Variant Multi Product Transhipment Model · Variant Multi Product Transhipment Model Sreenuvasalu Y1, Suresh BabuC2, Madhu Mohan Reddy P3, Sundara Murthy M4 1 Asst. Professor, Dept.

© 2016, IJARCSSE All Rights Reserved Page | 116

Volume 6, Issue 7, July 2016 ISSN: 2277 128X

International Journal of Advanced Research in Computer Science and Software Engineering Research Paper Available online at: www.ijarcsse.com

Variant Multi Product Transhipment Model Sreenuvasalu Y

1, Suresh BabuC

2, Madhu Mohan Reddy P

3, Sundara Murthy M

4

1 Asst. Professor, Dept. of Mathematics, Narayana Engineering College, Nellore, India

2, 3 Asst. Professor, Dept. of Mathematics, Siddharth Institute of Engineering and Technology, Puttur, India

4 Professors (Rtd.) Dept. of Mathematics, SV University, Tirupati, India

Abstract: ‘Transportation’ the term which plays a vital role in shaping the nation’s destiny though it’s a common

phenomenon. In which one of the important aspects is bulk transportation. The goods transported from source to

destination there be some hauling place is more natural. So, it drives me ahead to study on bulk transshipment

models. In this paper, we take some sources and destinations. Sources and destinations maintain some multiple

products. Each destination should get its own multiple requirements from unique source only. The distance/cost

between sources and destinations are to be known. All the sources and destinations need not be performed that means

some of the given sources and destinations should satisfied by the transshipment schedule. The objective is to find the

minimum cost/distance subject to the given constraints.

Keywords: Mathematical Formulation,Lexi-Search Algorithm, Pattern recognition technique, Combinatorial

structure. multipleproducts & optimal solution.

I. INTRODUCTION

There are various types of transportation models and the simplest of them was developed by[1, 3]. Several

extensions of transportation models and methods have been subsequently developed. In these models, the given cost of

transportation is for a unit quantity of item or commodity or product. But in Bulk Transportation Problem, the bulk

transportation cost (independent of quantity transported) from a source i to destination j is taken into consideration.The

problem of Bulk Transportation was first investigated by [4] who presented an algorithm. Later [12] also offered an

algorithm and [15] give an algorithms.Both the algorithms are based on Branch and Bound procedure to solve the Bulk

transportation problems.

A set of logic problems to solve multi-index transportation problems by [2] also conducted a detailed investigation

regarding the characteristics of multi-index transportation problem model. [9]used multi-index transportation problem

model to solve the shipping scheduling suggested that the employment of such transportation efficiency but also optimize

the integral system. These references are only a single objective model and its constraints are not fuzzy member. In the

case of cost-time trade- off bulk transportation problem [7] studied multi objective models and presented efficient

heuristic for this multi objective bulk transportation problem.

Furthermore [5, 10, 11, 13] have studied a variety of the problems.These investigations come under single

commodity bulk transportation problems. [8]Presenteda Multi-Product Bulk Transportation Problem to minimize the

total cost of the bulk transportation.[16]Presented three dimensional time minimization bulk transportation problem to

minimize the total time of goods transportation.A more generalized variation of the bulk TP problems studied by [14]

called A Variant Bulk Transportation Problem with Multiple Bulk Cost Constraint, where the products are supplied

simultaneously to the destinations according to their requirement from the sources. [6] Studied two stage multi objective

variant bulk transportation problem.Both of them presented Lexi Search algorithms and mentioned that the efficiency of

this algorithm over branch and bound algorithm.

II. PROBLEM DESCRIPTION

Let there be m sources with capacity S(i,1) and S(i,2) of products k=2 in ith

source . Let there be n destinations

cities with requirement of productsDR (i, 1) and DR (i, 2). Let there be r cities DT= {𝛼1, 𝛼2 ,−−−−𝛼𝑟} ⊂ 𝐷𝑅 which are

subset of destinations. Each of the destination cities in DT can supply to another rdestination cities, which is called as

transshipment. Let C(i,j) is the bulk supply cost from ith source to j

th destination for items (k=1,2) . Each destination

should get its supply of k products from one source only subject to availability. In this case of transshipment also this

condition that the product of the destinations should be from the same source only. In this transshipment model we do

not consider supply of the products from destinations to sources and sources to sourcesas they are not natural. The

objective of the problem is to find supply schedule for supplying the requirements of multiple products from the sources

to destinations subject to its availability and subject to the conditions such that the total bulk cost of supply products is

minimum.

Let a set S ={ 1,2,-------m} where m is the number of sources and S(i,1) and S(i,2) are the availability of k=2

products in ith

source. Let another set DR= {1, 2-------n} where n is the number destinations with requirement of

products. DR(i,1) and DR(i,2) are the requirement of two products in ith

destination where i∈ 𝐷𝑅.We denote a total

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Sreenuvasalu et al., International Journal of Advanced Research in Computer Science and Software Engg. 6(7),

July- 2016, pp. 116-125


number of sources in solution as m0 and a total number ofdestinations as no .Let C (i,j) be the bulk cost for supplying

from ith

source to jth

destinationfor all products k=1,2. The objective of the problem is to find minimum supply

schedulen0(<n)requirement of both products from m0 < m sources to destinations subject to its availability and subject

to the conditions such that the total bulk cost of supply products is minimum.

III. SPECIALITY OF THE MODEL

Present this model is a variant bulk transshipment model withmultiple products. A subject of destinations DT⊂ 𝐷𝑅

can supply to r destinations the bulk cost for k products is same as C (i, j) i.e the cost of supply as each of k products

from ith to j

th destinations. In this case of transshipment i∈ 𝐷𝑇, we consider only m0<m sources and n0<n destinations for

supply sehudele. In the above situations this model is useful and the same as specialty of the model.

IV. MATHEMATICAL FORMULATION

Minimum z(x) = 𝐶 𝑖, 𝑗 𝑋(𝑖, 𝑗)𝑗 𝑖 K------------ (1)

Where i∈ 𝑆1, j∈ 𝐷𝑅&𝑆1 =S∪ 𝐷𝑇 𝐷𝑅(𝑖. 𝑘) ≤ 𝑆 𝑖, 𝑘 𝑚

𝑖=1𝑛𝑖=1 ∀K, where k ∈ 𝐾------------ (2)

𝑆 (𝑖,𝑘) ≥ 𝐷𝑅 ( j, k) where j ∉ 𝐷𝑇& k∈ 𝐾-----------(3)

𝑠 (𝑖, 𝑘) ≥ 𝐷𝑅 (j,k) where j ∈ 𝐷𝑇& k∈ ------------ (4)

X (i, j) =0 or 1 ------------ (5)

Equation (1) represents the objective of the problem. i.e., to find minimum cost supply to the n0 destinations for its

requirement of two products.Constraint(2) represents the total requirement at the destination cities for each product of k

should be less than or equal to the total availability at the sources for each product of k.Equation (3) states that we

consider a destination j∉ 𝐷𝑇 then the requirement of DR (j, k) should be less than or equal to the availability of

kthproduct in i

th source at that point of supply i.e. denoted by𝑆 (𝑖,𝑘).Equation (4) denotes, if a destination j ∈ 𝐷𝑇 then the

requirement of kth

product in the jthdesignation along its supply to other destination should be less than or equal to

it‟ssources of theproduct k which is denoted by𝑆 (𝑖, 𝑘).Equation (5) indicates there are number of order pairs connecting

to sources to destinations or destinations to destinations.

V. NUMERICAL ILLUSTRATION:

The concepts and algorithm developed will be illustrated by a suitable numerical example for which set of

number of sources S={1,2,3,4} and set of number of destinations DR ={1,2,3,4,5,6,7} . Here m=4,n=7, m0 =3, and n0

=6, k=2 & DT=(1,5,7) ⊂ DR. 𝑆1=S∪ 𝐷𝑇=(1,2,3,4) ∪ (1D, 5D, 7D)= (1,2,3,4,1D,5D,7D) then the cost matrix C(i, j)

between the sources and destinations is as follows .

Table -1

C(i,j)=

In above table- 1, suppose C (4, 6) =21 means that the cost of the supply from source city 4 to destination city 6 is 21

units for both products k=1, 2.By the transshipment consider some of the destinations among seven destinations (i.e 1,

5&7) treated as sources. So the we add 1, 5& 7 destinations to the sources. Hence the total number of sources increases

from 4 to 7.∞ indicates irrelevant between the destinations. Consider two products are available at the sources. Let them

are S(i,1) and S(i,2). Then the availability of the given products at sources S (i,α) are as follows . Where ∝𝑖 𝑖𝑠 the type

of the product.

Table-2 (source product available table)

Sources 1 2 3 4

Product 1

capacity 60 60 60 60

Product 2

capacity 70 60 60 60

By the table-2, we have S (1, 2) =70 means source 1 has capacity of product 2 is 70 units. Consider two products are

required at each and every destination .Let them are DR(j.k) where k= 1,2,3,4,5,6 & 7.Then the required of the product at

destination DR(j,α ) are as follows . Where α is the type of the product.

Table-3 (Destination product requirement table)

SN 1 2 3 4 5 6 7

1 5 12 10 9 8 19 11

2 15 18 14 20 16 3 17

3 19 1 21 13 2 22 17

4 7 20 16 23 22 21 3

1D ∞ 23 6 18 24 9 25

5D 24 7 26 10 ∞ 25 26

7D 14 12 13 8 27 4 ∞


July- 2016, pp. 116-125


Destinations 1D 2 3 4 5D 6 7D

Product 1 Requirement 20 20 20 20 30 40 20

Product 2 requirement 20 30 10 20 20 20 30

By the table-3, we have DR (1, 2) =20 means first product requirement of second destination is 20 units. For this

numerical example the objective is the supply of the requirement of both products to n0 destinations frommosources with

minimum total cost subject to the conditions suchthat the total bulk cost of supply is minimum. Where transshipment

consider destination to destination.

VI. FEASIBLE SOLUTION

Feasible solution is a solution, which satisfies all the constraints in the problem. The constraints are discussed in the

mathematical formulation. Consider an ordered pair set {(4,1D) , (4,1D) , (1D,3) , (ID,3) , (4,7D) , (4,7D) , (2,6) , (2,6) ,

(3,5D) , (3,5D) , (3,2) , (3,2) } represents a feasible solution.In the following figure-1, squarerepresents sources with

capacity of products andcircle represents are destination cities. The value in square indicates availability of products

(product 1 numerator and product 2 denominator) number of that particular source. The value in circles indicates

requirement of products (product 1 numerator and product 2 denominator number) of that particular destination. A value

below and above the connectivity of the source city and destination city or destination city and destination city represents

cost between the particular source and destination cities.Consider a set of order pairs{ (4,1D) , (4,1D) , (1D,3) , (ID,3) ,

(4,7D) , (4,7D) , (2,6) , (2,6) , (3,5D) , (3,5D) , (3,2) , (3,2) } represents a feasible solution. FIGURE-1

1 1D 6 3

6

7

4 7

3 7D

6

3

5D 4

2

3 2 2

1

1

2 3 6

3

60

70

20

20

20

10

20

30

20

20

60

60

30

20

60

60

40

20

60

60

20

30

In above figure 1, source 4 supplies 20 units of product 1, 20 units of product 2 to the destination 1D with cost of

7 units .From there to supplies 20 units of product 1 , 10 units of product 2 to the destination 3 with the cost of 6 units.

Source 4 supplies 20 units of product 1, 30 units of product 2 to the destination 7D with cost of 3 units. source 3 supplies

20 units of product 1, 30 units of product 2 to the destination 2 with cost of 1 units .source 3 supplies 30 units of product

1 , 20 units of product 2 to the destination 5D with the cost of 2 units. Source2 supplies 40 units of product 1, 20 units

of product 2 to the destination 6 with cost of 3 units. Also 6 destinations among seven get their requirements of the

different products from given 3 sources among 4. Here we leave one source and one destination for truncation. So the

truncation conditions satisfied hence we called the above solution is feasible solution.Then the corresponding

transshipment cost of the above allocations of figure-1, as follows

The Solution is Z = 2(4,1D) X(4,1D) + 2(1D,3) X(1D,3) + 2(4,7D) X(4,7D) + 2(2,6) X(2,6) + 2 (3,5D)

X(3,5D) + 2(3,2) X(3,2) =14+12+6+6+4+2 = 44 units.


July- 2016, pp. 116-125


VII. CONCEPTS AND DEFINITIONS

A tour is a feasible trip - schedule for the transshipment. Trip-schedule can be represented by an approximate n x n x

r indicator array X = {x (i, j,k); x (i, j ,k) = 0 or 1} in which x (i, j,k ) = 1 indicates that the vehicle visits city j from city

i in kth season

A) .Definition of an Alphabet –Ttable:

There are mxmordered pairs in thetwo-dimensional array C. For convenience these are arranged in ascending order

of their corresponding cost are indexed from 1, 2, 3…. . (SundaraMurrty -1979) .Let SN=( 1, 2, 3…. ) be the set of

indices. Let C be the corresponding array of cost.. If a, bSN and a < b then C(a) C(b).. Also let the arrays R, C& C

be the array of row, column and cost indices. For our convience we used same notationC for column and cost indices.

CD be an array of cumulative sum of the elements of C. The arrays SN, C, CD, R, and C for the numerical example are

given in the table-6. If aSN then (R(a),C(a),c(a) ) is the ordered triple and D(a) = D (R(a), C(a)) is the value of the

ordered pair and CD (a) = 𝑐(𝑖)𝑎𝑖=1 Then the alphabet of the given cost matrix (table-6) is as follows.

Table-6 (Alphabet table)

S N C CD R C

1 1 1 3 2

2 2 3 3 5D

3 3 6 2 6

4 3 9 4 7D

5 4 13 7D 6

6 5 18 1 1D

7 6 24 1D 3

8 7 31 4 1D

9 7 38 5D 2

10 8 46 1 5D

11 8 54 7D 4

12 9 63 1 4

13 9 72 1D 6

14 10 82 1 3

15 10 92 5D 4

16 11 103 1 7D

17 ∞ 114 5D 5

18 12 126 1 2

19 12 138 7D 2

20 13 151 3 4

21 13 164 7D 3

22 14 178 2 3

23 14 192 7D 1D

24 15 207 2 1D

25 ∞ 222 1D 1

26 16 238 2 5D

27 16 254 4 3

28 17 271 2 7D

29 17 288 3 7D

30 18 306 2 2

31 18 324 1D 4

32 19 343 1 6

33 19 362 3 1D

34 20 382 2 4

35 20 402 4 2

36 21 423 3 3

37 21 444 4 6

38 22 466 3 6

39 22 488 4 5D

40 23 511 4 4

41 23 534 1D 2

42 24 558 1D 5D

43 24 582 5D 1D

44 25 607 1D 7D

45 25 632 5D 6


July- 2016, pp. 116-125


46 26 658 5D 3

47 26 684 5D 7D

48 27 711 7D 5D

49 ∞ 738 7D 7D

Let us consider 15ЄSN. It represents the ordered (R (15), C (15)) = (5D, 4). Then C (15) =C (5D, 4) = 10 and CD

(15) =92

B) Definition of a Word:

Let SN= (1,2...) be a set of indices, let c be an array of cost ,CD is an array of cumulative sum of elements in C.

Let arrays R, C and C be the row, column and cost indices of the ordered triples. Let C be the array of cost. Let Lk={a1,

a2,......,ak}, ai €SN be an ordered sequence of k indices from SN. The pattern represented by the ordered triples whose

indices are given by Lk is independent of the order of ai in the sequence. Hence for uniqueness the indices are arranged

in the increasing order such that ai ≤ai+1, i=1,2,........k-1. The set SN is defined as the “Alphabet-Table” with alphabet

order as (1,2,....,) and the ordered sequence Lk is defined as a “word” of length k. A word Lk is called a “sensible word”.

If ai ≤ai+1 , for i=1,2,....k-1 and if this condition is not met it is called a “insensible word”. A word Lkhas at least one

feasible word or equivalently the partial pattern represented by Lk has at least one feasible word or equivalently the

partial pattern represented by Lk is said to be feasible if the block of words represented by Lk has at least one feasible

word or, equivalently the partial pattern represented by Lk should not have any inconsistency.

In the partial word Lk any of the letters in SN can occupy the first place.Since the words of length greater than n-1

are necessarily infeasible,as any feasible pattern can have only n unit entries in it. LK is called a partial word if k<n-1, and

it is a full length word if k=n-1,or simply a word.A partial word Lk represents, a block of words with Lk as a leader i.e., as

it‟s first k letters. A leader is said to be feasible, if the block of word, defined by it has at least one feasible word.

C) Value of the Word:

The value of the( partial) word Lk, V(Lk) is defined recursively as V(Lk)=V(Lk-1)+D(ai) with V(L0)=0, where D(ai) is

the cost array arranged sush that D(ak)<D(ak+1).For our convince k be the value of bulk unit of D(ak). V(Lk) and V(X)

the values of the pattern X will be the same, since X is the (partial) pattern represented by Lk, (Sundara Murthy –

1979).Consider the partial word L4 = {(1(a), 1(b),2(a),2(b))}Then V (L4) =1+1+2+2=6.

D) Lower Bound of a Partial Word LB (Lk):

A lower bound LB (Lk) for the values of the block of words represented by Lk= (a1,a2, .....ak) can be defined as

follows. 𝐿𝐵(𝐿𝑘) = 𝑉(𝐿𝑘) + 𝐶𝐷(𝑎𝑘 + 𝑛 + 2 − 1 − 𝑘) − 𝐶𝐷(𝑎𝑘). Consider the partial word L4 = (1(a), 2(a), 3(a), 5(a))

and V (L4) =1+2+3+3=10. Then LB (L4) =V (L4) +CD (a4+n+2- 1-4) - CD (a4); LB (4) = 10+CD (5+7+1-4) - CD (5)

= 10+CD (9)- CD (5) =10+38-13 =35 units

VII. ALGORITHMS

The recursive algorithm for checking the feasibility of a partial word Lp is given as follows. In the algorithm first we

equate IX = 0, at the end if IX = 1 then the partial word is feasible, otherwise it is infeasible. For this algorithm we have

RA=R (ak+1) and CA=C (ak+1).

A) Arrays and Notations:

m = Number of Sources, n = Number of Destinations, k= For First Product is 1 and For Second Product is 2,

SA(i,j)= Capacity of two Products at Particular Source, DR (i,j) = Requirement of two Products at Particular

Destinations, DX (i) = Destination cities act as sources or List of Sub Sources, CDR (i,j) = Requirement of two Products

at current position.

B) Algorithm-1: (Feasible Checking) STEP0 : IX = 0 IF YES GO TO 1 & IF NO GO TO 6

STEP1 : IS (IC [CA] = 1) IF YES GO TO 6 &IF NO GO TO 2

STEP2 : IS (DX (RA) =1) IF YES GO TO 3 & IF NO GO TO 4

STEP3 : IS CDR (RA, 1) = DR (RA, 1) +DR (CA, 1)

IS CDR (RA, 2) = DR (RA, 2) +DR (CA, 2) IF YES GO TO 5

STEP4 : IS SA (RA, 1) ≥DR (CA, 1)

IS SA (RA, 2) ≥DR (CA, 2) IF YES GO TO 5 & IF NO GO TO 6

STEP5 : IX=1

STEP6 : STOP.

We start with the partial word L1 = (a1) = (1). A partial word Lk is constructed as Lk = Lk-1 * (ak). Where * indicates

chain formulation. We will calculate the values of V (Lk) and LB (Lk) simultaneously. Then two situations arises one for

branching and other for continuing the search.

1. LB (Lk) < VT. Then we check whether Lk is feasible or not. If it is feasible we proceed to consider a partial word

of under (k+1). Which represents a sub-block of the block of words represented by Lk? If Lk is not feasible then

consider the next partial word p by taking another letter which succeeds ak in the position. If all the words of

order p are exhausted then we consider the next partial word of order (k-1).


July- 2016, pp. 116-125


2. LB (Lk) > VT. In this case we reject the partial word Lk. We reject the block of word with Lk as leader as not

having optimum feasible solution and also reject all partial words of order p that succeeds Lk.To find Optimal

Feasible word a Lexi-Search Algorithm using PRT is developed and is given in below.

C) Algorithm-II: (Lexi-search algorithm):

STEP 0 : Initialization The arrays SN, IC, R, C, SS, DRS, SW, SA, DR, DC, RA, CA, L, V, LB and values

Q, m, n, k are made available. The values I=1, J=0, VT=9999 and Max= m×n–n

STEP 1 : J = J + 1, IS [IF J > Max IF YES GOTO 15& IF NO GOTO 2

STEP2 : V [I] = V [I-1] + C [J]; V [0] = 0

LB [I] = 2[V (I) + DC (J + n – I) – DC (J) ]

IS (LB [I] ≥ VT)IF YES GO TO 14 & IF NO GO TO 3

STEP : RA = R [J], CA = C [J] IF YES GOTO 4

STEP4 : Check Feasibility using algorithm 1

IS (IX = 1) IF YES GO TO 5& IF NO GO TO 1

STEP5 : IS (I = n-1) IF YES GO TO 10& IF NO GO TO 6

STEP6 : IS DX (RA=1) IF YES GOTO 7& IF NO GO TO 8

STEP7 : DR (RA, 1) = CDR (RA, 1)

DR (RA, 2) = CDR (RA, 2) IF YES GO TO 9

STEP8 : SA (RA.1) = SA (RA, 1)-DR (CA, 1)

SA (RA.2) = SA (RA, 2)-DR (CA, 2) IF YES GO TO 9

STEP9 : IC (CA) =1, L(I)=J, I-I+1, Max=Max+1 IF YES GO TO 1

STEP10 : VT = V [I], L [I] = J, L [I] is full length word and is feasible

Record L [I] and VT IF YES GO TO 11

STEP11 : I = I – 1, J = L [I], RA = R [J], CA = C [J]

IC [CA] = 0 & IS[DX (I)] =1 IF YES GO TO 12& IF NO GO TO 13

STEP12 : DR (RA.1) =DR (RA, 1)-CDR (RA,1)

DR (RA.2) =DR (RA,2)-CDR (RA,2) IF YES GO TO 1

STEPP13 : SA (RA.1) =SA (RA,1)+DR (RA,1)

SA (RA.2) =SA (RA,2)+DR (RA,2) IF YES GO TO 1

STEP14 : IS IF YES GOTO 15& IF NO GO TO 11

STEP15 : STOP

The current value of VT at the end of the search is the value of the optimal feasible word. At the end if VT = 999 it

indicates that there is no feasible solution.

VIII. SEARCH TABLE

The working details of getting an optimal word using the above algorithm for the illustrative numerical example is

given in the following table-6. The columns named (1), (2),(3)............gives the letters in the first, second, third and so on

places respectively. The columns V, LB, R, C and C letters indicates Value of that word ,lower bound of the word , row,

column and cost indices of accepted for connectivity. The last column denotes the remarks regarding the acceptability

of the partial word. In the following table „A‟ indicates ACCEPT and „R‟ indicates REJECT.

Table-7 (Search table)

SN 1 2 3 4 5 6 7 8 9 10 11 12 V LB R C RE

1 1(a) 1 36 3 2 A

2 1(b) 2 36 3 2 A

3 2(a) 4 36 3 5D A

4 2(b) 6 36 3 5D A

5 3(a) 9 36 2 6 A

6 3(b) 12 36 2 6 A

7 4(a) 15 36 4 7D A

8 4(b) 18 36 4 7D A

9 5(a) 22 7D 6 R

10 5(b) 22 7D 6 R

11 6(a) 23 1 1D R

12 6(b) 23 1 1D R

13 7(a) 24 44 1D 3 A

14 7(b) 30 44 1D 3 A

15 8(a) 37 44 4 1D A


July- 2016, pp. 116-125


16 8(b) 44 44 4 1D A=VT

17 8(b) 37 44 4 1D R,=VT

18 8(a) 31 45 4 1D R,>VT

19 7(b) 24 45 1D 3 R,>VT

20 5(a) 19 7D 6 R

21 5(b) 19 7D 6 R

22 6(a) 20 44 1 1D R,=VT

23 4(b) 15 39 4 7D A

24 5(a) 19 7D 6 R

25 5(b) 19 7D 6 R

26 6(a) 20 44 1 1D R,=VT

27 5(a) 16 7D 6 R

28 5(b) 16 45 7D 6 R,>VT

29 4(a) 12 39 4 7D A

30 4(b) 15 39 4 7D A

31 5(a) 19 7D 6 R

32 5(b) 19 7D 6 R

33 6(a) 20 44 1 1D R,=VT

34 5(a) 16 7D 6 R

35 5(b) 16 45 7D 6 R,>VT

36 4(b) 12 42 4 7D A

37 5(a) 16 7D 6 R

38 5(b) 16 45 7D 6 R,>VT

39 5(a) 13 46 7D 6 R,>VT

40 3(b) 9 39 2 6 A

41 4(a) 12 39 4 7D A

42 4(b) 15 39 4 7D A

43 5(a) 19 7D 6 R

44 5(b) 19 7D 6 R

45 6(a) 20 44 1 1D R,=VT

46 5(a) 16 7D 6 R

47 5(b) 16 45 7D 6 R,>VT

48 4(b) 12 42 4 7D A

49 5(a) 16 7D 6 R

50 5(b) 16 45 7D 6 R,>VT

51 5(a) 13 46 7D 6 R,>VT

52 4(a) 9 42 4 7D A

53 4(b) 12 42 4 7D A

54 5(a) 16 42 7D 6 A

55 5(b) 20 42 7D 6 A

56 6(a) 25 42 1 1D A

57 6(b) 30 42 1 1D A

58 7(a) 36 42 1D 3 A

59 7(b) 42 42 1D 3 A,=VT

60 7(b) 36 43 1D 3 R,>VT

61 7(a) 31 44 1D 3 R,>VT

62 6(b) 25 44 1 1D R,>VT

63 6(a) 21 45 1 1D R,>VT

64 5(b) 16 45 7D 6 R,>VT

65 5(a) 13 46 7D 6 R,>VT

66 4(b) 9 46 4 7D R,>VT

67 3(a) 7 40 2 6 A

68 3(b) 10 40 2 6 A


July- 2016, pp. 116-125


69 4(a) 13 40 4 7D A

70 4(b) 16 40 4 7D A

71 5(a) 20 7D 6 R

72 5(b) 20 42 7D 6 R,=VT

73 5(a) 17 43 7D 6 R,>VT

74 4(b) 13 43 4 7D R,>VT

75 4(a) 10 43 4 7D R,>VT

76 3(b) 7 43 2 6 R,>VT

77 2(b) 4 40 3 5D A

78 3(a) 7 40 2 6 A

79 3(b) 10 40 2 6 A

80 4(a) 13 40 4 7D A

81 4(b) 16 40 4 7D A

82 5(a) 20 7D 6 R

83 5(b) 20 42 7D 6 R,=VT

84 5(a) 17 43 7D 6 R,>VT

85 4(b) 13 43 4 7D R,>VT

86 4(a) 10 43 4 7D R,>VT

87 3(b) 7 43 2 6 R,>VT

88 3(a) 5 43 2 6 R,>VT

89 2(a) 3 41 3 5D A

90 2(b) 5 41 3 5D A

91 3(a) 8 41 2 6 A

92 3(b) 11 41 2 6 A

93 4(a) 14 41 4 7D A

94 4(b) 17 41 4 7D A

95 5(a) 21 7D 6 R

96 5(b) 21 43 7D 6 R,>VT

97 5(a) 18 44 7D 6 R,>VT

98 4(b) 14 44 4 7D R,>VT

99 4(a) 11 44 4 7D R,>VT

100 3(b) 8 44 2 6 R,>VT

101 3(a) 6 45 2 6 R,>VT

102 2(b) 3 45 3 5D R,>VT

103 1(b) 1 41 3 2 A

104 2(a) 3 41 3 5D A

105 2(b) 5 41 3 5D A

106 3(a) 8 41 2 6 A

107 3(b) 11 41 2 6 A

108 4(a) 14 41 4 7D A

109 4(b) 17 41 4 7D A

110 21 7D 6 R

111 21 43 7D 6 R,>VT

112 5(a) 18 44 7D 6 R,>VT

113 4(b) 14 44 4 7D R,>VT

114 4(a) 11 44 4 7D R,>VT

115 3(b) 8 44 2 6 R,>VT

116 3(a) 6 44 2 6 R,>VT

117 2(b) 3 45 3 5 R,>VT

118 2(a) 2 46 2 6 R,>VT

At the end of the search table-7, the current value of VT is 42 and it is the value of the optimal feasible word L12

={(1(a),1(b),2(a),2(b),4(a),4(b),5(a),5(b),6(a),6(b),7(a),7(b))}.It is given in the 59th row of the search table and the


July- 2016, pp. 116-125


corresponding order paires are(3,2),(3,2),(3,5D),(3,5D),(4,7D),(4,7D),(7D,6),(7D,6) (1,1D),(1,1D),(1D,3),(1D,3) .Then

the following figure-3 represents the optimal solution. FIGURE-3

1 5 1D 6 3

5 6

2 2 4

1

3 1 5D

2

2

4 3 7D 6

4

3 4

60

70

20

20

20

10

30

20

60

60

20

20

60

60

20

30

60

60

20

30

40

20

In above figure 3, Source 1 supplies 20 units of product 1, 20 units of product 2 to the destination 1D with cost of

5 units .From there to supplies 20 units of product 1 , 10 units of product 2 to the destination 3 with the cost of 6 units.

Source 3 supplies 20 units of product 1, 30 units of product 2 to the destination 2 with cost of 1 units .Source 3

supplies to 30 units of product 1 , 20 units of product 2 to the destination 5D with the cost of 2 units. Source 4

supplies 20 units of product 1, 30 units of product 2 to the destination 7D with cost of 3 units .From there to supplies 40

units of product 1 , 20 units of product 2 to the destination 6 with the cost of 4 units. 6 destinations among seven get

their requirement of the different products from given sources 3 among 4. Here we leave one source and one destination

for truncation. So all the conditions satisfied hence we called the above solution is optimal solution. Then the optimal

solution is

Z = 2(1,1D) X(1,1D) + 2 (1D, 3) X(1D,3)+ 2(3, 2) X(3,2) + 2 (3,5D) X(3,5D) + 2 (4,7D) X(4,7D)+ 2 (7D, 6)

X(7D,6)= 10+12+2+4+6+8 = 42 units.

IX. CONCLUSION

In thispaper,first the model is formulated as a transshipment problem. Write the corresponding mathematical

formulation and Developed Lexi-search Algorithm by using pattern recognition technique for solving this model.The

concepts and steps involved in the algorithm are explained with a suitable numerical example for understand.we find an

optimum solution with the given all constraints are satisfied.Based on this experience we can say that this algorithm is

efficient. The following References are used for making the model.

REFERENCES

[1] Dantzig G.B., and Ramser, J.H. (1959): “The Truck Dispatching Problem,” Management Science, Vol. 6, No.

1, pp.80-91.

[2] Junginger w.(1993) European journal of operational research 66,353-371.

[3] Koopmans, Tjalling, C. (1949): “Optimum utilization of the transportation system,”Econometrica 17

(Supplement) 136–146.

[4] Maio, A. D., Roveda, C. (1971): “An all zero-one algorithm for a certain class of transportation problems,”

Operations Research, 19, 1406-1418.

[5] Naganna B. Operations Research. Ph. D thesis, S. V. University, Tirupati, India. (2007)

[6] P. Guravaraju, purusotham S, Suresh Babu C and Madhu Mohan Reddy P: two stage multi objective variant

bulk TP model - International Journal of advanced research in computer science and software engineering -

ISSN 2277-128X Volume 5 (10), October (2015), pp. 115-125.


July- 2016, pp. 116-125


[7] Prakash, S., Kumar, P., Prasad, B.V.N.S., and Gupta, A. (1986): “Pareto optimal solutions of a cost– Time

trade-off bulk transportation problem,”

[8] Purusotham, S., Sundara Murthy, M. (2011): “An Exact Algorithm for Multi- Product Bulk Transportation

Problem”, International Journal on Computer Science and Engineering, Vol. 3(9), 3222-3236.

[9] Rautman .C.A. Ried R.A and Ryder E.E(1993) operational research 41,459-469.

[10] Sangeetham Prasad, Suresh Babu C, Purusotham S and Sundara Murthy M: A Variant Constrained Bulk

Transshipment Problem - Global Journal of Pure and Applied Mathematics ISSN 0973-1768 Volume 4 (2014)

[11] SobhanBabu, K. And Sundara Murthy, M. (2010): “An Efficient Algorithm for Variant Bulk Transportation

Problem,” International Journal of Engineering Science and Technology, vol: 2(7), pp: 2595-2600.

[12] Srinivasan, V and Thompson, G.L. (1973). An algorithm for assigning uses to sources in a special class of

transportation problem. Operations Research, 21, 284-295.

[13] Sundara Murthy, M - (1976). A Bulk Transportation Problem Opsearch, 13, pp. 143-155.

[14] Suresh Babu, C. (2013): “Lexi-Search Exact Algorithm for Variant TSP, Bulk TP and Minimum Spanning

Models” Ph.D thesis, S. V. University, Tirupati, India.

[15] VanitaVerma, Puri, M.C. (1996): “A branch and bound procedure for cost minimizing bulk transportation

problem,”Opsearch, Vol. 33(3), 145-161.

[16] Vidhyullatha, A. (2012): “Pattern Recognition Lexi – Search Exact Algorithms for Variant ASP and Bulk TP

Models”, a Ph. D. thesis, Sri Venkateswara University, Tirupati, India.

Variant Multi Product Transhipment Model · Variant Multi Product Transhipment Model Sreenuvasalu Y1, Suresh BabuC2, Madhu Mohan Reddy P3, Sundara Murthy M4 1 Asst. Professor, Dept.

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