*3.3. Example*

Consider the following matrix:

$$A = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1/2 & 1 & 5 & 6 \\ 1/3 & 1/5 & 1 & 7 \\ 1/4 & 1/6 & 1/7 & 1 \end{pmatrix}$$

The steps for the calculation of both measures are followed in Tables 3 and 4. It can be seen that for this size of matrix the calculation of the *T*-*GCI* is less complicated. 7

The steps for the calculation of the *GCI* are (Table 3): (i) the calculation of ∏*j aij* for each row; (ii) the calculation of priorities *wi*; (iii) the calculation of the errors *eij*; (iv) the calculation of the squared logarithm of the errors; and (v) the value of the *GCI*.


**Table 3.** Calculation of the *GCI*.

Table 4 shows the calculation of the product and squared logarithms for each of the four triads or cycles of length 3. The *T*-*GCI* is obtained by adding these values and dividing by *n*(*n* − <sup>1</sup>)(*n* − 2)/6.


**Table 4.** Calculation of the *T*-*GCI*.

#### **4. Inconsistency Measures Based on Cycles**

In general, if *A* is consistent, it holds [9] that

$$a\_{i\_1 i\_2} a\_{i\_2 i\_3} \dots a\_{i\_l i\_1} = 1 \qquad \forall i\_1, i\_2, \dots, i\_l \tag{14}$$

An inconsistency measure can therefore be defined through the intensities of cycles of any length *l* (> 2).

**Definition 2.** *Given a pairwise comparison matrix, <sup>A</sup>*(*nxn*) = *aij with aijaji* = 1 *and aij* > 0*, the l*−*Cycles Consistency Index is defined as*

$$I\_l(A) = \frac{\sum\_{i\_1 \neq i\_2 \neq \dots \neq i\_l} \log^2 \left( a\_{i\_1 i\_2} a\_{i\_2 i\_3} \dots a\_{i\_l i\_1} \right)}{l! V\_{n,l}} \tag{15}$$

*where Vn*,*<sup>l</sup> are the number of l-variations of n elements.*

**Remark 2.** *Clearly, the <sup>T</sup>*-*GCI*(*A*) = *<sup>I</sup>*3(*A*)*.*

A cycle of length greater than 3 can be easily broken down into product of cycles of length 3. For example (Figures 1 and 2), when *l* = 4:

$$a\_{ij}a\_{jk}a\_{kl}a\_{li} = a\_{ij}a\_{jk}\\\left(a\_{ki}a\_{ik}\right)a\_{kl}a\_{li} = \left(a\_{ij}a\_{jk}a\_{ki}\right)\left(a\_{ik}\\\left.a\_{kl}a\_{li}\right)$$

 **Figure 1.** Cycle of length 4.

**Figure 2.** Cycles of length 3.

Therefore, a measure of divergence *dijkl* = *f*(*aijajk aklali*) can be obtained as *dijkl* = *f* (*dijk*, *dikl*) and a global measure can be considered as a function of the intensities of the cycles of length 3.

The following result not only allows the expression of the general index *Il*(*A*) to be obtained in terms of cycles of length 3, but also proves that all indexes *Il*(*A*) provide the same value regardless of the value of *l*.

**Theorem 2.** *Given a pairwise comparison matrix, <sup>A</sup>*(*nxn*)= *aij with aijaji* = 1 *and aij* > 0*, it holds that*

$$I\_l(A) = \mathcal{GCI}(A) \tag{16}$$

**Proof.** See Appendix A.

These cycle-based indices are equivalent, regardless of the cycle length that is considered. Given that the complexity of calculation increases as *l* increases, it is sufficient to consider the *T*-*GCI* when the objective is to measure the inconsistency from a judgements cycles-based approach.

It can be easily verified that the calculation of the index *Il*(*A*) has complexity *<sup>o</sup>*(*n<sup>l</sup>*), with *n* being the order of the matrix *A*. From a computational point of view, its calculation for *n* > 3 will not be of interest. Since all these measures are equal, it is faster to obtain the value of *<sup>I</sup>*3(*A*) = *<sup>T</sup>*-*GCI*(*A*).

## **5. Conclusions**

The evaluation of the consistency of decision makers when incorporating their preferences in the AHP, in other words, when eliciting their judgements in pairwise comparison matrices, is one of the most outstanding characteristics of this multi-criteria technique. The initial measures of inconsistency, including the two most widely employed in the scientific literature (*CR* and *GCI*), were associated with indexes derived for the prioritisation method used (EV and RGM, respectively). Along with this approach, another method, based on triads, is developed. It evaluates the inconsistency using the definition of a consistent pairwise comparison matrix, or equivalently, in terms of the cycles of length 3, comprising the elements of the matrix.

The current paper introduces a new indicator based on triads (*T*-*GCI*) that links the two approaches followed in the evaluation of the inconsistency of AHP. As demonstrated in the paper, this indicator coincides with the *GCI*, a measure of inconsistency proposed for the RGM. The relationship between the two indicators can take advantage of the properties and characteristics of both to exploit their potential with regards to calculation.

The work does not analyse the relationship between consistency and representativeness of the priority vector derived from the pairwise comparison matrix. The main objective is to propose a new indicator based on triads, the *T*-*GCI*, which links the two approaches considered when defining inconsistency measures and is more efficient than the *GCI* for matrices with fewer than eight alternatives. This fact is of grea<sup>t</sup> significance for simulation studies that require reiterating on numerous occasions the calculation of the inconsistency measure. It is also necessary to emphasise the fact that our work does not deal with obtaining the *T*-*GCI* for incomplete matrices. Studies analysing the estimation of inconsistency for incomplete matrices can be seen in [65,66].

The relationship between the *T*-*GCI* and the *GCI*, and the similarities with different inconsistency measures already proved for the *GCI*, can be used to transfer the results obtained for it to other inconsistency measures, based on triads or the priority vectors; in particular, the setting of thresholds to make those inconsistency indices without thresholds operational. Finally, it has been shown that the generalisation of the *T*-*GCI* to cycles of any length can be expressed in terms of cycles of length 3, which facilitates calculation. Future extensions that are under consideration include how to obtain thresholds for the *T*-*GCI* that would be calculated directly, or obtained in a similar way to the methods for other related indicators.

**Author Contributions:** The paper was elaborated jointly by the four authors. Conceptualization, J.M.M.-J.; Data curation, M.T.E. and A.T.; Formal analysis, J.A., M.T.E. and J.M.M.-J.; Investigation, J.A., M.T.E. and A.T.; Methodology, J.M.M.-J.; Software, J.A.; Visualization, A.T.; Writing—review & editing, J.A., M.T.E. and J.M.M.-J. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research was supported by the Grupo Decisión Multicriterio Zaragoza research group (S35-17R, Government of Aragon) and FEDER funds.

**Acknowledgments:** The authors would like to acknowledge the work of English translation professional David Jones in preparing the final text, and also the excellent review process carried out by one of the three reviewers.

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **Appendix A. Proofs**

**Proof of Theorem 1.** Consider the definition of the index *<sup>T</sup>*-*GCI*(*A*), where the denominator of that measure is denoted as *D* = <sup>3</sup>*n*(*n* − <sup>1</sup>)(*n* − <sup>2</sup>). It can be seen that the sum extends, without problems, to all subscripts *i*, *j*, *k*.

$$\begin{aligned} \text{T-GCI}(A) &= \begin{array}{c} \frac{1}{D'} \sum\_{i,j,k} \log^2 \left( a\_{ij} a\_{jk} a\_{ki} \right) = \frac{1}{D'} \sum\_{i,j,k} \left( \log a\_{ij} + \log a\_{jk} + \log a\_{ki} \right)^2 =\\ & \sum\_{i,j,k} \left( \log a\_{ij}^2 + \log a\_{jk}^2 + \log a\_{ij}^2 + 2 \log a\_{ij} \log a\_{jk} + 2 \log a\_{ij} \log a\_{ki} + 2 \log a\_{jk} \log a\_{ki} \right) \end{array} \end{aligned}$$

By symmetry, the first three terms are identical and can be written as

$$\sum\_{i,j,k} \log a\_{ij}^2 = n \sum\_{i,j} \log a\_{ij}^2$$

On the other hand, the last three addends are also equal to each other. Thus, the index can be expressed as:

$$\begin{aligned} \label{eq:SDAR-T} \left( \begin{array}{c} T-\text{GCI}(A) \\ \hline D' \end{array} \right) &= \frac{1}{D'} \left( 3 \sum\_{i,j} \log a\_{ij}^2 + 6 \sum\_{i,j,k} \log a\_{ij} \log a\_{jk} \right) = \\ &= \frac{1}{3n(n-1)(n-2)} 3 \left( n \sum\_{i,j} \log a\_{ij}^2 + 2 \sum\_{i,j,k} \log a\_{ij} \log a\_{jk} \right) = \\ &= \frac{1}{n(n-1)(n-2)} \left( n \sum\_{i,j} \log a\_{ij}^2 + 2 \sum\_{i,j,k} \log a\_{ij} \log a\_{jk} \right) \end{aligned} \tag{A1}$$

Now, consider the expression of *GCI*:

$$GCI = \frac{\sum\_{i,j} \log^2 \varepsilon\_{ij}}{(n-1)(n-2)} = \frac{\sum\_{i,j} \log^2 a\_{ij} w\_j / w\_i}{(n-1)(n-2)}$$

where *eij* = *aij<sup>ω</sup>j*/*<sup>ω</sup>i* is the error obtained when the ratio of priorities *<sup>ω</sup>i*/*<sup>ω</sup>j* is approximated by *aij* and *w* = (*wi*) is the priority vector obtained with the RGM method: *ωi* = ∏*nk*=<sup>1</sup> *a*1/*<sup>n</sup> ik* .

Combining it, and naming *D* = (*n* − <sup>1</sup>)(*n* − <sup>2</sup>):

*GCI* = 1*D* ∑*i*,*j* log<sup>2</sup> *aij <sup>ω</sup>j ωi* = 1*D* ∑*i*,*j* log<sup>2</sup> *aij* ∏*nk*=<sup>1</sup> *a*1/*<sup>n</sup> jk* ∏*nk*=<sup>1</sup> *a*1/*<sup>n</sup> ik* = = 1 *D* ∑*i*,*j* log<sup>2</sup> ⎛⎝*aij* ∏*nk*=<sup>1</sup> *a*1/*<sup>n</sup> jk* ∏*nk*=<sup>1</sup> *a*1/*<sup>n</sup> ik* ⎞⎠ = 1*D* ∑*i*,*j* log *aij* + 1*n n*∑*k*=1 log *ajk* − 1*n n*∑*k*=1 log *a*1/*<sup>n</sup> ik* 2 = = 1 *n*2*D* ∑*i*,*j n* log *aij* + *n*∑*k*=1 log *ajk* + *n*∑*k*=1 log *aki*2 = = 1 *n*2*D* ∑*i*,*j n* log *aij* + *n*∑*k*=1 log *ajk* + *n*∑*k*=1 log *aki n* log *aij* + *n*∑*l*=1 log *ajl* + *n*∑*l*=1 log *ali* = = 1 *Dn*<sup>2</sup> ⎛⎝*n*2 ∑*i*,*j* log<sup>2</sup> *aij* + *n* ∑*i*,*j*,*l* log *aij* log *ajl* + *n* ∑*i*,*j*,*l* log *aij* log *ali*+ + *n* ∑ *i*,*j*,*k* log *aij* log *ajk* + ∑ *i*,*j*,*k*,*l* log *ajk* log *ajl* + ∑ *i*,*j*,*k*,*l* log *ajk* log *ali* + + *n* ∑ *i*,*j*,*k* log *aij* log *aki* + ∑ *i*,*j*,*k*,*l* log *aki* log *ajl* + ∑ *i*,*j*,*k*,*l* log *aki* log *ali*⎞⎠

It is clear that the second, third, fourth, and seventh terms are the same, represented as *n* ∑*<sup>i</sup>*,*j*,*<sup>k</sup>* log *aij* log *ajk*. Note that the fifth term:

$$\sum\_{i,j,k,l} \log a\_{jk} \log a\_{jl} = \sum\_{i} \sum\_{j,k,l} \log a\_{jk} \log a\_{jl} = -n \sum\_{j,k,l} \log a\_{lj} \log a\_{jk}$$

The same with the ninth addend:

$$\sum\_{i,j,k,l} \log a\_{ki} \log a\_{li} = \sum\_{j} \sum\_{i,k,l} \log a\_{ki} \log a\_{li} = -n \sum\_{i,k,l} \log a\_{li} \log a\_{ik}$$

The two addends are identical and represented by changing subindices to −*n* ∑*<sup>i</sup>*,*j*,*<sup>k</sup>* log *aij* log *ajk* The sixth and eighth addends can also be simplified:

$$\sum\_{i,j,k,l} \log a\_{jk} \log a\_{li} = \sum\_{j,k} \log a\_{jk} \sum\_{i,l} \log a\_{li} = 0 \times 0 = 0$$

Therefore,

$$\begin{split} \begin{array}{rcl} \text{GCI} &=& \frac{1}{\operatorname{Dn^2}} \left( n^2 \sum\_{l,j} \log^2 a\_{ij} + n \sum\_{l,j,k} \log a\_{il} \log a\_{jk} + n \sum\_{l,j,k} \log a\_{il} \log a\_{jk} + 1 \right) \\ &+ n \sum\_{l,j,k} \log a\_{il} \log a\_{jk} - n \sum\_{l,j,k} \log a\_{il} \log a\_{jk} + 0 + \\ &\quad + n \sum\_{l,j,k} \log a\_{il} \log a\_{jk} + 0 - n \sum\_{l,j,k} \log a\_{il} \log a\_{jk} \right) = \\ &=& \frac{1}{\operatorname{Dn^2}} \left( n^2 \sum\_{l,j} \log^2 a\_{ij} + 2n \sum\_{l,j,k} \log a\_{il} \log a\_{jk} \right) \end{split}$$

Finally, replacing the denominator *D* and simplifying:

$$\begin{aligned} \text{GCI} &=& \frac{1}{n^2(n-1)(n-2)} \left( n^2 \sum\_{i,j} \log^2 a\_{ij} + 2n \sum\_{i,j,k} \log a\_{ij} \log a\_{jk} \right) = \\ &=& \frac{1}{n(n-1)(n-2)} \left( n \sum\_{i,j} \log^2 a\_{ij} + 2 \sum\_{i,j,k} \log a\_{ij} \log a\_{jk} \right) \end{aligned} \tag{A2}$$

Equations (A1) and (A2) coincide, thus it is proved that *<sup>T</sup>*-*GCI*(*A*) = *GCI*(*A*).

**Proof of Theorem 2.** This is similar to the proof of Theorem 1. Commencing with the definition of the index *Il*(*A*) to prove that it matches Equation (A2).

$$\begin{aligned} I\_l(A) &= \begin{array}{c} \frac{1}{lV\_{n,l}} \sum\_{i\_1 \neq i\_2 \neq \dots \neq i\_l} \log^2 \left( a\_{i\_1 i\_2} a\_{i\_2 i\_3} \dots a\_{i\_l i\_1} \right) = \\ \frac{1}{lV\_{n,l}} \sum\_{i\_1 \neq \dots \neq i\_l} \left( \log a\_{i\_1 i\_2} + \log a\_{i\_2 i\_3} \dots + \log a\_{i\_l i\_1} \right)^2 \end{array} \end{aligned}$$

By developing the summation:

$$\begin{split} &\sum\_{i\_1 \neq \dots \neq i\_l} \left( \log a\_{i\_1 i\_2} + \log a\_{i\_2 i\_3} \dots + \log a\_{i\_l i\_1} \left( \log a\_{i\_1 i\_2} + \log a\_{i\_2 i\_3} \dots + \log a\_{i\_l i\_1} \right) \right) \\ &= \sum\_{i\_1 \neq \dots \neq i\_l} \left( \log^2 a\_{i\_1 i\_2} + \log^2 a\_{i\_2 i\_3} \dots + \log^2 a\_{i\_l i\_1} \right) + \\ &+ \sum\_{i\_1 \neq \dots \neq i\_l} \left( \log a\_{i\_1 i\_2} \log a\_{i\_2 i\_3} \dots + \log a\_{i\_{l-1} i\_l} \log a\_{i\_l i\_1} \right) + \\ &+ \sum\_{i\_1 \neq \dots \neq i\_l} \left( \dots \log a\_{i\_l i\_{l+1}} \log a\_{i\_l i\_{l+1}} \dots \right) \end{split}$$

In the first summation, squared logarithms of the judgements are grouped. It should be noted that, by symmetry, each of the terms must produce the same sum:

$$\begin{aligned} &\sum\_{i\_1 \neq \dots \neq i\_l} \left(\log^2 a\_{i\_1 i\_2} + \log^2 a\_{i\_2 i\_3} \dots + \log^2 a\_{i\_l i\_1}\right) = \\ &= \sum\_{i\_1 \neq \dots \neq i\_l} \log^2 a\_{i\_1 i\_2} = l \sum\_{i\_3 \neq \dots \neq i\_l} \sum\_{i\_1 \neq i\_2} \log^2 a\_{i\_1 i\_2} \\ &= \left. l(n-2)(n-3) \dots \left(n-l+1\right) \sum\_{i\_1 \neq i\_2} \log^2 a\_{i\_1 i\_2} \right| \end{aligned}$$

The second summation groups the products of the logarithms of judgements that are adjacent in the cycle, i.e. those that have a common subscript. Again, by symmetry, they are all the same and can be expressed as:

$$\begin{aligned} 2\sum\_{i\_1 \neq \dots \neq i\_l} \left( \log a\_{i\_1 i\_2} \log a\_{i\_2 i\_3} \dots + \log a\_{i\_{l-1} i\_l} \log a\_{i\_l i\_1} \right) &= \\ 2l \sum\_{i\_1 \neq \dots \neq i\_l} \log a\_{i\_1 i\_2} \log a\_{i\_2 i\_3} &= 2l \sum\_{i\_4 \neq \dots \neq i\_l} \sum\_{i\_1 \neq i\_2 \neq i\_3} \log a\_{i\_1 i\_2} \log a\_{i\_2 i\_3} = \\ = 2l(n-3) \cdot \dots \cdot (n-l+1) \sum\_{i\_1 \neq i\_2 \neq i\_3} \log a\_{i\_1 i\_2} \log a\_{i\_2 i\_3} &= \end{aligned}$$

Finally, the third sum contains products of logarithms of judgements that have no subscript in common. That each of them is equal to zero can be verified:

$$\sum\_{i\_1 \neq \dots \neq i\_l} \left( \log a\_{i\_r i\_{r+1}} \log a\_{i\_s i\_{s+1}} \right) = \sum\_{i\_1 \neq \dots \neq i\_l} \sum\_{i\_r \neq i\_{r+1}} \sum\_{i\_s \neq i\_{s+1}} \log a\_{i\_r i\_{r+1}} \log a\_{i\_s i\_{s+1}} = 1$$

$$= \sum\_{i\_1 \neq \dots \neq i\_l} \left( \sum\_{i\_r \neq i\_{r+1}} \log a\_{i\_r i\_{r+1}} \right) \left( \sum\_{i\_s \neq i\_{s+1}} \log a\_{i\_s i\_{s+1}} \right) = \sum\_{i\_1 \neq \dots \neq i\_l} 0 \times 0 = 0$$

Taking everything to the definition of the measure of inconsistency:

$$\begin{aligned} I\_l(A) &= \frac{l(n-2)(n-3)\cdots(n-l+1)\sum\_{i\_1 \neq i\_2} \log^2 a\_{i\_1 i\_2}}{\ln(n-1)\cdots(n-l+1)} + \\ &+ \frac{2l(n-3)\cdots(n-l+1)\sum\_{i\_1 \neq i\_2 \neq i\_3} \log a\_{i\_1 i\_2} \log a\_{i\_2 i\_3}}{\ln(n-1)\cdots(n-l+1)} = 0 \\ &= \frac{(n-2)\sum\_{i\_1 \neq i\_2} \log^2 a\_{i\_1 i\_2} + 2\sum\_{i\_1 \neq i\_2 \neq i\_3} \log a\_{i\_1 i\_2} \log a\_{i\_2 i\_3}}{n(n-1)(n-2)} \end{aligned}$$

Renaming the subscripts:

$$I\_l(A) = \frac{(n-2)\sum\_{i\neq j} \log^2 a\_{ij} + 2\sum\_{i\neq j\neq k} \log a\_{ij}\log a\_{jk}}{n(n-1)(n-2)}\tag{A3}$$

This expression is not exactly equivalent to Equation (A2), but if the latter is slightly modified:

$$GCI = \frac{1}{n(n-1)(n-2)} \left( n \sum\_{i,j} \log^2 a\_{ij} + 2 \sum\_{i,j,k} \log a\_{ij} \log a\_{jk} \right)$$

Summations are made that allow repetitions of subscripts. Repetitions are removed to compare this expression with the one that was previously obtained.

It is clear that:

$$\sum\_{i,j} \log^2 a\_{ij} = \sum\_{i \neq j} \log^2 a\_{ij} \text{ since } a\_{ii} = 1$$

Regarding the second term:

$$\begin{aligned} \sum\_{i,j,k} \log a\_{ij} \log a\_{jk} &= \sum\_{i \neq j,k} \log a\_{ij} \log a\_{jk} + \sum\_{i,k} \log a\_{ii} \log a\_{ik} = \\ &= \sum\_{i \neq j,k} \log a\_{ij} \log a\_{jk} + 0 = \\ &= \sum\_{i \neq j \neq k} \log a\_{ij} \log a\_{jk} + \sum\_{i \neq j,k=i} \log a\_{ij} \log a\_{jk} + \sum\_{i \neq j,k=i} \log a\_{ij} \log a\_{jk} = \\ &= \sum\_{i \neq j \neq k} \log a\_{ij} \log a\_{jk} + \sum\_{i \neq j} \log a\_{ij} \log a\_{ji} + \sum\_{i \neq j} \log a\_{ij} \log a\_{ij} = \\ &= \sum\_{i \neq j \neq k} \log a\_{ij} \log a\_{jk} - \sum\_{i \neq j} \log a\_{ij}^{2} + 0 \end{aligned}$$

Grouping everything:

$$\begin{aligned} GCI &=& \frac{n \sum\_{i \neq j} \log^2 a\_{ij} + 2 \left( \sum\_{i \neq j \neq k} \log a\_{ij} \log a\_{jk} - \sum\_{i \neq j} \log a\_{ij}^2 \right)}{n(n-1)(n-2)} \\ &=& \frac{(n-2) \sum\_{i \neq j} \log^2 a\_{ij} + 2 \sum\_{i \neq j \neq k} \log a\_{ij} \log a\_{jk}}{n(n-1)(n-2)} \end{aligned}$$

This now matches Equation (A3) developed from *Il*(*A*).
