of coefficients x such that:

abs((a2.')*x) -> min

abs ((a1.')*x) >= condition

under the conditions:

- Dimensions of vectors a1, a2 and x from 1 to p

- elements a1, a2 and x are complex numbers

- abs(a1_i)>0, abs(a2_i)>0, abs(x_i) = 1, where i =1..p

[Torsten]

So you want to maximize abs(a1.'*x). And you want to minimize abs(a2.'*x).

How do you want to combine these two objectives in one ?

[reincornator]

It is necessary to find a vector x such that:

abs ((a1.')*x) -> max - goal

abs((a2.')*x) < tol, wheh tol->0 - limitations

[Torsten]

Seems you get a quadratic objective with quadratic constraint. Try fmincon.

[Matt J]

abs(x_i) = 1, where i =1..p

Are you sure it's not abs(x_i) <= 1 or sum(abs(xi).^2)=1? It would be hard to satisfy the constraint as posted.

For example, if a2=[1;0;0;0] it would be impossible to satisfy abs(a2.'*x)-->0, because abs(a2.'*x)=abs(x1) and abs(x1) is constrained to 1.

[reincornator]

Torsten, thanks. The "fmincon" function only works with real numbers. I use complex numbers.

[reincornator]

Matt J, thanks. You are right, there is another constraint: abs(a1_i)>0, abs(a2_i)>0, i=1..p.

[Torsten]

@reincornator

With each optimization software you use, you will have to split in real and imaginary part and thus handle two real-valued numbers simultaneously.

[Matt J]

You are right, there is another constraint: abs(a1_i)>0, abs(a1_i)>0, i=1..p.

I assume you meant abs(a1_i)>0, abs(a2_i)>0.

That's still not enough. Consider a2=[2,1].'. Then if abs(x1)=abs(x2)=1,

abs(a2.'*x)=abs(2*x1+x2)
            >=abs( 2*abs(x1) -abs(x2))
            =1

So, the constraint abs(a2.'*x)=0 can never be satisfied.

[reincornator]

@Matt J, thank you. Corrected the condition.

abs((a2^T)*x) < tol, tol-> 0

[Matt J]

My counter-example still applies. There is little guarantee that a solution will exist under those constraints.

[Matt J]

If you had abs(a2_i)=1 for all i=1...p. You might be able to guarantee existence of a solution for p>1.

[reincornator]

@Torsten, in real numbers, the condition abs(x_i)=1, can be written as Re(x_i)=sqrt(1-Im(x_i)^2). How to write such a condition for vector x in constraints?

[Matt J]

How to write such a condition for vector x in constraints?

Your problem wouldn't be written in terms of x_i in what Torsten is recommending. Your objective and constraints would be written in terms of independent variables (u_i, v_i) where u_i stands in for the real component of x_i and v_i stands in for the imaginary component. The constraint, written differentiably, would be,

u_i^2 + v_i^2 = 1

[reincornator]

@Torsten @Matt J

How to set 2 restrictions in the "fmincon" function?

[Matt J]

c and ceq can be vectors.

Answer 1

[Matt J]

N=null(a2.');

a3=a1.'*N;

[~,idx]=max(abs(a3));


x=N(:,idx);

[reincornator]

No, it doesn't work. In addition, no restrictions are set: abs(x) == 1.

randcomplex = @(x,y) rand(x,y).*exp(2i*pi*rand(x,y));
a1 = randcomplex(1,10);
a2 = randcomplex(1,10);

N=null(a2.')

N =

  1×0 empty double row vector

a3=a1.'*N

a3 =

  10×0 empty double matrix

[~,idx]=max(abs(a3))

idx =

  1×0 empty double row vector

x=N(:,idx)

x =

  1×0 empty double row vector

[Matt J]

randcomplex = @(x,y) rand(x,y).*exp(2i*pi*rand(x,y));
a1 = randcomplex(10,1);
a2 = randcomplex(10,1);

N=null(a2.');
a3=a1.'*N;
[maxval,idx]=max(abs(a3));
x=N(:,idx)*conj(a3(idx))/abs(a3(idx))

x = 
   0.1546 - 0.3938i
   0.0998 + 0.0391i
   0.0617 - 0.0385i
   0.0246 + 0.1474i
   0.1035 + 0.0910i
   0.0625 + 0.0216i
  -0.5747 - 0.6314i
   0.0177 + 0.0661i
   0.0371 - 0.0283i
   0.0319 + 0.1476i

a2.'*x

ans = -8.3267e-17 + 5.5511e-17i

maxval,

maxval = 0.7671

abs(a1.'*x)

ans = 0.7671

[reincornator]

@Matt J, thank you. The graph shows that such a solution does not satisfy the condition abs(x_i)=1

randcomplex = @(x,y) rand(x,y).*exp(2i*pi*rand(x,y));
a1 = randcomplex(40,1);
a2 = randcomplex(40,1);

N=null(a2.');
a3=a1.'*N;
[maxval,idx]=max(abs(a3));
x=N(:,idx)*conj(a3(idx))/abs(a3(idx));
a2.'*x;
[a1.';a2.']*x

ans = 
   0.9866 + 0.0000i
   0.0000 - 0.0000i

plot(abs(x))

[Matt J]

The graph shows that such a solution does not satisfy the condition abs(x_i)=1

Yes, because as I've said above, I don't think you will be able to find such a solution.

[reincornator]

@Matt J, I found one solution. It is implemented using a special algorithm and shows a good result. But I'm looking for a better way.

[Torsten]

I thought a2'*x should be 0...