ON FUZZY TRANSPORTATION PROBLEM USING TRIANGULAR FUZZY NUMBERS WITH MODIFIED REVISED SIMPLEX METHOD

ON FUZZY TRANSPORTATION

PROBLEM USING TRIANGULAR

FUZZY NUMBERS WITH MODIFIED

REVISED SIMPLEX METHOD

B. Ramesh Kumar1

Department of Mathematics,

Sree sowdambika college of Engineering, Aruppukottai [email protected]

S.Murugesan1

Department of Mathematics, Sri Ramasamy Naidu Memorial College, Sattur

[email protected]

Abstract:

In this paper, we present a Modified Revised Simplex Method for minimizing fuzzy transportation problem of a Triangular Fuzzy Numbers. In which the supplies and demands are triangular fuzzy numbers. A parametric approach is used to obtain a fuzzy basic feasible solution with the help of Revised Simplex Method. We get optimal solution of fuzzy transportation problem in which number of constraints equal to number of occupied cells.

Key Words: Triangular Fuzzy Number, Fuzzy Transportation Problem, Modified Revised Simple Method, Optimal Solution,



-level set

1 Introduction

The Transportation problem refers to a special class of linear programming problems. In such problems product is to be transformed from several sources to numerous location at minimum cost, suppose there are “m” warehouses when a commodity is stocked and “n” markets where it’s needed. Let the supply available in the warehouses be a1,a2... a_mand the demands at the markets be b1,b2...bn respectively. In addition there is a

penalty

C

ij associated with transporting unit if product from the source i to destination j. This penalty may be cost (or) delivery time (or) safety of delivery etc. Various

x

_ij represents the unknown quantity to be shipped from source i to destination j.

The basic transportation problem was originally stated and latter discussed in [8]. A linear programming problem using L-R fuzzy number was given in9. An operator theory of parametric programming for general transportation problem was presented by10. A fuzzy transportation problem (FTP) is a transportation problem in which the transportation costs, supply and demand quantities are fuzzy quantities. The objective of the fuzzy transportation problem is to determine the shipping schedule that minimizes the total fuzzy transportation cost while satisfying fuzzy supply and fuzzy demand limits to deal quantitatively with imprecise information in making decisions, Bellman and Zadeh [3] introduced the notion of fuzziness. In the literature many researchers [6, 7] have developed various algorithms to solve FTP with equality constraints. The aim is to find the optimal transportation cost by Modified Revised Simplex Method.

2 Fuzzy Basics

L.A.Zadeh advanced the fuzzy theory in 1965; the theory proposes mathematical techniques for dealing with the concepts and problems that have much possible solution. In year 1974 concept of mathematical programming on a general level was first proposed by Tanaka et al. in the frame work of fuzzy decision. Now we summarize the definitions and results which are needed in sequal.

2.1 Definition

If X is a collection of objects denoted generically by X then the fuzzy subset

A

of X is defined as a set of ordered pairs

A



(x,A(x)/xX



Where



_A

(

x

)

is called the membership function for the fuzzy set

A

. The membership function maps each element of X to a membership grade (or) membership value between 0 and 1

2.2 Definition

The



cut (or)



- level set of fuzzy set

A

is a set consisting of those elements of the universe X whose membership values exceed the threshold level



that is

A

_





x

/



A

(

x

)







The Height of a fuzzy set is the supremum (maximum) of the membership grades of A. h(A) sup _A(x)

X x





 [h

(A) =1], In other words, there is an X for which



A

(

x

)



1

.Any set is not normal is called subnormal.

2.3 Definition

A real fuzzy number

a

is a fuzzy subset of the set of real numbers R with membership function



_a satisfying the conditions.

(i)



a is continuous from R to the closed interval [0,1] (ii)



_a is strictly increasing and continuous

[

a

a

]

(iii)



_a is strictly decreasing and continuous

[

a

a

]

where

a

,

a

,

a

are real numbers and the fuzzy number denoted by a =

[

a

,

a

,

a

]

is called Triangular fuzzy number.

2.4 Definition

Let

a



[

a

a

a

]

and

b



[

b

b

b

]

be two triangular fuzzy numbers then the arithmetic operations of

a and b are defined as

Addition :

a



b





[

a



b

]

[

a



b

]

[

a



b

]



Subtraction :

a



b





[

a



b

]

[

a



b

]

[

a



b

]



Multiplication:

(i) If a and b are positive:

a



b





[

a

b



b

a

]

[

ab

]

[

a

b



ab

]



(ii) If a or b is negative in any sense

a



[

a



a

a

]

and

b



[

b

b

b

]

a



b





[

b

a



a

b

]

[



ab

]

[

b

a



a

b

]



(iii) If a and b are negative in any sense

a



[

a



a

a

]

and

b



[

b



b

b

]

Example: Let

A



(

2

,

4

,

6

)

and

B



(

1

,

2

,

7

)

are two fuzzy numbers .Then

(i)

A



B



(

3

,

6

,

12

)

(ii)

A



B



(



5

,

2

,

5

)

, (A>B)

2.5 Definition

A New operation for addition and subtraction on Triangular fuzzy number

2.5.1 Subtraction

Let a[a a a] and b[bbb] then

a



b





[

a



b

]

[

a



b

]

[

a



b

]



. The new subtraction operation exists only if the following condition is satisfied

DP

(

a

)



DP

(

b

)

Where

2 2

)

(a a a and a a

DP    , DP (a)

denotes the Difference Point of a Triangular Fuzzy Number.

2.5.2 Addition

Let

a



[

a

a

a

]

and

b



[

b

b

b

]

then

a



b





[

a



b

]

[

a



b

]

[

a



b

]



. The new addition

operation exists only if the following condition is satisfied

MP

(

a

)



MP

(

b

)

,

2 2

)

(a a a and a a

MP    ,

where MP (a) denotes the Midpoint of a Triangular Fuzzy Number.

2.6 Definition

Fuzzy transportation problem formulate as the linear programming problem is



 

m

i n

j ij ij

x

c

1 1

n

j

m

i

,..

2

,

1

,..

2

,

1





, Subject to

0

1 1













 

ij

j m

i ij

i n

j ij

x

b

x

a

x

This is the linear programming in m x n decision variables m+n functional constraints, and m x n non negative constraints.

C

_ij

,

a

_i

,

b

_j are the Triangular Fuzzy Numbers.

3 Modified Revised Simplex Algorithm

In the view of the characteristic of L.P.P and the way of simplex method works, it can be easily being shown that a significant amount of redundant information is generated at each step.

The RSM is an implementation of the standard simplex method that uses an abbreviated table, reconstructing only, that data from the complete simplex table absolutely necessary to perform the steps of the simplex table. The following data is necessary for a simplex table.



 the list of basic variable

x

B



 the inverse matrix

S



B

1



 the RHS values

b



s

b



 the shadow prices

y



C

T_B

S



 the objective function value

f



C

_BT

b



y

_b

 

Step (0): An initial basis inverse 1

B is given with

b



B

1

b



0

. The columns of B are

[

A

j1

,

A

j2

...

A

jm

]

And

1





c

B

y

B is the vector of simplex multipliers.

Step (1): The coefficients of

c

for the nonbasic variables

x

_j are computed by pricing out the

original data

A

_j (ie) _ij

m

i i j j j

j c yA c ya

c





   

1

for j – non basic if all

c

_j



0

,

then stop, we are optimal. If we continue then there exists some

c

j



0

Step (2): Choose the variable to introduce the basis by

c

_s



max



c

_j

/

c

_j



0



Compute

A

S

B

A

s 1





if

A

S



0

then stop, the problem is unbounded if we continuous,

there exists

a

ij



0

for some

i



1

,

2

,..

m

Step (3): Choose the variable to drop from the basis by the minimum ratio

_{min , 0,} 

       

rs r

is i

rs r

a b a

b a

b

Step (4): Using Minimum Ration Value, we get the new basis inverse

B

1, the new RHS vector

b

and the

new vector of simplex multipliers

y



c

B

B

1 by the minimum ratios



0

rs r

a

b

Step (5): The initial basis in step (0) usually is composed of slack variables and artificial variables constituting an identity matrix, so that B = I and B-1 = I, also the more general statement of algorithm is given, since, after a problem has been solved once a good starting feasible basis is generally known, and it’s there fore unnecessary to start with identity basis.

4 Numerical Example

To solve the following fuzzy transportation problem of minimal cost with the initial fuzzy basic feasible solution obtained by Modified Revised Simplex Method whose fuzzy cost and fuzzy requirement table is given below.

Supply

Demand

Since





 



n

j j m

i

i

b

a

1 1

its balanced fuzzy transportation problem and then supply and demand costs are

non symmetric fuzzy triangular numbers. We reduce the above table and then consider the



level set .Fixed the constant value



=0.5, then we get the supply and demand value of the



parameter is

5 . 5 3

, 5 . 5

3a1 a2  1.5b13,4.5b2 5.5,0.5b31.5 Consider the supply value a₁ 4,a₂ 4,b₁2,b₂ 5,b₃ 1

The given problem reduced to fuzzy transportation table is.

4.1 Table

This Transportation problem formulated as L.P.P as following

23 22 21 13 12

11 3 7 2 6 3

2x x x x x x

MinZ      

Subject to

1

5

2

4

4

23 13

22 12

21 11

23 22 21

13 12 11

























x

x

x

x

x

x

x

x

x

x

x

x

An equivalent formulation to derive an optimal solution of the transportation scheme for the above example, apply the Modified Revised Simplex method and convert the inequality constraints into equality constraints.

23 22 21 13 12

11 3 7 2 6 3

2x x x x x x

MinZ      

Subject to

1

5

2

4

4

5 23 13

4 22 12

3 21 11

2 23 22 21

1 13 12 11



































s

x

x

s

x

x

s

x

x

s

x

x

x

s

x

x

x

The above problem can be written as the matrix form 0 _{, 0}

0 1

                  

 

X b X Z

A c

4.2 Table

The initial basic feasible solution is

1



B

Var. in

basis

Solu.

Var(b)

B

0

B

1

B

2

B

3

B

4

B

5 cjzj

x

11

x

12

x

x1312

21

x

x

₂₂

x

₂₃

Z 0 1 0 0 0 0 0 -2 -3 -7 -2 -6 -3

1

s

4 0 1 0 0 0 0 1 1 1 0 0 0

2

s

4 0 0 1 0 0 0 0 0 0 1 1 1

3

s

2 0 0 0 1 0 0 1 0 0 1 0 0

4

s

5 0 0 0 0 1 0 0 1 0 0 1 0

5

s

1 0 0 0 0 0 1 0 0 1 0 0 1

To select Vector corresponding to non-basic variables to enter into the basis











( 2, 3, 7, 2, 6, 3)



max

)( 1

( max

6 .. 2 , 1 , 0 ) (

max

1 1

       

 

     



i st

j j k

k

B in basis the not is colum B

row

i z

c z

c

The Vector

x

₁₃ is selected to enter into the basis, 1

b

b

2

b

3 Supply

1

a

2 3 7 4

2

a

2 6 3 4

                       1 0 0 0 1 7 13 1 1 ) 1 ( 13 B x

y                        1 5 2 4 4 0 ) ( 1 1 1 b B x_B

After having the selected non basis variables

x

₁₃ to enter into the basis. We shall calculate the

minimum ratio is selected the basic variable to leave the basis. ₁

1 1 , 0 5 , 0 2 , 0 4 , 1 4 min 1 13 1         y

xB . The vector

s

₅ is

selected to leave the basis.

The initial basic feasible solution is now updated by replacing the variable

s

5 with thevariable

x

13 in the basis.

4.3 Table

The initial basic feasible solution is

1 

B

Var. in basis Solu. var(b ) 0

B

B

₁

B

₂

B

₃

B

₄

B

₅ c_jz_j

x

₁₁

x

₁₂

s

₅

12

x

21

x

x

₂₂

x

₂₃

Z 1 1 0 0 0 0 1 -2 -3 0 -2 -6 -3

1

s

3 0 1 0 0 0 -1 1 1 0 0 0 0

2

s

3 0 0 1 0 0 -1 0 0 0 1 1 1

3

s

1 0 0 0 1 0 -1 1 0 0 1 0 0

4

s

4 0 0 0 0 1 -1 0 1 0 0 1 0

13

x

1 0 0 0 0 0 1 0 0 1 0 0 1

To select vector corresponding the non-basic variable to enter the basis







2

,

3

,

1

,

2

,

6

,

2



6

max

6

..

2

,

1

,

0

)

(

max



















z

c

z

i

c

_{k k j j}

                       0 1 0 1 0 6 22 1 1 ) 1 (

22 B x

y                        1 3 0 2 2 2 ) ( 1 1 1 b B xB

After having the selected non-basic variable

x

₂₂ to enter into the basis. We shall calculate the

minimum ratio is selected the basic variable to leave the basis ₂ 0 1 , 1 3 , 0 0 , 1 2 , 0 2 min 1 22 1         y

xB _.

The vector

s

2 is selected to leave the basis. The initial basic feasible solution is now updated by

4.4 Table

The initial basic feasible solution is

1



B

Var.

in basis

Solu.

var(b

B

0

B

1

B

2

B

3

B

4

B

5 cjzj

x

11

x

12

s

x512

21

x

s

2

x

23

Z 4 1 0 1 0 0 0 -2 -3 0 -2 0 -3

1

s

3 0 2 -1 0 0 -1 1 1 0 0 0 0

22

x

3 0 0 1 0 0 -1 0 0 0 1 1 1

3

s

1 0 0 -1 1 0 -3 1 0 0 1 0 0

4

s

1 0 0 -1 0 1 0 0 1 0 0 0 0

13

x

1 0 0 -1 0 0 3 0 0 1 0 0 1







2,3,0,1, 1,2



3 max

6 .. 2 , 1 , 0 ) (

max

  

    

z c z i

c_{k k j j}

The Vector

x

₁₂ is selected to enter into the basis, and

       

 

       

 



 

0 1 0 0 2 3

12 1 1 ) 1 ( 12 B x

y

       

 

       

 

  

 

0 2 6 2 2 7

) (

1 1 1

b B

xB

After having the selected non-basic variable

x

12 to enter into the basis. We shall calculate the

minimum ratio is selected the basic variable to leave the basis 1 2 0 , 1

2 , 0

6 , 0 2 , 2 2 min 1 12 1

    



  



y

xB _.

The vector

s

1 is selected to leave the basis.

x

12 be the entering variable.

4.5 Table

The initial basic feasible solution is

1



B

Var.

in basis

Solu.

Var

B

0

B

1

B

2

B

3

B

4

B

5 cjzj

x

11

s

1

s

x512

21

x

s

₂

x

₂₃

Z 3 1 -2/3 4/3 0 0 1/3 -2 0 0 -2 0 -3

12

x

0 0 8/3 -7/3 0 0 -4/3 1 1 0 0 0 0

22

x

0 0 2/3 -1/3 0 0 -4/3 0 0 0 1 1 1

3

s

1 0 2/3 -16/3 16 0 -37/3 1 0 0 1 0 0

4

s

1 0 2/3 -12/3 0 4 -1/3 0 0 0 0 0 0

13





3

8

3

4

,

3

4

,

3

2

,

3

1

,

3

2

,

3

8

max

6

..

2

,

1

,

0

)

(

max

































z

c

z

i

c

_{k k j j}

The Vector

x

11 is selected to enter into the basis, and

                       3 / 2 3 / 2 3 / 50 3 / 2 3 / 8 3 / 8 11 1 1 ) 1 ( 11 B x

y                          3 / 35 3 / 1 3 / 11 3 / 4 3 / 4 3 / 10 ) ( 1 1 1 b B xB

After having the selected non-basic variable

x

₁₁ to enter into the basis. We shall calculate the

minimum ratio is selected the basic variable to leave the basis 11/50 2 35 , 2 1 , 50 11 , 2 4 , 8 4 min 1 11 1          y xB

The vector

s

3 is selected to leave the basis.

x

11 be the entering variable.

4.6 Table

The initial basic feasible solution is





111 211 111 211 , 111 196 , 111 144 , 3 2 , 111 80 , 37 48 max 6 .. 2 , 1 , 0 ) ( max              

z c z i

c_{k k j j}

The Vector

x

₂₃ is selected to enter into the basis, and

                           6 7 . 1 6 . 0 1 . 1 1 . 3 9 . 1 23 1 1 ) 1 (

23 B x

y                           27 . 6 07 . 0 9 . 0 5 . 0 5 . 0 7 . 16 ) ( 1 1 1 _{B b}

x_B

After having the selected non-basic variable

x

₂₃ to enter into the basis. We shall calculate the

minimum ratio is selected the basic variable to leave the basis min



0.16,0.5,1.5,0.14,1.03



0.14 1 23 1   y

xB _.

1 

B

Var. in basis Solution

Var (b)

B

0 1

B B2 B3 B4

B

₅ c_jz_j

s

₃

s

₁

s

₅

12

x

21

x

s

2

x

23

Z 108/37 1 -80/3 196/111 -48/37 0 -2/3 0 0 0 -2 0 -3

12

x

0 0 302/3 -307/111 48/37 0 -1/3 0 1 0 0 0 0

22

x

0 0 80/111 -85/111 48/37 0 -1/3 0 0 0 1 1 1

11

x

1 0 -2/37 16/37 -48/37 0 -1 1 0 0 1 0 0

4

s

1 0 80/111 192/111 48/37 0 0 0 0 0 0 0 0

13

The vector

s

4 is selected to leave the basis. The initial basic feasible solution is now updated by replacing the

variable

s

₄ with the variable

x

₂₃ in the basis.

Table 4.7

The initial basic feasible solution is

Vari.in bsisi

Solu.

vari

B

0

B

1

B

2

B

3

B

4

B

5 cjzj

s

3

s

1

s

5 21

x

2

s

s

₄

z 4.81 1 5.84 10.57 10.5 0 0.67 0 0 0 -2 0 0

12

x 7.79 0 5.86 9.59 10.5 0 -0.33 0 1 0 0 0 0

22

x 7.79 0 5.86 11.59 10.5 0 -0.33 0 0 0 1 1 0

11

x 7.63 0 5.09 12.79 7.92 0 -1 1 0 0 1 0 0

23

x 7.71 0 5.14 12.36 9.21 0 0 0 0 0 0 0 1

13

x 8.79 0 5.86 17.26 10.5 0 5.43 0 0 1 0 0 0









0

)

0

,

57

.

10

,

07

.

19

,

67

.

0

,

84

.

5

,

5

.

10

(

max

6

..

2

,

1

,

0

)

(

max





















k k

j j k

k

z

c

i

z

c

z

c

The current solution is optimal solution

11

x

=7.63,

x

12=7.79,

x

13=8.79,

x

22=7.79,

x

23=7.71, z = 4.81

23 22 21 13 12

11 3 7 2 6 3

2x x x x x x

MinZ      

Min Z=170.03 --- (*)

4.8 RESULTS AND DISCUSSION

Comparing the result that is obtained in Table (4.7), we can easily deduce that the fuzzy transportation problem using



- level set and modified simplex algorithm easily render by optimal value. A decision maker is always concerned with the minimization of the utilization of the resources.

A Comparative evaluation of the solutions of linear programming problem (crisp transportation problem) using modified revised simplex method is obtained as follows: The finial iteration of the solution of linear programming problem using the revised simplex method:

Var.in bsisi

Solu.

vari

B

0

B

1

B

2

B

3

B

4

B

5 cjzj

s

1

s

4

s

5 21

x

2

s

x23

z 10.7 4

1 4.55 -0.82 0 2.33 0.50 0 0 0 -2 0 -3

11

x 7.36 0 4.55 -1.05 0 1.91 -0.77 1 0 0 0 0 0

22

x 7.38 0 4.55 -0.48 0 1.49 1.10 0 0 0 1 1 1

3

s

5.98 0 4.55 -1.28 0 1.49 -1.12 0 0 0 1 0 0

12

x 8.54 0 4.55 -1.22 0 2.33 -0.90 0 1 0 0 0 1

13

x 8.74 0 4.55 -1.28 0 1.47 -1.98 0 0 1 0 0 1









52 . 2

) 59 . 0 , 82 . 0 , 52 . 2 , 5 . 0 , 33 . 2 , 55 . 4 ( max

6 .. 2 , 1 , 0 ) (

max

 

   



     

k k

j j k

k

z c

i z

c z

The Vector

x

₂₁ is selected to enter into the basis, and

       

 

       

 



 

29 . 19

52 . 36

96 . 32

86 . 32

56 . 40

44 . 62

21 1 1 ) 1 ( 21 B x

y

       

 

       

 

     



 

28 . 1

22 . 1

28 . 0

48 . 0

05 . 1

82 . 2

) (

1 1 1 _{B b}

x_B

Since the condition

c

_k



z

_k



0

is not satisfied, the solution is not optimum. Even though

x

₂₁ enters the basis

based on the most negative of all the ratios

min(

)

B B

y

x

are negative. Hence we have an unbounded solution for

the given problem. But, we are using



-level set in fuzzy transportation problem with modified revised simplex method obtained the optimal solution is obtained in (from Table 4.7)

11

x

=7.63,

x

₁₂=7.79,

x

₁₃=8.79,

x

₂₂=7.79,

x

₂₃=7.71, z = 4.81

23 22 21 13 12

11 3 7 2 6 3

2x x x x x x

MinZ      

Min Z=170.03

This method of approach is more efficient in terms of utilization of resources and optimization of the objective function.

4.9 Conclusion:

In this paper, we obtained an optimal solution for fuzzy transportation with triangular membership functions. The new arithmetic operations of triangular fuzzy numbers are employed to get the fuzzy optimal solutions. The optimal solution obtained by using the Modified Revised Simplex method can also be verified in the triangular membership function, which would be a new attempt in solving the transportation problem in fuzzy environment. The same approach for solving the fuzzy problems may also be utilized in future studies in operational research. The authors value great comments and suggestion of referees.

References

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