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04/20/23http://
numericalmethods.eng.usf.edu 1
Gauss-Siedel Method
Industrial 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.
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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
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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
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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 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: Production Optimization
To find the number of toys a company should manufacture per day to optimally use their injection-molding machine and the assembly line, one needs to solve the following set of equations. The unknowns are the number of toys for boys, x1, number of toys for girls, x2, and the number of unisexual toys, x3.
0
1260
756
00.000.105.1
3333.06667.01667.0
6667.01667.03333.0
3
2
1
x
x
x
Find the values of x1, x2,and x3 using the Gauss-Seidel Method
Example: Production Optimization
The system of equations is:
Initial Guess: Assume an initial guess of
100
1000
1000
3
2
1
x
x
x
0
1260
756
00.000.105.1
3333.06667.01667.0
6667.01667.03333.0
3
2
1
x
x
x
Example: Production Optimization
Rewriting each equation
0
1260
756
00.000.105.1
3333.06667.01667.0
6667.01667.03333.0
3
2
1
x
x
x
3333.0
6667.01667.0756 321
xxx
6667.0
3333.01667.01260 312
xxx
0
)00.1(05.10 213
xxx
Example: Production Optimization
0
)00.1(05.10 213
xxx
The Equation for x3 is divided by 0 which is undefined. Therefore the order of the equations will need to be
reordered. Equation 3 and equation 1 will be switched. By switching equations 3 and 1, the matrix will also become
diagonally dominant.
The system of equations becomes:
756
1260
0
666701667033330
333306667016670
000001051
3
2
1
x
x
x
...
...
...
Example: Production Optimization
Rewriting each equation
756
1260
0
666701667033330
333306667016670
000001051
3
2
1
x
x
x
...
...
...
051
00010 321 .
xx.x
6667.0
1667.03333.0756 213
xxx
6667.0
3333.01667.01260 312
xxx
Example: Production Optimization
Applying the initial guess and solving for xi
Initial Guess
When solving for x2, how many of the initial guess values were used?
100
1000
1000
3
2
1
x
x
x
38.95205.1
10001000)00.1(0x1
8.16016667.0
1003333.038.9521667.01260x2
32.2576667.0
8.16011667.038.9523333.0756x3
Example: Production Optimization
Finding the absolute relative approximate error
100
newi
oldi
newi
ia x
xx
At the end of the first iteration
The maximum absolute relative approximate error
is 61.138%
%..
.a 00005100
38952
1000389521
%..
.a 57037100
81601
1000816012
%138.6110032.257
10032.2573
a
32257
81601
38952
3
2
1
.
.
.
x
x
x
Example: Production Optimization
Iteration 2
Using from Iteration 1,
the values of xi are found:
32257
81601
38952
3
2
1
.
.
.
x
x
x
51525
051
3225708160100101 .
.
...x
8137966670
3225733330515251667012602 .
.
....x
2952666670
813791667051525333307563 .
.
....x
Example: Production Optimization
Finding the absolute relative approximate error
At the end of the second iteration
The maximum absolute relative approximate error
is 878.59%
%570.371005.1525
38.9525.15251
a
%085.161008.1379
8.16018.13792
a
%59.878100295.26
32.257295.263
a
29526
81379
51525
3
2
1
.
.
.
x
x
x
Iteration x1 x2 x3
123456
952.381525.51314.11474.51406.01451.9
537.57016.08510.8744.86863.1618
1601.81379.81548.21476.31524.51501.9
37.57016.08510.8744.86863.16181.5021
257.3226.29589.87627.69449.86332.554
61.138878.5970.743224.5344.45953.170
Example: Production Optimization
Repeating more iterations, the following values are obtained%1a %
2a %3a
! Notice – After six iterations, the absolute relative approximate errors are decreasing, but are still
high.
Iteration x1 x2 x3
2021
1439.81439.8
0.000642760.00034987
1511.81511.8
0.000349870.00019257
36.11536.114
0.00914950.0049578
Example: Production Optimization
Repeating more iterations, the following values are obtained%1a %
2a %3a
The value of closely approaches the true value of
11336
81511
81439
3
2
1
.
.
.
x
x
x
11436
8.1511
81439
3
2
1
.
.
x
x
x
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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
jj
ijaa
i1
ii
n
ijj
ijii aa1
for all ‘i’ and for at least one ‘i’
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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.
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?
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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
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
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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
100.00240.6180.23621.5464.53910.74307
4.90003.71533.16443.02813.00343.0001
100.0031.88917.4084.49960.824990.10856
3.09233.81183.97083.99714.00014.0001
67.66218.8764.00420.657720.0743830.00101
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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 .
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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
21.000−196.15−1995.0−201492.0364×105
−2.0579×105
95.238110.71109.83109.90109.89109.89
0.8000014.421−116.021204.6−121401.2272×105
100.0094.453112.43109.63109.92109.89
50.680−462.304718.1−476364.8144×105
−4.8653×106
98.027110.96109.80109.90109.89109.89
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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.
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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
9432 321 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?
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