Solving Two Stage Fully Interval Integer Transportation Problems

(1)

Solving Two Stage Fully Interval Integer

Transportation Problems

V. E Sobana, D. Anuradha, Kaspar S

Abstract: A split and bind method has been proposed in this article to find all the possible solutions for two-stage fully interval integer transportation problem (TSFIITP). It is supported with a numerical example in medication logistics to determine the relevance of the proposed method.

Index Terms: Interval integer transportation problem, Optimal solution, Two-stage interval integer transportation problem. ——————————  ——————————

1. INTRODUCTION

The transportation problem (TP) is one of the best optimization methods that can be applied to different areas of human being activity. TP handles for optimizing the cost of goods shipped from a number of origins to various destinations. Destinations are incapable to obtain the amount in surplus of their smallest requirement, due to storage restrictions in some circumstances. They are prepared to obtain the surplus in the second stage, after they have used up part of the entire first shipment. According to Pandian and Natarajan [15], the manufactured goods shipped to the destination have two stages in such circumstances. Only the sufficient quantity of the manufactured goods is transported in the first stage, so that the smallest demands of the destinations are met and, once this has been done, the remaining quantities in the origins transport to the destinations according to the cost consideration. In these two stages the transportation of the product from the origins to the destinations is carried out in simultaneously. The goal is to minimize the sum of the costs of transportation in the two stages. Many researchers [8, 9, 11, 17, 2] have proposed two-stage time minimization problem and fuzzy two-stage TPs for single and multi-objective problems for solving them. A bi-criteria multi-stage TP was solved by Ellaimony et al. [6] without any transportation restrictions on the intermediate stages by using a heuristic procedure. Pandian and Natarajan [13] obtained an optimal solution for a TP based on the zeropoint method (ZPM). Several effective algorithms have beendeveloped for solving TPs with the hypothesis of accurate supply, demand, and penalty factors. When problems arise in actual life, these situations cannot always be met. To treat with inaccurate coefficients in TPs, various approaches have been used to solve interval programming approach and imprecise data such as [3, 4, 5, 20, 12, 14, 18]. Akilbasha et al. [1] have developed an advanced method for pharmaceutical sciences called the mid-width method for obtaining an optimal solution to fully interval integer TPs. Prabhjot Kaur et al. [16] and Sharma et al. [19]

presented two-stage interval time minimization TPs. In this paper, we proposed split and bind method for finding all possible solutions for the TSFIITP. This algorithm helps decision-making people to select a suitable solution based on their present circumstances and the same is illustrated by a numerical example in medication logistics to determine the relevance of the proposed method.

2 F

ULLY

I

NTERVAL

I

NTEGER

T

RANSPORTATION

P

ROBLEM

(F

IITP

)

Consider a FIITP is shown as below:

(G1) _ _



_  _{ } _

m n

1 2 ij ij

i=1 j=1

Minimize z ,z = c ,dij t ,sij

Subject to

1

, , , 1,2,..., n

ij ij i i j

t s a p i m



   _ _ 

 



(1)

1

, , , 1,2,...,

m

ij ij j j i

t s b q j n



        



(2)

tij0,sij0, for all i and j are integers (3)

Where c_ij and d_ij are positive real numbers for all i and j, a_i

and p_i are positive real numbers for all i and b_jandq_j are

positive real numbers for all j.

Upper and Lower bound ITP of the problem (G1) is given below:

Upper bound ITP (UBITP) of FIITP :

(G3) ₂

1 1

Minimize

m n

ij ij i j

z d s

 





Subject to

1

, 1,2,..., n

ij i j

s p i m



 



1

, 1,2,...,

m

ij j i

s q j n



 



0, and are integers ij

s  i j

The optimal objective of (G3)

Lower bound ITP (LBITP) of FIITP :

(G2) ₁

1 1 Minimize

m n

ij ij i j

z c t

 





Subject to

1

, 1,2,..., n

ij i j

t a i m



 



1

, 1,2,...,

m

ij j i

t b j n



 



0, and are integers ij

t  i j

The optimal objective of ————————————————

 V. E Sobana , Department of Mathematics, Vellore Institute of Technology, Vellore, Tamilnadu.

E-mail: [email protected]

 D. Anuradha, Department of Mathematics, Vellore Institute of Technology, Vellore, Tamilnadu.

E-mail: [email protected]

 Kaspar S , Department of Mathematics, Vellore Institute of Technology, Vellore, Tamilnadu.

(2)

2406

is ₂0 0

1 1

m n

ij ij i j

z d s

 





(4) (G2)is

m n

0 0

1 i=1 j=1

ij ij

z 



c t (5)

The definitions of the arithmetic operators, partial ordering of closed bounded intervals, feasible and optimal solutions of the interval can be found in [7, 10, 14].

3 TWO

STAGE

FULLY

INTERVAL

INTEGER

TRANSPORTATION

PROBLEM

When there is a storage problem at the destinations due to maximum storage conditions the problem (G1) can be converted as TSFIITP.

We define the same as follows

(P1) Minimize _z z₁, ₂   _ _z₃ z z₄, ₅ z₆_

Subject to

3 5

1 1

, ,

m n

ij ij ij ij i j

c e d g z z

 

   _ _

 



4 6

1 1

, ,

m n

ij ij ij ij i j

c f d h z z

 

   _ _

 



1

, , , 1,2,..., n

ij ij i i j

e g a p i m



   _ _ 

 



1

, , , 1,2,...,

m

ij ij j j i

e g l k j n



        



1 1 1

, ,

n n n

ij ij i ij i ij

j j j

f h a e p g

  

 

  _   _

  _ _



1

, , , 1,2,...,

m

ij ij j j j j i

f h b l q k j n



          



e f g hij, ,ij ij, ij0, for all and are integersi j

Where c_ij and d_ij is the unit transportation cost from origin i

to destination j,a_iand p_i are the source at the ith origin, b_j

and q_j are the demand at jth destination,l_j and k_jare the

maximum storage capacity of the jth destinations, e_ij,g_ijand

ij

f ,h_ij are the amount transported from ith origin to jth destinations in first stage and second stage.

Upper and Lower bound two stage ITP of the problem (P1) is given below:

Upper bound two-stage ITP (TSUBITP):

(P3) Minimize z₂ z₅z₆ Subject to

5

1 1

m n

ij ij i j

d g z

 





6

1 1

m n

ij ij i j

d h z

 





Lower bound two-stage ITP (TSLBITP):

(P2) Minimize z₁ z₃z₄ Subject to

3

1 1

m n

ij ij i j

c e z

 





4

1 1

m n

ij ij i j

c f z

 





1

, 1,2,..., n

ij i j

g p i m



 



1

, 1,2,...,

m

ij j i

g k j n



 



1 1

, 1,2,...,

n n

ij i ij

j j

h p g i m

 

  



1

, 1,2,...,

m

ij j j i

h q k j n



  



, 0, and are integers. ij ij

g h  i j

where k_jis the positive real number for all j.

1

, 1,2,..., n

ij i j

e a i m



 



1

, 1,2,...,

m

ij j i

e l j n



 



1 1

, 1,2,...,

n n

ij i ij

j j

f a e i m

 

  



1

, 1,2,...,

m

ij j j i

f b l j n



  



, 0, and are integers. ij ij

e f  i j

where j

l is the positive real number for all j.

4 SPLIT

AND

BIND

METHOD

We now propose a split and bind method for obtaining all the possible solutions to TSFIITP (P1). Pandian and Natarajan have proved the existence of the optimal solutions to the FIITP and two-stage TPs in [14, 15]. In a similar manner, we prove the existence of the optimal solution to the TSFIITP in the following theorem which will be used in the proposed method.

4.1 Theorem

An optimal solution of the problem (P1) = [(P2), (P3)] can be obtained from an optimal solution of the problem (G1) = [(G2), (G3)]. Further the respective optimal objective values of the problem (P1) = [(P2), (P3)] and (G1) = [(G2), (G3)] are the same.

Proof

Let (G1) = [(G2), (G3)] be the given FIITP.

Let



0



, and ij

t i j and



0



, and ij

s i j are the optimal solutions of the problems (G2) and (G3) with objective value

0

v

and

w

0respectively.

This implies{_t0_ij,s , and }0_ij _ i j is an optimal solution of the

problem (G1) with objective value _{v w}0_, 0

 .

Let (P1) = [(P2), (P3)] be the TSFIITP constructed from (G1) = [(G2), (G3)].

In the problem (P1), let us consider the TSLBITP problem (P2).

Let



0



, and ij

e i j and



0



, and ij

f i j be two sets of positive real numbers, obtained from the set



0



, and ij

t i j such that 0

1

, 1,2,..., n

ij i j

e a i m



 



0 1

, 1,2,...,

m

ij j i

e l j n



 



and 0 0 0

, and ij ij ij

f  t e i j .

Let 0 0

3

1 1

m n ij ij i j

z c e

 





, 0 0

4

1 1

m n ij ij i j

z c f

 





and 0 0 0

3 4

v z z . Clearly, 0 0 0 0

3 4

(3)

2407 problem (P2).

Now consider 0 0 0

3 4

v z z

= 0 0

1 1 1 1

m n m n

ij ij ij ij i j i j

c e c f

   





= 0 0

1 1

( )

 





m n

ij ij ij i j

c e f

= 0

1 1

(t ) m n

ij ij i j

c

 



= 0

1

z by

(5).

This implies 0 0

1

v z . (6)

Thus, 0 0

{e_ijf_ij, and }i j is an optimal solution to the problem (P2) with objective value 0 0 0

3 4

v z z .

In a similar way starting with the TSUBITP problem one can

prove that 0 0 0 0

2 5 6

w z z z (7)

Combining equations (6) and (7) we

get 0 0 0 0 0 0

3 4 5 6

[ ,v w ] [ z z z, z ] which proves the statement of theorem.

The split and bind method proceeds as follows:

Step 1: Construct the problem (P3) from the problem (P1). Step 2: Obtain an optimal solution for the problem (P3) according to any transportation algorithm without taking into account the maximum capacity of the destinations. Let { 0

ij

s , and i j} be an optimal solution for the UBITP.

Step 3: Obtain various possible split ups, for each allocated cell column wise taking into account the feasibility of capacity and optimal solutions present in that column.

Step 4: In stage I, consider same possible solution combinations obtained from the earlier split ups say 0

ij g .

Step 5: In stage 2,

a) Modify each supply as ith row supply is equal to

(old supply) - 0

ij i fixed

g

 

 







b) Similarly modify each demand as jth column demand is equal to ( ) ( )q_j  k_j

c) Modify each allotted cell ash_ij0is equal

to(s ) ( )0_ij  g_ij0

Step 6: Repeat the Step 4 and Step 5 to obtain all possible optimal solutions to the problem (P3).

Step 7: Construct the problem (P2) from the problem (P1). Step 8: Obtain an optimal solution for the problem (P2) according to any transportation algorithm without taking into account the maximum capacity of the destinations. Let

{t_ij0, and i j} be an optimal solution for the LBITP

with 0 0 ij ij

t s ,  and i j.

Step 9: Repeat Step 3 to Step 5 to obtain all possible

optimal solutions to the problem (P2).

Step 10: The optimal solution of the problem (P1) is obtained from the optimal solutions of the problem [(P2), (P3)] (by theorem 4.1)

The split and bind method for solving a TSFIITP is shown below using an example.

Consider a medication- manufacturing company consisting three factories and four warehouses. The maximum capacity of the destinations of the medication- manufacturing company

isgivenas[ , ] [5,5]l k₁ ₁  ,[ , ] [2,4]l k₂ ₂  ,[ , ] [10,10]l k₃ ₃  ,

4 4

[ , ] [12,14]l k  respectively. The below table shows the cost of transportation, supplies and demands are in the form of intervals.

Warehouses

W1 W2 W3 W4 Supply

Factory 1 _[2,4] [2,6] [10,18] [8,16] [7,9] Factory 2 _{[1, 2]} [7,10] [2,6] [3,5] [17, 21] Factory 3 _[7,9] [7,11] [3,5] [5,7] [16,18]

Demand _[10,12] [2, 4] [13,15] [15,17]

Now, using step 1 the TSUBITP to the given TSFIITP is given below

Warehouses

W1 W2 W3 W4 Supply

Factory 1 4 6 18 16 9

Factory 2 2 10 6 5 21

Factory 3 9 11 5 7 18

Demand 12 4 15 17

with maximum capacity of the upper bound destinations arek₁5,k₂4, k₃10,k₄14respectively.

Now, as in Step 2, using the ZPM [13], the optimal solution to the UBITP is 0

11 5

s  , 0

12 4

s  , 0

21 7

s  , 0

24 14

s  , 0

33 15

s  , 0

34 3

s  with the total minimum transportation cost as 224.

Now, as in Step 3, we consider the split ups of the above problem is given below

Warehouses

W1 W2 W3 W4 Supply

Factory 1

5(0,1,2,3,4 ,5)

4(4) 9

Factory 2

7(5,4,3,2,1 ,0)

14(14,13,12, 11)

21 Factory

3

15(10 )

3(0,1,2,3) 18 Demand 12(5) 4(4) 15(10

)

17(14)

Now, using Step 4 to Step 6, we obtain all possible optimal solutions with minimum total transportation costs at 224 of the TSUBITP which is shown below.

S.

No Stage-I UB optimal solutions Stage-II UB optimal solutions

1

01

11 0

g  ,g₁₂014,g₂₁015,g₂₄0114,g₃₃0110,g₃₄010

with total transportation cost is 154.

01

11 5

h  ,h₁₂010,h₂₁012,h₂₄010,h₃₃015,h₃₄013

with total transportation cost is 70. 2

02

11 1

g  ,g₁₂024,g₂₁02 4,g₂₄02 14,g₃₃02 10,g₃₄02 0

02

11 4

h  , h₁₂02 0,h₂₁02 3,h₂₄020,h₃₃025,h₃₄023

(4)

2408 3

03

11 2

g  ,g₁₂03 4,g₂₁033,g₂₄0314,g₃₃03 10,g₃₄03 0

03

11 3

h  ,h₁₂030,h₂₁034,h₂₄030,h₃₃035,h₃₄033

04

11 0

g  ,g₁₂04 4,g₂₁04 5,g₂₄0413,g₃₃04 10,g₃₄041

04

11 5

h  ,h₁₂040,h₂₁04 2,h₂₄041,h₃₃045,h₃₄04 2

05

11 1

g  , g₁₂05 4,g₂₁054,g₂₄0513,g₃₃0510,g₃₄05 1

05

11 4

h  ,h₁₂050,h₂₁053,h₂₄051,h₃₃055,h₃₄05 2

06

11 2

g  ,g₁₂06 4,g₂₁06 3,g₂₄0613,g₃₃06 10,g06₃₄ 1

06

11 3

h  ,h₁₂060,h₂₁064,h₂₄061,h₃₃065,h₃₄06 2

07

11 0

g  ,g₁₂07 4,g₂₁07 5,g₂₄0712,g₃₃07 10,g₃₄07 2

07

11 5

h  ,h₁₂07 0,h₂₁07 2,h₂₄07 2,h₃₃07 5,h₃₄071

08

11 1

g  ,g₁₂084,g₂₁08 4,g₂₄08 12,g₃₃08 10,g₃₄082

08

11 4

h  ,h₁₂080,h₂₁083,h₂₄082,h₃₃08 5,h₃₄081

09

11 2

g  ,g₁₂09 4,g₂₁09 3,g₂₄09 12,g₃₃0910,g₃₄092

09

11 3

h  ,h₁₂090,h₂₁094,h₂₄092,h₃₃095,h₃₄09 1

010

11 0

g  ,g₁₂010 4,g₂₁0105,g₂₄01011,g₃₃010 10,g₃₄010 3

010

11 5

h  ,h₁₂0100,h₂₁010 2,h₂₄010 3,h₃₃010 5,h₃₄010 0

011

11 1

g  ,g₁₂0114,g₂₁011 4,g₂₄01111,g₃₃01110,g₃₄0113

011

11 4

h  ,h₁₂0110,h₂₁0113,h₂₄0113,h₃₃011 5,h₃₄011 0

012

11 2

g  ,g₁₂0124,g012₂₁ 3,g₂₄01211,g₃₃01210, 012

34 3

g  with total transportation cost is 164.

012

11 3

h  ,h₁₂0120,h₂₁012 4,h₂₄012 3,h₃₃012 5,h₃₄012 0

013

11 5

g  ,g₁₂013 4,g013₂₁ 0,g₂₄01314,g₃₃01310,g₃₄013 0

013

11 0

h  ,h₁₂013 0,h₂₁013 7,h₂₄013 0,h₃₃013 5,h₃₄0133

014

11 4

g  ,g₁₂0144,g₂₁014 1,g₂₄01414,g₃₃014 10, g₃₄014 0

014

11 1

h  ,h₁₂014 0,h₂₁014 6,h₂₄014 0,h₃₃014 5,h₃₄014 3

015

11 3

g  ,g₁₂015 4,g015₂₁ 2,g₂₄01514,g₃₃01510,g₃₄015 0

015

11 2

h  ,h₁₂015 0,h₂₁015 5,h₂₄015 0,h₃₃0155,h₃₄0153

016

11 5

g  ,g₁₂016 4,g₂₁016 0,g₂₄01613,g₃₃016 10,g₃₄016 1

016

11 0

h  ,h₁₂016 0,h₂₁0167,h₂₄0161,h₃₃016 5,h₃₄016 2

017

11 4

g  ,g₁₂0174,g₂₁017 1,g₂₄01713,g₃₃017 10,

g

₃₄017



1

017

11 1

h  ,h₁₂017 0,h₂₁017 6,h₂₄017 1,h₃₃017 5,h₃₄017 2

018

11 3

g  ,g₁₂018 4,g018₂₁ 2,g₂₄01813,g₃₃01810,g₃₄018 1

018

11 2

h  , h₁₂0180,h₂₁0185,h₂₄0181,h₃₃018 5,h₃₄018 2

019

11 5

g  ,g₁₂019 4,g₂₁019 0,g₂₄01912,g₃₃019 10, 019

34 2

g  with total transportation cost is 168.

019

11 0

h  ,h₁₂019 0,h₂₁0197,h₂₄019 2,h₃₃019 5,h₃₄0191

020

11 4

g  ,g₁₂0204,g₂₁020 1,g₂₄02012,g₃₃020 10, g₃₄020 2

020

11 1

h  ,h₁₂020 0,h₂₁020 6,h₂₄020 2,h₃₃020 5,h₃₄0201

021

11 3

g  ,g₁₂0214,g021₂₁ 2,g₂₄02112,g₃₃02110,g₃₄0212

021

11 2

h  ,h₁₂0210,h₂₁0215,h₂₄0212,h₃₃021 5,h₃₄0211

022

11 5

g  ,g₁₂022 4,g₂₁022 0,g₂₄02211,g₃₃022 10,g₃₄022 3

022

11 0

h  ,h₁₂022 0,h₂₁0227,h₂₄022 3,h₃₃022 5,h₃₄0220

023

11 4

g  ,g₁₂023 4,g023₂₁ 1,g₂₄023 11,g023₃₃ 10,g₃₄023 3

023

11 1

h  ,h₁₂0230,h₂₁023 6,h₂₄023 3,h₃₃023 5,h₃₄023 0

024

11 3

g  ,g₁₂0244,g₂₁024 2,g₂₄02411,g₃₃02410,

g

₃₄024



3

024

11 2

h  ,h₁₂024 0,h₂₁024 5,h₂₄024 3,h₃₃024 5,h₃₄024 0

with total transportation cost is 58. Now, using Step 7 the TSLBITP to the given TSFIITP is given

below

Warehouses

W1 W2 W3 W4 Supply

Factory 1 2 2 10 8 7

Factory 2 1 7 2 3 17

Factory 3 7 7 3 5 16

Demand 10 2 13 15

with maximum capacity of the lower bound destinations are

1 5

l  ,l₂2,l₃10 and l₄12respectively.

Now, as in Step 8, using the ZPM [13], the optimal solution to the LBITP is 0

11 5

t  , 0

12 2

t  , 0

21 5

t  , 0

24 12

t  , 0

33 13

t  and 0

34 3

t  with the total minimum transportation costs as 109.

(5)

2409

S.No Stage-I LB optimal solutions Stage-II LB optimal solutions

1

01

11 0

e  , e₁₂012,e01₂₁5,e₂₄0112,e₃₃0110,e₃₄010

01

11 5

f  , f₁₂010, f₂₁010, f₂₄010, f₃₃01 3, f₃₄013

02

11 1

e  , e₁₂022,e₂₁024,e₂₄0212,e₃₃0210,e₃₄020

02

11 4

f  , f₁₂020, f₂₁021, f₂₄02 0, f₃₃023,f₃₄02 3

03

11 2

e  , e₁₂032,e₂₁033,e₂₄0312,e3303 10, 03

34 0

e  with total transportation cost is 77.

03

11 3

f  , f₁₂030, f₂₁032, f₂₄030, f33033, 03

34 3

f  with total transportation cost is 32

4

04

11 0

e  , e₁₂042,e₂₁045,e₂₄04 11,e₃₃04 10, e₃₄041

04

11 5

f  , f₁₂040, f₂₁040, f₂₄04 1, f₃₃043, f₃₄042

05

11 1

e  , e₁₂052,e₂₁054,e₂₄05 11,e₃₃0510,e₃₄051

05

11 4

f  , f₁₂050, f₂₁051, f₂₄05 1, f₃₃053,f₃₄052

06

11 2

e  , e₁₂062,e₂₁063,e₂₄06 11,e₃₃0610,e₃₄061

06

11 3

f  , f₁₂060, f₂₁062, f₂₄06 1, f₃₃063, f₃₄06 2

07

11 0

e  , e₁₂072,e₂₁075,e₂₄0710,e₃₃0710,e₃₄072

with total transportation cost is 79

07

11 5

f  , f₁₂070, f₂₁07 0, f₂₄07 2, f₃₃07 3, f₃₄071

with total transportation cost is 30 8

08

11 1

e  , e₁₂082,e₂₁084,e₂₄0810,e₃₃0810,e₃₄082

08

11 4

f  , f₁₂080, f₂₁081, f₂₄082, f₃₃083,f₃₄081

09

11 2

e  , e₁₂092,e₂₁093,e₂₄0910,e₃₃0910,e₃₄09 2

09

11 3

f  , f₁₂090, f₂₁092, f₂₄092, f₃₃093, f₃₄091

010

11 0

e  ,e₁₂010 2, 010

21 5

e  ,e₂₄0109,e₃₃01010,e₃₄010 3

010

11 5

f  , f₁₂0100, f₂₁0100, f₂₄010 3, f₃₃0103, f340100

011

11 1

e  , e₁₂0112,e011₂₁ 4,e₂₄0119,e₃₃01110,e₃₄0113

011

11 4

f  , f₁₂0110, f₂₁0111, f₂₄0113,f₃₃0113,f₃₄0110

012

11 2

e  , e₁₂012 2,e012₂₁ 3,e₂₄0129,e33012 10, 012

34 3

012

11 3

f  , f₁₂0120, f₂₁0122,f₂₄012 3, f330123, 012

34 0

f  with total transportation cost is 26.

13

013

11 5

e  , e₁₂013 2,e₂₁0130,e₂₁01312,e₃₃01310,e₃₄013 0

013

11 0

f  , f₁₂0130, f₂₁0135,f₂₄013 0, f₃₃0133, f₃₄0133

014

11 4

e  , e₁₂014 2,e014₂₁ 1,e₂₄01412,e₃₃01410,e₃₄014 0

014

11 1

f  , f₁₂014 0, f₂₁014 4, f₂₄014 0, f₃₃014 3, f₃₄014 3

015

11 3

e  , e₁₂015 2,e₂₁0152,e₂₄01512,e₃₃015 10,e₃₄0150

015

11 2

f  , f₁₂0150, f₂₁0153, f₂₄0150, f₃₃0153, f₃₄0153

016

11 5

e  , e₁₂016 2,e₂₁0160,e₂₄01611,e3301610, 016

34 1

016

11 0

f  , f₁₂0160, f₂₁0165,f₂₄016 1, f33016 3, 016

34 2

17

017

11 4

e  , e₁₂017 2,e017₂₁ 1,e₂₄01711,e₃₃01710,e₃₄017 1

017

11 1

f  , f₁₂017 0, f₂₁0174, f₂₄017 1, f₃₃017 3, f₃₄017 2

018

11 3

e  , e₁₂018 2,e₂₁0182,e₂₄01811,e₃₃01810,e₃₄0181

018

11 2

f  , f₁₂0180, f₂₁0183, f₂₄018 1, f₃₃0183, f₃₄018 2

019

11 5

e  , e₁₂019 2,e₂₁0190,e₂₄01910,e₃₃01910,e₃₄0192

019

11 0

f  , f₁₂0190, f₂₁0195,f₂₄019 2, f₃₃0193, f₃₄0191

020

11 4

e  , e₁₂020 2,e020₂₁ 1,e₂₄02010,e3302010, 020

34 2

020

11 1

f  , f₁₂020 0, f₂₁020 4, f₂₄020 2, f33020 3, 020

34 1

21

021

11 3

e  , e₁₂0212,e₂₁0212,e₂₄02110,e3302110, 021

34 2

021

11 2

f  , f₁₂0210, f₂₁0213, f₂₄0212,f330213, 021

34 1

22

022

11 5

e  , e₁₂022 2,e₂₁0220,e₂₄022 9,e022₃₃ 10,e₃₄022 3

022

11 0

f  , f₁₂0220, f₂₁0225, f₂₄022 3, f₃₃0223,f₃₄022 0

023

11 4

e  , e₁₂023 2,e023₂₁ 1,e₂₄023 9,e₃₃02310,e₃₄023 3

023

11 1

f  , f₁₂023 0, f₂₁0234, f₂₄023 3,f₃₃023 3, f₃₄0230

024

11 3

e  , e₁₂024 2,e₂₁0242,e₂₄0249,e3302410, 024

34 3

024

11 2

f  , f₁₂0240, f₂₁0243,f₂₄024 3, f33024 3, 024

34 0

(6)

2410 S.No Stage-I optimal solutions Stage-II optimal solutions

1 01 01

11 11

[e ,g ][0, 0], [e₁₂01,g₁₂01][2, 4],

01 01

21 21

[e ,g ][5, 5],[e₂₄01,g₂₄01][12,14],

01 01

33 33

[e ,g ][10,10], [e₃₄01,g₃₄01][0, 0]

with total transportation cost is [75,154].

01 01

11 11

[f ,h ][5, 5], [f₁₂01,h₁₂01][0, 0],

01 01

21 21

[f ,h ][0, 2],[f₂₄01,h₂₄01][0, 0],

01 01

33 33

[f ,h ][3, 5], [f₃₄01,h₃₄01][3, 3]

2 02 02

11 11

[e ,g ][1,1], [e₁₂02,g₁₂02][2, 4],

02 02

21 21

[e ,g ][4, 4],[e₂₄02,g₂₄02][12,14],

02 02

33 33

[e ,g ][10,10],[e₃₄02,g₃₄02][0, 0]

02 02

11 11

[f ,h ][4, 4], [f₁₂02,h₁₂02][0, 0],

02 02

21 21

[f ,h ][1, 3],[f₂₄02,h₂₄02][0, 0],

02 02

33 33

[f ,h ][3, 5], [f₃₄02,h₃₄02][3, 3]

3 03 03

11 11

[e ,g ][2, 2], [e₁₂03,g₁₂03][2, 4],

03 03

21 21

[e ,g ][3, 3],[e₂₄03,g₂₄03][12,14],

03 03

33 33

[e ,g ][10,10],[e₃₄03,g₃₄03][0, 0]

03 03

11 11

[f ,h ][3, 3], [f₁₂03,h₁₂03][0, 0],

03 03

21 21

[f ,h ][2, 4],[f₂₄03,h₂₄03][0, 0],

03 03

33 33

[f ,h ][3, 5], [f₃₄03,h₃₄03][3, 3]

4 04 04

11 11

[e ,g ][0, 0], [e₁₂04,g₁₂04][2, 4],

04 04

21 21

[e ,g ][5, 5],[e₂₄04,g₂₄04][11,13],

04 04

33 33

[e ,g ][10,10], [e₃₄04,g₃₄04][1,1]

04 04

11 11

[f ,h ][5, 5], [f₁₂04,h₁₂04] [0, 0] ,

04 04

21 21

[f ,h ] [0, 2] [f2404,h2404] [1,1] ,

04 04

33 33

[f ,h ][3, 5], [f₃₄04,h₃₄04][2, 2] with total transportation cost is [32,68].

5 05 05

11 11

[e ,g ][1,1], [e₁₂05,g₁₂05][2, 4],

05 05

21 21

[e ,g ][4, 4],[e₂₄05,g₂₄05][11,13],

05 05

33 33

[e ,g ][10,10], [e₃₄05,g₃₄05][1,1] with total transportation cost is [78,158].

05 05

11 11

[f ,h ] [4, 4] , [f₁₂05,h₁₂05] [0, 0] ,

05 05

21 21

[f ,h ] [1, 3] ,[f₂₄05,h₂₄05] [1,1] ,

05 05

33 33

[f ,h ][3,5], [f₃₄05,h₃₄05] [2, 2] , with total transportation cost is [31,66].

6 06 06

11 11

[e ,g ] [2, 2] , [e₁₂06,g₁₂06][2, 4],

06 06

21 21

[e ,g ][3, 3],[e₂₄06,g₂₄06] [11,13] ,

06 06

33 33

[e ,g ][10,10], [e₃₄06,g₃₄06] [1,1] with total transportation cost is [79,160].

06 06

11 11

[f ,h ] [3,3] , [f₁₂06,h₁₂06] [0, 0] ,

06 06

21 21

[f ,h ][2, 4] 06 06

24 24

[f ,h ] [1,1] ,

06 06

33 33

[f ,h ][3, 5], [f₃₄06,h₃₄06] [2, 2] with total transportation cost is [30,64].

7 07 07

11 11

[e ,g ][0, 0], [e₁₂07,g₁₂07][2, 4],

07 07 21 21

[e , g ] [5, 5]



,[e₂₄07,g₂₄07][10,12],

07 07

33 33

[e ,g ][10,10],[e₃₄07,g₃₄07][2, 2] with total transportation cost is [79,158].

07 07

11 11

[f ,h ] [5,5] , [f₁₂07,h₁₂07] [0, 0] ,

07 07

21 21

[f ,h ][0, 2] 07 07

24 24

[f ,h ] [2, 2] ,

07 07

33 33

[f ,h ][3, 5], [f₃₄07,h₃₄07] [1,1] with total transportation cost is [30,66].

8 08 08

11 11

[e ,g ][1,1], [e₁₂08,g₁₂08][2, 4],

08 08

21 21

[e ,g ][4, 4],[e₂₄08,g₂₄08] [10,12] ,

08 08

33 33

[e ,g ][10,10],[e₃₄08,g₃₄08] [2, 2] with total transportation cost is [80,160].

08 08

11 11

[f ,h ] [4, 4] , [f₁₂08,h₁₂08] [0, 0] ,

08 08

21 21

[f ,h ] [1, 3] ,[f₂₄08,h₂₄08] [2, 2] ,

08 08

33 33

[f ,h ][3,5], [f₃₄08,h₃₄08] [1,1] with total transportation cost is [29,64].

9 09 09

11 11

[e ,g ] [2, 2] , [e₁₂09,g₁₂09][2, 4],

09 09

21 21

[e ,g ][3, 3],[e₂₄09,g₂₄09][10,12],

09 09

33 33

[e ,g ][10,10],[e₃₄09,g₃₄09] [2, 2] with total transportation cost is [81,162].

09 09

11 11

[f ,h ] [3,3] , [f₁₂09,h₁₂09] [0, 0] ,

09 09

21 21

[f ,h ][2, 4][f2409,h2409] [2, 2] ,

09 09

33 33

[f ,h ][3, 5], [f₃₄09,h₃₄09] [1,1] with total transportation cost is [28,62].

10 010 010

11 11

[e ,g ] [0, 0] , [e₁₂010,g₁₂010][2, 4],

010 010

21 21

[e ,g ][5,5][e24010,g24010] [9,11] ,

010 010

33 33

[e ,g ][10,10],[e₃₄010,g₃₄010] [3, 3]with total transportation cost is [81,160].

010 010

11 11

[f ,h ] [5, 5] , [f₁₂010,h₁₂010][0, 0],

010 010

21 21

[f ,h ] [0, 2],[f₂₄010,h₂₄010] [3, 3] ,

010 010

33 33

[f ,h ][3, 5], [f₃₄010, 010

34 ] [0, 0]

h 

11 011 011

11 11

[e ,g ][1,1], [e₁₂011,g₁₂011] [2, 4] ,

011 011

21 21

[e ,g ] [4, 4] 011 011

24 24

[e ,g ] [9,11] ,

011 011

33 33

[e ,g ][10,10], [e₃₄011,g₃₄011][3, 3] with total transportation cost is [82,162].

011 011

11 11

[f ,h ] [4, 4] , [f₁₂011,h₁₂011] [0, 0] ,

011 011

21 21

[f ,h ][1,3],[f₂₄011,h₂₄011] [3,3] ,

011 011

33 33

[f ,h ] [3,5] , [f₃₄011, h34011][0, 0]

(7)

2411

12 012 012

11 11

[e ,g ] [2, 2] , [e₁₂012,g₁₂012][2, 4],

012 012

21 21

[e ,g ] [3,3] [e24012,g24012] [9,11] ,

012 012

33 33

[e ,g ] [6,10] , [e₃₄012,g₃₄012] [3,3] with total transportation cost is [83,164].

012 012

11 11

[f ,h ] [3, 3] , [f₁₂012,h₁₂012][0, 0],

012 012

21 21

[f ,h ] [2, 4],[f₂₄012,h₂₄012][3, 3],

012 012

33 33

[f ,h ][3, 5], [f₃₄012, 012

34 ] [0, 0]

h 

13 013 013

11 11

[e ,g ] [5,5] , [e₁₂013,g₁₂013] [2, 4] ,

013 013

21 21

[e ,g ][0, 0][e24013,g24013] [12,14] ,

013 013

33 33

[e ,g ][10,10], [e₃₄013,g₃₄013] [0, 0] with total transportation cost is [80,164].

013 013

11 11

[f ,h ][0, 0], [f₁₂013,h₁₂013] [0, 0] ,

013 013

21 21

[f ,h ][5,7],[f₂₄013,h₂₄013] [0, 0] ,

013 013

33 33

[f ,h ][3, 5], [f₃₄013, 013

34 ] [3, 3]

h 

14 014 014

11 11

[e ,g ] [4, 4] , [e₁₂014,g₁₂014][2, 4],

014 014

21 21

[e ,g ][1,1] 014 014

24 24

[e ,g ] [12,14] ,

014 014

33 33

014 014

11 11

[f ,h ][1,1], [f₁₂014,h₁₂014][0, 0],

014 014

21 21

[f ,h ] [4,6],[f₂₄014,h₂₄014] [0, 0] ,

014 014

33 33

[f ,h ][3, 5], [f₃₄014, 014

34 ] [3, 3]

h 

15 015 015

11 11

[e ,g ][3,3], [e₁₂015,g₁₂015][2, 4],

015 015

21 21

[e ,g ][2, 2][e24015,g24015] [12,14] ,

015 015

33 33

[e ,g ] [10,10] , [e₃₄015,g₃₄015] [0, 0] with total transportation cost is [78,160].

015 015

11 11

[f ,h ] [2, 2] , [f₁₂014,h₁₂014][0, 0],

014 014

21 21

[f ,h ] [3,5],[f₂₄014,h₂₄014] [0, 0] ,

014 014

33 33

[f ,h ][3, 5], [f₃₄014, 014

34 ] [3, 3]

h 

16 016 016

11 11

[e ,g ] [5,5] , [e₁₂016,g₁₂016][2, 4],

016 016

21 21

[e ,g ][0, 0][e24016,g24016] [11,13] ,

016 016

33 33

016 016

11 11

[f ,h ][0, 0], [f₁₂016,h₁₂016][0, 0],

016 016

21 21

[f ,h ] [5,7],[f₂₄016,h₂₄016][1,1],

016 016

33 33

[f ,h ][3, 5], [f₃₄016, 016

34 ] [2, 2]

h 

17 017 017

11 11

[e ,g ] [4, 4] , [e₁₂017,g₁₂017][2, 4],

017 017

21 21

[e ,g ][1,1] 017 017

24 24

[e ,g ] [11,13] ,

017 017

33 33

017 017

11 11

[f ,h ][1,1], [f₁₂017,h₁₂017][0, 0],

017 017

21 21

[f ,h ][4,6],[f₂₄017,h₂₄017][1,1],

017 017

33 33

[f ,h ][3, 5], [f₃₄017, h₃₄017][2, 2]

18 018 018

11 11

[e ,g ][3,3], [e₁₂018,g₁₂018][2, 4],

018 018

21 21

[e ,g ][2, 2] 018 018

24 24

[e ,g ] [11,13] ,

018 018

33 33

018 018

11 11

[f ,h ] [2, 2] , [f₁₂018,h₁₂018][0, 0],

018 018

21 21

[f ,h ][3,5],[f₂₄018,h₂₄018] [1,1] ,

018 018

33 33

[f ,h ][3, 5], [f₃₄018, 018

34 ] [2, 2]

h 

19 019 019

11 11

[e ,g ] [5,5] , [e₁₂019,g₁₂019][2, 4],

019 019

21 21

[e ,g ][0, 0][e24019,g24019] [10,12] ,

019 019

33 33

019 019

11 11

[f ,h ][0, 0], [f₁₂019,h₁₂019][0, 0],

019 019

21 21

[f ,h ] [5,7],[f₂₄019,h₂₄019] [2, 2] ,

019 019

33 33

[f ,h ][3, 5], [f₃₄019, 019

34 ] [1,1]

h 

20 020 020

11 11

[e ,g ] [4, 4] , [e₁₂020,g₁₂020][2, 4],

020 020

21 21

[e ,g ][1,1][e24020,g24020] [10,12] ,

020 020

33 33

020 020

11 11

[f ,h ][1,1], [f₁₂020,h₁₂020][0, 0],

020 020

21 21

[f ,h ] [4,6],[f₂₄020,h₂₄020] [2, 2] ,

020 020

33 33

[f ,h ][3, 5], [f₃₄020, 020

34 ] [1,1]

h 

21 021 021

11 11

[e ,g ] [3,3] , [e₁₂021,g₁₂021] [2, 4] ,

021 021

21 21

[e ,g ][2, 2] 021 021

24 24

[e ,g ] [10,12] ,

021 021

33 33

021 021

11 11

[f ,h ] [2, 2] , [f₁₂021,h₁₂021] [0, 0] ,

021 021

21 21

[f ,h ] [3,5],[f₂₄021,h₂₄021] [2, 2] ,

021 021

33 33

[f ,h ] [3,5] , [f₃₄021, h34021][1,1]

22 022 022

11 11

[e ,g ] [5,5] , [e₁₂022,g₁₂022][2, 4],

022 022

21 21

[e ,g ] [0, 0] [e24022,g24022] [9,11] ,

022 022

33 33

022 022

11 11

[f ,h ][0, 0], [f₁₂022,h₁₂022][0, 0],

022 022

21 21

[f ,h ] [5,7],[f₂₄022,h₂₄022][3, 3],

022 022

33 33

[f ,h ][3, 5], [f₃₄022, 022

34 ] [0, 0]

h 

23 023 023

11 11

[e ,g ][4, 4], [e₁₂023,g₁₂023][2, 4],

023 023

21 21

[e ,g ][1,1],[e₂₄023,g023₂₄ ] [9,11] ,

023 023

11 11

[f ,h ] [1,1] , [f₁₂023,h₁₂023] [0, 0] ,

023 023

21 21

(8)

2412

023 023

33 33

023 023

33 33

[f ,h ][3, 5], [f₃₄023, 023

34 ] [0, 0]

h 

24 024 024

11 11

[e ,g ][3,3],[e₁₂024,g₁₂024][2, 4],

024 024

21 21

[e ,g ][2, 2] 024 024

24 24

[e ,g ] [9,11] ,

024 024

33 33

[e ,g ] [6,10] , [e₃₄024,g₃₄024] [3,3] with total transportation cost is [84,166].

024 024

11 11

[f ,h ] [2, 2] , [f₁₂024,h₁₂024][0, 0],

024 024

21 21

[f ,h ] [3,5],[f₂₄024,h₂₄024][3, 3],

024 024

33 33

[f ,h ][3, 5], [f₃₄024, h₃₄024][0, 0]

5 CONCLUSION

In this paper, we presented a split and bind method for TSFIITP. The proposed method provides all the possible solutions and thus it can be served as an essential tool for the decision-making persons to select a suitable solution based on their present circumstances in medication problems of logistics.

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