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Transcript of 10/26/2015 1 Gauss-Siedel Method Civil Engineering Majors Authors: Autar Kaw Transforming.
04/21/23http://
numericalmethods.eng.usf.edu 1
Gauss-Siedel Method
Civil Engineering Majors
Authors: Autar Kaw
http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM
Undergraduates
Gauss-Seidel Method
http://numericalmethods.eng.usf.edu
http://numericalmethods.eng.usf.edu
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 iteration to check if error is within a pre-specified tolerance.
http://numericalmethods.eng.usf.edu
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 initial guess can be made, decreasing the number of iterations needed.
http://numericalmethods.eng.usf.edu
Gauss-Seidel MethodAlgorithm
A set of n equations and n unknowns:
11313212111 ... bxaxaxaxa nn
2323222121 ... bxaxaxaxa n2n
nnnnnnn bxaxaxaxa ...332211
. . . . . .
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 MethodAlgorithm
Rewriting each equation
11
131321211 a
xaxaxacx nn
nn
nnnnnnn
nn
nnnnnnnnnn
nn
a
xaxaxacx
a
xaxaxaxacx
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
http://numericalmethods.eng.usf.edu
Gauss-Seidel MethodAlgorithm
General Form of each equation
11
11
11
1 a
xac
x
n
jj
jj
22
21
22
2 a
xac
x
j
n
jj
j
1,1
11
,11
1
nn
n
njj
jjnn
n a
xac
x
nn
n
njj
jnjn
n a
xac
x
1
http://numericalmethods.eng.usf.edu
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?
http://numericalmethods.eng.usf.edu
Gauss-Seidel MethodSolve for the unknowns
Assume 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 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 MethodCalculate the Absolute Relative Approximate Error
100
newi
oldi
newi
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.
Example: Cylinder Stresses
To find the maximum stresses in a compounded cylinder, the following four simultaneous linear equations need to solved.
0
007.0
0
10887.7
106057.3102857.400
15384.05.615384.05.6
104619.5102857.4104619.5102857.4
00102307.9102857.4 3
57
5757
57
4
3
2
1
c
c
c
c
Example: Cylinder StressesIn the compound cylinder, the inner cylinder has an internal radius of a = 5”, and outer radius c = 6.5”, while the outer cylinder has an internal radius of c = 6.5” and outer radius, b=8”. Given E = 30×106 psi, ν = 0.3, and that the hoop
stress in outer cylinder is given by
2432
11
1 rcc
E
find the stress on the inside radius of the outer cylinder. Find the values of c1, c2, c3 and c4 using Gauss-Seidel Method.
Assume an initial guess of and conduct two iterations.
03.0
0002.0
001.0
005.0
4
3
2
1
c
c
c
c
Example: Cylinder Stresses
Rewriting each equation
0
007.0
0
10887.7
106057.3102857.400
15384.05.615384.05.6
104619.5102857.4104619.5102857.4
00102307.9102857.4 3
57
5757
57
4
3
2
1
c
c
c
c
7
43253
1 1028574
001023079108877
.
ccc..c
5
45
37
17
2 1046195
1046195102857410285740
.
c.c.c.c
56
153840153840560070 4213 .
c.c.c..c
5
37
214 1060573
1028574000
.
c.ccc
Example: Cylinder Stresses
47
53
1 10624911028574
00101023079108877
.
.
...c
35
5747
2 10556911046195
0301046195000201028574106249110285740
.
.
......c
434
3 104125256
03015384010556911538401062491560070
..
.......c
25
47
4 10867521060573
104125210285740
.
.
..c
Iteration 1
Substituting initial guesses into the equations
03.0
0002.0
001.0
005.0
4
3
2
1
c
c
c
c
Example: Cylinder Stresses
Finding the absolute relative approximate error
%.
..
%.
..
%.
..
%.
..
c
cc
a
a
a
a
newi
oldi
newi
ia
6223.41001086752
0301086752
098.171001041252
00201041252
770.351001055691
00101055691
1.29771001062491
00501062491
100
2
2
4
4
4
3
3
3
2
4
4
1
At the end of the first iteration
The maximum absolute relative approximate error
is 2977.1%
2
4
3
4
4
3
2
1
108675.2
104125.2
105569.1
106249.1
c
c
c
c
Example: Cylinder Stresses
Iteration 2
Using
2
4
3
4
4
3
2
1
108675.2
104125.2
105569.1
106249.1
c
c
c
c
4
7
353
1
105050.1
102857.4
10559.1)102307.9(10887.7c
2
5
47
4
4
234
3
3
5
254747
2
103643.2
106057.3
109892.1102857.40c
109892.1
5.6
108675.215384.0100639.2)15384.0()105050.1()5.6(007.0c
100639.2
104619.5
108675.2104619.5104125.2)102857.4()105050.1(102857.40c
Example: Cylinder Stresses
Finding the absolute relative approximate error for the second iteration
%.
..
%.
..
%.
..
%.
..
a
a
a
a
281.211001036432
10867521036432
281.211001098921
10412521098921
44.1751001006392
10556911006392
9702.71001050501
10624911050501
2
22
4
4
44
3
3
33
2
4
44
1
Example: Cylinder Stresses
At the end of the second iteration:
The maximum absolute relative approximate error
is 175.44%
2
4
3
4
4
3
2
1
103643.2
109892.1
100639.2
105050.1
c
c
c
c
At the end of the second iteration the stress on the inside radius of the outer cylinder is calculated.
psi21439
5.6
3.01103643.23.01109892.1
3.01
1030
11
1
224
2
6
232 4
rcc
E
Iteration c1 c2 c3 c4
1
2
3
4
5
6
−1.6249×10−4
−1.5050×10−4
−2.2848×10−4
−3.9711×10−4
−8.0755×10−4
−1.6874×10−3
2977.1
7.9702
34.132
42.464
50.825
52.142
1.5569×10−3
−2.0639×10−3
−9.8931×10−3
−2.8949×10−2
−6.9799×10−2
−1.7015×10−1
35.770
175.44
79.138
65.826
58.524
58.978
2.4125×10−4
1.9892×10−4
5.4716×10−5
−1.5927×10−4
−9.3454×10−4
−2.0085×10−3
17.098
21.281
263.55
134.53
82.957
53.472
2.8675×10−2
2.3643×10−2
6.5035×10−3
−1.8931×10−2
−1.1108×10−1
−2.3873×10−1
4.6223
21.281
263.55
134.35
82.957
53.472
Example: Cylinder Stresses
Conducting more iterations, the following values are obtained
%1a %
2a %3a %
4a
Notice: The absolute relative approximate errors are not decreasing
http://numericalmethods.eng.usf.edu
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
jj
ijaa
i1
ii
n
ijj
ijii aa1
for all ‘i’ and for at least one ‘i’
http://numericalmethods.eng.usf.edu
Gauss-Seidel Method: Pitfall
116123
14345
3481.52
A
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.
1293496
55323
5634124
]B[
Which coefficient matrix is diagonally dominant?
Most physical systems do result in simultaneous linear equations that have diagonally dominant coefficient matrices.
Example: Cylinder Stresses
Examination of the coefficient matrix reveals that it is not diagonally dominant and cannot be
rearranged to become diagonally dominant
57
5757
57
106057.3102857.400
15384.05.615384.05.6
104619.5102857.4104619.5102857.4
00102307.9102857.4
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
http://numericalmethods.eng.usf.edu
Gauss-Seidel Method: Example 2
Given the system of equations
15312 321 x- x x
2835 321 x x x 761373 321 x x x
1
0
1
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?
http://numericalmethods.eng.usf.edu
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
http://numericalmethods.eng.usf.edu
Gauss-Seidel Method: Example 2
76
28
1
1373
351
5312
3
2
1
a
a
a
Rewriting each equation
12
531 321
xxx
5
328 312
xxx
13
7376 213
xxx
With an initial guess of
1
0
1
3
2
1
x
x
x
50000.0
12
150311
x
9000.4
5
135.0282
x
0923.3
13
9000.4750000.03763
x
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Gauss-Seidel Method: Example 2
The 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 equations
After Iteration #2
0923.3
9000.4
5000.0
3
2
1
x
x
x
http://numericalmethods.eng.usf.edu
Gauss-Seidel Method: Example 2
Iteration #2 absolute relative approximate error
%61.24010014679.0
50000.014679.01a
%889.311007153.3
9000.47153.32a
%874.181008118.3
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?
Iteration a1 a2 a3
123456
0.500000.146790.742750.946750.991770.99919
100.00240.6180.23621.5464.5391
0.74307
4.90003.71533.16443.02813.00343.0001
100.0031.88917.4084.4996
0.824990.10856
3.09233.81183.97083.99714.00014.0001
67.66218.8764.0042
0.657720.0743830.00101
http://numericalmethods.eng.usf.edu
Gauss-Seidel Method: Example 2
Repeating 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
xThe solution obtained is close to the exact solution of .
http://numericalmethods.eng.usf.edu
Gauss-Seidel Method: Example 3
Given the system of equations
761373 321 xxx
2835 321 xxx
15312 321 xxx
With an initial guess of
1
0
1
3
2
1
x
x
x
Rewriting the equations
3
13776 321
xxx
5
328 312
xxx
5
3121 213
xxx
Iteration a1 A2 a3
123456
21.000−196.15−1995.0−20149
2.0364×105
−2.0579×105
95.238110.71109.83109.90109.89109.89
0.8000014.421
−116.021204.6−12140
1.2272×105
100.0094.453112.43109.63109.92109.89
50.680−462.304718.1−47636
4.8144×105
−4.8653×106
98.027110.96109.80109.90109.89109.89
http://numericalmethods.eng.usf.edu
Gauss-Seidel Method: Example 3
Conducting 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.
http://numericalmethods.eng.usf.edu
Gauss-Seidel MethodNot every system of equations can be rearranged to have a diagonally dominant coefficient matrix.
Observe the set of equations
3321 xxx
9432 321 xxx
97 321 xxx
Which equation(s) prevents this set of equation from having a diagonally dominant coefficient matrix?
http://numericalmethods.eng.usf.edu
Gauss-Seidel MethodSummary
-Advantages of the Gauss-Seidel Method
-Algorithm for the Gauss-Seidel Method
-Pitfalls of the Gauss-Seidel Method
http://numericalmethods.eng.usf.edu
Gauss-Seidel Method
Questions?
Additional ResourcesFor 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_seidel.html