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DAFFODIL INTERNATIONAL UNIVERSITY JOURNAL OF SCIENCE AND TECHNOLOGY, VOLUME 4, ISSUE 1, JANUARY 2009 28

A COMPARATIVE STUDY OF THE METHODS OF

SOLVING NON-LINEAR PROGRAMMING PROBLEM

Bimal Chandra Das

Department of Textile Engineering

Daffodil International University, Dhaka, BangladeshE-mail: [email protected].

Abstract: The work present in this paper is based

on a comparative study of the methods of solving

Non-linear programming (NLP) problem. We

know that Kuhn-Tucker condition method is an

efficient method of solving Non-linear

programming problem. By using Kuhn-Tucker

conditions the quadratic programming (QP)

problem reduced to form of Linear

programming(LP) problem, so practically simplex

type algorithm can be used to solve the quadratic

programming problem (Wolfe’s Algorithm).Wehave arranged the materials of this paper in

following way. Fist we discuss about non-linear

programming problems. In second step we discuss

Kuhn- Tucker condition method of solving NLP

problems. Finally we compare the solution

obtained by Kuhn- Tucker condition method with

other methods. For problem so consider we use

MATLAB programming to graph the constraints

for obtaining feasible region. Also we plot the

objective functions for determining optimum

points and compare the solution thus obtained

with exact solutions.

Keywords: Non-linear programming, objective function ,convex-region, pivotal element, optimal

solution.

1 IntroductionFor decision making optimization plays thecentral role. Optimization is the synonym of

the word maximization/minimization. It

means choose the best. In our time to takeany decision, we use most modern scientific

methods best on computer implementations.Modern optimization theory based on

computing and we can select the best

alternative value of the objective function.The optimization problems have two major divisions. One is linear programming

problem and other is non-linear programming

problem [1]. But the modern game theory,dynamic programming problem, integer

programming problem also part of the

optimization theory having wide range of

application in modern science, economics

and management. Linear and non-linear

programming problem optimizes an objective

function subject to a class of linear and non-

linear equality or inequality conditions called constraints, usually subject to non-negativity

restrictions of the variables. It was introduced

by [2]. In the present work we tried tocompare some methods of non-linear

programming problem. We know that, for

solving a non-linear programming problemvarious algorithms can be used, but only few

of the methods will be affective for solving

problems. Usually none of the algorithmshave no relations with others and there is no

universal algorithm like simplex method in

linear programming for solving a non-linear

programming problem. In the work we willapply the methods for solving problem rather

than its theoretical descriptions.

2 Non-linear programmingLike linear programming, non-linear

programming is a mathematical technique for determining the optimal solutions to many

business problems. The problem of

maximizing or minimizing a given function

⎪⎭

⎪⎬

⎫

≥≤

=

)b(b) x(g

) x f( Z

iii

Subject to

(1)

is called to general non-linear programming

problem, if the objective function i.e. the

function )(−

x f or any one of the constraints

function )( xgi is non-linear ; or both are

non-linear for −

x nonnegative. In a word, the

non-linear programming problem is that of

choosing nonnegative values of certainvariables, so as to maximize or minimize a

given non-linear function subject to a givenset of linear or non-linear inequality

constraints; or maximize or minimize a linear

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function subject to a given set of non-linear inequality. The problem (2.1) can be re

written as:

Maximize Z= )x.....,,,,( n321 x x x f

Subject to

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

≥≥≥

≥≤

≥≤

≥≤

000

,

..................................................

..................................................

,

,

21

321

223212

113211

n

mmnm

n

n

x , .... , x x

)b(b) x , .... ,x ,x(xg

)b(b) x , .... ,x ,x(xg

)b(b) x , .... ,x ,x(xg

(2)

Any vector −

x satisfying the constraints &

non-negativity restrictions will be called afeasible solution for the problem.

Geometrically, each of the n- no negativity

restrictions ,0≥ j x n j .....3,2,1= defines a

half- space of non-negative values & theintersection of all such half-spaces is the non-

negative orthant, a subset of Euclidian n-space. In E2 the non-negative orthant is the

non-negative first quadrant. Each of the n-

inequality constraints and it was introduced

by [3].

m1,2,....,i ) b( b )( ii-

=≥≤ xgi (3)

also defines a set of points in Euclidian n-

space and the intersection of these m-sets

with non-negative orthant is called asopportunity set i.e.

X= 0; b)g( :E --

n

≥≤∈− x x x (4)So, geometrically a non-linear programming

problem is that of finding a point or a set of points in the opportunity set at which the

highest contour of the objective function is

attained. We know that the optimum solution

of linear programming problem does occur atan extreme point of the convex opportunity

set. However in case of non-linear

programming problem the solution can existat the boundary or in the interior of the

opportunity set. (Weierstrass theorem : If

the non-empty set X is compact (i.e. closed

and bounded ) and objective function F )(−

x is

continuous on X then F )(−

x has a global

maximum either in X in the interior or on the boundary of X2.[5])

3 Kuhn-Tucker Theory

The impetus for generalizations of this theory

is contained in the material presented where

we observed that under certain conditions, a point at which f(x) takes on a relative

maximum for points satisfying

( ) ,,.....,2,1 , mib xgii == is a saddle point

of the Lagrangian function F(x,λ ).The

material presented was originally developed

By H.W. Kuhn and A.W.Tucker [5]. Thetheorem has been of fundamental importancein developing a numerical procedure for

solving quadratic programming problems.

A function F(x,λ ), x-being an n-component

and λ an m-component vector, is said to have

a saddle point at [ ]DD

λ , x if

( ) ( ) ( ) ,,, λ λ λ DDDD

xF xF xF ≤≤ (5)

holds for all x in an ∈-neighborhood of λ°. If

(5) holds for all x and λ , then F(x,λ ) is said

have a saddle point in the large on a global

saddle point at ( )DD

λ , x . We shall find it

convenient to specialize the definition of asaddle point to cases where certain

components of x and λ are restricted to be

non-negative, others are to be non-positive,

and a third category is to be unrestricted insign.

( )

( )

( )

( )6

,.....,1 ,0,

,.....,1 ,0,

,.....,1 ,0,

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

+==∂

∂

+=≥∂

∂

=≤∂

∂

nt j xF x

t s j xF x

s j xF x

j

j

j

DD

DD

DD

λ

λ

λ

;,.....,1

,0 ;,.....,1 ,0

t s j

xs j x j j

+=

≤=≥DD

( )7 ;,.....1 , nt jed unrestrict x j +=D

( ) ( )8;,.....,1 ,0, n j xF x

x j

j ==∂

∂DD

D

λ

( ) ui xF i

,.....,1 ,0, =≥∂

∂DD

λ λ

( ) ( )9 ,.....,1 ,0, vui xF i

+=≤∂

∂DD

λ λ

( ) mvi xF i

,.....,1 ,0, +==∂∂

DDλ

λ

vuiui ii ,...,1 ,0 ;,...,1 ,0 +=≤=≥DD

λ λ

( )10 ,...,1 , mvied unrestrict i +=D

λ

( ) ( )11 .,....,1 ,0, mi xF i

i ==∂

∂DD

D

λ λ

λ

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DAS: A COMPARATIVE STUDY OF THE METHODS OF SOLVING NON-LINEAR PROGRAMMING PROBLEM30

Equations (6) through (12) represent a set of

necessary conditions which [ ]DD

λ , x must

satisfy if ( )λ , xF has a saddle point at

[ ]DD

λ , x for [ ]λ , x ∈ W , provided that

./cF ∈

Now the sufficient conditions can be stated asfollows. Let [ ]

DDλ , x be a point satisfying (6)

through (12). Then if there exists an ∈-

neighborhood about [ ]DD

λ , x such that for

points [ ]D

λ , x ∈ W , in this neighborhood

( ) ( ) ( )( ) ( )13 ,,,DDDDDD

x-x xF xF xF x

λ λ λ ∇+≤

( ) ( ) ( )( ) ( )14 ,,,DDDDDD

λ λ λ λ λ λ - xF xF xF ∇+≥

then ( )λ , xF has a saddle point at [ ]DD

λ , x

for [ ]λ , x ∈ W . If (13) and (14) hold for all

,, 21 W W x ∈∈ λ it follows that has a global

saddle point at [ ]DD λ , x for [ ]λ , x ∈ W .From the above we can deduce Kuhn-Tucker’s conditions as follows:

Let us consider the non-linear programming

problem as:

( )( )

.0 ≥

≤

=

x

b xgtosubject

x f Z Maximize

To obtain Kuhn-Tucker necessary conditions,

let us convert the inequality constrains of the

above problem to equality constraints byadding a vector of m slack (surplus)

variables:

( ) x f Z Max =

( )

( ) ( ).,.....,, 21 m

T sssswhere xgbsor

bs xgtosubject

=−=

=+

Now the Lagrangian function for the problem

takes the form, ( ) ( ) { })(, xgb x f x L −+= λ λ

Assumed that ,, /cg f ∈ first order

necessary conditions can be obtained bytaking first order derivatives with respect to

λ , x of the Lagrangian function

( )

( )( ) .0

0

0

0

=−≡∂

∂∴

≥−≡∂

∂

=⎟ ⎠

⎞⎜⎝

⎛

∂

∂−

∂

∂≡

∂

∂∴

≤∂

∂−∂

∂≡∂

∂

xgbF

xgbF

and

x x

g

x

f x

x

F

xg

x f

xF

λ λ

λ

λ

λ

Rewriting the Lagrangian form we get

( ) ( ) ( )( ) .,.....,1 ,, mi xgb x f xF i

iii=−+≡ ∑ λ λ

Now if ( ) x f takes constraints local optimum

at* x x = it is necessary that a vector

*λ exists

such that

( ) ( )( ) ( )∑ ≤∇−∇≡∇i

i xi x x xg x f xF 0,.1 ***** λ λ

( )[ ] [ ( )

] 00)(

)(,.21

***

=≤∇−

∇≡∇

∗∗∗

∗

=

∑

∑

ji x

i

i

x

n

j

T

x

x xg

x f x xF

λ

λ

( ) ( ) ( ) iiii b xgif xgb xF ≤≥−≡∇ **** 0,.3 λ λ

( ) ( )( ) ( )

iiii

iiii

b xgif xgb

b xgif xgb

==−≡

≥≤−≡**

**

0

0

( )[ ] ( )[ ] .0,.4***** ∑ =−≡∇

i

iii ggb xF λ λ λ λ

4 Comparison of solutions by Kuhn-

Tucker condition and others

In this part we want to show that various NLP problems can be solved by differentmethods. Our aim is to show the affective

ness of the methods considered:

Example: Let us consider the problem

Maximize2

121 32 x x x Z −+=

Subject to the constraints:

0,

42

21

21

≥

≤+

x x

x x

First we want to solve above problem by

graphical solution method.The given problem can be rewriting as:

Maximize Z ( ) ⎟ ⎠

⎞⎜⎝

⎛ ++−−=

3

131 2

2

1 x x

Subject to the constraints

0,

42

21

21

≥

≤+

x x

x x

We observe that our objective function is a

parabola with vertex at (1, -1/3) and

constraints are linear. To solve the problem

graphically, first we constract the graph of the constraint in the first quadrant since

01 ≥ x and 02 ≥ x by considering the

inequation to equation. 42 21 =+ x x

and it was introduced by [6].Each point has co-ordinates of the

tupe ),( 21 x x

and conversely every ordered pair ),( 21 x x or

real numbers determines a point in the plane.

Thus our search for the number pair

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),( 21 x x is restricted to the points of the first

quadrant only

Fig. 1 Optimum solution by graphical method

We get the convex region OAB as

opportunity set. Since our search is for such a pair (x1 ,x2) which gives a maximum value of

2

121 32 x x x −+ and lies in the convex

region. The desire point will be that point of the region at which a side of the convex

region is tangent to the parabola. For this

proceed as follows:

Differentiating the equation of the parabola,we get

( ) 15 3

22

0232

1

1

2

1121

x

dx

dx

dx xdxdx

−−=⇒

=−+

Now differentiating the equation of the

constraint, we get

)16(2102

1

221 −=⇒=+ dx

dxdxdx

Solving equation (15) and (16), we get

4

1 1 = x

8

15xand 2 = .This shows that the

parabola ( ) ⎟ ⎠

⎞⎜⎝

⎛ ++−−=

3

131 2

2

1 x x z has a

tangent to it the line 42 21 =+ x x at the

point .8

15,

4

1⎟ ⎠

⎞⎜⎝

⎛ Hence we get the maximum

value of the objective function at this point.Therefore,

Zmax2

121 32 x x x −+=

.8

15x,

4

1

16

9721 === xat

Let us solve the above problem by using [7]

Kuhn-Tucker Conditions. The Lagrangianfunction of the given problem is

( ) ( ).2432,, 21

2

12121 x x x x x x xF −−+−+≡ λ λ

By Kuhn-Tucker conditions, we obtain

( ) ( ) ( )

( ) ( ) 0.with024F

02322

024F

)(

023F

,022 )(

21

211

2

2

1

1

21

2

1

1

≥=−−≡∂

∂

=−+−−≡∂∂+

∂∂

≥−−≡∂

∂

≤−≡∂

∂≤−−≡

∂

∂

λ λ λ

λ

λ λ

λ

λ λ

x xd

x x x x

F x

x

F xc

x xb

x x

x

F a

Now there arise the following cases:

Case (i) : Let 0=λ , in this case we get

from

0023x

F and 022

2

1

1

≤⋅−≡∂

∂≤−≡

∂

∂ x

x

F

03 ≤⇒ which is impossible and this

solution is to be discarded and it was

introduced by [8].Case (ii): Let 0≠λ . In this case we get

from

(17) 42 024

0)24(

2121

21

=+⇒=−−

=−−

x x x x

x xλ

Also from 022 1

1

≤−−≡∂

∂λ x

x

F

0232

≤−≡∂

∂λ

x

F 022 1 ≥−+∴ λ x

and 2

3032 ≥⇒≥− λ λ

If we take

2

3=λ , then

2

12 1 ≥ x

If we consider 2

12 1 = x then

4

11 = x .

Now putting the value of x1 in (4 .3),

we get8

15 x 2 =

( ) .2

3,

8

15,

4

1,, 21 ⎟

⎠

⎞⎜⎝

⎛ =∴ λ x x for this solution

satisfied 002

3F

satisfied 008

150

4

1x

satisfied 04

15116

8

15

24

1

4

F

satisfied 02

323

F

satisfied 02

314

2

3

4

122

2

2

1

1

4

2

1

=⋅≡∂

∂

=×+×≡∂

∂+

∂

∂

=

−−

=/⋅/−−≡∂

∂

=/

⋅/−≡∂

∂

=−−

=−⋅−≡∂

∂

λ λ

λ

x

F x

x

F

x

x

F

Thus all the Kuhn-Tucker necessary

conditions are satisfied at the point .8

15,

4

1⎟ ⎠

⎞⎜⎝

⎛

0 5 1 0 1 5 2 0

0

5

1 0

1 5

2 0

x 1

x 2

1 7 .

6 .

- 3 .1

- 3 .1g 1

g 2 g 3

O p t i m u m

0

A

B

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Hence the optimum (maximum) solution tothe given NLP problem is

2

121max 32 x x x Z −+=

.8

15,4

1at x 16

97

4

1

8

153

4

12

21

2

===

⎟ ⎠

⎞⎜⎝

⎛ −⋅+⋅=

x

Let us solve the problem by Beale’s method.

Maximize ( ) 2

121 32 x x x x f −+=

Subject to the constraints:

0,

42

21

21

≥

≤+

x x

x x

Introducing a slack variables s, the constraint

becomes

0,

42

21

21

≥

=++

x x

s x x

since there is only one constraint, let s be a basic variable. Thus we have by [9]

( ) ( ) 21 ,x x x ,s x NB B

== with 4=s

Expressing the basic x B and the objectivefunction in terms of non-basic x NB, we have

s =4-x1-2x2 and f =2x1+3x2-x12.

We evaluated the partial derivatives of f

w.r.to non-basic variables at x NB=0, we get

( )

3

202222

0

0

0

2

1

1

=⎟⎟

⎠

⎞⎜⎜

⎝

⎛

∂

∂

=⋅−=−=⎟⎟ ⎠

⎞⎜⎜⎝

⎛

∂

∂

=

=

=

NB

NB

NB

x

x

x

x

f

x x

f

since both the partial derivatives are positive,

the current solution can be improved. As

2 x

f

∂

∂gives the most positive value, x2 will

enter the basis. Now, to determine the leaving basic variable,

we compute the ratios:

203,

24min

,min,min22

20

32

30

=⎪⎭⎪⎬⎫

⎪⎩⎪⎨⎧

−=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

α

γ

α

α

γ

γ

α

α

kk

ko

hk

ho

since the minimum occurs for ,30

30

α

α s will

leave the basis and it was introduced by [10].

Thus expressing the new basic variable, x2 aswell as the objective function f

in terms of the new non-basic variables ( x1 and s) we have:

2

11

2

11

1

12

2

3

26

22232

222

xs x

xs x

x f and

s x x

−−+=

−⎟ ⎠

⎞⎜⎝

⎛ −−+=

−−=

we, again, evaluate the partial derivates of f

w. r. to the non-basic variables:

.2

3

2

12

2

1

0

010

1

1

−=⎟ ⎠

⎞⎜⎝

⎛

∂

∂

=⎟ ⎠

⎞⎜⎝

⎛ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂

∂

=

==

NB

NB

x

x x

s

f

x x

f

since the partial derivatives are not all

negative, the current solution is not optimal,clearly, x1 will enter the basis.

For the next Criterion, we compute the ratios.

4

3

2

2/1,

2/1

2,min

11

10

21

20 =⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

γ

γ

α

α

since the minimum of these ratios correspond

to

11

10

γ

γ , non-basic variables can be

removed. Thus we introduce a free variable,

u1 (unrestricted), as an additional non-basicvariable, defined by

11

1

14

12

2

1

2

1

2

1 x x

x

f u −=⎟

⎠

⎞⎜⎝

⎛ −=

∂

∂=

Note that now the basis has two basic

variables x2 and x1 (just entered). That is, wehave

( ) ( )211 ,, x x xand us x B NB ==

Expressing the basic x B in terms of non-basic

x NB , we have, 11 4

1u x −=

( ) .2

1

2

1

8

154

2

1 1312 su x x xand −+=−−=

The objective function, expressing in terms

of x NB is,

.2

3

16

97

4

1

2

1

2

1

8

153

4

12

2

1

2

111

us

usuu f

−−=

⎟ ⎠

⎞⎜⎝

⎛ −−⎟

⎠

⎞⎜⎝

⎛ −++⎟

⎠

⎞⎜⎝

⎛ −=

Now,2

3

0 0

−=⎟ ⎠

⎞⎜⎝

⎛

∂

∂

==

u NB xs

F ; 02 1

1

00

=−=⎟⎟ ⎠

⎞⎜⎜⎝

⎛

==

uu

f

u NB x

δ

δ

since 01

≤∂

∂

u

f for all x j in x NB and ,0=

∂

∂

u

f

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the current solution is optimal. Hence theoptimal basic feasible solution to the given

problem is:

16

97*,

8

15 ,

4

121 == Z x x

Let us solve the given problem by [3] using

Wolfe’s algorithm2

121 32 x x x Z Max −+=

subject to the constraints

0,

42

21

21

≥

≤+

x x

x x

First we convert the inequality constraints to

equality by introducing slack variables s 0≥

( ) 18 42 21 =++ s x x

Therefore, the Lagrangian function of the

problem is

( ) ( )21

2

121 2432, x x x x x x L −−+−+≡ λ λ

( )

( )4 b ,3

2C

,2

1

A ,2 1,00

01-

DHereT

=⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

==⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

= A

We know by Wolfe’s algorithm:

C V A DX T −=+− λ 2

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +⎟⎟

⎠

⎞⎜⎜⎝

⎛ −⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −⇒

3

2

2

1

0 0

0 12

2

1

2

1

v

v

x

x λ

( )

( )20 32

19 22

32

22

2

11

2

11

=−

=−+⇒

−=+−

−=+−−⇒

v

v x

v

v x

λ

λ

λ

λ

Introducing two artificial variables a1 and a2

in equations (4. 5) and (4. 6), we get

( )

( )22 32

21 22

22

111

=+−

=+−+

av

av x

λ

λ

The quadratic programming problem isequivalent to

( ) 21

21

..

aa Z Maxei

aa Z Min

−−=−

+=

subject to the constraints

0,0,0

0,,,,,,,

32

22

42

2211

212121

22

111

21

===

≥

=+−=+−+

=++

sv xv xwith

aasvv x x

av

av x

s x x

λ

λ

λ

λ

The solution of the problem can be shown in

the following modified simplex tableau: Itwas Introduced by [10]. Here we get three

constraints equations with eight unknown.

So, for basic solution of the system alwaysthree variables will take non-zero values and

others are zero. We will find out non zero

values for x1 , x2 &λ or s In our case initial

basic solution is 04,3,2 21 =⇒=== zsaa .

Our next goal is to improve the value of Z. In

tableau –1, we put constraints system as:

Table 1 Constraints system

x1 x2 λ v1 v2 s a1 a2 c

12

0

2

0

0

0

1

2

0

-1

0

0

0

-1

1

0

0

0

1

0

0

0

1

4

2

3

Now taking x1 as leading variables. Then weget min (4/1, 2/2) is 2/2. So, second element

of the 1st column is the pivotal element and

the corresponding column is the pivotalcolumn. So, x1 enters in basis. Reducing the

pivotal element to unity by dividing allelements of the pivotal row by 2, we get thetable 2.

Table 2 Reducing first pivotal element to unity.

x1 x2 λ v1 v2 s a1 a2 c

11

0

2

0

0

01/2

2

0-1/2

0

0

0

-1

1

0

0

01/2

0

0

0

1

4

1

3

Reducing zero all elements of the pivotal

column except pivotal one, we get the table 3

Table 3 Reducing zero all elements of 1st pivotalcolumn except pivotal one.

x1 x2 λ v1 v2 s a1 a2 c

0

10

2

00

-1/2

1/22

1/2

-1/20

0

0-1

1

00

-1/2

1/20

0

01

3

13

Now taking x2 as our next leading variable.

Then we get 1st element of the second column is our next pivotal element. Reducing

it to unity by dividing all elements of the

pivotal now by 2 and next taking λ as our

next leading variable. Then we get

min 2

3,

2/1

1

⎭⎬⎫

⎩⎨⎧ is

23 . So, 2 is our next pivotal

element. Reducing the pivotal element tounity, we get, the tableau-5.

Table 4 Reducing 3rd pivotal element to unity.

x1 x2 λ v1 v2 s a1 a2 c

0

1

0

1

0

0

-1/4

1/2

1

1/4

-1/2

0

0

0

-1/2

1/2

0

0

-1/4

1/2

0

0

0

1/2

3/2

1

3/2

7/27/2019 4352-16011-1-PB.pdf



Making zero all elements of the pivotalcolumn except pivotal one, we get the table 5.

Table 5 Optimal solution

From T-5 we obtain the optimal solution as

.2

3 ,

8

15 ,

4

121 === λ x x

Thus for the optimal solution for the given

QP problem is

( ) .8

15,

4

1,

16

97

4

1

8

153

4

12

32

21

2

2

121

⎟ ⎠

⎞⎜⎝

⎛ ==

⎟ ⎠

⎞⎜⎝

⎛ −⋅+⋅=

−+=

∗∗ x xat

x x x Z Max

which is same as we obtain by Kuhn-Tuker condition method. For all kinds of non-linear

programming problem, we can show that the

optimal solution by Kuhn-tucker condition issame as any other method we considered

Therefore the solution obtained by graphical

solution method, Kuhn-Tucker conditionsand Wolfe’s method are same.

5 ConclusionTo obtain an optimal solution to the non-linear programming problem, we observe that

Kuhn-Tucker conditions are more useful thanany other methods of solving NLP problem.

Because in a NLP problem particular problems are solve by particular method.

There is no universalism in the methods of solving NLP problem. But we have shown

that by Kuhn-Tucker conditions all kinds of

NLP problems can be solved. The NLP

problem involving two variables can easilysolve by graphical solution method, but the

problem involving more variables cannotsolve by graphical solution method. Besides,

for all NLP problems, graphical solution

method do not gives always-optimal solution.Only quadratic programming problems can

solve by Wolfe’s method.

Therefore, from the above discussion, we can

say that kuhn-Tucker conditions are the bestmethod for solving only NLP problem.

References[1] Greig, D. M: “Optimization”. Lougman- Group

United , New York (1980).[2] Gupta, P. K. Man Mohan: “Linear programming

and theory of Games” Sultan Chand & sons,

New Delhi, India

[3] Honhendalken, B. Von: “SimplicialDecomposition in Non-linear Programming

procedure”.

[4] G. Hadley: “Non-linear and dynamic programming”.

[5] M. S. Bazaraa & C. M. Shetty: “Non-linear

programming theory and algorithms”.[6] Mittal Sethi: “Linear Programming ’’Pragati

Prakashan, Meerut India 1997.[7] Adadie. J: “On the Kuhn-Tucker Theory,” in Non-

Linear Programming , J. Adadie. (Ed) 1967b.

[8] Hildreth. C: “A Quadratic Programming

procedure”.

Naval Research Logistics Quarterly.

[9] D. M.Himmeldlau: Applied Non-linear

programming.

[10] G. R. Walsh: “Methods of optimization” Jon Wiley

and sons ltd , 1975, Rev. 1985.

Bimal Chandra Das has

completed M. Sc in pure

Mathematics and B. Sc (Hons)

in Mathematics from Chittagong

University. Now he is working

as a Lecturer under theDepartment of Textile Engineering in Daffodil

International University. His area of research is

Operation Research.

x1 x2 λ

v1 v2 s a1 a2 c

0

10

1

00

0

01

1/4

-1/20

-1/8

1/4-1/2

1/2

00

-1/4

00

1/8

-1/41/2

15/8

1/43/2

4352-16011-1-pb.pdf

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