Low Complexity Polyhedral Calculation using Yalmip (LMI)

[Himanshu Nagpal]

Hi there,

I have been using Algorithm 1 in pic1 to compute low-complexity polyhedral invariant sets in Example 2 given in pic2. I am getting different answers when I use mosek and SeDuMi solvers and both of them are not matching with the correct answer. 

clc

clear all

A = [0.858, -0.053, 0.021, 0.193, -0.029, 0.027;

    -0.053, 0.716, 0.039, 0.051, 0.179, 0.162;

     0.021, 0.039, 0.698, 0.031, 0.135, -0.206;

     0.193, 0.051, 0.031, 0.567, 0.053, -0.014;

     -0.029, 0.179, 0.135, 0.053, 0.658, -0.031;

     0.027, 0.162, -0.206, -0.014, -0.031, 0.504];

 

B = [-0.438, -1.607, -0.474;

    -0.074, -1.479, 1.389;

    -1.647, 1.011, 0.204;

    0.866, 1.869, 0.821;

    0.633, 0.004, -1.364;

    -0.997, 1.185, -0.554];

u = [1,1,1]';

nx = size(A,2);

nu = size(B,2);

Q = sdpvar(nx,nx);

Y = sdpvar(nu,nx);

alpha = 1/sqrt(nx);

X = sdpvar(nu,nu);

obj = -log(det(Q));

umax = norm(u,inf);

LMI1 = Q;

LMI2 = [X, Y;

        Y', Q];

LMI3 = [alpha^2*Q, (A*Q+B*Y)';

        A*Q+B*Y,    Q];

% const = [LMI1 >= 0; LMI2 >= 0; LMI3 >= 0, X(1,1) <= u(1)^2, X(2,2) <= u(2)^2, X(3,3) <= u(3)^2];

const = [LMI1 >= 0; LMI2 >= 0; LMI3 >= 0, X(1,1) <= umax^2, X(2,2) <= umax^2, X(3,3) <= umax^2];

% ops = sdpsettings('solver','SeDuMi');

ops = sdpsettings('solver','mosek');

diagnostics = optimize(const, obj, ops);

P = (1/(alpha^2))*inv(value(Q));

M = det(P);..

Any suggestions please.

Thanks.

[Johan Löfberg]

Your problem is not well-posed as the objective can get arbitrary large (neither mosek nor sedumi terminates with a solution, so you are looking at crap)

You can easily see this if you solve

diagnostics = optimize([const,trace(Q)<=100], obj, sdpsettings('solver','mosek'))

when you change the constant 100 to 1000 or 10000, the objective simply gets larger with the trace constraint being active, and there is no upper limit

[Johan Löfberg]

You are missing (14) in your model.

[Johan Löfberg]

and I hope you understand that LMI1 is completely redundant as it is a consequence of LMI2 and LMI3 

[Himanshu Nagpal]

Thank you Johan for your quick response. 

Yes, (14) represents the state constraint which is not mentioned in the problem. 

[Himanshu Nagpal]

and sorry I didn't understand that problem is ill-posed. Is it my program or the problem itself?

[Johan Löfberg]

Without (14) the problem is simply not well-defined.

Since the discrete-time system is asymptotically stable (all eigenvalues of A in unit-cricle), I can use the control law u = 0*x, and thus R^n is invariant, i.e., any invariant region without state constraints is arbitrrily large.

[Johan Löfberg]

Ill-posed because there is no optimal solution to the problem. The largest possible log(det(Q)) is infinity