Relaxation Method 2012

7/31/2019 Relaxation Method 2012

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RELAXATION METHODS

Presenters : Saleem Abdool 11/0935/1430Michal Dhani 10/0937/1299Quincy Chester 10/0937/2244Kevin Tucker 10/0937/2516

Adeye Horatio 09/0935/1430


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Outline of presentationIntroductionTypes of Relaxation Methods

- Jacobi method

- Gauss-Seidel method

- Successive over relaxation(SOR)

Summary

Reference

Worksheet

Program solving relaxation Methods


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IntroductionWhat is Relaxation Method:

It is a method of solving simultaneous equations by guessing asolution and then reducing the errors that result by successiveapproximations until all the errors are less than some specifiedamount.

Why Relaxation Method:

Because relaxation methods may be applied to any system oflinear equations to interactively improve an approximation to theexact solution.

In principle, relaxation methods which are the basis of the Jacobi,Gauss-Seidel, Successive Over Relaxation methods may be appliedto any system of linear equations to interatively improve anapproximation to the exact solution.

One may solve these equation either (1) direct or (2) iterative

methods.


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IntroductionDirect Method

They are three such methods

Solution by determinants (Cramer rules)

Solution by inverse matrix and

Solution by successive elimination.

The first two methods are not practical in solving large systemsof equations. Even the third one may sometimes be toocomputer memory demanding that one needs to resort to the

iterative alternative.

Additionally, direct methods may leave round-off error problemsthat may result in solutions that are incorrect on computerssupporting or not supporting sufficient precision.


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Introduction

Iterative Method

Because of the limit in precision in mostcomputations representation of numbers, it isunlikely that one will have the exact solution evenwith the direct method.

Iterative methods do not produce exact solution,theoretically, in finite number of iteration.

However, given the imprecise nature of numberrepresentation on computers, iteration method mayhave some advantages over direct methods. Inreality, for large systems of equation, iterativemethods are the ones to choose


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Objectives

To identify the iteration methods

To show the application of iterative methodin engineering

To determine the convergence of eachmethod


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Why use Iterative Techniques? The method of solving simultaneous linearalgebraic equations using Gaussian Elimination andthe Gauss-Jordan Method. These techniques areknown as direct methods. Problems can arise from

round-off errors and zero on the diagonal.

One means of obtaining an approximate solution tothe equations is to use an educated guess.


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Iterative Methods

We will look at three iterativemethods:

Jacobi Method Gauss-Seidel Method

Successive over Relaxation (SOR)


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Jacobi methodIn numerical linear algebra, the Jacobi method is analgorithm for determining the solutions of a systemof linear equations with largest absolute values ineach row and column dominated by the diagonal

element. Each diagonal element is solved for, and anapproximate value plugged in. The process is theniterated until it converges. This algorithm is astripped-down version of the Jacobi transformation

method of matrix diagonalization. The method isnamed after German mathematician Carl GustavJakob Jacobi.


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Jacobi method

The technique solves for the entire set of

x values for each iteration.

The problem does not update the values

until an iteration is completed


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ExampleA linear system of the form with initial estimate is given by

We use the equation , described above, to estimate . First, we

rewrite the equation in a more convenient form , where

and . Note that where and are the strictly lower

and upper parts of . From the known values


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we determine as

Further, C is found as


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With T and C calculated, we estimate as :

The next iteration yields

This process is repeated until convergence (i.e., until is small). The solution after

25 iterations is


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The Gauss-Seidel Method

This is an iterative method used to solve alinear system of equations. It is named afterthe German mathematicians Carl FriedrichGauss and Philipp Ludwig von Seidel, and is

similar to the Jacobi method. Though it canbe applied to any matrix with non-zeroelements on the diagonals, Convergence issure if the matrix is diagonally dominant or

symmetrical and positive definite.


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How this method works:

It use absolute relative approximate errorafter each iteration to check if an error iswithin a prespecified tolerance byassuming an initial guess solution arraythen algebraically solving each linearequation for xi. After which the iterationmethod is repeated to check if the error iswithin the prespecified tolerance.


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Gauss-Seidel Method

http://numericalmethods.eng.usf.edu

AlgorithmRewriting each equation

11

13132121

1a

xaxaxacx nn

nn

nnnnnn

n

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 Method


AlgorithmGeneral Form of each equation

11

11

11

1a

xac

x

n

jj

jj

22

21

22

2a

xac

x

j

n

jj

j

1,1

11

,11

1

nn

n

njj

jjnn

na

xac

x

nn

n

njj

jnjn

na

xac

x

1


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Gauss-Seidel Method

n

-n

2

x

x

x

x

1

1


Solve for the unknowns

Assume an initial guess for [X] Use rewritten equations to solve foreach value of xi.

Important: Remember to use themost recent value of xi. Whichmeans to apply values calculated tothe calculations remaining in thecurrent iteration.


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Gauss-Seidel Method


Calculate the Absolute Relative Approximate Error

100

newi

old

i

new

i

ia x

xx

So when has the answer been found?

The iterations are stopped when the absolute relativeapproximate error is less than a prespecified tolerance for allunknowns.


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Gauss-Seidel Method: Example 1

Time, Velocity

5 106.8

8 177.2

12 279.2


The upward velocity of a rocketis given at three different times

The velocity data is approximated by a polynomial as:

12.t5,322

1 atatatv

st m/sv

Table 1 Velocity vs. Time data.


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3

2

1

3

2

3

2

2

2

1

2

1

1

1

1

v

v

v

a

a

a

tt

tt

tt

3

2

1

2.279

2.177

8.106

112144

1864

1525

3

2

1

a

a

a


Using a Matrix template of the form

The system of equations becomes

Initial Guess: Assume an initial guess of

5

2

1

3

2

1

a

a

a


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2.279

2.177

8.106

112144

1864

1525

3

2

1

a

a

a


Rewriting each equation

25

58.106 321

aaa

8

642.177 312

aaa

1

121442.279 213

aaa


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Applying the initial guess and solving for ai

5

2

1

3

2

1

a

a

a 6720.325

)5()2(58.106a1

8510.7

8

56720.3642.177a 2

36.155

1

8510.7126720.31442.279a3

Initial Guess

When solving for a2, how many of the initial guess values were used?


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36.155

8510.76720.3

3

2

1

a

a

a


100

new

i

old

i

new

i

ia x

xx

%76.721006720.3

0000.16720.31a

x

%47.1251008510.7

0000.28510.72a

x

%22.10310036.155

0000.536.1553a

x

Finding the absolute relative approximate error

At the end of the first iteration

The maximum absolute

relative approximate error is125.47%


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36.155

8510.7

6720.3

3

2

1

a

a

a


Iteration #2Using

056.1225

36.1558510.758.1061

a

882.54

8

36.155056.12642.1772

a

34.798

1

882.5412056.121442.2793

a

from iteration #1

the values of ai are found:


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Finding the absolute relative approximate error

%543.69100056.12

6720.3056.121a

x

%695.85100x

882.54

8510.7882.542

a

%540.8010034.79836.15534.798

3a

x

At the end of the second iteration

54.798

882.54

056.12

3

2

1

a

a

a

The maximum absoluterelative approximate error is

85.695%


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Iteration a1 a2 a3

1

23

4

5

6

3.6720

12.05647.182

193.33

800.53

3322.6

72.767

69.54374.447

75.595

75.850

75.906

7.8510

54.882

255.51

1093.4

4577.2

19049

125.47

85.69578.521

76.632

76.112

75.972

155.36

798.34

3448.9

14440

60072

249580

103.22

80.54076.852

76.116

75.963

75.931


0857.1

690.19

29048.0

a

a

a

3

2

1


Repeating 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


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Gauss-Seidel Method: Pitfall


Even though done correctly, the answer is not converging to thecorrect answer

This example illustrates a pitfall of the Gauss-Siedel method: not allsystems of equations will converge.

One class of system of equations always converges: One with a diagonallydominantcoefficient matrix.

Diagonally dominant: [A] in [A] [X] = [C] is diagonally dominant if:

n

j

j

ijaa

i

1

ii

n

ij

j

ijii aa1

for all i and for at least one i


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Given the system of equations

15312 321 x-xx

2835 321 xxx

761373 321 xxx

1

0

1

3

2

1

x

x

x

With an initial guess of

The coefficient matrix is:

1373

351

5312

A


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76

28

1

1373

351

5312

3

2

1

a

a

a

1

0

1

3

2

1

x

x

x


Rewriting each equation

12

531 321

xxx

5328 312 xxx

13

7376 213

xxx

With an initial guess of

50000.0

12

150311

x

9000.45 135.0282

x

0923.3

13

9000.4750000.03763

x


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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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8118.3

7153.3

14679.0

3

2

1

x

x

x

0923.3

9000.4

5000.0

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 theequations

After Iteration #2


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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%


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Successive Over relaxation Method

This method is based on to the Gauss-SeidelMethod and is specially formulated to give

you a more accurate solution in lessrepetition.

This is done with the use of a relaxationfactor().

http://numericalmethods.eng.usf.edu38


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This method is based on the general Formula

(+)

()

()

Where

()is the residual of

Hence from a rearranged for of the Gauss-

Seidel Method which is

(+)

()

()

()+

Where we can see the Residual bracketed



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The SOR formula then becomes:

(+)

()

()

()

+



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Example

Solve for values of x in the following system where:

2 1

2 3 0

23 1 Which becomes the matrix form

2 1 0

1 2 1

0 1 2

3

10

1

Where for

(+)

==



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Example contd Rewriting the actual solution

2 0 1 0 0 0 1

Using the equation:

(+) () ()

Taking to be 1 in this case



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Example contd (+) 0 1

And it continues as shown below


K 1 0.5 0.25 0.6252 0.625 0.625 0.8125

10 0.9985 0.9985 0.9926


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Example contd At different values of can see checking a value

each variable after 5 cycles (k=5)


3

0.5 0.651 0.545 0.704

1.0 0.953 0.953 0.976

1.99 0.995 0.995 1.94


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The key to successfully using this method is obtaining anappropriate value for each time.

This allows convergence to be faster than the previousmethods mentioned

For 0 < < 2 the system will converge.

But at < < 1 convergence is slower than the Gauss-

Seidel Method. While at 1 the system becomes Gauss-Seidel

Method.

And for 1 < < 2 the system converges fastest and in

the least steps. This does not mean that for the highest possible value of

we get the fastest convergence but an appropriatevalue of can be obtained by trial and error.



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The End

Relaxation Method 2012

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