Weak transitivity of interval-valued fuzzy relations

Weak transitivity of interval-valued fuzzy relations

Knowledge-Based Systems 63 (2014) 24–32 Contents lists available at ScienceDirect Knowledge-Based Systems journal homepage: www.elsevier.com/locate/...

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Knowledge-Based Systems 63 (2014) 24–32

Contents lists available at ScienceDirect

Knowledge-Based Systems journal homepage: www.elsevier.com/locate/knosys

Weak transitivity of interval-valued fuzzy relations Yejun Xu a,b, Huimin Wang a,b,⇑, Dejian Yu c a

State Key Laboratory of Hydrology-Water Resources and Hydraulic Engineering, Hohai University, Nanjing, Jiangsu 210098, PR China Business School, Hohai University, Nanjing, Jiangsu 211100, PR China c School of Information, Zhejiang University of Finance & Economics, 18 Xueyuan Street, Hangzhou 310018, PR China b

a r t i c l e

i n f o

Article history: Received 13 June 2013 Received in revised form 4 March 2014 Accepted 5 March 2014 Available online 28 March 2014 Keywords: Interval-valued fuzzy relations Weak transitivity Cycles Rationality Consistency

a b s t r a c t In this paper, we define and study the weak transitivity of interval-valued fuzzy relations (IVFRs). We propose the weak transitivity index (WTI) to measure the transitivity consistency degree of an IVFR, which is to count the cycles of length 3 in the digraph. Afterwards, an algorithm is proposed to compute the WTI and to locate each cycle, as well as to find the inconsistent judgments in an IVFR. In order to resolve the intransitivities of an IVFR, another algorithm is developed to find and remove all the 3-cycles in the digraph. Our method can not only repair the weak intransitivity for an IVFR, but also preserve the initial preference information as much as possible. Finally, two examples are shown to illustrate the proposed method. Ó 2014 Elsevier B.V. All rights reserved.

1. Introduction Interval-valued fuzzy set (IVFS) theory [23] is an extension of fuzzy theory. The membership degree of each element on an IVFS is defined on a closed subinterval of [0, 1]. IVFSs have been used in a number of different fields: image processing [34], intervalvalued logic [35,36], approximate reasoning [4,7,23] and so on. Transitivity is a fundamental notion in decision theory. It is most universally assumed in disciplines of decision theory and generally accepted in a principle of rationality. Yet, it is often violated in actual choice, particularly in pairwise choices. A first task for decision science is thus the resolution of intransitivities [27]. The weak transitivity is the usual transitivity condition that a logical and consistent person should use if he/she does not want to express inconsistent opinions, and therefore it is the minimum requirement condition that a consistent fuzzy preference relation should verify [31]. Weak transitivity is in fact acyclic about the alternatives ranking, i.e., if an alternative A is preferred or equivalent to B, and B is preferred or equivalent to C, then A must be preferred or equivalent to C. The transitivity assumption can be used to check for the judgment consistency of a decision maker (DM). If a DM provides a preference relation does not possess transitivity (i.e., inconsistency problems exist), the ranking result of alternatives is misleading [25,26,29]. ⇑ Corresponding author at: State Key Laboratory of Hydrology-Water Resources and Hydraulic Engineering, Hohai University, Nanjing, Jiangsu 210098, PR China. E-mail addresses: [email protected] (Y. Xu), [email protected] (H. Wang), [email protected] (D. Yu). http://dx.doi.org/10.1016/j.knosys.2014.03.003 0950-7051/Ó 2014 Elsevier B.V. All rights reserved.

Transitivity of a fuzzy preference relation has been received greatly attention in the past decades [2,3,9–16,18,20,28,29], such as weak transitivity (or called weak stochastic transitivity) [8,15,16,18,31,32,40,41], max–min transitivity [19,25,31,33,37], max-max transitivity [25,31], restricted max–min transitivity (or moderate stochastic transitivity) [8,15,16,25,31–33,37], restricted max-max transitivity (or strong stochastic transitivity) [8,15,16,18,25,31,32], multiplicative consistency [8,25] and additive consistency [8,25]. It should be pointed out that, strictly speaking, additive consistency is not a type of transitivity [17]. Gonzales-del-Campo et al. [22] proposed an algorithm to compute the transitive closure for an IVFR. IVFRs are also common fuzzy relations which experts express their comparison information for alternatives. The comparison information are not exact numerical values but interval numbers. Thus, the transitivity is also an important problem for IVFRs. However, little work has been done on the transitivity problem of IVFRs. Therefore, it is important to pay attention to this problem. This is the subject of the present paper. In this paper, we give a definition of the weak transitivity for IVFRs. We propose the weak transitivity index (WTI) to measure the consistency degree of an IVFR, which is to count the 3-cyles in the digraph. A procedure is proposed to compute the WTI and locate each cycle, as well as to find the inconsistent judgments in an IVFR. If an IVFR is not weakly transitive, another algorithm is developed to find and remove all the 3-cycles in the digraph. Moreover, our improving method can preserve the initial preference information as much as possible. The rest of the paper is set out as follows. Section 2 gives the basic concepts related to IVFSs. We give the definition of weak

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transitivity for an IVFR. Section 3 proposes a procedure to judge whether an IVFR is weakly transitive, as well as to find the inconsistent judgments in an IVFR. Section 4 develops a method to repairing the intransitivities for an IVFR. Section 5, two examples are illustrated to show the effectiveness and validity of the proposed methods. The conclusions, some characteristics and advantages of the proposed methods and future research are presented in Section 6.

generally needs to provide his/her preferences for each pair of alternatives, and perhaps it is possible that he/she is not so sure about it. Thus, it is very suitable to express the DM’s preference values with IVFVs rather than exact numerical values, and then constructs an interval-valued fuzzy relation (IVFR), which can be defined as follows.

2. Preliminaries

Definition 5 [43]. An IVFR R on the set X is represented by a matrix R = (rij)nn  X  X, where

In the following, we introduce some basic concepts related to interval-valued fuzzy sets.

r ii ¼ ½0:5; 0:5; for all i;j ¼ 1; 2; .. . ;n

h i h i r ij ¼ rij ; rþij ; rji ¼ r ji ; r þji ; r ij þ rþji ¼ r þij þ r ji ¼ 1; r þij P r ij P 0; ð5Þ

Definition 1. Let Y be a universe of discourse, then a fuzzy set:

A ¼ fhy; lA ðyÞijy 2 Yg

ð1Þ

defined by Zadeh [45] is characterized by a membership function:

lA:Y ? [0, 1], where lA(y) denotes the degree of membership of the element y to the set A. We will denote with L([0, 1]) the set of all closed subintervals of the closed interval [0, 1]. That is,

Lð½0; 1Þ ¼ fx ¼ ½x ; xþ jðx ; xþ Þ 2 ½0; 12 and x 6 xþ g Definition 2 [5]. An interval-valued fuzzy set (IVFS) A, on the universe U – ;, is a set such that

A ¼ fðu; AðuÞ ¼ ½A ðuÞ; Aþ ðuÞÞju 2 Ug

ð2Þ

where the function A: U ? L([0, 1]) is called the membership function. For convenience, we call a = [a, a+] an interval-valued fuzzy value (IVFV), where a 2 [0, 1], a+ 2 [0, 1], a 6 a+. Based on the concepts of score function and accuracy degree of intuitionistic fuzzy values, in the following, we define the corresponding concepts for IVFVs, which are used to compare two IVFVs. +





Definition 3. Let a = [a , a ] be an IVFV, where a 2 [0, 1], a+ 2 [0, 1], a 6 a+. The score of a can be evaluated by the score function s shown as

sðaÞ ¼ a þ aþ  1

ð3Þ

where s(a) 2 [1, 1]. The larger the score s(a), the greater the IVFV a. An accuracy function h to evaluate the degree of accuracy of a can be expressed as:

hðaÞ ¼ a þ 1  aþ

ð4Þ

where h(a) 2 [0, 1]. The lager the value of h(a), the more the degree of accuracy of a.

rij is interpreted as the preference degree of the alternative xi over þ xj: (1) rij = [0.5, 0.5] (i.e. r  ij ¼ r ij ¼ 0:5) denotes indifference between xi and xj (xi  xj); (2) [0.5, 0.5] < rij 6 [1, 1] denotes xi is strictly preferred to xj(xi  xj). Especially, rij = [1, 1] denotes that xi is definitely preferred to xj; (3) [0, 0] 6 rij < [0.5, 0.5] denotes that xj is strictly preferred to xi(xj  xi). Especially, rij = [0, 0] denotes that xj is definitely preferred to xi. In the following, we define the weak transitivity for IVFRs.

h i þ Definition 6. Let R = (rij)nn be an IVFR, where rij ¼ r  ij ; r ij ; i; j ¼ 1; 2; . . . ; n, for all i, j, k = 1, 2, . . . , n, i – j–k (1) if rik > [0.5, 0.5] and rkj P [0.5, 0.5], we have rij > [0.5, 0.5]; (2) if rik P [0.5, 0.5] and rkj > [0.5, 0.5], we have rij > [0.5, 0.5]; (3) if rik = [0.5, 0.5] and rkj = [0.5, 0.5], we have rij = [0.5, 0.5]. then R is weakly transitive. h i þ Definition 7. Let R = (rij)nn be an IVFR, where rij ¼ r  ij ; r ij ; i; j ¼ 1; 2; . . . ; n, for all i, j, k = 1, 2, . . . , n, i – j – k, if rik > [0.5, 0.5], rkj > [0.5, 0.5], we have rij > [0.5, 0.5], then R is strict weakly transitive. In the following, we will discuss the weak transitivity of the IVFR from the graph theory point of view. Some basic theory of digraph is presented as follows.

h i þ Definition 8. Let R = (rij)nn be an IVFR, where rij ¼ r  ij ; r ij ; i; j ¼ 1; 2; . . . ; n, we define the adjacency matrix E = (eij)nn of R as follows:

 eij ¼



+



+

1; rij P ½0:5; 0:5; i – j 0;

otherwise

ð6Þ



Definition 4. Let a = [a , a ], b = [b , b ] be two IVFVs, s(a) = a + a+  1 and s(b) = b + b+  1 be the scores of a and b, respectively, and let h(a) = a + 1  a+ and h(b) = b + 1  b+ be the accuracy degrees of a and b, respectively, then  If s(a) < s(b), then a is smaller than b, denoted by a < b.  If s(a) = s(b), then (1) If h(a) = h(b), then a and b represent the same information, denoted by a = b. (2) If h(a) < h(b), then a is smaller than b, denoted by a < b. For a decision making problem, let X = {x1, x2, . . . , xn} be a discrete set of alternatives. In the process of decision making, a DM

Let R = (rij)nn be an IVFR, we can construct the digraph G = (V, A) of R, where V = {v1, v2, . . . , vn} denotes the node set, A = {(vi, vj)ji – j, rij P [0.5, 0.5]} denotes the arc set. That is, if i – j, rij > [0.5, 0.5], then there is a directed arc in G from vi to vj, it is denoted by (vi, vj) or vi ? vj. rij is called the weight of the arc (vi, vj). Therefore, if rij = [0.5, 0.5] (i – j), then there is an arc from vi to vj, and also an arc from vj to vi. A directed path q in a graph G is a sequence of arcs v i1 ; v i2 ; v i3 ; . . . :. in G, where the nodes v ik are different. The length of a directed path is the number of successive arcs in the directed path. A cycle is a directed path that begins and ends at the same node. Proposition 1. Let R = (rij)nn be an IVFR, then there exists a directed path q of length n  1 in the digraph G of R.

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Proof. When n = 2, it holds obviously. Assume that when n = k, the proposition holds. Then, when n = k + 1,





Rk

a

b

½0:5; 0:5



h i where a ¼ ðri;kþ1i Þk1 ; ri;kþ1 ¼ ri;kþ1 ; rþi;kþ1 ; b ¼ ðbkþ1;i Þ1k ; bkþ1;i ¼ h  1  rþ i;kþ1 ; 1  r i;kþ1 , i = 1, 2, . . . , k. Rk is a k  k IVFR. By the assumption, we know that there is a directed path ðv i1 ; v i2 ; . . . ; v ik Þ, the length of the path is k  1. If r i1 ;kþ1 6 ½0:5; 0:5, then there is a directed path ðv kþ1 ; v i1 ; v i2 ; . . . ; v ik Þ, the length is k. If r i1 ;kþ1 > ½0:5; 0:5, then we check r i2 ;kþ1 . If r i2 ;kþ1 6 ½0:5; 0:5, then there exists a directed path ðv i1 ; v kþ1 ; v i2 ; . . . ; v ik Þ of length k. Otherwise, ri2 ;kþ1 > ½0:5; 0:5, we check ri3 ;kþ1 . Assume that the process continues to check the value rik ;kþ1 . At this moment, if rik ;kþ1 6 ½0:5; 0:5, then there is a directed path ðv i1 ; v i2 ; . . . ; v kþ1 ; v ik Þ, Otherwise, rik ;kþ1 > ½0:5; 0:5, then there is a directed path ðv i1 ; v i2 ; . . . ; v ik ; v kþ1 Þ. Thus, the proposition holds for n = k + 1. Therefore, the proposition holds for all n, which completes the proof of Proposition 1. h Theorem 1. Let R = (rij)nn be an IVFR, R is strict weakly transitive (i.e. rij–[0.5, 0.5], i – j) if and only if there is only one longest directed path of length n  1 in G. Proof. Necessity. By Proposition 1, there exists a directed path ðv i1 ; v i2 ; . . . ; v in Þ of length n  1 in G. Sufficiency. Assume that there exists another directed path ðv j1 ; v j2 ; . . . ; v jn Þ of length n  1 in G. Now we prove that these two directed paths are same. If they are not same, there must exist 1 6 s 6 n, such that i1 = j1, i2 = j2, . . . , is1 = js1, is–js. In the directed path ðv j1 ; v j2 ; . . . ; v jn Þ, then there must exist a sub-path from v js to v is , according to the Definition 8, we have r js is > ½0:5; 0:5. For the same reason, in the directed path ðv i1 ; v i2 ; . . . ; v in Þ, there must exist a sub-path from v is to v js , then we have ris js > ½0:5; 0:5, this is in þ contradiction with the condition r is js þ r js is ¼ 1 of an IVFR. Therefore, R is strict weakly transitive if and only if there is only one longest directed path of length n  1 in G, which completes the proof of Theorem 1. h

3. The judgment of weak transitivity According to the Definition 6 of weak transitivity of an IVFR R, if R is not weakly transitive, then there exist some inconsistent judgment elements in R, which satisfy one of the following: (a) rik P [0.5, 0.5], rkj > [0.5, 0.5], but rij 6 [0.5, 0.5]; (b) rik > [0.5, 0.5], rkj P [0.5, 0.5], but rij 6 [0.5, 0.5];

(c) rik = [0.5, 0.5], rkj = [0.5, 0.5], but rij –[0.5, 0.5]. In the above each case, there is a directed cycle of length 3 (simplified 3-cycle) (vi ? vk ? vj ? vi) in the digraph G of R. Theorem 2. Let R be an IVFR, G is the digraph of R, if there exists a cycle of length k(k P 4) in G, then there must exist a cycle of length 3 in G. Proof. If there exists a cycle of length 4, assume that the cycle is

v1 ? v2 ? v3 ? v4 ? v1, for the nodes v1, v3, if v1 ? v3, then there exists a cycle v1 ? v3 ? v4 ? v1, its length is 3; If v3 ? v1, then there exists a cycle v1 ? v2 ? v3 ? v1, its length is 3. If there exists a cycle which length is more than 4, we can prove it similarly, which can be seen from Fig. 1. Theorem 2 denotes that if there exist cycles in the digraph of an IVFR, then we only need to find all the 3-cycles in the digraph. That is to say, the inconsistent judgments of an IVFR could be determined by 3-cycles in G. Based on the Definition 6 and Theorem 2, we have the following result easily. h Theorem 3. Let R = (rij)nn be an IVFR, there is a directed 3-cycle (vi ? vk ? vj ? vi) in the digraph G of R, if and only if there exist the elements rik, rkj, rij(i – j–k), which satisfy one of the followings: (a) rik > [0.5, 0.5], rkj P [0.5, 0.5], or rik P [0.5, 0.5], rkj > [0.5, 0.5], but rij 6 [0.5, 0.5]; (b) rik = [0.5, 0.5], rkj = [0.5, 0.5], but rij – [0.5, 0.5]; (c) rik = [0.5, 0.5], rjk = [0.5, 0.5], rij = [0.5, 0.5].

Remark 1. Theorem 3 shows that each of the above cases would result a directed 3-cycle (vi ? vk ? vj ? vi). When the judgment elements rik, rkj, rij satisfy the third case (c) of Theorem 3, these judgment elements are reasonable, because xi, xk, xj are indifference (xi  xk  xj), but they would be two 3-cycles (vi ? vj ? vk ? vi, vi ? vk ? vj ? vi) in the digraph G. Therefore, the two 3-cycles do not result weak intransitivity. Only the first and second cases would lead to intransitivity, and the number of 3-cycles in the first and second cases reflect the inconsistency degree for an IVFR R. In the following, we will provide a method to compute the total number of 3-cycles in the digraph G. Theorem 4. Let R = (rij)nn be an IVFR, G is the digraph of R, E = (eij)nn is the adjacency matrix of R, which is defined by Eq. (6), then ð3Þ the trace (sum of diagonal elements) of E3 ¼ ðeij Þ is 3 times the number of 3-cycles in G, where E3 is the 3rd power of E. P P ð3Þ Proof. 8i 2 N; eii ¼ nk¼1 nj¼1 eik ekj eji , From the Definition 8 of adjacency matrix, eikekjeji = 1 if and only if eik = ekj = eji = 1, this means that there is a 3-cycle

v

v

v v

v

v

v

vk

v

… v v (a) 4-cycle

v (b) 5-cycle Fig. 1. Cycles in digraph.

vk (c) k-cycle

Y. Xu et al. / Knowledge-Based Systems 63 (2014) 24–32 ð3Þ

(vi ? vk ? vj ? vi) in G. Therefore, eii is the number of 3-cycles of P ð3Þ node vi, and thus the trace of E3 (i.e. ni¼1 eii Þ is 3 times the number of 3-cycles in G. h

Theorem 5. Let R = (rij)nn be an IVFR, E = (eij)nn is the adjacency matrix of R, which is defined by Eq. (6). Let B = (bij)nn = E2 ET, that is the Hadamard product of E2 and ET, then bij is the number of Pn ð3Þ 3-cycles that includes the directed arc (vj, vi), and j¼1 bij ¼ eii ; Pn Pn i¼1 j¼1 bij is 3 times the number of 3-cycles in G. P Proof. Since B = (bij)nn = E2 ET, then bij ¼ nk¼1 eik ekj eji , where k runs for all points in the graph. According to the Definition 8 of adjacency matrix, we have eikekjeji = 1 if and only if there exists a directed cycle (vi ? vk ? vj ? vi) in G, that is k satisfies the condition as shown in Fig. 2. bij is the number of 3-cycles which contains P the directed arc (vj,vi). Therefore, nj¼1 bij is the number of 3-cycles ð3Þ of node vi, then by Theorem 4, eii is also the number of 3-cycles of P P P Pn ð3Þ ð3Þ node vi, thus, nj¼1 bij ¼ eii , and ni¼1 nj¼1 bij ¼ i¼1 eii , which is 3 times the number of 3-cycles in G. Of course, if eji = 0, we have bij = 0, and bii = 0. Therefore, if the elements on ith row of matrix B are all zero, the node vi is not involved in any cycle. If the jth element on ith row (i.e. bij) is n, node vi has n arcs on cycles incoming from node vj. h

Remark 2. For an IVFR R = (rij)nn, if rik = [0.5, 0.5], rkj = [0.5, 0.5], rji = [0.5, 0.5], then there are two 3-cycles (vi ? vk ? vj ? vi,vj ? vk ? vi ? vj) in G. In fact, these cycles do not lead to weak intransitivity. Thus, the inconsistency degree of an IVFR is the 3-cycles exclude the ones of (c) in Theorem 3. From Theorems 4 and 5, we can get the following result easily.

ð3Þ

Theorem 6. Let R,E,B as before, if the trace of E3 ¼ ðeij Þ is equal to zero or the sum of all the elements of B is equal to zero P P P P P (i.e. ni¼1 nk¼1 nj¼1 eik ekj eji ¼ 0 or ni¼1 nj¼1 bij ¼ 0), then R is weakly transitive. Pn Pn Pn Pn Pn Proof. If i¼1 k¼1 j¼1 eik ekj eji ¼ 0 or i¼1 j¼1 bij ¼ 0, then there is no 3-cycle in the digraph of R, then R is weakly transitive. h In order to measure the inconsistency degree of an IVFR, we give the following definition. Definition 9. Let R, E, B be as before, then we call

Pn Pn WTI ¼

j¼1 bij

i¼1

3

l

ð7Þ

the weak transitivity index (WTI) of R, where l is the number of 3-cycles which satisfies the condition (c) of Theorem 3. Theorem 7. Let R be an IVFR, R is weakly transitive if and only if WTI = 0.

v

27

Proof. If R is weakly transitivity, according to Definition 6, there exists no judgments which satisfy: (a) rik > [0.5, 0.5], rkj P [0.5, 0.5], or rik P [0.5, 0.5], rkj > [0.5, 0.5], but rij 6 [0.5, 0.5], (b) rik = [0.5, 0.5], rkj = [0.5, 0.5], but rij – [0.5, 0.5] and by Theorem 3, the number of all 3-cycles in digraph G is equal to the number of 3-cycles which satisfies the condition (c) of Theorem 3, then we have WTI = 0. On the contrary, if WTI = 0, and by Theorem 3, there exist no 3cycles which satisfy the condition (a) and (b) of Theorem 2, thus, R is weakly transitivity, which completes the proof of Theorem 7. h Remark 3. Theorem 6 shows that if the trace of E3 is equal to zero, or all the elements of B is equal to zero, then R is weakly transitivity. However, if R is weakly transitivity, it does not mean that the trace of E3 is equal to zero, or all the elements of B is equal to zero, as noted in Remark 1, there are 3-cycles which satisfy the condition (c) of Theorem 3, these cycles do not cause intransitivity. That is why we present WTI to measure the inconsistency degree of an IVFR R. Thus, by Theorem 7, we can get the following result easily. Theorem 8. Let R be a strict pairwise comparison IVFR (i.e. rij – [0.5, 0.5],i – j), then R is weakly transitivity if and only if the numP P ber of 3-cycles is 0 (i.e. ni¼1 nj¼1 bij =3 ¼ 0). Theorem 9. [38]. Let R, E, B be as before, d1,d2, d3 2 {1, 2, . . . , n} and d1 < d2 < d3, B[d1, d2, d3] is a 3  3 principal sub-matrix whose entries are d1, d2, d3 th row and d1, d2, d3 th column of B, then there exists a 3-cycle which contains nodes v d1 ; v d2 ; v d3 in the digraph G of R if and only if there are no zero row and zero column in the principal sub-matrix B[d1, d2, d3]. Proof. In the case without loss of generality. Suppose that d1 = 1, d2 = 2, d3 = 3. Necessity If there exists a 3-cycle which contains the nodes v1, v2, v3 in digraph G of R, by the symmetry, let (v1 ? v2 ? v3 ? v1) be the 3-cycle. By Theorem 5, we have b21 – 0, b32 – 0, b13 – 0, thus there are no zero row and zero column in the sub-matrix B [1, 2, 3]. Sufficiency Suppose that there are no zero row and zero column in the principle sub-matrix B [1, 2, 3], then the entries of B [1, 2, 3] should satisfy: when i – j, there is at least one of the elements bij, bji not equal to zero. In the following, we prove this conclusion. In fact, if i – j, without loss of generality, suppose that i = 1, j = 2, such that b12 = b21 = 0, because there are no zero row and zero 2 3 b11 b12 b13 4 column in the sub-matrix B½1; 2; 3 ¼ b21 b22 b23 5 and bii = 0, b31 b32 b33 for i = 1, 2, 3, therefore, we have b13, b23, b31, b32 are not equal to zero. By Theorem 5, there exist the direct arcs v3 ? v1,v3 ? v2, v1 ? v3, v2 ? v3. And at the same time there exists an arc between different nodes in the digraph G, that is, there exists the arc v1 ? v2 or v2 ? v1, thus, there exists a 3-cycle v1 ? v2 ? v3 ? v1 or v1 ? v3 ? v2 ? v1, it denotes that b12 – 0 or b21 – 0, this is contradict with b12 = b21 = 0. Thus, there is at least one of the elements bij,bji(i – j) not equal to zero. In the following, we propose a method to search the 3-cycle which contains the nodes v1,v2,v3 in the digraph G of R.

ekj =1

eik =1 v eji =1

v

Fig. 2. Condition of forming a cycle.

(1) If b12 – 0, b23 – 0, b31 – 0, then there exists a 3-cycle v1 ? v3 ? v2 ? v1 in digraph G; (2) If b12 – 0, b23 – 0, b31 = 0. By Theorem 9, there are no zero row and zero column in the sub-matrix, we have b13 – 0, b21 – 0, b32 – 0, thus, there exists a 3-cycle v1 ? v2 ? v3 ? v1.

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(3) If b12 – 0, b23 = 0, we have b13 – 0, b21 – 0, b32 – 0, then there exists a 3-cycle v1 ? v2 ? v3 ? v1. (4) If b12 = 0, we have b21 – 0, b13 – 0, b32 – 0, then there exists a 3-cycle v1 ? v2 ? v3 ? v1. h By Theorem 9, we propose a procedure to compute the WTI, and search all the inconsistency entries in an IVFR.

experts dk (k = 1, 2, 3) compare these four alternatives and con  ðkÞ struct the IVFRs Rk ¼ r ij ðk ¼ 1; 2; 3Þ, respectively, shown as 44

follows (adapted from [44]):

3 ½0:5; 0:5 ½0:2; 0:6 ½0:5; 0:6 ½0:7; 0:9 6 ½0:4; 0:8 ½0:5; 0:5 ½0:3; 0:5 ½0:4; 0:5 7 7 6 R1 ¼ 6 7 4 ½0:4; 0:5 ½0:5; 0:7 ½0:5; 0:5 ½0:8; 0:8 5 2

½0:1; 0:3 ½0:5; 0:6 ½0:2; 0:2 ½0:5; 0:5 Algorithm 1. Let R = (rij)nn be an IVFR, G be the digraph of R, E be the adjacency matrix of G. Step 1. Compute B = (bij)nn = E2 ET, check all the 3  3 principal sub-matrices of B, then find all the 3-cycles in digraph G. We use C to denote the set of all the 3-cycles in digraph G. Step 2. Compute the number of all the 3-cycles in the digraph G P P (i.e. compute ni¼1 nj¼1 bij =3Þ, compute l, which is the number of 3-cycles which satisfy the condition (c) of Theorem 3, then we obtain WTI by Eq. (7). If l – 0, then we find all the 3-cycles which satisfy the condition (c) of Theorem 3. Step 3. We delete the 3-cycles which satisfy the condition (c) of Theorem 3 from the set C, then we get the inconsistency entries in the IVFR.

Example 3. A DM (potential buyer) intends to buy a house. He has three alternatives (houses) X = {x1, x2, x3} to choose. Tanking into consideration various factors such as price, size, distance to work, environmental characteristics, the DM constructs an IVFR R as follows (adapted from [21]).

2

½0:5; 0:5 ½0:2; 0:4 ½0:6; 0:6

2

½0:5; 0:5 6 ½0:4; 0:7 6 R2 ¼ 6 4 ½0:5; 0:6 ½0:3; 0:4 2

3 0 0 1 1 61 0 0 07 7 6 E1 ¼ 6 7; 40 1 0 15

3

3 0 1 0 1 60 0 1 07 7 6 E3 ¼ 6 7 41 0 0 05 2

0 0 0 0

0 1 1 0

The digraphs of Rk(k = 1, 2, 3), respectively, are shown in Fig. 3. In the following, we utilize Algorithm 1 to compute the WTI and find the cycles in each digraph.

Step 1: By the adjacency matrices Ek(k = 1, 2, 3), we have

2

3 2 3 2 3 2 1 0 0 0 0 0 0 2 0 1 0 60 2 0 17 60 0 0 17 60 1 0 17 6 7 3 6 7 3 6 7 E31 ¼ 6 7; E ¼ 6 7; E ¼ 6 7; 41 0 1 15 2 40 0 0 05 3 40 1 2 05 0 0 1 1

3

2

0 2 0 0

0 0 0 0 3

2

0 0 0 0

1 1 0 1 3

2

0 0 2 0

3

60 0 1 17 60 0 0 07 61 0 0 07 6 6 6 7 7 7 B1 ¼ 6 7; B2 ¼ 6 7; B3 ¼ 6 7: 41 0 0 05 40 0 0 05 40 1 0 15 1 0 0 0

3

2

0 0 0 6 7 E ¼ 4 0 0 0 5; 0 0 0

3

0 0 0 6 7 B ¼ 40 0 05 0 0 0

3

3

2

0 1 0 0

Then we have

j¼1 bij

3 0 0 0 1 61 0 1 17 7 6 E2 ¼ 6 7; 41 0 0 15

2

0 0 0

i¼1

½0:5; 0:5 ½0:8; 0:9 ½0:3; 0:6 ½0:6; 0:6

The adjacency matrices of Rk(k = 1,2,3), respectively, are:

6 7 E ¼ 41 0 15

P3 P3

½0:3; 0:5 ½0:2; 0:3 ½0:5; 0:5

½0:4; 0:4 ½0:5; 0:6 ½0:7; 0:7 ½0:5; 0:5

The adjacency matrix of R is:

2

½0:5; 0:5 ½0:4; 0:7 ½0:5; 0:7 7 7 7 ½0:3; 0:6 ½0:5; 0:5 ½0:7; 0:8 5

6 ½0:1; 0:2 ½0:5; 0:5 ½0:5; 0:7 ½0:4; 0:5 7 7 6 R3 ¼ 6 7 4 ½0:4; 0:7 ½0:3; 0:5 ½0:5; 0:5 ½0:3; 0:3 5

½0:4; 0:4 ½0:1; 0:3 ½0:5; 0:5

0 0 1

3

3

6 7 R ¼ 4 ½0:6; 0:8 ½0:5; 0:5 ½0:7; 0:9 5

2

½0:3; 0:6 ½0:4; 0:5 ½0:6; 0:7

0 0 0 0 Pn

1 0 0 0

ð3Þ eii

We can verify that it satisfies j¼1 bij ¼ for each IVFR. For the IVFR R1, all the 3  3 principal sub-matrices of B1 are:

P3 ¼

ð3Þ i¼1 eii

3

¼0

Thus, R is weakly transitive. Furthermore, by E, we can easily get x2  x1  x3, which is the same ranking obtained by Gong et al. [21]. However, Gong et al. [21] constructed a goal programming model to get the priority vector of the IVFR, and finally got the ranking order of the alternatives. Their method is so complicated and needs huge computation. For the decision making, the DM finally wants to rank the alternatives and then to find the best one, therefore, our method is simpler and more effective from the point of view of ranking alternatives. Example 4. For a group decision making problem, there are four alternatives xi (i = 1, 2, 3, 4), and three experts dk(k = 1, 2, 3). The

There are no zero row and zero column in the principal sub-matrices B1 [1, 2, 3], B1 [1,2,4]. Therefore, there are two 3-cycles in the digraph G. Then C = {v1 ? v3 ? v2 ? v1, v1 ? v4 ? v2 ? v1}.

v

v

v

v

v

v

(a) digraph of R1

v

v

v

v

(b) digraph of R2 Fig. 3. Digraph of Example 4.

v

v

(c) digraph of R3

29

Y. Xu et al. / Knowledge-Based Systems 63 (2014) 24–32

P4 P4

Step 2.

j¼1 bij =3

P4

ð3Þ

¼ i¼1 eii =3 ¼ 2, then there are two 3-cycles, and both the two 3-cycles do not satisfy the condition (c) of Theorem 3, thus l = 0, WTI = 2. The sets of inconsistent entries are {r13, r32, r21}, {r14, r42, r21}. i¼1

In the digraph of R1, we also can find there are two 3-cycles. Furthermore, b12 = 2 denotes that r21 appears 2 times in the inconsistent entries, b23 = 1 denotes that r13 appears once in the inconsistent entries, and so on. Similarly, we have R2 is weakly transitive, as well as there is no cycle in the digraph of R2. There are two 3-cycles in digraph of R3, and C = {v1 ? v4 ? v3 ? v1, v1 ? v2 ? v3 ? v1}, the weak transitivity index of R3 is WTI = 2. The inconsistent judgments are {r14, r43, r31}, {r12, r23, r31}. According to our above analysis, we know that, there are two experts who provide their preference relation information is unreasonable, that is there are two experts whose preference values are contradictory. Xu and Yager [44] used these IVFRs for consensus analysis in group decision making, therefore, it is unreliable for the final result.

Example 5. Suppose that there an expert compares four alternatives xi (i = 1, 2, 3, 4) and constructs the following IVFR.

2

½0:5; 0:5 ½0:6; 0:8 ½0:5; 0:5 ½0:5; 0:5

3

6 ½0:2; 0:4 ½0:5; 0:5 ½0:2; 0:4 ½0:2; 0:4 7 7 6 R¼6 7 4 ½0:5; 0:5 ½0:6; 0:8 ½0:5; 0:5 ½0:5; 0:5 5 ½0:5; 0:5 ½0:6; 0:8 ½0:5; 0:5 ½0:5; 0:5 The adjacency matrix of R is

2

0 1 1 1

3

60 0 0 07 7 6 E¼6 7 41 1 0 15 1 1 1 0 Then its corresponding digraph G is shown in Fig. 4. Then, we have

2

2 4 3 3

3

2

0 0 1 1

3

60 0 0 07 60 0 0 07 6 6 7 7 E3 ¼ 6 7; B ¼ 6 7 43 4 2 35 41 0 0 15 3 4 3 2 1 0 1 0 P4 P4

j¼1 bij

i¼1

3

4. A method to repair weak intransitivity for an IVFR In the above, we have proposed how to judge whether an IVFR is weakly transitive or not. We also propose a method to find the inconsistent judgments when an IVFR is not weakly transitive. In the following, we study how to repair weak intransitivity for an IVFR. For the IVFR R1 in Example 4, we know that there are two 3-cycles (C = {v1 ? v3 ? v2 ? v1, v1 ? v4 ? v2 ? v1}) in the digraph G of R1, and the inconsistent entries of R1 are {r13, r32, r21}, {r14, r42, r21}. If we repair these inconsistent entries randomly, we cannot obtain an weakly transitive IVFR quickly. For example, if we change the inconsistent entry r14 = [0.7, 0.9] into r14 = [0.1, 0.3], then the IVFR R1 becomes:

2

½0:5; 0:5 ½0:2; 0:6 ½0:5; 0:6 ½0:1; 0:3

3

7 6 6 ½0:4; 0:8 ½0:5; 0:5 ½0:3; 0:5 ½0:4; 0:5 7 7 6 R1 ¼ 6 7 6 ½0:4; 0:5 ½0:5; 0:7 ½0:5; 0:5 ½0:8; 0:8 7 5 4 ½0:7; 0:9 ½0:5; 0:6 ½0:2; 0:2 ½0:5; 0:5 Then we have

2

0 61 6 E¼6 40 1

0 1 0

3

0 0 07 7 7; 1 0 15 1 0 0

2

2 60 6 E3 ¼ 6 41 0

1 0

0

3

1 0 17 7 7; 0 2 05 1 1 1

2

3 0 1 0 1 60 0 1 07 6 7 B¼6 7 42 0 0 05 0 0 1 0

Using Algorithm 1, we can get there are no zero row and zero column in the principle sub-matrices B [1, 2, 3], B [1,3,4], therefore, there are still two 3-cycles in the digraph G. Then, C = {v1 ? v3 ? v2 ? v1,v1 ? v3 ? v4 ? v1}, and both the two 3-cycles do not satisfy the condition (c) of Theorem 3, and thus we have WTI = 2. Although the 3-cycle v1 ? v4 ? v2 ? v1 is removed, the new one v1 ? v3 ? v4 ? v1 appears. Therefore, we cannot modify the inconsistent entries randomly. In order to solve this problem, we propose the following rules. Rule 1. If one inconsistent arc vi ? vj (its corresponding entry is rij P [0.5, 0.5]) appears most frequently in the 3-cycles, then the inconsistent entry rij (rij P [0.5, 0.5], arc vi ? vj) should be repaired firstly.

thus

trðE3 Þ ¼ 3

C = {v1 ? v3 ? v4 ? v1,v1 ? v4 ? v3 ? v1}, and both the two 3-cycles satisfy the condition (c) of Theorem 3, and thus we have l = 2, WTI = 0. Then the IVFR R is weakly transitive although there are two 3-cycles in the digraph G.

¼2

Then there are two 3-cycles. All the 3  3 principal sub-matrices of B are:

Rule 1 denotes that we should give priority to repair the inconsistent arc vi ? vj (its corresponding entry is rij) which appears most frequently in the 3-cycles, it can remove the cycles maximally. Genh i h i þ 0  þ erally, we can set r0ij ¼ r  ji ; r ji ; r ji ¼ r ij ; r ij , which denotes that we only reverse the direction of the arc (vi ? vj), and the preference degree between xi and xj is not changed. If rij = [0.5, 0.5], we set r0ij ¼ ½0:4; 0:4; r 0ji ¼ ½0:6; 0:6.

There are no zero row and zero column in the principal sub-matrix B [1,3,4], therefore, there are 3-cycles in the digraph G. Then,

v

v

v

v

Fig. 4. Digraph G of Example 5.

Rule 2. If two or more inconsistent arcs appear the same times in the 3-cycles, then the inconsistent entry which is closer to [0.5, 0.5] should be repaired firstly. If the distances between the inconsistent entries and [0.5, 0.5] are the same, then we choose any inconsistent entry (arc) to repair. The judgment rij is closer to [0.5, 0.5], which denotes that the decision maker is more difficult to determine which one is better between xi and xj. And this would be easier to cause the intransitivity, thus, we give priority to modify these judgments.

30

Y. Xu et al. / Knowledge-Based Systems 63 (2014) 24–32

For simplicity, we can use the Hamming distance to measure the distance between rij and [0.5, 0.5], i.e., dðrij ; ½0:5; 0:5Þ ¼          rij  0:5 þ r þ ij  0:5.

Let k = k + 1 and return to Step 2. Step 4. Output k, E(k), B(k), WTI(k) and R(k). R(k) is the modified IVFR. Step 5. End.

Based on the Rules 1 and 2 and Algorithm 1, the following algorithm is proposed to improve the weak transitivity of an IVFR.

5. Numerical examples

Algorithm 2. Let R = (rij)nn be an IVFR, G is the digraph of R, E = (eij)nn is the adjacency matrix of R, B = (bij)nn = E2 ET.

Example 6. The IVFR R1 in Example 4. For the IVFR R1 in Example 4, we know that there are two 3cycles (C = {v1 ? v3 ? v2 ? v1,v1 ? v4 ? v2 ? v1}) in the digraph G of R1, and the inconsistent entries of R1 are {r13, r32, r21}, {r14, r42, r21}. The arc v2 ? v1 (which denotes the judgment r21) appears twice in the cycles while others appear only once. Therefore, we give priority to modify the judgment r21. Let ð1Þ ð1Þ r21 ¼ ½0:2; 0:6; r12 ¼ ½0:4; 0:8; k ¼ 1, then

Step 1. Let R(0) = R = (rij)nn, E(0) = E, B(0) = B, k = 0. Step 2. Using Algorithm 1 to compute the WTI(k) of an IVFR R(k). If WTI(k) = 0, then R(k) is weakly transitive, go to Step 4. If WTI(k) – 0, then we get the set C(k) of all the 3-cycles in the digraph G(k), and also get all the inconsistent judgments of R(k). ðkÞ Step 3. Search the inconsistent judgment rij which appears most frequently in the 3-cycles (i.e. find the element n o ðkÞ maxi;j bji ). If there are two or more inconsistent entries,

2 ð1Þ R1

then we choose the one is closer to [0.5, 0.5]. If their distances between [0.5, 0.5] are still same, then we choose

and

3

7 6 6 ½0:2; 0:6 ½0:5; 0:5 ½0:3; 0:5 ½0:4; 0:5 7 7 6 ¼6 7 6 ½0:4; 0:5 ½0:5; 0:7 ½0:5; 0:5 ½0:8; 0:8 7 5 4 ½0:1; 0:3 ½0:5; 0:6 ½0:2; 0:2 ½0:5; 0:5

any inconsistent judgment to repair. Let Rðkþ1Þ ¼   ðkþ1Þ r ij , where nn 8h i ðkÞ þðkÞ ðkÞ ðkÞ > rji ; r ji ; r ij is inconsistent entry and r ij – ½0:5; 0:5 > > > < ðkþ1Þ rij ¼ ½0:4; 0:4; rijðkÞ is inconsistent entry and r ðkÞ ij ¼ ½0:5;0:5 > h i > > > ðkÞ þðkÞ : r ;r ; otherwise ij

½0:5; 0:5 ½0:4; 0:8 ½0:5; 0:6 ½0:7; 0:9

And

2

0 1 1 1

3

2

7 6 60 0 0 07 ð1Þ 7; E1 ¼ 6 7 6 40 1 0 15 0 1 0

ij

0 0 0 0

3

7 6 60 0 0 07 ð1Þ 7 B1 ¼ 6 7 6 40 0 0 05

0

0 0 0 0

By Algorithm 1, we know that there is no 3-cycle in the digraph G of ð1Þ ð1Þ R1 . R1 has weak transitivity.

8h i ðkÞ þðkÞ ðkÞ ðkÞ > > rij ; r ij ; r ji is inconsistent entry and r ji – ½0:5;0:5 > > < ðkþ1Þ rji ¼ ½0:6; 0:6; rjiðkÞ is inconsistent entry and r ðkÞ ji ¼ ½0:5;0:5 > h i > > > : rðkÞ ; r þðkÞ ; otherwise ji ji

Example 7. Suppose that there an expert compares five alternatives xi (i = 1, 2, . . . , 5) and constructs the following IVFR.

Table 1 The repairing process for the IVFR R in Example 7.

R(k )

k

E (k )

⎡[0.5, 0.5] ⎢[0.4, 0.4] ⎢ ⎢[0.7, 0.8] ⎢ ⎢[0.6, 0.6] ⎢⎣[0.3, 0.4]

[0.6, 0.6] [0.5, 0.5] [0.1, 0.2] [0.5, 0.5] [0.2, 0.3]

[0.2, 0.3] [0.8, 0.9] [0.5, 0.5] [0.5, 0.5] [0.1, 0.2]

[0.4, 0.4] [0.5, 0.5] [0.5, 0.5] [0.5, 0.5] [0.6, 0.8]

[0.6, 0.7]⎤ [0.7, 0.8]⎥⎥ [0.8, 0.9] ⎥ ⎥ [0.2, 0.4]⎥ [0.5, 0.5] ⎥⎦

⎡0 ⎢0 ⎢ ⎢1 ⎢ ⎢1 ⎢⎣0

1 0 0 1 0

0 1 0 1 0

0 1 1 0 1

1⎤ 1 ⎥⎥ 1⎥ ⎥ 0⎥ 0 ⎥⎦

⎡0 ⎢2 ⎢ ⎢0 ⎢ ⎢0 ⎢1 ⎣

⎡[0.5, 0.5] ⎢[0.4, 0.4] ⎢ ⎢[0.7, 0.8] ⎢ ⎢[0.6, 0.6] ⎢⎣[0.3, 0.4]

[0.6, 0.6] [0.5, 0.5] [0.1, 0.2] [0.5, 0.5] [0.2, 0.3]

[0.2, 0.3] [0.8, 0.9] [0.5, 0.5] [0.5, 0.5] [0.1, 0.2]

[0.4, 0.4] [0.5, 0.5] [0.5, 0.5] [0.5, 0.5] [0.2, 0.4]

[0.6, 0.7]⎤ [0.7, 0.8]⎥⎥ [0.8, 0.9] ⎥ ⎥ [0.6, 0.8]⎥ [0.5, 0.5] ⎥⎦

⎡0 ⎢0 ⎢ ⎢1 ⎢ ⎢1 ⎢⎣0

1 0 0 1 0

0 1 0 1 0

0 1 1 0 0

1⎤ 1 ⎥⎥ 1⎥ ⎥ 1⎥ 0 ⎥⎦

⎡0 ⎢ ⎢2 ⎢0 ⎢ ⎢0 ⎢0 ⎣

2

⎡[0.5, 0.5] ⎢[0.6, 0.6] ⎢ ⎢[0.7, 0.3] ⎢ ⎢[0.6, 0.6] ⎣⎢[0.3, 0.4]

[0.4, 0.4] [0.5, 0.5] [0.1, 0.2] [0.5, 0.5] [0.2, 0.8]

[0.2, 0.3] [0.8, 0.9] [0.5, 0.5] [0.5, 0.5] [0.1, 0.2]

[0.4, 0.4] [0.5, 0.5] [0.5, 0.5] [0.5, 0.5] [0.2, 0.4]

[0.6, 0.7]⎤ [0.7, 0.8]⎥⎥ [0.8, 0.9] ⎥ ⎥ [0.6, 0.8]⎥ [0.5, 0.5] ⎥⎦

⎡0 ⎢1 ⎢ ⎢1 ⎢ ⎢1 ⎣⎢0

0 0 0 1 0

0 1 0 1 0

0 1 1 0 0

1⎤ 1 ⎥⎥ 1⎥ ⎥ 1⎥ 0 ⎥⎦

⎡0 ⎢0 ⎢ ⎢0 ⎢ ⎢0 ⎢0 ⎣

0 0 0 0 1 0

3

⎡[0.5, 0.5] ⎢[0.6, 0.6] ⎢ ⎢[0.7, 0.8] ⎢ ⎢[0.6, 0.6] ⎢⎣[0.3, 0.4]

[0.4, 0.4] [0.5, 0.5] [0.1, 0.2] [0.5, 0.5] [0.2, 0.3]

[0.2, 0.3] [0.8, 0.9] [0.5, 0.5] [0.6, 0.6] [0.1, 0.2]

[0.4, 0.4] [0.5, 0.5] [0.4, 0.4] [0.5, 0.5] [0.2, 0.4]

[0.6, 0.7]⎤ [0.7, 0.8]⎥⎥ [0.8, 0.9] ⎥ ⎥ [0.6, 0.8]⎥ [0.5, 0.5] ⎥⎦

⎡0 ⎢1 ⎢ ⎢1 ⎢ ⎢1 ⎢⎣0

0 0 0 1 0

0 1 0 1 0

0 1 0 0 0

1⎤ 1 ⎥⎥ 1⎥ ⎥ 1⎥ 0 ⎥⎦

⎡0 ⎢0 ⎢ ⎢0 ⎢ ⎢0 ⎢⎣0

0 0 0 0 0

0

1

WTI ( k )

B(k )

v1 → v2 → v3 → v1

0 1 2 0 ⎤ 0 0 2 0 ⎥⎥ 2 0 1 0 ⎥ ⎥ 1 1 0 3⎥ 1 1 0 0 ⎥⎦

6

0⎤ ⎥ 0⎥ 0⎥ ⎥ 0⎥ 0 ⎥⎦

3

0 0⎤ 1 0 ⎥⎥ 0 0⎥ ⎥ 0 1 0 0⎥ 0 0 0 0 ⎥⎦

1

0⎤ 0 ⎥⎥ 0⎥ ⎥ 0⎥ 0 ⎥⎦

0

0 1 1 0 2 1 0

0 0 1 0

0 0 0 0 0

1 0 0 0

0 0 0 0 0

C (k ) v1 → v2 → v4 → v1 v1 → v5 → v4 → v1 v2 → v3 → v4 → v2 v2 → v5 → v4 → v2 v3 → v5 → v4 → v3

v1 → v2 → v3 → v1 v1 → v2 → v4 → v1 v2 → v3 → v4 → v2

v2 → v3 → v4 → v2

Y. Xu et al. / Knowledge-Based Systems 63 (2014) 24–32

2

½0:5; 0:5 6 6 ½0:4; 0:4 6 6 R¼6 6 ½0:7; 0:8 6 6 ½0:6; 0:6 4 ½0:3; 0:4

½0:6; 0:6 ½0:2; 0:3 ½0:4; 0:4 ½0:6; 0:7 ½0:5; 0:5 ½0:8; 0:9 ½0:5; 0:5 ½0:1; 0:2 ½0:5; 0:5 ½0:5; 0:5 ½0:5; 0:5 ½0:5; 0:5 ½0:5; 0:5 ½0:2; 0:3 ½0:1; 0:2 ½0:6; 0:8

3

7 ½0:7; 0:8 7 7 7 ½0:8; 0:9 7 7 7 ½0:2; 0:4 7 5 ½0:5; 0:5

According to the Algorithm 2 given in the previous section, we can repair the IVFR R. After three times of repairing, we obtain the modified IVFR R with weak transitivity. The values k, R(k), E(k), B(k), WTI(k), C(k) are listed in Table 1. For the initial IVFR (i.e.k = 0), there are six 3-cycles in the digraph of R. Since n o ð0Þ ð0Þ ð0Þ max bji ¼ b45 ¼ 3, we select r 54 to modify first. Let ð1Þ

ð1Þ

r 54 ¼ ½0:2; 0:4; r45 ¼ ½0:6; 0:8; k ¼ 1,

then go to next iteran o ð1Þ ð1Þ ð1Þ ¼ b21 ¼ b32 ¼ 2. When k = 1, we have max bji    ð1Þ ð1Þ Because d r12 ; ½0:5; 0:5 ¼ j0:6  0:5j þ j0:6  0:5j ¼ 0:2 < d r 23 ;

tion.

ð1Þ

½0:5; 0:5Þ ¼ j0:8  0:5j þ j0:9  0:5j ¼ 0:7, we choose r12 to repair. Let

ð2Þ r 12

ð2Þ ½0:4; 0:4; r21 ¼ ½0:6; 0:6, k = n o ð2Þ ð2Þ ð2Þ ¼ 2; max bji ¼ b24 ¼ b32

2, then go to next iteration.  ð2Þ ð2Þ When k ¼ b43 ¼ 1. Because d r 42 ;     ð2Þ ð2Þ ½0:5; 0:5Þ ¼ d r 34 ; ½0:5; 0:5 ¼ 0 < d r 23 ; ½0:5; 0:5 ¼ 0:7, we can ¼

31

is not weakly transitive, then the results obtained by these methods would not be reliable. (3) If an IVFR is not weakly transitive, we propose an algorithm to repair the intransitivities of an IVFR. Our algorithm eliminates 3-cycles in a digraph most effectively. It is showed that the proposed method can not only improve the transitivity of an IVFR, but also preserve the initial preference information as much as possible. We do not change the comparison intensity between different alternatives, and only reverse the direction of the arc which causes the cycles. Furthermore, our method can be used for an IVFR not only with strict comparison information, but also with non-strict comparison information. In the future, we will investigate the application of the method to the consensus process [30] of group decision making problem, and other types of preference relations [1,6,24,39]. Acknowledgements

½0:4; 0:4; ¼ ½0:6; 0:6; k ¼ 3, then go to next iteration. When k = 3, WTI = 0, there is no 3-cycle. At this time, we obtain the modified IVFR R(3) with weak transitivity. Furthermore, by E(3), we can get the ranking of five alternatives is: x2  x4  x3  x1  x5.

The authors are very grateful to the Editor-in-Chief, Associate Editor and the two anonymous reviewers for their constructive comments and suggestions that have helped to improve the quality of this paper. This work was partially supported by the National Natural Science Foundation of China (Nos. 71101043 and 71301142), the major project of the National Social Science Fund of China (No. 12&ZD214), Program for Excellent Talents in Hohai University, and Zhejiang Natural Science Foundation of China (No. LQ13G010004).

6. Conclusions

References

Transitivity is a fundamental notion in decision theory. It is almost universally assumed in disciplines of decision theory and generally accepted as a principle of rationality. In this paper, we study the weak transitivity of an IVFR. We propose the definition of the weak transitivity of an IVFR. For a complete IVFR, we prove that if an IVFR is not weakly transitive, we only need to find the cycles of length 3 in its digraph. We also propose WTI to measure the consistency degree of an IVFR. Afterwards, an algorithm is proposed to compute WTI and to locate each cycle, as well as to find the inconsistent judgments in an IVFR. In order to resolve the intransitivities of an IVFR, another algorithm is developed to find and remove all the 3-cycles in the digraph. Finally, two examples are shown to illustrate the proposed method. Our proposed approach has the following distinct characteristics and advantages.

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ð2Þ

choose either r 42

ð2Þ

or r34

ð3Þ

to repair. For instance, let r 34 ¼

ð3Þ r43 (3)

(1) Multiple attribute decision making (MADM) consists of selecting the most desirable alternative from a given alternative set according to a set of alternatives. In order to make the right decision, the preference information which is provided by the DM should be rationality, that is, the DM should not provide the contradictory pairwise comparison judgments. Weak transitivity is the minimum requirement for the relational choice in decision making process. Our method can judge whether an IVFR is weakly transitive or not. (2) If an IVFR is weakly transitive, we can rank the alternatives directly. Therefore, from the point of view of ranking alternatives, our method can simplify the computation process, while other methods (for example Gong et al. [21], Xu [42]) usually need to compute the weight vector, all these methods are so complicated and need huge computation. Furthermore, these methods do not judge whether the DM’s preference information is consistent or not. If an IVFR

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