Transcript Gauss - Seidel Method - Transforming Numerical Methods

Gauss-Siedel Method
Major: All Engineering Majors
Authors: Autar Kaw
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates
7/7/2015
http://numericalmethods.eng.usf.edu
1
Gauss-Seidel Method
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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.
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Gauss-Seidel Method
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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Gauss-Seidel Method
Algorithm
A set of n equations and n unknowns:
a11x1  a12 x2  a13 x3  ...  a1n xn  b1
a21 x1  a22 x2  a23 x3  ... a2n xn  b2
.
.
.
.
.
.
an1x1  an2 x2  an3 x3  ...  ann xn  bn
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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Gauss-Seidel Method
Algorithm
Rewriting each equation
x1 
c1  a12 x2  a13 x3   a1n xn
a11
From Equation 1
c2  a21 x1  a23 x3   a2 n xn
a22
From equation 2
x2 
xn 1 
xn 



cn 1  an 1,1 x1  an 1, 2 x2   an 1,n  2 xn  2  an 1,n xn
From equation n-1
an 1,n 1
cn  an1 x1  an 2 x2    an ,n 1 xn 1
From equation n
ann
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Gauss-Seidel Method
Algorithm
General Form of each equation
n
c1   a1 j x j
x1 
j 1
j 1
a11
cn1 
xn1 
n
a
j 1
j  n 1
n 1, j
xj
an1,n1
n
c n   a nj x j
n
c2   a2 j x j
x2 
j 1
j 2
a 22
xn 
j 1
j n
a nn
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Gauss-Seidel Method
Algorithm
General Form for any row ‘i’
n
ci   aij x j
xi 
j 1
j i
aii
, i  1,2,, n.
How or where can this equation be used?
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Gauss-Seidel Method
Solve for the unknowns
Assume an initial guess for [X]
 x1 
x 
 2
  
 
 xn-1 
 xn 
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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Gauss-Seidel Method
Calculate the Absolute Relative Approximate Error
a i
xinew  xiold

100
new
xi
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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Gauss-Seidel Method: Example 1
The upward velocity of a rocket
is given at three different times
Table 1 Velocity vs. Time data.
Time, t s Velocity v m/s
5
106.8
8
177.2
12
279.2
The velocity data is approximated by a polynomial as:
vt   a1t 2  a2t  a3 , 5  t  12.
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Gauss-Seidel Method: Example 1
Using a Matrix template of the form
The system of equations becomes
Initial Guess: Assume an initial guess of
t12
2
t 2
t32

t1
t2
t3
1  a1   v1 

1 a2   v2 
1  a3  v3 
 25 5 1  a1  106.8 
 64 8 1 a   177.2 

  2 

144 12 1  a3  279.2
 a1   1 
a   2
 2  
 a 3  5 
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Gauss-Seidel Method: Example 1
Rewriting each equation
 25 5 1  a1  106.8 
 64 8 1 a   177.2 

  2 

144 12 1  a3  279.2
106 .8  5a 2  a 3
a1 
25
177 .2  64 a1  a3
a2 
8
279 .2  144 a1  12 a 2
a3 
1
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Gauss-Seidel Method: Example 1
Applying the initial guess and solving for ai
 a1   1 
a   2
 2  
 a 3  5 
Initial Guess
a1 
106 .8  5(2)  (5)
 3.6720
25
177 .2  643.6720   5
a2 
 7.8510
8
a3 
279 .2  144 3.6720   12 7.8510 
 155 .36
1
When solving for a2, how many of the initial guess values were used?
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Gauss-Seidel Method: Example 1
Finding the absolute relative approximate error
a i
xinew  xiold

100
new
xi
a 1 
a
2
3.6720 1.0000
x100  72.76%
3.6720
 7.8510 2.0000

x100  125.47%
 7.8510
a 3 
At the end of the first iteration
 a1   3.6720 
a    7.8510
 2 

 a 3    155.36
The maximum absolute
relative approximate error is
125.47%
 155.36  5.0000
x100  103.22%
 155.36
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Gauss-Seidel Method: Example 1
Using
 a1   3.6720 
a    7.8510
 2 

 a3    155.36
Iteration #2
the values of ai are found:
a1 
106 .8  5 7.8510   155 .36
 12.056
25
a2 
177 .2  6412.056   155 .36
 54.882
8
from iteration #1
a3 
279 .2  144 12.056   12 54.882 
 798 .34
1
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Gauss-Seidel Method: Example 1
Finding the absolute relative approximate error
12.056 3.6720
At the end of the second iteration
a 1 
x100  69.543%
12.056
 a   12.056 
1
a 2 
a 3 
 54.882  7.8510
x100  85.695%
 54.882
 798.34   155.36
x100  80.540%
 798.34
a    54.882
 2 

 a3   798.54
The maximum absolute
relative approximate error is
85.695%
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Gauss-Seidel Method: Example 1
Repeating more iterations, the following values are obtained
Iteration
1
2
3
4
5
6
a1
3.6720
12.056
47.182
193.33
800.53
3322.6
a 1 %
72.767
69.543
74.447
75.595
75.850
75.906
a2
−7.8510
−54.882
−255.51
−1093.4
−4577.2
−19049
a 2 %
125.47
85.695
78.521
76.632
76.112
75.972
a3
−155.36
−798.34
−3448.9
−14440
−60072
−249580
a 3 %
103.22
80.540
76.852
76.116
75.963
75.931
Notice – The relative errors are not decreasing at any significant rate
 a 1  0.29048
Also, the solution is not converging to the true solution of   

a

19
.
690
 2 

a 3   1.0857 
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Gauss-Seidel Method: Pitfall
What 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 all
systems 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
n
aii   aij
j 1
j i
for all ‘i’
and
aii   aij
for at least one ‘i’
j 1
j i
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Gauss-Seidel Method: Pitfall
Diagonally dominant: The coefficient on the diagonal must be at least
equal to the sum of the other coefficients in that row and at least one row
with a diagonal coefficient greater than the sum of the other coefficients
in that row.
Which coefficient matrix is diagonally dominant?
 2 5.81 34
A    45 43 1 
123 16
1 
124 34 56 
[ B]   23 53 5 
 96 34 129
Most physical systems do result in simultaneous linear equations that
have diagonally dominant coefficient matrices.
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Gauss-Seidel Method: Example 2
Given the system of equations
12x1  3x2- 5x3  1
x1  5x2  3x3  28
3x1  7 x2  13x3  76
With an initial guess of
 x1  1 
 x   0 
 2  
 x 3  1 
The coefficient matrix is:
12 3  5
A   1 5 3 
 3 7 13 
Will the solution converge using the
Gauss-Siedel method?
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Gauss-Seidel Method: Example 2
Checking if the coefficient matrix is diagonally dominant
12 3  5
A   1 5 3 
 3 7 13 
a11  12  12  a12  a13  3   5  8
a22  5  5  a21  a23  1  3  4
a33  13  13  a31  a32  3  7  10
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
Rewriting each equation
With an initial guess of
12 3  5  a1   1 
 1 5 3  a   28

  2  
 3 7 13   a3  76
1  3 x 2  5 x3
x1 
12
28  x1  3x3
x2 
5
76  3 x1  7 x 2
x3 
13
 x1  1 
 x   0 
 2  
 x 3  1 
x1 
1  30   51
 0.50000
12
28  0.5  31
x2 
 4.9000
5
76  30.50000   74.9000 
x3 
 3.0923
13
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Gauss-Seidel Method: Example 2
The absolute relative approximate error
0.50000 1.0000
a 1 
100  100.00%
0.50000
a
a
2
3
4.9000 0

100  100.00%
4.9000
3.0923 1.0000

100  67.662%
3.0923
The maximum absolute relative error after the first iteration is 100%
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Gauss-Seidel Method: Example 2
After Iteration #1
 x1  0.5000
 x    4.9000
 2 

 x3   3.0923
Substituting the x values into the
equations
x1 
1  34.9000   53.0923 
 0.14679
12
x2 
28  0.14679   33.0923 
 3.7153
5
x3 
76  30.14679   74.900 
 3.8118
13
After Iteration #2
 x1  0.14679
 x    3.7153 
 2 

 x3   3.8118 
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Gauss-Seidel Method: Example 2
Iteration #2 absolute relative approximate error
a 1 
a
a
2
3
0.14679 0.50000
100  240.61%
0.14679
3.7153 4.9000

100  31.889%
3.7153
3.8118 3.0923

100  18.874%
3.8118
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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Gauss-Seidel Method: Example 2
Repeating more iterations, the following values are obtained
Iteration
a1
a 1 %
a2
a 2 %
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
a 3 %
67.662
18.876
4.0042
0.65772
0.074383
0.00101
 x1  0.99919
The solution obtained  x    3.0001  is close to the exact solution of
 2 

 x3   4.0001
 x1  1 
 x   3 .
 2  
 x 3   4 
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Gauss-Seidel Method: Example 3
Given the system of equations
3x1  7 x2  13x3  76
Rewriting the equations
x1  5x2  3x3  28
76  7 x2  13 x3
x1 
3
12x1  3x2  5x3  1
With an initial guess of
 x1  1 
 x   0 
 2  
 x3  1 
28  x1  3 x3
x2 
5
1  12 x1  3 x 2
x3 
5
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Gauss-Seidel Method: Example 3
Conducting six iterations, the following values are obtained
Iteration
a1
1
2
3
4
5
6
21.000
−196.15
−1995.0
−20149
2.0364×105
−2.0579×105
a 1 %
A2
95.238
0.80000
110.71
14.421
109.83
−116.02
109.90
1204.6
109.89
−12140
109.89 1.2272×105
a 2 %
a3
a 3 %
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
The values are not converging.
Does this mean that the Gauss-Seidel method cannot be used?
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Gauss-Seidel Method
The Gauss-Seidel Method can still be used
The coefficient matrix is not
diagonally dominant
But this is the same set of
equations used in example #2,
which did converge.
 3 7 13 
A   1 5 3 
12 3  5
12 3  5
A   1 5 3 
 3 7 13 
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 Method
Not every system of equations can be rearranged to have a
diagonally dominant coefficient matrix.
Observe the set of equations
x1  x2  x3  3
2 x1  3x2  4 x3  9
x1  7 x2  x3  9
Which equation(s) prevents this set of equation from having a
diagonally dominant coefficient matrix?
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Gauss-Seidel Method
Summary
-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 Resources
For all resources on this topic such as digital audiovisual
lectures, primers, textbook chapters, multiple-choice
tests, worksheets in MATLAB, MATHEMATICA, MathCad
and MAPLE, blogs, related physical problems, please
visit
http://numericalmethods.eng.usf.edu/topics/gauss_seid
el.html
THE END
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