A direct consistency test and improvement method for the ......of perfect consistency and acceptable consistency. Saaty (1977) also developed an eigenvector method to derive priorities

Fuzzy Optimization and Decision Making (2020) 19:359–388https://doi.org/10.1007/s10700-020-09323-y

A direct consistency test and improvement method for theanalytic hierarchy process

Kang Xu1,2 · Jiuping Xu1

Published online: 7 April 2020© Springer Science+Business Media, LLC, part of Springer Nature 2020

AbstractThe consistency test is a vital component of pairwise comparison matrices if mean-ingful results are to be guaranteed, and it has been studied extensively since theanalytic hierarchy process was developed by Saaty. However, when using the exist-ing methods, it is imperative to carry out matrix operations, which are usuallynot intuitional. In this paper, a direct method, from the perspective of 3 tuples,is proposed to test and improve the consistency of a pairwise comparison matrix,which is far more intuitional and easier to understand. In the proposed 3 tuplesiterative method, only simple mathematical calculations are needed, without theneed for matrix operations. Furthermore, the calculation of Saaty’s consistencyratio is only conducted to verify the proposed method. Some related theorems andpropositions are proved mathematically or verified by random simulations. Theproposed method is applied to some published examples to verify its effective-ness and practicability. Finally, we provide an improved iterative method, whichis based on the 3 tuples iterative method. The improved iterative method andthe 3 tuples iterative method are closely linked and each has its own differentperspective. Moreover, they are applied on different occasions and reinforce oneanother.

Keywords Multiple criteria decision making · Analytical hierarchy process ·Pairwise comparison matrix · Consistency test · 3 tuples

B Jiuping [email protected]

Kang [email protected]

1 Business School, Sichuan University, Chengdu 610065, People’s Republic of China

2 School of Economics and Management, Hubei University of Automotive Technology, Shiyan442002, Hubei, People’s Republic of China

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360 K. Xu, J. Xu

1 Introduction

The analytical hierarchy process (AHP), developed by Saaty (1977, 1978, 1979,1980), is a powerful tool inmultiple criteria decisionmaking (MCDM)when resolvingproblems that combine qualitative and quantitative analysis. In recent years, the AHPhas become a highly important research topic (Goyal and Kaushal 2018; Krejčí et al.2017), and the scope of its application is very extensive.

The AHP is a powerful tool for dealing with multiple criteria decision problems inwhich the criteria (or objectives) are compared pairwise with respect to their impor-tance. Moreover, the judgements regarding the paired comparisons of objectives arecollected using pairwise comparison matrices.

Consistency is a basic requirement for pairwise comparison matrices if meaningfulresults are to be guaranteed, and checking the consistency of pairwise comparisonmatrices is a crucial step to avoid misleading solutions (Zhu et al. 2016). Since aninconsistency has a serious impact on the results of a priority vector, many studieshave focused on the consistency test, and how to improve it, for a number of decades(Chiclana et al. 2009; Ramík 2018; Saaty 1986, 1990; Wu et al. 2019; Yager andAlajlan 2015).

However, some of the existing methods are complicated and difficult to use whenrevising the inconsistent comparison matrix, whilst others make it difficult to preservemost of the original comparison information, as a new matrix has to be constructedto replace the original one (Ergu et al. 2011). Ergu et al’s method has overcomethe aforementioned problems. They proposed a method, which is simpler and moreefficient and accurate than the previous approaches, to improve the consistency ratioof the pairwise comparison matrix (Ergu et al. 2011). The latest research, such asZhang et al. (2018), proposed there were two paradoxes in Saaty’s consistency testin the pairwise comparison method using Saaty’s scale. They then investigated theconsistency by dividing an n × n pairwise comparison matrix into 3× 3 submatrices.

However, in Ergu et al’s method, it is still imperative to carry out matrix oper-ations, which are usually not intuitional. Furthermore, the existing methods cannot respond to certain problems directly. For example, assume a 3 × 3 positivepairwise comparison matrix A = (ai j )3×3; according to our current understand-ing, it can be found that a13 = a12a23 or a13/(a12a23) = 1 means thatA is of perfect consistency. Moreover, consider the following questions. Whatvalues should a13/(a12a23) take when it means that A is of acceptable consis-tency? Does it matter whether or not a13/(a12a23) = 0.99 or a13/(a12a23) =1.01 means that A is of acceptable consistency? What about a13/(a12a23) =0.9 or a13/(a12a23) = 1.1? Or even a13/(a12a23) = 0.5 or a13/(a12a23) =2?

The objective of this paper is to provide a direct consistency test and improvementmethod for the pairwise comparison matrix in AHP, and subsequently reply to thequestions above directly. As it does not carry out matrix operations, the proposedmethod is only for testing and improving the consistency according to the 3 tuples(aik, akj , ai j ) in the pairwise comparison matrix.

The contributions of this paper are as follows. First, a direct method, from theperspective of 3 tuples (aik, akj , ai j ), rather than carrying out matrix operations,

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A direct consistency test and improvement method for the… 361

is provided to test the consistency of a pairwise comparison matrix. We divide ann × n pairwise comparison matrix into 3 tuples (aik, akj , ai j ). Moreover, the inter-nal quantitative relationships between the consistency ratio of the comparison matrixand the 3 tuples are ascertained. Second, this paper presents a simpler algorithm forimproving the consistency of a pairwise comparison matrix. The proposed methodis far more intuitional and easy to understand, only simple mathematical calcula-tions are needed, and there are no matrix operations. Furthermore, the calculationof Saaty’s consistency ratio is only conducted to verify the proposed method. Third,the proposed method can also retain most of the information provided by the orig-inal comparison matrix, indicate the modification direction and provide the optimalvalues; using the proposed method, the inconsistent elements in the matrix can befound accurately and rapidly. Finally, we provide an improved iterative method,which is based on the 3 tuples iterative method. The improved iterative methodand the 3 tuples iterative method are closely linked and each has its own differ-ent perspective. Moreover, they are applied on different occasions and reinforce oneanother.

The rest of this paper is organized as follows. In Sect. 2, some concepts and prop-erties associated with the topic are briefly reviewed. In Sect. 3, a direct acceptableconsistency test is proposed, sufficient and approximately equivalent conditions foran acceptable consistency are developed and some valuable theorems and propo-sitions are proved mathematically or verified by random simulations. In Sect. 4,a 3 tuples iterative method for improving the consistency process is proposed. Inorder to compare the proposed method with those suggested by other researchers,it is applied to some published examples, and the results are compared. In Sect. 5,an improved iterative method, which is based on the 3 tuples iterative method,and is convenient for preference relations with high orders. Some examples arecalculated by employing the improved iterative method, and an actual case studyis used in order to demonstrate the effectiveness of the proposed approach. Inthe final section, conclusions are drawn and future research directions are dis-cussed.

2 Preliminaries

In this section, some concepts and properties associated with the topic are brieflyreviewed.

2.1 The 1–9 ratio scale and pairwise comparisonmatrix

Saaty (1980) proposed a 1–9 ratio scale (shown in Table 1) as a basis for decisionmakers (DMs) to provide judgements over the paired comparisons of objectives.Then the judgements are collected by multiplicative preference relations, where reci-procity is required, which are called pairwise comparison matrices, or simply calledcomparison matrices. A pairwise comparison matrix A = (ai j )n×n can be shownas:

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362 K. Xu, J. Xu

Table 1 The 1–9 ratio scale

Scale Meaning

1 Equally preferred

3 Moderately preferred

5 Strongly preferred

7 Very strongly preferred

9 Extremely preferred

Other values between 1 and 9 Intermediate values used to represent compromise

A =

⎡⎢⎢⎢⎢⎢⎢⎢⎣

1 a12 a13 · · · a1n1 a23 · · · a2n... 1

...... 1/ai j

.... . .

...... 1

⎤⎥⎥⎥⎥⎥⎥⎥⎦

where A = (ai j )n×n (n ≥ 2), ai j a ji = 1, ai j ∈{ 19 ,

18 ,

17 , . . . ,

12 , 1, 2, . . . 7, 8, 9

}, i,

j = 1, 2, . . . , n.

2.2 The consistency of a pairwise comparisonmatrix

Saaty (1980) presented a consistency ratio in theAHP, and also developed the conceptsof perfect consistency and acceptable consistency. Saaty (1977) also developed aneigenvector method to derive priorities from comparison matrices, and then defined aconsistency index to measure their consistency degrees.

For a set of objectives X = {x1, x2, . . . , xn} and a constructed comparison matrixA = (ai j )n×n , the eigenvector method is based on solving the following equation:

Aω = λmaxω,n∑

i=1ωi = 1 (1)

where λmax is the maximum eigenvalue of A, and ω is the priority vector of theobjectives.

Let

C I = λmax − nn − 1 (2)

then the consistency index is defined as

CR = C IRI

(3)

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Table 2 The random index (RI )by Saaty (2001) n 1 2 3 4 5 6 7 8 9 10

RI 0 0 0.52 0.89 1.11 1.25 1.35 1.40 1.45 1.49

where RI is a random consistency index, which is the average value of C I obtainedfrom 10,000 random pairwise comparison matrices, whose entries were randomlygenerated using the 1 to 9 scale. Saaty considers that a value of CR under 0.10indicates the pairwise comparison matrix is of acceptable consistency. Table 2 givesvalues of RI for different matrix orders (n). Some cases require the consistency ratioto be less than 5% for n = 3 and less than 8% for n = 4 (Saaty 2001).Definition 1 Saaty (1980) A reciprocal judgement matrix A = (ai j )n×n is of perfectconsistency if for any positive integers i, j, k (≤ n), the following result holds:

ai j = aikak j (4)

3 A direct acceptable consistency test

To check the consistency of pairwise comparison matrices using the existing meth-ods, it is imperative to carry out matrix operations, which are usually not intuitional.Moreover, as outlined in the introduction, the existing methods cannot resolve certainproblems directly. A direct acceptable consistency test method is provided to test theconsistency of a pairwise comparison matrix in this paper. 3× 3 pairwise comparisonmatrices are investigated first.

3.1 A direct acceptable consistency test for 3× 3 pairwise comparisonmatrices

In this section a correspondence between the consistency ratio CR and the ratioa13/(a12a23) is proposed.

Theorem 1 For a 3 × 3 pairwise comparison matrix A = (ai j )3×3 (a ji =1/ai j , i, j = 1, 2, 3), ai j ∈

{ 19 ,

18 ,

17 , . . . ,

12 , 1, 2, . . . 7, 8, 9

}, there is a correspon-

dence between Saaty’s consistency ratio and the ratio a13/(a12a23).

(1) When the threshold value of the consistency ratio is 0.05,

CR < 0.05 ⇔ 0.505 < a13a12a23

< 1.979

where ‘⇔’ means being equivalent, namely, C R < 0.05 is equivalent to 0.505 <a13/(a12a23) < 1.979.

(2) When the threshold value of the consistency ratio is 0.1,

C R < 0.1 ⇔ 0.382 < a13a12a23

< 2.620

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364 K. Xu, J. Xu

where ‘⇔’ means being equivalent, namely, C R < 0.1 is equivalent to 0.382 <a13/(a12a23) < 2.620.

Hereby 0.505, 1.979, 0.382, 2.620 all are approximate values, which are numbers tothree decimal places.

Proof Since ai j ∈{ 19 ,

18 ,

17 , . . . ,

12 , 1, 2, . . . 7, 8, 9

}, there are only 17 possible values

from 19 to 9 for a12, a23 or a13. By using the multiplication principle of combinatorialmathematics, there are 17× 17× 17 = 4, 913 different combinations of a12, a23, a13.Furthermore, due to a ji = 1/ai j (i, j = 1, 2, 3), there are only 4913 combinations ofai j . Therefore, Theorem 1 can be proved by enumerating all 4913 cases one by one.It is completed by using MATLAB (The MATLAB codes are listed in “Appendix 1”).

The proof of Theorem 1 can be included with that of Theorem 2. In other words, ifTheorem 2 holds, then Theorem 1 also holds.

Theorem 2 When ai j (a ji = 1/ai j , i, j = 1, 2, 3) is continuous, i.e., for arbitraryai j ∈ [1/9, 9], the conclusions in Theorem 1 also hold.Proof Suppose that a pairwise comparison matrix

Ȧ =⎡⎣

1 a c1/a 1 b1/c 1/b 1

⎤⎦

where a, b, c ∈ [1/9, 9] andc

ab= k (5)

for brevity, hereby we let a12 = a, a23 = b, a13 = c, then we calculate the character-istic polynomial of Ȧ as follows:

∣∣λE − Ȧ∣∣ =∣∣∣∣∣∣

λ − 1 −a −c−(1/a) λ − 1 −b−(1/c) −(1/b) λ − 1

∣∣∣∣∣∣

= (λ − 1)3 − 3(λ − 1) −(ab

c+ c

ab

)

where E denotes a 3× 3 identity matrix and λ denotes the eigenvalues; i.e., the rootsof the characteristic equation

∣∣λE − Ȧ∣∣ = 0, namely,

(λ − 1)3 − 3(λ − 1) −(ab

c+ c

ab

)= 0 (6)

According to Eq. (5), Eq. (6) is equivalent to

(λ − 1)3 − 3(λ − 1) −(1

k+ k

)= 0 (7)

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Fig. 1 The correspondence between CR and k

Fig. 2 h(x) = x3 − 3x − t

Let

x = λ − 1, t = 1k

+ k

then Eq. (6) is equivalent to

x3 − 3x − t = 0 (8)

According to Eq. (7), it can be found that the roots of Eq. (7), or equivalent Eqs.(6) and (8) are only dependent on the parameter k, and thus the consistency ratio CRis only dependent on the parameter k. Figure 1 shows a corresponding line chart ofCR and k. The data for Fig. 1 is omitted for the purposes of brevity.

In order to find the roots of Eq. (8), suppose that

h(x) = x3 − 3x − t

where t is a parameter.

As shown in Fig. 2, when t increases, the largest root of h(x) = 0 also increases.h is continuous and differentiable everywhere, the derivative of h is

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366 K. Xu, J. Xu

dh

dx= 3x2 − 3 = 3(x2 − 1)

When x > 1, dhdx > 0, and since it is continuous everywhere, h is strictly increasingon the interval [1,+∞) and due to

h(1) = −2 − t < 0(t = k + 1

k≥ 2

)

limx→+∞ h(x) = +∞

therefore, function h has just one zero point on the interval [1,+∞), i.e., Eq. (8) hasjust one root on the interval [1,+∞); accordingly the root on the interval [1,+∞)is the largest root of Eq. (8), so in the following discussion we only consider theinterval [1,+∞). Suppose the largest root of Eq. (8) is xmax, since we have supposedthat x = λ − 1, the corresponding λ, namely, λmax = xmax + 1, is also the largesteigenvalue according to Eq. (6) or (7). According to Eq. (8), we have

t = x3 − 3x

It is obvious that t = x3−3x is also strictly increasing on the interval [1,+∞). Noticethat we now only consider the interval [1,+∞); according to Eqs. (2) and (3) andTable 2, we have

CR < 0.1 ⇔ C IRI

< 0.1

⇔ λmax − n0.52(n − 1) =

λmax − 30.52 × 2 < 0.1

⇔ (xmax + 1) − 30.52 × 2 < 0.1

⇔ xmax < 2.104⇔ t = x3 − 3x < 2.1043 − 3 × 2.104⇔ k + 1

k< 2.1043 − 3 × 2.104

(t = k + 1

k

)

⇔ 0.382 < k < 2.620

where ‘⇔’ means being equivalent. t = k+ 1k is decreasing on the interval (0, 1], andincreasing on the interval [1,+∞), solving the equation

k + 1k

= 2.1043 − 3 × 2.104 (9)

the approximations of the two roots are obtained:

k1 = 0.382,

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Table 3 Values of CR (thethreshold value is 0.05) fordifferent k

k < 0.505 0.505 (0.505, 1.979) 1.979 > 1.979

CR > 0.05 0.05 < 0.05 0.05 > 0.05

Table 4 Values of CR (thethreshold value is 0.1) fordifferent k

k < 0.382 0.382 (0.382,2.620) 2.620 > 2.620

CR > 0.1 0.1 < 0.1 0.1 > 0.1

k2 = 2.620

which are numbers to three decimal places. The domain of k is k1 < k < k2. Thiscompletes the proof of case (2); the proof of the other case is similar.

Notice that in this paper 0.505, 1.979, 0.382, 2.620 are all approximate values,which are numbers to three decimal places. If it is necessary, more accurate approx-imations or even the exact values can be found by solving Eq. (9), but are omittedhere for brevity; this is the same for other parts of the paper. Using the exact roots tosimulate, from 1,000,000 random pairwise comparison matrices, Theorem 2 is foundto be satisfied. The exact roots of Eq. (9) and the Matlab codes are listed in “Appendix2”. The random simulation is also further proof of Theorem 2.

As shown in Tables 3 and 4, if the threshold value of CR is 0.05, then k ∈(0.505, 1.979) means that Ȧ is of acceptable consistency; if the threshold value ofCR is 0.1, then k ∈ (0.382, 2.620) means that Ȧ is of acceptable consistency.Example 1 If a12 = 3, a23 = 4, take 0.1 as the threshold value of the consistency ratio,then

0.382 × 3 × 4 < a13 < 2.620 × 3 × 4

where a13 ∈ [1/9, 9], namely,

4.58 < a13 ≤ 9

means that A is of acceptable consistency. If ai j ∈{ 19 ,

18 , . . . ,

12 , 1, 2, . . . , 8, 9

}is

needed, then a13 = 5, 6, . . . , 9.

3.2 A direct consistency test method for higher order pairwise comparisonmatrices

In this section, we investigate the approximate condition for an acceptable consistencyin the environment of a 4 × 4 and higher order pairwise comparison matrix.Proposition 1 Suppose a pairwise comparison matrix A,

A = (ai j )n×n (n ≥ 3), ai j a ji = 1, ai j ∈{1

9,1

8,1

7, . . . ,

1

2, 1, 2, . . . 7, 8, 9

}

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368 K. Xu, J. Xu

θki j =ai j

aikak j(1 ≤ i < k < j ≤ n),

considering that if θki j > 1 (i = 1, 2, 3, 4), there is an upward deviation from aperfect consistency, while if θki j < 1, there is a downward deviation. In order to avoida situation that the upward and downward deviations cancel each other out, supposethat the adjusted θki j , i.e.,

†θki j ={1/θki j , i f θ

ki j < 1;

θki j , else

then we have

6

n(n − 1)(n − 2)n−2∑i=1

n∑j=i+2

j−1∑k=i+1

†θki j < 2.620approximately⇔ CR(A) < 0.1 (10)

whereapproximately⇔ means that in formula (10), the former is approximately equivalent

to the latter with a very small percentage error. 2.620 is an approximate value, whichis a number to three decimal places; the total number of all †θki j is C

3n = n(n−1)(n−2)6

and 6n(n−1)(n−2)∑n−2

i=1∑n

j=i+2∑ j−1

k=i+1 †θki j is the average of all †θ

ki j .

Proof Proposition 1 involves some approximately equivalent conditions, which can beverified by random simulations. The MATLAB codes for 4× 4 and 5× 5 comparisonmatrices are listed in “Appendix 3”, the rest are similar in nature.

Proposition 2 When ai j ∈[ 19 , 9

], i, j = 1, 2, . . . , n, i.e., when ai j is continuous, the

conclusions in Proposition 1 including formula (10) also hold.

Proof Similar to the verification of Theorem 2; the MATLAB codes are omitted.

According to Eq. (9), the two roots k1, k2 are the inverse of each other, namely,0.382 and 2, 620 are almost the inverse of each other, hence †θki j < 2.620 is equivalent

to 0.382 < θki j < 2.620. Thus a simple and direct consistency test method can beobtained:

Definition 2 Let A be as before, and

6

n(n − 1)(n − 2)n−2∑i=1

n∑j=i+2

j−1∑k=i+1

†θki j < 2.620

then A is considered to be of an acceptable consistency.

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4 A 3 tuples iterativemethod for improving the consistency process

In this section,we investigate the consistency improvement process.A3 tuples iterativemethod is proposed for improving the consistency process.

4.1 A 3 tuples iterative method

According to the aforementioned theory, we propose a 3 tuples iterative method toadjust a pairwise comparison matrix so that it is of acceptable consistency.

For convenience, (aik, akj , ai j ) (1 ≤ i < k < j ≤ n) is called a 3 tuple. Further,if

0.382 <ai j

aikak j< 2.620

then (aik, akj , ai j ) is called a 3 tuple satisfying an acceptable consistency, otherwiseit is called an inconsistent 3 tuple.

Using the 3 tuples iterative method to improve the consistency of the comparisonmatrix A, the general steps are as follows.

Step 1 Calculating. As mentioned in Proposition 1, calculate the ratios

θki j =ai j

aikak j(1 ≤ i < k < j ≤ n),


ki j < 1;

θki j , else

When calculating Saaty’s consistency ratio, if CR(A) < 0.1 is satisfied, then thematrix A is of an acceptable consistency. This completes the calculation. Otherwise, ifCR(A) < 0.1 is not satisfied, or 6n(n−1)(n−2)

∑n−2i=1

∑nj=i+2

∑ j−1k=i+1 †θ

ki j >> 2.620

is satisfied (by Definition 2), then the matrix A is not of an acceptable consistency;therefore go to Step 2.

Step 2 Identifying. Find all the inconsistent 3 tuples, which satisfy †θki j ≥ 2.620.Particularly for some fairly large values in †θki j , their corresponding 3 tuples are seri-ously inconsistent, and should be adjusted first.

Step 3 Adjusting. Adjust the elements in the inconsistent 3 tuples. Then go to Step1.When the elements in the inconsistent 3 tuples are adjusted, the following principlesshould be obeyed as far as possible.

Principle 1Adjust as few elements in the comparisonmatrix as possible. In thiswaythe original information from decision makers can be preserved as much as possible.

Principle 2 If an element appears in a tuplewith themaximumof all ratios †θki j (1 ≤i < k < j ≤ n), and this element simultaneously appears in several other seriouslyinconsistent 3 tuples, then this element should be adjusted first.

If there are 3 tuple judgments with more than one common element that are out ofthe acceptable consistency scope, then one must adjust the element with the maximumdeviation (ratios) and the highest frequencyof occurrence in these inconsistent 3 tuples.

Step 4 Finish.

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370 K. Xu, J. Xu

Note 1 According to the reciprocity (ai j a ji = 1), in the pairwise comparisonmatrix A, only the elements in the upper triangulation matrix of A are consideredby the authors. Similar processing can be found in Petra and Lidija (2012). In theconsistency improvement process, when we adjust one element ai j (i < j) in theupper triangulation matrix of A without an explanation, the reciprocity is alwayssatisfied; i.e., the element a ji is adjusted simultaneously to let ai j a ji = 1 still hold.This applies elsewhere in the paper.

In some special cases, if we adjust individual elements in the submatrix of a com-parison matrix A, then the submatrix has a perfect consistency; at this moment, it isnecessary to firstly adjust the elements in the submatrix of the comparison matrix A.

Note 2 Due to the convergence of the monotone bounded series of numbers, theiterative algorithm above is obviously convergent.

4.2 Some illustrative examples and comparisons

In order to compare our method with those of others, the proposed method in thispaper is applied to some published examples.

Example 2 The 4 × 4 pairwise comparison matrix A is inconsistent with CR =1.0242 > 0.1.

A =

⎡⎢⎢⎣1 2 4 1/8

1 2 41 2

1

⎤⎥⎥⎦

Step 1 Calculating the ratios,

θ213 =a13

a12a23= 4

2 × 2 = 1,

θ324 =a24

a23a34= 4

2 × 2 = 1,

θ214 =a14

a12a24=

18

2 × 4 =1

64, †θ214 =

1

θ214= 64 > 2.620,

θ314 =a14

a13a34=

18

4 × 2 =1

64, †θ314 =

1

θ314= 64 > 2.620,

Step 2 Therefore the 3 tuples (a12, a23, a13) and (a23, a34, a24) are of perfect consis-tency, while the 3 tuples (a12, a24, a14) and (a13, a34, a14) are seriously inconsistent.

†θ213 + †θ324 + †θ214 + †θ3144

= 1 + 1 + 64 + 644

= 32.5 � 2.620

means that the comparison matrix A is seriously inconsistent.

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Step 3 Adjust the 3 tuples (a12, a24, a14) and (a13, a34, a14), out of which a14 is acommon element. Therefore, a14 should be adjusted first. Let

a14 = a12a24 = 2 × 4 = 8,

or

a14 = a13a34 = 4 × 2 = 8

the adjusted comparison matrix A is obtained.

Aad justed =

⎡⎢⎢⎣1 2 4 8

1 2 41 2

1

⎤⎥⎥⎦

CR(Aad justed) = 0,

which completes Example 2.

Notice that in Example 2, when we let 0.382a12a24 < a14 < 2.620a12a24 or0.382a13a34 < a14 < 2.620a13a34, i.e., 3.056 < a14 < 20.96, and due to 19 ≤ a14 ≤9, we have a14 ∈ (3.056, 9], or even a14 varies in a larger domain, A is still of anacceptable consistency. By using formula (10), we can get an approximate interval(1.887, 9] for a14. The calculation process is as follows.

†θ213 + †θ324 + †θ214 + †θ3144

< 2.620

(1) When a14 < 8,

θ214 =a14

a12a24= a14

8< 1, θ314 =

a14a13a34

= a148

< 1,

†θ214 =1

θ214= 8

a14, †θ314 =

1

θ314= 8

a14,

†θ213 = θ213 = 1, †θ324 = θ324 = 1,†θ213 + †θ324 + †θ214 + †θ314

4= 1 + 1 +

8a14

+ 8a144

< 2.620,

a14 >16

8.48≈ 1.887,

thus 1.887 < a14 ≤ 8 is obtained.(2) When a14 ≥ 8,

θ214 =a14

a12a24= a14

8≥ 1, θ314 =

a14a13a34

= a148

≥ 1,†θ213 + †θ324 + †θ214 + †θ314

4= 1 + 1 +

a148 + a148

4< 2.620,

then a14 < 33.92 is obtained.

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372 K. Xu, J. Xu

Since a14 ∈{ 19 ,

18 ,

17 , . . . ,

12 , 1, 2, . . . 7, 8, 9

}, finally the domain making the com-

parison matrix A be of an acceptable consistency is

a14 = 2, 3, 4, . . . , 9(when a14 = 1, CR(A) = 0.2184; when a14 = 2, CR(A) = 0.0933; . . . ; whena14 = 9, CR(A) ≈ 0).

Example 2 has appeared in Ergu et al. (2011); compared with their methods, theresults are the same, but the proposed method is much more intuitional and easyto understand, as only simple mathematical calculations are needed, without matrixoperations. The calculation of the consistency ratio is only conducted to verify theproposed method.

In the proposed method, the identification process of the inconsistent elements isintuitional, rapid and efficient. The elements in the comparison matrix are adjusted aslittle as possible, that is only a pair of elements are adjusted, and after that the pairwisecomparison matrix A is of an acceptable consistency, or even of perfect consistency.

In the proposed method, the elements needing to be adjusted are identified accu-rately, and furthermore the specific domain of the adjusted element is also ascertained.

Example 3 The 8 order pairwise comparison matrix A was first introduced inKwiesielewicz and Uden (2002) as an example, then Ergu et al. (2011) also usedthis matrix as an example, and compared their method with that of Kwiesielewicz andUden (2002). Now we will use this matrix as an example to facilitate a comparativeanalysis.

A =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 2 1/2 2 1/2 2 1/2 21 4 1 1/4 1 1/4 1

1 4 1 4 1 41 1/4 1 1/4 1

1 4 1 41 1/4 1

1 41

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

CR(A) = 0.1055Step 1Calculating the ratios. This is a rather special kind ofmatrix, so first we considerthe 4 × 4 submatrix of A,

Asub4×4 =

⎡⎢⎢⎣1 2 1/2 2

1 4 11 4

1

⎤⎥⎥⎦ , CR(Asub4×4) = 0.4071 > 0.1,

θ213 =a13

a12a23=

12

2 × 4 =1

8, †θ213 =

1

θ213= 8 > 2.620,

θ324 =a24

a23a34= 1

4 × 4 =1

16, †θ324 =

1

θ324= 16 > 2.620,

123


θ214 =a14

a12a24= 2

2 × 1 = 1, †θ214 = θ214 = 1 < 2.620,

θ314 =a14

a13a34= 21

2 × 4= 1, †θ314 = θ314 = 1 < 2.620

Step2Thus the3 tuples (a12, a24, a14), (a13, a34, a14) are of an acceptable consistency,or even of a perfect consistency, while the 3 tuples (a12, a23, a13), (a23, a34, a24) areseriously inconsistent; out of which a23 is a common element. Thus a23 should beadjusted first.Step 3 Let

a23 = a13a12

=12

2= 1

4,

or

a23 = a24a34

= 14

= 14,

then

Asub-ad justed4×4 =

⎡⎢⎢⎣1 2 1/2 2

1 1/4 11 4

1

⎤⎥⎥⎦ , CR(A

sub-ad justed4×4 ) = 0 < 0.1

Meanwhile,

Aad justed =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 2 1/2 2 1/2 2 1/2 21 1/4 1 1/4 1 1/4 1

1 4 1 4 1 41 1/4 1 1/4 1

1 4 1 41 1/4 1

1 41

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

, CR(Aad justed) = 0 < 0.1

Example 4 The following pairwise comparison matrix is obtained by adding one rowand one column with a random value to the comparison matrix in Example 3, and itis also from Ergu et al. (2011).

123

374 K. Xu, J. Xu

A =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 2 1/2 2 1/2 2 1/2 2 1/31 4 1 1/4 1 1/4 1 1/4

1 4 1 4 1 4 1/71 1/4 1 1/4 1 1/6

1 4 1 4 61 1/4 1 1/3

1 4 71 1/2

1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

As with Example 3, adjust a23 from 4 to 14 , then the submatrix from the first 8 rowsand the first 8 columns in the matrix A, i.e.,

Asub-ad justed8×8 =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 2 1/2 2 1/2 2 1/2 21 1/4 1 1/4 1 1/4 1

1 4 1 4 1 41 1/4 1 1/4 1

1 4 1 41 1/4 1

1 41

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

is of perfect consistency, CR(Asub-ad justed8×8 ) = 0 < 0.1.CR(Aad justed) = 0.1335 > 0.1, at this moment, it is no longer necessary to adjust

the elements in the submatrix Asub-ad justed8×8 , only the elements from the ninth row andthe ninth column in Aad justed need to be adjusted.Step 1 Calculating the ratios. Since Asub-ad justed8×8 has been of perfect consistency, itis not necessary to consider the 3 tuples in Asub-ad justed8×8 . The rest of the 3 tuples arecalculated as follows. Using the formulas

θki j =ai j

aikak j(1 ≤ i < k < j ≤ n),


ki j < 1;

θki j , else

the calculation results are listed in Table 5.Step 2After observation, it can be found that out of all †θki j , †θ

739 = 49 is themaximum,

and †θ539 = 42, †θ839 = 14, †θ719 = 10.5, †θ749 = 10.5, †θ639 = 9.34, †θ519 = 9,†θ549 = 9 are also quite large ratios; therefore the corresponding 3 tuples need to beadjusted, where the element a39 appears with the highest frequency. Thus a39 shouldbe adjusted first.

123


Table 5 Values of †θki j

i, k, j 129 139 149 159 169 179 189 239 249 259

†θki j 1.5 4.67 1 9 2 10.5 3 7 1.5 6

i, k, j 269 279 289 349 359 369 379 389 459 469

†θki j 1.33 7 2 4.67 42 9.34 49 14 9 2

i, k, j 479 489 569 579 589 679 689 789

†θki j 10.5 3 4.5 1.17 3 5.25 1.5 3.5

Table 6 Adjusted a39 and the corresponding consistency ratio

adjusted a39 ... 1/5 1/4 1/3 1/2 1 2

CR(Aad justed ) ... 0.1103 0.0980 0.0852 0.0721 0.0597 0.0566

adjusted a39 3 4 5 6 7 8 9

CR(Aad justed ) 0.0586 0.0618 0.0654 0.0654 0.0732 0.0771 0.0810

Step 3 Since the adjusted element a39, i.e.,

aad justed39 ∈{1

9,1

8, . . . ,

1

2, 1, 2, . . . , 8, 9

}

which can be enumerated one by one; the calculation results are listed in Table 6.Out of which, a39 = 2 (a93 = 12 ) can maximize the consistent degree of A, at this

moment,

Aad justed =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 2 1/2 2 1/2 2 1/2 2 1/31 1/4 1 1/4 1 1/4 1 1/4

1 4 1 4 1 4 21 1/4 1 1/4 1 1/6

1 4 1 4 61 1/4 1 1/3

1 4 71 1/2

1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

CR(Aad justed) = 0.0566 < 0.1

The adjusted matrix A has been of an acceptable consistency, and in the whole pro-cess only two pairs of elements in the matrix A are adjusted; therefore compared tothe published methods, the calculation process of the proposed method is far moreintuitional and simple, without the need for matrix operations.

Example 5 The following pairwise comparison matrix was first introduced in Xu andWei (1999) as an example, then was also used by Saaty (2003) as an example todescribe the method embedded in the Expert Choice Software, which was used to

123

376 K. Xu, J. Xu

Table 7 Values of †θki j

i, k, j 128 138 148 158 168 178 127 137 147 157

†θki j 2.86 2.4 3.5 4 4 1.5 3 54 3 3.6

i, k, j 167 126 136 146 156 125 135 145 124 134

†θki j 3.6 2.5 2 3.43 2 2.5 1.5 2.57 3.57 6

i, k, j 123 238 248 258 268 278 237 247 257 267

†θki j 1.8 2.14 4.37 3.5 3.5 1.43 10 3.57 3 3

i, k, j 236 246 256 235 245 234 348 358 368 378

†θki j 2.25 2.4 2 3 1.8 6 3.75 2.5 3.33 15

i, k, j 347 357 367 346 356 345 458 468 478 457

†θki j 7 10 7.50 2.67 2.67 1.5 2.25 3 1.75 2.14

i, k, j 467 456 568 578 567 678

†θki j 2.86 1.5 2 1.67 2 1.67

detect the inconsistencies, and also by Cao et al. (2008) as an inconsistent pairwisecomparison matrix to test their proposed heuristic method (Ergu et al. 2011). Later,Ergu et al. (2011) also used this matrix to illustrate their method. Now we will alsoapply this matrix to test the proposed method.

A =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 5 3 7 6 6 1/3 1/41 1/3 5 3 3 1/5 1/7

1 6 3 4 6 1/51 1/3 1/4 1/7 1/8

1 1/2 1/5 1/61 1/5 1/6

1 1/21

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

CR(A) = 0.1703 > 0.1

Step 1 Calculating the ratios. Using the formulas

θki j =ai j

aikak j(1 ≤ i < k < j ≤ n),


ki j < 1;

θki j , else

the calculation results are listed in Table 7.The average of all †θki j is 4.32 > 2.620, by Proposition 3, the comparison matrix

is inconsistent (CR(A) = 0.1703 > 0.1). This is also a verification of Proposition 3.Step 2After observation, it can be found that out of all †θki j , †θ

317 = 54 is themaximum,

and †θ738 = 15, †θ327 = 10, †θ537 = 10, †θ637 = 7.5, †θ437 = 7 are also quite large ratios,therefore the corresponding 3 tuples

123


Table 8 Adjusted a37 and the corresponding consistency ratio

adjusted a37 1/9 1/8 1/7 1/6 1/5 1/4

CR(Aad justed ) 0.1044 0.1007 0.0969 0.0933 0.0897 0.0864

adjusted a37 1/3 1/2 1 2 3 ...

CR(Aad justed ) 0.0838 0.0828 0.0886 0.1058 0.1231 ...

(a13, a37, a17), (a23, a37, a27), (a37, a78, a38), (a35, a57, a37), (a36, a67, a37),

(a34, a47, a37),

are seriously inconsistent; out of which a37 is a common element, so a37 should beadjusted first.Step 3 Adjust the element a37. Since the adjusted element a37, i.e.,

aad justed37 ∈{1

9,1

8, . . . ,

1

2, 1, 2, . . . , 8, 9

}

which can be enumerated one by one; the calculation results are listed in Table 8.Therefore, a37 = 17 , 16 , . . . , 12 , 1 can make A be of an acceptable consistency, out ofwhich a37 = 12 (a73 = 2) can maximize the consistent degree of A, at this moment,CR(Aad justed) = 0.0828.

In order to facilitate a comparative analysis, we used several of the same examplesas in Ergu et al. (2011). Ergu et al. compared their method with some other meth-ods and their comparison is quite sufficient. They pointed out that in Xu and Wei’smethod (1999), many elements in the original comparison matrix have been changed.Based on the original inconsistent matrix, a consistent matrix is generated by an auto-adaptive process instead of revising single elements. A similar problem occurs inCao et al. (2008). Compared with the other two methods, Saaty’s method is easier touse because it is based on the ratio of priorities and designed for the Perron Eigen-value Method (EM) (1977) and AHP. However, the ‘precise’ number recommendedby Saaty’s method is really an approximated value. Ergu et al. (2011) concluded thatall aforementioned three methods are based on the priority vector ratios, which arecalculated by the inconsistent comparison matrix. Apart from the EM, different meth-ods, have been proposed to derive a priority vector with a given pairwise comparisonmatrix, including the methods in Barzilai (1997), Saaty (1980), Chu et al. (1979),Crawford and Williams (1985), Bryson (1995).

Ergu et al. (2011) also argued that in their method, all values provided by experts inthe original comparison matrix have been retained, except the inconsistent elements,a37 and a73. Furthermore, Ergu et al’s method does not violate the scale [1, 9], needsfewer computations than the methods of Xu andWei and Cao et al., and also preservesmore original comparison information than these two methods. Ergu et al’s methodcan also show the modification direction and provide the optimal values. All of theaforementioned advantages are also the advantages of ourmethod.The computations in

123

378 K. Xu, J. Xu

our method are simple, intuitional and easy to understand. Only simple mathematicalcalculations are needed, without matrix operations. Furthermore, the calculation ofSaaty’s consistency ratio is only conducted to verify the proposedmethod. Ourmethodcan also show the modification direction and provide the optimal values; moreover,most of the information provided by the original comparison matrix can be retainedin our method.

In the current method, it is still imperative to carry out matrix operations, whichare usually not intuitional. The proposed method in this paper has preserved the afore-mentioned advantages, while overcoming the disadvantages mentioned above.

The latest literature, such as Zhang et al. (2018) has some similar ideas to ourmethod, as they proposed two paradoxes regarding Saaty’s consistency test in thepairwise comparison method with Saaty’s scale. They also investigated the consis-tency by dividing an n × n pairwise comparison matrix into 3× 3 submatrices, whichis similar to our method. However, they investigated the consistency test in the pair-wise comparison method with a fixed numerical scale; thus there is a large differencebetween their research direction and ours. Furthermore, they divided an n×n pairwisecomparison matrix into 3 × 3 submatrices, while we divide an n × n pairwise com-parison matrix into 3 tuples (aik, akj , ai j ). Moreover, we have ascertained the internalquantitative relationships between the consistency ratio of the comparison matrix andthe 3 tuples, which are intuitional and easy to understand.

5 An improved iterativemethod for preference relations with highorders

The proposed 3 tuples method is intuitional, easy to understand, and more suitablefor manual calculations, especially for preference relations with low orders and somespecific preference relations with high orders. In this section, we provide an improvediterative method, which is based on the 3 tuples iterative method, and is convenientfor preference relations with high orders.

5.1 An improved iterative method to increase the computational efficiency forpreference relations with high orders

We have the following proposition

Proposition 3 Suppose the comparison matrix A = (ai j )n×n is of a perfect consis-tency, then the following Eqs. (11), (12), (13), and (14) all hold.

nai j =n∑

k=1aikak j (11)

nA = A2 (12)n∑

k=1aikak j

nai j= 1 (13)

123


bi jnai j

= 1 (14)

Proof According to Definition 1 and Eq. (4), it is known that if a comparison matrixA = (ai j )n×n is of a perfect consistency, then for any positive integers i, j, k (≤ n),ai j = aikak j holds. Therefore, let k = 1, 2, . . . , n, then we have

ai j = ai1a1 jai j = ai2a2 j· · · · · ·ai j = ainanj

Add these formulas together, and Eq. (11) is obtained.Notice that

∑nk=1 aikak j is the element in the i th row and j th column of the matrix

A2, in other words, suppose that

A2 = (bi j )n×nthen we have

bi j =n∑

k=1aikak j

According to the arbitrariness of i, j , in fact Eq. (11) is equivalent to Eq. (12).Since all the elements are positive, i.e., for any positive integers i, j = 1, 2, . . . , n,

ai j > 0, hence Eq. (11) or Eq. (12) is equivalent to Eq. (13) or Eq. (14), whichcompletes the proof process.

Let

ci j = max(

bi jnai j

,nai jbi j

), i, j = 1, 2, . . . n (15)

then in all the elements ci j , the largest one (suppose it is ci0 j0 ) corresponds to themost serious inconsistency. According to the aforementioned Principle 1, adjust asfew elements in the comparison matrix as possible; at this moment, adjusting ai0 j0has the highest efficiency. Therefore, the corresponding element in the matrix A, i.e.,ai0 j0 should be adjusted first. Thus we have the following improved iterative method,which is convenient for preference relations with high orders.

Step 1 Calculating. When calculating Saaty’s consistency ratio (CR), if it is foundthat the consistency ratio of the matrix A satisfies CR(A) < 0.1, then the matrix Ais of an acceptable consistency and the computations are completed; therefore go toStep 4. Otherwise, go to Step 2.

Step 2 Identifying. Calculate B = A2 = (bi j )n×n and nA, let C ′ = (c′i j )n×n =(bi jnai j

)n×n ; C = (ci j )n×n , ci j = max

(c′i j , 1c′i j ,

), namely, ci j = max

(bi jnai j

,nai jbi j

),

calculate the matrix C , ascertain the largest out of all the elements in the matrix C .

123

380 K. Xu, J. Xu

Suppose the largest one is ci1 j1 , i.e., the element in the i1th row and j1th column ofthe matrix C , then go to Step 3.

Step 3 Adjusting. Adjust the corresponding element ai1 j1 in the matrix A. Becauseai j ∈

{ 19 ,

18 ,

17 , . . . ,

12 , 1, 2, . . . 7, 8, 9

}, all the values of ai j , which can make A be

of an acceptable consistency, can be ascertained by enumerating one by one; out ofwhich the optimal value of ai j , which can maximize the consistent degree of A, canalso be ascertained. Then go to Step 1.

Step 4 Finish.Note 3 In theory, Eqs. (11), (12), (13), (14) are just necessary conditions, which can

help in identifying the most inconsistent element, but are not sufficient conditions forA being of a perfect consistency.When Eqs. (11), (12), (13), (14) are satisfied, whetheror not A is of a perfect consistency or even an acceptable consistency, this needs tobe checked further. If the answer is ‘not’, further improvement can be achieved byemploying the proposed 3 tuples method.

Example 6 By reusing A in Example 5, i.e., let

A = (ai j )8×8 =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 5 3 7 6 6 1/3 1/41/5 1 1/3 5 3 3 1/5 1/71/3 3 1 6 3 4 6 1/51/7 1/5 1/6 1 1/3 1/4 1/7 1/81/6 1/3 1/3 3 1 1/2 1/5 1/61/6 1/3 1/4 4 2 1 1/5 1/63 5 1/6 7 5 5 1 1/24 7 5 8 6 6 2 1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

Step 1 Calculate Saaty’s consistency ratio. CR(A) = 0.1703 > 0.1, so A is not ofan acceptable consistency, go to Step 2.

Step 2 Calculate B = A2 = (bi j )n×n and nA, let C = (ci j )n×n , ci j =max

(bi jnai j

,nai jbi j

), calculate the matrix C .

By using MATLAB, it is found that

B = A2 = (bi j )n×n

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

8 27.8167 13.6389 103.3333 53.5000 46.9167 23.5667 4.85603.3968 8 4.5976 36.9429 17.7238 13.1405 4.6667 2.127422.0905 42.6000 8 97.9333 58.2000 53.2000 15.3683 5.82861.4071 3.3980 1.6510 8 4.5881 4.2548 1.7400 0.51632.2897 5.4333 2.7694 15.5667 8 7.0833 3.3841 1.04762.6548 5.8833 3.3111 21.0667 10.0833 8 3.3270 1.406011.7222 33.7333 17.5833 100.0000 58.8333 50.9167 8 5.039320.2095 64.6000 29.5000 165.0000 96.6667 92.0000 40.2762 8

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

123


C ′ = (c′i j )n×n =(

bi jnai j

)

n×n

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 0.6954 0.5683 1.8452 1.1146 0.9774 8.8375 2.42802.1230 1 1.7241 0.9236 0.7385 0.5475 2.9167 1.86158.2839 1.7750 1 2.0403 2.4250 1.6625 0.3202 3.64291.2312 2.1238 1.2382 1 1.7205 2.1274 1.5225 0.51631.7173 2.0375 1.0385 0.6486 1 1.7708 2.1151 0.78571.9911 2.2062 1.6556 0.6583 0.6302 1 2.0794 1.05450.4884 0.8433 13.1875 1.7857 1.4708 1.2729 1 1.25980.6315 1.1536 0.7375 2.5781 2.0139 1.9167 2.5173 1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

Let ci j = max(c′i j , 1c′i j ,

), namely, C = (ci j )n×n , ci j = max

(bi jnai j

,nai jbi j

), then we

have

C =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 1.4380 1.7597 1.8452 1.1146 1.0231 8.8375 2.42802.1230 1 1.7241 1.0828 1.3541 1.8264 2.9167 1.86158.2839 1.7750 1 2.0403 2.4250 1.6625 3.1233 3.64291.2312 2.1238 1.2382 1 1.7205 2.1274 1.5225 1.93701.7173 2.0375 1.0385 1.5418 1 1.7708 2.1151 1.27271.9911 2.2062 1.6556 1.5190 1.5868 1 2.0794 1.05452.0474 1.1858 13.1875 1.7857 1.4708 1.2729 1 1.25981.5834 1.1536 1.3559 2.5781 2.0139 1.9167 2.5173 1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

It can be ascertained that the largest element in the matrix C is c73 = 13.1875, whichmeans that there is a most serious inconsistency in the corresponding position of thematrix A; therefore, the corresponding element a73 in the matrix A should be adjustedfirst. The remaining steps are similar to Example 5; please refer to Step 3 and Table 8in Example 5.

It is obvious that adjusting a73 and a37 lead to the same results, and thus the twomethods, i.e., the 3 tuples iterative method and the improved iterative method, lead tothe same results.

Example 7 Consider the following 16 × 16 matrix,A = (ai j )16×16

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 1 2 4 1/4 3 1/8 1/9 1/9 5 1/9 1/7 1/3 1/2 1/6 1/51 1 2 3 1/5 3 1/9 1/9 1/9 4 1/2 1/8 1/4 1/3 1/7 1/61/2 1/2 1 2 1/6 1 1/9 1/9 1/9 3 1/3 1/9 1/5 1/4 1/8 1/71/4 1/3 1/2 1 1/8 1 1/9 1/9 1/9 2 1/5 1/9 1/7 1/6 1/9 1/94 5 6 8 1 7 1/4 1/5 1/6 9 3 1/3 1 2 1/2 11/3 1/3 1 1 1/7 1 1/9 8 1/9 2 1/4 1/9 1/6 1/5 1/9 1/88 9 9 9 4 9 1 1 1/2 9 7 1 5 6 2 39 9 9 9 5 1/8 1 1 1 9 8 2 6 7 3 49 9 9 9 6 9 2 1 1 1/7 9 3 7 8 4 51/5 1/4 1/3 1/2 1/9 1/2 1/9 1/9 7 1 1/6 1/9 1/8 1/7 1/9 1/99 2 3 5 1/3 4 1/7 1/8 1/9 6 1 1/6 1/2 1 1/5 1/47 8 9 9 3 9 1 1/2 1/3 9 6 1 4 5 1 23 4 5 7 1 6 1/5 1/6 1/7 8 2 1/4 1 1 1/3 1/22 3 4 6 1/2 5 1/6 1/7 1/8 7 1 1/5 1 1 1/4 1/36 7 8 9 2 9 1/2 1/3 1/4 9 5 1 3 4 1 15 6 7 9 1 8 1/3 1/4 1/5 9 4 1/2 2 3 1 1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

123

382 K. Xu, J. Xu

Step 1 Calculate Saaty’s consistency ratio. CR(A) = 0.4473 > 0.1, so A is not ofan acceptable consistency, go to Step 2.

Step 2 Calculate B = A2 = (bi j )n×n and nA, let C = (ci j )n×n ,ci j = max

(bi jnai j

,nai jbi j


By using MATLAB, it is found that

C =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1.0000 1.0952 1.2643 1.7729 1.4555 1.5625 1.3856 14.7147 20.70021.1076 1.0000 1.3587 1.4218 1.6590 1.6946 1.4022 14.5683 16.68711.6039 1.4901 1.0000 1.4343 1.5426 1.1707 1.0262 5.2808 12.40582.4419 1.7681 1.5289 1.0000 1.7269 1.1836 1.3341 5.0653 8.32031.1214 1.3801 1.1222 1.0115 1.0000 1.0489 1.8252 19.3072 25.557715.3431 15.2308 5.2815 5.5761 18.8524 1.0000 5.2391 7.5379 12.80491.4848 1.1406 1.5417 2.1806 1.4208 1.8273 1.0000 5.2919 9.27891.5874 1.4013 1.8168 2.5460 1.4322 154.8750 1.1707 1.0000 4.69911.9012 1.7121 2.2121 3.0074 1.3918 2.5778 1.4333 5.5459 1.000022.0632 17.5877 13.5650 9.3964 24.9921 9.1244 8.3466 6.5703 7.71113.8918 1.0410 1.1096 1.0766 1.8174 1.0927 2.0476 17.8936 25.52721.3665 1.0098 1.2431 1.7847 1.3613 1.5039 1.1996 10.3205 13.56501.0935 1.4628 1.2276 1.1626 1.2720 1.1800 1.8270 19.7363 26.29841.1212 1.4468 1.2961 1.3249 1.2383 1.3038 1.7467 19.1101 26.11581.2635 1.1200 1.1113 1.4323 1.1646 1.2104 1.3830 15.1118 17.69191.1804 1.2554 1.0086 1.1323 1.3126 1.0768 1.6868 17.7696 21.6856

1.6061 6.1556 1.3994 1.3799 1.1445 1.4301 1.45811.3895 1.2532 1.4491 1.6768 1.5971 1.5189 1.59721.6680 1.4629 1.1796 1.6274 1.6407 1.3023 1.42781.6216 2.0329 1.0672 1.9078 2.0399 1.1933 1.52721.1568 1.4482 1.5931 1.3230 1.1394 1.2914 1.25782.9269 17.5081 9.8690 19.4811 19.0666 14.5763 17.21302.6394 1.1614 1.2636 1.3417 1.2177 1.2162 1.40102.9472 1.0593 1.3155 1.2889 1.1364 1.5106 1.5262218.4375 1.0457 1.6680 1.2127 1.0486 1.6816 1.56421.0000 25.4536 12.4058 26.0801 26.2620 16.5308 20.72451.1873 1.0000 2.0086 1.5716 1.0361 1.9789 1.92962.2121 1.2916 1.0000 1.3800 1.3079 1.2844 1.18621.0084 1.2589 1.6823 1.0000 1.3241 1.5127 1.25571.1387 1.2119 1.6664 1.2735 1.0000 1.5851 1.46321.8162 1.4085 1.2225 1.3329 1.3544 1.0000 1.30011.4637 1.4773 1.3330 1.1479 1.3156 1.2415 1.0000

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

It can be ascertained that the largest element in the matrix C is c9,10 = 218.4375,which means that there is a most serious inconsistency in the corresponding positionof the matrix A; therefore, the corresponding element a9,10 in the matrix A should beadjusted first.

Step 3 Adjust a9,10 to a9,10−ad justed = 9 (a10,9−ad justed = 1/9). At this point,CR(A) = 0.2598. Furthermore, in a same way, the second iteration is carried out, a8,6is adjusted to a8,6−ad justed = 9 (a6,8−ad justed = 1/9) with CR(A) = 0.0623 < 0.1;at this moment, the adjusted matrix A is of an acceptable consistency.

In Example 7, the required computing time is very short and the calculation effi-ciency is quite high.

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5.2 An actual case study

In this section, an actual case study is used for demonstrating the effectiveness of theproposed approach.

Example 8 Many companies have recently tried to select suitable suppliers in orderto improve product quality. Therefore, supplier selection has become a key issue thatboth enterprises and scholars are paying attention to. Supplier selection is particularlyimportant for seafood companies. Accordingly, in this case study a seafood companyneeds to select a supplier. There are ten companies, x1, x2, ..., x9 and x10, which canbe selected (the companies’ names will not be released due to confidentiality).

To select the best supplier, the seafood company employs a consultancy firm toevaluate the ten competing suppliers. The experts provide their preference informationregarding the alternatives as follows:

A = (ai j )10×10 =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 4 5 8 5 6 1/3 1/5 3 1/21/4 1 6 5 6 3 1/5 1/8 4 1/41/5 1/6 1 1/3 1/3 1/4 1/6 1/6 1 1/81/8 1/5 3 1 1/3 1/4 1/7 1/7 1 1/51/5 1/6 3 3 1 1/2 1/5 1/6 6 1/61/6 1/3 4 4 2 1 1/5 1/7 1 1/33 5 6 7 5 5 1 1/2 8 1/35 8 6 7 6 7 2 1 6 61/3 1/4 1 1 1/6 1 1/8 1/6 1 1/72 4 8 5 6 3 3 1/6 7 1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

Step 1 It is found that CR(A) = 0.1336, so A is not of an acceptable consistencyand should be adjusted.

Step 2 Calculate B = A2 = (bi j )n×n and nA, let C = (ci j )n×n, ci j =max

(bi jnai j

,nai jbi j


By using MATLAB, it is found that the largest element in the matrix C is c95 =5.1893, hence there is a most serious inconsistency, and a95 should be adjusted first.

Step 3 Adjust a95 to a95-ad justed = 1 (a59-ad justed = 1). At this point, CR(A) =0.1157. Furthermore, in a sameway, the second iteration is carried out, a10,8 is adjustedto a10,8-ad justed = 1 (a8,10-ad justed = 1) with CR(A) = 0.0993 < 0.1; at thismoment, the adjusted matrix A is of an acceptable consistency.

Step 4 By using the row geometric mean method (RGMM) developed by Crawfordand Williams (1985), the weight vector for x1, x2, . . . , x9 and x10 is obtained:

(1.85, 1.13, 0.28, 0.35, 0.49, 0.67, 2.71, 3.92, 0.44, 3.22)

therefore these companies can be sorted as follows:

x8 � x10 � x7 � x1 � x2 � x6 � x5 � x9 � x4 � x3which indicates that x8 is the most desirable according to the consultancy firm.

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384 K. Xu, J. Xu

5.3 Comparisons and discussion

When comparing the two methods, i.e., the 3 tuples iterative method and the improvediterativemethod, one can find that the 3 tuples iterativemethod can help in ascertainingthe internal quantitative relationships between the consistency ratio of the comparisonmatrix and the 3 tuples. Furthermore, the 3 tuples iterative method is intuitional, easyto understand, and more suitable for manual calculations, especially for preferencerelations with low orders and some specific preference relations with high orders;e.g. Examples 2, 3 and 4. The improved iterative method is also intuitional, easy tounderstand, and easy to use for preference relations with high orders, e.g. Example 7.However, the matrix operation utilizes the power of the algebraic operation.

The 3 tuples iterative method utilizes the ratios max(

ai jaikak j

,aikak jai j

)(i, j, k =

1, 2, . . . , n, i < k < j); while the improved iterative method utilizes the ratios

max

(nai j∑n

k=1 aikak j,

∑nk=1 aikak jnai j

)(i, j, k = 1, 2, . . . , n). In summary, the two methods

are closely linked and each has its own different perspective; moreover, they can beapplied on different occasions and reinforce one another.

It should be noted that since in Ergu et al’s method (2011) they reached the conclu-sion that A2 − nA = 0, which is equivalent to our conclusion A2 = nA, the proposedimproved iterative method and Ergu et al’s method have a similar theoretical basis.However, the latter’s method is based on the principle that the elements in the matrixA2 − nA are near to 0 if A is close to being of a perfect consistency; while our pro-posed improved iterative method is based on the principle that the ratios between thecorresponding elements in the matrices A2 and nA should be near to 1 if A is close tobeing of a perfect consistency. Moreover, the proposed improved iterative method isalso more intuitional and easy to understand than their method.

6 Conclusions

In this paper, an intuitional method has been proposed to illustrate the consistencyproblem from the perspective of 3 tuples (aik, akj , ai j ), which does not require matrixoperations. We have found that in a pairwise comparison matrix, if all the 3 tuples(aik, akj , ai j ) satisfy 0.382 <

ai jaikak j

< 2.620, then the pairwise comparison matrixis of an acceptable consistency (Saaty’s consistency ratio is less than 0.1). Somerelated theorems and propositions have been proved mathematically or verified byrandom simulations. The effectiveness of the method has also been demonstratedthrough a comparison with several published examples. Furthermore, an improvediterative method has been proposed, which is based on the proposed 3 tuples method.The proposed improved iterative method is also intuitional, easy to understand, andconvenient for preference relations with high orders.

Compared to the existing methods, the proposed method in this paper has thefollowing advantages and features:

(1) The internal quantitative relationships between the consistency ratio of the pair-wise comparison matrix and the 3 tuples have been ascertained. A new method,

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from the perspective of 3 tuples, is provided to test the consistency of a pairwisecomparison matrix. We divide an n × n pairwise comparison matrix into 3 tuples(aik, akj , ai j ).

(2) This paper presents a simpler algorithm for the consistency test. The proposed 3tuples iterative method is more intuitional and easy to understand, as only simplemathematical calculations are needed, without the need for matrix operations. Thecalculation of Saaty’s consistency ratio is only conducted to verify the proposedmethod.

(3) The proposed method can also retain most of the information provided by theoriginal comparison matrix, show the modification direction and provide the opti-mal values. At the same time, by using the proposed method, the inconsistentelements in the matrix can be found accurately and rapidly.

(4) The proposed improved iterative method is also intuitional, easy to understand,and convenient for preference relations with high orders.

The two methods are closely linked and each has its own different perspective;moreover, they are applied on different occasions and reinforce one another.

In future research, some similar methods may be applied to other related issuesregarding the pairwise comparison matrix; for example,the consensus of the pairwisecomparison matrix.

Appendix 1

S=0,v=[1/9 ,1/8 ,1/7 ,1/6 ,1/5 ,1/4 ,1/3 ,1/2 ,1 ,2 ,3 ,4 ,5 ,6 ,7 ,8 ,9];

for i=1:length(v) for j=1:length(v) for m=1:length(v)

x12=v( i ) ,x13=v( j ) ,x23=v(m) ,x21=1/x12,x31=1/x13,x32=1/x23;B=[1,x12,x13;x21,1 ,x23;x31,x32,1] ; [x,y]=eig (B) ,eigenvalue=diag(y) ;Lamda=eigenvalue(1) ,Lamda1=Lamda−3,CI=Lamda1/2 , CR=CI/0.52 ,

i f 0.505

386 K. Xu, J. Xu

Appendix 2

>> syms xx=solve (1. /x+x==power(2.104,3)−3∗2.104)x =

3379975011117215/2251799813685248 − (3∗705958741651544450761739614969^(1/2))/2251799813685248(3∗705958741651544450761739614969^(1/2))/2251799813685248 + 3379975011117215/2251799813685248

>> 3379975011117215/2251799813685248 − (3∗705958741651544450761739614969^(1/2))/2251799813685248ans = 0.3816

>> (3∗705958741651544450761739614969^(1/2))/2251799813685248 + 3379975011117215/2251799813685248ans = 2.6204

>> S=0,N=1000000;for i=1:N;

x12=(9−(1/9))∗rand(1 ,1)+(1/9) , x13=(9−(1/9))∗rand(1 ,1)+(1/9) ,x23=(9−(1/9))∗rand(1 ,1)+(1/9) ,x21=1/x12,x31=1/x13, x32=1/x23,

B=[1,x12,x13;x21,1 ,x23;x31,x32,1] ; [x,y]=eig (B) ,eigenvalue=diag(y) ;Lamda=eigenvalue(1) ,Lamda1=Lamda−3,CI=Lamda1/2 , CR=CI/0.52 ,

m=3379975011117215/2251799813685248−(3∗705958741651544450761739614969^(1/2))/2251799813685248n=(3∗705958741651544450761739614969^(1/2))/2251799813685248+3379975011117215/2251799813685248i f m


i f (max(x13/(x12∗x23) ,(x12∗x23) /x13)+max(x24/(x23∗x34) ,(x23∗x34) /x24)+max(x14/(x12∗x24) ,(x12∗x24) /x14)+max(x14/(x13∗x34) ,(x13∗x34) /x14) ) S

S=0,N=10000000;for i=1:N;a=[1/9 ,1/8 ,1/7 ,1/6 ,1/5 ,1/4 ,1/3 ,1/2 ,1 ,2 ,3 ,4 ,5 ,6 ,7 ,8 ,9];c=randperm(numel(a) ) ;b=a(c(1:10) ) ;x12=b(1 ,1) ,x13=b(1 ,2) ,x14=b(1 ,3) ,x15=b(1 ,4) ,x23=b(1 ,5) ,x24=b(1 ,6) ,x25=b(1 ,7) ,x34=b(1 ,8) ,x35=b(1 ,9) ,x45=b(1,10) ,

x21=1/x12,x31=1/x13,x41=1/x14,x51=1/x15,x32=1/x23,x42=1/x24,x52=1/x25, x43=1/x34, x53=1/x35, x54=1/x45

B=[1,x12,x13,x14,x15;x21,1 ,x23,x24,x25;x31,x32,1 ,x34,x35;x41,x42,x43,1 ,x45;x51,x52,x53,x54,1][x,y]=eig (B) ,eigenvalue=diag(y) ; lamda=eigenvalue(1) ,CI=(lamda−5)/4 ,CR=CI/1.11 ,

m=3379975011117215/2251799813685248−(3∗705958741651544450761739614969^(1/2))/2251799813685248n=(3∗705958741651544450761739614969^(1/2))/2251799813685248+3379975011117215/2251799813685248i f (max(x13/(x12∗x23) ,(x12∗x23) /x13)+max(x24/(x23∗x34) ,(x23∗x34) /x24)+max(x14/(x12∗x24) ,(x12∗x24) /x14)+max(x14/(x13∗x34) ,(x13∗x34) /x14)+max(x15/(x12∗x25) ,(x12∗x25) /x15)+max(x15/(x13∗x35) ,(x13∗x35) /x15)+max(x15/(x14∗x45) ,(x14∗x45) /x15)+max(x25/(x23∗x35) ,(x23∗x35) /x25)+max(x25/(x24∗x45) ,(x24∗x45) /x25)+max(x35/(x34∗x45) ,(x34∗x45) /x35) ) S

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A direct consistency test and improvement method for the analytic hierarchy processAbstract1 Introduction2 Preliminaries2.1 The 1–9 ratio scale and pairwise comparison matrix2.2 The consistency of a pairwise comparison matrix

3 A direct acceptable consistency test3.1 A direct acceptable consistency test for 3 times3 pairwise comparison matrices3.2 A direct consistency test method for higher order pairwise comparison matrices

4 A 3 tuples iterative method for improving the consistency process4.1 A 3 tuples iterative method4.2 Some illustrative examples and comparisons

5 An improved iterative method for preference relations with high orders5.1 An improved iterative method to increase the computational efficiency for preference relations with high orders5.2 An actual case study5.3 Comparisons and discussion

6 ConclusionsAppendix 1Appendix 2References

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A direct consistency test and improvement method for the ......of perfect consistency and acceptable consistency. Saaty (1977) also developed an eigenvector method to derive priorities