jacobi and gauss-seidel iteration methods, use of software packages

7/29/2019 Jacobi and Gauss-Seidel Iteration Methods, Use of Software Packages

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Outlines

Jacobi and Gauss-Seidel Iteration Methods, Useof Software Packages

Mike Renfro

February 20, 2008

Mike Renfro Jacobi and Gauss-Seidel Iteration Methods, Use of Software P


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OutlinesPart I: Review of Previous LecturePart II: Jacobi and Gauss-Seidel Iteration Methods, Use of Softw

Review of Previous Lecture



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OutlinesPart I: Review of Previous LecturePart II: Jacobi and Gauss-Seidel Iteration Methods, Use of Softw

Jacobi and Gauss-Seidel Iteration Methods, Use ofSoftware Packages

Jacobi Iteration MethodIntroduction

ExampleNotes on Convergence Criteria

Gauss-Seidel Iteration MethodIntroductionExample

Use of Software PackagesMATLABExcel



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Part I




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Cramers Rule

Gauss Elimination


J bi It ti M th d


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Jacobi Iteration MethodGauss-Seidel Iteration Method

Use of Software Packages

Part II

Jacobi and Gauss-Seidel Iteration Method, Use ofSoftware Packages


Jacobi Iteration Method Introduction


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IntroductionExampleNotes on Convergence Criteria

Jacobi Iteration Method: Introduction

Consider a system of equations in algebraic form

a11x1 + a12x2 + a13x3 + a14x4 + + a1nxn =b1

a21x1 + a22x2 + a23x3 + a24x4 + + a2n

xn

=b2a31x1 + a32x2 + a33x3 + a34x4 + + a3nxn =b3

a41x1 + a42x2 + a43x3 + a44x4 + + a4nxn =b4...

an1x1 + an2x2 + an3x3 + an4x4 + + annxn =bn




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Jacobi Iteration Method: Introduction

These n equations can be rewritten to isolate an unknown on oneside of each equation:

x1 =1

a11(b1 a12x2 a13x3 a14x4 a1nxn)

x2 =

1

a22 (b2 a21x1 a23x3 a24x4 a2nxn)

x3 =1

a33(b3 a31x1 a32x2 a34x4 a3nxn)

x4 =1

a44

(b4 a41x1 a42x2 a43x3 a4nxn)

...

xn =1

ann(bn an1x1 an2x2 an3x3 an,n1xn1)




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So Whats the Point?

It looks like all weve done at this point is some useless algebraicmanipulations. But if we substitute some assumed starting valuesfor x1, x2, , xn on the right hand side of each of the rewrittenequations, well get a new set of xi values on the left hand side. Ifwe repeat the process, substituting the just-calculated x valuesinto the right hand side of the equations, we should get closer and

closer to the actual values of x that solve the equations.




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Example

Example 3.19 (p.184) Solve the following system of equationsusing the Jacobi iteration method with an initial guess of xi = 0:

5x1 x2 + 2x3 =1

2x1 + 6x2 3x3 =2

2x1 + x2 + 7x3 =32




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Example

Step 1: reformat the equations, solving the first one for x1, thesecond for x2, and the third for x3:

x1 = 15

(1 + x2 2x3)

x2 =1

6(2 2x1 + 3x3)

x3 =

1

7 (32 2x1 x2)




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Gauss-Seidel Iteration MethodUse of Software Packages


Example

Step 2: Substitute the initial guesses for xi into the right-hand sideof the equations:

x1 = 1

5(1 + 0 2 (0))

x2 = 16

(2 2 (0) + 3 (0))

x3 =1

7(32 2(0) (0))

x(2)1 = 0.2000

x(2)2 =0.3333

x(2)3 =4.5714




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Example

Step 3: Substitute the calculated values for for xi into theright-hand side of the equations:

x1 = 1

5(1 + 0.3333 2 (4.5714))

x2 = 16

(2 2 (0.2000) + 3 (4.5714))

x3 =1

7(32 2 (0.2000) (0.3333))

x(3)1 =1.5619

x(3)2 =2.6857

x(3)3 =4.5810




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Example

Step 4, 5, : Continue substituting xi values into the right-handside of the equations and watch for them to converge to finalvalues:

Iteration number x1 x2 x31 0.0000 0.0000 0.00002 -0.2000 0.3333 4.57143 1.5619 2.6857 4.58104 1.0952 2.1032 3.74155 0.8760 1.8390 3.9580

6 1.0154 2.0204 4.05847 1.0193 2.0241 3.99278 0.9923 1.9899 3.99109 0.9984 1.9981 4.0037

10 1.0018 2.0023 4.0007


Jacobi Iteration MethodG S id l I i M h d

IntroductionE l


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Notes on Convergence Criteria

Like many of the iterative root-finding methods in Chapter 2, theJacobi iteration method is not guaranteed to converge on the exactanswer in every possible case and every possible initial guess.

However, if the equations in the system are diagonally dominant,then the Jacobi iteration method is guaranteed to convergeregardless of the starting guess for x.Diagonal dominance is defined as the condition where the

coefficient along the diagonal on any row is larger in absolute valuethan the sum of the absolute values of the other coefficients on thesame row.


Jacobi Iteration MethodG S id l It ti M th d

IntroductionE l


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Is Example 3.19 Diagonally Dominant?

5x1 x2 + 2x3 =1

2x1 + 6x2 3x3 =2

2x1 + x2 + 7x3 =32


Jacobi Iteration MethodGauss Seidel Iteration Method

IntroductionExample


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Is Example 3.19 Diagonally Dominant?

5 1 22 6 32 1 7

x1x2x3

=

12

32

| 5| > | 1| + |2|

|6| > |2| + | 3|

|7| > |2| + |1|

Yes, this system of equations is diagonally dominant. We willconverge to the exact solution regardless of the values of xi westart with.


Jacobi Iteration MethodGauss Seidel Iteration Method

Introduction


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Example

Gauss-Seidel Iteration: Introduction

Gauss-Seidel iteration is similar to Jacobi iteration, except thatnew values for xi are used on the right-hand side of the equationsas soon as they become available. It improves upon the Jacobimethod in two respects:

Convergence is quicker, since you benefit from the newer,more accurate xi values earlier.

Memory requirements are reduced by 50%, since you onlyneed to keep track of one set of xi values, not two sets.



Introduction


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Example

Example

Example 3.20 (p.186) Solve the following system of equationsusing the Gauss-Seidel iteration method with an initial guess of

xi = 0:

5x1 x2 + 2x3 =1

2x1 + 6x2 3x3 =2

2x1 + x2 + 7x3 =32



IntroductionE l


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Gauss Seidel Iteration MethodUse of Software Packages

Example

Example

Step 1: reformat the equations, solving the first one for x1, thesecond for x2, and the third for x3:

x1 = 15

(1 + x2 2x3)

x2 =1

6(2 2x1 + 3x3)

x3 =

1

7 (32 2x1 x2)



IntroductionE l


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Use of Software PackagesExample

Example

Step 2a: Substitute the initial guesses for xi into the right-handside of the first equation:

x(2)1 =

1

5(1 + 0 2(0)) = 0.2000

Step 2b: Substitute the new x1 value with the initial guess for x3into the second equation:

x(2)2 =

1

6(2 2 (0.2000) + 3 (0)) = 0.4000

Step 2c: Substitute the new x1 and x2 values into the thirdequation:

x(2)3 =

1

7(32 2 (0.2000) (0.4000)) = 4.5714



IntroductionExample


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Use of Software PackagesExample

Example

Step 3, 4, : Repeat step 2 and watch for the xi values toconverge to an exact solution.

Iteration number x1 x2 x3

1 0.0000 0.0000 0.00002 -0.2000 0.4000 4.57143 1.5486 2.1029 3.82864 0.9109 1.9440 4.03355 1.0246 2.0085 3.9918

6 0.9950 1.9975 4.00187 1.0012 2.0005 3.99968 0.9997 1.9999 4.0001



MATLABExcel


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Use of Software PackagesExcel

Example 3.27

Find the solution to the following system of equations:

4 1 1 01 4 0 11 0 4 10 1 1 4

x1x2x3x4

=

200400

0200



MATLABExcel


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MATLAB Setup

>> A =[ -4 1 1 -0; 1 -4 0 1; 1 0 -4 1; 0 1 1 -4]

A =

-4 1 1 0

1 -4 0 1

1 0 -4 1

0 1 1 -4

> > b = [ -2 00; -400; 0; -200]

b =

-200

-4000

-200



U f S f P k

MATLABExcel


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2 MATLAB Solutions

> > x = in v (A )* b

x =

1 0 0 . 0 0 0 0

1 5 0 . 0 0 0 0

5 0 . 0 0 0 01 0 0 . 0 0 0 0

> > x =A \b

x =

1 0 0 . 0 0 0 0

1 5 0 . 0 0 0 05 0 . 0 0 0 0

1 0 0 . 0 0 0 0



U f S ft P k

MATLABExcel


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Example 3.29 (Corrected)

Use MATLAB to find the inverse of the matrix [A] given by

[A] =

1 1213

1n

1

2

1

3

1

4 1

n+113

14

15

1n+2

......

......

1n

1n+1

1n+2

12n1

, n = 50

and find the error in the inverse matrix by calculating[C] = [A][A]1 and summing up the absolute values of the matrixelements. If there was no error at all, the sum would be n.




MATLABExcel


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MATLAB Solution 1

> > A = h i lb ( 50 );

> > r es u lt = i nv ( A ) * A ;

W ar ni ng : M at ri x is c lo se to s in gu la r or b ad ly s ca le d .R es ul ts m ay be i na cc ur at e . R CO ND = 1 .6 03 36 6 e - 02 0.

> > e rr = s u m ( a bs ( r e su lt ( :) ))

err =

7 . 3 8 8 6 e + 0 1 1




MATLABExcel


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MATLAB Solution 2

> > A = h i lb ( 50 );

> > r es u lt = A \ A ;

W ar ni ng : M at ri x is c lo se to s in gu la r or b ad ly s ca le d .

R es ul ts m ay be i na cc ur at e . R CO ND = 1 .6 03 36 6 e - 02 0.> > e rr = s u m ( a bs ( r e su lt ( :) ))

err =

1 9 2 . 9 8 5 6

>>

The error on this version is much, much lower. This is why wedont use the inv() function on ill-conditioned matrices.




MATLABExcel


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Excel

Microsoft Excel can also do matrix inversion, but its a bitcumbersome. However, if its the only tool you have at somepoint, its better than nothing. See the the matrix inversion.xlsfile located with the lecture slides on the web.


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