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7/23/2011 http://numericalmethods.eng.usf.edu 1

Gauss-Siedel Method

Major: All Engineering Majors

Authors: Autar Kaw

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

Undergraduates

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Gauss-Seidel Method

http://numericalmethods.eng.usf.edu

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Gauss-Seidel Method An iterative method.

Basic Pr ocedur e:

- Algebr aically solve each linear equation f or xi

- Assume an initial guess solution arr ay

-Solve f or each xi and r epeat

-Use absolute r elative appr oximate err or af ter each iter ationto check if err or is within a pr e-specif ied toler ance.

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Gauss-Seidel MethodWhy?

The Gauss-Seidel Method allows the user to contr ol r ound-off err or .

Elimination methods such as Gaussian Elimination and LU

Decomposition ar e pr one to pr one to r ound-off err or .

Also: If the physics of the pr oblem ar e under stood, a close initialguess can be made, decr easing the number of iter ations needed.

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Gauss-Seidel Method Algor ithm

A set of n equations and n unknowns:

11313212111 ... b xa xa xa xa nn !

2323222121 ... b xa xa xa xa n2n !

nnnnnnn b xa xa xa xa ! ...332211

. .

. .

. .

If : the diagonal elements ar e

non-zer o

Rewr ite each equation solving

f or the corr esponding unknown

ex:

Fir st equation, solve f or x1

Second equation, solve f or x2

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Rewr iting each equation

11

13132121

1

a

xa xa xac x nn

!

--

nn

nnnnnn

n

nn

nnnnnnnnnn

nn

a

xa xa xac x

a xa xa xa xac x

a

xa xa xac x

11,2211

1,1

,122,122,111,111

22

232312122

!

!

!

--

--

///

--

Fr om Equation 1

Fr om equation 2

Fr om equation n-1

Fr om equation n

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Gener al For m of each equation

11

11

11

1a

xac

x

n

j j

j j

§{!

!

22

21

22

2a

xac

x

j

n

j j

j

§{!

!

1,1

11

,11

1

{!

§

!

nn

n

n j j

j jnn

na

xac

x

nn

n

n j j jnjn

na

xac

x§{!

!

1

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Gener al For m f or any r ow µi¶

.,,2,1,

1

nia

xac

xii

n

i j j

jiji

i -!

!

§{!

How or wher e can this equation be used?

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Gauss-Seidel MethodCalculate the Absolute Relative Appr oximate Err or

100v

!� new

i

old

i

new

i

ia

x

x x

So when has the answer been f ound?

The iter ations ar e stopped when the absolute r elative

appr oximate err or is less than a pr especif ied toler ance f or all

unknowns.

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Gauss-Seidel Method: Example 1The upwar d velocity of a r ocket

is given at thr ee diff er ent times

Time, Velocity

5 106.8

8 177.2

12 279.2

The velocity data is appr oximated by a polynomial as:

12.t5,32

2

1 ee! at at at v

s t m/s v

Table 1 Velocity vs. Time data.

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Gauss-Seidel Method: Example 1

¼

¼¼

½

»

¬

¬¬«

!

¼

¼¼

½

»

¬

¬¬«

¼

¼¼

½

»

¬

¬¬«

3

2

1

323

2

2

2

1

2

1

1

1

1

v

v

v

a

a

a

t t

t t

t t

3

2

1Using a Matr ix template of the f or m

The system of equations becomes

¼¼¼

½

»

¬¬¬«

!

¼¼¼

½

»

¬¬¬«

¼¼¼

½

»

¬¬¬«

2.279

2.177

8.106

112144

1864

1525

3

2

1

a

a

a

Initial Guess: Assume an initial guess of

¼¼¼

½

»

¬¬¬«

!

¼¼¼

½

»

¬¬¬«

5

2

1

3

2

1

a

a

a

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¼¼¼

½

»

¬¬¬«

!

¼¼¼

½

»

¬¬¬«

¼¼¼

½

»

¬¬¬«

2.279

2.177

8.106

112144

1864

1525

3

2

1

a

a

a

25

58.106 32

1

aaa

!

8

642.177 31

2

aaa

!

1

121442.279 21

3

aaa

!

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Gauss-Seidel Method: Example 1 Applying the initial guess and solving f or ai

¼¼

¼

½

»

¬¬

¬«

!

¼¼

¼

½

»

¬¬

¬«

52

1

3

2

1

aa

a 6720.325

)5()2(58.106a1 !

!

8510.7

8

56720.3642.177a 2 !

!

36.155

1

8510.7126720.31442.279a

3!

!

Initial Guess

When solving f or a2, how many of the initial guess values wer e used?

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%76.721006720.3

0000.16720.31a !

!� x

%47.1251008510.7

0000.28510.72a !

!� x

%22.10310036.155

0000.536.1553a !

!� x

Finding the absolute r elative appr oximate err or

100v

!�new

i

old

i

new

i

ia

x

x x At the end of the f ir st iter ation

The maximum absolute

r elative appr oximate err or is 125.47%

¼¼¼

½

»

¬¬¬«

!

¼¼¼

½

»

¬¬¬«

36.155

8510.7

6720.3

3

2

1

a

a

a

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Gauss-Seidel Method: Example 1Iter ation #2

Using

¼¼

¼

½

»

¬¬

¬«

!

¼¼

¼

½

»

¬¬

¬«

36.1558510.7

6720.3

3

2

1

aa

a

056.1225

36.1558510.758.1061 !

!a

882.54

8

36.155056.12642.1772 !

!a

34.798

1

882.5412056.121442.2793 !

!a

fr om iter ation #1

the values of ai ar e f ound:

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Gauss-Seidel Method: Example 1Finding the absolute r elative appr oximate err or

%543.69100056.12

6720.3056.121a !

!� x

%695.85100x

882.54

8510.7882.542

!

!�a

%540.8010034.798

36.15534.7983a !

!� x

At the end of the second iter ation

¼¼¼

½

»

¬¬¬«

!

¼¼¼

½

»

¬¬¬«

54.798

882.54

056.12

3

2

1

a

a

a

The maximum absolute

r elative appr oximate err or is

85.695%

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Iteration a 1 a 2 a 3

12

3

4

5

6

3.672012.056

47.182

193.33

800.53

3322.6

72.76769.543

74.447

75.595

75.850

75.906

í7.8510í54.882

í255.51

í1093.4

í4577.2

í19049

125.4785.695

78.521

76.632

76.112

75.972

í155.36í798.34

í3448.9

í14440

í60072

í249580

103.2280.540

76.852

76.116

75.963

75.931

http://numerica lmethods.eng.usf.edu

Ga uss-Seidel Method: Exa mple 1

¼¼¼

½

»

¬¬¬«

!

¼¼¼

½

»

¬¬¬«

0857.1

690.19

29048.0

a

a

a

3

2

1

Repeating mor e iter ations, the f ollowing values ar e obtained

%1a

� %2a

� %3a�

Notice ± The r elative err or s ar e not decr easing at any signif icant r ate

Also, the solution is not converging to the tr ue solution of

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Gauss-Seidel Method: Example 2Given the system of equations

15312 321 x- x x !

2835 321

x x x !

761373 321 ! x x x

¼¼¼

½

»

¬¬¬«

!

¼¼¼

½

»

¬¬¬«

1

01

3

2

1

x

x x

With an initial guess of

The coeff icient matr ix is:

? A

¼¼

¼

½

»

¬¬

¬«

!

1373

351

5312

A

Will the solution converge using the

Gauss-Siedel method?

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? A

¼¼¼

½

»

¬¬¬

«

!

1373

351

5312

A

Checking if the coeff icient matr ix is diagonally dominant

43155 232122 !!u!! aaa

10731313 323133 !!u!! aaa

8531212 131211 !!u!! aaa

The inequalities ar e all tr ue and at least one r ow is strictly gr eater than:

Ther ef or e: The solution should converge using the Gauss-Siedel Method

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¼¼¼

½

»

¬¬¬«

!

¼¼¼

½

»

¬¬¬«

¼¼¼

½

»

¬¬¬«

76

28

1

1373

351

5312

3

2

1

a

a

a


12

531 32

1

x x x

!

5328 31

2 x x x

!

13

7376 21

3

x x x

!


¼¼¼

½

»

¬¬¬«

!

¼¼¼

½

»

¬¬¬«

1

0

1

3

2

1

x

x

x

50000.0

12

150311 !

! x

9000.45

135.0282 !! x

0923.3

13

9000.4750000.03763 !

! x

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Gauss-Seidel Method: Example 2The absolute r elative appr oximate err or

%00.10010050000.0

0000.150000.01

!v

!�a

%00.1001009000.4

09000.42a !v

!�

%662.671000923.3

0000.10923.33a !v

!�

The maximum absolute r elative err or af ter the f ir st iter ation is 100%

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¼

¼¼

½

»

¬

¬¬«

!

¼

¼¼

½

»

¬

¬¬«

8118.3

7153.3

14679.0

3

2

1

x

x

x

Af ter Iter ation #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 Af ter Iter ation #2

¼

¼¼

½

»

¬

¬¬«

!

¼

¼¼

½

»

¬

¬¬«

0923.3

9000.4

5000.0

3

2

1

x

x

x

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Gauss-Seidel Method: Example 2Iter ation #2 absolute r elative appr oximate err or

%61.24010014679.0

50000.014679.01a !v

!�

%889.311007153.3

9000.47153.32a !v

!�

%874.18100

8118.3

0923.38118.33a !v

!�

The maximum absolute r elative err or af ter the f ir st iter ation is 240.61%

This is much larger than the maximum absolute r elative err or obtained in

iter ation #1. Is this a pr oblem?

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Iteration a1 a2 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

67.662

18.876

4.0042

0.65772

0.074383

0.00101


Gauss-Seidel Method: Example 2Repeating mor e iter ations, the f ollowing values ar e obtained

%1a�

%2a� %

3a�

¼¼¼

½

»

¬¬¬«

!

¼¼¼

½

»

¬¬¬«

4

3

1

3

2

1

x

x

x

¼¼¼

½

»

¬¬¬«

!

¼¼¼

½

»

¬¬¬«

0001.4

0001.3

99919.0

3

2

1

x

x

x

The solution obtained is close to the exact solution of .

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Gauss-Seidel Method: Example 3Given the system of equations

761373 321 ! x x x

2835 321 ! x x x

15312 321 ! x x x


¼¼¼

½

»

¬¬¬«

!

¼¼¼

½

»

¬¬¬«

1

0

1

3

2

1

x

x

x

Rewr iting the equations

3

13776 321

x x x

!

5

328 312

x x x

!

5

3121 21

3

!

x x x

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Iteration a1 A2 a3

1

2

3

4

5

6

21.000

í196.15

í1995.0

í20149

2.0364×105

í2.0579×105

95.238

110.71

109.83

109.90

109.89

109.89

0.80000

14.421

í116.02

1204.6

í12140

1.2272×105

100.00

94.453

112.43

109.63

109.92

109.89

50.680

í462.30

4718.1

í47636

4.8144×105

í4.8653×106

98.027

110.96

109.80

109.90

109.89

109.89


Gauss-Seidel Method: Example 3Conducting six iter ations, the f ollowing values ar e obtained

%1a

� %2a

� %3a

�

The values ar e 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 coeff icient matr ix is not

diagonally dominant? A

¼¼

¼

½

»

¬¬

¬«

!

5312351

1373

A

But this is the same set of

equations used in example #2,

which did converge.? A

¼¼¼

½

»

¬¬¬«

!

1373

351

5312

A

If a system of linear equations is not diagonally dominant, check to see if

r earr anging the equations can f or m a diagonally dominant matr ix.

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Gauss-Seidel MethodNot every system of equations can be r earr anged to have a

diagonally dominant coeff icient matr ix.

Observe the set of equations

3321 ! x x x

9432 321 ! x x x

97321

! x x x

Which equation(s) pr events this set of equation fr om having a

diagonally dominant coeff icient matr ix?

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Gauss-Seidel MethodSummary

- Advantages of the Gauss-Seidel Method- Algor ithm f or the Gauss-Seidel Method

-Pitf alls of the Gauss-Seidel Method

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Gauss-Seidel Method

Questions?

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Additional ResourcesFor all resources on this topic such as digital audiovisuallectures, primers, textbook chapters, multiple-choicetests, worksheets in MATLAB, MATHEMATICA, MathCadand MAPLE, blogs, related physical problems, pleasevisit

http://numericalmethods.eng.usf.edu/topics/gauss_seid

el.html

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