# How can i solve this cost function?

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Volcano on 29 Jun 2022
Commented: Sam Chak on 29 Jun 2022
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,
Volcano on 29 Jun 2022
@Sam Chak sorry, but what is LMI?
There should be optimal q vector which minimize J.

Matt J on 29 Jun 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));
Solving problem using quadprog. Minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the value of the optimality tolerance, and constraints are satisfied to within the value of the constraint tolerance.
q=sol.q
q = 4×1
50.7877 74.3710 90.2892 124.5521
Volcano on 29 Jun 2022
Thanks a lot @Matt J

Sam Chak on 29 Jun 2022
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)
Local minimum found. Optimization completed because the size of the gradient is less than the value of the optimality tolerance.
q = 1×4
70.5372 99.4606 70.5396 99.4624
fval = 0.0457
Sam Chak on 29 Jun 2022
Thanks @Matt J for the explanations. I learned something new today.