Demonstratie - metoda SOR

Cuprins
Consideram sistemul cu matricea

Initializare

A = [2,1;5,7];
b = [4;19];
xn = [2;1];
omega = 1.11001; %SOR
%omega =1; %Gauss-Seidel

Pregatirea matricelor metodei

M=1/omega*diag(diag(A))+tril(A,-1)
M = 2×2
1.801785569499374 0 5.000000000000000 6.306249493247809
N = M -A
N = 2×2
-0.198214430500626 -1.000000000000000 0 -0.693750506752191

Prima iteratie

xv = xn;
xn = M\(N*xv+b);
ea = norm(xn-xv,inf);
xn, ea
xn = 2×1
1.444995000000000 1.757189357178571
ea = 0.757189357178571

A doua iteratie

xv = xn;
xn = M\(N*xv+b);
ea = norm(xn-xv,inf);
xn, ea
xn = 2×1
1.085807220869107 1.958678117933273
ea = 0.359187779130893

Dupa 20 de iteratii

for k=1:19
xv = xn;
xn = M\(N*xv+b);
ea = norm(xn-xv,inf);
end
xn
xn = 2×1
1.000000000000000 2.000000000000000
ea
ea = 2.220446049250313e-16
er =ea/norm(xn,inf)
er = 1.110223024625157e-16