Demonstratie - metoda SOR
Contents
Consideram sistemul
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) N = M -A
M = 1.801785569499374 0 5.000000000000000 6.306249493247809 N = -0.198214430500626 -1.000000000000000 0 -0.693750506752191
Prima iteratie
xv = xn; xn = M\(N*xv+b); ea = norm(xn-xv,inf); xn, ea
xn = 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 = 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 ea er =ea/norm(xn,inf)
xn = 1.000000000000000 2.000000000000000 ea = 2.220446049250313e-16 er = 1.110223024625157e-16