1. ITERATIVE METHODS Prepared by, Dr. L. Benedict Michael Raj
Associate Professor
Department of Mathematics,
St. Joseph’s College(Autonomous),
Trichy

2. Iterative Methods
This method is also called as indirect method. There are two types of iterative methods.
(1)Gauss Jacobi Method
(2)Gauss Seidel Method
The condition for convergence Gauss Jacobi & Gauss Seidel methods is given by the following rule.
The process of iteration will converge if each equation of the system, the absolute value of the largest coefficient is greater than the sum of the absolute values of all the remaining coefficients.

3. Gauss Jacobi Method Problem
solve the equations by Jacobi iteration method the system.
8x-3y+2z=20
6x+3y+12z=35
4x+11y-z=33
Solution
consider the given system as
We write the equations in the form
x=1/8(20+3y-2z)
y=1/11(33-4x+z)
z=1/12(35-6x-3y)

4. We start from an approximation x = y = x I.NO
X=1/8(20+3y-2z)
Y=1/11(33-4x+z)
Z=1/12(35-6x-3y) 1. 2. 3. 4. 5. 6. 7. 8.
2.5
2.8958
3.1543
3.0419
3.0168
3.0104
3.0157
3.0169 3
2.3560
2.0303
1.9329
1.9696
1.9859
1.9885
1.9865
2.9166
0.9166
0.8797
0.8319
0.9127
0.9158
0.9149
0.9116

5. I.NO
X=1/8(20+3y-2z)
Y=1/11(33-4x+z)
Z=1/12(35-6x-3y) 9.
10.
11.
12.
3.0170
3.0167
1.9857
1.9858
0.9115
0.9116
0.9118
Therefore, from the 11th &12th iterations, the values x, y, z are same correct to four decimal places. Stopping at this stage, we get x= y= z=0.9118

6. Gauss Seidel Method Problem
* This is modification of Gauss Jacobi method.
* The current values of the unknowns at each stage of iteration are used in preceeding to the next stage of iteration, this method is more rapid in convergence than Gauss Jacobi method.
Problem
Solve the equations by Gauss Seidel method.
8x-3y+2z=20
6x+3y+12z=35
4x+11y-z=33

7. Solution
consider the given system as
8x-3y+2z=20
4x+11y-z=33
6x+3y+12z=35
we write the equations in the form
x=1/8(20+3y-2z)
y=1/11(33-4x+z)
z=1/12(35-6x-3y)
x = y = z = 0

8. I.NO X=1/8(20+3y-2z) Y=1/11(33-4x+z)
Z=1/12(35-6x-3y) 1. 2. 3. 4. 5. 6. 7.
2.5
2.9981
3.0266
3.0165
3.0166
3.0167
2.0909
2.0137
1.9825
1.9856
1.9859
1.9858
1.1439
0.9141
0.9077
0.9120
0.9118

9. since at the 6th & 7th iterations the values of x, y, z are the same
since at the 6th & 7th iterations the values of x, y, z are the same. They are x= y= z=0.9118

10. THANK YOU