Help in solving a weird error with yalmip!

[David]

Hello all,

I have the following code

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N = 50;

h = 1/N;

H1 = [ones(1,N-10) zeros(1,10)].';

H2 = 5*[ones(1,N-10) zeros(1,10)].';

x = sdpvar(N,1);

y = sdpvar(N,1);

Objective = 0.5*h*(sum(log(x))+sum(log(1-y.^2)));

Constraints = set(sum(h*abs(H2).*sqrtm(x).*y.*sqrt(abs(H1).^2+1/var))-0.5*sum(h*abs(H2).^2.*(x-1))-sum(h*real(conj(H2).*H1)) == 0);

Constraints = Constraints + set(sum(h*x) == 1);

Constraints = Constraints + set(-1<=y<=1)+set(x>=0);

assign(x,1)

assign(y,1)

solvesdp(Constraints,-Objective)

solx = double(x);

soly= double(y);

Objective = 0.5*h*(sum(log(solx.*(1-soly.^2))));

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Now, using this code, I obtained that the maximal objective value is equal to 0.0819. However, if I simply choose the following solution

solx = ones(N_partition,1);

soly = H1./sqrt(abs(H1).^2+1);

which satisfy all the constraints, then the obtained objective is bigger = 0.2773 !! 

1. Where is my mistake?

2. Why when I changed the objective function from 

Objective = 0.5*h*(sum(log(x))+sum(log(1-y.^2)))

to

Objective = 0.5*h*(sum(log(x.*(1-y.^2))))

which are essentially the same, I obtained different solutions?

Thanks a lot!

[David]

For completeness, the problem that I wish to solve is mathematically given by

MAX 0.5*h*(sum(log(x))+sum(log(1-y.^2)))

{x,y}

SUBJECT TO

sum(h*abs(H2).*sqrtm(x).*y.*sqrt(abs(H1).^2+1/var))-0.5*sum(h*abs(H2).^2.*(x-1))-sum(h*real(conj(H2).*H1)) == 0

&&

sum(h*x) == 1

&&

x>=0

&&

-1<=y<=1.

[Johan Löfberg]

Are you saying that you can generate a better solution than what is obtained by the solver? That is of absolutely no surprise, as the problem looks nastily non-convex, hence a local solver will only give you *some* locally optimal point, if it even finds a feasible solution.

BTW: var is undefined so I cannot run the code

BTW2: SET is obsolete, use concatenation  

Constraints = [Constraints, sum(h*x) == 1];

[Johan Löfberg]

You are assigning initial values to x and y, but these will not be used since you haven't told the solver to use these

solvesdp(Constraints,-Objective,sdpsettings('usex0',1))

[David]

> Are you saying that you can generate a better solution than what is obtained by the solver? That is of absolutely no surprise, as the problem looks nastily non-convex, hence a local solver will only give you *some* locally

> optimal point, if it even finds a feasible solution.

Yes, exactly. What about changing the first equality constraint to inequality?

> BTW: var is undefined so I cannot run the code

Ops, just take var=1.

> BTW2: SET is obsolete, use concatenation  

OK, thanks!

[Johan Löfberg]

I think you are making some sign error on the objetive.

You want to minimize -Objective. The solution I get (using var=1) with fmincon is 0.08

If I compute the value with you solution, -Objective equals  0.27726 which is worse than the computed one. Hence, fmincon computes a better solution.

(in other words, -0.08 is bigger than -0.27 which is good since you want to maximize that number)

[David]

I want to obtain the following quantity

R = 

-1 * MAX 0.5*h*(sum(log(x))+sum(log(1-y.^2)))

{x,y}

SUBJECT TO

sum(h*abs(H2).*sqrtm(x).*y.*sqrt(abs(H1).^2+1/var))-0.5*sum(h*abs(H2).^2.*(x-1))-sum(h*real(conj(H2).*H1)) == 0

&&

sum(h*x) == 1

&&

x>=0

&&

-1<=y<=1.

Thus, I use solvesdp(Constraints,-Objective) to solve a minimization problem instead. 

[David]

Sorry, please Ignore my last comment.

[Johan Löfberg]

Well, it boils the curvature of sqrt(x)*y, which is neither convex nor concave, so it will never be a convex  constraint

[X,Y] = meshgrid(0:0.01:2,-1:0.01:1);

mesh((X.^.5).*Y)