Demonstratie - metoda SOR

Contents

Consideram sistemul

$$\left[ \begin{array} [c]{cc}
2 & 1\\
5 & 7
\end{array} \right]x = \left[
\begin{array}
[c]{c}
4\\
19
\end{array}
\right]
  $$

Initializare

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

Pregatirea matricelor metodei

$$A = D-L-U$$

$$M = \frac{1}{\omega}D - L, N = M - A$$

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