Factor relation analysis for sustainable recycling partner evaluation using probabilistic linguistic DEMATEL

Abstract

Evaluating and selecting suitable sustainable recycling partners is a key work in sustainable supply chain management. In order to deal with the probabilistic linguistic influence relations between criteria and obtain the key factors that influence the evaluation results of sustainable recycling partners, we propose a new decision-making trial and evaluation laboratory (DEMATEL) method. First, we propose a new generalized weighted ordered weighted averaging (GWOWA) operator and discuss its properties. Second, we use probabilistic linguistic term sets (PLTSs) to aggregate the experts’ hesitant fuzzy linguistic decision-making information and develop a novel method of transforming PLTSs into triangular fuzzy numbers (TFNs) based on the proposed GWOWA operator and the characteristics of PLTSs. Furthermore, we propose a method of making criteria relation analysis based on DEMATEL with TFNs. With the method, we not only access the importance weights of criteria but also obtain the influence relation among the criteria and cluster the criteria into two groups: cause group and effect group. Finally, we apply our method to a real case of sustainable recycling partner selection.

Appendix

Theorem 1

Let $a_{1} ,a_{2} , \ldots ,a_{n}$ be a series of positive numbers. If $a_{1} = a_{2} = \cdots = a_{n} = a$, then

$$ f_{GWOWA}^{P,W} (a_{1} ,a_{2} , \ldots ,a_{n} ) = a. $$

Proof

Since $a_{1} = a_{2} = \cdots = a_{n} = a$, we have $f_{GWOWA}^{P,W} (a_{1} ,a_{2} , \ldots ,a_{n} ) = (\sum\nolimits_{i = 1}^{n} {v_{i} a_{{}}^{\lambda } } )^{{\frac{1}{\lambda }}}$. Owing to $\sum\nolimits_{i = 1}^{n} {v_{i} } = 1$, we can obtain $(\sum\nolimits_{i = 1}^{n} {v_{i} a_{{}}^{\lambda } } )^{{\frac{1}{\lambda }}} = a$.□

Theorem 2

Let $a_{1} ,a_{2} , \ldots ,a_{n}$ and $b_{1} ,b_{2} , \ldots ,b_{n}$ be two arbitrary series of positive numbers. If $a_{j} \ge b_{j}$($j = 1,2, \ldots ,n$), then $f_{GWOWA}^{P,W} (a_{1} ,a_{2} , \ldots ,a_{n} ) \ge f_{GWOWA}^{P,W} (b_{1} ,b_{2} , \ldots ,b_{n} )$.

Proof

Since $a_{j} \ge b_{j}$($j = 1,2, \ldots ,n$), we have $a_{j}^{\lambda } \ge b_{j}^{\lambda }$. According to Torra (1997), we can obtain $f_{WOWA}^{P,W} (a_{1}^{\lambda } ,a_{2}^{\lambda } , \ldots ,a_{n}^{\lambda } ) \ge f_{WOWA}^{P,W} (b_{1}^{\lambda } ,b_{2}^{\lambda } , \ldots ,b_{n}^{\lambda } )$. Therefore, we have $[f_{WOWA}^{P,W} (a_{1}^{\lambda } ,a_{2}^{\lambda } , \ldots ,a_{n}^{\lambda } )]^{{\frac{1}{\lambda }}} \ge [f_{WOWA}^{P,W} (b_{1}^{\lambda } ,b_{2}^{\lambda } , \ldots ,b_{n}^{\lambda } )]^{{\frac{1}{\lambda }}}$. In other words, $f_{GWOWA}^{P,W} (a_{1} ,a_{2} , \ldots ,a_{n} ) \ge f_{GWOWA}^{P,W} (b_{1} ,b_{2} , \ldots ,b_{n} )$ holds.□

Theorem 3

Let $a_{1} ,a_{2} , \ldots ,a_{n}$ be a series of positive numbers, then

$$ \hbox{min} \{ a_{1} ,a_{2} , \ldots ,a_{n} \} \le f_{GWOWA}^{P,W} (a_{1} ,a_{2} , \ldots ,a_{n} ) \le \hbox{max} \{ a_{1} ,a_{2} , \ldots ,a_{n} \} . $$

Proof

Let $\hbox{min} \{ a_{1} ,a_{2} , \ldots ,a_{n} \} = a^{ - }$ and $\hbox{max} \{ a_{1} ,a_{2} , \ldots ,a_{n} \} = a^{ + }$. We can easily obtain $f_{GWOWA}^{P,W} (a_{1} ,a_{2} , \ldots ,a_{n} ) \le (\sum\nolimits_{i = 1}^{n} {v_{i} (a^{ + } )^{\lambda } } )^{{\frac{1}{\lambda }}} = a^{ + }$. Similarly, we can obtain $f_{GWOWA}^{P,W} (a_{1} ,a_{2} , \ldots ,a_{n} ) \ge (\sum\nolimits_{i = 1}^{n} {v_{i} (a^{ - } )^{\lambda } } )^{{\frac{1}{\lambda }}} = a^{ - }$.

Therefore, $\hbox{min} \{ a_{1} ,a_{2} , \ldots a_{n} \} \le f_{GWOWA}^{P,W} (a_{1} ,a_{2} , \ldots ,a_{n} ) \le \hbox{max} \{ a_{1} ,a_{2} , \ldots ,a_{n} \}$.□

Theorem 4

(1) If $\lambda = 1$, then the GWOWA operator reduces to the WOWA operator.

Furthermore, if $p_{1} = p_{2} = \cdots = p_{n} = \frac{1}{n}$, then the GWOWA operator reduces to the OWA operator; and if $w_{1} = w_{2} = \cdots = w_{n} = \frac{1}{n}$, then the GWOWA operator reduces to the weighted average operator.

(2) If $\lambda \to + \infty$, then the GWOWA operator reduces to the max operator, that is,

$$ f_{GWOWA}^{P,W} (a_{1} ,a_{2} , \ldots ,a_{n} ) = \mathop {\hbox{max} }\limits_{j \in N} \{ a_{j} \} . $$

Proof

(1) When $\lambda = 1$, then $f_{GWOWA}^{P,W} (a_{1} ,a_{2} , \ldots ,a_{n} ) = (\sum\nolimits_{i = 1}^{n} {v_{i} b_{i}^{\lambda } } )^{{\frac{1}{\lambda }}} = \sum\nolimits_{i = 1}^{n} {v_{i} b_{i} } = f_{WOWA}^{P,W} (a_{1} ,a_{2} , \ldots ,a_{n} )$.

Furthermore, when $p_{1} = p_{2} = \cdots = p_{n} = \frac{1}{n}$, then $v_{i} = w^{*} (\sum\nolimits_{j = 1}^{i} {p_{\sigma (j)} } ) - w^{*} (\sum\nolimits_{j = 1}^{i - 1} {p_{\sigma (j)} } ) = \sum\nolimits_{j \ge i}^{{}} {w_{j} } - \sum\nolimits_{j < i}^{{}} {w_{j} } = w_{i}$. Therefore, the GWOWA operator reduces to the OWA operator.

When $w_{1} = w_{2} = \cdots = w_{n} = \frac{1}{n}$, then $w^{*} (x) = x$. Therefore, we have $v_{i} = w^{*} (\sum\nolimits_{j = 1}^{i} {p_{\sigma (j)} } ) - w^{*} (\sum\nolimits_{j = 1}^{i - 1} {p_{\sigma (j)} } ) = p_{\sigma (i)}$.

Then the GWOWA operator reduces to the weighted average operator.

(2) When $\lambda \to + \infty$, $f_{GWOWA}^{P,W} (a_{1} ,a_{2} , \ldots ,a_{n} ) = \mathop {\lim }\nolimits_{\lambda \to \infty } (\sum\nolimits_{i = 1}^{n} {v_{i} (b_{i} )^{\lambda } )^{{\frac{1}{\lambda }}} } = \mathop {\hbox{max} }\nolimits_{j \in N} \{ a_{j} \}$.□

Theorem 5

Let $a_{1} ,a_{2} , \ldots ,a_{n}$ be a series of positive numbers, $\eta$ be an arbitrary sequence in $N = \{ 1,2,3 \ldots ,n\}$, and $f_{\eta } (a_{1} ,a_{2} , \ldots ,a_{n} ) = (\sum\nolimits_{j \in N}^{{}} {v_{j} a_{{_{\eta (j)} }}^{\lambda } } )^{{\frac{1}{\lambda }}}$. If $w_{1} \ge w_{2} \ge \cdots \ge w_{n}$, then

$$ f_{\eta } (a_{1} ,a_{2} , \ldots ,a_{n} ) \le f_{GWOWA}^{P,W} (a_{1} ,a_{2} , \ldots ,a_{n} ) \le \left( {nw_{1} \sum\limits_{i = 1}^{n} {p_{i} a_{i}^{\lambda } } } \right)^{{\frac{1}{\lambda }}} . $$

Proof

(1) Without loss of generality we assume that $a_{1} \ge a_{2} \ge \cdots \ge a_{n}$. Given two elements $a_{\eta (i)}$ and $a_{\eta (i + 1)}$, we assume that $a_{\eta (i)} \le a_{\eta (i + 1)}$ in sequence $\eta$. We interchange the two elements $a_{\eta (i)}$ and $a_{\eta (i + 1)}$, and denote this by sequence $\eta^{'}$. We easily obtain $f_{\eta } (a_{1} , \ldots ,a_{n} ) - f_{{\eta^{'} }} (a_{1} , \ldots ,a_{n} ) = v_{i} a_{{^{{_{\eta (i)} }} }}^{\lambda } + v_{i + 1} a_{\eta (i + 1)}^{\lambda } - v_{i}^{'} a_{\eta (i + 1)}^{\lambda } - v_{i + 1}^{'} a_{{^{{_{\eta (i)} }} }}^{\lambda } = a_{\eta (i + 1)}^{\lambda } (v_{i + 1} - v_{i}^{'} ) - a_{{^{{_{\eta (i)} }} }}^{\lambda } (v_{i + 1}^{'} - v_{i} )$.

Because $v_{i} + v_{i + 1} = v_{i}^{'} + v_{i + 1}^{'}$ holds, we have $v_{i + 1} - v_{i}^{'} = v_{i + 1}^{'} - v_{i}$.

Because $w_{1} \ge w_{2} \ge \cdots \ge w_{n}$, function $w^{*}$ is concave. Then we obtain $\frac{{v_{i + 1} }}{{p_{\eta (i + 1)} }} \le \frac{{v_{i}^{'} }}{{p_{\eta (i + 1)} }}$. Because $p_{\eta (i + 1)} > 0$, we obtain $v_{i + 1} - v_{i}^{'} \le 0$. Therefore, we have $f_{\eta } (a_{1} , \ldots ,a_{n} ) - f_{{\eta^{'} }} (a_{1} , \ldots ,a_{n} )_{{^{{_{\eta (i)} }} }}^{\lambda } \ge 0$ and $f_{\eta } (a_{1} ,a_{2} , \ldots ,a_{n} ) \le f_{GWOWA}^{P,W} (a_{1} ,a_{2} , \ldots ,a_{n} )$.

(2) Because function $w^{*}$ is concave, we have the slope $\frac{{w_{1} }}{{{\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 n}}\right.\kern-0pt} \!\lower0.7ex\hbox{$n$}}}} \ge \frac{{v_{i} }}{{p_{\sigma (i)} }}$. Hence,

$$ f_{GWOWA}^{P,W} (a_{1} ,a_{2} , \ldots a_{n} ) = \left( {\sum\limits_{i = 1}^{n} {v_{i} b_{i}^{\lambda } } } \right)^{{\frac{1}{\lambda }}} = \left( {\sum\limits_{i = 1}^{n} {v_{i} a_{\sigma (i)}^{\lambda } } } \right)^{{\frac{1}{\lambda }}} \le \left( {\sum\limits_{i = 1}^{n} {nw_{1} p_{\sigma (i)} a_{\sigma (i)}^{\lambda } } } \right)^{{\frac{1}{\lambda }}} = \left( {nw_{1} \sum\limits_{i = 1}^{n} {p_{i} a_{i}^{\lambda } } } \right)^{{\frac{1}{\lambda }}} . $$

Therefore, $f_{\eta } (a_{1} ,a_{2} , \ldots ,a_{n} ) \le f_{GWOWA}^{P,W} (a_{1} ,a_{2} , \ldots ,a_{n} ) \le (nw_{1} \sum\nolimits_{i = 1}^{n} {p_{i} a_{i}^{\lambda } } )^{{\frac{1}{\lambda }}}$ holds.□