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Lecture03 - Iterative Methods

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  • 8/9/2019 Lecture03 - Iterative Methods

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    Solution of Linear System of Equations

    Lecture 3:

    Iterative Methods

    MTH2212 Computational Methods and Statistics

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    Dr. M. HrairiDr. M. Hrairi MTH2212MTH2212 -- Computational Methods and StatisticsComputational Methods and Statistics 22

    Objectives

    Introduction

    Jacobi Method

    Gauss-Seidel Method

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    Introduction

    To solve the linear system Ax = b we may use either:

    Direct Methods

    - Gaussianelimination

    - PLU decomposition Iterative Methods

    - Jacobi Method

    - Gauss-Seidel Method

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    Iterative Methods

    Suppose we solve Ax = b for a given matrix A by finding the

    PLU decomposition

    Ifwe change the vectorb,we may continue to use the PLU

    Ifwe change A,we now have to re-compute the PLUdecomposition: expensive

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    Iterative Methods

    Instead, suppose we have solved the system

    Ax = b

    for a given matrix A

    Suppose we change A slightly, e.g., modify a single resistor

    in a circuit

    Ifwe call that new matrix Amod

    , is it possible to use the

    solution to Ax = b to solve Amod

    x = b?

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    Iterative Methods

    They provide an alternative to the elimination method.

    Let Ax = b be the set ofequations to be solved.

    The system Ax = b is reshaped by solving the first equation

    for x1, the second equation for x2, and the third for x3, and

    nth equation for xn.

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    Iterative Methods

    For ease ofcomputation, lets assume we have a 3x3

    system ofequations to solve.

    If the diagonal elements are all non-zero then:

    !!

    !

    3333232131

    2323222121

    1313212111

    bxaxaxa

    bxaxaxa

    bxaxaxa

    33

    23213133

    22

    32312122

    11

    31321211

    a

    xaxabx

    a

    xaxabx

    a

    xaxabx

    !

    !

    !

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    Dr. M. HrairiDr. M. Hrairi MTHMTH22122212 -- Computational Methods and StatisticsComputational Methods and Statistics 88

    Jacobi Iteration Method

    1. Assume all the xs are zero

    2. Substitute the zeros into the three equations to get:

    3. Repeat the procedure until the error criterion is satisfied:

    11

    1

    1 a

    bx !

    22

    2

    2 a

    bx !

    33

    3

    3 a

    bx !

    si

    j

    i

    j

    i

    jja

    xxx Iev

    !I

    %00,

    33

    23233

    3

    22

    32322

    2

    3322

    a

    xaxabx

    a

    xaxabx

    a

    xaxabx

    ii

    i

    iii

    ii

    i

    !

    !

    !

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    Gauss-Seidel Method

    It is the most commonly used iterative method.

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    Gauss-Seidel Procedure

    1. Assume all the xs are zero

    2. Substitute the zeros into the first equation i.e. equation (1)to give:

    3. Substitute the new value ofx1 and x3 = 0 into equation (2)to compute x2

    4. Substitute the value ofx1 and the new value ofx2 inequation (3) to estimate x3

    11

    11

    a

    bx !

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    Gauss-Seidel Procedure

    5. Return to equation (1) and repeat the entire procedure untilthe error criterion is satisfied:

    33

    1

    232

    1

    13131

    3

    22

    323

    1

    12121

    2

    11

    31321211

    1

    a

    aab

    a

    aab

    a

    aab

    ii

    i

    ii

    i

    ii

    i

    si

    j

    i

    j

    i

    jja

    xxx Iev

    I

    %00,

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    Dr. M. HrairiDr. M. Hrairi MTHMTH22122212 -- Computational Methods and StatisticsComputational Methods and Statistics 1212

    Example 1

    Use Gauss-Seidel method to solve the following set of

    linear equations:

    3x1 0.1x2 0.2x3 = 7.85 (1)

    0.1x1 + 7x2 0.3x3 = -19.3 (2)

    0.3x1 0.2x2 + 10x3 = 71.4 (3)

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    Example 1 - Solution

    First we have:

    10

    2.03.04.71

    7

    3.01.03.19

    3

    2.01.085.7

    213

    31

    2

    321

    xx

    x

    xx

    x

    xx

    x

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    Example 1 - Solution

    1st iteration

    Assumethat x2 = 0 and x3 = 0, weobtain

    Substitute x1 = 2.616667 and x3 = 0 intoequation (2)

    Substitute x1

    = 2.616667 and x2

    = -2.794524 intoequation (3)

    This completes the first iteration

    6 6667.23

    85.7!!x

    794524.27

    0)616667.2(1.03.192

    x

    0056 0.70

    )794524.2(2.0)6 6667.2(3.04.73 !

    !x

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    Example 1 - Solution

    2nd iteration

    990557.)005610.7(.0)79454.(1.085.7

    1 !

    !x

    499625.27

    )005610.7(3.0)990557.2(1.03.192 !

    !x

    000291.710

    )499625.2(2.0)990557.2(3.04.713 !

    !x

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    Dr. M. HrairiDr. M. Hrairi MTHMTH22122212 -- Computational Methods and StatisticsComputational Methods and Statistics 1616

    Example 1 - Solution

    Error estimate

    Forx1

    Forx2

    Forx3

    %12%12

    1221 !

    ! aI

    %8.11%100499625.2

    )794524.2(499625.22, !

    ! aI

    %% !

    ! aI

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    Convergence

    Gauss-Seidel is similar in spirit to the simple fixed-pointiteration.

    Gauss-Seidel will converge iffor every equation of the

    system,w

    e have:

    Such system is said to be diagonally dominant.

    This criterion is sufficient but not necessary forconvergence.

    {!

    "n

    ijj

    ijii aa1

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    Dr. M. HrairiDr. M. Hrairi MTHMTH22122212 -- Computational Methods and StatisticsComputational Methods and Statistics 1818

    Relaxation

    Designed to Enhance convergence.

    After each new value ofx is computed, that value ismodified using:

    Where is a weighting factor.

    The choice of is problem-specific and is often determinedempirically.

    oldi

    new

    i

    new

    ixxx PP! 1

    20 ePe

    P

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    Gauss-Seidel/Jacobi Iteration Methods

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    Dr. M. HrairiDr. M. Hrairi MTHMTH22122212 -- Computational Methods and StatisticsComputational Methods and Statistics 2020

    Gauss-Seidel iteration converges more rapidly than the

    Jacobi iteration does; since, it uses the latest updates.

    But there are some cases that Jacobi iteration does

    converge but Gauss-Seidel does not.

    Gauss-Seidel/Jacobi Iteration Methods

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    Assignment #1

    Computational Methods

    12.11, 12.30, 12.33

    Statistics

    2.2, 2.14, 2.22, 2.26, 2.28, 2.37, 2.45, 2.52, 2.65, 2.74

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