How can i solve this cost function?

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Volcano 2022-6-29

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评论： Sam Chak 2022-6-29

采纳的回答： Matt J

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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,

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Volcano 2022-6-29

@Sam Chak Yes, it is 2*2 matrix. Do you think there is a mistake in formulation?

Volcano 2022-6-29

@Sam Chak sorry, but what is LMI?

There should be optimal q vector which minimize J.

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采纳的回答

Matt J 2022-6-29

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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

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Volcano 2022-6-29

Thanks a lot @Matt J

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Sam Chak 2022-6-29

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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)
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

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.