How can i solve this cost function?
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Hi,
I have a cost function which includes vectors and matrices:
v = [(1 / (0.47 * 0.94 * 1500)^2) (1 / (0.47 * 0.94 * 1700)^2) (1 / (0.47 * 0.94 * 2000)^2) (1 / (0.47 * 0.94 * 2200)^2)];
delta = diag(v);
nu = transpose([120 340]);
K = [((1.95 / (2*0.47)) .* [-1 1 -1 1]);...
1 1 1 1 ];
zeta = [2 0; 0 1];
%J = transpose(q) * delta * q + (transpose(K*q - nu)) * zeta * (K*q - nu) %Cost function
%q = transpose([X Y Z T]) %Output of cost function
I am trying to find q vector without any constraint. What kind of methods can be used to solve related equation?
Thanks,
8 commentaires
Réponse acceptée
Matt J
le 29 Juin 2022
v = [(1 / (0.47 * 0.94 * 1500)^2) (1 / (0.47 * 0.94 * 1700)^2) (1 / (0.47 * 0.94 * 2000)^2) (1 / (0.47 * 0.94 * 2200)^2)];
delta = diag(v);
nu = transpose([120 340]);
K = [((1.95 / (2*0.47)) .* [-1 1 -1 1]);...
1 1 1 1 ];
zeta = [2 0; 0 1];
q=optimvar('q',4,1);
J=transpose(q) * delta * q + (transpose(K*q - nu)) * zeta * (K*q - nu);
sol=solve(optimproblem('Objective',J));
q=sol.q
Plus de réponses (1)
Sam Chak
le 29 Juin 2022
Hi @Volcano
I converted the matrix equation into a scalar equation. Since there is no constraint, fminunc() is used and the local minimum is found.
v = [(1 / (0.47 * 0.94 * 1500)^2) (1 / (0.47 * 0.94 * 1700)^2) (1 / (0.47 * 0.94 * 2000)^2) (1 / (0.47 * 0.94 * 2200)^2)];
delta = diag(v);
nu = transpose([120 340]);
K = [((1.95 / (2*0.47)) .* [-1 1 -1 1]); 1 1 1 1];
zeta = [2 0; 0 1];
% Cost function
J = @(q) v(1)*q(1).^2 + v(2)*q(2).^2 + v(3)*q(3).^2 + v(4)*q(4).^2 + zeta(1,1)*(K(1,1)*q(1) + K(1,2)*q(2) + K(1,3)*q(3) + K(1,4)*q(4) - nu(1)).^2 + zeta(2,2)*(K(2,1)*q(1) + K(2,2)*q(2) + K(2,3)*q(3) + K(2,4)*q(4) - nu(2)).^2;
% Initial guess of q solution
q0 = [1 1 1 1];
% Minimize unconstrained multivariable function
[q, fval] = fminunc(J, q0)
4 commentaires
Matt J
le 29 Juin 2022
Modifié(e) : Matt J
le 29 Juin 2022
It doesn't seem possible that it is a local minimum, because it is a convex problem. Possibly, fminunc's finite difference approximations to the gradient, or maybe the optimoption defaults, led to incomplete convergence. quadprog doesn't rely on approximate gradients, so it wouldn't have that difficulty to overcome.
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