Demonstratie - metoda relaxarii (SOR)

Consideram sistemul cu matricea

Contents

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

Initializare

A = [2,1;5,7];
b = [4;19];
xn = [2;1];

Pregatirea matricelor metodei

$$M = diag(A)/\omega-L, N = M - A$$

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