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Iterative Solution Methods

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Iterative Solution Methods. Starts with an initial approximation for the solution vector (x 0 ) At each iteration updates the x vector by using the sytem Ax=b During the iterations A, matrix is not changed so sparcity is preserved Each iteration involves a matrix-vector product - PowerPoint PPT Presentation
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1 Iterative Solution Methods Starts with an initial approximation for the solution vector (x 0 ) At each iteration updates the x vector by using the sytem Ax=b During the iterations A, matrix is not changed so sparcity is preserved Each iteration involves a matrix-vector product If A is sparse this product is efficiently done
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Page 1: Iterative Solution Methods

1

Iterative Solution Methods Starts with an initial approximation for the

solution vector (x0)

At each iteration updates the x vector by using the sytem Ax=b

During the iterations A, matrix is not changed so sparcity is preserved

Each iteration involves a matrix-vector product

If A is sparse this product is efficiently done

Page 2: Iterative Solution Methods

2

Iterative solution procedure Write the system Ax=b in an equivalent formx=Ex+f (like x=g(x) for fixed-point iteration)

Starting with x0, generate a sequence of approximations {xk} iteratively byxk+1=Exk+f

Representation of E and f depends on the type of the method used

But for every method E and f are obtained from A and b, but in a different way

Page 3: Iterative Solution Methods

3

Convergence As k, the sequence {xk} converges to

the solution vector under some conditions on E matrix

This imposes different conditions on A matrix for different methods

For the same A matrix, one method may converge while the other may diverge

Therefore for each method the relation between A and E should be found to decide on the convergence

Page 4: Iterative Solution Methods

4

Different Iterative methods

Jacobi Iteration Gauss-Seidel Iteration Successive Over Relaxation (S.O.R)

SOR is a method used to accelerate the convergence

Gauss-Seidel Iteration is a special case of SOR method

Page 5: Iterative Solution Methods

5

Jacobi iteration

nnnnnn

nn

nn

bxaxaxa

bxaxaxa

bxaxaxa

2211

22222121

11212111

0

02

01

0

nx

x

x

x

)(1 0

102121

11

11 nnxaxaba

x

1

1 1

1 1 i

j

n

ij

kjij

kjiji

ii

ki xaxab

ax

)(1 0

11022

011

1 nnnnnn

nnn xaxaxaba

x

)(1 0

20323

01212

22

12 nnxaxaxaba

x

Page 6: Iterative Solution Methods

6

xk+1=Exk+f iteration for Jacobi method

A can be written as A=L+D+U (not decomposition)

000

00

0

00

00

00

0

00

000

23

1312

33

22

11

3231

21

333231

232221

131211

a

aa

a

a

a

aa

a

aaa

aaa

aaa

n

ij

kjij

i

j

kjiji

ii

ki xaxab

ax

1

1

1

1 1 xk+1=-D-1(L+U)xk+D-

1b

E=-D-1(L+U)

f=D-1b

Ax=b (L+D+U)x=b

Dxk+1 =-(L+U)xk+b

kk UxLxDxk+1

Page 7: Iterative Solution Methods

7

Gauss-Seidel (GS) iteration

nnnnnn

nn

nn

bxaxaxa

bxaxaxa

bxaxaxa

2211

22222121

11212111

0

02

01

0

nx

x

x

x

1

1 1

11 1 i

j

n

ij

kjij

kjiji

ii

ki xaxab

ax)(

1 01

02121

11

11 nnxaxaba

x

)(1 1

11122

111

1 nnnnnn

nnn xaxaxaba

x

)(1 0

20323

11212

22

12 nnxaxaxaba

x

Use the latestupdate

Page 8: Iterative Solution Methods

8

x(k+1)=Ex(k)+f iteration for Gauss-Seidel

1

1 1

11 1 i

j

n

ij

kjij

kjiji

ii

ki xaxab

ax

xk+1=-(D+L)-1Uxk+(D+L)-1b

E=-(D+L)-1U

f=-(D+L)-1b

Ax=b (L+D+U)x=b

(D+L)xk+1 =-Uxk+b

k1k UxLx Dxk+1

Page 9: Iterative Solution Methods

9

Comparison

Gauss-Seidel iteration converges more rapidly than the Jacobi iteration since it uses the latest updates

But there are some cases that Jacobi iteration does converge but Gauss-Seidel does not

To accelerate the Gauss-Seidel method even further, successive over relaxation method can be used

Page 10: Iterative Solution Methods

10

Successive Over Relaxation Method

GS iteration can be also written as follows

ki

ki

ki

i

j

n

ij

kjij

kjiji

ii

ki

ki

xx

xaxaba

xx

1

1

1

11 1

Correction term

0ix

1ix

2ix

3ix

0i

1i

2i

0i

1i

2i

1Multiply with

Faster convergence

Page 11: Iterative Solution Methods

11

SOR

1

1 1

11

1

1

11

1

1)1(

1

i

j

n

ij

kjij

kjiji

ii

ki

ki

i

j

n

ij

kjij

kjiji

ii

ki

ki

ki

ki

ki

xaxaba

xx

xaxaba

xx

xx

1<<2 over relaxation (faster convergence)0<<1 under relaxation (slower convergence)There is an optimum value for Find it by trial and error (usually around 1.6)

Page 12: Iterative Solution Methods

12

x(k+1)=Ex(k)+f iteration for SOR

1

1 1

11 1)1(

i

j

n

ij

kjij

kjiji

ii

ki

ki xaxab

axx

Dxk+1=(1-)Dxk+b-Lxk+1-Uxk

(D+ L)xk+1=[(1-)D-U]xk+b

E=(D+ L)-1[(1-)D-U]

f= (D+ L)-1b

Page 13: Iterative Solution Methods

13

The Conjugate Gradient Method

iiii

iTi

iTi

i

iiii

iTi

iTi

i

drd

rr

rr

Adxx

Add

rr

Axbrd

111

111

1

000

• Converges if A is a symmetric positive definite matrix

• Convergence is faster

Page 14: Iterative Solution Methods

14

Convergence of Iterative Methods

x̂Define the solution vector as

Define an error vector as ke

xex kk ˆSubstitute this into fExx kk 1

kkk EefxEfxeExe ˆ)ˆ(ˆ1

0)1(211 eEEEEeEEeEee kkkkk

Page 15: Iterative Solution Methods

15

Convergence of Iterative Methods

0)1(0)1(1 eEeEe kkk

Convergence condition

0Lim0Lim )1(1

k

k

k

kEe if

iteration power

The iterative method will converge for any initial iteration vector if the following condition is satisfied

Page 16: Iterative Solution Methods

16

Norm of a vector

A vector norm should satisfy these conditions

yxyx

αxαx

xx

xx

scalar for

vectorzero a is iff

vector nonzeroevery for

0

0

Vector norms can be defined in different forms as long as the norm definition satisfies these conditions

Page 17: Iterative Solution Methods

17

Commonly used vector norms

nxxxx 211

Sum norm or ℓ1 norm

Euclidean norm or ℓ2 norm

222

212 nxxxx

Maximum norm or ℓ norm

ii xx max

Page 18: Iterative Solution Methods

18

Norm of a matrix

A matrix norm should satisfy these conditions

BABA

αAαA

AA

A

scalar for

matrix zero a is iff

0

0

Important identitiy

vectora is xxAAx

Page 19: Iterative Solution Methods

19

Commonly used matrix norms

Maximum column-sum norm or ℓ1 norm

Spectral norm or ℓ2 norm

Maximum row-sum norm or ℓ norm

m

iij

njaA

111max

AAA T of eigenvalue maximum2

n

jij

miaA

11max

Page 20: Iterative Solution Methods

20

Example

Compute the ℓ1 and ℓ norms of the matrix

186

427

593 17

13

15

A

16 19 10

1A

Page 21: Iterative Solution Methods

21

Convergence condition

0lim0lim )1(1

k

k

k

kEe if

Express E in terms of modal matrix P and :Diagonal matrix with eigenvalues of E on the diagonal

1)1()1(

111)1(

1

PPE

PPPPPPE

PPE

kk

k

1

12

11

1

kn

k

k

k

,...,n,i

PPE

iki

k

k

k

k

k

k

k

21for 10lim

0lim0lim0lim

1

)1(1)1()1(

Page 22: Iterative Solution Methods

22

Sufficient condition for convergence

If the magnitude of all eigenvalues of iteration matrixE is less than 1 than the iteration is convergent

ExExxEEx

xEx

xEx

It is easier to compute the norm of a matrix than to compute its eigenvalues

1E is a sufficient condition for convergence

Page 23: Iterative Solution Methods

23

Convergence of Jacobi iteration

E=-D-1(L+U)

0

0

0

11

11

1

22

2

22

23

22

21

11

1

11

12

nn

nn

nn

n

nn

nn

n

n

a

a

a

aa

a

a

a

a

a

a

aa

a

a

a

E

Page 24: Iterative Solution Methods

24

Convergence of Jacobi iteration

n

jij

ijii

n

jij ii

ij

aa

a

aE

1

1

11

n1,2,...,ifor

Evaluate the infinity(maximum row sum) norm of E

Diagonally dominant matrix

If A is a diagonally dominant matrix, then Jacobi iteration converges for any initial vector

Page 25: Iterative Solution Methods

25

Stopping Criteria

Ax=b

At any iteration k, the residual term is

rk=b-Axk

Check the norm of the residual term

||b-Axk||

If it is less than a threshold value stop

Page 26: Iterative Solution Methods

26

Example 1 (Jacobi Iteration)

15

21

7

512

184

114

3

2

1

x

x

x

0

0

00x

5

215

8

421

4

7

02

011

3

03

011

2

03

021

1

xxx

xxx

xxx

0.35

15

625.28

21

75.14

7

7395.262

0 Axb

0452.102

1 Axb

Diagonally dominant matrix

Page 27: Iterative Solution Methods

27

Example 1 continued...

5

215

8

421

4

7

12

112

3

13

112

2

13

122

1

xxx

xxx

xxx

225.45

625.275.1215

875.38

375.1421

65625.14

3625.27

7413.6

2

2 Axb

8875.25

875.365625.1215

98125.38

225.465625.1421

6625.14

225.4875.37

33

32

31

x

x

x

9534.12

2 Axb

Matrix is diagonally dominant, Jacobi iterations are converging

Page 28: Iterative Solution Methods

28

Example 2

7

21

15

114

184

512

3

2

1

x

x

x

0

0

00x 7395.26

2

0 Axb

02

01

13

03

011

2

03

021

1

47

8

421

2

515

xxx

xxx

xxx

0.7

625.28

21

5.72

15

The matrix is not diagonally dominant

8546.542

1 Axb

Page 29: Iterative Solution Methods

29

Example 2 continued...

625.39625.25.747

25.08

75.7421

3125.112

75625.215

13

12

11

x

x

x

3761.2082

2 Axb

The residual term is increasing at each

iteration, so the iterations are diverging.

Note that the matrix is not diagonally

dominant

Page 30: Iterative Solution Methods

30

Convergence of Gauss-Seidel iteration

GS iteration converges for any initial vector if A is a diagonally dominant matrix

GS iteration converges for any initial vector if A is a symmetric and positive definite matrix

Matrix A is positive definite if

xTAx>0 for every nonzero x vector

Page 31: Iterative Solution Methods

31

Positive Definite Matrices

A matrix is positive definite if all its eigenvalues are positive

A symmetric diagonally dominant matrix with positive diagonal entries is positive definite

If a matrix is positive definite All the diagonal entries are positive

The largest (in magnitude) element of the whole matrix must lie on the diagonal

Page 32: Iterative Solution Methods

32

Positive Definitiness Check

5225

21512

251220

2525

21512

51220

Not positive definiteLargest element is not on the diagonal

Not positive definiteAll diagonal entries are not positive

2525

21512

51220 Positive definiteSymmetric, diagonally dominant, all diagonal entries are positive

Page 33: Iterative Solution Methods

33

Positive Definitiness Check

2528

21512

51220

A decision can not be made just by investigating the matrix.

The matrix is diagonally dominant and all diagonal entries are positive but it is not symmetric.

To decide, check if all the eigenvalues are positive

Page 34: Iterative Solution Methods

34

Example (Gauss-Seidel Iteration)

15

21

7

512

184

114

3

2

1

x

x

x

0

0

00x

5

215

8

421

4

7

12

111

3

03

111

2

03

021

1

xxx

xxx

xxx

0.35

5.375.1215

5.38

75.1421

75.14

7

7395.262

0 Axb

0414.32

1 Axb

Diagonally dominant matrix

0452.102

1 Axb

Jacobi iteration

Page 35: Iterative Solution Methods

35

Example 1 continued...

5

215

8

421

4

7

22

212

3

13

212

2

13

122

1

xxx

xxx

xxx

9625.25

9375.3875.1215

9375.38

3875.1421

875.14

35.37

4765.02

2 Axb

7413.62

2 Axb

Jacobi iteration

When both Jacobi and Gauss-Seidel iterations converge, Gauss-Seidel converges faster

Page 36: Iterative Solution Methods

36

Convergence of SOR method

If 0<<2, SOR method converges for any initial vector if A matrix is symmetric and positive definite

If >2, SOR method diverges

If 0<<1, SOR method converges but the convergence rate is slower (deceleration) than the Gauss-Seidel method.

Page 37: Iterative Solution Methods

37

Operation count The operation count for Gaussian

Elimination or LU Decomposition was 0 (n3), order of n3.

For iterative methods, the number of scalar multiplications is 0 (n2) at each iteration.

If the total number of iterations required for convergence is much less than n, then iterative methods are more efficient than direct methods.

Also iterative methods are well suited for sparse matrices


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