Lecture Week8 Q

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  • 8/10/2019 Lecture Week8 Q

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    Tutorial - week9

    To minimiz2 2 2

    -

    with the starting poi nt ( ) ( ) by

    , ,

    , , ,

    ,

    ,0 0 0

    -

    x y z 0 0 0

    ( )1 Steepest Descent Method

    ( )2 Conjugate Gradient ethM od

  • 8/10/2019 Lecture Week8 Q

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    calculating by:( ) , , , FirstlyT

    f x y zSolution1

    - -

    - -( )- -

    , , , ,

    , , , ,T

    0 0 0 0 0 0 0 0 0 0 0

    - -

    r f x y z 2x - y 1 2y x z 2z - y 1

    .- -, ,

    T1 0 1

    finding which ( )byminimizes

    ,Next 0 00 f x r

    x

    - - x r 0 0 0 1 0 1

    , ,- - 0

    ( ) ( ) (-( )

    -

    ) ( )-

    - - - - - -

    , , , , , ,1 1 1 0 0 0 0

    2 2 2 2

    0 f x y z f x y z r f q

    -

    ( ) ( ).

    --

    2q 2 2

    4 2d 1

    d

    0

    2d d

  • 8/10/2019 Lecture Week8 Q

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    - - - -

    ( )

    Therefore

    1 1 11 - con.

    , take the negative gradient to find the next search direction:

    Thirdly

    2 2 2

    - -( - -) , , , ,1 1 1 1 1 1 1 1 1 1

    r f x y z 2x - y 1 2y x z 2z - y

    T

    1

    1

    ( ) ( )- ( )- ( ) ( )- - - - .-, , , ,

    T2 - 1 2 2 - 1 0 1 0

    1 1 1 10 0 0 -

    2 2 2 2

    , update the iteration formula byThen

    ( ) ( -) -, , , , ,, , ,

    2 2 2 1 1 1 1x y z x y z r 0 0 1 02 2

    -

    -- ., , 2

    -2

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    we( have) ,Thus1 - con.

    ( ) (- - ) ( ).( ) ( ), , , , , ,2 2 2 1 1 1 1x y z f x y z r f q2 2

    -

    ( ) (- ) (- )( ( )) ( ) ( )(- ) (- ) (- )-2 2 2q 1 -

    2 2 2 2 2

    - - -

    .- -2

    2

    ( )- -

    ( )-Set

    2

    1

    ddq 12 1 2 0

    ( , ,2 2x y () -) -, , , , , ,

    2 1 1 1 11

    1 1

    z x y z r 0 0 1

    1

    - 0

    .-- -, , 1 1 1

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    the negative gradient to find( ) Take1 - con.

    ( ) - -- -, , , ,2 2 2 2 2 2 2 2 2 2 2T

    r f x y z 2x - y 1 2y x z 2z - y 1

    ( ) ( ) ( )- (- - - - - )- - - -( ) ( ) ( ), ,

    2 -

    2 2 2 2 2 2

    1 2 2 - 1

    2

    - - ., ,

    10

    2

    1

    2

    update the iter,Again ation formula by

    ( ) - - - - -)( , , , , , , , ,

    3 3 3 2 2 2 2 x y z x y z r 02 2 2 2 2

    - ( ) - - ( .), ,

    1 1 1

    1 12 2

    2

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    ( ) , we obtainThus1 - con.

    - ( ) - - ( ) ( ).( ), , , ,

    3 3 3f x y z f 1 q2

    1 2 2

    - ( ) - ( )( ) ( ( )) ( )- -

    2

    2

    q 1 -

    1 1

    1 1

    1

    2 2

    1

    2 2

    - -- --

    21 1 1

    1 1 1 1

    2

    2 2 212

    3 1

    -2

    2 4

    1 3 1

    ( )-

    - . 2d

    1 1

    q 2 2

    04

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    ( )( ) ( ), , , ,Thus 3 3 3 2 2 2 22x y z x y z r1 - con.

    - - -- - - - - ., , , , , ,

    1 3 1 32 4

    1 1 1 1 102 2 2 2 2 2 4

    the next search direction:FindT

    -- - -, , , ,3 3 3 3 3 3 3 3 3 33- -

    - - - - - - - -- -

    T

    2 - 1 2 2 - 13 1 1 3 3 3 1

    T

    14 2 2 4 4 4 2

    -, , , , , ,

    4 4 4 3 3 3 32

    3 1 3 1 313

    -- - - - ., , , , , , , ,

    4 4 424 2 4 2 44

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    by( ) Find2 2 2

    31 - con.

    .-, ,

    4 4 4 4 4 4 4 4 4 4 4

    2

    21 1

    -

    3 3

    - - - -

    2

    q -2

    4 4

    ( ) - - -- ( )-

    3 3 3

    4 4 4

    1 1

    2

    ( ) - -21 3 9

    1

    ( ) - - 21 3 9

    1 d

    .- 3d

    22 4d

    0

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    ( ) ( ) ( ), , , , ,Thus 4 4 4 3 3 3 331 x y z x y z r- con.

    -- -- - ., , , , , ,

    0 04 2 4 2 42 4 4

    the next search direction:

    - ---

    FindT

    - -

    -

    , , , ,4 4 4 4 4 44 4 4 4 4

    2

    - - - -- -- - -, ,

    T3 3 33 3

    - 1 2 2 - 13 3

    T 4 4 44 4

    1 14 4

    , , , , , ,

    5 5 5 4 4 4 4

    4 4

    - - - ( ) .- ( ) -, , , , , ,

    0 34 4 4 4 4

    4

    344

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    .

    ( ) by

    -, ,

    Find2 2 2

    4

    x z x x 1 - z z z

    1

    - con.

    ( ) ( ( )) ( )- ( ) - ( ) - -

    2

    2q 1 -3 31 1

    3 3

    2

    1 1 13

    - - --- 4 4 4

    1 7

    4

    51

    - -8 8

    16

    1 7 51

    ( )- -

    - .

    4

    q 1 18 8 16

    0

    1

    d

    - - - - - .

    Thus3 3 3 1 1 7

    x 01 7 3

    4 2 84 4 84 4 4

  • 8/10/2019 Lecture Week8 Q

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    the next search direction:( ) FindT

    1 - con.

    - -

    ( ) ( ) ( )- ( )- ( ) ( ) ( )- - -

    - -

    - - .- - --

    , , , ,

    , , , ,

    5 5 5 5 5 5 5 5 5 5 5T T- -

    7 7 12 -

    7 71 2 2 - 1 0 0

    3 3 3

    ( ) ( ) -, , , ,6 6 6 5 5 5 57

    x y z x y rz

    - - -, , , ,

    3 7 1

    0 0

    - - .- ( ), ,

    7

    8

    1 3

    7

    84

    by ( ) ( ) ( ).-, ,Find 2 2 26 6 6 6 6 6 6 6 6 6 6 52

    f x y z x x 1 - y y y z z z q

    - -( ) ( ( )) ( )- ( ) - ( )

    28 8

    q 1 3 34 4

    -

    - ( ) - - -- ( () .- - )

    27 7 7 1 7 4938 8 8 16 16 32

    4

    3 1

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    - - 21 7 49

    ( )( -) q 1 116 16 32 08 16

    dd d

    1 - con.

    ( ) ( ), , , ,6 6 6 5 5 5 55 5x y z x y z r1

    2

    - - - ., , , , , ,

    7 3 7 1 7 7 7 0 0

    8 4 8 4 8 8 8

    1

    2

    ( ) is no, ,Since 6 6 6x y z t very close to ( ) ( ), , , , ,x y z - 1 -1 -1

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    1-

    6 2.

    - -( ) ( ), , , , , , , ,

    7 7 7 6 6 6 6 6x y z x y z r 08 8 8 8 8

    2

    - ., ,

    15 7 15

    16 8 16

    - - -( ) ( ) -, , , , , , , 8 8 8 7 7 77 7 715 7 15

    x y z x y z r 0 , 1

    0

    - ( ) -- ., , 15 151

    7

  • 8/10/2019 Lecture Week8 Q

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    by ( )( ) , ,Find 8 8 87 f x y z1 - con.

    ( )- ( )- ( ) ( )- - - ., , 2 f q 716 64 64 128

    16

    78

    -( -)( )

    2

    7

    7q

    64 64 128 0

    d 1

    d

    ( ) - - - --, , , , , , ,

    7

    15 7 15 1 15x y z 0 0

    1 .- -,

    15 15

    we think that ( ) is very close to ( ) ( ) , , , , , , ,If 8 8 8x y z x y z - 1 -1 -1

    we can check the by evaluating

    - -( ), , , ,

    convergence criteria

    88 8 8 88 8 88 82y x2x - y 1f x y z 2z -z y 1

    ,1 16 0 ( ) ( ) ( ) ., ,2 2 21 161 16 1 16 2 16 0 0880 4

    may take it to be close enough to solution in

    .

    We Example 2.

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    :( )Soluti2 on

    we can rewr te to a qua ract c orm y, , ,rst y

    T T

    x y z1

    , ,

    2

    ( ), , , , ,

    x y y -1 2 -1 y 1 0 1 y2 z z0 -1 2

    2 -1 0- -

    -1,

    0 -1 2 .

    -1

  • 8/10/2019 Lecture Week8 Q

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    the begining of iterations, we calculate the residual

    vector from

    ( ) At

    r x

    2 - con.

    .- 0 0 0

    -1 2 -1 0 0 -1 r b - Ax r 0 -1 2 -1 0 0

    - - -

    we use as our i i in t,Next0

    r al search direction and compute,0 0

    p

    , ,

    T

    0 0

    --1 0 -1 0

    -1r r 2 1

    .

    , ,

    T

    0 0

    0

    2 -1 0 -1p Ap 4-1 0 -1 -1 2 -1 0

    2

    can now compu e ( ) ( )t , , , ,We1 1 1 0 00 0 0

    - -

    x y z x y z p

    ( ) .- - - -, , , , , ,

    10

    1 10

    2 2 20 0 1 0 1

    resu comp e es e rs era on.s

  • 8/10/2019 Lecture Week8 Q

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    we can compute( ) So-1 2 -1 0 -1 0

    2 - con.

    ,

    1 0 00 r r - Ap 0 - -1 2 -1 0 -1-1 0 -1

    2 2 -1 0

    computing the scalar and the next search direction, .Next0 1

    p

    TT - -0

    and,, , , ,

    , , , ,

    1

    0

    1 1

    TT

    0 0T

    r r -1 0 -1 -1 0 -1 2

    ., , , , , , 0

    T T

    1 1 0 p r p 0 -1 0 -1 0 -1 - -1 -2 2 2

    .

    , , , ,

    1 1

    TT

    1 1 0 -1 0 0 -1 0r r

    1

    , ,

    T

    1

    1

    12 -1 0p Ap 1 1

    - -1 - -1 2 -1

    .

    , ,

    T

    - -1 -11 1

    -

  • 8/10/2019 Lecture Week8 Q

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    ( ) 2 - con.

    we can find ( ) by, , ,Using 2 2 21 x y z

    , , , ,2 2 2 1 1 1 11

    1 1 1 1

    - - - - - - - - ., , , , , , 2 2 2 2

    ( ) is the point of ( ).minimum, , , ,2 2 2x y z f x y z , , .