'error message Numerical problems (SeDuMi-1.3) when try to find the controller parameter'

[zia]

Why when i simulate this simple code, got message numerical problems??Its means that my code is not feasible??

Code:

opt = sdpsettings;

opt = sdpsettings(opt,'verbose',0);

z=tf('z');

%% System A'=A+B2F, B'=B1, C'=F, D=0

A=[0.7 0 0; 0 0 1; 0 -1 2];

B1=[1; 0 ; 1];

B2=[-1; 0; -1];

D=0;

L=sdpvar(1,3,'full');

na=size(A,1);

nc1=size(L,1); % size of the row of matrix L

nb=size(B1,2); % size of the 'column' of matrix B

X= sdpvar(na,na,'symmetric'); % set the variable X

Gm = sdpvar(1,1);

%L=K*X

%% Linear approximation

Mh=[-X                A*X+B2*L      B1            zeros(na,nc1);

   X*A'+L'*B2'        -X           zeros(na,nb)     L';

    B1'            zeros(nb,na)    -Gm*eye(nb)       0;

    zeros(nc1,na)     L              0            -Gm*eye(nc1)];

    F1 = [ Mh <=0,  X>=0] 

    solution = solvesdp(F1,Gm,opt)

    REPORT = solution.problem  

    Gm1=double(Gm)

    w=double(L)

    x=double(X)

    Fs=w/x

Result:

++++++++++++++++++++++++++++++++

|   ID|              Constraint|

++++++++++++++++++++++++++++++++

|   #1|   Matrix inequality 8x8|

|   #2|   Matrix inequality 3x3|

++++++++++++++++++++++++++++++++

solution = 

  struct with fields:

    yalmiptime: 0.1512

    solvertime: 0.5118

          info: 'Numerical problems (SeDuMi-1.3)'

       problem: 4

Really appreciate your reply. Thanks.

[Johan Löfberg]

No, it simply means that sedumi runs into numerical problems as you have posed a problem where most likely the optimal X tends to infinity (or is unreasonably large numerically). The computed solution is feasible, hence the problem is obviously feasible

>> check(F1)

 

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

|   ID|          Constraint|   Primal residual|   Dual residual|

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

|   #1|   Matrix inequality|        9.3443e-06|      9.5987e-11|

|   #2|   Matrix inequality|           0.53147|      1.6339e-10|

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

 

>> value(X)

ans =

   1.0e+05 *

    1.9766   -0.0009   -0.0009

   -0.0009    2.9621    2.9588

   -0.0009    2.9588    2.9555

Putting some regularization on X might give a better problem

solution = solvesdp(F1,Gm + .0001*norm(X))

or avoid the ill-posed optimal solution by noting that the optimal gamma appears to be 1, solve a feasibility problem instead where you aim for 5% sub-optimal at most. Most often, the solver will steer towards a better conditioned solution

solution = solvesdp([F1,Gm <= 1.05])