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1/10/2010 http://numericalmethods.eng.usf.edu 1
Gauss-Siedel Method
Computer Engineering Majors
Authors: Autar Kaw
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Gauss-Seidel Method
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Gauss-Seidel MethodAn 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 iterationto check if error is within a pre-specified tolerance.
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Gauss-Seidel MethodWhy?
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 initialguess can be made, decreasing the number of iterations needed.
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Gauss-Seidel MethodAlgorithmA set of n equations and n unknowns:
11313212111 ... bxaxaxaxa nn =++++
2323222121 ... bxaxaxaxa n2n =++++
nnnnnnn bxaxaxaxa =++++ ...332211
. .
. .
. .
If: the diagonal elements arenon-zero
Rewrite each equation solvingfor the corresponding unknown
ex:
First equation, solve for x1
Second equation, solve for x2
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Gauss-Seidel MethodAlgorithmRewriting each equation
11
13132121
1a
xaxaxacx nn
=
nn
nnnnnn
n
nn
nnnnnnnnnn
nn
a
xaxaxacx
axaxaxaxacx
a
xaxaxacx
11,2211
1,1
,122,122,111,111
22
232312122
=
=
=
From Equation 1
From equation 2
From equation n-1
From equation n
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Gauss-Seidel MethodAlgorithmGeneral Form of each equation
11
11
11
1a
xac
x
n
jj
jj=
=
22
21
22
2a
xac
x
j
n
jj
j
==
1,1
11
,11
1
=
=nn
n
njj
jjnn
na
xac
x
nn
n
njj
jnjn
na
xac
x
=
=1
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Gauss-Seidel MethodAlgorithm
General Form for any row i
.,,2,1,1
nia
xac
xii
n
ijj
jiji
i =
=
=
How or where can this equation be used?
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Gauss-Seidel MethodSolve for the unknownsAssume an initial guess for [X]
n
-n
2
x
x
x
x
1
1
Use rewritten equations to solve for
each value of xi.
Important: Remember to use themost recent value of xi. Which
means to apply values calculated to
the calculations remaining in the
current iteration.
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Gauss-Seidel MethodCalculate the Absolute Relative Approximate Error
100
=
newi
old
i
new
i
ia x
xx
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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Example: Surface Shape DetectionTo infer the surface shape of an object from images taken of a surfacefrom three different directions, one needs to solve the following set ofequations
=
239
248
247
9428.02357.02357.0
9701.02425.00
9701.002425.0
3
2
1
x
x
x
The right hand side values are the light intensities from the middle ofthe images, while the coefficient matrix is dependent on the light
source directions with respect to the camera. The unknowns are theincident intensities that will determine the shape of the object.
Find the values of x1, x2, and x3 use the Gauss-Seidel method.
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Example: Surface Shape DetectionThe system of equations is:
Initial Guess:Assume an initial guess of
=
239
248
247
9428.02357.02357.0
9701.02425.00
9701.002425.0
3
2
1
x
x
x
=
10
1010
3
2
1
x
xx
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Example: Surface Shape DetectionRewriting each equation
2425.0
)9701.0(0247 321
xxx
=
2425.0
)9701.0(0248 312
xxx
=
9428.0
)2357.0()2357.0(239 213
=
xxx
=
239248
247
9428.02357.02357.09701.02425.00
9701.002425.0
3
2
1
xx
x
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Example: Surface Shape DetectionIteration 1Substituting initial guesses into the equations
=
10
10
10
3
2
1
x
x
x
( ) 6105824250
10970101002471 ..
.x ==
( )71062
24250
10970106105802482 .
.
..x =
=
( ) ( )81783
94280
710622357061058235702393 .
.
....x =
=
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Example: Surface Shape DetectionFinding the absolute relative approximate error
( )%.
.
.
%..
.
%..
.
x
xx
a
a
a
new
i
old
i
new
i
ia
2810110081783
1081783
0599910071062
1071062
0559910061058
1061058
100
3
2
1
=
=
=
=
=
=
=
At the end of the first iteration
The maximum absolute
relative approximate error is101.28%
=
81783
71062561058
3
2
1
.
..
x
xx
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Example: Surface Shape DetectionIteration 2Using
=
81.783
7.1062
6.1058
x
x
x
3
2
1
( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )98803
94280
92112235700211723570239
9211224250
8178397010021170248
0211724250
8178397010710620247
3
2
1
..
....x
..
...x
..
...x
=
=
==
=
=
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Example: Surface Shape DetectionFinding the absolute relative approximate error for the second iteration
( )
( )
( ) %..
..
%..
..
%.
.
..
a
a
a
4919710098803
8178398803
3015010092112
7106292112
0015010002117
6105802117
3
2
1
==
=
=
=
= At the end of the first
iteration
The maximum absoluterelative approximate error is
197.49%
=
98.803
9.2112
0.2117
3
2
1
x
x
x
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Iteration x1
x2
x3
1
2
3
4
5
6
1058.6
2117.0
4234.8
8470.1
16942
33888
99.055
150.00
149.99
150.00
149.99
150.00
1062.7
2112.9
4238.9
8466.0
16946
33884
99.059
150.30
149.85
150.07
149.96
150.01
783.81
803.98
2371.9
3980.5
8725.7
16689
101.28
197.49
133.90
159.59
145.62
152.28
Example: Surface Shape DetectionRepeating more iterations, the following values are obtained%
1a %
2a %
3a
Notice: The absolute relative approximate errors are not decreasing.
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Gauss-Seidel Method: PitfallWhat went wrong?Even though done correctly, the answer is not converging to the
correct answer
This example illustrates a pitfall of the Gauss-Siedel method: not allsystems of equations will converge.
Is there a fix?
One class of system of equations always converges: One with a diagonally
dominant coefficient matrix.
Diagonally dominant: [A] in [A] [X] = [C] is diagonally dominant if:
=
n
j
j
ijaa
i
1
ii
=
>n
ij
j
ijii aa1
for all i and for at least one i
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Gauss-Seidel Method: PitfallDiagonally Dominant: In other words.
For every row: the element on the diagonal needs to be equal than orgreater than the sum of the other elements of the coefficient matrix
For at least one row: The element on the diagonal needs to be greaterthan the sum of the elements.
What can be done? If the coefficient matrix is not originally diagonallydominant, the rows can be rearranged to make it diagonally dominant.
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Example: Surface Shape DetectionExamination of the coefficient matrix reveals that it is not diagonallydominant and cannot be rearranged to become diagonally dominant
This particular problem is an example of a system of linear
equations that cannot be solved using the Gauss-Seidel method.
Other methods that would work:
1. Gaussian elimination 2. LU Decomposition
9428.02357.02357.0
9701.02425.009701.002425.0
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Gauss-Seidel Method: Example 2Given the system of equations
15312321
x-xx =+
2835 321 xxx =++
761373 321 =++ xxx
=
1
01
3
2
1
x
x
x
With an initial guess of
The coefficient matrix is:
[ ]
=
1373
351
5312
A
Will the solution converge using the
Gauss-Siedel method?
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Gauss-Seidel Method: Example 2
[ ]
=1373
351
5312
A
Checking if the coefficient matrix is diagonally dominant
43155 232122 =+=+== aaa
10731313 323133 =+=+== aaa
8531212 131211 =+=+== aaa
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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Gauss-Seidel Method: Example 2
=
76
28
1
1373
351
5312
3
2
1
a
a
a
Rewriting each equation
12
531 321
xxx
+=
5328 312 xxx =
13
7376 213
xxx
=
With an initial guess of
=
1
0
1
3
2
1
x
x
x
( ) ( )50000.0
12
150311 =
+=x
( ) ( ) 9000.45 135.0282 =
=x
( ) ( )0923.3
13
9000.4750000.03763 =
=x
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Gauss-Seidel Method: Example 2The absolute relative approximate error
%00.10010050000.0
0000.150000.01
=
=a
%00.1001009000.4
09000.42a
=
=
%662.671000923.3
0000.10923.33a
==
The maximum absolute relative error after the first iteration is 100%
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Gauss-Seidel Method: Example 2
=
8118.3
7153.3
14679.0
3
2
1
x
x
x
After Iteration #1
( ) ( )14679.0
12
0923.359000.4311 =
+=x
( ) ( )7153.3
5
0923.3314679.0282 =
=x
( ) ( )8118.3
13
900.4714679.03763 =
=x
Substituting the x values into the
equationsAfter Iteration #2
=
0923.3
9000.4
5000.0
3
2
1
x
x
x
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Gauss-Seidel Method: Example 2Iteration 2 absolute relative approximate error
%61.24010014679.0
50000.014679.01a
=
=
%889.311007153.3
9000.47153.32a
=
=
%874.18100
8118.3
0923.38118.33a
=
=
The maximum absolute relative error after the first iteration is 240.61%
This is much larger than the maximum absolute relative error obtained in
iteration #1. Is this a problem?
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Iteration a1
a2
a3
1
2
3
4
5
6
0.50000
0.14679
0.74275
0.94675
0.99177
0.99919
100.00
240.61
80.236
21.546
4.5391
0.74307
4.9000
3.7153
3.1644
3.0281
3.0034
3.0001
100.00
31.889
17.408
4.4996
0.82499
0.10856
3.0923
3.8118
3.9708
3.9971
4.0001
4.0001
67.662
18.876
4.0042
0.65772
0.074383
0.00101
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Gauss-Seidel Method: Example 2Repeating more iterations, the following values are obtained%
1a %
2a %
3a
=
4
3
1
3
2
1
x
x
x
=
0001.4
0001.3
99919.0
3
2
1
x
x
x
The solution obtained is close to the exact solution of .
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Gauss-Seidel Method: Example 3Given the system of equations
761373 321 =++ xxx
2835 321 =++ xxx15312 321 =+ xxx
With an initial guess of
=
1
0
1
3
2
1
x
x
x
Rewriting the equations
313776 321 xxx =
5
328 312
xxx
=
5
3121 213
=
xxx
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Iteration a1
A2
a3
1
2
3
4
5
6
21.000
196.15
1995.0
20149
2.0364 105
2.0579 105
95.238
110.71
109.83
109.90
109.89
109.89
0.80000
14.421
116.02
1204.6
12140
1.2272 105
100.00
94.453
112.43
109.63
109.92
109.89
50.680
462.30
4718.1
47636
4.8144 105
4.8653 106
98.027
110.96
109.80
109.90
109.89
109.89
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Gauss-Seidel Method: Example 3Conducting six iterations, the following values are obtained%
1a %
2a %
3a
The values are not converging.
Does this mean that the Gauss-Seidel method cannot be used?
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Gauss-Seidel MethodThe Gauss-Seidel Method can still be used
The coefficient matrix is not
diagonally dominant
[ ]
=
5312
351
1373
A
But this is the same set of
equations used in example #2,
which did converge. [ ]
=
1373
351
5312
A
If a system of linear equations is not diagonally dominant, check to see if
rearranging the equations can form a diagonally dominant matrix.
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Gauss-Seidel MethodNot every system of equations can be rearranged to have a
diagonally dominant coefficient matrix.
Observe the set of equations
3321 =++ xxx
9432321
=++ xxx
97 321 =++ xxx
Which equation(s) prevents this set of equation from having a
diagonally dominant coefficient matrix?
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Gauss-Seidel MethodSummary
-Advantages of the Gauss-Seidel Method
-Algorithm for the Gauss-Seidel Method
-Pitfalls of the Gauss-Seidel Method
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Gauss-Seidel Method
Questions?
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Additional ResourcesFor all resources on this topic such as digital audiovisuallectures, primers, textbook chapters, multiple-choicetests, worksheets in MATLAB, MATHEMATICA, MathCad
and MAPLE, blogs, related physical problems, pleasevisit
http://numericalmethods.eng.usf.edu/topics/gauss_seid
el.html
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