Basic HTML version of Foils prepared March 15 00
| You can show that for Jacobi, best choice is ? = 1 i.e. that relaxation strategy is not very useful |
| For Gauss-Seidel, then in simple one dimensional case with N grid points, one can show that there is an optimal choice ? = 2 ( 1 - ? / N) i.e. almost equals 2 |
| Then the number of iterations needed by optimal SOR to get a given error can be shown to be proportional to N not N2 as in Jacobi or Gauss Seidel. So Ratio of Iterations is: |
| Jacobi 2 N2/ ?2 |
| Gauss-Seidel N2/ ?2 |
| SOR on Gauss-Seidel N/ ? |