A little Much about the Successive Over Relaxation (SOR) method which is the iterative method for solving the linear equations.

1. Successive over Relaxation Method
Iterative Method (Numerical Analysis)
Khushdil Ahmad
Government College University Lahore
Department of Mathematics
December 7, 2018
Khushdil Ahmad Successive over Relaxation Method

2. Abstracts
In this Talk I will Discussed the following topics:
1 Formula of Successive Over Relaxation
2 Some Important
3 Proof of SOR
4 Examples
Khushdil Ahmad Successive over Relaxation Method

3. Formula of Successive Over Relaxation
x
(k+1)
i = x
(k)
i +
w
aii
[bi −
i−1
j=1
aij x
(k+1)
j −
n
j=1
aij x(k)
] (1)
OR
x
(k+1)
1 = x
(k)
1 + w
a11
[b1 − a11x
(k)
1 − a12x
(k)
2 − a13x
(k)
3 ]
x
(k+1)
2 = x
(k)
2 + w
a22
[b2 − a21x
(k+1)
1 − a22x
(k)
2 − a23x
(k)
3 ]
x
(k+1)
3 = x
(k)
3 + w
a33
[b3 − a31x
(k+1)
1 − a32x
(k+1)
2 − a33x
(k)
3 ]
Khushdil Ahmad Successive over Relaxation Method

4. Some Important Terms
Here w is the Relaxation Perimeter.
1 If w = 1 then this method will reduced to Gauss Seidal
Method.
2 If w > 1 then this method is called Over Relaxation.
3 If w <1 then this method is called under Relaxation.
4 If w Lies between 0< w <2 then the method converges
otherwise diverges.
5 We can find ”w” by the formula:
w =
2
1 − p(tj )2
(2)
and p(tj ) = D−1(L + U)
Khushdil Ahmad Successive over Relaxation Method

5. PROOF
We Prove this formula from Gauss Seidel Iterative Formula. As
Generalized form of Gauss Siedel is :
x
(k+1)
i = 1
aii
[bi − i−1
j=1 aij x
(k+1)
i − n
j=1 aij x(k)]
Now adding and Subtracting xk
i on both sides of equation
x
(k+1)
i = x
(k)
i +
1
aii
[bi −
i−1
j=1
aij x
(k+1)
i −
n
j=1
aij x(k)
] (3)
Replace ”1” by ”w” in equation (3) then we get
x
(k+1)
i = x
(k)
i +
w
aii
[bi −
i−1
j=1
aij x
(k+1)
i −
n
j=1
aij x(k)
] (4)
Khushdil Ahmad Successive over Relaxation Method

6. PROOF
Equation (4) Is the Requires equation of Successive over
Relaxation (SOR) Method
If we expand the ” ” and put the values of i = 1,2,3,. . . ,n then
we get the Equations
x
(k+1)
1 = x
(k)
1 + w
a11
[b1 − a11x
(k)
1 − a12x
(k)
2 − a13x
(k)
3 ]
x
(k+1)
2 = x
(k)
2 + w
a22
[b2 − a21x
(k+1)
1 − a22x
(k)
2 − a23x
(k)
3 ]
x
(k+1)
3 = x
(k)
3 + w
a33
[b3 − a31x
(k+1)
1 − a32x
(k+1)
2 − a33x
(k)
3 ]
Khushdil Ahmad Successive over Relaxation Method

7. Example 01
Question: Solve the following equations by SOR with w = 1.45
5x1 + 2x2 − x3 = 6
x1 + 6x2 − 3x3 = 4
2x1 + x2 + 4x3 = 7
Khushdil Ahmad Successive over Relaxation Method

8. Solution:
As Formula of SOR
x
(k+1)
1 = x
(k)
1 + w
a11
[b1 − a11x
(k)
1 − a12x
(k)
2 − a13x
(k)
3 ]
x
(k+1)
2 = x
(k)
2 + w
a22
[b2 − a21x
(k+1)
1 − a22x
(k)
2 − a23x
(k)
3 ]
x
(k+1)
3 = x
(k)
3 + w
a33
[b3 − a31x
(k+1)
1 − a32x
(k+1)
2 − a33x
(k)
3 ]
Here the Equation are in Matrix form are:


5 2 −1
1 6 −3
2 1 4




x1
x2
x3

 =


6
4
7


and 

a11 a12 a13
a21 a22 a23
a31 a32 a33

 =


5 2 −1
1 6 −3
2 1 4


Khushdil Ahmad Successive over Relaxation Method

9. Solution:
Here
a11 = 5, a12 = 2, a13 = −1. a21 = 1, a22 = 6, a23 = −3. a31 =
2, a32 = 1, a33 = 4.b1 = 6, b2 = 4 and b3 = 7.
For Initial Estimate
Take x0
1 = x0
2 = x0
3 = 0
First Iteration
Put k=0 and w = 1.45 in General Formula’s of SOR,
∗1 For x1
1
x
(1)
1 = x
(0)
1 + w
a11
[b1 − a11x
(0)
1 − a12x
(0)
2 − a13x
(0)
3 ]
By Putting Values
x
(1)
1 = 0 + 1.45
5 [6 − 5(0) − 2(0) − (−1)(0)]
x
(1)
1 = 1.74
Khushdil Ahmad Successive over Relaxation Method

10. Solution:
∗2 For x
(1)
2
x
(1)
2 = x
(0)
2 + w
a22
[b2 − a21x
(1)
1 − a22x
(0)
2 − a23x
(0)
3 ]
By Putting Values
x
(1)
2 = 0 + 1.45
6 [4 − 1(1.74) − 6(0) − (−3)(0)]
x
(1)
2 = 0.5462
∗3 For x
(1)
3
x
(1)
3 = x
(0)
3 + w
a33
[b3 − a31x
(1)
1 − a32x
(1)
2 − a33x
(0)
3 ]
By Putting Values
x
(1)
3 = 0 + 1.45
4 [7 − 2(1.74) − 1(0.5462) − (4)(0)]
x
(1)
3 = 1.0780
Khushdil Ahmad Successive over Relaxation Method

11. Solution:
Second Iteration
Put k=1 , and w = 1.45 in General Formula’s of SOR, and
x
(1)
1 = 1.74, x
(1)
2 = 0.5462 x
(1)
3 = 1.0780
∗1 For x2
1
x
(2)
1 = x
(1)
1 + w
a11
[b1 − a11x
(1)
1 − a12x
(1)
2 − a13x
(1)
3 ]
By Putting Values
x
(2)
1 = 1.74 + 1.45
5 [6 − 5(1.74) − 2(0.5462) − (−1)(1.0780)]
x
(2)
1 = 0.9528
∗2 For x
(2)
2
x
(2)
2 = x
(1)
2 + w
a22
[b2 − a21x
(2)
1 − a22x
(1)
2 − a23x
(1)
3 ]
By Putting Values
x
(2)
2 = 0.5462+ 1.45
6 [4−1(0.9528)−6(0.5426)−(−3)(1.0780)]
Khushdil Ahmad Successive over Relaxation Method

12. Solution:
x
(2)
2 = 1.7519
∗3 For x
(2)
3
x
(2)
3 = x
(1)
3 + w
a33
[b3 − a31x
(2)
1 − a32x
(3)
2 − a33x
(1)
3 ]
By Putting Values
x
(2)
3 = 1.0780 + 1.45
4 [7 − 2(0.9528) − 1(1.7519) − 4(1.0780)]
x
(2)
3 = 0.7266
Khushdil Ahmad Successive over Relaxation Method

13. Solution:
Similarly
iteration No (k) x
(k)
1 x
(k)
2 x
(k)
3
3 0.5064 0.5827 1.6322
4 1.6475 1.6710 0.8707
What happen when we change the value of w?
If we change the value of w towards ”2” then answer will converge
quickly than of near to ”1” and if w = 1 then it reduces to Gauss
Siedel Method and if we put w < 1 then it will take more time to
get the better approximation.
If w > ”2” then the Method will diverges.
Khushdil Ahmad Successive over Relaxation Method

14. Thank You
THANK YOU
Khushdil Ahmad Successive over Relaxation Method