Problem with minimizing a LQR problem with Genetic Algorithm
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Gilad Shaul
am 15 Okt. 2022
Beantwortet: Sam Chak
am 17 Okt. 2022
I am trying to solve a LQR problem using Genetic Algorithm.
The values of the LQR matrices are not importent to me, I am just trying to understand the way that it is work.
First I solve the system using the Matlab's LQR algorithm and them I try to solve using the Genetic Algorithm and compare the results.
In the code attach I am trying to compare between P to P2, but I dont get a match.
Also when I run the code, I get the following Error: Optimization terminated: average change in the fitness value less than options
%% State Space
A = [1 2; 3 4];
B = [1; 1];
Q = [1 0;0 1];
R = 1;
%% Slove the LQR
[K,P1] = lqr(A,B,Q,R);
P2 = norm(P1);
%% Solve the Genetic Algorithm
P = ga(@cost_fun,2);
function P = cost_fun(K1)
A = [1 2; 3 4];
B = [1; 0];
Q = [1 0;0 1];
R = 1;
A_i = A - B*K1;
Q_i = - Q - K1'*R*K1;
P_i = lyap(A_i,Q_i);
P = norm(P_i);
end
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Sam Chak
am 15 Okt. 2022
Hi @gilad shaul
Minimizing the norm(P) does not work because there are infinite K that can make A'*P + P*A - K'*R*K + Q exactly 0. If you want to solve using genetic algorithm, ga(), then you should directly minimize the Finite-time Performance Index J (a scalar) instead:
Else, if you want to find the elements in the positive-definite matrix
that minimizes J, then you should use gamultiobj() instead (because there are multiple fitness functions). In 2nd-order systems, there are three fitness functions involving
,
,
because the positive-definite matrix
looks like this:
% ------- Computation of K via LQR (solving Algebraic Riccati Matrix Equation) -------
% State Space
A = [0 1; 0 -1];
B = [0; 1];
Q = [1 0; 0 1];
R = 1;
% LQR
K = lqr(A, B, Q, R)
% [K, P, E] = lqr(A, B, Q, R)
% Riccati Equation
% A'*P + P*A - (P*B)*inv(R)*(B'*P) + Q
% QQ = - (P*B)*inv(R)*(B'*P) + Q;
% QQ = - (P*B)*K + Q;
% QQ = - K'*R*K + Q;
% PP = lyap(A', QQ) % returns the same positive-definite matrix P
% ------- Computation of K via GA (minimizing Finite-time Performance Index J) -------
fitfun = @costfun;
PopSize = 100;
MaxGens = 200;
nvars = 2;
A = -eye(nvars);
b = zeros(nvars, 1);
Aeq = [];
beq = [];
lb = [];
ub = [];
nonlcon = [];
options = optimoptions('ga', 'PopulationSize', PopSize, 'MaxGenerations', MaxGens);
[KK, fval, exitflag, output] = ga(fitfun, nvars, A, b, Aeq, beq, lb, ub, nonlcon, options)
function J = costfun(param)
% Parameters
A = [0 1; 0 -1];
B = [0; 1];
Q = [1 0; 0 1];
R = 1;
K = [param(1) param(2)];
% State-Space
odefcn = @(t, x) (A - B*K)*x;
tspan = [0 20]; % Finite-time
x0 = [1 0];
[t, x] = ode45(odefcn, tspan, x0);
% Performance Index
u = - K(1)*x(:,1) - K(2)*x(:,2);
itgrd = Q(1,1)*x(:,1).^2 + Q(2,2)*x(:,2).^2 + R*u.^2; % integrand
J = trapz(t, itgrd);
end
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Sam Chak
am 17 Okt. 2022
Hi @gilad shaul
To find
through
, you need to expand the Continuous-time Algebraic Riccati Equation (CARE) in terms of
, and then use gamultiobj() to solve the root-finding problem.
If you like the alternative solution, please consider voting 👍 the Answer as a little token of appreciation.
%% ----- Expanding the CARE -----
syms p11 p12 p22
A = sym('A', [2 2]); % state matrix
B = sym('B', [2 1]); % input matrix
P = sym('P', [2 2]); % positive definite matrix
A = [sym('0') sym('1');
sym('0') sym('-1')];
B = [sym('0'); sym('1')];
P = [p11 p12;
p12 p22];
Q = sym(eye(2));
R = sym('1');
CARE = A.'*P + P*A - P*B/R*B.'*P + Q
%% ----- Solving using gamultiobj -----
fitfun = @multiobjfcn;
PopSize = 15;
MaxGen = 30;
nvars = 3;
A = -eye(nvars);
b = zeros(nvars, 1);
Aeq = [];
beq = [];
lb = [0 0 0];
ub = [3 3 3];
nonlcon = [];
options = optimoptions('gamultiobj', 'ConstraintTolerance', 1e-6, 'FunctionTolerance', 1e-6, 'HybridFcn', @fgoalattain);
[p, fval] = gamultiobj(fitfun, nvars, A, b, Aeq, beq, lb, ub, nonlcon, options);
popt = p(end, :);
R = 1;
B = [0; 1];
P = [popt(1) popt(2); popt(2) popt(3)]
Kopt = R\(B'*P)
%% ----- Multi-objective Functions -----
function y = multiobjfcn(p)
y = zeros(3, 1);
y(1) = (1 - p(2)^2)^2; % Remember to square the CARE
y(2) = (- p(3)^2 - 2*p(3) + 2*p(2) + 1)^2;
y(3) = (p(1) - p(2) - p(2)*p(3))^2;
end
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