Transcript PPT - Math For College

Gauss-Siedel Method
Electrical Engineering Majors
Authors: Autar Kaw
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates
4/13/2015
http://numericalmethods.eng.usf.edu
1
Gauss-Seidel Method
http://numericalmethods.eng.usf.edu
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
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.
http://numericalmethods.eng.usf.edu
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
http://numericalmethods.eng.usf.edu
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
http://numericalmethods.eng.usf.edu
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
http://numericalmethods.eng.usf.edu
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?
http://numericalmethods.eng.usf.edu
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.
http://numericalmethods.eng.usf.edu
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.
http://numericalmethods.eng.usf.edu
Example: Unbalanced three phase load
Three-phase loads are common in AC systems. When the system
is balanced the analysis can be simplified to a single equivalent circuit
model. However, when it is unbalanced the only practical solution
involves the solution of simultaneous linear equations. In a model the
following equations need to be solved.
0.7460  0.4516
0.4516 0.7460

0.0100  0.0080

0.0080 0.0100
0.0100  0.0080

0.0080 0.0100
0.0100  0.0080 0.0100  0.0080  I ar   120 
0.0080 0.0100 0.0080 0.0100   I ai   0.000 
0.7787  0.5205 0.0100  0.0080  I br   60.00
  

0.5205 0.7787 0.0080 0.0100   I bi    103.9 
0.0100  0.0080 0.8080  0.6040  I cr   60.00
  

0.0080 0.0100 0.6040 0.8080   I ci   103.9 
Find the values of Iar , Iai , Ibr , Ibi , Icr , and Ici using the Gauss-Seidel
method.
Example: Unbalanced three phase load
Rewrite each equation to solve for each of the unknowns
I ar 
120 .00   0.4516 I ai  0.0100 I br   0.0080 I bi  0.0100 I cr   0.0080 I ci
0.7460
I ai 
0.000  0.4516 I ar  0.0080 I br  0.0100 I bi  0.0080 I cr  0.0100 I ci
0.7460
I br 
 60.00  0.0100 I ar   0.0080 I ai   0.5205 I bi  0.0100 I cr   0.0080 I ci
0.7787
I bi 
 103.9  0.0080 I ar  0.0100 I ai  0.5205 I br  0.0080 I cr  0.0100 I ci
0.7787
I cr 
 60.00  0.0100 I ar   0.0080 I ai0.0100 I br   0.0080 I bi   0.6040 I ci
0.8080
I ci 
103.9  0.0080 I ar  0.0100 I ai  0.0080 I br  0.0100 I bi  0.6040 I cr
0.8080
Example: Unbalanced three phase load
For iteration 1, start with an initial guess value
Initial Guess:
 I ar  20
 I  20
 ai   
 I br  20
  
 I bi  20
 I cr  20
   
 I ci  20
Example: Unbalanced three phase load
Substituting the guess values into the first equation
120  0.4516I ai  0.0100I br   0.0080I bi  0.0100I cr   0.0080I ci
0.7460
 172.86
I ar 
Substituting the new value of Iar and the remaining guess values into
the second equation
0.00  0.4516I ar  0.0080I br  0.0100I bi  0.0080I cr  0.0100I ci
0.7460
 105.61
I ai 
Example: Unbalanced three phase load
Substituting the new values Iar , Iai , and the remaining guess values into
the third equation
 60.00  0.0100I ar   0.0080I ai   0.5205I bi  0.0100I cr   0.0080I ci
0.7787
 67.039
I br 
Substituting the new values Iar , Iai , Ibr , and the remaining guess values into
the fourth equation
 103.9  0.0080I ar  0.0100I ai  0.5205I br  0.0080I cr  0.0100I ci
0.7787
 89.499
I bi 
Example: Unbalanced three phase load
Substituting the new values Iar , Iai , Ibr , Ibi , and the remaining guess
values into the fifth equation
 60.00  0.0100I ar   0.0080I ai0.0100I br   0.0080I bi   0.6040I ci
0.8080
 62.548
I cr 
Substituting the new values Iar , Iai , Ibr , Ibi , Icr , and the remaining guess
value into the sixth equation
103.9  0.0080I ar  0.0100I ai  0.0080I br  0.0100I bi  0.6040I cr
0.8080
 176.71
I ci 
Example: Unbalanced three phase load
At the end of the first iteration, the solution matrix is:
 I ar   172.86 
 I    105.61
 ai  

I br   67.039
 

I

89
.
499
 bi  

 I cr    62.548
  

I
176
.
71

 ci  
How accurate is the solution? Find the absolute relative
approximate error using:
a i
xinew  xiold

100
new
xi
Example: Unbalanced three phase load
Calculating the absolute relative approximate errors
172.86  20
100  88.430%
172.86
a 5 
 62.548 20
100  131.98%
 62.548
a 2 
 105.61 20
100  118.94%
 105.61
a 6 
176.71 20
100  88.682%
176.71
a 3 
 67.039 20
100  129.83%
 67.039
a 1 
a
4
 89.499 20

100  122.35%
 89.499
The maximum error after
the first iteration is:
131.98%
Another iteration is needed!
Example: Unbalanced three phase load
Starting with the values obtained in iteration #1
 I ar   172.86 
 I    105.61
 ai  

 I br   67.039
 

I

89
.
499
bi
  

 I cr    62.548
  

I
176
.
71

 ci  
Substituting the values from Iteration 1 into the first equation
120.00   0.4516I ai  0.0100I br   0.0080I bi  0.0100I cr   0.0080I ci
I ar 
0.7460
 99.600
Example: Unbalanced three phase load
Substituting the new value of Iar and the remaining values from
Iteration 1 into the second equation
0.00  0.4516I ar  0.0080I br  0.0100I bi  0.0080I cr  0.0100I ci
0.7460
 60.073
I ai 
Substituting the new values Iar , Iai , and the remaining values from
Iteration 1 into the third equation
 60.00  0.0100I ar   0.0080I ai   0.5205I bi  0.0100I cr   0.0080I ci
0.7787
 136.15
I br 
Example: Unbalanced three phase load
Substituting the new values Iar , Iai , Ibr , and the remaining values from
Iteration 1 into the fourth equation
 103.9  0.0080I ar  0.0100I ai  0.5205I br  0.0080I cr  0.0100I ci
0.7787
 44.299
I bi 
Substituting the new values Iar , Iai , Ibr , Ibi , and the remaining values
From Iteration 1 into the fifth equation
 60.00  0.0100I ar   0.0080I ai0.0100I br   0.0080I bi   0.6040I ci
0.8080
 57.259
I cr 
Example: Unbalanced three phase load
Substituting the new values Iar , Iai , Ibr , Ibi , Icr , and the remaining
value from Iteration 1 into the sixth equation
103.9  0.0080I ar  0.0100I ai  0.0080I br  0.0100I bi  0.6040I cr
0.8080
 87.441
I ci 
The solution matrix at the end of
the second iteration is:
 I ar   99.600 
 I    60.073
 ai  

 I br    136.15 
 

I

44
.
299
 bi  

 I cr   57.259 
  

 I ci   87.441 
Example: Unbalanced three phase load
Calculating the absolute relative approximate errors for
the second iteration
a 1 
a 2 
a 3 
a 4 
99.600 172.86
100  73.552%
99.600
 60.073 (105.61)
100  75.796%
 60.073
 136.35  (67.039)
100  50.762%
 136.35
 44.299 (89.499)
100  102.03%
 44.299
a 5 
57.259 (62.548)
100  209.24%
57.259
a 6 
87.441 176.71
100  102.09%
87.441
The maximum error after
the second iteration is:
209.24%
More iterations are needed!
Example: Unbalanced three phase load
Repeating more iterations, the following values are obtained
Iteration
1
2
3
4
5
6
Iteration
1
2
3
4
5
6
Iar
Iai
Ibr
Ibi
Icr
Ici
172.86
99.600
126.01
117.25
119.87
119.28
−105.61
−60.073
−76.015
−70.707
−72.301
−71.936
−67.039
−136.15
−108.90
−119.62
−115.62
−116.98
−89.499
−44.299
−62.667
−55.432
−58.141
−57.216
−62.548
57.259
−10.478
27.658
6.2513
18.241
176.71
87.441
137.97
109.45
125.49
116.53
a 1 %
a 2 %
a 3 %
a 4 %
a 5 %
a 6 %
88.430
73.552
20.960
7.4738
2.1840
0.49408
118.94
75.796
20.972
7.5067
2.2048
0.50789
129.83
50.762
25.027
8.9631
3.4633
1.1629
122.35
102.03
29.311
13.053
4.6595
1.6170
131.98
209.24
646.45
137.89
342.43
65.729
88.682
102.09
36.623
26.001
12.742
7.6884
Example: Unbalanced three phase load
After six iterations,
the solution matrix is
 I ar   119.28 
 I   71.936
 ai  

 I br    116.98
 

I
57
.
216
 bi  

 I cr   18.241 
  

 I ci   116.53 
The maximum error after
the sixth iteration is:
65.729%
The absolute relative approximate error is still high, but allowing for
more iterations, the error quickly begins to converge to zero.
What could have been done differently to allow for a faster
convergence?
Example: Unbalanced three phase load
Repeating more iterations, the following values are obtained
Iteration
32
33
Iar
119.33
119.33
Iai
Ibr
Ibi
Icr
−57.432
−57.432
13.940
13.940
Ici
−71.973
−71.973
−116.66
−116.66
a 2 %
a 3 %
a 4 %
a 5 %
a 6 %
119.74
119.74
Iteration
a 1 %
32
3.0666×10−7
3.0047×10−7
4.2389×10−7
5.7116×10−7
2.0941×10−5
1.8238×10−6
33
1.7062×10−7
1.6718×10−7
2.3601×10−7
3.1801×10−7
1.1647×10−5
1.0144×10−6
Example: Unbalanced three phase load
After 33 iterations, the solution matrix is
 I ar   119.33 
 I   71.973
 ai  

 I br    116.66
 

I

57
.
432
 bi  

 I cr   13.940 
  

 I ci   119.74 
The maximum absolute relative approximate error is 1.1647×10−5%.
Gauss-Seidel Method: Pitfall
Even though done correctly, the answer may not converge to the
correct answer.
This is 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
http://numericalmethods.eng.usf.edu
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.
http://numericalmethods.eng.usf.edu
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?
http://numericalmethods.eng.usf.edu
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
http://numericalmethods.eng.usf.edu
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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