Fletcher Reeves conjugate method
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Hello,
My program is giving the right solution for the problem, but I believe it is doing unecessary steps. For a problem with initial point at [4 6], my code using conjugate method is doing more steps than when I try to solve the same problem using the steepest descent method.
-> Main function:
function [x_opt,f_opt,k] = conjugate_gradient (fob,g_fob,x0,tol_grad);
c0 = feval(g_fob,x0); % evaluate gradient at initial point
k = 0;
if norm(c0) < tol_grad
x_opt = x0; % optimum point
f_opt = feval(fob,x_opt); % cost function value
else
d0= -c0; % search direction
alfa0 = equal_interval_line_search(x0,d0,fob,0.5,1e-6); %line search (step size)
x1= x0+ alfa0*d0;
c1 = feval(g_fob,x1);
while norm(c1) > tol_grad
beta = (norm(c1)/norm(c0))^2;
d1= -c1+beta*d0;
alfa1 = equal_interval_line_search(x1,d1,fob,0.5,1e-6);
x2= x1+alfa1*d1;
c0=c1;
c1= feval(g_fob,x2);
d0=d1;
x1=x2;
k=k+1;
end
x_opt = x1;
f_opt = feval(fob,x_opt);
end
Cost function:
function f = fob_8_58(x);
f = 8*x(1)^2 + 8*x(2)^2 - 80*((x(1)^2+x(2)^2-20*x(2)+100)^0.5)- 80*((x(1)^2+x(2)^2+20*x(2)+100)^0.5)-5*x(1)-5*x(2);
->Gradient fuction:
function g = grad_fob_8_58(x)
g(1) = 16*x(1) - 80*x(1)/((x(1)^2+x(2)^2-20*x(2)+100)^0.5)- 80*x(1)/((x(1)^2+x(2)^2+20*x(2)+100)^0.5)-5;
g(2) =16*x(2) - 80*(x(2)-10)/((x(1)^2+x(2)^2-20*x(2)+100)^0.5)- 80*(x(2)+10)/((x(1)^2+x(2)^2+20*x(2)+100)^0.5)-5;
Answers (1)
Matt J
on 17 Aug 2017
0 votes
One reason might be that you are not doing any restarts in your conjugate gradient implementation. In non-quadratic problems, the sequence of directions, d, will typically lose conjugacy as the iterations progress and you need to restart with d=-gradient from time to time.
5 Comments
Matt J
on 17 Aug 2017
Also, Fletcher-Reeve may not be optimal for some non-quadratic problems. Other choices of beta selection can be found here
I've always preferred Polak-Ribiere.
Matt J
on 17 Aug 2017
how can I do this?
Set beta to 0 at periodic intervals.
Matheus
on 17 Aug 2017
Matheus
on 17 Aug 2017
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