Page 329 - MATLAB an introduction with applications

314 ——— MATLAB: An Introduction with Applications

Example E5.30: Repeat problem E5.30 using Fletcher-Reeves method.
Solution:
2
2
Given f(x , x ) = 4x + 3x – 5x x – 8x 1
1 2
2
1
2
1
Here gradient df = [8x – 5x – 8; 6x – 5x ] T
1
2
1
2
Complete program is given below:
% CONJUGATE GRADIENT METHOD
global X V
X=[0;0];
N=length(X);
g0=–feval(‘df’,X);
V=g0;
n=50;
for i=1:n
[s fmin]=fminbnd(‘fline’,–10,10);
X=X+s*V
g1=–feval(‘df’,X);
gamma=(norm(g1)/norm(g0))^2;
V=g1+gamma*V;
g0=g1;
end
fprintf(‘Minimum point for this function is \n’);
fprintf(‘%f\n%f\n’,X(1),X(2));

This requires the following function files
%LINEAR FUNCTION OF s
function z=fline(s)
global V X
p1=X(1)+s*V(1);
p2=X(2)+s*V(2);
z=4*p1^2+3*p2^2–5*p1*p2–8*p1;
%GRADIENT FUNCTION
function y=df(X)
y=[8*X(1)–5*X(2)–8;6*X(2)–5*X(1)];

The output is as follows:
Minimum point for this function is
2.086957
1.739130
The actual output is [2.087 1.7392] T