Kuhn Tucker Condition
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Optimization Methods: Optimization using Calculus – Kuhn-Tucker Conditions 1
Module - 2 Lecture Notes – 5
Kuhn-Tucker Conditions
Introduction
In the previous lecture the optimization of functions of multiple variables subjected to
equality constraints using the method of constrained variation and the method of Lagrange
multipliers was dealt. In this lecture the Kuhn-Tucker conditions will be discussed with
examples for a point to be a local optimum in case of a function subject to inequality
constraints.
Kuhn-Tucker Conditions
It was previously established that for both an unconstrained optimization problem and an
optimization problem with an equality constraint the first-order conditions are sufficient for a
global optimum when the objective and constraint functions satisfy appropriate
concavity/convexity conditions. The same is true for an optimization problem with inequality
constraints.
The Kuhn-Tucker conditions are 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
belong to a class called the convex programming problems.
Consider the following optimization problem:
Minimize f(X) subject to gj(X) ≤ 0 for j = 1,2,…,p ; where X = [x1 x2 . . . xn]
Then the Kuhn-Tucker conditions for X* = [x1* x2
* . . . xn*] to be a local minimum are
10 1, 2,...,
0 1, 2,...,
0 1, 2,...,
0 1, 2,...,
m
jji i
j j
j
j
f g i nx x
g j
g j
m
m
j m
λ
λ
λ
=
∂ ∂+ = =
∂ ∂
= =
≤ =
≥ =
∑
(1)
D Nagesh Kumar, IISc, Bangalore
M2L5
Optimization Methods: Optimization using Calculus – Kuhn-Tucker Conditions 2
In case of minimization problems, if the constraints are of the form gj(X) 0, then ≥ jλ have
to be nonpositive in (1). On the other hand, if the problem is one of maximization with the
constraints in the form gj(X) ≥ 0, then jλ have to be nonnegative.
It may be noted that sign convention has to be strictly followed for the Kuhn-Tucker
conditions to be applicable.
Example 1
Minimize 2 21 22 3 2
3f x x x= + + subject to the constraints
1 1 2 3
2 1 2 3
2 12 3
g x x xg x x x= − − ≤= + − ≤
28
using Kuhn-Tucker conditions.
Solution:
The Kuhn-Tucker conditions are given by
a) 1 21 2 0
i i i
g gfx x x
λ λ∂ ∂∂+ + =
∂ ∂ ∂
i.e.
1 1 2
2 1 2
3 1 2
2 0 (2) 4 2 0 (3)6 2 3 0
xx
x
λ λλ λλ λ
+ + =− + =− − = (4)
b) 0 j jgλ =
i.e.
1 1 2 3
2 1 2 3
( 2 12) 0 (5)( 2 3 8) 0 (6)x x xx x x
λλ
− − − =+ − − =
c) 0 jg ≤
D Nagesh Kumar, IISc, Bangalore
M2L5
Optimization Methods: Optimization using Calculus – Kuhn-Tucker Conditions 3
i.e.,
1 2 3
1 2 3
2 12 0 (7)2 3 8 0 (8)
x x xx x x− − − ≤+ − − ≤
d) 0jλ ≥
i.e.,
1
2
0 (9)0 (10)
λλ≥≥
From (5) either 1λ = 0 or, 1 2 32 12x x x− − − = 0
Case 1: 1λ = 0
From (2), (3) and (4) we have x1 = x2 = 2 / 2λ− and x3 = 2 / 2λ .
Using these in (6) we get 22 2 28 0, 0 orλ λ λ+ = ∴ = −8
From (10), 2 0 λ ≥ , therefore, 2λ =0, X* = [ 0, 0, 0 ], this solution set satisfies all of (6) to (9)
Case 2: 1 2 32 12 0x x x− − − =
Using (2), (3) and (4), we have 1 2 1 2 1 22 2 3 12 02 4 3
λ λ λ λ λ λ− − − +− − − = or,
1 217 12 144λ λ+ = − . But conditions (9) and (10) give us 1 0λ ≥ and 2 0λ ≥ simultaneously,
which cannot be possible with 1 217 12 144λ λ+ = − .
Hence the solution set for this optimization problem is X* = [ 0 0 0 ]
D Nagesh Kumar, IISc, Bangalore
M2L5
Optimization Methods: Optimization using Calculus – Kuhn-Tucker Conditions 4
Example 2
Minimize 2 21 2 60 1f x x x= + + subject to the constraints
1 1
2 1 2
80 0120 0
g xg x x= − ≥= + − ≥
using Kuhn-Tucker conditions.
Solution
The Kuhn-Tucker conditions are given by
a) 31 21 2 3 0
i i i i
gg gfx x x x
λ λ λ ∂∂ ∂∂+ + + =
∂ ∂ ∂ ∂
i.e.
1 1 2
2 2
2 60 0 (11)2 0 (12)
xx
λ λλ
+ + + =+ =
b) 0 j jgλ =
i.e.
1 1
2 1 2( 120) 0 (14)x x( 80) 0 (13)xλ
λ + − =− =
80 0 (15)x − ≥
c) 0 jg ≤
i.e.,
1
1 2 120 0 (16)x x+ + ≥
d) 0jλ ≤
i.e.,
D Nagesh Kumar, IISc, Bangalore
M2L5
Optimization Methods: Optimization using Calculus – Kuhn-Tucker Conditions 5
1
2
0 (17)0 (18)
λλ≤≤
From (13) either 1λ = 0 or, 1( 80) 0x − =
Case 1: 1λ = 0
From (11) and (12) we have 21 302x λ= − − and 2
2 2x λ= −
Using these in (14) we get ; ( )2 2 150 0λ λ − = 2 0 150orλ∴ = −
Considering 2 0 λ = , X* = [ 30, 0].
But this solution set violates (15) and (16)
For 2 150 λ = − , X* = [ 45, 75].
But this solution set violates (15) .
Case 2: 1 80) 0x − =(
Using in (11) and (12), we have 1 80x =
2 2
1 2
22 220 (19)
xx
λλ
= −= −
Substitute (19) in (14), we have
( )2 22 40x x− − 0=
0=
.
For this to be true, either 2 20 40x or x= −
For , 2 0x = 1 220λ = − . This solution set violates (15) and (16)
D Nagesh Kumar, IISc, Bangalore
M2L5
Optimization Methods: Optimization using Calculus – Kuhn-Tucker Conditions 6
For ,2 40 0x − = 1 2140 80andλ λ= − = − . This solution set is satisfying all equations from
(15) to (19) and hence the desired. Therefore, the solution set for this optimization problem
is X* = [ 80 40 ].
References / Further Reading:
1. Rao S.S., Engineering Optimization – Theory and Practice, Third Edition, New Age
International Limited, New Delhi, 2000.
2. Ravindran A., D.T. Phillips and J.J. Solberg, Operations Research – Principles and
Practice, John Wiley & Sons, New York, 2001.
3. Taha H.A., Operations Research – An Introduction, Prentice-Hall of India Pvt. Ltd., New
Delhi, 2005.
4. Vedula S., and P.P. Mujumdar, Water Resources Systems: Modelling Techniques and
Analysis, Tata McGraw Hill, New Delhi, 2005.
D Nagesh Kumar, IISc, Bangalore
M2L5