# SOR

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```					More on iterative solution of Ax  b

Iterative solution

Jacobi’s method

Gauss’-Seidel’s
Approach

SOR method

SOR method. Successive Over-Relaxation Method

Consider the basic framework for an iterative
solution of Ax  b through an example.

4 x1  3 x2          24
3x1  4 x2  x3  30
 x2  4 x3  24

Its solution is x  3 4  5t .

We begin with a starting solution x( 0 )  1 1 1t .

Gauss-Seidel frames the equations in the following
way:
x1 k )  0.75 x2k 1 )  6
(              (

x2k )  0.75 x1 k )  0.25 x3 k 1 )  7.5
(             (             (

(             (
x3 k )  0.25 x2k )  6

For Gauss’-Seidel,

1  i 1             n
( k 1 )      
xi( k )                  (k )
  aij x j   aij x j         bi 
aii  j 1
              j i 1               


In SOR, we begin right about here. We take ( 1   )
units of the previous estimate of xi and take  units
of the currently available estimate to formulate the
iterative solution:

xi( k )  ( 1   )xi( k 1 ) 
          i 1              n                             

(k )
 aij x j               aij x(j k 1 )    bi 
aii 
       j 1            j i 1                         


The value of  is usually less than 2.0. When   1
we get the Gauss’-Seidel method; for 0    1, we
get Under-Relaxation Method. For 1    2 , we
have the Over-Relaxation Method.
If we take   1.5 , our SOR equations appear as

x1 k )  0.5 x1 k 1 ) 
(             (            1.5
4
 3x2k 1 )  24
(

1.5
x2k )  0.5 x2k 1 ) 
(            (
( 3x1 k )  4 x2k 1 )  x3 k 1 )  30 )
(          (          (
4

….

Assignment. Write a program that would iteratively
obtain SOR solutions given A and b .

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 views: 15 posted: 3/8/2012 language: English pages: 3