Efficient approach for solving type-1 intuitionistic fuzzy

Int J Syst Assur Eng Manag DOI 10.1007/s13198-014-0274-x

ORIGINAL ARTICLE

Efficient approach for solving type-1 intuitionistic fuzzy transportation problem Sujeet Kumar Singh • Shiv Prasad Yadav

Received: 15 November 2013 / Revised: 5 March 2014  The Society for Reliability Engineering, Quality and Operations Management (SREQOM), India and The Division of Operation and Maintenance, Lulea University of Technology, Sweden 2014

Abstract In today’s real world problems such as in corporate or in industrial, the experts and the decision makers have to suffer with uncertainty as well as with hesitation usually, due to the complexity of the situations. The main reasons behind these complexities are lack of good communications with all involved persons, error in data’s, understanding of markets, unawareness of customers etc. So, In this paper, we consider a transportation problem having uncertainty and hesitation in supply and demand. We formulate the problem and utilize triangular intuitionistic fuzzy numbers (TIFNs) to deal with uncertainty and hesitation. We propose intuitionistic fuzzy methods to find starting basic feasible solution in terms of TIFNs. Intuitionistic fuzzy modified distribution method has been proposed to find optimal solution. The shortcomings of the existing methods have been pointed out. The proposed method is illustrated by numerical examples. Keywords Triangular intuitionistic fuzzy number  Intuitionistic fuzzy transportation problem  Basic feasible solution  Optimal solution

1 Introduction The fuzzy set theory introduced by Zadeh (1965) has been applied successfully in various fields. The concept of

S. K. Singh (&)  S. P. Yadav Department of Mathematics, Indian Institute of Technology Roorkee, Roorkee, India e-mail: [email protected] S. P. Yadav e-mail: [email protected]

intuitionistic fuzzy set (IFS) proposed by Atanassov (1986) is found to be highly useful to deal with vagueness. The major advantage of IFS over fuzzy set is that IFS separates the degree of membership (belongingness) and the degree of non-membership (non-belongingness) of an element in the set. With the help of IFS theory decision maker (DM) can decide about the degree of acceptance, degree of nonacceptance and degree of hesitation for some quantity. In case of transportation problem, the DM can decide about the level of acceptance and non-acceptance for the transportation cost. Due to this, the application of IFS theory became very popular in decision making problems. The concept of fuzzy mathematical programming was firstly introduced by Bellman and Zadeh (1970). Then many authors have used fuzzy set theory in various optimization problems (Bector and Chandra 2002; Ganesan and Veeramani 2006; Ebrahimipur et al. 2011; Kapur et al. 2011; Nagoorgani and Ponnalagu 2012; Kheirfam and Verdegay 2013). Nagoorgani and Razak (2006) presented a two stage cost minimizing fuzzy transportation problem (FTP) in which supplies and demands are trapezoidal fuzzy numbers. Dinager and Palanivel (2009) investigated a method to solve FTP by taking trapezoidal fuzzy numbers. Pandian and Natarajan (2010) presented a new algorithm for finding a fuzzy optimal solution for FTP. Mohideen and Kumar (2010) did a comparative study on transportation problem in fuzzy environment. Kaur and Kumar (2012) solved FTP taking generalized trapezoidal fuzzy numbers. In various real world transportation problems, the demands as well as availabilities are not known exactly. These are uncertain quantities with hesitation due to various factors. This may be due to lack of good communications with all involved persons, error in data’s, understanding of markets, unawareness of customers and many more. In such

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situations the DM cannot predict exactly. He may hesitate. Thus to represent the uncertainty with hesitation, IFS theory seems to be more reliable than the general fuzzy set theory. In this paper, we utilize triangular intuitionistic fuzzy numbers (TIFN) to represent demands and availabilities. Some authors have solved transportation problem by using intuitionistic fuzzy numbers (IFNs) (Hussain and Kumar 2012; Nagoorgani and Abbas 2013). In the existing methods for solving IFTP (Hussain and Kumar 2012; Nagoorgani and Abbas 2013) authors have proposed ranking procedures for ordering the IFNs. In both the procedures the used ranking functions are not universal. These ranking functions cannot be used for ordering all IFNs, hence these ranking cannot be utilized for solving a general IFTP. To overcome this shortcoming, we developed a new ordering procedure using accuracy function of TIFN and used this ordering to develop an algorithm for finding optimal solution of IFTP. The paper is organized as follows: Sect. 2 deals with some definitions from literature and arithmetic operations on TIFNs (Atanassov 1986; Zimmerman 1996; Hussain and Kumar 2012). In Sect. 3, a new ordering of TIFNs is given using accuracy function and the shortcomings of the ranking functions used in Hussain and Kumar (2012) and Nagoorgani and Abbas (2013) are discussed. Section 4 deals with IFTP of type-1. Section 5 deals with solution procedure. In Sect. 6, a numerical example is given to illustrate the method to get optimal solution. In Sect. 7, result and discussion of the obtained transportation cost is given followed by conclusion in Sect. 8.

2 Some Definitions Definition 2.1 Let X be a universal set. Then a fuzzy set A~ in Xis defined by   A~ ¼ ðx; lA~ðxÞÞ : x 2 X ; where lA~ : X ! ½0; 1: Definition 2.2 Let X be a universe of discourse. Then an IFS A~I in X is defined by a set of ordered triples   A~I ¼ \x; lA~I ðxÞ; tA~I ðxÞ [ : x 2 X ; where lA~I ; tA~I : X ! ½0; 1 are functions such that 0  lA~I ðxÞ þ tA~I ðxÞ  1; 8x 2 X: lA~I ðxÞ represents the degree of membership and tA~I ðxÞ represents the degree of non- membership of the element x 2 X being in A~I . The degree of hesitation for the element x 2 X being in A~I is given by hðxÞ ¼ 1  lA~I ðxÞ  tA~I ðxÞ  1 8x 2 X: ~I

Definition 2.3 An intuitionistic fuzzy subset A ¼   \x; lA~I ðxÞ; tA~I ðxÞ [ : x 2 X of the real line R is called an IFN if the following hold:

123

1 2

There exists m2R such that lA~I ðmÞ ¼ 1 and tA~I ðmÞ ¼ 0 (m is called the mean value of A~I ). lA~I and tA~I are piecewise continuous mappings from Rto the closed interval [0, 1] and the relation 0  lA~I ðxÞ þ tA~I ðxÞ  1; 8x 2 R holds.

e I is The membership and non-membership functions of A of the following form: 8 t1 ðxÞ; m  a  x  m; > > < 1; x ¼ m; lA~I ðxÞ ¼ h ðxÞ; m  x  m þ b; > 1 > : 0; otherwise; where t1 and h1 are piecewise continuous; strictly increasing and strictly decreasing functions in [m - a, m] and [m, m ? b] respectively. 8 t2 ðxÞ; m  a0  x  m; 0  t1 ðxÞ þ t2 ðxÞ  1; > > < 0; x ¼ m; tA~I ðxÞ ¼ h2 ðxÞ; m  x  m þ b0 ; 0  h1 ðxÞ þ h2 ðxÞ  1; > > : 1; otherwise; where t2 and h2 are piecewise continuous; strictly decreasing and strictly increasing functions in [m - a’, m] and [m, m ? b’] respectively. Here m is the mean value of A~I ; a and b are left and right 0 spreads of membership function lA~I ðxÞ respectively; a and 0 b are left and right spreads of non- membership function tA~I ðxÞ respectively. The IFN A~I is represented by e I ¼ ðm; a; b; a0 ; b0 Þ. A Definition 2.4 A TIFN A~I is an IFS in R with the following membership function lA~I and non-membership function mA~I : 8 xa 1 > ; a1 \x  a2 > > > a  a 2 1 > < a3  x lA~I ðxÞ ¼ ; a2  x\a3 > > a3  a2 > > > : 0; otherwise 8 a2  x 0 > a1 \x  a2 0 ; > > > a2  a1 > < x  a2 0 mA~I ðxÞ ¼ ; a2  x\a3 0 > > a  a 2 > > 3 > : 1; otherwise; 0

0

where a1 B a1 \ a2 \ a3 B a3. This TIFN is denoted by e I ¼ ða1 ; a2 ; a3 ; a0 ; a2 ; a0 Þ Fig. 1. A 1 3 Arithmetic operations on TIFNs    e I ¼ a1 ; a2 ; a3 ; a0 ; a2 ; a0 and B e I ¼ b1 ; b2 ; b3 ; b01 ; Let A 1 3  0 eI  B e I ¼ a1 þ b1 ; a2 þ b2 ; a3 þ b3 ; a01 b2 ; b3 Þ Addition: A 0 0 0 e I HB e I ¼ ð a1  b3 ; þb1 ; a2 þ b2 ; a3 þ b3 Þ Subtraction: A

Int J Syst Assur Eng Manag

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðxl ðA~I ÞÞ2 þ ðyl ðA~I ÞÞ2 RankðA~I Þ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðxm ðA~I ÞÞ2 þ ðym ðA~I ÞÞ2

Fig. 1 Membership and non-membership functions of TIFN 0

0

0

Here it is clear that the defuzzified value of TIFN A~I does not depend only on the components of the number but also depend on DM as the values yl ðA~I Þ and ym ðA~I Þ are independent from the components of TIFN. These are chosen by the DM. Thus the defuzzified value will not give a unique representation of a TIFN. It will give different value for different DM. Also it is not possible always to find defuzzified value for all TIFNs by this ranking method, which is clear by following example.

0

a2  b2 ; a3  b1 ; a1  b3 ; a2  b2 ; a3  b1 Þ Multiplication:   e I B e I ¼ l1 ; l2 ; l3 ; l01 ; l2 ; l03 ; A where l1 ¼ minfa1 b1 ; a1 b3 ; a3 b1 ; a3 b3 g; n 0 0 0 0 0 0 0 0o 0 l1 ¼ min a1 b1 ; a1 b3 ; a3 b1 ; a3 b3 l3 ¼ maxfa1 b1 ; a1 b3 ; a3 b1 ; a3 b3 g; n 0 0 0 0 0 0 0 0o 0 l3 ¼ max a1 b1 ; a1 b3 ; a3 b1 ; a3 b3 l 2 ¼ a2 b2 : Scalar multiplication:   e I ¼ ka1 ; ka2 ; ka3 ; ka0 ; ka2 ; ka0 : k [ 0 1 kA 1 3   e I ¼ ka3 ; ka2 ; ka1 ; ka0 ; ka2 ; ka0 : k\0: 2 kA 3 1

3 Ordering of TIFNs 3.1 The existing ordering Let A~I ¼ ða1 ; a2 ; a3 ; a01 ; a2 ; a03 Þ be a TIFN. Then Hussain and Kumar (2012) defined the ranking for A~I as follows:

e I ¼ ð1; 1; 2; 0; 1; 3Þ and B e I ¼ ð1; 2; A

eI 2; 0; 2; 2Þ, Here we cannot find the values of xl A  I e as the denumerator values a2 - a1 and a03 - a2 and xm B reduce to zero respectively. Thus we cannot determine the defuzzified values. Hence we cannot rank these numbers by the ranking function given by Hussain and Kumar (2012). Thus this ranking method is not universal; we cannot order all TIFNs by this ranking. This is not an efficient ranking for all TIFNs. Therefore this ranking method cannot be utilized for solving a general IFTP. Example

1

Let

3.2 The existing ordering   e I ¼ a1 ; a2 ; a3 ; a0 ; a2 ; a0 be a TIFN. Then NagoLet A 1 3 organi and Abbas (2013) have given the graded mean   e I ¼ a 1 ; a2 ; a3 ; a0 ; a2 ; a0 integration representation of A 1 3 h

i e I ; Pb A e I ; where where Pa A eI ¼ e I ¼ Pa A as P A l m l  h

0 0 0 2bða1 a3 Þþa2 þ2a3 aða1 a3 Þ b eI þa3 þ 2a2 3 and Pm A ¼ . 3 e I ¼ ða1 ; a2 ; a3 ; They said that for any two TIFNs A

  0 0 eI e I ¼ b1 ; b2 ; b3 ; b01 ; b2 ; b03 ; Pal A a1 ; a2 ; a3 Þ and B

 3   3  1 2 3 1 2 3

6ða2 a1 Þ 2a2  3a2 a1 þ a1 þ 6ða3 a2 Þ 2a3  3a2 a3 þ 2a2 I e xl A ¼ a3 a1 2

0      0 2 3

0 2 0 3 3 1 3 1 ð2a 2a  3a a þ2 a Þ  3a a3 þa2 þ 0 0 2 2 2 1 1 3 6ða2 a1 Þ 6ða3 a2 Þ



eI ¼ xm A

yl ðA~I Þ ¼ 13 given by

0

0

a3 a1 2

and ym ðA~I Þ ¼ 23 : Then ranking of A~I is



 I  I e I \B e I \Pb B e and \Pbm A e )A eI: \Pal B m

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e I ¼ ð2; 3; 6; 1; 3; 8Þ and B eI ¼ Example 2 Let A ð2; 4; 5; 1; 4; 7Þ: Then we have 4a þ 12 14b þ 19 eI ¼ eI ¼ ; Pbm A ; Pal A 3 3  I  3a þ 13  I  10b þ 16 e ¼ e ¼ ; Pbm B ; Pal B 3 3

a = 0, b = 1

eI ¼ 4 Pal A

e I ¼ 1:66 Pbm A  I e ¼ 4:33 Pal B   b eI Pm B ¼ 2

 I e I \Pa B e Pal A l

 I e I \Pb B e Pbm A m

a = 0.5, b = 0.5

e I ¼ 3:33 Pal A

eI ¼ 4 Pbm A  I e ¼ 3:83 Pal B   b eI Pm B ¼ 3:66

 I e I \Pa B e Pal A l

 I e I [ Pb B e Pbm A m

a = 1, b = 0

e I ¼ 2:66 Pal A

e I ¼ 6:33 Pbm A  I e ¼ 3:33 Pal B   b eI Pm B ¼ 5:33

 I e I \Pa B e Pal A l

 I e I [ Pb B e Pbm A m

Thus it is clear that in this case we cannot order TIFNs I ~ A and B~I : From Example 2, we get that the ranking (ordering) used by Nagoorgani and Abbas (2013) is not universal. By this ranking we cannot order all TIFNs. Hence we cannot solve a general IFTP using this ordering. 3.3 Proposed ordering using accuracy function   e I ¼ a1 ; a2 ; a3 ; a0 ; a2 ; a0 be a TIFN. The score Let A 1 3 function for the membership function lA~I is denoted by

SðlA~I Þ and is defined by S leI ¼ a1 þ2a42 þa3 . The score A function for the non-membership function mA~I is denoted by



0 0 a þ2a þa S meI and is defined by S meI ¼ 1 42 3 . A A

e I is denoted by f A e I and The accuracy function of A defined by

eI

f A







S leI þ S meI A A ¼ 2  0 0  ða1 þ 2a2 þ a3 Þ þ a1 þ 2a2 þ a3 ¼ : 8

We define the ordering of TIFNs as follow using accuracy function, which totally depends on membership and non-membership of a TIFN.   e I ¼ a1 ; a2 ; a3 ; a0 ; a2 ; a0 eI ¼ Let A and B 1 3   0 0 b1 ; b2 ; b3 ; b1 ; b2 ; b3 be two TIFNs. Then (a)

f ðA~I Þ  f ðB~I Þ ) A~I  B~I

(b)

f ðA~I Þ  f ðB~I Þ ) A~I  B~I

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(c)

f ðA~I Þ ¼ f ðB~I Þ ) A~I ¼ B~I

(d)

MinðA~I ; B~I Þ ¼ A~I if A~I  B~I or B~I  A~I

(e)

MaxðA~I ; B~I Þ ¼ A~I if A~I  B~I or B~I  A~I

4 Type-1 intuitionistic fuzzy transportation problem (IFTP-1) In transportation problem the DM or the expert hesitates due to many factors from both sides that is from supplier side and demand side. Sometimes a DM is not sure that how much quantity of a particular product is available at his warehouse at a particular time by different reason. Such that, he has not good communications to his fellows or he is not sure that how much quantity of particular product can be produced by the available row materials by that particular time. Similarly, he may hesitate from demand side. Suppose some new product is to be launch in a market then he cannot decide exactly that how much quantity of this product should transport to a particular destination. This may be due to unawareness of the customers about this product or difference in price and utility of the product to the similar one. We utilize IFNs to deal with hesitation and uncertainty. Let us consider a transportation problem with m origins and n destinations. Let cij be the cost of transporting one unit of the product from the ith origin to the jth destination.  0 0 Let e a Ii ¼ ai1 ; ai2 ; ai3 ; ai1 ; ai2 ; ai3 be the IF quantity available at the ith origin.

0 0 e b Ij ¼ b1j ; b2j ; b3j ; bj1 ; b2j ; bj3 be the IF quantity needed

0 0 at the jth destination. e x Iij ¼ xij1 ; xij2 ; xij3 ; xij1 ; xij2 ; xij3 be the IF quantity transported from the ith origin to the jth destination. Then the balanced IFTP-1 is given by Xm Xn eI ¼ c e x Iij Min Z i¼1 j¼1 ij Xn e s:t: a Ii ; i ¼ 1; 2; . . .; m xI ¼ e j¼1 ij Xm ex I ¼ e b Ij ; j ¼ 1; 2; . . .; n i¼1 ij e 0; i ¼ 1; 2; . . .m; j ¼ 1; 2; . . .; n: x Iij  e 5 Proposed methods to find starting basic feasible solution (BFS) In this section we introduce some methods to find starting BFS of the given balanced IFTP. 5.1 Intuitionistic fuzzy north west corner method (IFNWCM) for IFTP-1 The following steps are involved in this method to find staring BFS.

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Step1: Select the North West Corner Cell (NWCC) of

b Ij , using the IFTP table (IFTPT) and find the min e a Ii ; e accuracy function. There may arise the following three cases:

b Ij ¼ e a Ii in a Ii , then allocate ex Iij ¼ e Case-1. If min e a Ii ; e the NWCC of m 9 n IFTPT. Delete the ith row to obtain a I

a Ii in new IFTPT of order (m-1) 9 n. Replace e b Ij by bj He the obtained IFTPT and then go to Step 2.

Case-2. If min e a Ii ; e b Ij ¼ e b Ij in b Ij then allocate, ex Iij ¼ e the NWCC of m 9 n IFTPT. Delete the jth column to obtain a new IFTPT of orderm 9 (n-1). Replace e a Ii by e a Ii He b Ij in the obtained IFTPT and then go to the Step 2. b I , then either follow Case 1 or Case 2 Case-3. If e aI ¼ e i

j

but not both together. Step2: Repeat Step 1 for obtained IFTPT until the IFTPT reduces into a table of order 1 9 1. Step3: Starting BFS and IF transportation cost (IFTC) PP are given by e x Iij and cij e x Iij respectively. 5.2 Intuitionistic fuzzy least cost method (IFLCM) for IFTP-1 The following steps are involved in this method: Step 1: Determine the smallest cost in the IFTPT. Let it

be cij. Find e x Iij ¼ min a~Ii ; b~Ij using accuracy function. The

resultant in front of the row on right. This is the penalty for the first row. Similarly, compute the penalties for each row and write them in front of the corresponding row. In similar fashion find the penalties for the columns and write them in the bottom of the IFTPT below corresponding columns. Step 2: Select the highest penalty and observe the row or column for which this corresponds. Determine the smallest cost in the selected row or column. Let it be cij. Find aIi ; b~Ij Þ. Again there may arise three cases:x~Iij ¼ minð~

Case-1. If min a~Ii ; b~Ij ¼ a~Ii , then allocate x~Iij ¼ a~Ii in the (i,j)th cell of m 9 n IFTPT. Delete the ith row to obtain a aIi in new IFTPT of order (m-1) 9 n. Replace b~Ij by b~Ij H~ the obtained IFTPT and then go to the Step 3. Case-2. If minð~ aIi ; b~Ij Þ ¼ b~Ij , then allocate x~Iij ¼ b~Ij in the (i,j)th cell of m 9 n IFTPT. Ignore the jth column to obtain a new IFTPT of order m 9 (n-1). Replace a~Ii by a~Ii Hb~Ij in the obtained table and then go to the Step 3. Csase-3. If a~Ii ¼ b~Ij , then either follow Case 1 or Case 2 but not both together. Step 3: Calculate the penalties for the reduced table obtained in Step 1. Repeat Step 2 until the table is reduced into 1 9 1. Step 4: The starting BFS and IFTC are given by x~Iij and PP cij x~Iij respectively.

following three cases may arise:

b Ij ¼ e x iij ¼ e a Ii in a Ii , then allocate e Case-1. If min e a Ii ; e

5.4 Intuitionistic fuzzy modified distribution method (IFMODIM) to find optimal solution of IFTP-1

the (i,j)th cell of m 9 n IFTPT. Ignore the ith row to obtain aIi in a new IFTPT of order(m-1) 9 n. Replace b~Ij by b~Ij H~ the obtained IFTPT and then go to the Step 2. Case-2. If minð~ aIi ; b~Ij Þ ¼ b~Ij , then allocate e x Iij ¼ e b Ij in the (i,j)th cell of m 9 n IFTPT. Ignore the jth column to obtain a new IFTPT of order m 9 (n-1). Replace a~Ii by a~Ii Hb~Ij in the obtained table and then go to the Step 2. Case-3. If a~Ii ¼ b~Ij , then either follow Case 1 or Case 2 but not both together. Step 2: Repeat Step1 for the obtained IFTPT until it reduces to an 1 9 1 IFTPT. Step 3: The starting BFS and IFTC are given by x~Iij and PP x Iij respectively. cij e

In this section IFMODIM is introduced for optimal solution of IFTP-1. The following steps are to be taken while finding the optimal solution. Step 1: Find the starting BFS using IFNWCM, IFLCM or IFVAM. Step 2: Define dual variables ui and vj corresponding to the ith row and thejth column respectively such that ui ? vj = cij for a basic cell (i,j). Step 3: Define zij = ui ? vj. Find zij-cij for all nonbasic variables and write them in the right lower corner of the concerned cell. Any one of the following two cases may arise: Case-1: If zij - cij B 0, for all i,j. Then the starting BFS is optimal stop there. Case-2: If there exists at least one zij - cij s.t. zij cij [ 0, then the BFS is not optimal. Go to Step 4. Step 4: In IFTPT choose that zij - cij which is the most positive. Step 5: Assign h~I quantity in that cell for which zij - cij is the most positive and make a loop as follows:

5.3 Intuitionistic fuzzy vogel’s approximation method (IFVAM) for IFTP-1 The following steps are involved in this method:Step 1: Take the first row and choose its smallest entry and subtract it from the next smallest entry. Write the

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Rule for making the loop—start from h~I -cell and move horizontally and vertically to the nearest basic cell with the restriction that the end of the loop must not lie in any nonbasic cell except h~I -cell. In this way return to h~I -cell to complete the loop. Step 6: Add and subtract h~I in concerned cell of the loop maintaining feasibility and define the value of h~I as the minimum of x~Iij from which h~I is subtracted, using accuracy function. Step 7: Inserting the value of h~I the next BFS is obtained which improves the IFTC. While inserting value of h~I one cell assumes 0~I value, i.e., this cell become non basic. This gives the improved BFS. Step 8: Repeat step 1–7 until all zij - cij B 0Vi, j. Step 9: The obtained IF optimal solution and IFTC are PP given by x~Iij and cij x~Iij respectively, i = 1, 2, 3,…, m and j = 1, 2, 3,…, n.

6 Numerical examples Example 1 Hussain and Kumar (2012) There are four sources and four destinations, all the sources are connected to all the destinations by roads and the goods are transported by trucks. The transportation cost per unit is known as in Table 1 below. The availability and demand of goods are not known exactly, which are given in terms of TIFNs as below. Find the optimal allocation which minimizes total transportation cost. Solution by proposed method: Here we have, (3,4,6; 1,4,8)  (2,5,7;1,5,8)  (10,15,20;8,15,22)  (2,3,5;1,3,6) = (2,4,5;1,4,6)  (4,6,8;3,6,9)  (3,7,12;2,7,13)  (8,10,13; 5,10,16) = (17,27,38;11,27,44). Hence the problem is balanced. To compare the above mentioned IFNs we find the accuracy functional values of a~Ii ‘s and b~Ij ‘s as under: 30 48 58 82 aI2 Þ ¼ ; f ð~ aI3 Þ ¼ ; f ð~ aI4 Þ ¼ ; f ðb~I1 Þ f ð~ aI1 Þ ¼ ; f ð~ 8 8 8 8 34 38 120 26 I I I ; f ðb~4 Þ ¼ : ¼ ; f ðb~2 Þ ¼ ; f ðb~3 Þ ¼ 8 8 8 8

Table 1 IFTP

subtracted is (2,3,5;1,3,6) because (2,3,5;1,3,6) \ (3,7,12;2,7,13). Hence allocating h~I = (2, 3, 5; 1, 3, 6), we get the new improved solution-1 given in Table 4. In Table 4, z12 - c12 = 8 which is the most positive and (2,4,5;1,4,6) \ (2,5,7;1,5,8) so allocating h~I = (2,4,5;1, 4,6), we get the improved solution-2 given in Table 5. In Table 5, z31 - c31 = 5 which is the most positive and (-2,4,10;-3,4,11) \ (0,4,9;-4,4,13), therefore allocating h~I = (-2,4,10;-3,4,11), we have the improved solution-3 given in Table 6. In Table 6, z41 - c41 = 3 which is the most positive and (-10, 0, 11;-15, 0, 16) \ (-8, 10, 29;-17, 10, 38). Therefore allocating h~I = (-10, 0, 11;-15, 0, 16), we get the improved solution-4 given in Table 7. In Table 7, all zij - cij B 0 so, the obtained solution is optimal. The optimal solution is x~I12 = (2, 4, 5; 1, 4, 6), x~I22 = (-3, 1, 5; -5, 1, 7), x~I23 = (-19,5,29; -30,5,40), x~I31 = (-2,4,10;-3,4,11), x~I34 = (2,3,5;1,3,6), x~I41 ¼ ð10; 0; 11; 15; 0; 16Þ; x~I43 ¼ ð19; 10; 39; 33; 10; 53Þ: Table 2 The starting BFS 16

1

8

13 10

(2,4,5;1,4,6) 11

4

7

(2,4,5;1,4,6)

(2,5,7;1,5,8)

(-7,1,8;-12,1,13)

8

15

9

2

(3,7,12;2,7,13) 6

12

5

14

(-10,7,24;-18,7,32)

(2,3,5;1,3,6)

D1

D2

D3

D4

a~Ii

S1

16

1

8

13

(2,4,5;1,4,6)

S2 S3

11 8

4 15

7 9

10 2

(4,6,8;3,6,9) (3,7,12;2,7,13)

S4 b~I

6

12

5

14

(3,4,6;1,4,8)

(2,5,7;1,5,8)

(10,15,20;8,15,22)

(2,3,5;1,3,6)

(8,10,13;5,10,16) P I P I a~ ¼ b~

j

123

Now to find the starting BFS we can apply any one of the methods discussed in Sect. 5. Here we have applied IFNWCM to find the starting BFS. The BFS obtained is given in Table 2. Now we will apply IFMODIM to test the optimality of the obtained starting BFS. In Table 3, manyzij - cij [ 0, so solution is not optimal so we form a loop as shown above and improve the solution. z34 - c34 = 16 is the most positive, allocating h~I quantity there, we get the smallest value among which h~I

i

j

Int J Syst Assur Eng Manag Table 3 Optimality test of the starting BFS

16 1 (2,4,5;1,4,6) 8 11 4 u2 = 0 (-2,0,4;-5,0,7) (2,5,7;1,5,8) 8 15 u3 = 2 5 6 12 u4 =-2 3 v1=11 v2=4

8

13

u1 = 5

4

8

9 (-7,1,8;-12,1,13)

10

9 (3,7,12;2,7,13)

2

6

θI

16 14 (2,3,5;1,3,6)

5 (-10,7,24;-18,7,32) v3=7

v4=16

Table 4 Improved solution-1

u1 = 5 u2 = 0 u3 = 2 u4 = -2

16 (2,4,5;1,4,6)

1

11 (-2,0,4;-5,0,7) 8 5 6 3

4 (2,5,7;1,5,8) 15

8

θI

12

v1 =11

13 4

8

v2 = 4

7 (-7,1,8;-12,1,13) 9 (-2,4,10;-3,4,11) 5 (-8,10,29;-17,10,38)

10 2 (2,3,5;1,3,6) 14

v3 = 7

v4 = 0

8

13

7 (-7,1,8;-12,1,13) 9 (-2,4,10;-3,4,11)

10

Table 5 Improved solution-2

16 u1 = -3 u2 = 0 u3 = 2

11 (0,4,9;-4,4,13) 8

θI

1 (2,4,5;1,4,6) 4 (-3,1,5;-5,1,7) 15

5

6

12

u4 = -2

3 v1 = 11

v2 = 4

5 (-8,10,29;-17,10,38) v3 =7

2 (2,3,5;1,3,6) 14 v4 =0

Table 6 Improved solution-3

16 u1 = -3 u2 = 0 u3 = -3

11 (-10,0,11;-15,0,16) 8 (-2,4,10;-3,4,11) 6

u4 = -2

θ

1 (2,4,5;1,4,6) 4 (-3,1,5;-5,1,7) 15 12

I

v1 = 11

3 v2 = 4

8

13

7 (-9,5,18;-15,5,24) 9

10

5 (-8,10,29;-17,10,38) v3 = 7

2 (2,3,5;1,3,6) 14 v4 = 5

123

Int J Syst Assur Eng Manag Table 7 Improved solution-4 (optimal solution)

16 u1 = -3 11 u2 = 0 u3 = 0 u4 = -2

8 (-2,4,10;-3,4,11) 6 (-10,0,11;-15,0,16) v1= 8

1 (2,4,5;1,4,6) 4 (-3,1,5;-5,1,7) 15 12

8

13

7 (-19,5,29;-30,5,40) 9

10

5 (-19,10,39;-33,10,53) v3 = 7

v2 = 4

I The IFTC Z~opt = (2,4,5;1,4,6)  4(-3,1,5;-5,1,7)  7(-19,5,29;-30,5,40)  8(-2,4,10;-3,4,11)  2(2,3,5; 1,3,6)  6(-10,0,11;-15,0,16)  5(-19,10,39;-33,10, 53) = (-310,131,579;-506,131,775). Defuzzified value of the transportation cost I ~ f ðZopt Þ ¼ 310þ262þ579506þ262þ775 ¼ 132:75 8 Solution by existing methods: The optimal solution by I ¼ ð76; 131; 345; existing methods is given by Z~ex I I I ¼ Z~ex . 173; 131; 442Þ, f ðZ~ex Þ ¼ 132:75. Here we got Z~opt So, the optimal solution in this case is likely equal to the optimal solution obtained by the proposed method, but the ranking used in existing methods are not universal.

Example 2 Let us consider the following problem as in (Nagoorgani and Abbas 2013).

D1

D2

D3

D4

a~Ii

S1

2

3

11

7

(4,6,9;2,6,10)

S2

1

0

6

1

(0.5,1,3;0,1,5)

S3 b~I

5

8

15

9

(6,7,9; 5,7,11)

(4,5,7; 3,5,8)

(2,3,5; 1.5,3,6)

(1,2,3; 0.5,2,4)

(8.5,10,12;8,10,14) P I P I a~ ¼ b~

j

i

j

I Solving by proposed approach we have, Z~opt = I ~ (-928.5, 100, 1138.5;-1831, 100, 2048), f ðZopt Þ = 103.375. I And that of by existing approaches is given by Z~ex = I (-67, 100, 278.5; -185,100,402), f ðZ~ex Þ ¼ 103:5625: I I \Z~ex ; thus sometimes our approach Here we got Z~opt gives better optimal solution than the existing approaches.

7 Result and discussion I The IFTC Z~opt obtained in Numerical Example 1 is I ~ Zopt = (-310,131,579;-506,131,775) …(1)

123

2 (2,3,5;1,3,6) 14 v4 = 2

The result in (1) can be explained (Refer to Fig. 2) as follows: Assuming that lZ~I ðcÞ is membership and mZ~I ðcÞ is nonmembership value at c. Then the degree of acceptance (in percentage) of the transportation cost c for the DM is 100:lZ~I ðcÞ and the degree of non-acceptance of the transportation cost c for the DM is 100:mZ~I ðcÞ. The degree of hesitation for the acceptance of c is given by 100ð1  lZ~I ðcÞ  mZ~I ðcÞÞ. Values of lZ~I ðcÞ and mZ~I ðcÞ at different values of c can be evaluated using Eq. (1a) and (1b) respectively. 8 c þ 310 > > > < 441 ; 310\c  131; ... ð1aÞ lZ~I ðcÞ ¼ 579  c ; 131  c\579; > > > 448 : 0 otherwise 8 131  c > > > < 637 ; 506\c  131; ::: ð1bÞ mZ~I ðcÞ ¼ c  131 ; 131  c\775; > > > : 644 1; otherwise: Similarly the result obtained in Example 2 can be discussed.

1

0

-506

-310

131

Fig. 2 Intuitionistic fuzzy transportation cost

579

775

Int J Syst Assur Eng Manag

8 Conclusions In this paper new methods are proposed to find the starting BFS and optimal solution of IFTP-1 in which availabilities and demands of goods are TIFNs. The algorithm is very easy to understand and to apply in solving IFTPs because these methods are similar to the methods which are applied in solving crisp transportation problems. Also the shortcomings in existing methods proposed by Hussain and Kumar (2012), Nagoorgani and Abbas (2013) are noticed. The ranking functions used by the existing approaches are not universal for ordering TIFNs. By using the existing ranking functions, ordering of all TIFNs is not possible. Hence existing approaches could not be utilized to solve a general IFTP. These approaches only can be utilized in solving particular types of IFTPs. Our ranking approach is universal, it can be utilize for ordering all TIFNs. Also we observed that our approach sometimes gives better solution than existing approaches. In future this approach can be applied in solving transportation problems having uncertainty and hesitation in availabilities and demands of commodities. In future our algorithm can be extended for solving transportation problems having all parameters as IFNs. Acknowledgments The authors gratefully acknowledge the critical comments given by the learned reviewer which helped us to improve the manuscript. The first author gratefully acknowledges the financial support given by the Ministry of Human Resource and Development (MHRD), Govt. of India, India.

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