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 ≤

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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