Problem in cvx optimization


(Elham) #1

I WORK on code as

F=[0.95 0.05  ;
   0 1
   ];
H=[1 0];
P0=100*eye(2);
x=[1;-1];
xhat=[1;-1];

   cvx_begin sdp 
   variable L(2,1) 
   variable P(2,2) symmetric
   minimize (trace (P))
   
   c=[eye(2) -L]*[10^4*eye(2)  zeros(2,1);zeros(1,2) 0.0001];
   
  [P       (F-L*H)*P0  c
 
 P0*(F-L*H)'    P0    zeros(2,3)
 c'              zeros(3,2) [10^4*eye(2)  zeros(2,1);zeros(1,2) 0.0001]
 ]>=0 
   cvx_end

L is a vector which is optimization variable bu all the time i have L(2,1)=0 it means just L(1,1) arrange why??


(Mark L. Stone) #2

You provided one optimization problem, and its optimal solution has L(2,1) = 0. So that’s the optimal solution. I believe (but have nott definitively checked) that any solution not having L(2,1) = 0 is not optimal. I have no idea what your optimization problem represents, so I have no idea why an optimal value L(2,1) = 0 should be surprising or disconcerting.