Lecture 11BUE

8/14/2019 Lecture 11BUE

1/28

ScheduleWeek Date Topic Classification of Topic

1 9 Feb. 2010 Introduction toNumerical Methods

and Type of Errors

Measuring errors, Binary representation,Propagation of errors and Taylor series

2 14 Feb. 2010 Nonlinear Equations Bisection Method

3 21 Feb. 2010 Newton-Raphson Method

4 28 Feb. 2010 Interpolation Lagrange Interpolation

5 7 March 2010 Newton's Divided Difference Method

6 14 March 2010 Differentiation Newton's Forward and BackwardDivided Difference

7 21 March 2010 Regression Least squares

8 28 March 2010 Systems of LinearEquations

Gauss-Jacobi

9 11 April 2010 Gauss- Seidel

10 18 April 2010 Integration Composite Trapezoidal and SimpsonRules

11 25 April 2010 Ordinary DifferentialEquations

Euler's Method

12 2 May 2010 Runge-Kutta 2nd and4th order Method


2/28

Gauss-Jacobi

Iterative or approximate methodsprovide an alternative to theelimination methods. The Gauss-Jacobimethod is the most commonly used

iterative method.

The system [A]{X}={B} is reshapedby solving the first equation forx1, the

second equation for x2, and the thirdforx3, and n

th equation forxn.


3/28

Gauss-Jacobi Method

Algorithm

A set of nequations and nunknowns:

11313212111 ... bxaxaxaxa nn

2323222121 ... bxaxaxaxa n2n

nnnnnnn bxaxaxaxa ...332211

. .

. .

. .

If: the diagonal elements arenon-zero

Rewrite each equation solvingfor the corresponding unknown

ex:

First equation, solve for x1

Second equation, solve for x2


4/28


5/28

Gauss-Jacobi 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 themost recent value of xi. Which

means to apply values calculated to

the calculations remaining in the

current iteration.


6/28

3

333

2

212

1

111

equations33For

c

zbxadz

b

zcxady

a

zcybdx

Now we can start the solution process by

choosing guesses for the xs. A simple way toobtain initial guesses is to assume that they

are zero. These zeros can be substituted into

x1equation to calculate a newx

0=d

1/a

1.


7/28

First iteration

3

333

2

222

1

111

equations33For

c

ybxadz

b

zcxady

a

zcybdx

2

030331

3

020221

1

010111

equations33For

b

ybxadz

c

zcxady

a

zcybdx

Applying the initial guess and solving for x, y,

and z

0

0

0

z

y

x

z

y

x


8/28

Gauss-Jacobi 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 relative approximate error is less

than a prespecified tolerance for all

unknowns.


9/28

Convergence Criterion for Gauss-

Jacobi Method The Gauss-Jacobi method has two

fundamental problems as any iterative

method:

It is sometimes nonconvergent, and

If it converges, converges very slowly.


10/28

Gauss-Jacobi Method

Why?

The Gauss-Jacobi 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.


11/28

Gauss-Jacobi Method: Example 1

The upward velocity of a rocket is given at three different

times

Time Velocity

5 106.8

8 177.2

12 279.2

The velocity data is approximated by a polynomial as:

12.t5,322

1 atatatv

st m/sv

Table 1 Velocity vs. Time data.


12/28


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

1Using a Matrix template of the form

The system of equations becomes

2.279

2.177

8.106

112144

1864

1525

3

2

1

a

a

a

Initial Guess: Assume an initial guess of

5

2

1

3

2

1

a

a

a


13/28

Gauss-Jacobi Method: Example 1Rewriting each equation

2.279

2.177

8.106

112144

1864

1525

3

2

1

a

a

a

25

58.106 321

aaa

8

642.177 312

aaa

1

121442.279 213

aaa


14/28


Applying the initial guess and solving for ai

5

2

1

3

2

1

a

a

a 6720.325

)5()2(58.1061

a

525.13

8

51642.1772

a

2.111

1

21211442.2793

a

Initial Guess


15/28


%76.721006720.3

0000.16720.31a

x

%212.85100525.13

0000.2525.132a

x

%5.951002.111

0000.52.1113a

x

Finding the absolute relative approximate error

100

new

i

old

i

new

i

ia x

xx At the end of the first iteration

The maximum absolute

relative approximate error is95.5%

2.111

525.136720.3

3

2

1

a

aa


16/28


Iteration #2

Using

2.111

525.13

6720.3

3

2

1

a

a

a

881.225

2.111525.1358.106

a1

126.21

8

2.111672.3642.177a2

868.411

1

525.1312672.31442.279a3

from iteration #1

the values of ai are found:


17/28


Finding the absolute relative approximate error

%455.227100881.2

6720.3881.21a

x

%02.164100126.21

525.13126.212a

x

%05.127100686.411

2.111868.4113a

x

At the end of the second iteration

868.411

126.21

881.2

3

2

1

a

a

a

The maximum absolute

relative approximate error is

227.455%


18/28

Iteration a1 a2 a3

1

23

4

3.6720

-2.8818.2972

-42.545

72.767

227.455134.722

119.502

13.525

21.12644.573

162.67

85.212

164.02147.396

127.403

111.2

-411.868947.576

-1450.47

95.5

127.05143.465

165.329


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


19/28

Gauss-Jacobi Method: PitfallWhat went wrong?

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

correct answer

This example illustrates a pitfall of the Gauss-Jacobi method: not allsystems 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

ij

j

ijii aa1

n

j

j

ijaa

i

1

iifor all i and

for at least one i


20/28

Gauss-Jacobi Method: Pitfall

116123

14345

3481.52

A

Diagonally dominant: The coefficient on the diagonal must be at leastequal 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.


21/28

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

20238 zyx

zyx 33114

351236 zyx

0

00

z

yx

With an initial guess of

The coefficient matrix is:

1236

1114

238

A

Will the solution converge using the

Gauss-Jacobi method?


22/28

Gauss-Jacobi Method: Example 2Checking if the coefficient matrix is diagonally dominant

5141111 232122 aaa

9361212 323133 aaa

52388 131211 aaa

The inequalities are all true and at least one row is strictlygreater than:

Therefore: The solution should converge using the Gauss-Jacobi Method

12361114

238

A


23/28


3533

20

zy

x

12361114

238

Rewriting each equation

8

2320 zyx

11433 zxy

12

3635 yxz

With an initial guess of

0

0

0

z

y

x

5.2

12

020320x1

311 00433y1

9166667.2

13

030635z1


24/28

Gauss-Jacobi Method: Example 2The absolute relative approximate error

%00.1001005.2

0.05.21

a

%00.1001000.3

0.00.32

a

%100100916666.2

0.0916666.23

a

The maximum absolute relative error after the first iteration is 100%


25/28


916666.0

3560606.2

895833.2

2

2

2

z

y

x

After Iteration #1

895833.2

8

9166667.2233202

x

3560606.2

11

9166667.25.24332

y

916666.0

12

335.26352

z

Substituting the values into

the equationsAfter Iteration #2

9166667.2

0000.3

5000.2

1

1

1

z

y

x

111 ,, zyx


26/28

Gauss-Jacobi Method: Example 2Iteration #2 absolute relative approximate error

%669055.13100895833.2

50000.2895833.21a

%33119.271003560606.2

0.33560606.22a

%181.218100

916667.0

916667.2916667.03a

The maximum absolute relative error after the second iteration is 218.181%

This is much larger than the maximum absolute relative error obtained in

iteration #1. Is this a problem?


27/28

Iteration x y z

1

23

4

5

2.5

2.895833.1543

3.04143

3.01687

100.00

13.66908.1942

3.7110

0.8140

3

2.35602.0303

1.9329

1.9696

100.00

27.331116.041

5.0390

1.8633

2.9166

0.916660.8797

0.83191

0.9127

82.193

218.1814.2014

5.7446

8.8911

Gauss-Jacobi Method: Example 2Repeating more iterations, the following values are obtained%

1a %

2a %

3a

9118312.0

9858987.1

0167424.3

z

y

xThe solution obtained is close to the exact solution of .

9118.0

985.1

0167.3

z

y

x


28/28

Gauss-Jacobi MethodThe Gauss-Jacobi 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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Lecture 11BUE

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