Solution of system of equations by Gauss-seidel method.

1
Gauss-Seidel Method

• Solution of System of Equations

2
Gauss-Seidel Method
Iterative or approximate methods provide an
alternative to the elimination methods. The
Gauss-Seidel method is the most commonly used
iterative method
The system AXB is reshaped by solving the
first equation for x1, the second equation for
x2, and the third for x3, and nth equation for
xn. For conciseness, we will limit ourselves to a
3x3 set of equations.
Consider the system of equations
The above three equations can be solved by
Gauss-Seidel method as follow.
3
Algorithm
Step-1-
Solve the given three equations for x1,x2,x3
respectively as
Step-2-
Take the initial approximations as x10, x20,
x30.
Step-3-
In each iteration find the values of x1,x2 and x3
using equation-1,2,3 in step-1 with available
updated values of x1,x2,x3.
Step-4-
Repeat the process until given stopping criteria
will met.
4
Summary of Gauss-Seidel Method
Zeroth approximation
1st iteration
5
Summary of Gauss-Seidel Method
Zeroth approximation
1st iteration
6
Summary of Gauss-Seidel Method
Zeroth approximation
1st iteration
2nd iteration
7
Summary of Gauss-Seidel Method
Zeroth approximation
1st iteration
2nd iteration
8
Summary of Gauss-Seidel Method
Zeroth approximation
1st iteration
2nd iteration
9
Convergence Criterion for Gauss-Seidel Method-
The Gauss-Seidel method has two fundamental
problems as any iterative method
1) It is sometimes no convergent,.
2) If it converges, converges very slowly
10
Note-
If the given equations doesnt satisfy these
conditions, but satisfy after re-arranging them,
then we can re-arrange the equations in such a
way that the can satisfy these conditions. This
will make easier to solve
These conditions are sufficient for convergence
but not necessary.
11
Example
12
Ans-
13
Ans-
14
Approximated values in different iterations
Iteration Unknown Value Max.error
1 x1 2.7 100 100
1 x2 8.9 100 100
1 x3 -6.62 100 100
2 x1 0.258 946.5 946.5
2 x2 7.91433 12.45 946.5
2 x3 -5.93446 11.5 946.5
3 x1 0.523687 50.73 50.73
3 x2 8.010001 1.19 50.73
3 x3 -6.00674 1.20 50.73
4 x1 0.497326 5.30 5.30
4 x2 7.999091 0.14 5.30
4 x3 -5.99928 0.12 5.30
5 x1 0.500253 0.59 0.59
5 x2 8.000112 0.01 0.59
5 x3 -6.0007 0.01 0.59
15
Thank You ALL

• Presented by
• Boina Anil Kumar
• Asst.Prof in Mathematics
• MITS, Rayagada.
• Odisha, India
• E-mail anil.anisrav_at_gmail.com
• Visit at http//www.mits.edu.in/academics.phpfac
  ulty_BSH-tab

16
Thank You ALL
Make Presentation much more fun