Cyclic sequential process of pairwise comparisons with application to multi-criteria decision making

Abstract

The technique of paired comparisons is commonly used for finding an optimal solution to multi-criteria decision-making (MCDM) problems. The process of comparing alternatives is worth investigations due to the limitation and complexity of human cognition. In this paper, we propose a cyclic sequential process of pairwise comparisons to produce a real-valued preference relation without reciprocal property. The non-reciprocal property characterizes the uncertainty experienced by the decision maker (DM). The concepts of consistency and approximate consistency are defined by considering the inherent property of the derived uncertain preference relation. An optimization model is given to elicit the priority vector from uncertain preference relations. A novel yet effective possibility degree formula is established to rank interval numbers. A new decision making model is constructed and illustrated by carrying out numerical examples. As compared to the existing works, a novel process of pairwise comparisons is proposed to generate an uncertain preference relation and cope with the uncertainty in a decision making problem. The proposed model can be used to reduce the workload of providing pairwise comparisons for the DM and reach an acceptable decision.

Appendix A

The proof procedures of Theorems 1 and 2 are provided below:

Proof of Theorem 1. (1) Since $1/9\le a_{nk}\le 9,$ $1/9\le a_{kn}\le 9$ and further according to Assumption 1, it gives $1/9<a_{nk}a_{kn}<9$ and $\left| a_{nk}a_{kn}-1\right| <8.$ Therefore, we have the following result:

$$\begin{aligned} 0\le \frac{1}{8(n-1)}\sum _{k=1}^{n}\left| a_{nk}a_{kn}-1\right| <\frac{1}{8(n-1)}\sum _{k=1}^{n-1}8=1, \quad n\ge 2, \end{aligned}$$

meaning that $CI_{c}\in [0,1).$

(2) On the one hand, if $\textbf{A}_{r}$ and $\textbf{A}_{c}$ are consistent, it follows $a_{nk}a_{kn}=1$ $(k\in I_{n})$ according to Definition 2. The application of Eq. (5) leads to $CI_{c}=0.$ On the other hand, when $CI_{c}=0,$ it is natural to give $a_{nk}a_{kn}=1$ $(k\in I_{n}).$ In terms of Definition 2, $\textbf{A}_{r}$ and $\textbf{A}_{c}$ are consistent.

Proof of Theorem 2. (1) We first consider M2 in Eq. (8) and construct the following Lagrangian function:

$$\begin{aligned} f(w_{1},w_{2},\cdots ,w_{n},\lambda )=\sum _{k=1}^{n-1}\left( a_{kn}-\frac{w_{k}}{w_{n}}\right) ^{2}+\lambda (\sum _{k=1}^{n}w_{k}-1). \end{aligned}$$

It is easy to give:

$$\begin{aligned}{} & {} \frac{\partial f}{\partial w_{k}}=2(a_{kn}-\frac{w_{k}}{w_{n}})(-\dfrac{1}{w_{n}})+\lambda =0, \quad k=1, 2, \cdots , n-1, \end{aligned}$$

(23)

$$\begin{aligned}{} & {} \frac{\partial f}{\partial w_{n}}=2\sum _{k=1}^{n-1}\left( a_{kn}-\frac{w_{k}}{w_{n}}\right) \dfrac{w_{k}}{w_{n}^{2}}+\lambda =0. \end{aligned}$$

(24)

Then we have

$$\begin{aligned} \sum _{k=1}^{n-1}\frac{\partial f}{\partial w_{k}}=2\sum _{k=1}^{n-1}\left( a_{k1}-\frac{w_{k}}{w_{n}}\right) \left( -\dfrac{1}{w_{n}}\right) +(n-1)\lambda =0. \end{aligned}$$

If $\lambda >0$, then one can get

$$\begin{aligned} a_{kn}-\frac{w_{k}}{w_{n}}>0 ,\quad k=2,3,\cdots ,n. \end{aligned}$$

This means that

$$\begin{aligned} \frac{\partial f}{\partial w_{n}} =2\sum _{k=1}^{n-1}\left( a_{kn}-\frac{w_{k}}{w_{n}}\right) \dfrac{w_{k}}{w_{n}^{2}}+\lambda >0. \end{aligned}$$

The result is in contradiction with the observation in (24).

Similarly, when $\lambda <0,$ a contradictory result can be reduced. Therefore, we have $\lambda =0,$ and

$$\begin{aligned} a_{nk}=\frac{w_{k}}{w_{n}},\quad k=1,2,3,\cdots ,n. \end{aligned}$$

Since

$$\begin{aligned} \sum _{k=1}^{n}w_{k} =1,\quad \sum _{k=1}^{n}a_{kn}=a_{nn}+\sum _{k=1}^{n-1}a_{kn}=1+ \sum _{k=1}^{n-1}\frac{w_{k}}{w_{n}} =\frac{1}{w_{n}}, \end{aligned}$$

it follows

$$\begin{aligned} w_{i}^{(2)}=\displaystyle a_{in}\left( \sum _{k=1}^{n}a_{kn}\right) ^{-1}, \quad F_{2}=0, \quad i\in I_{n}. \end{aligned}$$

(2) For M1 in Eq. (7), we can construct the following two functions:

$$\begin{aligned} g_{1}(w_{1},w_{2},\cdots , w_{n})= \sum _{k=1}^{n-1}\left( a_{nk}-\frac{w_{n}}{w_{k}}\right) ^{2}, \\ g_{2}(w_{1},w_{2},\cdots ,w_{n})= \sum _{k=1}^{n-1}\left( b_{nk}-\frac{w_{k}}{w_{n}}\right) ^{2}, \end{aligned}$$

where $b_{nk}=\dfrac{1}{a_{nk}}.$ On the one hand, it is found that

$$\begin{aligned} g_{1}(w_{1},w_{2},\cdots ,w_{n})&=\sum _{k=1}^{n-1}\left( \dfrac{1}{b_{nk}}-\frac{1}{\frac{w_{k}}{w_{n}}}\right) ^{2}\nonumber \\&=\sum _{k=1}^{n-1}\left( \frac{b_{nk}-\frac{w_{k}}{w_{n}}}{b_{nk}\frac{w_{k}}{w_{n}}}\right) ^{2}\nonumber \\&\le \frac{g_{2}(w_{1},w_{2},\cdots ,w_{n})}{C^{2}}, \end{aligned}$$

(25)

where

$$\begin{aligned} C = \min \left\{ {b_{{n1}} \frac{{w_{1} }}{{w_{n} }},b_{{n2}} \frac{{w_{2} }}{{w_{n} }}, \cdots ,b_{{n(n - 1)}} \frac{{w_{{n - 1}} }}{{w_{n} }}} \right\}. \end{aligned}$$

Thus, $\forall \textbf{w}=\{w_{1},w_{2},\cdots ,w_{n}\}, \exists C>0,$ such that $0\le g_{1}(\textbf{w})\le \frac{1}{C^{2}}g_{2}(\textbf{w}).$ According to case (1), when $\textbf{w}^{*}=\left( w_{1}^{*}, w_{2}^{*}, \cdots , w_{n}^{*}\right)$ with

$$\begin{aligned} w_{i}^{*}=\displaystyle \frac{1}{a_{ni}}\left( \sum _{k=1}^{n}\dfrac{1}{a_{nk}}\right) ^{-1}, \end{aligned}$$

it gives $g_{2}(\textbf{w}^{*})=0.$ In other words, $0\le g_{1}(\textbf{w}^{*})\le \frac{1}{C_{1}^{2}}g_{2}(\textbf{w}^{*})=0,$ where

$$\begin{aligned} C_{1}=\min _{k=1,2,\cdots n-1}\left\{ b_{nk}\frac{w_{k}^{*}}{w_{n}^{*}}\right\} . \end{aligned}$$

On the other hand, $\forall \textbf{w}\ne \textbf{w}^{*}$, $g_{1}(\textbf{w}^{*})>0.$ In summary, one has

$$\begin{aligned} w_{i}^{(1)}=\displaystyle \frac{1}{a_{ni}}\left( \sum _{k=1}^{n}\dfrac{1}{a_{nk}}\right) ^{-1},\quad F_{1}=0, \quad i\in I_{n}. \end{aligned}$$