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D Nagesh Kumar, IISc Optimization Methods: M2L51
Optimization using Calculus
Kuhn-Tucker Conditions
D Nagesh Kumar, IISc Optimization Methods: M2L52
Introduction
Optimization with multiple decision variables and equality
constraint : Lagrange Multipliers.
Optimization with multiple decision variables and inequality
constraint : Kuhn-Tucker (KT) conditions
KT condition: Both necessary and sufficient if the objective
function is concave and each constraint is linear or each constraint
function is concave, i.e. the problems belongs to a class called the
convex programming problems.
D Nagesh Kumar, IISc Optimization Methods: M2L53
Kuhn Tucker Conditions: Optimization Model
Consider the following optimization problem
Minimize f(X)
subject to
gj(X) 0 for j=1,2,…,p
where the decision variable vector
X=[x1,x2,…,xn]
D Nagesh Kumar, IISc Optimization Methods: M2L54
Kuhn Tucker Conditions
1
0 1, 2,...,
0 1, 2,...,
0 1, 2,...,
0 1, 2,...,
m
jji i
j j
j
j
f gi n
x x
g j m
g j m
j m
Kuhn-Tucker conditions for X* = [x1* x2
* . . . xn*] to be a local minimum are
D Nagesh Kumar, IISc Optimization Methods: M2L55
Kuhn Tucker Conditions …contd.
j
In case of minimization problems, if the constraints are
of the form gj(X) 0, then have to be non-positive
On the other hand, if the problem is one of maximization
with the constraints in the form gj(X) 0, then have to
be nonnegative.
j
D Nagesh Kumar, IISc Optimization Methods: M2L56
Example (1)
Minimize
subject to
2 2 21 2 32 3f x x x
1 1 2 3
2 1 2 3
2 12
2 3 8
g x x x
g x x x
D Nagesh Kumar, IISc Optimization Methods: M2L57
Example (1) …contd.
1 21 2 0
i i i
g gf
x x x
1 1 2
2 1 2
3 1 2
2 0 (2)
4 2 0 (3)
6 2 3 0 (4)
x
x
x
0 j jg 1 1 2 3
2 1 2 3
( 2 12) 0 (5)
( 2 3 8) 0 (6)
x x x
x x x
0 jg 1 2 3
1 2 3
2 12 0 (7)
2 3 8 0 (8)
x x x
x x x
0j 1
2
0 (9)
0 (10)
Kuhn – Tucker Conditions
D Nagesh Kumar, IISc Optimization Methods: M2L58
Example (1) …contd.
From (5) either = 0 or ,
Case 1
From (2), (3) and (4) we have x1 = x2 = and x3 =
Using these in (6) we get
From (10), , therefore, =0,
Therefore, X* = [ 0, 0, 0 ]
This solution set satisfies all of (6) to (9)
1 1 2 32 12 0x x x
2 / 2 2 / 222 2 28 0, 0 8or
2 0 2
D Nagesh Kumar, IISc Optimization Methods: M2L59
Example (1) …contd.
Case 2:
Using (2), (3) and (4), we have
or
But conditions (9) and (10) give us and simultaneously,
which cannot be possible with
Hence the solution set for this optimization problem is X* = [ 0 0 0 ]
1 2 32 12 0x x x 1 2 1 2 1 22 2 3
12 02 4 3
1 217 12 144
1 0 2 0
1 217 12 144
D Nagesh Kumar, IISc Optimization Methods: M2L510
Example (2)
Minimize
subject to
2 21 2 160f x x x
1 1
2 1 2
80 0
120 0
g x
g x x
D Nagesh Kumar, IISc Optimization Methods: M2L511
Example (2) …contd.
1 21 2 0
i i i
g gf
x x x
0 j jg
0 jg
0j
Kuhn – Tucker Conditions
1 1 2
2 2
2 60 0 (11)
2 0 (12)
x
x
1 1
2 1 2
( 80) 0 (13)
( 120) 0 (14)
x
x x
1
1 2
80 0 (15)
120 0 (16)
x
x x
1
2
0 (17)
0 (18)
D Nagesh Kumar, IISc Optimization Methods: M2L512
Example (2) …contd.
From (13) either = 0 or ,
Case 1
From (11) and (12) we have and
Using these in (14) we get
Considering , X* = [ 30, 0]. But this solution set violates (15)
and (16)
For , X* = [ 45, 75]. But this solution set violates (15)
1 1( 80) 0x
21 302x 2
2 2x
2 2 150 0
2 0 150or
2 0
2 150
D Nagesh Kumar, IISc Optimization Methods: M2L513
Example (2) …contd.
Case 2:
Using in (11) and (12), we have
Substitute (19) in (14), we have
For this to be true, either
1( 80) 0x
1 80x
2 2
1 2
2
2 220 (19)
x
x
2 22 40 0x x
2 20 40 0x or x
D Nagesh Kumar, IISc Optimization Methods: M2L514
Example (2) …contd.
For ,
This solution set violates (15) and (16)
For ,
This solution set is satisfying all equations from (15) to (19) and
hence
the desired
Thus, the solution set for this optimization problem is X* = [ 80 40 ].
2 0x 1 220
2 40 0x 1 2140 80and
D Nagesh Kumar, IISc Optimization Methods: M2L515
Thank you
