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Gauss – Jacobi Iteration Method
Gauss - Seidal Iteration 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.
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Introduction:
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
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Introduction (continued..)
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
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nnnnnn
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
2211
22222121
11212111 - - - - - - - - (1)
- - - - - - - - (2)
- - - - - - - - (3)
has a unique solution.
Gauss – Jacobi Iteration Method
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Second assumption:
761373 321 xxx2835 321 xxx10312 321 xxx
10312 321 xxx
2835 321 xxx761373 321 xxx
Gauss– Jacobi Iteration Method
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n
j1j
ijaa
i
ii
To begin the Jacobi method ,solve
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Gauss– Jacobi Iteration Method
nnnnnn
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
2211
22222121
11212111
Gauss– Jacobi Iteration Method
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(7)
Gauss– Jacobi Iteration Method
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8
Gauss– Jacobi Iteration Method
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9
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Gauss– Jacobi Iteration Method
Gauss– Jacobi Iteration Method
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nnnnnn
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
2211
22222121
11212111- - - -(1)
- - - -(2)
- - - -(3)
Gauss– Jacobi Iteration Method
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Gauss– Jacobi Iteration Method
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Gauss– Jacobi Iteration Method
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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.
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7162727 131211 aaa
Gauss– Jacobi Iteration Method
From the given equation we have
17
27
685 321
xxx
15
2672 312
xxx
54
110 213
xxx
(1)
Gauss– Jacobi Iteration Method
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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
Gauss– Jacobi Iteration Method
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Gauss– Jacobi Iteration Method
The results are tabulated
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Gauss– Jacobi Iteration Method
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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
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Gauss –Seidal Iteration Method
Consider a system of strictly diagonally dominant equation as
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nnnnnn
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
2211
22222121
11212111- - - - -(1)
- - - - - (2)
- - - - - (3)
Gauss –Seidal Iteration Method
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Gauss –Seidal Iteration Method
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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
Gauss –Seidal Iteration Method
Soln: From the given equation ,we have
- - - - - - - (1)
- - - - - - -(2)
- - - - - - - (3)
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27
685 321
xxx
15
2672 312
xxx
54
110 213
xxx
Gauss –Seidal Iteration Method
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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
Gauss –Seidal Iteration Method
1st Iteration:
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91317.13
)1(
54074.32
)1(
14815.31
)1(
x
x
x
Gauss –Seidal Iteration Method
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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
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
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..)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
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
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