Download - Lecture Week9 Q KKT
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Tutorial - week 9 points of the following two problemsFind KKT
- - -( )( ) ,21 1 2 f x 4x x x 2 Minimize 51
( ) ,
- -1 2 1 1h x = x 1.5x 2x 1 = 0
h x =
.- -2 2 x 2x 2x 4.25 = 0
- -- ,2 Minimize x 4x x x 2.52
( ) + 2 22 1 1
subject toc x = 4.25 x 2x 2x 0
( -) -2 22 1 1h x = x 1.5x 2x 1 = 0.
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( ) : define theWeSol iotu n Lagrange function1
, (
- - - - - -
) ( ) ( )( )
-1 1 2 2
2 2 2 2 2
L x f x h x h x
the conditions will be
.
,Thus
1 1 2 1 2 1 1 2 2 1 1 . . .
KKT ,
-( )
-(1 11
L x 8x 1 2
x ) ( )-1 2 13x 4x 2 0
, -
- -
1 2 2 22 2 2
1 2 x 2 x 0 x
--
2 22 1 1
x 2x 2x 4.25 0
1 2 .
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we solve( ) ,Next -0.549 1.597 1 - con.
( , )- - ,
2 22 1 1 -0.549 - 1.597 x 1.5x 2x 1 0 x x
- ,-
( , )2 1 1 x x x
1.695 - 1.3
. . . 5
86 we need to check the multiplier ,Thirdly Lagrange s
( , ) - - -
- 1 1 1 2 1
1 2 2 2-0.549 1.597
x x x 1 2 x 2 x 0
- -.1 2 1
3.647 4.196 5.392 0 0.855
- -1 2 23.194 3.194
1 0 0.5
42
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- - -
( ) ( )( ) 1 1 1 2 1
8x 1 2 3x 4x 2 01 - con.
( ) -- -
,1 2 2 2
1 2 1
-0.549 -1.597
3.647 4.196 5.392 0 0.520 .
-1 2 2
3.194 3.194 1 0 0.833
-
--1 1 1 2 1
( , )
- 1 2 2 2 1.695 1.386 1 2 x 2 x 0
- - .- -
1 2 1
1 2 2
3.085 4.78 12.56 0 1.8152.792 2.792 1
0 1.456
(- - ) ( )
-
1 1 1 2 18x 1 2 3x 4x 2 0 - ,1 2 2 2 . - .
- -1 2 13.085 4.78 12.56 0 1.379 .-1 2 22.792 2.792 1 0 1. 3
7
7
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( )we have( ) , ,Then 1 2 f x x 1 - con.
( ) ( ) ( )( ) ( ) ( ) ( )( ) - - - -
- - - , ,
2 f 4 2.5
f -0.-0.549 1.5
549 - 1.597 4 -0.549 -0.549 -1.597 2.597 -0.549 -0.549 1.597 2.3
0.8516 424
( )( - -( -) () ) , 21.695 1.386 1.695 1.695 1.3 f 4 6 8
- - -( ( ) ( ) ( )) , 21.695 - 1.386 1.695 1
2.5
f 4
5.9
2.5
111
.695 -1.386 8.6831
:Conclusion
( )
- ( ,
, ,
-0.54
9 1.597 2.342 f f 4 ) ,-0.549 - 1.597 0.8516
( ) optimal soluhence is th( e) .tio f n o ,
, , ,
, ,
1 2
. - . .-0.549 1.
. . x x f 59
.7
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( , )-
-0.549 1.597 =
( )Figure - 1
( , ) 1.695 1.386 f = 5.9111
( )1h x = 0
( )2h x = 0
( , ) 1.695 - 1.386 f = 8.6831
( , ) -0.549 - 1.597 f = 0.8516
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: we have( ) ,Solution Firstly Lagrange function2
- - -( - - ) ( - - ) .
, , 1 1 22 2 2 22 1 1 2 1 1
x x x x . x 2x 2x 4.25 1.5x 2x 1
the conditions will be , , ,
Thus L x
KKT
- - -1 11
x x x
( )-
, ,1 x
L x
- -
2 22
2 22 1 1
x x 2x 2x 4.25 0
- - 2 22 1 1 x 1.5x 2x 1 0
- -
- -2 1 1
2 2
x x x .
x 1.5x 2x 1 0
.and free 0
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we solve( ) ,Next-0.549 1.597
2 - con.
( )(
- - )
, ,
2 22 1 1 -0.549 - 1.597 x 1.5x 2x 1 0 x x
- -
( )
,
,
2 1 1 . x x x .
1.695 - 1.386
.
we check the multipliers in case ,Then Lagrange 2 s:
( ) and free ( )and free
a > 0 activ cb 0
c
ei ivnact e
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and free.( )( ) > 0 a2 - con.
need to check the multipliers
- - -
We Lagrange
( ) - , 1 1 1
2 2-0.549 1.597 1 2 x 2 x 0
- --
3.647 4.196 5.392 0 0.3.194
3.194 0
1
( ).
- !855
0.542 0Ok
Violation
- - -( ) ( )1 1 18x 1 4x 2 2 3x 0
( ) -
,2 2 -0.549 -1.597 1 2 x 2 x 0
- -.
- ( !) . . . .
3.194 3.19
4 1 0 0 Violation .833
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( )a - con.
- - -
( ) ( ) 1 1 1
x 1 4x 2 2 3x 0
( ) -
- -
,2 2 1.695 1.386
3.085 4.78 Ok 12.56 0 1.815
.- - ( !)Violati 2.792 2.792 1 0 1.456 on
( ) ( )-- -1 1 1 x 1 4x 2 2 3x 0 - 1.695 -1.386
- - ( )3.085 4.78 12.56 0 1.379 Ok .- ( !)2.792 2.792 1 0 1.737 Violation
.ase
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and free.( )( ) 0 b2 - con.
con t ons w e( , ) e
L x
- -- -
-( ),
1 11
1 2 1 2
x 16x x 3x 2x 2 0 L x
-
2
2
x
-22 .
- -
1 1
4 3 21 2 1 2
.
16x x 3x 2x 2 0 - -
--2 2 1 1 1 12 1 1 x 1.5x 2x 1 0
( )
( )
-( )
..
, ,
,1 2
Ok
Ok
. x x
. - .
0.50 1 0
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( ) and free.( .) b 0 2 - con
( ) ( )we check at and by( )- -
, , ,Now2 2
0.2291 - 0.7878c x c x x 2 x 2x 4.25
0 1
( ) ( ) ( )( -) ,2 2
0.2291 - 0.7878 -0.7878 0.2291 0.2291c = 2 2
- -
( ) - - - (( ) ( ) ( ) ) , 2 2 . . .
c = 2 2 4.25 = 3.25 0.0 1 1 0 0 Ok
both and are th( ) e Poi( t) sn , , , . Thus 0.2291 - 0.7878 0 1 KKT
( -) , ,
2
. - .0.2294 1 - ( )- -2.5 1.73130.2291 0.7878
- - - -( ) optimawe can choose as the of l solu t oi( n)
, , , ,1 2
.0
. x x f 1 .
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( ,-
( ))
-0.549 1
a .597
f = 2.3424(2)Figure -( , )( )
-0 1
f b
3.5
, )) (( 1.695 1.386 f = 5
a .9111
( )h x = 0
( )c x = 0
( , )( ) 1.695 - 1.386 f = 8.6831
a
( , )( )
-0.549 - 1.597 a f = 0.8516 -)
,(
. - .b f 1.7313