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Gauss-Siedel Method

Civil Engineering Majors

Authors: Autar Kaw

http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM

Undergraduates

Gauss-Seidel Method

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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.

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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

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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.

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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

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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

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.

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

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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

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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

100.00240.6180.23621.5464.5391

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

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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

21.000−196.15−1995.0−20149

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 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

THE END

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