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;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; 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; optimization, conjugate method, fletcher reeves MATLAB Answers — New Questions

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