Demonstratie - metoda relaxarii (SOR)
Consideram sistemul cu matricea
Contents
Initializare
A = [2,1;5,7]; b = [4;19]; xn = [2;1];
Pregatirea matricelor metodei
omega = 1; %optim 1.11
D = diag(diag(A)); L=-tril(A,-1);
M = 1/omega*D-L
N = M -A
M = 2 0 5 7 N = 0 -1 0 0
Prima iteratie
xv = xn; xn = M\(N*xv+b); ea = norm(xn-xv,inf); xn, ea
xn = 1.5000e+00 1.6429e+00 ea = 6.4286e-01
A doua iteratie
xv = xn; xn = M\(N*xv+b); ea = norm(xn-xv,inf); xn, ea
xn = 1.1786e+00 1.8724e+00 ea = 3.2143e-01
Dupa 25 de iteratii
for k=1:23 xv = xn; xn = M\(N*xv+b); ea = norm(xn-xv,inf); end xn ea er =ea/norm(xn,inf)
xn = 1.0000e+00 2.0000e+00 ea = 1.6690e-11 er = 8.3448e-12