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Lecture 8 - Optimization With Inequality Constraints, Lecture notes of Economics

Optimization With Inequality Constraints

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2018/2019

Uploaded on 08/10/2019

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9/7/2018 Prepared by Nachrowi 1
Optimization with Inequality Constraints
Applications:
(i) Max U U (x1 , x2, …, x n ); S.t.: P1 x1 + P2 x2+...+ Pn x n B; x1, x2,,x n 0
(ii) Min C = PK K + PL L; S.t.: K L Q 0 ( Q 0 > 0); K , L 0
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Download Lecture 8 - Optimization With Inequality Constraints and more Lecture notes Economics in PDF only on Docsity!

Optimization with Inequality Constraints

Applications:

(i) Max U  U ( x

, x

x

n

); S.t.: P

x

+ P

x

+...+ P

n

x

n

 B; x

, x

, x

n

(ii) Min C = P

K

K + P

L

L; S.t.: K

L

≥ Q

( Q

> 0); K , L ≥ 0

In general, profit maximization and cost minimization are written in

the following forms:

Max  = f ( x 1 , x 2 , (^) …, xn ); there are n products and m constraints

S.t. : g

1 ( x 1 , x (^) 2……… xn )  r (^1)

g

2 ( x 1 , x (^) 2……… xn )  r (^2)

g

m ( x 1 , x (^) 2………xn )  r (^) m

Min C = f ( x 1 , x (^) 2…………… x (^) n )

S.t. : g

1 ( x 1 , x2…………… xn )  r (^1)

g

2 ( x 1 , x2…………… xn )  r (^2)

g

m ( x 1 , x2…………… xn )  r (^) m

9/7/2018 Prepared by Nachrowi^4

Finding the Solution to an NLP Problem Graphically

Example (1): Min C = ( x 1 - 4 )

2

+ ( x 2 - 4 )

2

S.t. 2 x 1 + 3 x 2  6

x 2  3 x 1  2 x 2   12

x 1 , x 2  0

Where’s the solution?

a circle with its center at ( 4 , 4 )

2 x 1 + 3 x 2 = 6

3 x 1 + 2 x 2 = 12

x 1

The solution is located at the tangential point between the line 3 x

+2 x

=12 and the circle with

its center at (4,4).

Why: (a) C is minimum when x

= 4; x

= 4. However this point not feasible.

(b) The point in the feasible region with the closest radius to the tangential point.

( ii ) Find the solution from the 2 equations: the slope and the constraint line

2 x 1

  • 3 x 2

= – 4

3 x 1 + 2 x 2 = 12

6 x 1

  • 9 x 2

= – 12 ; 4 x 1

  • 6 x 2

= – 8

6 x 1 + 4 x 2 = 24 9 x 1 + 6 x 2 = 36

13 x (^) 2 = 36 13 x 1 = 2 8

x (^) 2

=

/ 13 x (^) 1

=

/ 13  C

= 4

/ 13

+

9/7/2018 Prepared by Nachrowi 8

Observations on the Feasible Area:

  1. This area is a convex set.
  2. The solution acquired is the global solution.

Example (2)

Min C = ( x (^) 1 – 4 )

2

  • ( x (^) 2 – 4 )

2

S.t. : x 1 + x (^2)  5

- x (^1)  – 6 - 2 x (^2)  – 11

x 1 , x 2  0

x 1 = 6

x 1 + x 2 = 5

5 x 1

x 2 = 5

1 / 2

Example (3):

Max  = 2 x 1 + x 2

S.t. :  x (^) 1

2

  • 4 x 1  x 2  0

2 x 1 + 3 x 2  12

x 1 , x 2  0

x (^2)

4 -x (^1)

**2

  • 4 x 1**  x 2 = 0

F 1

2 x 1 + 3 x 2 = 12

F 2 P 3

2 4 6 x (^1)

P 2

P 1

Observations:

( i ) Feasible set : F 1  F 2 is not convex.

( ii ) P 1 is optimum for F 1

However  point  F 2 is better than P 1  P 1 is optimum relative.

( iii ) P 3 with x 1

= 6 and x 2

= 0 is the global solution;  (^2)

= 12.

Notes:

( i ) If a feasible set is not convex  the solution may not be global nor unique.

( ii ) If a feasible set is convex  there exists a global solution.

9/7/2018 Prepared by Nachrowi^13

Non-Negative Constraints

 Max  = f ( x 1 ) ; S.t. x (^) 1  0

 Possible solutions:

(i) Optimum solution A : interior solution; f  ( x A ) = 0

(ii) Optimum solution B : boundary solution; f ( xB ) = 0

(iii) Optimum solutions C , D : non-stationary; f( xC ) < 0 ; f( xD ) < 0

A
B
A
D
  C

x 1 x 1 x 1

Observations:

x (^) 1

is the local maximum for  if one of the following 3 conditions is fulfilled:

( i )   ( x 1

) = 0 and x 1

0 ( point A )

( ii )   ( x 1

) = 0 and x 1

= 0 ( point B )

( iii )   ( x 1

) < 0 and x 1

= 0 ( points C & D )

Mathematically, they can be summarized: f ( x 1

)  0; x 1

≥ 0 and x 1

. f ( x 1

) = 0

This condition is the FONC local maximum with non-negative constraints.

INEQUALITY CONSTRAINTS (3 Variables and 2 Constraints)

Max.  = f ( x 1 , x 2 , x 3 )

S.t. : g

1 ( x 1 , x 2 , x 3 )  r 1

g

2 ( x 1 , x 2 , x 3 )  r 2

x 1 , x 2 , x (^) 3 ≥ 0

 Max.  =  ( x 1 , x 2 , x 3 )

S.t. : g

1 ( x 1 , x 2 , x 3 ) + s 1 = r 1

g

2 ( x 1 , x 2 , x 3 ) + s 2 = r 2

x 1 , x 2 , x 3 , s 1 , s 2 ≥ 0

Ignoring the non-negative constraints:
Lagrangian Function: Z
= f( x 1 , x 2 , x 3 )+ y 1  r 1 – g

1

( x 1 , x 2 , x 3 )– s 1  + y 2  r 2 – g

2

( x 1 , x 2 , x 3 ) - s 2 
FONC: 0

y

Z

y

Z

s

Z

s

Z

x

Z

x

Z

x

Z

1 2 3 1 2 1 2

Due to the following requirements: x j ≥ 0 ; s i ≥ 0 ;
FONC: (i)  0 ; x j ≥ 0 and x j. = 0
(ii)  0 ; s i ≥ 0 and s i. = 0
(iii) = 0 ; i = 1 , 2 and j = 1 , 2 , 3.

j

x

Z

j

x

Z

i

s

Z

i

y

Z

i

s

Z

 The FONC can be expressed as:

( i ) = fj  ( y 1. gj

1 + y 2. gj

2 )  0 ; x j ≥ 0 ; x (^) j. = 0

( ii ) r (^) i  g

i ( x (^) 1 , x (^) 2 , x (^) 3 ) ≥ 0 ; y (^) i ≥ 0 and y (^) i [ r (^) i  g

i ( ) ] = 0

 This condition is also known as K K T condition (a version of K K T ):

Max.  = f

s.t : g

1 ( )  r (^1)

g

2 ( )  r (^2)

x 1 , x (^) 2 , x (^) 3 ≥ 0

j

x
Z

j

x

Z

Another Method to Find the K K T Condition

 Max. f ; st. g

1  r 1 ; g

2  r 2 ; x ≥ 0

 F. L Z = f + y (^) 1 [ r 1  g

1 ] + y 2 [ r 2  g

2 ]

( i ) = fj  ( y 1. gj

1

  • y 2. gj

2 )  0 ; x j ≥ 0 ; x (^) j. = 0

( ii ) = r (^) i  g

i () ≥ 0 y (^) i ≥ 0 ; y (^) i. = 0

This is the K K T condition that must be remembered.

j

x

Z

j

x

Z

i

y

Z

i

y

Z