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Mehar method for solving unbalanced generalized interval-valued trapezoidal fuzzy number transportation problems
Chapter 18
Mehar method for solving unbalanced generalized intervalvalued trapezoidal fuzzy number transportation problems Akansha Mishraa, Amit Kumara, S.S. Appadoob a
School of Mathematics, Thapar Institute of Engineering & Technology, Patiala, India, Department of Supply Chain Management, University of Manitoba, Winnipeg, MB, Canada
b
1 Introduction Ebrahimnejad (2016) pointed out that several methods have been proposed for solving fuzzy/fully fuzzy transportation problems (transportation problems in which each parameter is represented by a triangular/trapezoidal fuzzy number) (Ebrahimnejad, 2014, 2015a,b,c, 2016; Oheigeartaigh, 1982; Chanas et al., 1984, 1993; Chanas and Kuchta, 1996; Jimenez and Verdegay, 1998, 1999; Liu and Kao, 2004; Dinagar and Palanivel, 2009; Pandian and Natarajan, 2010; Kumar and Kaur, 2010, 2011a,b, 2014; Gupta et al., 2012; Shanmugasundari and Ganesan, 2013; Sudhagar and Ganesan, 2012; Kaur and Kumar, 2012; Chiang, 2005; Gupta and Kumar, 2012). However, no method has been proposed to find the solution of generalized interval-valued fuzzy transportation problems (transportation problems in which each parameter is represented by a triangular/trapezoidal fuzzy number). To fill this gap, Ebrahimnejad (2016) proposed the following methods: 1. A method to transform an unbalanced generalized IVTrFTP into a balanced generalized IVTrFTP. 2. A linear programming method to find the solution of a balanced generalized IVTrFTP. Ebrahimnejad (2016) considered an unbalanced generalized IVTrFNTP and applied the methods, proposed by him, to transform it into a balanced generalized IVTrFNTP as well as to find its solution.
Maritime Transport and Regional Sustainability. https://doi.org/10.1016/B978-0-12-819134-7.00019-8 © 2020 Elsevier Inc. All rights reserved.
311
312 PART | V Other key issues
In this paper, it is shown that on applying Ebrahimnejad’s method for transforming an unbalanced generalized IVTrFNTP into a balanced generalized IVTrFNTP (Ebrahimnejad, 2016), the obtained dummy supply and/or dummy demand is not a generalized interval-valued trapezoidal fuzzy number (IVTrFN), and therefore this method is not valid. In addition, a new method (known as the Mehar method) is proposed to transform an unbalanced generalized IVTrFNTP into a balanced generalized IVTrFNTP. Furthermore, the validity of the proposed Mehar method is discussed. This paper is organized as follows:
2 Ebrahimnejad’s method for transforming an unbalanced generalized IVTrFNTP into a balanced generalized IVTrFNTP Ebrahimnejad proposed the following method to transform an unbalanced generalized IVTrFNTP into a balanced generalized m
n
∑a ≠ ∑b
IVTrFNTP, i.e., to transform
i
i =1
j
m
n
∑a = ∑b , where, m
into
i
j =1
j
i =1
j =1
represents number of sources, n represents number of destinations, m m m m m m m m m i = a L , a L , a L , a L ; ω L , aU , aU , aU , aU ; ω U a ∑ ∑ ∑ ∑ ∑ i1 ∑ i 2 ∑ i 3 i4 i1 ∑ i 2 ∑ i 3 i4 i =1 i =1 i =1 i =1 i =1 i =1 i =1 i =1 i =1 represents total interval-valued fuzzy supply, and n n n n n n n n n b j = ∑b Lj1 , ∑b Lj 2 , ∑b Lj 3 , ∑b Lj 4 ; ω L , ∑bUj1 , ∑bUj 2 , ∑bUj 3 , ∑bUj 4 ; ω U ∑ j =1 j =1 j =1 j =1 j =1 j =1 j =1 j =1 j =1 represents total interval-valued fuzzy demand. The generalized IVTrFN ⟨(ai1L, U U U U ai2L, ai3L, ai4L; ωL), (ai1 , ai2, ai3, ai4; ωU)⟩ represents the supply of the product at ith U U U U source Si. The generalized IVTrFN ⟨(bj1L, bj2L, bj3L, bj4L; ωL), (bj1 , bj2, bj3, bj4; ωU)⟩ represents the demand of the product at jth destination Dj. Case 1: If
m
∑a
L i1
i =1
m
∑a i =1
U i1
n
≤ ∑bUj1 ,
dummy
j =1
m
n
≤ ∑b Lj1 , j =1
∑a
U i2
i =1
source
m
∑a
L i2
i =1
n
≤ ∑bUj 2 , j =1
m
n
≤ ∑b Lj 2 , j =1
∑a i =1
Sm + 1
U i3
m
∑a
n
≤ ∑bUj 3 , j =1
having
i =1 m
L i3
∑a i =1
n
≤ ∑b
U i4
L 3
j =1 n
,
m
∑a i =1
L i4
n
≤ ∑b Lj 4 , j =1
≤ ∑bUj 4 , then add a j =1
dummy
n m m n m n L m L n L L L L L L L ∑b j1 − ∑ai1 , ∑b j 2 − ∑ai 2 , ∑b j 3 − ∑ai 3 , ∑b j 4 − ∑ai 4 ; ω , j =1 i =1 i =1 j =1 i =1 j =1 i =1 j =1 n U m U n U m U n U m U n U m U U ∑b j1 − ∑ai1 , ∑b j 2 − ∑ai 2 , ∑b j 3 − ∑ai 3 , ∑b j 4 − ∑ai 4 ; ω i =1 j =1 i =1 j =1 i =1 j =1 i =1 j =1
supply
Trapezoidal fuzzy number transportation problems Chapter | 18 313
by considering the cost for supplying the unit quantity of the product from the dummy source Sm + 1 to all the destinations as a generalized IVTrFN 0 = ( ( 0, 0, 0, 0; 1) , ( 0, 0, 0, 0; 1) . n m n
Case 2: If n
∑b
U j1
j =1
then
∑b j =1
L j1
≤ ∑a , n
≤ ∑aiU1 ,
∑b
i =1
a
L i1
i =1
m
add
)
m
U j2
j =1
n
∑b
L j2
j =1
m
n
m
L j3
j =1
i =1
≤ ∑aiL3 , i =1
n
∑b
L j4
j =1 n
m
m
≤ ∑aiL4 , i =1 m
≤ ∑aiU2 ,
∑b
≤ ∑aiU3 ,
∑b
≤ ∑aiU4 ,
destination
Dn + 1
having
dummy
demand
U j3
j =1
i =1
dummy
∑b
≤ ∑aiL2 ,
U j4
j =1
i =1
i =1
m n n m n m L n L m L L L L L L L ∑ai1 − ∑b j1 , ∑ai 2 − ∑b j 2 , ∑ai 3 − ∑b j 3 , ∑ai 4 − ∑b j 4 ; ω , i =1 j =1 j =1 i =1 j =1 i =1 j =1 i =1 m n m n m n m n U U U U U U U U U ∑ai1 − ∑b j1 , ∑ai 2 − ∑b j 2 , ∑ai 3 − ∑b j 3 , ∑ai 4 − ∑b j 4 ; ω j =1 i =1 j =1 i =1 j =1 i =1 j =1 i =1 by considering the cost for supplying the unit quantity of the product from the dummy destination Dn + 1 to all the sources as a generalized IVTrFN, 0 = ( ( 0, 0, 0, 0;1) , ( 0, 0, 0, 0;1) .
)
Case 3: If neither Case 1 nor Case 2 is satisfied, then carry out the following: (i) Add a dummy source Sm + 1 having the dummy supply (A(Lm +1)1 , A(Lm +1) 2 , U A(Lm +1)3, A(Lm +1) 4 ) , (A(Um +1)1 , A(Um +1) 2 , A( m +1)3, A(Um +1) 4 ) by considering the cost for supplying the unit quantity of the product from all the dummy sources Sm + 1 to all the dummy destinations as generalized IVTrFN 0 = ( ( 0, 0, 0, 0;1) , ( 0, 0, 0, 0;1)
)
A(Lm +1)1 =
n
m
∑b − ∑a j =1
U j1
i =1
U i1
m n + max 0,∑b Lj1 − ∑aiL1 , j =1 i =1
n m m n A(Lm +1)2 = A(Lm +1)1 + max 0, ∑b Lj 2 − ∑b Lj1 − ∑aiL2 − ∑aiL1 , j =1 i =1 j =1 i =1
n m m n A(Lm +1)3 = A(Lm +1)2 + max 0, ∑b Lj 3 − ∑b Lj 2 − ∑aiL3 − ∑aiL2 , j =1 i =1 j =1 i =1 n m n m A(Lm +1) 4 = A(Lm +1)3 + max 0, ∑b Lj 4 − ∑b Lj 3 − ∑aiL4 − ∑aiL3 , j =1 i =1 i =1 j =1 m n A(Um +1)1 = max 0,∑bUj1 − ∑aiU1 , j =1 i =1
314 PART | V Other key issues
n m m n A(Um +1)2 = ∑bUj1 − ∑aiU1 + max 0,∑bUj1 − ∑aiU1 j =1 i =1 i =1 j =1 n m m n U + max 0, ∑b j 2 − ∑bUj1 − ∑aiU2 − ∑aiU1 , j =1 i =1 i =1 j =1 n m m n A(Um +1)3 = A(Um +1)2 + max 0, ∑bUj 3 − ∑bUj 2 − ∑aiU3 − ∑aiU2 , j =1 i =1 j =1 i =1 n m n m A(Um +1) 4 = A(Um +1)3 + max 0, ∑bUj 4 − ∑bUj 3 − ∑aiU4 − ∑aiU3 . j =1 i =1 i =1 j =1 L L (ii) Add a dummy destination Dn + 1 having the dummy demand (B( n +1)1 , B( n +1) 2, B(Ln +1)3 , B(Ln +1) 4 ) , (B(Un +1)1 , B(Un +1) 2, B(Un +1)3 , B(Un +1) 4 ) by considering the cost for supplying the unit quantity of the product from all the sources to the dummy destination Dn + 1 as generalized IVTrFN 0 = ( ( 0, 0, 0, 0;1) , ( 0, 0, 0, 0;1) . n m = ∑b − ∑a + max 0,∑aiL1 − ∑b Lj1 , j =1 i =1 j =1 i =1 n
L ( n +1)1
B
m
U j1
U i1
m n m n B(Ln +1)2 = B(Ln +1)1 + max 0, ∑aiL2 − ∑aiL1 − ∑b Lj 2 − ∑b Lj1 , i =1 j =1 j =1 i =1 m n m n B(Ln +1)3 = B(Ln +1)2 + max 0, ∑aiL3 − ∑aiL2 − ∑b Lj 3 − ∑b Lj 2 , i =1 j =1 j =1 i =1 m n n m B(Ln +1) 4 = B(Ln +1)3 + max 0, ∑aiL4 − ∑aiL3 − ∑b Lj 4 − ∑b Lj 3 , i =1 j =1 j =1 i =1 n m B(Un +1)1 = max 0,∑aiU1 − ∑bUj1 , i =1 j =1 n m + max 0,∑aiU1 − ∑bUj1 i =1 j =1 j =1 i =1 n m U m U n U + max 0, ∑ai 2 − ∑ai1 − ∑b j 2 − ∑bUj1 , i =1 j =1 j =1 i =1
B(Un +1)2 =
m
n
∑a − ∑b U i1
U j1
)
Trapezoidal fuzzy number transportation problems Chapter | 18 315
m n m n B(Un +1)3 = B(Un +1)2 + max 0, ∑aiU3 − ∑aiU2 − ∑bUj 3 − ∑bUj 2 , i =1 j =1 j =1 i =1 m m n n B(Un +1) 4 = B(Un +1)3 + max 0, ∑aiU4 − ∑aiU3 − ∑bUj 4 − ∑bUj 3 . i =1 j =1 j =1 i =1
3 Flaws of Ebrahimnejad’s method for transforming an unbalanced generalized IVTrFNTP into a balanced generalized IVTrFNTP In this section, some numerical values are considered to show that Ebrahimnejad’s method for transforming an unbalanced generalized IVTrFNTP into a balanced generalized IVTrFNTP (Ebrahimnejad, 2016) is not valid. (1) Let
and
m m m m m L m L m L m L U U L U U U ∑ai1 , ∑ai 2 , ∑ai 3 , ∑ai 4 ; ω , ∑ai1 , ∑ai 2 ,∑ai 3 , ∑ai 4 ; ω i =1 i =1 i =1 i =1 i =1 i =1 i =1 i =1 = (1, 2, 7,11;1) , ( 0, 3, 8, 7;1)
n L n L n L n L n U n U n U n U U L ∑b j1 , ∑b j 2 , ∑b j 3 , ∑b j 4 ; ω , ∑b j1 , ∑b j 2 , ∑b j 3 , ∑b j 4 ; ω j =1 j =1 j =1 j =1 j =1 j =1 j =1 j =1 = ( 4, 8, 9,13;1) , (1, 6,10,18;1) . m
Since
∑a i =1
m
∑a i =1
U i1
n
L i1
≤ ∑bUj1 , j =1
n
m
≤ ∑b Lj1 ,
∑a
j =1
m
∑a i =1
U i2
L i2
i =1
n
≤ ∑bUj 2 , j =1
n
≤ ∑b Lj 2 , j =1
m
∑a i =1
U i3
m
∑a
L i3
i =1
n
≤ ∑bUj 3 , j =1
m
n
≤ ∑b Lj 3 , j =1
∑a i =1
U i4
m
∑a i =1
L i4
n
≤ ∑b Lj 4 , j =1
n
≤ ∑bUj 4 , i.e., Case 1 of j =1
Ebrahimnejad’s method (Ebrahimnejad, 2016), discussed in Section 2, is satisfied. So, according to (Ebrahimnejad’s (2016) method, the dummy supply will be ⟨(4 − 1, 8 − 2, 9 − 7, 13 − 11; 1), (1 − 0, 6 − 3, 10 − 8, 18 − 7; 1)⟩ = ⟨(3, 6, 2, 2; 1), (1, 3, 2, 11; 1)⟩. However, it is not a generalized IVTrFN for the following reason: It can be easily verified from the graphical representation of generalized IVTrFN (Ebrahimnejad, 2016) as well as from the existing definition of a generalized IVTrFN (Ebrahimnejad, 2016) that in a generalized IVTrFN ⟨(a1L, a2L, a3L, a4L; ωL), (a1U, a2U, a3U, a4U; ωU)⟩ the condition a1U ≤ a1L ≤ a2U ≤ a2L ≤ a3L ≤ a3U ≤ a4L ≤ a4U should always be satisfied. While, for the obtained dummy supply ⟨(3, 6, 2, 2; 1), (1, 3, 2, 11; 1)⟩, this condition is not satisfied. Hence, Ebrahimnejad’s method (Ebrahimnejad, 2016) to obtain the dummy supply is not correct.
316 PART | V Other key issues
(2) Let
and
m m m m m L m L m L m L U U L U U U a , a , a , a ; ω , a , a , a , ∑ i1 ∑ i 2 ∑ i 3 ∑ i 4 ∑ i1 ∑ i 2 ∑ i 3 ∑ai 4 ; ω i =1 i =1 i =1 i =1 i =1 i =1 i =1 i =1 = ( 4, 8, 9,13;1) , (1, 6,10,18;1)
n n n L n L n L n L n U n U U U U L ∑b j1 , ∑b j 2 , ∑b j 3 , ∑b j 4 ; ω , ∑b j1 , ∑b j 2 ,0∑b j 3 , ∑b j 4 ; ω j =1 j =1 j =1 j =1 j =1 j =1 j =1 j =1 = (1, 2, 7,11;1) , ( 0, 3, 8, 7;1) .
Since n
∑b
U j1
n
m
j =1 m
i =1 n
n
≤ ∑aiU1 ,
∑b
U j2
m
∑b Lj 2 ≤ ∑aiL2 ,
∑b Lj1 ≤ ∑aiL1 ,
j =1
m
i =1
n
n
m
∑b Lj 3 ≤ ∑aiL3 , j =1
m
i =1
n
≤ ∑aiU2 , ∑bUj 3 ≤ ∑aiU3 ,
∑b
U j4
n
m
j =1
i =1
∑b Lj 4 ≤ ∑aiL4 ,
m
≤ ∑aiU4 ,
j =1 i =1 j =1 i =1 j =1 i =1 j =1 i =1 i.e., Case 2 of Ebrahimnejad’s method (Ebrahimnejad, 2016), discussed in Section 2, is satisfied. So, according to Ebrahimnejad’s method (Ebrahimnejad, 2016), the dummy demand will be ⟨(4 − 1, 8 − 2, 9 − 7, 13 − 11; 1), (1 − 0, 6 − 3, 10 − 8, 18 − 7; 1)⟩ = ⟨(3, 6, 2, 2; 1), (1, 3, 2, 11; 1)⟩.
However, it is not a generalized IVTrFN for the following reason. It can be easily verified from the graphical representation of generalized IVTrFN (Ebrahimnejad, 2016) as well as from the existing definition of a generalized IVTrFN (Ebrahimnejad, 2016) that in a generalized IVTrFN ⟨(a1L, a2L, a3L, a4L; ωL), (a1U, a2U, a3U, a4U; ωU)⟩ the condition a1U ≤ a1L ≤ a2U ≤ a2L ≤ a3L ≤ a3U ≤ a4L ≤ a4U should always be satisfied. While, for the obtained dummy demand ⟨(3, 6, 2, 2; 1), (1, 3, 2, 11; 1)⟩, this condition is not satisfied. Hence, Ebrahimnejad’s method (Ebrahimnejad, 2016) to obtain the dummy demand is not correct. (3) Let
m m m m m L m L m L m L U U L U U U ∑ai1 , ∑ai 2 , ∑ai 3 , ∑ai 4 ; ω , ∑ai1 , ∑ai 2 , ∑ai 3 , ∑ai 4 ; ω i =1 i =1 i =1 i =1 i =1 i =1 i =1 i =1 2 = 110,150,160,180; , (100,140,170,190; 1) 3
n n n n n n n n and ∑b Lj1 , ∑b Lj 2 , ∑b Lj 3 , ∑b Lj 4 ; ω L , ∑bUj1 , ∑bUj 2 , ∑bUj 3 , ∑bUj 4 ; ω U j =1 j =1 j =1 j =1 j =1 j =1 j =1 j =1 2 = 90,120,140, 200; , ( 75,105,155, 215;1) . 3 n
Since n
∑b j =1
U j1
∑b
L j1
j =1
m
≤ ∑aiU1 , i =1
n
m
∑b
≤ ∑aiL1 ,
j =1
i =1
n
∑b j =1
U j2
L j2
m
n
≤ ∑aiL2 ,
∑b
m
i =1
m
n
m
i =1
j =1
i =1
≤ ∑aiU2 , ∑bUj 3 ≤ ∑aiU3 ,
j =1
∑a i =1
i =1
∑a i =1
m
m
≤ ∑aiL3 ,
L j3
U i4
n
L i4
n
≤ ∑b Lj 4 , j =1
≤ ∑b , i.e., Case 3 of j =1
U j4
Trapezoidal fuzzy number transportation problems Chapter | 18 317
Ebrahimnejad’s method (Ebrahimnejad, 2016), discussed in Section 2, is satisfied. So, according to Ebrahimnejad’s method (Ebrahimnejad, 2016), 2 the dummy demand will be . 45, 55, 55, 55; 3 , ( 25, 60, 60, 60; 1) However, the obtained dummy demand is not a generalized IVTrFN for the following reason. It can be easily verified from the graphical representation of generalized IVTrFN (Ebrahimnejad, 2016) as well as from the existing definition of a generalized IVTrFN (Ebrahimnejad, 2016) that in a generalized IVTrFN ⟨(a1L, a2L, a3L, a4L; ωL), (a1U, a2U, a3U, aU4 ; ωU)⟩ the condition a1U ≤ a1L ≤ a2U ≤ a2L ≤ a3L ≤ a3U ≤ a4L ≤ a4U should always be satisfied. While, for the obtained 2 , this condition is not 45, 55, 55, 55; 3 , ( 25, 60, 60, 60;1) satisfied. Hence, Ebrahimnejad’s method (Ebrahimnejad, 2016) to obtain the dummy demand is not correct. dummy demand
4 Proposed Mehar method for transforming an unbalanced generalized IVTrFNTP into a balanced generalized IVTrFNTP In this section, a new method (known as the Mehar method) is proposed to transform an unbalanced generalized IVTrFNTP into a balanced generalized IVTrFNTP. Let us consider an unbalanced generalized IVTrFNTP having m sources and n destinations such that the availability of the product and the demand of the product at ith source (Si) and jth destination (Dj) be represented by generalized IVTrFNTP a i = aiL1 , aiL2 , aiL3 , aiL4 ; ω L , aiU1 , aiU2 , aiU3 , aiU4 ; ω U and
(
b j =
(b
L j1
)(
)(
, b Lj 2 , b Lj 3 , b Lj 4 ; ω L , bUj1 , bUj 2 , bUj 3 , bUj 4 ; ω U
)
)
, respectively. Then, this un-
balanced generalized IVTrFNTP can be transformed into a balanced generalized IVTrFNTP as follows: Case 1: If m
∑a
U i2
i =1 m
m
m
∑a i =1
n
i =1
i =1
m
j =1
j =1
m
n
n
m
i =1 m
i =1
j =1
m
L i1
i =1
m
L i4
n
m
∑a − ∑a ∑a
i =1
∑a
n
≤ ∑bUj1 ,
− ∑a ≤ ∑b − ∑b , L i1
∑a − ∑a L i3
U i1
L i2
U j2
L j1
≤ ∑b Lj 3 − ∑b Lj 2 , j =1
j =1
n
n
− ∑aiU3 ≤ ∑b Lj 4 − ∑bUj 3 , i =1
j =1
j =1
L i2
i =1
U i1
n
j =1
m
i =1 m
∑a i =1
U i4
j =1
n
n
i =1
j =1
j =1
m
n
n
j =1 n
j =1 n
j =1
j =1
− ∑a ≤ ∑b Lj 2 − ∑bUj 2 , U i2
∑a − ∑a U i3
n
≤ ∑b Lj1 − ∑bUj1 ,
i =1 m
L i3
≤ ∑bUj 3 − ∑b Lj 3 ,
− ∑aiL4 ≤ ∑bUj 4 − ∑b Lj 4 then add a i =1
L ( m +1)1
dummy source Sm + 1 having dummy supply (A , A(Lm +1) 2 , A(Lm +1)3 , U L U U U A( m +1) 4 ) , (A( m +1)1 , A( m +1) 2 , A( m +1)3 , A( m +1) 4 ) by considering the cost for supplying the unit quantity of the product from the dummy source Sm + 1 to all the destinations Dn + 1 as a generalized IVTrFN 0 = ( 0, 0, 0, 0;1) , ( 0, 0, 0, 0;1) .
318 PART | V Other key issues n m n m n m A(Um +1)1 = ∑bUj1 − ∑aiU1 , A(Lm +1)1 = A(Um +1)1 + ∑b Lj1 − ∑bUj1 − ∑aiL1 − ∑aiU1 , j =1 i =1 j =1 i =1 j =1 i =1 n n m m A(Um +1)2 = A(Lm +1)1 + ∑bUj 2 − ∑b Lj1 − ∑aiU2 − ∑aiL1 , j =1 i =1 j =1 i =1 n n m m A(Lm +1)2 = A(Um +1)2 + ∑b Lj 2 − ∑bUj 2 − ∑aiL2 − ∑aiU2 , j =1 i =1 j =1 i =1 n m n m A(Lm +1)3 = A(Lm +1)2 + ∑b Lj 3 − ∑b Lj 2 − ∑aiL3 − ∑aiL2 , j =1 i =1 j =1 i =1 n n m m U L U L U L A( m +1)3 = A( m +1)3 + ∑b j 3 − ∑b j 3 − ∑ai 3 − ∑ai 3 , j =1 i =1 j =1 i =1 n m n m A(Lm +1) 4 = A(Um +1)3 + ∑b Lj 4 − ∑bUj 3 − ∑aiL4 − ∑aiU3 , j =1 i =1 j =1 i =1 n m n m A(Um +1) 4 = A(Lm +1) 4 + ∑bUj 4 − ∑b Lj 4 − ∑aiU4 − ∑aiL4 . j =1 i =1 j =1 i =1 n
m
∑bUj1 ≤ ∑aiU1 ,
Case 2: If
j =1
n
∑b
U j2
j =1 n
∑b
L j3
j =1 n
∑b j =1
L j4
n
i =1
m
n
n
j =1
j =1
m
n
− ∑b Lj1 ≤ ∑aiU2 − ∑aiL1 , j =1
i =1
i =1
n
m
m
m
m
∑b Lj1 − ∑bUj1 ≤ ∑aiL1 − ∑aiU1 , ∑b j =1
− ∑b Lj 2 ≤ ∑aiL3 − ∑aiL2 , j =1
i =1
i =1
n
m
m
j =1
i =1
i =1
L j2
− ∑bUj 3 ≤ ∑aiL4 − ∑aiU3 ,
n
∑b
U j3
j =1 n
∑b j =1
U j4
i =1
i =1
n
m
m
j =1
i =1
i =1
n
m
m
j =1
i =1
i =1
n
m
m
j =1
i =1
i =1
− ∑bUj 2 ≤ ∑aiL2 − ∑aiU2 , − ∑b Lj 3 ≤ ∑aiU3 − ∑aiL3 , − ∑b Lj 4 ≤ ∑aiU4 − ∑aiL4 then add a
L dummy destination Dn + 1 having dummy demand (B( n +1)1 , B(Ln +1) 2 , B(Ln +1)3 , U B(Ln +1) 4 ) , (B(Un +1)1 , B(Un +1) 2 , B( n +1)3 , B(Un +1) 4 ) by considering the cost for supplying the unit quantity of the product from all the sources to the dummy destination Dn + 1 as a generalized IVTrFN 0 = ( ( 0, 0, 0, 0; 1) , ( 0, 0, 0, 0; 1) .
)
m n m n m n B(Un +1)1 = ∑aiU1 − ∑bUj1 , B(Ln +1)1 = B(Un +1)1 + ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 , i =1 j =1 i =1 j =1 i =1 j =1 m n m n B(Un +1)2 = B(Ln +1)1 + ∑aiU2 − ∑aiL1 − ∑bUj 2 − ∑b Lj1 , i =1 j =1 i =1 j =1 m n m n B(Ln +1)2 = B(Un +1)2 + ∑aiL2 − ∑aiU2 − ∑b Lj 2 − ∑bUj 2 , i =1 j =1 i =1 j =1
Trapezoidal fuzzy number transportation problems Chapter | 18 319
m n m n B(Ln +1)3 = B(Ln +1)2 + ∑aiL3 − ∑aiL2 − ∑b Lj 3 − ∑b Lj 2 , i =1 j =1 i =1 j =1 m n m n B(Un +1)3 = B(Ln +1)3 + ∑aiU3 − ∑aiL3 − ∑bUj 3 − ∑b Lj 3 , i =1 j =1 i =1 j =1 m n m n B(Ln +1) 4 = B(Un +1)3 + ∑aiL4 − ∑aiU3 − ∑b Lj 4 − ∑bUj 3 , i =1 j =1 i =1 j =1 m n m n B(Un +1) 4 = B(Ln +1) 4 + ∑aiU4 − ∑aiL4 − ∑bUj 4 − ∑b Lj 4 . i =1 j =1 i =1 j =1
Case 3: If neither Case 1 nor Case 2 is satisfied, then carry out the following: L (1) Add a dummy source Sm + 1 having dummy supply (A(Lm +1)1 , A( m +1) 2 , U A(Lm +1)3 , A(Lm +1) 4 ) , (A(Um +1)1 , A(Um +1) 2 , A(Um +1)3 , A( m +1) 4 ) by considering the cost for supplying the unit quantity of the product from the dummy source Sm + 1 to all the destinations as a generalized IVTrFN 0 = ( ( 0, 0, 0, 0;1) , ( 0, 0, 0, 0;1) .
)
m n A(Um +1)1 = max 0, ∑bUj1 − ∑aiU1 , i =1 j =1 n m n m A(Lm +1)1 = A(Um +1)1 + max 0, ∑b Lj1 − ∑bUj1 − ∑aiL1 − ∑aiU1 , j =1 j =1 i =1 i =1 n m n m A(Um +1)2 = A(Lm +1)1 + max 0, ∑bUj 2 − ∑b Lj1 − ∑aiU2 − ∑aiL1 , j =1 j =1 i =1 i =1 n m n m A(Lm +1)2 = A(Um +1)2 + max 0, ∑b Lj 2 − ∑bUj 2 − ∑aiL2 − ∑aiU2 , j =1 j =1 i =1 i =1 n m n m A(Lm +1)3 = A(Lm +1)2 + max 0, ∑b Lj 3 − ∑b Lj 2 − ∑aiL3 − ∑aiL2 , j =1 i =1 i =1 j =1 n m n m A(Um +1)3 = A(Lm +1)3 + max 0, ∑bUj 3 − ∑b Lj 3 − ∑aiU3 − ∑aiL3 , j =1 i =1 i =1 j =1 n m n m A(Lm +1) 4 = A(Um +1)3 + max 0, ∑b Lj 4 − ∑bUj 3 − ∑aiL4 − ∑aiU3 , j =1 j =1 i =1 i =1 n m n m A(Um +1) 4 = A(Lm +1) 4 + max 0, ∑bUj 4 − ∑b Lj 4 − ∑aiU4 − ∑aiL4 j =1 j =1 i =1 i =1
320 PART | V Other key issues L L (2) Add a dummy destination Dn + 1 having the dummy demand (B( m +1)1 , B( m +1) 2 , U U L L U B( m +1)3 , B( m +1) 4 ) , (B( m +1)1 , B(Um +1) 2 , B( m +1)3 , B( m +1) 4 ) by considering the cost for supplying the unit quantity of the product from all the sources to the dummy destination Dn + 1 as a generalized IVTrFN 0 = ( ( 0, 0, 0, 0;1) , ( 0, 0, 0, 0;1) .
)
n m B(Un +1)1 = max 0, ∑aiU1 − ∑bUj1 , j =1 i =1 m n m n B(Ln +1)1 = B(Un +1)1 + max 0, ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 , i =1 j =1 j =1 i =1 m n n m B(Un +1)2 = B(Ln +1)1 + max 0, ∑aiU2 − ∑aiL1 − ∑bUj 2 − ∑b Lj1 , i =1 i =1 j =1 j =1
m n m n B(Ln +1)2 = B(Un +1)2 + max 0, ∑aiL2 − ∑aiU2 − ∑b Lj 2 − ∑bUj 2 , i =1 j =1 j =1 i =1 m n m n B(Ln +1)3 = B(Ln +1)2 + max 0, ∑aiL3 − ∑aiL2 − ∑b Lj 3 − ∑b Lj 2 , i =1 j =1 j =1 i =1 m n m n B(Un +1)3 = B(Ln +1)3 + max 0, ∑aiU3 − ∑aiL3 − ∑bUj 3 − ∑b Lj 3 , i =1 i =1 j =1 j =1 m n m n B(Ln +1) 4 = B(Un +1)3 + max 0, ∑aiL4 − ∑aiU3 − ∑b Lj 4 − ∑bUj 3 , i =1 j =1 j =1 i =1 m n m n B(Un +1) 4 = B(Ln +1) 4 + max 0, ∑aiU4 − ∑aiL4 − ∑bUj 4 − ∑b Lj 4 . i =1 j =1 j =1 i =1
5 Validity of the proposed Mehar method To prove the validity of the proposed Mehar method, it is sufficient to prove that: 1. The obtained dummy supply will be a generalized IVTrFN. 2. The obtainedm dummy demand will be a generalized IVTrFN. n 3. In Case 1, a i + A m +1 = b j .
∑ i =1
∑ j =1
Trapezoidal fuzzy number transportation problems Chapter | 18 321
4. In Case 2,
m
n
∑a = ∑b i
i =1
5. In Case 3,
i =1
n +1 . +B
j =1
m
∑a
j
n
i
n +1 . + A m +1 = ∑b j + B j =1
Therefore, the same is proved in this section.
5.1 The obtained dummy supply will be a generalized IVTrFN It is obvious from Case 1 of the proposed Mehar method that for the obtained dummy supply (A(Lm +1)1 , A(Lm +1) 2 , A(Lm +1)3 , A(Lm +1) 4 ) , (A(Um +1)1 , A(Um +1) 2 , A(Um +1)3 , A(Um +1) 4 ) , the L conditions A(Um +1)1 ≤ A(Lm +1)1 ≤ A(Um +1) 2 ≤ A(Lm +1) 2 ≤ A(Lm +1)3 ≤ A(Um +1)3 ≤ A( m +1) 4
≤ A(Um +1) 4 , will be satisfied. Therefore, the dummy supply (A(Lm +1)1 , A(Lm +1) 2 , A(Lm +1)3 , A(Lm +1) 4 ) , (A(Um +1)1 , A(Um +1) 2 , A(Um +1)3 , A(Um +1) 4 ) , obtained from Case 1 of the proposed Mehar method, will always be a generalized IVTrFN.
5.2 The obtained dummy demand will be a generalized IVTrFN It is obvious from Case 2 of the proposed Mehar method that for the obtained U U dummy demand (B(Ln +1)1 , B(Ln +1) 2 , B(Ln +1)3 , B(Ln +1) 4 ) , (B(Un +1)1 , B( n +1) 2 , B( n +1)3 , L ≤ BU ≤ BL ≤ BU ≤ B( n +1)3 B(Un +1) 4 ) , the conditions B(Un +1)1 ≤ B L ( n +1)1
( n +1) 2
( n +1) 2
( n +1) 3
L ( n +1) 4 L L ( n +1) 4 ( n +1) 3
≤ B ≤ B , will be satisfied. Therefore, the dummy demand (B(Ln +1)1 , U U L B( n +1) 2 , B ) , (B(Un +1)1 , B( n +1) 2 , B( n +1)3 , B(Un +1) 4 ) , obtained from the , B Case 2 of the proposed Mehar method, will always be a generalized IVTrFN. U ( n +1) 4
5.3 The obtained dummy supply and dummy demand will be a generalized IVTrFNs It is obvious from Case 3 of the proposed Mehar method that for the obtained L U dummy supply (A( m +1)1 , A(Lm +1) 2 , A(Lm +1)3 , A(Lm +1) 4 ) , (A(Um +1)1 , A(Um +1) 2 , A( m +1)3 , A(Um +1) 4 ) , the conditions A(Um +1)1 ≤ A(Lm +1)1 ≤ A(Um +1) 2 ≤ A(Lm +1) 2 ≤ A(Lm +1)3 ≤ A(Um +1)3 ≤ A(Lm +1) 4 ≤ A(Um +1) 4 , will be satisfied. In addition, for the obtained dummy demand (B(Ln +1)1 , B(Ln +1) 2 , B(Ln +1)3 , B(Ln +1) 4 ) , (B(Un +1)1 , B(Un +1) 2 , B(Un +1)3 , B(Un +1) 4 ) , U the conditions B(Un +1)1 ≤ B(Ln +1)1 ≤ B(Un +1) 2 ≤ B(Ln +1) 2 ≤ B(Ln +1)3 ≤ B( n +1)3 ≤ B(Ln +1) 4 ≤ L B(Un +1) 4 , will be satisfied. Therefore, the dummy supply (A( m +1)1 , A(Lm +1) 2 , A(Lm +1)3 , U A(Lm +1) 4 ) , (A(Um +1)1 , A(Um +1) 2 , A( m +1)3 , A(Um +1) 4 ) and the dummy demand (B(Ln +1)1 , B(Ln +1) 2 , B(Ln +1)3 , B(Ln +1) 4 ) , (B(Un +1)1 , B(Un +1) 2 , B(Un +1)3 , B(Un +1) 4 ) , obtained from Case 3 of the proposed Mehar method, will always be a generalized IVTrFNs.
322 PART | V Other key issues
5.4 Validity of the condition
m
∑ a i =1
n
i
+ A m +1 = ∑ b j j =1
m
n
∑a i + A m +1 = ∑b j i =1
⇒ ∑aiL1 , i =1 m
+
(A
m
m
m
j =1
∑a , ∑a , ∑a L i2
i =1
L ( m +1)1
i =1
L ( m +1) 2
,A
n = ∑b Lj1 , j =1
L i3
n
L i4
i =1
L ( m +1)3
,A n
; ω L , ∑aiU1 , i =1
L ( m +1) 4
,A n
∑b , ∑b , ∑b j =1
L j2
m
L j3
j =1
j =1
L j4
),( A
U ( m +1)1
m
m
m
∑a , ∑a , ∑a U i2
U i3
i =1
i =1
U ( m +1) 2
U ( m +1)3
,A
n ; ω L , ∑bUj1 , j =1
n
U i4
i =1
,A
n
; ωU
,A(Um +1) 4 n
)
∑b , ∑b , ∑b j =1
U j2
j =1
U j3
j =1
U j4
; ω U .
m m m m L L L L L L L L L ∑ai1 + A( m +1)1 , ∑ai 2 + A( m +1)2 , ∑ai 3 + A( m +1)3 , ∑ai 4 + A( m +1) 4 ; ω , i =1 i =1 i =1 ⇒ im=1 m m m U U U U U U U U U ∑ai1 + A( m +1)1 , ∑ai 2 + A( m +1)2 , ∑ai 3 + A( m +1)3 , ∑ai 4 + A( m +1) ; ω i =1 i =1 i =1 i =1 n n n n n L n L n L n L = ∑b j1 , ∑b j 2 , ∑b j 3 , ∑b j 4 ; ω L , ∑bUj1 , ∑bUj 2 , ∑bUj 3 , ∑bUj 4 ; ω U . j =1 j =1 j =1 j =1 j =1 j =1 j =1 j =1 m
n
⇒ ∑aiL1 + A(Lm +1)1 = ∑b Lj1 , i =1
m
∑a i =1
L i2
(1)
j =1
n
+ A(Lm +1)2 = ∑b Lj 2 ,
(2)
j =1
m
n
i =1
j =1
m
n
i =1
j =1
m
n
i =1
j =1
m
n
i =1
j =1
m
n
i =1
j =1
∑aiL3 + A(Lm +1)3 = ∑b Lj 3 , ∑aiL4 + A(Lm +1)4 = ∑b Lj 4 , ∑aiU1 + A(Um +1)1 = ∑bUj1 , ∑aiU2 + A(Um +1)2 = ∑bUj 2 , ∑aiU3 + A(Um +1)3 = ∑bUj 3 ,
(3) (4) (5) (6) (7)
Trapezoidal fuzzy number transportation problems Chapter | 18 323
m
(8)
n
∑a
+ A(Um +1) 4 = ∑bUj 4 .
U i4
i =1
j =1
m
∑a
It is obvious that in order to prove
n
i
i =1
+ A m +1 = ∑b j , there is a need to j =1
prove that Eqs. (1)–(8) are satisfied. Here, only the validity of Eq. (1) is proved. The validity of the remaining equations can be proved in the same manner. m n n L m L m U L L L U U a + A = a + A + b − b ∑ i1 ( m +1)1 ∑ i1 ( m +1)1 ∑ j1 ∑ j1 − ∑ai1 − ∑ai1 i =1 i =1 j =1 i =1 j =1 i =1 m n m n m n m = ∑aiL1 + ∑bUj1 − ∑aiU1 + ∑b Lj1 − ∑bUj1 − ∑aiL1 − ∑aiU1 i =1 j =1 i =1 j =1 i =1 j =1 i =1 m
n
= ∑b Lj1. j =1
m
n
+B n + 1 j
∑ a = ∑ b
5.5 Validity of the condition
i
i =1
m
j =1
n
∑a = ∑b i
i =1
j
n +1 +B
j =1
m m m m ⇒ ∑aiL1 , ∑aiL2 , ∑aiL3 , ∑aiL4 ; ω L , ∑aiU1 , ∑aiU2 , ∑aiU3 , ∑aiU4 ; ω U = = i 1=i 1 =i 1 =i 1 i 1 =i 1 =i 1 =i 1 n n n n n L n L n L n L = ∑b j1 , ∑b j 2 , ∑b j 3 , ∑b j 4 ; ω L , ∑bUj1 , ∑bUj 2 , ∑bUj 3 , ∑bUj 4 ; ω U = = j 1 =j 1 =j 1 =j 1 j 1 =j 1 =j 1 =j 1 m
+
m
( B(
L n +1)1
m
m
)(
, B(Ln +1) 2 , B(Ln +1)3 , B(Ln +1) 4 , B(Un +1)1 , B(Un +1) 2 , B(Un +1)3 , B(Un +1) 4
).
m m m m m m m m ⇒ ∑aiL1 , ∑aiL2 , ∑aiL3 , ∑aiL4 ; ω L , ∑aiU1 , ∑aiU2 , ∑aiU3 , ∑aiU4 ; ω U i =1 i =1 i =1 i =1 i =1 i =1 i =1 i =1 n n n n L L L L L L L L L ∑b j1 + B( n +1)1 , ∑b j 2 + B( n +1)2 , ∑b j 3 + B( n +1)3 , ∑b j 4 + B( n +1) 4 ; ω , j =1 j =1 j =1 j =1 . = n n n n U U U U U U U U U ∑b j1 + B( n +1)1 , ∑b j 2 + B( n +1)2 , ∑b j 3 + B( n +1)3 , ∑b j 4 + B( n +1) 4 ; ω j =1 j =1 j =1 j =1 m
n
i =1
j =1
⇒ ∑aiL1 = ∑b Lj1 + B(Ln +1)1 , m
∑a i =1
L i2
(9)
n
= ∑b Lj 2 + B(Ln +1)2 , j =1
(10)
324 PART | V Other key issues
m
∑a
L i3
i =1 m
L i4
i =1
U i1
i =1
U i2
i =1
n
= ∑bUj1 + B(Un +1)1 ,
(13)
n
= ∑bUj 2 + B(Un +1)2 ,
(14)
j =1
∑a
U i3
i =1 m
(12)
j =1
∑a m
n
= ∑b Lj 4 + B(Ln +1) 4 , j =1
∑a m
(11)
j =1
∑a m
n
= ∑b Lj 3 + B(Ln +1)3 ,
n
= ∑bUj 3 + B(Un +1)3 ,
(15)
j =1
∑a
U i4
i =1
n
= ∑bUj 4 + B(Un +1) 4 .
(16)
j =1
m
n
∑a = ∑b
j + B n +1 It is obvious that in order to prove i =1 , there is a need to j =1 prove that Eqs. (9)–(16) are satisfied. Here, only the validity of Eq. (9) is proved. The validity of the remaining equations can be proved in the same manner. i
n m n m n + B(Ln +1)1 = ∑b Lj1 + B(Un +1)1 + ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 j =1 j =1 i =1 j =1 i =1 j =1 m n m n m m n n = ∑b Lj1 + ∑aiU1 − ∑bUj1 + ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 = ∑aiL1 . j =1 i =1 j =1 i =1 j =1 i =1 j =1 i =1 n
∑b
L j1
m
5.6 Validity of the condition
∑ a i =1
m
∑a
n
i
n + 1 + A m + 1 = ∑ b j + B j =1
n
i
i =1
n +1 . + A m +1 = ∑b j + B j =1
m m m m m m m m ⇒ ∑aiL1 , ∑aiL2 , ∑aiL3 , ∑aiL4 ; ω L , ∑aiU1 , ∑aiU2 , ∑aiU3 , ∑aiU4 ; ω U i =1 i =1 i =1 i =1 i =1 i =1 i =1 i =1
+
(( A
L
L
L
L
( m +1)1 , A( m +1) 2 , A( m +1)3 , A( m +1) 4
) , ( A(
U m +1)1
, A(Um +1)2 , A(Um +1)3 , A(Um +1) 4
)
= ∑b Lj1 , ∑b Lj 2 , ∑b Lj 3 , ∑b Lj 4 ; ω L , ∑bUj1 , ∑bUj 2 , ∑bUj 3 , ∑bUj 4 ; ω U j =1 j =1 j =1 j =1 j =1 j =1 j =1 j =1 n
+
( B(
L n +1)1
n
n
n
n
)(
n
n
n
, B(Ln +1)2 , B(Ln +1)3 , B(Ln +1) 4 , B(Un +1)1 , B(Un +1)2 , B(Un +1)3 , B(Un +1) 4
).
Trapezoidal fuzzy number transportation problems Chapter | 18 325
m m m m L L L L L L L L L a A , a A , a A , + + + ∑ i1 ( m +1)1 ∑ i 2 ( m +1) 2 ∑ i 3 ( m +1)3 ∑ai 4 + A( m +1) 4 ; ω , i =1 i =1 i =1 ⇒ im=1 m m m U U U U U U U U U ∑ai1 + A( m +1)1 , ∑ai 2 + A( m +1)2 , ∑ai 3 + A( m +1)3 , ∑ai 4 + A( m +1) ; ω i =1 i =1 i =1 i =1 n n n n L L L L L L L L L ∑b j1 + B( n +1)1 ,∑b j 2 + B( n +1)2 , ∑b j 3 + B( n +1)3 , ∑b j 4 + B( n +1) 4 ; ω , j =1 j =1 j =1 j =1 . = n n n n U U U U U U U U U ∑b j1 + B( n +1)1 , ∑b j 2 + B( n +1)2 , ∑b j 3 + B( n +1)3 , ∑b j 4 + B( n +1) 4 ; ω j =1 j =1 j =1 j =1 m
n
i =1
j =1
⇒ ∑aiL1 + A(Lm +1)1 = ∑b Lj1 + B(Ln +1)1 , m
∑a
L i2
i =1 m
L i3
i =1
L i4
i =1 m
U i1
i =1
∑a
U i2
i =1
U i3
i =1 m
i =1
(19)
n
+ A(Lm +1) 4 = ∑b Lj 4 + B(Ln +1) 4 ,
(20)
n
+ A(Um +1)1 = ∑bUj1 + B(Un +1)1 ,
(21)
j =1 n
+ A(Um +1)2 = ∑bUj 2 + B(Un +1)2 ,
(22)
j =1
∑a ∑a
n
+ A(Lm +1)3 = ∑b Lj 3 + B(Ln +1)3 ,
j =1
∑a
m
(18)
j =1
∑a
m
n
+ A(Lm +1)2 = ∑b Lj 2 + B(Ln +1)2 , j =1
∑a m
(17)
U i4
n
+ A(Um +1)3 = ∑bUj 3 + B(Un +1)3 ,
(23)
j =1 n
+ A(Um +1) 4 = ∑bUj 4 + B(Un +1) 4 .
It is obvious that in order to prove
j =1 m
∑a i =1
(24) n
i
n +1 , there is a + A m +1 = ∑b j + B j =1
need to prove that Eqs. (17)–(24) are satisfied. Here, only the validity of Eq. (17) is proved. The validity of the remaining equations can be proved in the same manner. Putting the values of A(Lm +1)1 and B(Ln +1)1 in Eq. (17), it will be transformed into Eq. (25). m m n n n m + max 0, ∑bUj1 − ∑aiU1 + max 0, ∑b Lj1 − ∑bUj1 − ∑aiL1 − ∑aiU1 i =1 i =1 j =1 i =1 j =1 j =1 i =1 (25) n n m L m U n L m U n U L U = ∑b j1 + max 0, ∑ai1 − ∑b j1 + max 0, ∑ai1 − ∑ai1 − ∑b j1 − ∑b j1 j =1 j =1 i =1 j =1 j =1 i =1 i =1 m
∑a
L i1
326 PART | V Other key issues
There may be the following four cases: n Case 1: n m L m U and L U ∑b j1 − ∑b j1 ≥ ∑ai1 − ∑ai1 j =1 i =1 j =1 i =1 n m n m Case 2: ∑b Lj1 − ∑bUj1 ≤ ∑aiL1 − ∑aiU1 and j =1 i =1 j =1 i =1 n n L m L m U U Case 3: ∑b j1 − ∑b j1 ≤ ∑ai1 − ∑ai1 and j =1 i =1 j =1 i =1 n m m n L U L U b j1 − ∑b j1 ≥ ∑ai1 − ∑ai1 Case 4: ∑ j =1 j =1 i =1 and i =1
If
n n L m L m U U ∑b j1 − ∑b j1 ≥ ∑ai1 − ∑ai1 j =1 i =1 j =1 i =1
n m m n max 0, ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 = 0 , j =1 i =1 j =1 i =1
n
m
j =1
i =1
∑bUj1 ≥ ∑aiU1 n
∑b
U j1
j =1 n
∑b
U j1
j =1 n
∑b j =1
and
U j1
.
m
≤ ∑aiU1 . i =1 m
≥ ∑aiU1 . i =1 m
≤ ∑aiU1 . i =1
n
m
j =1
i =1
∑bUj1 ≥ ∑aiU1
then
n m max 0, ∑aiU1 − ∑bUj1 = 0 , j =1 i =1
m n n m m m n , n max 0, ∑b Lj1 − ∑bUj1 − ∑aiL1 − ∑aiU1 max 0, bU − aU = bU − aU j1 i1 j1 ∑ ∑ ∑ ∑ i1 j =1 i =1 j =1 i =1 i =1 i =1 j =1 j =1 n m n m = ∑b Lj1 − ∑bUj1 − ∑aiL1 − ∑aiU1 i j j = 1 = 1 = 1 i = 1
and therefore Eq. (25) will be satisfied. If n n L m L m U U ∑b j1 − ∑b j1 ≤ ∑ai1 − ∑ai1 j =1 i =1 j =1 i =1
m n n m max 0, ∑b Lj1 − ∑bUj1 − ∑aiL1 − ∑aiU1 = 0, i =1 j =1 i =1 j =1
and
n
m
∑bUj1 ≤ ∑aiU1 j =1
then
i =1
m max 0, ∑bUj1 − ∑aiU1 = 0, i =1 j =1 n
m n m n m m n n max 0, ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 = ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 , i =1 i =1 j =1 i =1 j =1 j =1 j =1 i =1
n n m and therefore Eq. (25) will be m max 0, ∑aiU1 − ∑bUj1 = ∑aiU1 − ∑bUj1 j =1 j =1 i =1 i =1
satisfied. If
L U L U ∑b j1 − ∑b j1 ≤ ∑ai1 − ∑ai1 j =1 i =1 j =1 i =1 n
n
m
m
m n n m max 0, ∑b Lj1 − ∑bUj1 − ∑aiL1 − ∑aiU1 = 0, j =1 i =1 j =1 i =1
and
n
m
j =1
i =1
∑bUj1 ≥ ∑aiU1
then
m n max 0, ∑bUj1 − ∑aiU1 = 0, i =1 j =1
Trapezoidal fuzzy number transportation problems Chapter | 18 327
n m m n m m n n max 0, ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 = ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 , i =1 j =1 j =1 i =1 i =1 j =1 j =1 i =1
n n and therefore Eq. (25) will be m m max 0, ∑aiU1 − ∑bUj1 = ∑aiU1 − ∑bUj1 j =1 j =1 i =1 i =1 satisfied. n m n m n m If ∑b Lj1 − ∑bUj1 ≥ ∑aiL1 − ∑aiU1 and ∑bUj1 ≤ ∑aiU1 then the values of j =1 i =1 j =1 i =1 j =1 i =1 m n max 0, ∑bUj1 − ∑aiU1 = 0, i =1 j =1
m n m n max 0, ∑b Lj1 − ∑bUj1 − ∑aiL1 − ∑aiU1 = 0, j =1 i =1 j =1 i =1
n m m n m m n n max 0, ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 = ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 , i =1 j =1 j =1 i =1 i =1 j =1 j =1 i =1
n n m m max 0, ∑aiU1 − ∑bUj1 = ∑aiU1 − ∑bUj1 and therefore Eq. (25) will be j =1 j =1 i =1 i =1 satisfied.
6 Invalidity of the existing result Ebrahimnejad (2016) considered a generalized IVTrFNTP having two sources, S1, S2, and three destinations, D1, D2, D3, such that: (i) The
generalized
IVTrF
supplies
2 70, 90, 90,100; 3 , (65, 85, 95,105; 1) ,
at
sources
S1
and
S2
are
2 40, 60, 70, 80; 3 , (35, 55, 75, 85; 1) ,
respectively. (ii) The generalized IVTrF demands at destinations D1, D2 and D3 are 2 30, 40, 50, 70; 3 , (25, 35, 55, 75; 1) ,
2 20, 30, 40, 50; , (15, 25, 45, 55;1) 3
and 40, 50, 50, 80; 2 , (35, 45, 55, 85; 1) , respectively. 3 Ebrahimnejad (2016) claimed that
2 2 70, 90, 90,100; 3 , (65, 85, 95,105; 1) + 40, 60, 70, 80; 3 , (35, 55, 75, 85; 1)
is not equal to
2 = 110,150,160,180; , (35, 55, 75, 85; 1)) 3 2 2 30, 40, 50, 70; 3 , (25, 35, 55, 75,1) + 20, 30, 40, 50, 3 , (15, 25, 45, 55; 1) 2 2 + 40, 50, 50, 80; , (35, 45, 55, 85; 1) = 190,120,140, 200; , (75,105,155, 215; 1) , 3 3
328 PART | V Other key issues
i.e., the considered generalized IVTrFNTP is an unbalanced generalized IVTrFNTP. Therefore, there is a need to add a dummy source S3 having a 2 dummy generalized IVTrF supply 25, 25, 35, 75; 3 , ( 0, 25, 45, 85; 1) and a dummy destination D4 having a dummy generalized IVTrF demand 2 45, 55, 55, 55; 3 , ( 25, 60, 60, 60; 1) in order to solve the considered gener alized IVTrFNTP. However, the dummy generalized IVTrF demand , obtained by Ebrahimnejad (2016), 2 45, 55, 55, 55; 3 , ( 25, 60, 60, 60; 1) is not a generalized IVTrFN for the following reason. It can be easily verified from the graphical representation of generalized IVTrFN (Ebrahimnejad, 2016) as well as from the existing definition of a generalized IVTrFN (Ebrahimnejad, 2016) that in a generalized IVTrFN ⟨(bL1, bL2, bL3, bL4; ωL), (bU1 , bU2 , bU3 , bU4 ; ωU)⟩, the condition bU1 ≤ bL1 ≤ bU2 ≤ b2L ≤ bL3 ≤ bU3 ≤ bL4 ≤ bU4 should always be satisfied. While it can be easily verified that if the generalized IVTrF demand 2 45, 55, 55, 55; 3 , ( 25, 60, 60, 60; 1) is compared with a generalized IVTrFN ⟨(bL1, bL2, bL3, bL4; ωL), (bU1 , bU2 , bU3 , bU4 ; ωU)⟩, then bL1 = 45, bL2 = 55, bL3 = 55, bL4 = 55, bU1 = 25, bU2 = 60, bU3 = 60, b4U = 60. It is obvious that bU2 ≥ bL2, i.e., the necessary condition bU2 ≤ bL2 is not satisfied. Therefore, the obtained dummy demand is not a generalized IVTrFN. Hence, the result of this problem, obtained by Ebrahimnejad (2016), is not correct.
7 Exact dummy supply and dummy demand for the existing generalized IVTrFNTP Ebrahimnejad (2016) solved the generalized IVTrFNTP, to illustrate the proposed method. Since, in the considered generalized IVTrFNTP, total supply 2 110,150,160,180; 3 , (100,140,170,190; 1) is not equal to the total de2 mand , Ebrahimnejad applied 90,120,140, 200; 3 , ( 75,105,155, 215; 1) the proposed method to transform this problem into a balanced generalized IVTrFNTP and claimed that there is a need to add a dummy source having a dummy generalized interval-valued trapezoidal fuzzy (IVTrF) supply 2 25, 25, 35, 75; 3 , ( 0, 25, 45, 85; 1) as well as a dummy destination having 2 a dummy generalized IVTrF demand 45, 55, 55, 55; , ( 25, 60, 60, 60; 1) . 3
Trapezoidal fuzzy number transportation problems Chapter | 18 329
However, as discussed in Section 3, the result of this problem, obtained by Ebrahimnejad (2016), is not correct. In this section, the proposed Mehar method is used to find the correct dummy supply and dummy demand. Using the proposed Mehar method, the exact dummy supply and dummy demand can be obtained as follows: 2
∑a i =1
i
2 2 2 2 2 2 2 2 = ∑aiL1 , ∑aiL2 , ∑aiL3 , ∑aiL4 ; ω L , ∑aiU1 , ∑aiU2 , ∑aiU3 , ∑aiU4 ; ω U i =1 i =1 i =1 i =1 i =1 i =1 i =1 i =1
2 = 110,150,160,180; , (100,140,170,190; 1) 3 and
3
∑b j =1
j
3 3 3 3 3 3 3 3 = ∑b Lj1 , ∑b Lj 2 , ∑b Lj 3 , ∑b Lj 4 ; ω L , ∑bUj1 , ∑bUj 2 , ∑bUj 3 , ∑bUj 4 ; ω U j =1 j =1 j =1 j =1 j =1 j =1 j =1 j =1
2 = 90,120,140, 200; , ( 75,105,155, 215;1) . 3 Case 3 of the proposed Mehar method is satisfied. Therefore, according to Case 3 of the proposed Mehar method, 3 2 3 2 U A33 = A33L + max 0, ∑bUj 3 − ∑b Lj 3 − ∑aiU3 − ∑aiL3 j =1 i =1 i =1 j =1
= 20 + max {0,15 − 10} = 25,
3 2 3 2 U A34L = A33 + max 0, ∑b Lj 4 − ∑bUj 3 − ∑aiL4 − ∑aiU3 j =1 i =1 i =1 j =1 = 25 + max {0,45 − 10} = 60, 3 2 3 2 U A34 = A34L + max 0, ∑bUj 4 − ∑b Lj 4 − ∑aiU4 − ∑aiL4 j =1 i =1 i =1 j =1 = 60 + max {0,15 − 10} = 65, 3 2 U B41 = max 0, ∑aiU1 − ∑bUj1 j =1 i =1
= max {0,100 − 75} = 25,
330 PART | V Other key issues 2 3 2 3 L U B41 = B41 + max 0, ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 , i =1 j =1 j =1 i =1 = 25 + max {0,10 − 15} = 25, 2 3 2 3 U L B42 = B41 + max 0, ∑aiU2 − ∑aiL1 − ∑bUj 2 − ∑b Lj1 i =1 j =1 j =1 i =1
= 25 + max {0,30 − 15} = 40,
2 3 3 2 L U B42 = B42 + max 0, ∑aiL2 − ∑aiU2 − ∑b Lj 2 − ∑bUj 2 i =1 i =1 j =1 j =1 = 40 + max {0,10 − 15} = 40, 2 3 2 3 L L B43 = B42 + max 0, ∑aiL3 − ∑aiL2 − ∑b Lj 3 − ∑b Lj 2 i =1 j =1 j =1 i =1 = 40 + max {0,10 − 20} = 40,
2 3 3 2 U L B43 = B43 + max 0, ∑aiU3 − ∑aiL3 − ∑bUj 3 − ∑b jL3 i =1 i =1 j =1 j =1 = 40 + max {0,10 − 15} = 40, 2 3 3 2 L U B44 = B43 + max 0, ∑aiL4 − ∑aiU3 − ∑b Lj 4 − ∑bUj 3 i =1 i =1 j =1 j =1 = 40 + max {0,10 − 45} = 40, 2 3 2 3 U L B44 = B44 + max 0, ∑aiU4 − ∑aiL4 − ∑bUj 4 − ∑b jL4 i =1 i =1 j =1 j =1
= 40 + max {0,10 − 15} = 40.
Therefore, the dummy fuzzy supply
( A(
L m +1)1
)(
, A(Lm +1)2 , A(Lm +1)3 , A(Lm +1) 4 ; ω L , A(Um +1)1 , A(Um +1)2 , A(Um +1)3 , A(Um +1) 4 ; ω U
2 = 5,10, 20,60; , ( 0, 5, 25, 65; 1) and the dummy fuzzy demand 3
)
Trapezoidal fuzzy number transportation problems Chapter | 18 331
( B(
L n +1)1
)(
, B(Ln +1)2 , B(Ln +1)3 , BF(Ln +1) 4 ; ω L , B(Un +1)1 , B(Un +1)2 , B(Un +1)3 , B(Un +1) 4 ; ω U
)
2 = 25, 40, 40, 40; , ( 25, 40, 40, 40; 1) . 3
8 Exact results of the existing generalized IVTrFNTP To find the solution of the generalized IVTrFNTP, presented in Ebrahimnejad (2016) has solved a crisp linear programming problem having 56 equality constraints. Out of these 56 equality constraints, 8 constraints correspond to dummy supply and 8 constraints correspond to dummy demand. However, as discussed in Section 3, the dummy supply and dummy demand, obtained by Ebrahimnejad (2016), is not correct. Therefore, the result of this problem, obtained by Ebrahimnejad (2016), is not correct. To find the exact result of this problem, the equality constraints C1 of the existing crisp linear programming problem have been replaced by the constraints C2.
L L L L x31,1 + x32,1 + x33,1 + x34,1 = 25, L L L L x31,2 + x32,2 + x33,2 + x34,2 = 25, L L L L x31,3 + x32,3 + x33,3 + x34,3 = 35, L L L L x31,4 + x32,4 + x33,4 + x34,4 = 75, U U U U x31,1 + x32,1 + x33,1 + x34,1 = 0, U U U U x31,2 + x32,2 + x33,2 + x34,2 = 25, U U U U x31,3 + x32,3 + x33,3 + x34,3 = 45, U U U U x31,4 + x32,4 + x33,4 + x34,4 = 85, ( C1 ) L L L x14,1 + x24,1 + x34,1 = 45, L L L x14,2 + x24,2 + x34,2 = 55, L L L x14,3 + x24,3 + x34,3 = 55, L L L x14,4 + x24,4 + x34,4 = 55, U U U x14,1 + x24,1 + x34,1 = 25, U U U x14,2 + x24,2 + x34,2 = 60, U U x14,3 + x2U4,3 + x34,3 60, = U U U 60. x14,4 + x24,4 + x34,4 =
332 PART | V Other key issues L L L L x31,1 + x32,1 + x33,1 + x34,1 = 5, L L L L x31,2 + x32,2 + x33,2 + x34,2 = 10, L L L L x31,3 + x32,3 + x33,3 + x34,3 = 20, L L L L x31,4 + x32,4 + x33,4 + x34,4 = 60, U U U U x31,1 + x32,1 + x33,1 + x34,1 = 0, U U U U x31,2 + x32,2 + x33,2 + x34,2 = 5, U U U U 25, x31,3 + x32,3 + x33 ,3 + x34,3 = U U U U x31,4 65, + x32,4 + x33,4 + x34,4 = ( C2 ) L L L x14,1 + x24,1 + x34,1 = 25, L L L x14,2 40, + x24,2 + x34,2 = L L L x14,3 + x24,3 + x34,3 = 40, L L L x14,4 40, + x24,4 + x34,4 = U U U x14,1 + x24,1 + x34,1 = 25, U U U x14,2 40, + x24,2 + x34,2 = U U U x14,3 + x24, 40, 3 + x34,3 = U U U 40. x14,4 + x24,4 + x34,4 =
On solving the existing crisp linear programming problem (Ebrahimnejad, 2016) with this modification, the obtained exact optimal solution and the optimal cost of the existing generalized IVTrFNTP, presented by Table 1, is 2 x 11 = 30, 40, 40, 50; , (25, 35, 45, 55;1) , 3 2 x 12 = 15, 20, 20, 20; , (15, 20, 20, 20;1) , 3 2 x 13 = 0, 0,0, 0; , (0, 0, 0, 0;1) , 3 2 x 14 = 25, 30, 30, 30; , (25, 30, 30, 30;1) 3 2 2 x 21 = 0, 0,10,10; , (0, 0,10,10;1) , x 22 = 0, 0, 0, 0; , (0, 0, 0, 0; 1) , 3 3 2 2 x 23 = 40, 50, 50, 50; , (35, 45, 55, 65; 1) , x 24 = 0,10,10,10; , (0,10,10,10; 1 , 3 3
2 2 x 31 = 0, 0, 0,10; , (0, 0, 0,10; 1) , x 32 = 5,10, 20, 30; , (0, 5, 25, 35; 1) , 3 3
Trapezoidal fuzzy number transportation problems Chapter | 18 333
2 2 x 33 = 0, 0, 0, 20; , (0, 0, 0, 20; 1) , x 34 = 0, 0, 0, 0; , (0, 0, 0, 0; 1) . 3 3
9 Conclusion In this chapter it has been shown that the existing method for transforming an unbalanced generalized IVTrFNTP into a balanced generalized IVTrFNTP is not valid. In addition, a new method (known as the Mehar method) was proposed for the same purpose and it was proved that this method is valid. Furthermore, the exact result of the existing unbalanced generalized IVTrFNTP was obtained.
References Chanas, S., Kuchta, D., 1996. A concept of the optimal solution of the transportation problem with fuzzy cost coefficients. Fuzzy Sets Syst. 82 (2), 299–305. Chanas, S., Kolodziejczyk, W., Machaj, A., 1984. A fuzzy approach to the transportation problem. Fuzzy Sets Syst. 13 (3), 211–221. Chanas, S., Delgado, M., Verdegay, J.L., Vila, M.A., 1993. Interval and fuzzy extensions of classical transportation problems. Transp. Plan. Technol. 17 (2), 203–218. Chiang, J., 2005. The optimal solution of the transportation problem with fuzzy demand and fuzzy product. J. Inf. Sci. Eng. 21, 439–451. Dinagar, D.S., Palanivel, K., 2009. The transportation problem in fuzzy environment. Int. J. Algorithms Comput. Math. 2 (3), 65–71. Ebrahimnejad, A., 2014. A simplified new approach for solving fuzzy transportation problems with generalized trapezoidal fuzzy numbers. Appl. Soft Comput. 19, 171–176. Ebrahimnejad, A., 2015a. A duality approach for solving bounded linear programming problems with fuzzy variables based on ranking functions and its application in bounded transportation problems. Int. J. Syst. Sci. 46 (11), 2048–2060. Ebrahimnejad, A., 2015b. Note on a fuzzy approach to transport optimization problem. Optim. Eng. https://doi.org/10.1007/s11081-015-9277-y. Ebrahimnejad, A., 2015c. An improved approach for solving transportation problem with triangular fuzzy numbers. J. Intell. Fuzzy Syst. 29 (2), 963–974. Ebrahimnejad, A., 2016. Fuzzy linear programming approach for solving transportation problems with interval-valued trapezoidal fuzzy numbers. Sadhana 41 (3), 299–316. Gupta, A., Kumar, A., 2012. A new method for solving linear multi-objective transportation problems with fuzzy parameters. Appl. Math. Model. 36, 1421–1430. Gupta, A., Kumar, A., Kaur, A., 2012. Mehar’s method to find exact fuzzy optimal solution of unbalanced fully fuzzy multiobjective transportation problems. Optim. Lett. 6, 1737–1751. Jimenez, F., Verdegay, J.L., 1998. Uncertain solid transportation problem. Fuzzy Sets Syst. 100 (13), 45–57. Jimenez, F., Verdegay, J.L., 1999. Solving fuzzy solid transportation problems by an evolutionary algorithm based parametric approach. Eur. J. Oper. Res. 117 (3), 485–510. Kaur, A., Kumar, A., 2012. A new approach for solving fuzzy transportation problems using generalized trapezoidal fuzzy numbers. Appl. Soft Comput. 12 (3), 1201–1213. Kumar, A., Kaur, A., 2010. Application of linear programming for solving fuzzy transportation problems. J. Appl. Math. Inform. 29 (3–4), 831–846. Kumar, A., Kaur, A., 2011a. A new method for solving fuzzy transportation problems using ranking function. Appl. Math. Model. 35 (12), 5652–5661. Kumar, A., Kaur, A., 2011b. Application of classical transportation methods to find the fuzzy optimal solution of fuzzy transportation problems. Fuzzy Inf. Eng. 3 (1), 81–99. Kumar, A., Kaur, A., 2014. Optimal way of selecting cities and conveyances for supplying coal in uncertain environment. Sadhana. https://doi.org/10.1007/s12046-013-0207-4.
334 PART | V Other key issues Liu, S.T., Kao, C., 2004. Solving fuzzy transportation problems based on extension principle. Eur. J. Oper. Res. 153 (3), 661–674. Oheigeartaigh, M., 1982. A fuzzy transportation algorithm. Fuzzy Sets Syst. 8 (3), 235–243. Pandian, P., Natarajan, G., 2010. A new algorithm for finding a fuzzy optimal solution for fuzzy transportation problems. Appl. Math. Sci. 4 (2), 79–90. Shanmugasundari, M., Ganesan, K., 2013. A novel approach for the fuzzy optimal solution of fuzzy transportation problem. Int. J. Eng. Res. Appl. 3 (1), 1416–1421. Sudhagar, S., Ganesan, K., 2012. A fuzzy approach to transport optimization problem. Optim. Eng. https://doi.org/10.1007/s11081-012-9202-6.
Further reading Kumar, A., Kaur, A., 2012. Methods for solving unbalanced fuzzy transportation problems. Oper. Res. 12 (3), 287–316.
Mehar method for solving unbalanced generalized intervalvalued trapezoidal fuzzy number transportation problems Akansha Mishraa, Amit Kumara, S.S. Appadoob a
School of Mathematics, Thapar Institute of Engineering & Technology, Patiala, India, Department of Supply Chain Management, University of Manitoba, Winnipeg, MB, Canada
b
1 Introduction Ebrahimnejad (2016) pointed out that several methods have been proposed for solving fuzzy/fully fuzzy transportation problems (transportation problems in which each parameter is represented by a triangular/trapezoidal fuzzy number) (Ebrahimnejad, 2014, 2015a,b,c, 2016; Oheigeartaigh, 1982; Chanas et al., 1984, 1993; Chanas and Kuchta, 1996; Jimenez and Verdegay, 1998, 1999; Liu and Kao, 2004; Dinagar and Palanivel, 2009; Pandian and Natarajan, 2010; Kumar and Kaur, 2010, 2011a,b, 2014; Gupta et al., 2012; Shanmugasundari and Ganesan, 2013; Sudhagar and Ganesan, 2012; Kaur and Kumar, 2012; Chiang, 2005; Gupta and Kumar, 2012). However, no method has been proposed to find the solution of generalized interval-valued fuzzy transportation problems (transportation problems in which each parameter is represented by a triangular/trapezoidal fuzzy number). To fill this gap, Ebrahimnejad (2016) proposed the following methods: 1. A method to transform an unbalanced generalized IVTrFTP into a balanced generalized IVTrFTP. 2. A linear programming method to find the solution of a balanced generalized IVTrFTP. Ebrahimnejad (2016) considered an unbalanced generalized IVTrFNTP and applied the methods, proposed by him, to transform it into a balanced generalized IVTrFNTP as well as to find its solution.
Maritime Transport and Regional Sustainability. https://doi.org/10.1016/B978-0-12-819134-7.00019-8 © 2020 Elsevier Inc. All rights reserved.
311
312 PART | V Other key issues
In this paper, it is shown that on applying Ebrahimnejad’s method for transforming an unbalanced generalized IVTrFNTP into a balanced generalized IVTrFNTP (Ebrahimnejad, 2016), the obtained dummy supply and/or dummy demand is not a generalized interval-valued trapezoidal fuzzy number (IVTrFN), and therefore this method is not valid. In addition, a new method (known as the Mehar method) is proposed to transform an unbalanced generalized IVTrFNTP into a balanced generalized IVTrFNTP. Furthermore, the validity of the proposed Mehar method is discussed. This paper is organized as follows:
2 Ebrahimnejad’s method for transforming an unbalanced generalized IVTrFNTP into a balanced generalized IVTrFNTP Ebrahimnejad proposed the following method to transform an unbalanced generalized IVTrFNTP into a balanced generalized m
n
∑a ≠ ∑b
IVTrFNTP, i.e., to transform
i
i =1
j
m
n
∑a = ∑b , where, m
into
i
j =1
j
i =1
j =1
represents number of sources, n represents number of destinations, m m m m m m m m m i = a L , a L , a L , a L ; ω L , aU , aU , aU , aU ; ω U a ∑ ∑ ∑ ∑ ∑ i1 ∑ i 2 ∑ i 3 i4 i1 ∑ i 2 ∑ i 3 i4 i =1 i =1 i =1 i =1 i =1 i =1 i =1 i =1 i =1 represents total interval-valued fuzzy supply, and n n n n n n n n n b j = ∑b Lj1 , ∑b Lj 2 , ∑b Lj 3 , ∑b Lj 4 ; ω L , ∑bUj1 , ∑bUj 2 , ∑bUj 3 , ∑bUj 4 ; ω U ∑ j =1 j =1 j =1 j =1 j =1 j =1 j =1 j =1 j =1 represents total interval-valued fuzzy demand. The generalized IVTrFN ⟨(ai1L, U U U U ai2L, ai3L, ai4L; ωL), (ai1 , ai2, ai3, ai4; ωU)⟩ represents the supply of the product at ith U U U U source Si. The generalized IVTrFN ⟨(bj1L, bj2L, bj3L, bj4L; ωL), (bj1 , bj2, bj3, bj4; ωU)⟩ represents the demand of the product at jth destination Dj. Case 1: If
m
∑a
L i1
i =1
m
∑a i =1
U i1
n
≤ ∑bUj1 ,
dummy
j =1
m
n
≤ ∑b Lj1 , j =1
∑a
U i2
i =1
source
m
∑a
L i2
i =1
n
≤ ∑bUj 2 , j =1
m
n
≤ ∑b Lj 2 , j =1
∑a i =1
Sm + 1
U i3
m
∑a
n
≤ ∑bUj 3 , j =1
having
i =1 m
L i3
∑a i =1
n
≤ ∑b
U i4
L 3
j =1 n
,
m
∑a i =1
L i4
n
≤ ∑b Lj 4 , j =1
≤ ∑bUj 4 , then add a j =1
dummy
n m m n m n L m L n L L L L L L L ∑b j1 − ∑ai1 , ∑b j 2 − ∑ai 2 , ∑b j 3 − ∑ai 3 , ∑b j 4 − ∑ai 4 ; ω , j =1 i =1 i =1 j =1 i =1 j =1 i =1 j =1 n U m U n U m U n U m U n U m U U ∑b j1 − ∑ai1 , ∑b j 2 − ∑ai 2 , ∑b j 3 − ∑ai 3 , ∑b j 4 − ∑ai 4 ; ω i =1 j =1 i =1 j =1 i =1 j =1 i =1 j =1
supply
Trapezoidal fuzzy number transportation problems Chapter | 18 313
by considering the cost for supplying the unit quantity of the product from the dummy source Sm + 1 to all the destinations as a generalized IVTrFN 0 = ( ( 0, 0, 0, 0; 1) , ( 0, 0, 0, 0; 1) . n m n
Case 2: If n
∑b
U j1
j =1
then
∑b j =1
L j1
≤ ∑a , n
≤ ∑aiU1 ,
∑b
i =1
a
L i1
i =1
m
add
)
m
U j2
j =1
n
∑b
L j2
j =1
m
n
m
L j3
j =1
i =1
≤ ∑aiL3 , i =1
n
∑b
L j4
j =1 n
m
m
≤ ∑aiL4 , i =1 m
≤ ∑aiU2 ,
∑b
≤ ∑aiU3 ,
∑b
≤ ∑aiU4 ,
destination
Dn + 1
having
dummy
demand
U j3
j =1
i =1
dummy
∑b
≤ ∑aiL2 ,
U j4
j =1
i =1
i =1
m n n m n m L n L m L L L L L L L ∑ai1 − ∑b j1 , ∑ai 2 − ∑b j 2 , ∑ai 3 − ∑b j 3 , ∑ai 4 − ∑b j 4 ; ω , i =1 j =1 j =1 i =1 j =1 i =1 j =1 i =1 m n m n m n m n U U U U U U U U U ∑ai1 − ∑b j1 , ∑ai 2 − ∑b j 2 , ∑ai 3 − ∑b j 3 , ∑ai 4 − ∑b j 4 ; ω j =1 i =1 j =1 i =1 j =1 i =1 j =1 i =1 by considering the cost for supplying the unit quantity of the product from the dummy destination Dn + 1 to all the sources as a generalized IVTrFN, 0 = ( ( 0, 0, 0, 0;1) , ( 0, 0, 0, 0;1) .
)
Case 3: If neither Case 1 nor Case 2 is satisfied, then carry out the following: (i) Add a dummy source Sm + 1 having the dummy supply (A(Lm +1)1 , A(Lm +1) 2 , U A(Lm +1)3, A(Lm +1) 4 ) , (A(Um +1)1 , A(Um +1) 2 , A( m +1)3, A(Um +1) 4 ) by considering the cost for supplying the unit quantity of the product from all the dummy sources Sm + 1 to all the dummy destinations as generalized IVTrFN 0 = ( ( 0, 0, 0, 0;1) , ( 0, 0, 0, 0;1)
)
A(Lm +1)1 =
n
m
∑b − ∑a j =1
U j1
i =1
U i1
m n + max 0,∑b Lj1 − ∑aiL1 , j =1 i =1
n m m n A(Lm +1)2 = A(Lm +1)1 + max 0, ∑b Lj 2 − ∑b Lj1 − ∑aiL2 − ∑aiL1 , j =1 i =1 j =1 i =1
n m m n A(Lm +1)3 = A(Lm +1)2 + max 0, ∑b Lj 3 − ∑b Lj 2 − ∑aiL3 − ∑aiL2 , j =1 i =1 j =1 i =1 n m n m A(Lm +1) 4 = A(Lm +1)3 + max 0, ∑b Lj 4 − ∑b Lj 3 − ∑aiL4 − ∑aiL3 , j =1 i =1 i =1 j =1 m n A(Um +1)1 = max 0,∑bUj1 − ∑aiU1 , j =1 i =1
314 PART | V Other key issues
n m m n A(Um +1)2 = ∑bUj1 − ∑aiU1 + max 0,∑bUj1 − ∑aiU1 j =1 i =1 i =1 j =1 n m m n U + max 0, ∑b j 2 − ∑bUj1 − ∑aiU2 − ∑aiU1 , j =1 i =1 i =1 j =1 n m m n A(Um +1)3 = A(Um +1)2 + max 0, ∑bUj 3 − ∑bUj 2 − ∑aiU3 − ∑aiU2 , j =1 i =1 j =1 i =1 n m n m A(Um +1) 4 = A(Um +1)3 + max 0, ∑bUj 4 − ∑bUj 3 − ∑aiU4 − ∑aiU3 . j =1 i =1 i =1 j =1 L L (ii) Add a dummy destination Dn + 1 having the dummy demand (B( n +1)1 , B( n +1) 2, B(Ln +1)3 , B(Ln +1) 4 ) , (B(Un +1)1 , B(Un +1) 2, B(Un +1)3 , B(Un +1) 4 ) by considering the cost for supplying the unit quantity of the product from all the sources to the dummy destination Dn + 1 as generalized IVTrFN 0 = ( ( 0, 0, 0, 0;1) , ( 0, 0, 0, 0;1) . n m = ∑b − ∑a + max 0,∑aiL1 − ∑b Lj1 , j =1 i =1 j =1 i =1 n
L ( n +1)1
B
m
U j1
U i1
m n m n B(Ln +1)2 = B(Ln +1)1 + max 0, ∑aiL2 − ∑aiL1 − ∑b Lj 2 − ∑b Lj1 , i =1 j =1 j =1 i =1 m n m n B(Ln +1)3 = B(Ln +1)2 + max 0, ∑aiL3 − ∑aiL2 − ∑b Lj 3 − ∑b Lj 2 , i =1 j =1 j =1 i =1 m n n m B(Ln +1) 4 = B(Ln +1)3 + max 0, ∑aiL4 − ∑aiL3 − ∑b Lj 4 − ∑b Lj 3 , i =1 j =1 j =1 i =1 n m B(Un +1)1 = max 0,∑aiU1 − ∑bUj1 , i =1 j =1 n m + max 0,∑aiU1 − ∑bUj1 i =1 j =1 j =1 i =1 n m U m U n U + max 0, ∑ai 2 − ∑ai1 − ∑b j 2 − ∑bUj1 , i =1 j =1 j =1 i =1
B(Un +1)2 =
m
n
∑a − ∑b U i1
U j1
)
Trapezoidal fuzzy number transportation problems Chapter | 18 315
m n m n B(Un +1)3 = B(Un +1)2 + max 0, ∑aiU3 − ∑aiU2 − ∑bUj 3 − ∑bUj 2 , i =1 j =1 j =1 i =1 m m n n B(Un +1) 4 = B(Un +1)3 + max 0, ∑aiU4 − ∑aiU3 − ∑bUj 4 − ∑bUj 3 . i =1 j =1 j =1 i =1
3 Flaws of Ebrahimnejad’s method for transforming an unbalanced generalized IVTrFNTP into a balanced generalized IVTrFNTP In this section, some numerical values are considered to show that Ebrahimnejad’s method for transforming an unbalanced generalized IVTrFNTP into a balanced generalized IVTrFNTP (Ebrahimnejad, 2016) is not valid. (1) Let
and
m m m m m L m L m L m L U U L U U U ∑ai1 , ∑ai 2 , ∑ai 3 , ∑ai 4 ; ω , ∑ai1 , ∑ai 2 ,∑ai 3 , ∑ai 4 ; ω i =1 i =1 i =1 i =1 i =1 i =1 i =1 i =1 = (1, 2, 7,11;1) , ( 0, 3, 8, 7;1)
n L n L n L n L n U n U n U n U U L ∑b j1 , ∑b j 2 , ∑b j 3 , ∑b j 4 ; ω , ∑b j1 , ∑b j 2 , ∑b j 3 , ∑b j 4 ; ω j =1 j =1 j =1 j =1 j =1 j =1 j =1 j =1 = ( 4, 8, 9,13;1) , (1, 6,10,18;1) . m
Since
∑a i =1
m
∑a i =1
U i1
n
L i1
≤ ∑bUj1 , j =1
n
m
≤ ∑b Lj1 ,
∑a
j =1
m
∑a i =1
U i2
L i2
i =1
n
≤ ∑bUj 2 , j =1
n
≤ ∑b Lj 2 , j =1
m
∑a i =1
U i3
m
∑a
L i3
i =1
n
≤ ∑bUj 3 , j =1
m
n
≤ ∑b Lj 3 , j =1
∑a i =1
U i4
m
∑a i =1
L i4
n
≤ ∑b Lj 4 , j =1
n
≤ ∑bUj 4 , i.e., Case 1 of j =1
Ebrahimnejad’s method (Ebrahimnejad, 2016), discussed in Section 2, is satisfied. So, according to (Ebrahimnejad’s (2016) method, the dummy supply will be ⟨(4 − 1, 8 − 2, 9 − 7, 13 − 11; 1), (1 − 0, 6 − 3, 10 − 8, 18 − 7; 1)⟩ = ⟨(3, 6, 2, 2; 1), (1, 3, 2, 11; 1)⟩. However, it is not a generalized IVTrFN for the following reason: It can be easily verified from the graphical representation of generalized IVTrFN (Ebrahimnejad, 2016) as well as from the existing definition of a generalized IVTrFN (Ebrahimnejad, 2016) that in a generalized IVTrFN ⟨(a1L, a2L, a3L, a4L; ωL), (a1U, a2U, a3U, a4U; ωU)⟩ the condition a1U ≤ a1L ≤ a2U ≤ a2L ≤ a3L ≤ a3U ≤ a4L ≤ a4U should always be satisfied. While, for the obtained dummy supply ⟨(3, 6, 2, 2; 1), (1, 3, 2, 11; 1)⟩, this condition is not satisfied. Hence, Ebrahimnejad’s method (Ebrahimnejad, 2016) to obtain the dummy supply is not correct.
316 PART | V Other key issues
(2) Let
and
m m m m m L m L m L m L U U L U U U a , a , a , a ; ω , a , a , a , ∑ i1 ∑ i 2 ∑ i 3 ∑ i 4 ∑ i1 ∑ i 2 ∑ i 3 ∑ai 4 ; ω i =1 i =1 i =1 i =1 i =1 i =1 i =1 i =1 = ( 4, 8, 9,13;1) , (1, 6,10,18;1)
n n n L n L n L n L n U n U U U U L ∑b j1 , ∑b j 2 , ∑b j 3 , ∑b j 4 ; ω , ∑b j1 , ∑b j 2 ,0∑b j 3 , ∑b j 4 ; ω j =1 j =1 j =1 j =1 j =1 j =1 j =1 j =1 = (1, 2, 7,11;1) , ( 0, 3, 8, 7;1) .
Since n
∑b
U j1
n
m
j =1 m
i =1 n
n
≤ ∑aiU1 ,
∑b
U j2
m
∑b Lj 2 ≤ ∑aiL2 ,
∑b Lj1 ≤ ∑aiL1 ,
j =1
m
i =1
n
n
m
∑b Lj 3 ≤ ∑aiL3 , j =1
m
i =1
n
≤ ∑aiU2 , ∑bUj 3 ≤ ∑aiU3 ,
∑b
U j4
n
m
j =1
i =1
∑b Lj 4 ≤ ∑aiL4 ,
m
≤ ∑aiU4 ,
j =1 i =1 j =1 i =1 j =1 i =1 j =1 i =1 i.e., Case 2 of Ebrahimnejad’s method (Ebrahimnejad, 2016), discussed in Section 2, is satisfied. So, according to Ebrahimnejad’s method (Ebrahimnejad, 2016), the dummy demand will be ⟨(4 − 1, 8 − 2, 9 − 7, 13 − 11; 1), (1 − 0, 6 − 3, 10 − 8, 18 − 7; 1)⟩ = ⟨(3, 6, 2, 2; 1), (1, 3, 2, 11; 1)⟩.
However, it is not a generalized IVTrFN for the following reason. It can be easily verified from the graphical representation of generalized IVTrFN (Ebrahimnejad, 2016) as well as from the existing definition of a generalized IVTrFN (Ebrahimnejad, 2016) that in a generalized IVTrFN ⟨(a1L, a2L, a3L, a4L; ωL), (a1U, a2U, a3U, a4U; ωU)⟩ the condition a1U ≤ a1L ≤ a2U ≤ a2L ≤ a3L ≤ a3U ≤ a4L ≤ a4U should always be satisfied. While, for the obtained dummy demand ⟨(3, 6, 2, 2; 1), (1, 3, 2, 11; 1)⟩, this condition is not satisfied. Hence, Ebrahimnejad’s method (Ebrahimnejad, 2016) to obtain the dummy demand is not correct. (3) Let
m m m m m L m L m L m L U U L U U U ∑ai1 , ∑ai 2 , ∑ai 3 , ∑ai 4 ; ω , ∑ai1 , ∑ai 2 , ∑ai 3 , ∑ai 4 ; ω i =1 i =1 i =1 i =1 i =1 i =1 i =1 i =1 2 = 110,150,160,180; , (100,140,170,190; 1) 3
n n n n n n n n and ∑b Lj1 , ∑b Lj 2 , ∑b Lj 3 , ∑b Lj 4 ; ω L , ∑bUj1 , ∑bUj 2 , ∑bUj 3 , ∑bUj 4 ; ω U j =1 j =1 j =1 j =1 j =1 j =1 j =1 j =1 2 = 90,120,140, 200; , ( 75,105,155, 215;1) . 3 n
Since n
∑b j =1
U j1
∑b
L j1
j =1
m
≤ ∑aiU1 , i =1
n
m
∑b
≤ ∑aiL1 ,
j =1
i =1
n
∑b j =1
U j2
L j2
m
n
≤ ∑aiL2 ,
∑b
m
i =1
m
n
m
i =1
j =1
i =1
≤ ∑aiU2 , ∑bUj 3 ≤ ∑aiU3 ,
j =1
∑a i =1
i =1
∑a i =1
m
m
≤ ∑aiL3 ,
L j3
U i4
n
L i4
n
≤ ∑b Lj 4 , j =1
≤ ∑b , i.e., Case 3 of j =1
U j4
Trapezoidal fuzzy number transportation problems Chapter | 18 317
Ebrahimnejad’s method (Ebrahimnejad, 2016), discussed in Section 2, is satisfied. So, according to Ebrahimnejad’s method (Ebrahimnejad, 2016), 2 the dummy demand will be . 45, 55, 55, 55; 3 , ( 25, 60, 60, 60; 1) However, the obtained dummy demand is not a generalized IVTrFN for the following reason. It can be easily verified from the graphical representation of generalized IVTrFN (Ebrahimnejad, 2016) as well as from the existing definition of a generalized IVTrFN (Ebrahimnejad, 2016) that in a generalized IVTrFN ⟨(a1L, a2L, a3L, a4L; ωL), (a1U, a2U, a3U, aU4 ; ωU)⟩ the condition a1U ≤ a1L ≤ a2U ≤ a2L ≤ a3L ≤ a3U ≤ a4L ≤ a4U should always be satisfied. While, for the obtained 2 , this condition is not 45, 55, 55, 55; 3 , ( 25, 60, 60, 60;1) satisfied. Hence, Ebrahimnejad’s method (Ebrahimnejad, 2016) to obtain the dummy demand is not correct. dummy demand
4 Proposed Mehar method for transforming an unbalanced generalized IVTrFNTP into a balanced generalized IVTrFNTP In this section, a new method (known as the Mehar method) is proposed to transform an unbalanced generalized IVTrFNTP into a balanced generalized IVTrFNTP. Let us consider an unbalanced generalized IVTrFNTP having m sources and n destinations such that the availability of the product and the demand of the product at ith source (Si) and jth destination (Dj) be represented by generalized IVTrFNTP a i = aiL1 , aiL2 , aiL3 , aiL4 ; ω L , aiU1 , aiU2 , aiU3 , aiU4 ; ω U and
(
b j =
(b
L j1
)(
)(
, b Lj 2 , b Lj 3 , b Lj 4 ; ω L , bUj1 , bUj 2 , bUj 3 , bUj 4 ; ω U
)
)
, respectively. Then, this un-
balanced generalized IVTrFNTP can be transformed into a balanced generalized IVTrFNTP as follows: Case 1: If m
∑a
U i2
i =1 m
m
m
∑a i =1
n
i =1
i =1
m
j =1
j =1
m
n
n
m
i =1 m
i =1
j =1
m
L i1
i =1
m
L i4
n
m
∑a − ∑a ∑a
i =1
∑a
n
≤ ∑bUj1 ,
− ∑a ≤ ∑b − ∑b , L i1
∑a − ∑a L i3
U i1
L i2
U j2
L j1
≤ ∑b Lj 3 − ∑b Lj 2 , j =1
j =1
n
n
− ∑aiU3 ≤ ∑b Lj 4 − ∑bUj 3 , i =1
j =1
j =1
L i2
i =1
U i1
n
j =1
m
i =1 m
∑a i =1
U i4
j =1
n
n
i =1
j =1
j =1
m
n
n
j =1 n
j =1 n
j =1
j =1
− ∑a ≤ ∑b Lj 2 − ∑bUj 2 , U i2
∑a − ∑a U i3
n
≤ ∑b Lj1 − ∑bUj1 ,
i =1 m
L i3
≤ ∑bUj 3 − ∑b Lj 3 ,
− ∑aiL4 ≤ ∑bUj 4 − ∑b Lj 4 then add a i =1
L ( m +1)1
dummy source Sm + 1 having dummy supply (A , A(Lm +1) 2 , A(Lm +1)3 , U L U U U A( m +1) 4 ) , (A( m +1)1 , A( m +1) 2 , A( m +1)3 , A( m +1) 4 ) by considering the cost for supplying the unit quantity of the product from the dummy source Sm + 1 to all the destinations Dn + 1 as a generalized IVTrFN 0 = ( 0, 0, 0, 0;1) , ( 0, 0, 0, 0;1) .
318 PART | V Other key issues n m n m n m A(Um +1)1 = ∑bUj1 − ∑aiU1 , A(Lm +1)1 = A(Um +1)1 + ∑b Lj1 − ∑bUj1 − ∑aiL1 − ∑aiU1 , j =1 i =1 j =1 i =1 j =1 i =1 n n m m A(Um +1)2 = A(Lm +1)1 + ∑bUj 2 − ∑b Lj1 − ∑aiU2 − ∑aiL1 , j =1 i =1 j =1 i =1 n n m m A(Lm +1)2 = A(Um +1)2 + ∑b Lj 2 − ∑bUj 2 − ∑aiL2 − ∑aiU2 , j =1 i =1 j =1 i =1 n m n m A(Lm +1)3 = A(Lm +1)2 + ∑b Lj 3 − ∑b Lj 2 − ∑aiL3 − ∑aiL2 , j =1 i =1 j =1 i =1 n n m m U L U L U L A( m +1)3 = A( m +1)3 + ∑b j 3 − ∑b j 3 − ∑ai 3 − ∑ai 3 , j =1 i =1 j =1 i =1 n m n m A(Lm +1) 4 = A(Um +1)3 + ∑b Lj 4 − ∑bUj 3 − ∑aiL4 − ∑aiU3 , j =1 i =1 j =1 i =1 n m n m A(Um +1) 4 = A(Lm +1) 4 + ∑bUj 4 − ∑b Lj 4 − ∑aiU4 − ∑aiL4 . j =1 i =1 j =1 i =1 n
m
∑bUj1 ≤ ∑aiU1 ,
Case 2: If
j =1
n
∑b
U j2
j =1 n
∑b
L j3
j =1 n
∑b j =1
L j4
n
i =1
m
n
n
j =1
j =1
m
n
− ∑b Lj1 ≤ ∑aiU2 − ∑aiL1 , j =1
i =1
i =1
n
m
m
m
m
∑b Lj1 − ∑bUj1 ≤ ∑aiL1 − ∑aiU1 , ∑b j =1
− ∑b Lj 2 ≤ ∑aiL3 − ∑aiL2 , j =1
i =1
i =1
n
m
m
j =1
i =1
i =1
L j2
− ∑bUj 3 ≤ ∑aiL4 − ∑aiU3 ,
n
∑b
U j3
j =1 n
∑b j =1
U j4
i =1
i =1
n
m
m
j =1
i =1
i =1
n
m
m
j =1
i =1
i =1
n
m
m
j =1
i =1
i =1
− ∑bUj 2 ≤ ∑aiL2 − ∑aiU2 , − ∑b Lj 3 ≤ ∑aiU3 − ∑aiL3 , − ∑b Lj 4 ≤ ∑aiU4 − ∑aiL4 then add a
L dummy destination Dn + 1 having dummy demand (B( n +1)1 , B(Ln +1) 2 , B(Ln +1)3 , U B(Ln +1) 4 ) , (B(Un +1)1 , B(Un +1) 2 , B( n +1)3 , B(Un +1) 4 ) by considering the cost for supplying the unit quantity of the product from all the sources to the dummy destination Dn + 1 as a generalized IVTrFN 0 = ( ( 0, 0, 0, 0; 1) , ( 0, 0, 0, 0; 1) .
)
m n m n m n B(Un +1)1 = ∑aiU1 − ∑bUj1 , B(Ln +1)1 = B(Un +1)1 + ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 , i =1 j =1 i =1 j =1 i =1 j =1 m n m n B(Un +1)2 = B(Ln +1)1 + ∑aiU2 − ∑aiL1 − ∑bUj 2 − ∑b Lj1 , i =1 j =1 i =1 j =1 m n m n B(Ln +1)2 = B(Un +1)2 + ∑aiL2 − ∑aiU2 − ∑b Lj 2 − ∑bUj 2 , i =1 j =1 i =1 j =1
Trapezoidal fuzzy number transportation problems Chapter | 18 319
m n m n B(Ln +1)3 = B(Ln +1)2 + ∑aiL3 − ∑aiL2 − ∑b Lj 3 − ∑b Lj 2 , i =1 j =1 i =1 j =1 m n m n B(Un +1)3 = B(Ln +1)3 + ∑aiU3 − ∑aiL3 − ∑bUj 3 − ∑b Lj 3 , i =1 j =1 i =1 j =1 m n m n B(Ln +1) 4 = B(Un +1)3 + ∑aiL4 − ∑aiU3 − ∑b Lj 4 − ∑bUj 3 , i =1 j =1 i =1 j =1 m n m n B(Un +1) 4 = B(Ln +1) 4 + ∑aiU4 − ∑aiL4 − ∑bUj 4 − ∑b Lj 4 . i =1 j =1 i =1 j =1
Case 3: If neither Case 1 nor Case 2 is satisfied, then carry out the following: L (1) Add a dummy source Sm + 1 having dummy supply (A(Lm +1)1 , A( m +1) 2 , U A(Lm +1)3 , A(Lm +1) 4 ) , (A(Um +1)1 , A(Um +1) 2 , A(Um +1)3 , A( m +1) 4 ) by considering the cost for supplying the unit quantity of the product from the dummy source Sm + 1 to all the destinations as a generalized IVTrFN 0 = ( ( 0, 0, 0, 0;1) , ( 0, 0, 0, 0;1) .
)
m n A(Um +1)1 = max 0, ∑bUj1 − ∑aiU1 , i =1 j =1 n m n m A(Lm +1)1 = A(Um +1)1 + max 0, ∑b Lj1 − ∑bUj1 − ∑aiL1 − ∑aiU1 , j =1 j =1 i =1 i =1 n m n m A(Um +1)2 = A(Lm +1)1 + max 0, ∑bUj 2 − ∑b Lj1 − ∑aiU2 − ∑aiL1 , j =1 j =1 i =1 i =1 n m n m A(Lm +1)2 = A(Um +1)2 + max 0, ∑b Lj 2 − ∑bUj 2 − ∑aiL2 − ∑aiU2 , j =1 j =1 i =1 i =1 n m n m A(Lm +1)3 = A(Lm +1)2 + max 0, ∑b Lj 3 − ∑b Lj 2 − ∑aiL3 − ∑aiL2 , j =1 i =1 i =1 j =1 n m n m A(Um +1)3 = A(Lm +1)3 + max 0, ∑bUj 3 − ∑b Lj 3 − ∑aiU3 − ∑aiL3 , j =1 i =1 i =1 j =1 n m n m A(Lm +1) 4 = A(Um +1)3 + max 0, ∑b Lj 4 − ∑bUj 3 − ∑aiL4 − ∑aiU3 , j =1 j =1 i =1 i =1 n m n m A(Um +1) 4 = A(Lm +1) 4 + max 0, ∑bUj 4 − ∑b Lj 4 − ∑aiU4 − ∑aiL4 j =1 j =1 i =1 i =1
320 PART | V Other key issues L L (2) Add a dummy destination Dn + 1 having the dummy demand (B( m +1)1 , B( m +1) 2 , U U L L U B( m +1)3 , B( m +1) 4 ) , (B( m +1)1 , B(Um +1) 2 , B( m +1)3 , B( m +1) 4 ) by considering the cost for supplying the unit quantity of the product from all the sources to the dummy destination Dn + 1 as a generalized IVTrFN 0 = ( ( 0, 0, 0, 0;1) , ( 0, 0, 0, 0;1) .
)
n m B(Un +1)1 = max 0, ∑aiU1 − ∑bUj1 , j =1 i =1 m n m n B(Ln +1)1 = B(Un +1)1 + max 0, ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 , i =1 j =1 j =1 i =1 m n n m B(Un +1)2 = B(Ln +1)1 + max 0, ∑aiU2 − ∑aiL1 − ∑bUj 2 − ∑b Lj1 , i =1 i =1 j =1 j =1
m n m n B(Ln +1)2 = B(Un +1)2 + max 0, ∑aiL2 − ∑aiU2 − ∑b Lj 2 − ∑bUj 2 , i =1 j =1 j =1 i =1 m n m n B(Ln +1)3 = B(Ln +1)2 + max 0, ∑aiL3 − ∑aiL2 − ∑b Lj 3 − ∑b Lj 2 , i =1 j =1 j =1 i =1 m n m n B(Un +1)3 = B(Ln +1)3 + max 0, ∑aiU3 − ∑aiL3 − ∑bUj 3 − ∑b Lj 3 , i =1 i =1 j =1 j =1 m n m n B(Ln +1) 4 = B(Un +1)3 + max 0, ∑aiL4 − ∑aiU3 − ∑b Lj 4 − ∑bUj 3 , i =1 j =1 j =1 i =1 m n m n B(Un +1) 4 = B(Ln +1) 4 + max 0, ∑aiU4 − ∑aiL4 − ∑bUj 4 − ∑b Lj 4 . i =1 j =1 j =1 i =1
5 Validity of the proposed Mehar method To prove the validity of the proposed Mehar method, it is sufficient to prove that: 1. The obtained dummy supply will be a generalized IVTrFN. 2. The obtainedm dummy demand will be a generalized IVTrFN. n 3. In Case 1, a i + A m +1 = b j .
∑ i =1
∑ j =1
Trapezoidal fuzzy number transportation problems Chapter | 18 321
4. In Case 2,
m
n
∑a = ∑b i
i =1
5. In Case 3,
i =1
n +1 . +B
j =1
m
∑a
j
n
i
n +1 . + A m +1 = ∑b j + B j =1
Therefore, the same is proved in this section.
5.1 The obtained dummy supply will be a generalized IVTrFN It is obvious from Case 1 of the proposed Mehar method that for the obtained dummy supply (A(Lm +1)1 , A(Lm +1) 2 , A(Lm +1)3 , A(Lm +1) 4 ) , (A(Um +1)1 , A(Um +1) 2 , A(Um +1)3 , A(Um +1) 4 ) , the L conditions A(Um +1)1 ≤ A(Lm +1)1 ≤ A(Um +1) 2 ≤ A(Lm +1) 2 ≤ A(Lm +1)3 ≤ A(Um +1)3 ≤ A( m +1) 4
≤ A(Um +1) 4 , will be satisfied. Therefore, the dummy supply (A(Lm +1)1 , A(Lm +1) 2 , A(Lm +1)3 , A(Lm +1) 4 ) , (A(Um +1)1 , A(Um +1) 2 , A(Um +1)3 , A(Um +1) 4 ) , obtained from Case 1 of the proposed Mehar method, will always be a generalized IVTrFN.
5.2 The obtained dummy demand will be a generalized IVTrFN It is obvious from Case 2 of the proposed Mehar method that for the obtained U U dummy demand (B(Ln +1)1 , B(Ln +1) 2 , B(Ln +1)3 , B(Ln +1) 4 ) , (B(Un +1)1 , B( n +1) 2 , B( n +1)3 , L ≤ BU ≤ BL ≤ BU ≤ B( n +1)3 B(Un +1) 4 ) , the conditions B(Un +1)1 ≤ B L ( n +1)1
( n +1) 2
( n +1) 2
( n +1) 3
L ( n +1) 4 L L ( n +1) 4 ( n +1) 3
≤ B ≤ B , will be satisfied. Therefore, the dummy demand (B(Ln +1)1 , U U L B( n +1) 2 , B ) , (B(Un +1)1 , B( n +1) 2 , B( n +1)3 , B(Un +1) 4 ) , obtained from the , B Case 2 of the proposed Mehar method, will always be a generalized IVTrFN. U ( n +1) 4
5.3 The obtained dummy supply and dummy demand will be a generalized IVTrFNs It is obvious from Case 3 of the proposed Mehar method that for the obtained L U dummy supply (A( m +1)1 , A(Lm +1) 2 , A(Lm +1)3 , A(Lm +1) 4 ) , (A(Um +1)1 , A(Um +1) 2 , A( m +1)3 , A(Um +1) 4 ) , the conditions A(Um +1)1 ≤ A(Lm +1)1 ≤ A(Um +1) 2 ≤ A(Lm +1) 2 ≤ A(Lm +1)3 ≤ A(Um +1)3 ≤ A(Lm +1) 4 ≤ A(Um +1) 4 , will be satisfied. In addition, for the obtained dummy demand (B(Ln +1)1 , B(Ln +1) 2 , B(Ln +1)3 , B(Ln +1) 4 ) , (B(Un +1)1 , B(Un +1) 2 , B(Un +1)3 , B(Un +1) 4 ) , U the conditions B(Un +1)1 ≤ B(Ln +1)1 ≤ B(Un +1) 2 ≤ B(Ln +1) 2 ≤ B(Ln +1)3 ≤ B( n +1)3 ≤ B(Ln +1) 4 ≤ L B(Un +1) 4 , will be satisfied. Therefore, the dummy supply (A( m +1)1 , A(Lm +1) 2 , A(Lm +1)3 , U A(Lm +1) 4 ) , (A(Um +1)1 , A(Um +1) 2 , A( m +1)3 , A(Um +1) 4 ) and the dummy demand (B(Ln +1)1 , B(Ln +1) 2 , B(Ln +1)3 , B(Ln +1) 4 ) , (B(Un +1)1 , B(Un +1) 2 , B(Un +1)3 , B(Un +1) 4 ) , obtained from Case 3 of the proposed Mehar method, will always be a generalized IVTrFNs.
322 PART | V Other key issues
5.4 Validity of the condition
m
∑ a i =1
n
i
+ A m +1 = ∑ b j j =1
m
n
∑a i + A m +1 = ∑b j i =1
⇒ ∑aiL1 , i =1 m
+
(A
m
m
m
j =1
∑a , ∑a , ∑a L i2
i =1
L ( m +1)1
i =1
L ( m +1) 2
,A
n = ∑b Lj1 , j =1
L i3
n
L i4
i =1
L ( m +1)3
,A n
; ω L , ∑aiU1 , i =1
L ( m +1) 4
,A n
∑b , ∑b , ∑b j =1
L j2
m
L j3
j =1
j =1
L j4
),( A
U ( m +1)1
m
m
m
∑a , ∑a , ∑a U i2
U i3
i =1
i =1
U ( m +1) 2
U ( m +1)3
,A
n ; ω L , ∑bUj1 , j =1
n
U i4
i =1
,A
n
; ωU
,A(Um +1) 4 n
)
∑b , ∑b , ∑b j =1
U j2
j =1
U j3
j =1
U j4
; ω U .
m m m m L L L L L L L L L ∑ai1 + A( m +1)1 , ∑ai 2 + A( m +1)2 , ∑ai 3 + A( m +1)3 , ∑ai 4 + A( m +1) 4 ; ω , i =1 i =1 i =1 ⇒ im=1 m m m U U U U U U U U U ∑ai1 + A( m +1)1 , ∑ai 2 + A( m +1)2 , ∑ai 3 + A( m +1)3 , ∑ai 4 + A( m +1) ; ω i =1 i =1 i =1 i =1 n n n n n L n L n L n L = ∑b j1 , ∑b j 2 , ∑b j 3 , ∑b j 4 ; ω L , ∑bUj1 , ∑bUj 2 , ∑bUj 3 , ∑bUj 4 ; ω U . j =1 j =1 j =1 j =1 j =1 j =1 j =1 j =1 m
n
⇒ ∑aiL1 + A(Lm +1)1 = ∑b Lj1 , i =1
m
∑a i =1
L i2
(1)
j =1
n
+ A(Lm +1)2 = ∑b Lj 2 ,
(2)
j =1
m
n
i =1
j =1
m
n
i =1
j =1
m
n
i =1
j =1
m
n
i =1
j =1
m
n
i =1
j =1
∑aiL3 + A(Lm +1)3 = ∑b Lj 3 , ∑aiL4 + A(Lm +1)4 = ∑b Lj 4 , ∑aiU1 + A(Um +1)1 = ∑bUj1 , ∑aiU2 + A(Um +1)2 = ∑bUj 2 , ∑aiU3 + A(Um +1)3 = ∑bUj 3 ,
(3) (4) (5) (6) (7)
Trapezoidal fuzzy number transportation problems Chapter | 18 323
m
(8)
n
∑a
+ A(Um +1) 4 = ∑bUj 4 .
U i4
i =1
j =1
m
∑a
It is obvious that in order to prove
n
i
i =1
+ A m +1 = ∑b j , there is a need to j =1
prove that Eqs. (1)–(8) are satisfied. Here, only the validity of Eq. (1) is proved. The validity of the remaining equations can be proved in the same manner. m n n L m L m U L L L U U a + A = a + A + b − b ∑ i1 ( m +1)1 ∑ i1 ( m +1)1 ∑ j1 ∑ j1 − ∑ai1 − ∑ai1 i =1 i =1 j =1 i =1 j =1 i =1 m n m n m n m = ∑aiL1 + ∑bUj1 − ∑aiU1 + ∑b Lj1 − ∑bUj1 − ∑aiL1 − ∑aiU1 i =1 j =1 i =1 j =1 i =1 j =1 i =1 m
n
= ∑b Lj1. j =1
m
n
+B n + 1 j
∑ a = ∑ b
5.5 Validity of the condition
i
i =1
m
j =1
n
∑a = ∑b i
i =1
j
n +1 +B
j =1
m m m m ⇒ ∑aiL1 , ∑aiL2 , ∑aiL3 , ∑aiL4 ; ω L , ∑aiU1 , ∑aiU2 , ∑aiU3 , ∑aiU4 ; ω U = = i 1=i 1 =i 1 =i 1 i 1 =i 1 =i 1 =i 1 n n n n n L n L n L n L = ∑b j1 , ∑b j 2 , ∑b j 3 , ∑b j 4 ; ω L , ∑bUj1 , ∑bUj 2 , ∑bUj 3 , ∑bUj 4 ; ω U = = j 1 =j 1 =j 1 =j 1 j 1 =j 1 =j 1 =j 1 m
+
m
( B(
L n +1)1
m
m
)(
, B(Ln +1) 2 , B(Ln +1)3 , B(Ln +1) 4 , B(Un +1)1 , B(Un +1) 2 , B(Un +1)3 , B(Un +1) 4
).
m m m m m m m m ⇒ ∑aiL1 , ∑aiL2 , ∑aiL3 , ∑aiL4 ; ω L , ∑aiU1 , ∑aiU2 , ∑aiU3 , ∑aiU4 ; ω U i =1 i =1 i =1 i =1 i =1 i =1 i =1 i =1 n n n n L L L L L L L L L ∑b j1 + B( n +1)1 , ∑b j 2 + B( n +1)2 , ∑b j 3 + B( n +1)3 , ∑b j 4 + B( n +1) 4 ; ω , j =1 j =1 j =1 j =1 . = n n n n U U U U U U U U U ∑b j1 + B( n +1)1 , ∑b j 2 + B( n +1)2 , ∑b j 3 + B( n +1)3 , ∑b j 4 + B( n +1) 4 ; ω j =1 j =1 j =1 j =1 m
n
i =1
j =1
⇒ ∑aiL1 = ∑b Lj1 + B(Ln +1)1 , m
∑a i =1
L i2
(9)
n
= ∑b Lj 2 + B(Ln +1)2 , j =1
(10)
324 PART | V Other key issues
m
∑a
L i3
i =1 m
L i4
i =1
U i1
i =1
U i2
i =1
n
= ∑bUj1 + B(Un +1)1 ,
(13)
n
= ∑bUj 2 + B(Un +1)2 ,
(14)
j =1
∑a
U i3
i =1 m
(12)
j =1
∑a m
n
= ∑b Lj 4 + B(Ln +1) 4 , j =1
∑a m
(11)
j =1
∑a m
n
= ∑b Lj 3 + B(Ln +1)3 ,
n
= ∑bUj 3 + B(Un +1)3 ,
(15)
j =1
∑a
U i4
i =1
n
= ∑bUj 4 + B(Un +1) 4 .
(16)
j =1
m
n
∑a = ∑b
j + B n +1 It is obvious that in order to prove i =1 , there is a need to j =1 prove that Eqs. (9)–(16) are satisfied. Here, only the validity of Eq. (9) is proved. The validity of the remaining equations can be proved in the same manner. i
n m n m n + B(Ln +1)1 = ∑b Lj1 + B(Un +1)1 + ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 j =1 j =1 i =1 j =1 i =1 j =1 m n m n m m n n = ∑b Lj1 + ∑aiU1 − ∑bUj1 + ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 = ∑aiL1 . j =1 i =1 j =1 i =1 j =1 i =1 j =1 i =1 n
∑b
L j1
m
5.6 Validity of the condition
∑ a i =1
m
∑a
n
i
n + 1 + A m + 1 = ∑ b j + B j =1
n
i
i =1
n +1 . + A m +1 = ∑b j + B j =1
m m m m m m m m ⇒ ∑aiL1 , ∑aiL2 , ∑aiL3 , ∑aiL4 ; ω L , ∑aiU1 , ∑aiU2 , ∑aiU3 , ∑aiU4 ; ω U i =1 i =1 i =1 i =1 i =1 i =1 i =1 i =1
+
(( A
L
L
L
L
( m +1)1 , A( m +1) 2 , A( m +1)3 , A( m +1) 4
) , ( A(
U m +1)1
, A(Um +1)2 , A(Um +1)3 , A(Um +1) 4
)
= ∑b Lj1 , ∑b Lj 2 , ∑b Lj 3 , ∑b Lj 4 ; ω L , ∑bUj1 , ∑bUj 2 , ∑bUj 3 , ∑bUj 4 ; ω U j =1 j =1 j =1 j =1 j =1 j =1 j =1 j =1 n
+
( B(
L n +1)1
n
n
n
n
)(
n
n
n
, B(Ln +1)2 , B(Ln +1)3 , B(Ln +1) 4 , B(Un +1)1 , B(Un +1)2 , B(Un +1)3 , B(Un +1) 4
).
Trapezoidal fuzzy number transportation problems Chapter | 18 325
m m m m L L L L L L L L L a A , a A , a A , + + + ∑ i1 ( m +1)1 ∑ i 2 ( m +1) 2 ∑ i 3 ( m +1)3 ∑ai 4 + A( m +1) 4 ; ω , i =1 i =1 i =1 ⇒ im=1 m m m U U U U U U U U U ∑ai1 + A( m +1)1 , ∑ai 2 + A( m +1)2 , ∑ai 3 + A( m +1)3 , ∑ai 4 + A( m +1) ; ω i =1 i =1 i =1 i =1 n n n n L L L L L L L L L ∑b j1 + B( n +1)1 ,∑b j 2 + B( n +1)2 , ∑b j 3 + B( n +1)3 , ∑b j 4 + B( n +1) 4 ; ω , j =1 j =1 j =1 j =1 . = n n n n U U U U U U U U U ∑b j1 + B( n +1)1 , ∑b j 2 + B( n +1)2 , ∑b j 3 + B( n +1)3 , ∑b j 4 + B( n +1) 4 ; ω j =1 j =1 j =1 j =1 m
n
i =1
j =1
⇒ ∑aiL1 + A(Lm +1)1 = ∑b Lj1 + B(Ln +1)1 , m
∑a
L i2
i =1 m
L i3
i =1
L i4
i =1 m
U i1
i =1
∑a
U i2
i =1
U i3
i =1 m
i =1
(19)
n
+ A(Lm +1) 4 = ∑b Lj 4 + B(Ln +1) 4 ,
(20)
n
+ A(Um +1)1 = ∑bUj1 + B(Un +1)1 ,
(21)
j =1 n
+ A(Um +1)2 = ∑bUj 2 + B(Un +1)2 ,
(22)
j =1
∑a ∑a
n
+ A(Lm +1)3 = ∑b Lj 3 + B(Ln +1)3 ,
j =1
∑a
m
(18)
j =1
∑a
m
n
+ A(Lm +1)2 = ∑b Lj 2 + B(Ln +1)2 , j =1
∑a m
(17)
U i4
n
+ A(Um +1)3 = ∑bUj 3 + B(Un +1)3 ,
(23)
j =1 n
+ A(Um +1) 4 = ∑bUj 4 + B(Un +1) 4 .
It is obvious that in order to prove
j =1 m
∑a i =1
(24) n
i
n +1 , there is a + A m +1 = ∑b j + B j =1
need to prove that Eqs. (17)–(24) are satisfied. Here, only the validity of Eq. (17) is proved. The validity of the remaining equations can be proved in the same manner. Putting the values of A(Lm +1)1 and B(Ln +1)1 in Eq. (17), it will be transformed into Eq. (25). m m n n n m + max 0, ∑bUj1 − ∑aiU1 + max 0, ∑b Lj1 − ∑bUj1 − ∑aiL1 − ∑aiU1 i =1 i =1 j =1 i =1 j =1 j =1 i =1 (25) n n m L m U n L m U n U L U = ∑b j1 + max 0, ∑ai1 − ∑b j1 + max 0, ∑ai1 − ∑ai1 − ∑b j1 − ∑b j1 j =1 j =1 i =1 j =1 j =1 i =1 i =1 m
∑a
L i1
326 PART | V Other key issues
There may be the following four cases: n Case 1: n m L m U and L U ∑b j1 − ∑b j1 ≥ ∑ai1 − ∑ai1 j =1 i =1 j =1 i =1 n m n m Case 2: ∑b Lj1 − ∑bUj1 ≤ ∑aiL1 − ∑aiU1 and j =1 i =1 j =1 i =1 n n L m L m U U Case 3: ∑b j1 − ∑b j1 ≤ ∑ai1 − ∑ai1 and j =1 i =1 j =1 i =1 n m m n L U L U b j1 − ∑b j1 ≥ ∑ai1 − ∑ai1 Case 4: ∑ j =1 j =1 i =1 and i =1
If
n n L m L m U U ∑b j1 − ∑b j1 ≥ ∑ai1 − ∑ai1 j =1 i =1 j =1 i =1
n m m n max 0, ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 = 0 , j =1 i =1 j =1 i =1
n
m
j =1
i =1
∑bUj1 ≥ ∑aiU1 n
∑b
U j1
j =1 n
∑b
U j1
j =1 n
∑b j =1
and
U j1
.
m
≤ ∑aiU1 . i =1 m
≥ ∑aiU1 . i =1 m
≤ ∑aiU1 . i =1
n
m
j =1
i =1
∑bUj1 ≥ ∑aiU1
then
n m max 0, ∑aiU1 − ∑bUj1 = 0 , j =1 i =1
m n n m m m n , n max 0, ∑b Lj1 − ∑bUj1 − ∑aiL1 − ∑aiU1 max 0, bU − aU = bU − aU j1 i1 j1 ∑ ∑ ∑ ∑ i1 j =1 i =1 j =1 i =1 i =1 i =1 j =1 j =1 n m n m = ∑b Lj1 − ∑bUj1 − ∑aiL1 − ∑aiU1 i j j = 1 = 1 = 1 i = 1
and therefore Eq. (25) will be satisfied. If n n L m L m U U ∑b j1 − ∑b j1 ≤ ∑ai1 − ∑ai1 j =1 i =1 j =1 i =1
m n n m max 0, ∑b Lj1 − ∑bUj1 − ∑aiL1 − ∑aiU1 = 0, i =1 j =1 i =1 j =1
and
n
m
∑bUj1 ≤ ∑aiU1 j =1
then
i =1
m max 0, ∑bUj1 − ∑aiU1 = 0, i =1 j =1 n
m n m n m m n n max 0, ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 = ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 , i =1 i =1 j =1 i =1 j =1 j =1 j =1 i =1
n n m and therefore Eq. (25) will be m max 0, ∑aiU1 − ∑bUj1 = ∑aiU1 − ∑bUj1 j =1 j =1 i =1 i =1
satisfied. If
L U L U ∑b j1 − ∑b j1 ≤ ∑ai1 − ∑ai1 j =1 i =1 j =1 i =1 n
n
m
m
m n n m max 0, ∑b Lj1 − ∑bUj1 − ∑aiL1 − ∑aiU1 = 0, j =1 i =1 j =1 i =1
and
n
m
j =1
i =1
∑bUj1 ≥ ∑aiU1
then
m n max 0, ∑bUj1 − ∑aiU1 = 0, i =1 j =1
Trapezoidal fuzzy number transportation problems Chapter | 18 327
n m m n m m n n max 0, ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 = ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 , i =1 j =1 j =1 i =1 i =1 j =1 j =1 i =1
n n and therefore Eq. (25) will be m m max 0, ∑aiU1 − ∑bUj1 = ∑aiU1 − ∑bUj1 j =1 j =1 i =1 i =1 satisfied. n m n m n m If ∑b Lj1 − ∑bUj1 ≥ ∑aiL1 − ∑aiU1 and ∑bUj1 ≤ ∑aiU1 then the values of j =1 i =1 j =1 i =1 j =1 i =1 m n max 0, ∑bUj1 − ∑aiU1 = 0, i =1 j =1
m n m n max 0, ∑b Lj1 − ∑bUj1 − ∑aiL1 − ∑aiU1 = 0, j =1 i =1 j =1 i =1
n m m n m m n n max 0, ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 = ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 , i =1 j =1 j =1 i =1 i =1 j =1 j =1 i =1
n n m m max 0, ∑aiU1 − ∑bUj1 = ∑aiU1 − ∑bUj1 and therefore Eq. (25) will be j =1 j =1 i =1 i =1 satisfied.
6 Invalidity of the existing result Ebrahimnejad (2016) considered a generalized IVTrFNTP having two sources, S1, S2, and three destinations, D1, D2, D3, such that: (i) The
generalized
IVTrF
supplies
2 70, 90, 90,100; 3 , (65, 85, 95,105; 1) ,
at
sources
S1
and
S2
are
2 40, 60, 70, 80; 3 , (35, 55, 75, 85; 1) ,
respectively. (ii) The generalized IVTrF demands at destinations D1, D2 and D3 are 2 30, 40, 50, 70; 3 , (25, 35, 55, 75; 1) ,
2 20, 30, 40, 50; , (15, 25, 45, 55;1) 3
and 40, 50, 50, 80; 2 , (35, 45, 55, 85; 1) , respectively. 3 Ebrahimnejad (2016) claimed that
2 2 70, 90, 90,100; 3 , (65, 85, 95,105; 1) + 40, 60, 70, 80; 3 , (35, 55, 75, 85; 1)
is not equal to
2 = 110,150,160,180; , (35, 55, 75, 85; 1)) 3 2 2 30, 40, 50, 70; 3 , (25, 35, 55, 75,1) + 20, 30, 40, 50, 3 , (15, 25, 45, 55; 1) 2 2 + 40, 50, 50, 80; , (35, 45, 55, 85; 1) = 190,120,140, 200; , (75,105,155, 215; 1) , 3 3
328 PART | V Other key issues
i.e., the considered generalized IVTrFNTP is an unbalanced generalized IVTrFNTP. Therefore, there is a need to add a dummy source S3 having a 2 dummy generalized IVTrF supply 25, 25, 35, 75; 3 , ( 0, 25, 45, 85; 1) and a dummy destination D4 having a dummy generalized IVTrF demand 2 45, 55, 55, 55; 3 , ( 25, 60, 60, 60; 1) in order to solve the considered gener alized IVTrFNTP. However, the dummy generalized IVTrF demand , obtained by Ebrahimnejad (2016), 2 45, 55, 55, 55; 3 , ( 25, 60, 60, 60; 1) is not a generalized IVTrFN for the following reason. It can be easily verified from the graphical representation of generalized IVTrFN (Ebrahimnejad, 2016) as well as from the existing definition of a generalized IVTrFN (Ebrahimnejad, 2016) that in a generalized IVTrFN ⟨(bL1, bL2, bL3, bL4; ωL), (bU1 , bU2 , bU3 , bU4 ; ωU)⟩, the condition bU1 ≤ bL1 ≤ bU2 ≤ b2L ≤ bL3 ≤ bU3 ≤ bL4 ≤ bU4 should always be satisfied. While it can be easily verified that if the generalized IVTrF demand 2 45, 55, 55, 55; 3 , ( 25, 60, 60, 60; 1) is compared with a generalized IVTrFN ⟨(bL1, bL2, bL3, bL4; ωL), (bU1 , bU2 , bU3 , bU4 ; ωU)⟩, then bL1 = 45, bL2 = 55, bL3 = 55, bL4 = 55, bU1 = 25, bU2 = 60, bU3 = 60, b4U = 60. It is obvious that bU2 ≥ bL2, i.e., the necessary condition bU2 ≤ bL2 is not satisfied. Therefore, the obtained dummy demand is not a generalized IVTrFN. Hence, the result of this problem, obtained by Ebrahimnejad (2016), is not correct.
7 Exact dummy supply and dummy demand for the existing generalized IVTrFNTP Ebrahimnejad (2016) solved the generalized IVTrFNTP, to illustrate the proposed method. Since, in the considered generalized IVTrFNTP, total supply 2 110,150,160,180; 3 , (100,140,170,190; 1) is not equal to the total de2 mand , Ebrahimnejad applied 90,120,140, 200; 3 , ( 75,105,155, 215; 1) the proposed method to transform this problem into a balanced generalized IVTrFNTP and claimed that there is a need to add a dummy source having a dummy generalized interval-valued trapezoidal fuzzy (IVTrF) supply 2 25, 25, 35, 75; 3 , ( 0, 25, 45, 85; 1) as well as a dummy destination having 2 a dummy generalized IVTrF demand 45, 55, 55, 55; , ( 25, 60, 60, 60; 1) . 3
Trapezoidal fuzzy number transportation problems Chapter | 18 329
However, as discussed in Section 3, the result of this problem, obtained by Ebrahimnejad (2016), is not correct. In this section, the proposed Mehar method is used to find the correct dummy supply and dummy demand. Using the proposed Mehar method, the exact dummy supply and dummy demand can be obtained as follows: 2
∑a i =1
i
2 2 2 2 2 2 2 2 = ∑aiL1 , ∑aiL2 , ∑aiL3 , ∑aiL4 ; ω L , ∑aiU1 , ∑aiU2 , ∑aiU3 , ∑aiU4 ; ω U i =1 i =1 i =1 i =1 i =1 i =1 i =1 i =1
2 = 110,150,160,180; , (100,140,170,190; 1) 3 and
3
∑b j =1
j
3 3 3 3 3 3 3 3 = ∑b Lj1 , ∑b Lj 2 , ∑b Lj 3 , ∑b Lj 4 ; ω L , ∑bUj1 , ∑bUj 2 , ∑bUj 3 , ∑bUj 4 ; ω U j =1 j =1 j =1 j =1 j =1 j =1 j =1 j =1
2 = 90,120,140, 200; , ( 75,105,155, 215;1) . 3 Case 3 of the proposed Mehar method is satisfied. Therefore, according to Case 3 of the proposed Mehar method, 3 2 3 2 U A33 = A33L + max 0, ∑bUj 3 − ∑b Lj 3 − ∑aiU3 − ∑aiL3 j =1 i =1 i =1 j =1
= 20 + max {0,15 − 10} = 25,
3 2 3 2 U A34L = A33 + max 0, ∑b Lj 4 − ∑bUj 3 − ∑aiL4 − ∑aiU3 j =1 i =1 i =1 j =1 = 25 + max {0,45 − 10} = 60, 3 2 3 2 U A34 = A34L + max 0, ∑bUj 4 − ∑b Lj 4 − ∑aiU4 − ∑aiL4 j =1 i =1 i =1 j =1 = 60 + max {0,15 − 10} = 65, 3 2 U B41 = max 0, ∑aiU1 − ∑bUj1 j =1 i =1
= max {0,100 − 75} = 25,
330 PART | V Other key issues 2 3 2 3 L U B41 = B41 + max 0, ∑aiL1 − ∑aiU1 − ∑b Lj1 − ∑bUj1 , i =1 j =1 j =1 i =1 = 25 + max {0,10 − 15} = 25, 2 3 2 3 U L B42 = B41 + max 0, ∑aiU2 − ∑aiL1 − ∑bUj 2 − ∑b Lj1 i =1 j =1 j =1 i =1
= 25 + max {0,30 − 15} = 40,
2 3 3 2 L U B42 = B42 + max 0, ∑aiL2 − ∑aiU2 − ∑b Lj 2 − ∑bUj 2 i =1 i =1 j =1 j =1 = 40 + max {0,10 − 15} = 40, 2 3 2 3 L L B43 = B42 + max 0, ∑aiL3 − ∑aiL2 − ∑b Lj 3 − ∑b Lj 2 i =1 j =1 j =1 i =1 = 40 + max {0,10 − 20} = 40,
2 3 3 2 U L B43 = B43 + max 0, ∑aiU3 − ∑aiL3 − ∑bUj 3 − ∑b jL3 i =1 i =1 j =1 j =1 = 40 + max {0,10 − 15} = 40, 2 3 3 2 L U B44 = B43 + max 0, ∑aiL4 − ∑aiU3 − ∑b Lj 4 − ∑bUj 3 i =1 i =1 j =1 j =1 = 40 + max {0,10 − 45} = 40, 2 3 2 3 U L B44 = B44 + max 0, ∑aiU4 − ∑aiL4 − ∑bUj 4 − ∑b jL4 i =1 i =1 j =1 j =1
= 40 + max {0,10 − 15} = 40.
Therefore, the dummy fuzzy supply
( A(
L m +1)1
)(
, A(Lm +1)2 , A(Lm +1)3 , A(Lm +1) 4 ; ω L , A(Um +1)1 , A(Um +1)2 , A(Um +1)3 , A(Um +1) 4 ; ω U
2 = 5,10, 20,60; , ( 0, 5, 25, 65; 1) and the dummy fuzzy demand 3
)
Trapezoidal fuzzy number transportation problems Chapter | 18 331
( B(
L n +1)1
)(
, B(Ln +1)2 , B(Ln +1)3 , BF(Ln +1) 4 ; ω L , B(Un +1)1 , B(Un +1)2 , B(Un +1)3 , B(Un +1) 4 ; ω U
)
2 = 25, 40, 40, 40; , ( 25, 40, 40, 40; 1) . 3
8 Exact results of the existing generalized IVTrFNTP To find the solution of the generalized IVTrFNTP, presented in Ebrahimnejad (2016) has solved a crisp linear programming problem having 56 equality constraints. Out of these 56 equality constraints, 8 constraints correspond to dummy supply and 8 constraints correspond to dummy demand. However, as discussed in Section 3, the dummy supply and dummy demand, obtained by Ebrahimnejad (2016), is not correct. Therefore, the result of this problem, obtained by Ebrahimnejad (2016), is not correct. To find the exact result of this problem, the equality constraints C1 of the existing crisp linear programming problem have been replaced by the constraints C2.
L L L L x31,1 + x32,1 + x33,1 + x34,1 = 25, L L L L x31,2 + x32,2 + x33,2 + x34,2 = 25, L L L L x31,3 + x32,3 + x33,3 + x34,3 = 35, L L L L x31,4 + x32,4 + x33,4 + x34,4 = 75, U U U U x31,1 + x32,1 + x33,1 + x34,1 = 0, U U U U x31,2 + x32,2 + x33,2 + x34,2 = 25, U U U U x31,3 + x32,3 + x33,3 + x34,3 = 45, U U U U x31,4 + x32,4 + x33,4 + x34,4 = 85, ( C1 ) L L L x14,1 + x24,1 + x34,1 = 45, L L L x14,2 + x24,2 + x34,2 = 55, L L L x14,3 + x24,3 + x34,3 = 55, L L L x14,4 + x24,4 + x34,4 = 55, U U U x14,1 + x24,1 + x34,1 = 25, U U U x14,2 + x24,2 + x34,2 = 60, U U x14,3 + x2U4,3 + x34,3 60, = U U U 60. x14,4 + x24,4 + x34,4 =
332 PART | V Other key issues L L L L x31,1 + x32,1 + x33,1 + x34,1 = 5, L L L L x31,2 + x32,2 + x33,2 + x34,2 = 10, L L L L x31,3 + x32,3 + x33,3 + x34,3 = 20, L L L L x31,4 + x32,4 + x33,4 + x34,4 = 60, U U U U x31,1 + x32,1 + x33,1 + x34,1 = 0, U U U U x31,2 + x32,2 + x33,2 + x34,2 = 5, U U U U 25, x31,3 + x32,3 + x33 ,3 + x34,3 = U U U U x31,4 65, + x32,4 + x33,4 + x34,4 = ( C2 ) L L L x14,1 + x24,1 + x34,1 = 25, L L L x14,2 40, + x24,2 + x34,2 = L L L x14,3 + x24,3 + x34,3 = 40, L L L x14,4 40, + x24,4 + x34,4 = U U U x14,1 + x24,1 + x34,1 = 25, U U U x14,2 40, + x24,2 + x34,2 = U U U x14,3 + x24, 40, 3 + x34,3 = U U U 40. x14,4 + x24,4 + x34,4 =
On solving the existing crisp linear programming problem (Ebrahimnejad, 2016) with this modification, the obtained exact optimal solution and the optimal cost of the existing generalized IVTrFNTP, presented by Table 1, is 2 x 11 = 30, 40, 40, 50; , (25, 35, 45, 55;1) , 3 2 x 12 = 15, 20, 20, 20; , (15, 20, 20, 20;1) , 3 2 x 13 = 0, 0,0, 0; , (0, 0, 0, 0;1) , 3 2 x 14 = 25, 30, 30, 30; , (25, 30, 30, 30;1) 3 2 2 x 21 = 0, 0,10,10; , (0, 0,10,10;1) , x 22 = 0, 0, 0, 0; , (0, 0, 0, 0; 1) , 3 3 2 2 x 23 = 40, 50, 50, 50; , (35, 45, 55, 65; 1) , x 24 = 0,10,10,10; , (0,10,10,10; 1 , 3 3
2 2 x 31 = 0, 0, 0,10; , (0, 0, 0,10; 1) , x 32 = 5,10, 20, 30; , (0, 5, 25, 35; 1) , 3 3
Trapezoidal fuzzy number transportation problems Chapter | 18 333
2 2 x 33 = 0, 0, 0, 20; , (0, 0, 0, 20; 1) , x 34 = 0, 0, 0, 0; , (0, 0, 0, 0; 1) . 3 3
9 Conclusion In this chapter it has been shown that the existing method for transforming an unbalanced generalized IVTrFNTP into a balanced generalized IVTrFNTP is not valid. In addition, a new method (known as the Mehar method) was proposed for the same purpose and it was proved that this method is valid. Furthermore, the exact result of the existing unbalanced generalized IVTrFNTP was obtained.
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Further reading Kumar, A., Kaur, A., 2012. Methods for solving unbalanced fuzzy transportation problems. Oper. Res. 12 (3), 287–316.