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NUMERICAL METHODS -Iterative methods(indirect method)

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1 Gauss – Jacobi Iteration Method Gauss - Seidal Iteration Method
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Page 1: NUMERICAL METHODS -Iterative methods(indirect method)

1

Gauss – Jacobi Iteration Method

Gauss - Seidal Iteration Method

Page 2: NUMERICAL METHODS -Iterative methods(indirect method)

Iterative Method

Simultaneous linear algebraic equation occur in various fields of Science and Engineering.

We know that a given system of linear equation can be solved by applying Gauss Elimination Method and Gauss – Jordon Method.

But these method is sensitive to round off error.

In certain cases iterative method is used.

Iterative methods are those in which the solution is got by successive approximation.

Thus in an indirect method or iterative method, the amount of computation depends on the degree of accuracy required.

2

Introduction:

Page 3: NUMERICAL METHODS -Iterative methods(indirect method)

Iterative Method

Iterative methods such as the Gauss – Seidal method give the user control of the round off.

But this method of iteration is not applicable to all systems of equation. In order that the iteration may succeed, each equation of the system

must contain one large co-efficient. The large co-efficient must be attached to a different unknown in that

equation. This requirement will be got when the large coefficients are along the

leading diagonal of the coefficient matrix. When the equation are in this form, they are solvable by the method of

successive approximation. Two iterative method - i) Gauss - Jacobi iteration method

ii) Gauss - Seidal iteration method

3

Introduction (continued..)

Page 4: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss – Jacobi Iteration Method:

The first iterative technique is called the Jacobi method named after

Carl Gustav Jacob Jacobi(1804- 1851).

Two assumption made on Jacobi method:

1)The system given by

4

nnnnnn

nn

nn

bxaxaxa

bxaxaxa

bxaxaxa

2211

22222121

11212111 - - - - - - - - (1)

- - - - - - - - (2)

- - - - - - - - (3)

has a unique solution.

Page 5: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss – Jacobi Iteration Method

5

Second assumption:

761373 321 xxx2835 321 xxx10312 321 xxx

10312 321 xxx

2835 321 xxx761373 321 xxx

Page 6: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss– Jacobi Iteration Method

6

n

j1j

ijaa

i

ii

Page 7: NUMERICAL METHODS -Iterative methods(indirect method)

To begin the Jacobi method ,solve

7

Gauss– Jacobi Iteration Method

nnnnnn

nn

nn

bxaxaxa

bxaxaxa

bxaxaxa

2211

22222121

11212111

Page 8: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss– Jacobi Iteration Method

8

(7)

Page 9: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss– Jacobi Iteration Method

9

8

Page 10: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss– Jacobi Iteration Method

10

9

Page 11: NUMERICAL METHODS -Iterative methods(indirect method)

11

Gauss– Jacobi Iteration Method

Page 12: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss– Jacobi Iteration Method

12

nnnnnn

nn

nn

bxaxaxa

bxaxaxa

bxaxaxa

2211

22222121

11212111- - - -(1)

- - - -(2)

- - - -(3)

Page 13: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss– Jacobi Iteration Method

13

Page 14: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss– Jacobi Iteration Method

14

Page 15: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss– Jacobi Iteration Method

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Page 16: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss– Jacobi Iteration Method

Solution:

In the given equation , the largest co-efficient is attached to a

different unknown.

Checking the system is diagonally dominant .

Here

Then system of equation is diagonally dominant .so iteration method

can be applied.

16

7162727 131211 aaa

Page 17: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss– Jacobi Iteration Method

From the given equation we have

17

27

685 321

xxx

15

2672 312

xxx

54

110 213

xxx

(1)

Page 18: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss– Jacobi Iteration Method

18

27

)0()0(685)1(

1

x

15

)0(2)0(672

2

)1( x

54

)0()0(110

3

)1( x

=3.14815

=4.8

=2.03704

First approximation

Page 19: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss– Jacobi Iteration Method

19

Page 20: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss– Jacobi Iteration Method

The results are tabulated

20

Page 21: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss– Jacobi Iteration Method

21

Page 22: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss –Seidal Iteration Method

Modification of Gauss- Jacobi method,

named after Carl Friedrich Gauss and Philipp Ludwig Von Seidal.

This method requires fewer iteration to produce the same degree

of accuracy.

This method is almost identical with Gauss –Jacobi method except

in considering the iteration equations.

The sufficient condition for convergence in the Gauss –Seidal

method is that the system of equation must be strictly diagonally

dominant

22

Page 23: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss –Seidal Iteration Method

Consider a system of strictly diagonally dominant equation as

23

nnnnnn

nn

nn

bxaxaxa

bxaxaxa

bxaxaxa

2211

22222121

11212111- - - - -(1)

- - - - - (2)

- - - - - (3)

Page 24: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss –Seidal Iteration Method

24

Page 25: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss –Seidal Iteration Method

25

Page 26: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss –Seidal Iteration Method

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The successive iteration are generated by the scheme called

iteration formulae of Gauss –Seidal method are as

The number of iterations k required depends upon the desired degree of accuracy

Page 27: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss –Seidal Iteration Method

Soln: From the given equation ,we have

- - - - - - - (1)

- - - - - - -(2)

- - - - - - - (3)

27

27

685 321

xxx

15

2672 312

xxx

54

110 213

xxx

Page 28: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss –Seidal Iteration Method

28

91317.154

)54074.5()14815.3(110

3

)1(

x

14815.327

)0()0(685

1

)2(

x

54074.315

)0(2)14815.3(672

2

)1(

x

Page 29: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss –Seidal Iteration Method

1st Iteration:

29

91317.13

)1(

54074.32

)1(

14815.31

)1(

x

x

x

Page 30: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss –Seidal Iteration Method

30

For the second iteration,

91317.13

)1(

54074.32

)1(

14815.31

)1(

x

x

x

43218.227

6853

)1(

2

)1(

1

)2(

xx

x

57204.315

26723

)1(

1

)2(

2

)2(

xx

x

92585.154

1102

)2(

1

)2(

3

)2(

xx

x

Page 31: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss –Seidal Iteration Method

Thus the iteration is continued .The results are tabulated.

S.No Iteration or approximation

1 0 0 0 0

2 1 3.14815 3.54074 1.91317

3 2 2.43218 3.57204 1.92585

4 3 2.42569 3.57294 1.92595

5 4 2.42549 3.57301 1.92595

6 5 2.42548 3.57301 1.92595

31

..)3,2,,( ii1,

ix ..)3,2,,(2

iiix ..)3,2,,(3

iiix

4th and 5th iteration are practically the same to four places.So we stop iteration process.

Ans: 9260.13

;57301.32

;4255.21

xxx

Page 32: NUMERICAL METHODS -Iterative methods(indirect method)

Gauss –Seidal Iteration Method

Comparison of Gauss elimination and Gauss- Seidal Iteration methods:

Gauss- Seidal iteration method converges only for special systems of

equations. For some systems, elimination is the only course available.

The round off error is smaller in iteration methods.

Iteration is a self correcting method

32


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