Download - 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
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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
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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
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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
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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
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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
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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
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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
-
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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 , , .