# Finite time horizon sdp problem

I have the following sdp formulation in infinite time horizon

\begin{equation} \begin{split} \min_{P,Z} &\ trace(SXP)+trace(Z)\\ \text{sbj to } & \\ &\begin{bmatrix} XP-I & YP\\ P^T Y^T & XP \end{bmatrix} \succeq 0\\ &\begin{bmatrix} Z & R^{1/2}UP\\ P^T U^T R^{1/2} & XP \end{bmatrix}\succeq 0 \end{split} \end{equation}

that is in cvx:

       cvx_begin sdp
variable Z(1,1)
variable P(T,n)
minimize( trace(S*X*P)+ trace(Z) )
subject to
[Z, R^{1/2}*U*P; P'*U'*R^{1/2}, X*P] >= 0
[X*P-eye(n), Y*P; P'*Y', X*P] >= 0
cvx_end
K_cvx = -U*P*inv(X*P)


Now, I want to re-write it in finite horizon. Assuming my formulation is correct, it is the following

\begin{equation} \begin{split} \min_{\substack{P_k,Z_k \\k\in \{ 0,\dots, N-1\} }} &\ trace(S_f X P(N)) + \sum_{k=0}^{N-1} \Big(trace(SXP(k))+trace(Z_k)\Big) \\ \text{sbj to } & \\ &\begin{bmatrix} X P(k+1) -I & Y P(k)\\ (Y P(k))^T & X P(k) \end{bmatrix} \succeq 0\\ &\begin{bmatrix} Z_k & R^{1/2}U P(k)\\ P(k)^T U^T R^{1/2} & X P(k) \end{bmatrix}\succeq 0\\ & X P(0) \succeq I_n \end{split} \end{equation}

which in cvx is

for j = 1:N-1
cvx_begin sdp
variable Z(1,1,N)
variable P(T,n,N)
minimize( trace(S*X*P(:,:,j))+ trace(Z(:,:,j)) + trace(Sf*X*P(:,:,N)) );
subject to
[Z(:,:,j), R^(0.5)*U*P(:,:,j); P(:,:,j)'*U' *R^(0.5), X*P(:,:,j)] >= 0;
[X*P(:,:,j+1)-eye(n), Y*P(:,:,j); P(:,:,j)' *Y' , X*P(:,:,j)] >= 0;
X*P(:,:,1) >= eye(n);
cvx_end
K_cvx_fh{j} = -U*P(:,:,j)*inv(X*P(:,:,j));
end


however, for {N \rightarrow \infty}, the solution is not converging to the infinite horizon solution. Do you have any tips to correct it?

I won’t address what should or shouldn’t converge to what.

However, are you sure you are even specifying semidefinite constraints? Specifically, what guarantees that X*P is symmetric? You haven;t declared P to be symmetric (recall your previous thread LMI formulation using Yalmip vs CVX ), and I have no idea what (the numerical value of ) X is. Did you get a warning about asymmetry?

You mkay be better off at a different forum for a discussion of what should converge to what.

I already know the value of K (which it is obtained with Yalmip) and I want to obtain the same value using cvx. Solving the problem for infinite horizon formulation, K_{cvx} gives indeed K. I want also to write the sdp formulation in finite time, and to see if it is correct, I have chosen a big N so that the final K_{cvx_{fh}} converges to K_{cvx}. Unfortunately, there is still something missing because K_{cvx_{fh}} \neq K_{cvx} for N\rightarrow \infty .

Is it because I haven’t written properly in the code the \sum_{k=0}^{N-1} in the trace of the second formulation?

In each CVX invocation, the objective as you have written the code is the sum of 3 trace terms. You are solving N-1 separate CVX problems. if that is not what you want to do, you need to change your code. You can, for instance, use a for loop from 0 to N-1 inside one CVX invocation (cvx_begin … cvx_end) to build up an objective function summing over terms from 0 to N-1.

If you show successful YALMIP code, hopefully it will not be difficult to determine how to implement it using CVX.

you are right, I am solving N cvx problem, which should not be the case. I need to solve this objective function \min_{P(k), Z(k)} \ trace(S_f X P(N)) + \sum_{k=0}^{N-1} (trace(SX P(k)) + trace(Z(k))). So i need to move the loop inside cvx_begin and take the sum over k in the objective function. Do you have any tutorials to help me out with this formulation? How can I write something like

    minimize( sum(trace(SX P(k)) + trace(Z(k)) )


and also I believe that it is wrong to define Z and P as

     variable Z(1,1,N)
variable P(T,n,N)


since the loop is inside cvx and I need to solve just one objective function.

Fix this up to be what you want.

    Objective = 0
for k=1:N
Objective = Objective + trace(S*X*P(:,:,k)) + trace(Z(:,:,k))
end
minimize(Objective)


Constraints can also go inside a for loop if needed.

So there should be

     cvx_begin sdp
variable Z(1,1,N)
variable P(T,n,N)
obj = 0;
for j = 1:N-1
obj = obj + trace(S*X*P(:,:,j))+ trace(Z(:,:,j) )
subject to
[Z(:,:,j), R^(0.5)*U*P(:,:,j); P(:,:,j)'*U' *R^(0.5), X*P(:,:,j)] >= 0;
[X*P(:,:,j+1)-eye(n), Y*P(:,:,j); P(:,:,j)' *Y' , X*P(:,:,j)] >= 0;
X*P(:,:,1) >= eye(n);
end
minimize(obj + + trace(Sf*X*P(:,:,N))  )
cvx_end

K_cvx_fh = -U*P(:,:,N)*inv(X*P(:,:,N));


Still,

    Status: Failed
Optimal value (cvx_optval): NaN

Is this infeasible? Look at my first post of this thread and address that.

Hmm, well if n does not equal N, then P can not be symmetric. Nevertheless, if X*P(:,:,j) is not symmetric, those can not be valid SDPs.

yes this is the problem indeed, this is not clear to me how these 3D arrays work…
so that would be XP(k+1)-I_n where P(k+1) is size T\times n and X is a data vector of size n \times T and k=0, \dots ,N.

Considering the 3d array

X*P(:,:,j+1)-eye(n)

should I change the size of eye(n)?

I have no idea what you are doing. It is necessary, but not sufficient, that dimensions be conformal (consistent, as required in pre-2016b MATLAB).

P(N,N,K) symmetric
declares that each of the K 2-D slices, P(:,:,k) for k from 1 to N is symmetric. However, you have P(T,n,N), so this can not be declared symmetric unless T = n.

If you can not construct a symmetric matrix, then you can not use it in an SDP. Maybe you need to go back to the origins of the problem and make sure you have proper a mathematical formulation for specifying SDPs. Try to get the single time period problem correct before you venture into the multi-period problem.

This is the single time (asymptotic behavior )problem

   cvx_begin sdp
variable Z(1,1)
variable P(T,n)
minimize( trace(S*X*P)+ trace(Z) )
subject to
[Z, R^{1/2}*U*P; P'*U'*R^{1/2}, X*P] >= 0
[X*P-eye(n), Y*P; P'*Y', X*P] >= 0
cvx_end
K_cvx = -U*P*inv(X*P)


also in this case I have P(T,n) where T\neq n but it returns the correct value for K_{cvx}. Here there is no terminal penalty trace(Sf*X*P(:,:,N)) in the cost function and no initial condition X*P(:,:,1) >= eye(n)

Using the optimal X, is X*P symmetric psd? using optimal Z and X, are [Z, R^{1/2}*U*P; P'*U'*R^{1/2}, X*P] and [X*P-eye(n), Y*P; P'*Y', X*P] psd?

They are psd but not symmetric for both cases.

CVX deals with symmetric matrices in SDP (LMI) constraints. If the matrix is not symmetric, then you have not implemented anything meaningful in CVX. Perhaps you happen to get a correct or reasonable answer in some cases.

You need to formulate structurally symmetric matrices to be constrained to be semidefinite. If you can not ot do that, I think you are playing around in nonsense, whether you use CVX or YALMIP.