Satisfactory consistency judgement and inconsistency adjustment of linguistic judgement matrix

Fengxia Jin; Zhi Zhou; Yunxia Ma; Yonggang Chen

Dziennik i szczegóły wydania

Applied Mathematics and Nonlinear Sciences

Informacje o czasopiśmie

Format: Czasopismo
eISSN: 2444-8656
Pierwsze wydanie: 01 Jan 2016
Częstotliwość wydawania: 2 razy w roku
Języki: Angielski

In this study, first, the definitions of the preference relation matrix and the 0-1 permutation preference matrix of the linguistic judgement matrix are given. The method of judging the satisfactory consistency of the linguistic judgement matrix by the standard 0-1 arrangement matrix is obtained. This method not only solves the problem of satisfactory consistency when there are no equivalent objects but is also simple and effective. Then, the definition of the cyclic circle matrix of the three objects is given. According to the size of the preference value of the object line, the cyclic cycle and the adjusted language judgement matrix are obtained. Finally, the rationality and validity of the method are verified by examples.

Keywords

linguistic judgement matrix
preference relation matrix
satisfactory consistency
consistency modification

Artykuł

Introduction

In the decision analysis, due to the fuzziness, complexity and uncertainty of human thinking and objective things, decision-makers often cannot give accurate values when making judgements on things by pairwise comparison. Herrera et al. [1,2] noted the concept of using linguistic information to express evaluation results. The linguistic judgement matrix has become an important information form for decision-makers to compare objects. However, it is also important whether the linguistic judgement matrix is consistent before the decision-maker gives the judgement result. At present, the research in this field has attracted the extensive attention of scholars and achieved some results [3,4,5,6,7,8]. The number of illogical objects is found by constructing the directed graph, and then the satisfactory consistency degree of linguistic judgement is calculated [3]; this problem is solved by deriving the compatibility of matrix and weight matrix [4].

Many achievements on the consistency of linguistic judgement matrix have been obtained [9,10,11,12,13,14,15,16]. From the current research results, few scholars directly use the definition to determine the satisfactory consistency of the linguistic judgement matrix. Most scholars use graph theory knowledge or define a preference relation matrix to determine the satisfactory consistency of the linguistic judgement matrix. Whether the diagonal element of the reachable matrix is 0 was used to determine whether the linguistic judgement matrix has satisfactory consistency [9]. A method for determining the satisfactory consistency of the linguistic judgement matrix was established by using the concepts of the derived matrix and preference relation matrix [10]. The linguistic judgement matrix and graph theory were linked together to solve the problem of satisfactory consistency of linguistic judgement matrix with no difference between two objects [11]. Fan [12] used the concept of the induced matrix to transform the linguistic judgement matrix into a quantitative matrix to solve this problem. However in these methods, multiple matrices are required to be multiplied. Based on the former, this paper also uses the preference relation matrix to analyse the satisfactory consistency of the linguistic judgement matrix, but it is different. It proposes the concepts of the equivalent scheme, row preference value, column preference value, and standard 0-1 permutation matrix. This paper also presents a method to determine the satisfactory consistency of the linguistic judgement matrix. The method not only solves the problem of satisfactory consistency but is also simple and effective.

However, only judging whether the linguistic judgement matrix is consistent cannot meet the requirements of decision-makers. In recent years, some achievements have been made in other aspects of the linguistic judgement matrix [17,18,19,20,21,22,23,24,25,26]. Definitions of the consistency, acceptable incompleteness and complete consistency of the linguistic judgement matrix are given as well as a method for ranking alternatives [17]. A new proof method is given in reference [18]. It proves that if the judgement matrices given by each decision-maker were acceptable, then their weighted geometric average judgement matrices were also acceptable. For uncertain linguistic multi-attribute decision-making problems with incomplete weight coefficients, an optimisation model was established by transforming uncertain linguistic information into trapezoidal fuzzy numbers to obtain scheme weight coefficients [21]. The definition and properties of binary intuitionistic fuzzy language preference information are given, and a ranking method based on binary intuitionistic fuzzy language preference information is proposed [22]. Pang [23] compared the preference information of individuals and groups and measured the influence of the preference information of individuals and groups on decision-making from three evaluation indicators of consistency, closeness and monotonicity. Hsu [25] used the direct and indirect pairwise comparison rules to construct a multi-criteria incomplete language preference relationship model to solve the problem of increasing solutions and decreasing consistency.

This paper discusses the method to determine and adjust the satisfactory consistency of the linguistic judgement matrix with strict preference relation between two objects. In this paper, the satisfactory consistency of the linguistic judgement matrix can be modified by using whether the 0-1 permutation preference relation matrix of the linguistic judgement matrix is an upper triangular matrix, and then the three objects for forming a circular circle of judgement results are found by using the cyclic matrix. Using the size of the row preference value to modify the cyclic circle, the 0-1 type permutation preference relation matrix with satisfactory consistency is obtained. Finally the linguistic judgement matrix with satisfactory consistency is obtained.

Basic concepts

I = {1, 2, …, n} and U = {0, 1, 2, …, T} are two sets, where T is an even number. The preference information of pairwise comparison given by decision-makers can be described by a matrix P = (p_ij)_n×n. The objects in the matrix are selected from the linguistic term set S = {s_i|i ∈ U} as the evaluation results of x_i and x_j. The number of objects in the matrix is called the granularity of linguistic term set. For example, a linguistic term set with 13 granularity can be described as S={s₀=DD=absolute difference, s₁=VHD=quite poor, s₂=HD=very poor, s₃=MD=weak, s₄=LD=Poor, s₅=VLD=slightly poor, s₆=AS=equivalent, s₇=VLP=slightly better, s₈=LP=better, s₉=MP=good, s₁₀=HP=very good, s₁₁=VHP=quite good, s₁₂ = DP=absolutely good}, and should have the following properties:

Orderliness: when i<j, there is s_i ≺ s_j or s_j ≻ s_i, namely it is inferior s_i to s_j or s_j better than s_i;

Inverse operation neg: neg(s_i)=s_j, j = T −i;

Maximisation operation: if s_i ≥ s_j, max{s_i, s_j} = s_i;

Minimisation operation: if s_i ≥ s_j, min{s_i, s_j} = s_j;

$S^{L} = {s_{0}, s_{1}, \cdot \cdot \cdot, s_{\frac{T}{2} - 1}}$ {S^L} = \{{s_0},{s_1}, \cdot \cdot \cdot,{s_{{T \over 2} - 1}}\} , $S^{U} = {s_{\frac{T}{2} + 1}, \cdot \cdot \cdot, s_{T}}$ {S^U} = \{{s_{{T \over 2} + 1}}, \cdot \cdot \cdot,{s_T}\} , $S_{\frac{T}{2}}^{L} = {s_{0}, s_{1}, \cdot \cdot \cdot, s_{\frac{T}{2}}}$ S_{{T \over 2}}^L = \{{s_0},{s_1}, \cdot \cdot \cdot,{s_{{T \over 2}}}\} , $S_{\frac{T}{2}}^{U} = {s_{\frac{T}{2}}, \cdot \cdot \cdot, s_{T}}$ S_{{T \over 2}}^U = \{{s_{{T \over 2}}}, \cdot \cdot \cdot,{s_T}\} are the four linguistic term sets.

Definition 2.1

[11,12] A linguistic judgement matrix on the finite object set X = {x₁, x₂, …, x_n} for the linguistic term set S = {s_i |i ∈ U} is defined as P = (p_ij)_n×n, where (1) $p_{ij} \in S; p_{ii} = s_{\frac{T}{2}}; p_{ij} = s_{k}, p_{ji} = neg (s_{k});$ {p_{ij}} \in S;{p_{ii}} = {s_{{T \over 2}}};{p_{ij}} = {s_k},{p_{ji}} = neg({s_k}); for all i, j ∈ I.

Definition 2.2

In the linguistic judgement matrix, if $p_{ij} \in S_{\frac{T}{2}}^{U}$ {p_{ij}} \in S_{{T \over 2}}^U and $p_{ji} \in S_{\frac{T}{2}}^{U}$ {p_{ji}} \in S_{{T \over 2}}^U , that is $p_{ij} = s_{\frac{T}{2}}$ {p_{ij}} = {s_{{T \over 2}}} , then x_i and x_j are called equivalent objects and denoted as x_i ∼ x_j.

According to the meaning of the objects in the linguistic term set, we can know that the objects in the linguistic judgement matrix have the following properties:

If $p_{ij} = s_{\frac{T}{2}}$ {p_{ij}} = {s_{{T \over 2}}} , there is no difference between x_i and x_j, then x_i ∼ x_j;

If p_ij = s_k ∈ S^U, it means strictly that x_i better than x_j, then record as x_i ≻ x_j, and the greater k, the greater the degree of superiority;

If p_ij = s_k ∈ S^L, it means strictly x_i inferior to x_j, then record as x_i ≺ x_j, and the smaller k, the greater the degree of inferiority;

If $p_{ij} = s_{k} \in S_{\frac{T}{2}}^{U}$ {p_{ij}} = {s_k} \in S_{{T \over 2}}^U , it means x_i is not inferior to x_j, then record as x_i ≥ x_j;

If $p_{ij} = s_{k} \in S_{\frac{T}{2}}^{L}$ {p_{ij}} = {s_k} \in S_{{T \over 2}}^L , it means x_i is not better than x_j, then record as x_i ≤ x_j.

Definition 2.3

A preference relation matrix of the linguistic judgement matrix on the finite object set X = {x₁, x₂, …, x_n} for the linguistic term set S = {s_i |i ∈ U} is defined as Q = (q_ij)_n×n, where (2) $q_{ij} = {\begin{matrix} 1 & if p_{ij} \in S_{\frac{T}{2}}^{U} \\ 0 & if p_{ij} \in S^{L} \end{matrix}$ {q_{ij}} = \left\{{\matrix{1 & {if\,{p_{ij}} \in S_{{T \over 2}}^U} \cr 0 & {if\,{p_{ij}} \in {S^L}} \cr}} \right.

Definition 2.4

In the preference relation matrix Q = (q_ij)_n×n, $a_{i} = \sum_{j = 1}^{n} q_{ij}$ {a_i} = \sum\limits_{j = 1}^n {q_{ij}} is the row preference value of row i of the preference relation matrix; $b_{j} = \sum_{i = 1}^{n} q_{ij}$ {b_j} = \sum\limits_{i = 1}^n {q_{ij}} is the column preference value of column j of the preference relation matrix

Definition 2.5

A linguistic judgement matrix P = (p_ij)_n×n has satisfactory consistency if the relationship between the advantages and disadvantages of each scheme is transitive, and there is no circulation phenomenon except for the equivalent scheme.

For example, x₁ ≻ x₂ ≻ x₃ ≻ x₁ is a cyclic cycle.

Determination of satisfactory consistency and ranking of objects

In the last section, the definition of the linguistic judgement matrix is given, and the definitions of the equivalent scheme, the row preference value and the column preference value, and the preference relation matrix are proposed. On the basis of comparison of each object, the method to determine the satisfactory consistency of the linguistic judgement matrix is given, and the ranking method is also given.

3.1

Determination of satisfactory consistency

Definition 3.1

Let Q = (q_ij)_n×n be the preference relation matrix of the linguistic judgment matrix P = (p_ij)_n×n, and arrange the elements according to the number of preference values of the row. In order to ensure that the original preference relation remains unchanged, the columns are also adjusted accordingly. In this way, a new matrix R = (r_ij)_n×n is obtained, which is called the 0-1 permutation preference relation matrix of a linguistic judgement matrix.

For example: if $Q = \begin{array}{l} x_{1} \begin{matrix} x_{2} & x_{3} & x_{4} \end{matrix} \\ \begin{array}{l} \begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} & (\begin{matrix} 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 1 & 1 \end{matrix}) \end{array} \end{array}$ Q = \matrix{{{x_1}\matrix{{} & {{x_2}} & {{x_3}} & {{x_4}} \cr}} \hfill \cr {\matrix{{\matrix{{{x_1}} \hfill \cr {{x_2}} \hfill \cr {{x_3}} \hfill \cr {{x_4}} \hfill \cr}} \hfill & {\left({\matrix{1 & 0 & 1 & 0 \cr 1 & 1 & 1 & 1 \cr 0 & 0 & 1 & 0 \cr 1 & 0 & 1 & 1 \cr}} \right)} \hfill \cr}} \hfill \cr} , then $R = \begin{array}{l} x_{2} \begin{matrix} x_{4} & x_{1} & x_{3} \end{matrix} \\ \begin{array}{l} \begin{array}{l} x_{2} \\ x_{4} \\ x_{1} \\ x_{3} \end{array} & (\begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{matrix}) \end{array} \end{array}$ R = \matrix{{{x_2}\matrix{{} & {{x_4}} & {{x_1}} & {{x_3}} \cr}} \hfill \cr {\matrix{{\matrix{{{x_2}} \hfill \cr {{x_4}} \hfill \cr {{x_1}} \hfill \cr {{x_3}} \hfill \cr}} \hfill & {\left({\matrix{1 & 1 & 1 & 1 \cr 0 & 1 & 1 & 1 \cr 0 & 0 & 1 & 1 \cr 0 & 0 & 0 & 1 \cr}} \right)} \hfill \cr}} \hfill \cr} ; if $Q = \begin{array}{l} x_{1} \begin{matrix} x_{2} & x_{3} & x_{4} \end{matrix} \\ \begin{array}{l} \begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} & (\begin{matrix} 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \end{matrix}) \end{array} \end{array}$ Q = \matrix{{{x_1}\matrix{{} & {{x_2}} & {{x_3}} & {{x_4}} \cr}} \hfill \cr {\matrix{{\matrix{{{x_1}} \hfill \cr {{x_2}} \hfill \cr {{x_3}} \hfill \cr {{x_4}} \hfill \cr}} \hfill & {\left({\matrix{1 & 0 & 1 & 0 \cr 1 & 1 & 0 & 0 \cr 0 & 1 & 1 & 0 \cr 1 & 1 & 1 & 1 \cr}} \right)} \hfill \cr}} \hfill \cr} , then $R = \begin{array}{l} x_{4} \begin{matrix} x_{1} & x_{2} & x_{3} \end{matrix} \\ \begin{array}{l} \begin{array}{l} x_{4} \\ x_{1} \\ x_{2} \\ x_{3} \end{array} & (\begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{matrix}) \end{array} \end{array}$ \matrix{{\matrix{{\quad \quad {x_4}} & {{x_1}} & {{x_2}} & {{x_3}} \cr}} \hfill \cr {R = \matrix{{{x_4}} \cr {{x_1}} \cr {{x_2}} \cr {{x_3}} \cr} \left({\matrix{1 & 1 & 1 & 1 \cr 0 & 1 & 0 & 1 \cr 0 & 1 & 1 & 0 \cr 0 & 0 & 1 & 1 \cr}} \right)} \hfill \cr} . or $R = \begin{array}{l} x_{4} \begin{matrix} x_{3} & x_{2} & x_{1} \end{matrix} \\ \begin{array}{l} \begin{array}{l} x_{4} \\ x_{3} \\ x_{2} \\ x_{1} \end{array} & (\begin{matrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 1 & 0 & 1 \end{matrix}) \end{array} \end{array}$ \matrix{{\matrix{{\quad \quad {x_4}} & {{x_3}} & {{x_2}} & {{x_1}} \cr}} \hfill \cr {R = \matrix{{{x_4}} \cr {{x_3}} \cr {{x_2}} \cr {{x_1}} \cr} \left({\matrix{1 & 1 & 1 & 1 \cr 0 & 1 & 1 & 0 \cr 0 & 0 & 1 & 0 \cr 1 & 1 & 0 & 1 \cr}} \right)} \hfill \cr} .

According to the example above_St can be concluded that the 0-1 permutation preference relation matrix of the preference relation matrix is not unique.

Definition 3.2

Let A = (a_ij)_n×n be a standard 0-1 permutation matrix, if the elements of each row satisfy the following: (1) the number of 1s in row i is greater than or equal to the number of 1s in row i + 1; $a_{ij} = 1, a_{ij + 1} = 1, \cdot \cdot \cdot, a_{in} = 1; a_{ij - 1} = 0, a_{ij - 2} = 0, \cdot \cdot \cdot, a_{i 1} = 0 .$ {a_{ij}} = 1,{a_{ij + 1}} = 1, \cdot \cdot \cdot,{a_{in}} = 1;{a_{ij - 1}} = 0,{a_{ij - 2}} = 0, \cdot \cdot \cdot,{a_{i1}} = 0.

For example, the following matrices are the standard 0-1 permutation matrices: $A = (\begin{array}{l} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{array})$ A = \left({\matrix{1 & 1 & 1 & 1 \cr 0 & 1 & 1 & 1 \cr 0 & 0 & 1 & 1 \cr 0 & 0 & 0 & 1 \cr}} \right) ; $B = (\begin{array}{l} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{array})$ B = \left({\matrix{1 & 1 & 1 & 1 \cr 0 & 1 & 1 & 1 \cr 0 & 1 & 1 & 1 \cr 0 & 0 & 0 & 1 \cr}} \right) ; $C = (\begin{array}{l} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 \end{array})$ C = \left({\matrix{1 & 1 & 1 & 1 \cr 0 & 1 & 1 & 1 \cr 0 & 0 & 1 & 1 \cr 0 & 0 & 1 & 1 \cr}} \right) ; $D = (\begin{array}{l} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 \end{array})$ D = \left({\matrix{1 & 1 & 1 & 1 & 1 \cr 0 & 1 & 1 & 1 & 1 \cr 0 & 0 & 1 & 1 & 1 \cr 0 & 0 & 1 & 1 & 1 \cr 0 & 0 & 0 & 0 & 1 \cr}} \right) ; $E = (\begin{array}{l} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 \end{array})$ E = \left({\matrix{1 & 1 & 1 & 1 & 1 \cr 1 & 1 & 1 & 1 & 1 \cr 0 & 0 & 1 & 1 & 1 \cr 0 & 0 & 0 & 1 & 1 \cr 0 & 0 & 0 & 1 & 1 \cr}} \right) .

Theorem 3.1

A linguistic judgement matrix P = (p_ij)_n×n is satisfactory if and only if its 0-1 permutation preference relation matrix R = (r_ij)_n×n is a standard 0-1 permutation matrix.

Proof

First, we prove the necessity. According to definition 2.5, if P = (p_ij)_n×n has satisfactory consistency, the orders of the objects x₁,x₂,x₃, …, x_n can be expressed as x₍₁₎, x₍₂₎, …, x_(n). The 0-1 permutation preference relation matrix known from definition 3.1 satisfies that the number of 1s in the row i is greater than or equal to the number of 1s in the row i+1. Let x_(i) be one of the objects, it can to set x_(j), x_(j+1), …, x_(n) to be the objects which are not superior x_(i) according to the relationship between the advantages and disadvantages (it may exist equivalent schemes, so x_(j) is not necessarily the object x_(i+1)).

Then, $p_{(i) (j)}, p_{(i) (j + 1)}, \cdot \cdot \cdot, p_{(i) (n)} \in S_{\frac{T}{2}}^{U}, p_{(i) (j - 1)}, p_{(i) (j - 2)}, \cdot \cdot \cdot, p_{(i) (1)} \in S^{L} .$ {p_{(i)(j)}},{p_{(i)(j + 1)}}, \cdot \cdot \cdot,{p_{(i)(n)}} \in S_{{T \over 2}}^U,{p_{(i)(j - 1)}},{p_{(i)(j - 2)}}, \cdot \cdot \cdot,{p_{(i)(1)}} \in {S^L}.

From equation (2), it can be concluded that, $q_{(i) (j)}, q_{(i) (j + 1)}, \cdot \cdot \cdot, q_{(i) (n)} = 1, q_{(i) (j - 1)}, q_{(i) (j - 2)}, \cdot \cdot \cdot, q_{(i) (1)} = 0 .$ {q_{(i)(j)}},{q_{(i)(j + 1)}}, \cdot \cdot \cdot,{q_{(i)(n)}} = 1,{q_{(i)(j - 1)}},{q_{(i)(j - 2)}}, \cdot \cdot \cdot,{q_{(i)(1)}} = 0.

The corresponding 0-1 permutation preference relation matrix is r_ij, r_ij₊₁, …, r_in = 1,r_ij₋₁, r_ij₋₂, …, r_i₁ = 0.

It can be seen from the above that when the linguistic judgement matrix has satisfactory consistency, its 0-1 permutation preference relation matrix is the standard 0-1 permutation matrix.

Second, we prove the sufficiency. Let R be the 0-1 permutation preference relation matrix of the linguistic judgement matrix that satisfies the following conditions: r_ij, r_ij₊₁, …, r_in = 1, r_ij₋₁, r_ij₋₂, …, r_i₁ = 0,

Suppose the object x_(i) corresponds to row i, according to formula (2) the object x_(i) is not inferior to the objects x_(j), x_(j+1), …, x_(n) and is inferior to the objects x_(j−1), x_(j−2), …, x₍₁₎, and the number of 1s in the row i is more than the number of 1s in the row i + 1, then the relationship between the advantages and disadvantages of each object can be obtained. At the same time, the relationship has transitivity and no circulation phenomenon; therefore, P = (p_ij)_n×n has satisfactory consistency.

From the above proof, it can be concluded that the 0-1 permutation preference relation matrix is an upper triangular matrix when P has a strict preference relation. In this way, a method to judge whether the linguistic judgement matrix has satisfactory consistency can be obtained. The specific steps are as follows:

Step 1: Write the corresponding preference relation matrix according to the linguistic judgement matrix;

Step 2: Write the 0-1 permutation preference relation matrix according to the preference relation matrix;

Step 3: Judge whether the 0-1 permutation preference relation matrix is the standard 0-1 permutation matrix;

Step 4: If it is an upper triangular matrix, it has a strict preference relationship; otherwise, proceed to the next step;

Step 5: If it is a standard 0-1 permutation matrix, P has satisfactory consistency, otherwise it does not;

Step 6: Judge the end.

3.2

Ranking of objects

According to the above judgement method, we can judge whether a linguistic judgement matrix has satisfactory consistency or not. At the same time, we can also get a ranking method of objects. The specific steps are as follows:

Step 1: Judge whether the linguistic judgement matrix given by the decision-maker has satisfactory consistency;

Step 2: If there is no satisfactory consistency, it will be returned to the decision-maker for adjustment until satisfactory consistency is obtained;

Step 3: If there is satisfactory consistency, write the 0-1 permutation preference relation matrix of the linguistic judgement matrix, that is, the standard 0-1 permutation matrix;

Step 4: Judge the advantages and disadvantages of the scheme according to the standard 0-1 type permutation matrix;

Step 5: Get the sequence of advantages and disadvantages of the object.

Example 1

A decision-maker evaluates four types of fax machines and gives the following preference information: $P = (\begin{array}{l} AS & LD & MP & HD \\ LP & AS & HP & MP \\ MD & HD & AS & HD \\ HP & MD & HP & AS \end{array}) .$ P = \left({\matrix{{AS} & {LD} & {MP} & {HD} \cr {LP} & {AS} & {HP} & {MP} \cr {MD} & {HD} & {AS} & {HD} \cr {HP} & {MD} & {HP} & {AS} \cr}} \right).

Judge whether P is consistent or not, and if there is satisfactory consistency, the sequence of advantages and disadvantages of the objects is given.

The preference relation matrix obtained from formula (2) is as follows: $\begin{array}{l} \begin{array}{l} x_{1} & x_{2} & x_{3} & x_{4} \end{array} \\ Q = \begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} (\begin{array}{l} 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 1 & 1 \end{array}) \end{array} .$ \matrix{{\matrix{{\quad \quad {x_1}} & {{x_2}} & {{x_3}} & {{x_4}} \cr}} \hfill \cr {Q = \matrix{{{x_1}} \cr {{x_2}} \cr {{x_3}} \cr {{x_4}} \cr} \left({\matrix{1 & 0 & 1 & 0 \cr 1 & 1 & 1 & 1 \cr 0 & 0 & 1 & 0 \cr 1 & 0 & 1 & 1 \cr}} \right)} \hfill \cr}.

According to Definition 3.1, the 0-1 permutation preference relation matrix is $\begin{array}{l} \begin{array}{l} x_{2} & x_{4} & x_{1} & x_{3} \end{array} \\ R = \begin{array}{l} x_{2} \\ x_{4} \\ x_{1} \\ x_{3} \end{array} (\begin{array}{l} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{array}) \end{array} .$ \matrix{{\matrix{{\quad \quad {x_2}} & {{x_4}} & {{x_1}} & {{x_3}} \cr}} \hfill \cr {R = \matrix{{{x_2}} \cr {{x_4}} \cr {{x_1}} \cr {{x_3}} \cr} \left({\matrix{1 & 1 & 1 & 1 \cr 0 & 1 & 1 & 1 \cr 0 & 0 & 1 & 1 \cr 0 & 0 & 0 & 1 \cr}} \right)} \hfill \cr}

Obviously, it is a standard 0-1 permutation matrix, so it has satisfactory consistency, the sequence of advantages and disadvantages of the objects is x₂ ≻ x₄ ≻ x₁ ≻ x₃.

Example 2

Let the decision-maker give the linguistic judgement matrix of five alternatives X = {x₁, x₂, x₃, x₄, x₅} as follows: $P = (\begin{array}{l} AS & VLP & LP & AS & MP \\ VLD & AS & VLD & LP & DVL \\ LD & VLP & AS & AS & HP \\ AS & LD & AS & AS & VLD \\ MD & VLP & HD & VLP & AS \end{array}),$ P = \left({\matrix{{AS} & {VLP} & {LP} & {AS} & {MP} \cr {VLD} & {AS} & {VLD} & {LP} & {DVL} \cr {LD} & {VLP} & {AS} & {AS} & {HP} \cr {AS} & {LD} & {AS} & {AS} & {VLD} \cr {MD} & {VLP} & {HD} & {VLP} & {AS} \cr}} \right),

Judge whether P is consistent or not.

The preference relation matrix Q of Pi s $\begin{array}{l} \begin{array}{l} x_{1} & x_{2} & x_{3} & x_{4} & x_{5} \end{array} \\ Q = \begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{array} (\begin{array}{l} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 1 & 1 \end{array}) \end{array}$ \matrix{{\matrix{{\quad \quad {x_1}} \hfill & {{x_2}} \hfill & {{x_3}} \hfill & {{x_4}} \hfill & {{x_5}} \hfill \cr}} \hfill \cr {Q = \matrix{{{x_1}} \cr {{x_2}} \cr {{x_3}} \cr {{x_4}} \cr {{x_5}} \cr} \left({\matrix{1 & 1 & 1 & 1 & 1 \cr 0 & 1 & 0 & 1 & 0 \cr 0 & 1 & 1 & 1 & 1 \cr 1 & 0 & 1 & 1 & 0 \cr 0 & 1 & 0 & 1 & 1 \cr}} \right)} \hfill \cr}

and the 0-1 permutation preference relation matrix is $\begin{array}{l} \begin{array}{l} x_{1} & x_{3} & x_{4} & x_{5} & x_{2} \end{array} \\ R = \begin{array}{l} x_{1} \\ x_{3} \\ x_{4} \\ x_{5} \\ x_{2} \end{array} (\begin{array}{l} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \end{array}) \end{array} .$ \matrix{{\matrix{{\quad \quad {x_1}} \hfill & {{x_3}} \hfill & {{x_4}} \hfill & {{x_5}} \hfill & {{x_2}} \hfill \cr}} \hfill \cr {R = \matrix{{{x_1}} \cr {{x_3}} \cr {{x_4}} \cr {{x_5}} \cr {{x_2}} \cr} \left({\matrix{1 & 1 & 1 & 1 & 1 \cr 0 & 1 & 1 & 1 & 1 \cr 1 & 1 & 1 & 0 & 0 \cr 0 & 0 & 1 & 1 & 1 \cr 0 & 0 & 1 & 0 & 1 \cr}} \right)} \hfill \cr}.

Obviously, it is not the standard 0-1 permutation matrix, so the linguistic judgement matrix does not have satisfactory consistency. It is necessary to return it to the decision-maker for adjustment or modify the satisfactory consistency until a satisfactory consistent linguistic judgement matrix is obtained, which is not discussed here.

Satisfaction consistency determination and inconsistency adjustment

Theorem 4.1

The sufficient and necessary condition for the linguistic judgement matrix P = (p_ij)_n×n with strict preference relation to having satisfactory consistency is that the 0-1 permutation preference relation matrix of the linguistic judgement matrix R = (r_ij)_n×n is an upper triangular matrix.

Following Definition 3.2 and Theorem 3.1, the conclusion can be easily derived.

Definition 4.2

The secondary sub-formulas of the 0-1 permutation preference relation matrix of linguistic judgement matrix are called the cyclic circle matrix, if they have the following form: $\begin{array}{l} \begin{array}{l} x_{i} & x_{k} \end{array} \\ \begin{array}{l} x_{j} \\ x_{i} \end{array} (\begin{array}{l} 1 & 0 \\ 1 & 1 \end{array}) \end{array}$ \eqalign{& \matrix{{} \hfill & {{x_i}} \hfill & {{x_k}} \hfill \cr} \cr & \matrix{{{x_j}} \cr {{x_i}} \cr} \left({\matrix{1 & 0 \cr 1 & 1 \cr}} \right) \cr} ; $\begin{array}{l} \begin{array}{l} x_{k} & x_{i} \end{array} \\ \begin{array}{l} x_{i} \\ x_{j} \end{array} (\begin{array}{l} 1 & 1 \\ 0 & 1 \end{array}) \end{array}$ \eqalign{& \matrix{{} \hfill & {{x_k}} \hfill & {{x_i}} \hfill \cr} \cr & \matrix{{{x_i}} \cr {{x_j}} \cr} \left({\matrix{1 & 1 \cr 0 & 1 \cr}} \right) \cr} ; $\begin{array}{l} \begin{array}{l} x_{i} & x_{k} \end{array} \\ \begin{array}{l} x_{i} \\ x_{j} \end{array} (\begin{array}{l} 1 & 1 \\ 1 & 0 \end{array}) \end{array}$ \eqalign{& \matrix{{} \hfill & {{x_i}} \hfill & {{x_k}} \hfill \cr} \cr & \matrix{{{x_i}} \cr {{x_j}} \cr} \left({\matrix{1 & 1 \cr 1 & 0 \cr}} \right) \cr} ; $\begin{array}{l} \begin{array}{l} x_{k} & x_{i} \end{array} \\ \begin{array}{l} x_{j} \\ x_{i} \end{array} (\begin{array}{l} 0 & 1 \\ 1 & 1 \end{array}) \end{array}$ \eqalign{& \matrix{{} \hfill & {{x_k}} \hfill & {{x_i}} \hfill \cr} \cr & \matrix{{{x_j}} \cr {{x_i}} \cr} \left({\matrix{0 & 1 \cr 1 & 1 \cr}} \right) \cr} .

The cyclic circle matrix is the secondary sub-formula of the 0-1 permutation preference relation matrix, which is not necessarily true in reverse. There is only one element of the cyclic circle matrix in the main diagonal of the 0-1 permutation preference relation matrix, and the comparison relationship between the objects and the original linguistic judgement matrix remains unchanged. The comparison result of the three objects constituting the cyclic circle matrix is cyclic, and the cyclic circle matrix composed of the three objects can be the above four forms.

Theorem 4.2

The sufficient and necessary condition for the linguistic judgement matrix P = (p_ij)_n×n not to have satisfactory consistency is that there are cyclic circle matrices in the secondary sub-formula of the 0-1 permutation preference relation matrix R = (r_ij)_n×n.

Proof

First, we prove the sufficiency. If there is a cyclic circle matrix in the 0-1 permutation preference relation matrix of the linguistic judgement matrix, it shows that the comparison results of the three objects are a cycle circle. Therefore, the comparison results of the objects are not transitive, and the linguistic judgement matrix is not of satisfactory consistency.

Second, we prove the necessity. According to theorem 1, the linguistic judgement matrix has strict preference relation between two objects, then its 0-1 permutation preference relation matrix R = (r_ij)_n×n is an upper triangular matrix. On the contrary, we can conclude that the linguistic judgement matrix does not have satisfactory consistency. If r_ji = 1, then r_ij = 0 (in the 0-1 permutation preference relation matrix, the corresponding objects are x₍₁₎, x₍₂₎, …, x_(n) from the first row to the last row), it is only necessary to prove that there exists secondary sub formula is a cyclic circle matrix.

Discussing circular circle matrix of r_ji = 1 and r_ii = 1. Because r_ji = 1 is in the lower triangular matrix, the form of the cyclic circle matrix is $\begin{array}{l} \begin{array}{l} x_{(i)} & x_{(k)} \end{array} \\ \begin{array}{l} x_{(i)} \\ x_{(j)} \end{array} (\begin{array}{l} 1 & 1 \\ 1 & 0 \end{array}) \end{array}$ \eqalign{& \matrix{{} \hfill & {{x_{(i)}}} \hfill & {{x_{(k)}}} \hfill \cr} \cr & \matrix{{{x_{(i)}}} \cr {{x_{(j)}}} \cr} \left({\matrix{1 & 1 \cr 1 & 0 \cr}} \right) \cr} or $\begin{array}{l} \begin{array}{l} x_{(k)} & x_{(i)} \end{array} \\ \begin{array}{l} x_{(i)} \\ x_{(j)} \end{array} (\begin{array}{l} 1 & 1 \\ 0 & 1 \end{array}) \end{array}$ \eqalign{& \matrix{{} \hfill & {{x_{(k)}}} \hfill & {{x_{(i)}}} \hfill \cr} \cr & \matrix{{{x_{(i)}}} \cr {{x_{(j)}}} \cr} \left({\matrix{1 & 1 \cr 0 & 1 \cr}} \right) \cr} Suppose that the elements with 1 after the main diagonal of the row i are r_ip, …, r_iq, then the corresponding elements in the row r_ip, …, r_iq are r_jp, …, r_jq. Suppose r_jp, …, r_jq are all 1. If there is an element of 1 in front of the main diagonal of the row i, then set r_ik = 1. if r_jk= 0, then the formal cyclic circle matrix is $\begin{array}{l} \begin{array}{l} x_{(k)} & x_{(i)} \end{array} \\ \begin{array}{l} x_{(i)} \\ x_{(j)} \end{array} (\begin{array}{l} 1 & 1 \\ 0 & 1 \end{array}) \end{array}$ \eqalign{& \matrix{{} \hfill & {{x_{(k)}}} \hfill & {{x_{(i)}}} \hfill \cr} \cr & \matrix{{{x_{(i)}}} \cr {{x_{(j)}}} \cr} \left({\matrix{1 & 1 \cr 0 & 1 \cr}} \right) \cr} . If r_jk= 1 and r_ji = 1, r_jj = 1, there are more 1s in the row of j than in the row of i. This is not true, so there must be an element of 0 in the row r_jp, …, r_jq. If r_jl = 0, the form of cyclic circle matrix is $\begin{array}{l} \begin{array}{l} x_{(i)} & x_{(l)} \end{array} \\ \begin{array}{l} x_{(i)} \\ x_{(j)} \end{array} (\begin{array}{l} 1 & 1 \\ 1 & 0 \end{array}) \end{array}$ \eqalign{& \matrix{{} \hfill & {{x_{(i)}}} \hfill & {{x_{(l)}}} \hfill \cr} \cr & \matrix{{{x_{(i)}}} \cr {{x_{(j)}}} \cr} \left({\matrix{1 & 1 \cr 1 & 0 \cr}} \right) \cr} . If there is no element with 1s in front of the main diagonal of the row, no matter whether there is a 1 in front of r_ji = 1, then there must be an element of 0 in the row r_jp, …, r_jq. Otherwise, the number of 1s in the row j is more than the number of 1s in the row i, then the form of cyclic circle matrix is $\begin{array}{l} \begin{array}{l} x_{(i)} & x_{(l)} \end{array} \\ \begin{array}{l} x_{(i)} \\ x_{(j)} \end{array} (\begin{array}{l} 1 & 1 \\ 1 & 0 \end{array}) \end{array}$ \eqalign{& \matrix{{} \hfill & {{x_{(i)}}} \hfill & {{x_{(l)}}} \hfill \cr} \cr & \matrix{{{x_{(i)}}} \cr {{x_{(j)}}} \cr} \left({\matrix{1 & 1 \cr 1 & 0 \cr}} \right) \cr} .

Discussing the cyclic circle matrix of r_ji = 1 and r_jj = 1. r_ji = 1 and r_jj = 1 are in the same row, so they can only form the cyclic circle matrix of the form of $\begin{array}{l} \begin{array}{l} x_{(i)} & x_{(j)} \end{array} \\ \begin{array}{l} x_{(k)} \\ x_{(j)} \end{array} (\begin{array}{l} 0 & 1 \\ 1 & 1 \end{array}) \end{array}$ \eqalign{& \matrix{{} \hfill & {{x_{(i)}}} \hfill & {{x_{(j)}}} \hfill \cr} \cr & \matrix{{{x_{(k)}}} \cr {{x_{(j)}}} \cr} \left({\matrix{0 & 1 \cr 1 & 1 \cr}} \right) \cr} or $\begin{array}{l} \begin{array}{l} x_{(i)} & x_{(j)} \end{array} \\ \begin{array}{l} x_{(j)} \\ x_{(l)} \end{array} (\begin{array}{l} 1 & 1 \\ 0 & 1 \end{array}) \end{array}$ \eqalign{& \matrix{{} \hfill & {{x_{(i)}}} \hfill & {{x_{(j)}}} \hfill \cr} \cr & \matrix{{{x_{(j)}}} \cr {{x_{(l)}}} \cr} \left({\matrix{1 & 1 \cr 0 & 1 \cr}} \right) \cr} . Suppose that the element with 1 in column j are r_mj, … , r_nj and the element in the corresponding column i are r_mi, …, r_ni. In the 0-1 permutation preference relation matrix, the number of 1s in the upper row is always greater than or equal to the number of 1s in the following row; correspondingly, the number of 1s in the front column is always less than or equal to the number of 1s in the following column. Assuming that r_mi, …, r_ni are 1, if r_pi = 1 and r_pi = 0, then the formal circular matrix is $\begin{array}{l} \begin{array}{l} x_{(i)} & x_{(j)} \end{array} \\ \begin{array}{l} x_{(j)} \\ x_{(p)} \end{array} (\begin{array}{l} 1 & 1 \\ 0 & 1 \end{array}) \end{array}$ \eqalign{& \matrix{{} \hfill & {{x_{(i)}}} \hfill & {{x_{(j)}}} \hfill \cr} \cr & \matrix{{{x_{(j)}}} \cr {{x_{(p)}}} \cr} \left({\matrix{1 & 1 \cr 0 & 1 \cr}} \right) \cr} . If there are elements with 1 below the column and r_pi = 1, r_ji = 1, r_ii = 1, then the number of 1s in the column i is more than the number of 1s in the column j; this is not true, then there must be 0 elements in the column, and the formal cyclic circle matrix is $\begin{array}{l} \begin{array}{l} x_{(i)} & x_{(j)} \end{array} \\ \begin{array}{l} x_{(k)} \\ x_{(j)} \end{array} (\begin{array}{l} 0 & 1 \\ 1 & 1 \end{array}) \end{array}$ \eqalign{& \matrix{{} \hfill & {{x_{(i)}}} \hfill & {{x_{(j)}}} \hfill \cr} \cr & \matrix{{{x_{(k)}}} \cr {{x_{(j)}}} \cr} \left({\matrix{0 & 1 \cr 1 & 1 \cr}} \right) \cr} . If there are no elements of 1 below column j of the element of r_jj, and all of them are 1, then column i has one more element r_ji = 1 than column j. Then there must be 0 elements in the column r_mi, …, r_ni, and the formal circular matrix is $\begin{array}{l} \begin{array}{l} x_{(i)} & x_{(j)} \end{array} \\ \begin{array}{l} x_{(l)} \\ x_{(j)} \end{array} (\begin{array}{l} 0 & 1 \\ 1 & 1 \end{array}) \end{array}$ \eqalign{& \matrix{{} \hfill & {{x_{(i)}}} \hfill & {{x_{(j)}}} \hfill \cr} \cr & \matrix{{{x_{(l)}}} \cr {{x_{(j)}}} \cr} \left({\matrix{0 & 1 \cr 1 & 1 \cr}} \right) \cr} .

It is discussed that there is an element of 1 in the lower semi-triangular matrix, and the corresponding element is 0 in the upper triangular matrix of the 0-1 permutation preference relation matrix. Since 1 of the lower triangular matrix and 0 of the corresponding upper semi-triangular matrix are the results of the same pair of scheme comparison, the 0 element in the upper triangular matrix is no longer discussed. Theorem 2 is proved.

From Theorem 2, we get a method to judge whether the linguistic judgement matrix has satisfactory consistency. At the same time, we also find out the circle composed of three objects. If we modify the circle formed, we can get the order of the objects. Now, we give the criteria of correcting the circulation circle:

If the row preference values of the three objects are not equal, the one with the smallest row preference value is regarded as the worst object among the three objects;

If the row preference values of two objects are equal and less than that of the other object, the object corresponding to the row with element 1 in the lower triangular matrix is recorded as the worst object;

If the row preference values of the three objects are equal, the sum of the subscripts of the linguistic phrases in the linguistic judgement matrix is compared, and the one with the smallest subscript sum is regarded as the worst object among the three objects; if the sum of subscripts is equal, according to the comparison results with the optimal object, the worst one is marked as the worst one.

After finding the worst object, if the cyclic circle is x_(i) ≻ x_(k) ≻ x_(j) and x_(j) is the worst scheme, the corresponding 0-1 permutation preference relation matrix with satisfactory consistency will be obtained by r_ji = 1 changing to r_ji = 0 in the 0-1 permutation preference relation matrix.

The specific steps of judgement and correction are given as follows:

Step 1: Give the preference relation matrix of linguistic judgement matrix;

Step 2: 0-1 permutation preference relation matrix of linguistic judgement matrix is given;

Step 3: Judge whether the 0-1 permutation preference relation matrix is an upper triangular matrix; if the linguistic judgement matrix has satisfactory consistency, the judgement ends, otherwise proceed to the next step;

Step 4: Find the cyclic matrix in the 0-1 permutation matrix, and get the objects of forming the cyclic matrix;

Step 5: According to the principle of modified cycle, the first cycle is modified to obtain the modified 0-1 permutation preference relation matrix;

Step 6: Rearrange the 0-1 permutation preference relation matrix and turn to step 3;

Step 7: According to the principle of determining the comparison results, the modified linguistic judgement matrix is obtained;

Step 8: Get the ranking of the objects;

Step 9: Judge the end.

Example 1

The decision-maker makes the following evaluation for the four printers X = {x₁, x₂, x₃, x₄}, and gives the following preference information: $\begin{array}{l} \begin{matrix} x_{1} & x_{2} & x_{3} & x_{4} \end{matrix} \\ P = \begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} (\begin{array}{l} AS & LD & MP & MD \\ LP & AS & VLD & MD \\ MD & VLP & AS & HD \\ MP & MP & HP & AS \end{array}) \end{array}$ \matrix{{\matrix{{\,\,\,\,\,\,\,\,\,\,{x_1}} & {{x_2}} & {{x_3}} & {{x_4}} \cr}} \cr {{\bf{P}} = \matrix{{{x_1}} \cr {{x_2}} \cr {{x_3}} \cr {{x_4}} \cr} \left({\matrix{{AS} & {LD} & {MP} & {MD} \cr {LP} & {AS} & {VLD} & {MD} \cr {MD} & {VLP} & {AS} & {HD} \cr {MP} & {MP} & {HP} & {AS} \cr}} \right)} \cr}

According to the above steps, judge or revise the satisfactory consistency of P.

The preference relation matrix obtained from Definition 3.1 is as follows: $Q = {(q_{ij})}_{n \times n} = \begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} \begin{array}{l} \begin{array}{l} x_{1} & x_{2} & x_{3} & x_{4} \end{array} \\ (\begin{array}{l} 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \end{array}) \end{array} .$ {\bf{Q}} = {({q_{ij}})_{n \times n}} = \matrix{{{x_1}} \cr {{x_2}} \cr {{x_3}} \cr {{x_4}} \cr} \matrix{{\matrix{{{x_1}} & {{x_2}} & {{x_3}} & {{x_4}} \cr}} \cr {\left({\matrix{1 & 0 & 1 & 0 \cr 1 & 1 & 0 & 0 \cr 0 & 1 & 1 & 0 \cr 1 & 1 & 1 & 1 \cr}} \right)} \cr}.

According to Definition 3.1, the 0-1 permutation preference relation matrix of P is obtained: $R = {(r_{ij})}_{n \times n} = \begin{array}{l} x_{4} \\ x_{1} \\ x_{2} \\ x_{3} \end{array} \begin{array}{l} \begin{array}{l} x_{4} & x_{1} & x_{2} & x_{3} \end{array} \\ (\begin{array}{l} 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{array}) \end{array} .$ {\bf{R}} = {({r_{ij}})_{n \times n}} = \matrix{{{x_4}} \cr {{x_1}} \cr {{x_2}} \cr {{x_3}} \cr} \matrix{{\matrix{{{x_4}} & {{x_1}} & {{x_2}} & {{x_3}} \cr}} \cr {\left({\matrix{1 & 1 & 1 & 1 \cr 0 & 1 & 0 & 1 \cr 0 & 1 & 1 & 0 \cr 0 & 0 & 1 & 1 \cr}} \right)} \cr}.

From the 0-1 permutation preference relation matrix, we can get r₃₂ = 1 and r₄₃ = 1. According to the definition of the cyclic circle matrix, we can get the cyclic circle matrix which contains r₃₂ = 1 and r₄₃ = 1 as $\begin{array}{l} \begin{array}{l} x_{1} & x_{2} \end{array} \\ \begin{array}{l} x_{2} \\ x_{3} \end{array} (\begin{array}{l} 1 & 1 \\ 0 & 1 \end{array}) \end{array}$ \eqalign{& \matrix{{} \hfill & {{x_1}} \hfill & {{x_2}} \hfill \cr} \cr & \matrix{{{x_2}} \cr {{x_3}} \cr} \left({\matrix{1 & 1 \cr 0 & 1 \cr}} \right) \cr} and $\begin{array}{l} \begin{array}{l} x_{1} & x_{3} \end{array} \\ \begin{array}{l} x_{1} \\ x_{2} \end{array} (\begin{array}{l} 1 & 1 \\ 1 & 0 \end{array}) \end{array}$ \eqalign{& \matrix{{} \hfill & {{x_1}} \hfill & {{x_3}} \hfill \cr} \cr & \matrix{{{x_1}} \cr {{x_2}} \cr} \left({\matrix{1 & 1 \cr 1 & 0 \cr}} \right) \cr} .

It can be seen that it is a circular circle matrix composed of x₁, x₂, x₃, and the row preference values of the three objects are equal. If the subscript sum of the three objects is the smallest, then object x₃ is regarded as the worst object. The order of the three objects is x₂ ≻ x₁ ≻ x₃. The modified 0-1 permutation preference relation matrix is: $R = {(r_{ij})}_{n \times n} = \begin{array}{l} x_{4} \\ x_{2} \\ x_{1} \\ x_{3} \end{array} \begin{array}{l} \begin{array}{l} x_{4} & x_{2} & x_{1} & x_{3} \end{array} \\ (\begin{array}{l} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{array}) \end{array} .$ {\bf{R}} = {({r_{ij}})_{n \times n}} = \matrix{{{x_4}} \cr {{x_2}} \cr {{x_1}} \cr {{x_3}} \cr} \matrix{{\matrix{{{x_4}} & {{x_2}} & {{x_1}} & {{x_3}} \cr}} \cr {\left({\matrix{1 & 1 & 1 & 1 \cr 0 & 1 & 1 & 1 \cr 0 & 0 & 1 & 1 \cr 0 & 0 & 0 & 1 \cr}} \right)} \cr}.

Only by x₂ ≺ x₃ changing to x₂ ≻ x₃, the corresponding intermediate language phrase MP in the language judgement matrix is selected as the comparison result of the comparison result. The modified linguistic judgement matrix is obtained as follows: $P = \begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} \begin{array}{l} \begin{array}{l} x_{1} & x_{2} & x_{3} & x_{4} \end{array} \\ (\begin{array}{l} AS & LD & MP & MD \\ LP & AS & VLP & MD \\ MD & MP & AS & HD \\ MP & MP & HP & AS \end{array}) \end{array} .$ {\bf{P}} = \matrix{{{x_1}} \cr {{x_2}} \cr {{x_3}} \cr {{x_4}} \cr} \matrix{{\matrix{{{x_1}} & {{x_2}} & {{x_3}} & {{x_4}} \cr}} \cr {\left({\matrix{{AS} & {LD} & {MP} & {MD} \cr {LP} & {AS} & {VLP} & {MD} \cr {MD} & {MP} & {AS} & {HD} \cr {MP} & {MP} & {HP} & {AS} \cr}} \right)} \cr}.

So far, the determination and correction of satisfactory consistency have been completed, and the arrangement order of the schemes is x₄ ≻ x₂ ≻ x₁ ≻ x₃.

Conclusions

In this paper, we give a method to judge whether the linguistic judgement matrix with the not equivalent object have satisfactory consistency. Then, for the linguistic judgement matrix with strict preference relation of pairwise objects, we give a method to determine the satisfactory consistency of the linguistic judgement matrix. We use whether the 0-1 permutation preference relation matrix is an upper triangular matrix to determine the satisfactory consistency of the linguistic judgement matrix. This method is simple and intuitive. It does not need to establish models and complex operations but only needs to make simple transformation of the matrix. Then, three objects of forming a cyclic circle are found by using the cyclic matrix, and the modified linguistic judgement matrix is obtained by giving the correction principle. However, the modified method is only applicable to the matrix with strict preference relationship, and there are some defects in the matrix with an equivalent scheme. Further research is needed.

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Applied Mathematics and Nonlinear Sciences

Satisfactory consistency judgement and inconsistency adjustment of linguistic judgement matrix

Fengxia Jin,Fengxia Jin,Zhi Zhou,Zhi Zhou,Yunxia Ma orazYunxia Ma orazYonggang ChenYonggang Chen

Data publikacji: 15 Apr 2022

Tom & Zeszyt: AHEAD OF PRINT

Zakres stron: -

Otrzymano: 09 Dec 2020

Przyjęty: 24 Jan 2021

Applied Mathematics and Nonlinear Sciences

Keywords

Fengxia Jin,
Fengxia Jin,
Zhi Zhou,
Zhi Zhou,
Yunxia Ma oraz
Yunxia Ma oraz
Yonggang Chen
Yonggang Chen