The document describes the Gauss-Seidel method, an iterative method for solving systems of linear equations. It explains the basic procedure, which involves rewriting the equations to solve for each unknown in turn and repeating until the approximate error is within a tolerance. An example applying the method to a system of 3 equations and 3 unknowns is shown, calculating the solution over multiple iterations until convergence within a specified error.

1. Admission in India 2015
By:
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2. 2
Gauss-Siedel Method
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3. Gauss-Seidel Method
An iterative method.
Basic Procedure:
-Algebraically solve each linear equation for xi
-Assume an initial guess solution array
-Solve for each xi and repeat
-Use absolute relative approximate error after each iteration
to check if error is within a pre-specified tolerance.
http://numericalmethods.eng.usf.edu
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4. Gauss-Seidel Method
http://numericalmethods.eng.usf.edu
Why?
The Gauss-Seidel Method allows the user to control round-off error.
Elimination methods such as Gaussian Elimination and LU
Decomposition are prone to prone to round-off error.
Also: If the physics of the problem are understood, a close initial
guess can be made, decreasing the number of iterations needed.
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5. Gauss-Seidel Method
http://numericalmethods.eng.usf.edu
Algorithm
A set of n equations and n unknowns:
11 1 12 2 13 3 1 1 a x a x a x ... a x b n n + + + + =
21 1 22 2 23 3 2 a x a x a x ... a x b 2n n + + + + =
. .
. .
. .
n n n nn n n a x + a x + a x + ...+ a x = b 1 1 2 2 3 3
If: the diagonal elements are
non-zero
Rewrite each equation solving
for the corresponding unknown
ex:
First equation, solve for x1
Second equation, solve for x2
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6. Gauss-Seidel Method
http://numericalmethods.eng.usf.edu
Algorithm
Rewriting each equation
x c - a x - a x 
-
a = 1 12 2 13 3 1
n xn 1 a
11
x = c - a x - a x 
-
a x
2 21 1 23 3 2
  
c a x a x 
a x a x
n - 1 n - 1,1 1 n - 1,2 2 n - 1, n - 2 n - 2 n -
1,
n n
1, 1
c a x a x a x
n n n n n n
nn
n
n n
n
n n
a
x
a
x
a
1 1 2 2 , 1 1
1
22
2
- -
- -
-
- - - -
=
- - - -
=

From Equation 1
From equation 2
From equation n-1
From equation n
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7. Gauss-Seidel Method
n
c a x
n - 1 n -
1,
j j
n j j
- ¹ =
1
1
n
c a x
n nj j
n j j
http://numericalmethods.eng.usf.edu
Algorithm
General Form of each equation
å j j
c a x
1 1
1
1
1 a
11
x
n
j j
¹ =
-
=
å j
c a x
2 2
1
2
22
2 a
x
j
n
j j
¹ =
-
=
1, 1
1
- -
-
- å
=
n n
n a
x
nn
n a
x
å
¹ =
-
=
1
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8. Gauss-Seidel Method
http://numericalmethods.eng.usf.edu
Algorithm
General Form for any row ‘i’
, 1,2, , .
c -
a x
i ij j
1
i n
a
x
ii
n
i j j
=
= i 
å
¹ =
How or where can this equation be used?
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9. Gauss-Seidel Method
Solve for the unknowns
http://numericalmethods.eng.usf.edu
Assume an initial guess for [X]
ù
ú ú ú ú ú ú
û
é
ê ê ê ê ê ê
ë
x
1
2
x
n-
n
x
x
1

Use rewritten equations to solve for
each value of xi.
Important: Remember to use the
most recent value of xi. Which
means to apply values calculated to
the calculations remaining in the
current iteration.
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10. Gauss-Seidel Method
Calculate the Absolute Relative Approximate Error
100
http://numericalmethods.eng.usf.edu
x x
x
new
i
old
i
new
i
a i e = - ´
So when has the answer been found?
The iterations are stopped when the absolute relative
approximate error is less than a prespecified tolerance for all
unknowns.
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11. Gauss-Seidel Method: Example 2
12 3 5
ù
é -
=
1 5 3
http://numericalmethods.eng.usf.edu
Given the system of equations
12x 3x - 5x 1 1 2 3 + =
x 5x 3x 28 1 2 3 + + =
3x 7x 13x 76 1 2 3 + + =
With an initial guess of
ù
ú ú ú
û
1
é
=
ê ê ê
ë
ù
ú ú ú û
é
ê ê ê
ë
0
1
x
1
2
3
x
x
The coefficient matrix is:
[ ]
ú ú ú
û
ê ê ê
ë
3 7 13
A
Will the solution converge using the
Gauss-Siedel method?
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12. Gauss-Seidel Method: Example 2
12 12 3 5 8 11 12 13 a = = ³ a + a = + - =
http://numericalmethods.eng.usf.edu
[ ]
ù
ú ú ú
û
é -
=
12 3 5
ê ê ê
ë
1 5 3
3 7 13
A
Checking if the coefficient matrix is diagonally dominant
5 5 1 3 4 22 21 23 a = = ³ a + a = + =
13 13 3 7 10 33 31 32 a = = ³ a + a = + =
The inequalities are all true and at least one row is strictly greater than:
Therefore: The solution should converge using the Gauss-Siedel Method
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13. Gauss-Seidel Method: Example 2
ù
1
é
=
ù
é
0
x
1
2
x
1 x = - + =
2 x = - - =
x = 76 - 3 0.50000 - 7 4.9000
=
3 http://numericalmethods.eng.usf.edu
Rewriting each equation
ù
ú ú ú û
é
=
ê ê ê
ë
ù
ú ú ú
û
é
ê ê ê
ë
12 3 5
ù
ú ú ú
û
é -
ê ê ê
ë
1
28
76
1 5 3
3 7 13
a
1
2
a
x = - x + x
3
a
1 3 5 2 3
12
1
x = - x - x
28 3 1 3
5
2
x = - x - x
76 3 7 1 2
13
3
With an initial guess of
ú ú ú
û
ê ê ê
ë
ú ú ú
û
ê ê ê
ë
1
3
x
( ) ( ) 0.50000
12
1 3 0 5 1
( ) ( ) 4.9000
5
28 0.5 3 1
( ) ( ) 3.0923
13
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14. Gauss-Seidel Method: Example 2
The absolute relative approximate error
a 1 Î = - ´ =
100 67.662%
http://numericalmethods.eng.usf.edu
0.50000 1.0000
0.50000
a 2 Î = - ´ =
100 100.00%
4.9000 0
4.9000
a 3 Î = - ´ =
100 67.662%
3.0923 1.0000
3.0923
The maximum absolute relative error after the first iteration is 100%
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15. Gauss-Seidel Method: Example 2
ù
é
=
ù
é
0.5000
x
1
Substituting the x values into the equations After Iteration #2 ú ú ú
ù
ú ú ú
û
0.14679
é
=
ê ê ê
ë
ù
ú ú ú
û
é
ê ê ê
ë
3.7153
3.8118
x
1
2
x
x
3
http://numericalmethods.eng.usf.edu
After Iteration #1
4.9000
2
x
( ) ( ) 0.14679
x = 1 - 3 4.9000 + 5 3.0923
=
1 12
( ) ( ) 3.7153
x = 28 - 0.14679 - 3 3.0923
=
2 5
( ) ( ) 3.8118
x = 76 - 3 0.14679 - 7 4.900
=
3 13
û
ê ê ê
ë
ú ú ú
û
ê ê ê
ë
3.0923
3
x
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16. Gauss-Seidel Method: Example 2
Iteration #2 absolute relative approximate error
1 Î = - ´ = a
100 240.62%
http://numericalmethods.eng.usf.edu
0.14679 0.50000
0.14679
2 Î = - ´ = a
100 31.887%
3.7153 4.9000
3.7153
3 Î = - ´ = a
100 18.876%
3.8118 3.0923
3.8118
The maximum absolute relative error after the first iteration is 240.62%
This is much larger than the maximum absolute relative error obtained in
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17. Gauss-Seidel Method: Example 2
Repeating more iterations, the following values are obtained
a 2 e a 3 Iteration a e 1 a2 a3
http://numericalmethods.eng.usf.edu
a 1 e
123456
0.50000
0.14679
0.74275
0.94675
0.99177
0.99919
67.662
240.62
80.23
21.547
4.5394
0.74260
4.900
3.7153
3.1644
3.0281
3.0034
3.0001
100.00
31.887
17.409
4.5012
0.82240
0.11000
3.0923
3.8118
3.9708
3.9971
4.0001
4.0001
67.662
18.876
4.0042
0.65798
0.07499
0.00000
ù
ú ú ú
û
ù
é
=
ê ê ê
ë
0.99919
3.0001
ù
ú ú ú
û
é
ê ê ê
ë
1
3
4
x
1
2
3
x
x
ú ú ú
û
é
=
ê ê ê
ë
ù
ú ú ú
û
é
ê ê ê
ë
4.0001
x
1
2
x
3
x
The solution obtained
is close to the exact solution of admission.edhole.co
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