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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