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1/9/2010 http://numericalmethods.eng.usf.edu 1
Gauss-Siedel Method
Chemical Engineering Majors
Authors: Autar Kaw
http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM
Undergraduates
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
http://numericalmethods.eng.usf.edu
Gauss-Seidel MethodAlgorithm
Rewriting each equation
11
131321211 a
xaxaxacx nn−−−=
nn
nnnnnnn
nn
nnnnnnnnnn
nn
axaxaxac
x
axaxaxaxac
x
axaxaxacx
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
xx
xx
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 xxx
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: Liquid-Liquid ExtractionA liquid-liquid extraction process conducted in the ElectrochemicalMaterials Laboratory involved the extraction of nickel from the aqueousphase into an organic phase. A typical experimental data from thelaboratory is given below:
Ni aqueous phase, a (g/l) 2 2.5 3
Ni organic phase, g (g/l) 8.57 10 12
Assuming g is the amount of Ni in organic phase and a is the amount of Ni in the aqueous phase, the quadratic interpolant that estimates g is given by
5.32,322
1 ≤≤++= axaxaxg
Example: Liquid-Liquid ExtractionThe solution for the unknowns x1, x2, and x3 is given by
=
121057.8
13915.225.6124
3
2
1
xxx
Find the values of x1, x2,and x3 using the Gauss Seidel method. Estimate the amount of nickel in organic phase when 2.3 g/l is in
the aqueous phase using quadratic interpolation.
Initial Guess: Conduct two iterations.
=
111
3
2
1
xxx
Example: Liquid-Liquid Extraction
Rewriting each equation4257.8 32
1xxx −−
=
5.225.610 31
2xxx −−
=
13912 21
3xxx −−
=
=
121057.8
13915.225.6124
3
2
1
xxx
Example: Liquid-Liquid ExtractionIteration 1Applying the initial guess and solving for each xi
Initial Guess
When solving for x2, how many of the initial guess values were used?
3925.14
11257.81 =
−×−=x
11875.05.2
13925.125.6102 =
−×−=x
88875.01
11875.033925.19123 −=
×−×−=x
=
111
3
2
1
xxx
Example: Liquid-Liquid ExtractionFinding the absolute relative approximate error for Iteration 1.
100×−
=∈ newi
oldi
newi
ia xxx At the end of the Iteration 1
The maximum absolute relative approximate error is
742.11%.
%..
.a 18728100
39251139251
1=×
−=∈
%..
.a 11742100
1187501118750
2=×
−=∈
%..
.a 52212100
8887501888750
3=×
−−−
=∈
−=
88875.011875.03925.1
3
2
1
xxx
Example: Liquid-Liquid ExtractionIteration 2
Using from Iteration 1
the values of xi are found.
−=
88875.011875.03925.1
3
2
1
xxx
5245.41
)4078.1(33053.2912x
4078.15.2
)88875.0(3053.225.610x
3053.24
)88875.0(11875.0257.8x
3
2
1
−=−×−×−
=
−=−−×−
=
=−−×−
=
Example: Liquid-Liquid ExtractionFinding the absolute relative approximate error for Iteration 2.
At the end of the Iteration 1
The maximum absolute relative approximate error is
108.44%
%357.80
1005245.4
)88875.0(5245.4%44.108
1004078.1
11875.040778.1%596.39
1003053.2
3925.13053.2
3
2
1
=
×−
−−−=∈
=
×−
−−=∈
=
×−
=∈
a
a
a
−−=
5245.44078.1
3053.2
xxx
3
2
1
Iteration x1 x2 x3
123456
1.39252.30533.97757.058412.75223.291
28.186739.596042.04143.64944.64945.249
0.11875−1.4078−4.1340−9.0877−18.175−34.930
742.1053108.435365.94654.51049.99947.967
−0.88875−4.5245−11.396−24.262−48.243−92.827
212.5280.35760.29653.03249.70848.030
Example: Liquid-Liquid ExtractionRepeating more iterations, the following values are obtained
%1a∈ %
2a∈ %3a∈
Notice – The relative errors are not decreasing at any significant rate
Also, the solution is not converging to the true solution of
−=
55.827.2
14.1
xxx
3
2
1
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-Seidel 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
ijaai1
ii ∑≠=
⟩n
ijj
ijii aa1
for all ‘i’ and for at least one ‘i’
Gauss-Siedel Method: PitfallDiagonally 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.
Original (Non-Diagonally dominant) Rewritten (Diagonally dominant)
=
57.81012
12415.225.6139
3
2
1
xxx
=
121057.8
13915.225.6124
3
2
1
xxx
Example: Liquid-Liquid Extraction
Rewriting each equation
Iteration 1With an initial guess of
=
111
3
2
1
xxx
2589.21
3778.1288889.0457.8x
3778.15.2
188889.025.610x
88889.09
11312x
3
2
1
=×−×−
=
=−×−
=
=−×−
=
Example: Liquid-Liquid ExtractionThe absolute relative approximate error for Iteration 1 is
The maximum absolute relative error after the first iteration is 55.730%
%730.551002589.2
12589.2
%419.271003778.1
13778.1
%5.1210088889.0
188889.0
3
2
1
=×−
=∈
=×−
=∈
=×−
=∈
a
a
a At the end of Iteration 1
=
2589.23778.18889.0
3
2
1
xxx
Example: Liquid-Liquid ExtractionIteration 2
Using from Iteration 1
the values for xi are found
=
2589.23778.18889.0
3
2
1
xxx
0002.31
5387.1262309.0457.8x
5387.15.2
2589.2162309.025.610x
62309.09
2589.213778.1312x
3
2
1
=×−×−
=
=×−×−
=
=×−×−
=
Example: Liquid-Liquid ExtractionThe absolute relative approximate error for Iteration 2
The maximum absolute relative error after the first iteration is 42.659%
%709.241000002.3
2589.20002.3
%460.101005387.1
3778.15387.1
%659.4210062309.0
88889.062309.0
3
2
1
=×−
=∈
=×−
=∈
=×−
=∈
a
a
a At the end of Iteration 2
=
0002.35387.162309.0
xxx
3
2
1
Iteration x1 x2 x3
123456
0.888890.623090.487070.421780.394940.38890
12.542.65927.92615.4796.79601.5521
1.37781.53871.58221.56271.50961.4393
27.41910.4562.75061.25373.51314.8828
2.25893.00023.45723.75763.97104.1357
55.73024.70913.2207.99285.37473.9826
Example: Liquid-Liquid ExtractionRepeating more iterations, the following values are obtained
%1a∈ %
2a∈ %3a∈
After six iterations, the absolute relative approximate error seems to be decreasing. Conducting more iterations allows the absolute relative approximate error to decrease to an
acceptable level.
Iteration x1 x2 x3
199200
1.13351.1337
0.0144120.014056
−2.2389−2.2397
0.0348710.034005
8.51398.5148
0.0106660.010403
Example: Liquid-Liquid ExtractionRepeating more iterations, the following values are obtained
%1a∈ %
2a∈ %3a∈
−=
55.827.2
14.1
xxx
3
2
1
−=
5148.82397.2
1337.1
3
2
1
xxx
The value of closely approaches the true value of
Example: Liquid-Liquid Extraction
( ) ( ) ( )( ) ( )( ) 514882397213371 2
322
1
.a.a.
xaxaxag
+−+=
++=
( ) ( ) ( ) ( )g/l36089
514883223972321337132 2
.......g
=+×−+×=
Where g is grams of nickel in the organic phase and a is the grams/liter in the aqueous phase.
The polynomial that passes through the three data points is then
When 2.3g/l is in the aqueous phase, using quadratic interpolation, the estimated the amount of nickel in the organic phase
http://numericalmethods.eng.usf.edu
Gauss-Seidel Method: Example 3Given the system of equations
761373 321 =++ xxx
2835 321 =++ xxx15312 321 =−+ xxx
With an initial guess of
=
101
3
2
1
xxx
Rewriting the equations
313776 32
1xxx −−
=
5328 31
2xxx −−
=
53121 21
3 −−−
=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
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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 [ ]
−=
5312351
1373A
But this same set of equations will converge. [ ]
−=
13733515312
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