Using 'fminunc', I receive Optimization completed because the size of the gradient is less than the value of the optimality tolerance.
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I use 'fminunc' to find the value of B (2*4 matrix) that minimzes the difference between the corresponding elements in two vectors as indicated in the attached code. In other words, I want to find the B (2*4 matrix) that makes the elements of beta_d (1*4 vector) which is a function of B matrix, equal to the corresponding ones of a "given" beta_u (1*4 vector), i.e. beta_d(1,1) = beta_u(1,1) && beta_d(1,2) = beta_u(1,2) && beta_d(1,3) = beta_u(1,3) && beta_d(1,4) = beta_u(1,4).
However, I usually receive the following message without getting any result and the program seems to go on an infinite loop!
Local minimum found.
Optimization completed because the size of the gradient is less than
the value of the optimality tolerance.
<stopping criteria details>
the code:
clc;
clear;
% System paramters:
N = 2;
K = 4;
C_l = 4;
H = [-0.3208 -0.9784; -1.5994 -1.4689; -1.5197 -0.4568; -0.0993 -0.7667]; % 4*2 matrix
A = [-1 1; 0 1]; % 2*2 matrix
C = [-0.20 0.4 0.6 -0.2; -0.2 0.4 0.6 -0.2; 0.4 0.2 -0.2 0.4; 0.4 0.2 -0.2 0.4]; % 4*4 matrix
P = [250000 0 0 0; 0 250000 0 0; 0 0 250000 0; 0 0 0 250000]; % 4*4 matrix
beta_u = [50.2207 50.2207 20.3433 20.3433]; % 1*4 vector
beta_d = zeros(1,4); % intial value
B = zeros(2,4); %intial value
%store inputs to a struct for shorter syntax
s=struct;[s.H,s.A,s.C,s.P,s.C_l,s.N,s.K]=deal(H,A,C,P,C_l,N,K);
%fminunc optimization
while (sum(abs(beta_u-beta_d))>=0.1)
initial_guess = randn(2,4);
OLS = @(B_d,input_vars)sum((beta_u-myfun(B_d,input_vars)).^2);%ordinary least squares cost function
opts = optimoptions(@fminunc,'MaxIterations',10000,'MaxFunctionEvaluations',50000,'CheckGradients',true);
B = fminunc(OLS, initial_guess, opts,s);
input_vars = s;
[beta_d, D_d]=myfun(B,input_vars);
end
%calculate beta_d from B and the other inputs
function [beta_d, D_d]=myfun(B,input_vars)
%load parameters
s=input_vars;[H,A,C,P,C_l,N,K]=deal(s.H,s.A,s.C,s.P,s.C_l,s.N,s.K);
for j=1:1:N
d(j) = (B(j,:)*P*B(j,:)')/((2^(2*C_l))-(norm(A(:,j))^2));
end
D_d = diag(d);
for i=1:1:K
V_d(i)=C(i,:)*P*B'*H(i,:)'*inv(1+H(i,:)*(A'*D_d*A+B*P*B')*H(i,:)');
sigma_d(i)=norm((V_d(i)*H(i,:)*B-C(i,:))*(P^(1/2)))^2+(V_d(i)^2)*(1+H(i,:)*A'*D_d*A*H(i,:)');
beta_d(i)=((P(i,i))/sigma_d(:,i));
end
end
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Akzeptierte Antwort
Matt J
am 21 Mai 2020
Bearbeitet: Matt J
am 21 Mai 2020
If the problem doesn't have a solution satisfying the tolerance condition in your while loop
(sum(abs(beta_u-beta_d))<0.1)
then the loop has no way of stopping.
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