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Optimization With Inequality Constraints
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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
2
2
a circle with its center at ( 4 , 4 )
2 x 1 + 3 x 2 = 6
3 x 1 + 2 x 2 = 12
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
= – 4
3 x 1 + 2 x 2 = 12
6 x 1
= – 12 ; 4 x 1
= – 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:
Example (2)
Min C = ( x (^) 1 – 4 )
2
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
Max = 2 x 1 + x 2
S.t. : x (^) 1
2
2 x 1 + 3 x 2 12
x 1 , x 2 0
x (^2)
4 -x (^1)
**2
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
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
1
2
y
y
s
s
x
x
x
1 2 3 1 2 1 2
j
x
j
x
i
s
i
y
i
s
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
j
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
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
