Asymmetric normalized probabilistic linguistic term set based on prospect theory and its application to multi-attribute decision-making

Abstract

The probabilistic linguistic term set (PLTS) shows great superiority in expressing decision-makers’ opinions. The multi-attribute decision-making (MADM) problem under a PLTS environment has gained attention from numerous scholars. However, the majority of current studies are not precise enough in capturing information on PLTS. To address this problem, this paper presents a preference ranking organization method for enrichment of evaluations (PROMETHEE) based on the redefined PLTS and novel score function to solve MADM problems under a PLTS environment. First, an asymmetric normalized PLTS based on prospect theory (ANPLTSPT) is developed. Compared with the PLTS, ANPLTSPT offers a more realistic portrayal of decision-makers’ psychological state while ensuring the superiority of the PLTS. Second, regarding the structural complexity of ANPLTSPT, this paper attempts to simplify the computational process through a score function that can embody the characteristics of ANPLTSPT. Inspired by previously formulated score functions, a novel score function called Score-InInHe is developed, the corresponding definitions are given, and some further properties are discussed. With the support of the proposed Score-InInHe, the total score entropy is defined and an objective method to determine the attribute weights is proposed. Finally, the proposed approach is applied to the selection of a green supplier and the determination of air quality. The validity and realistic applicability of the proposed approach are demonstrated through comparative analyses and discussions.

Appendix

1.1 A.1 The proof of Theorem 1 (1)

Proof

For $\# L_{A} \left( p \right) = 1$, $F\left( {L_{A} \left( p \right)} \right) = 1 + \log_{2} \left( {1 + p^{\left( 1 \right)} } \right)$. Then, $\partial F\left( {L_{A} \left( p \right)} \right)/\partial \# L_{A} \left( p \right) = 0$.

For $\# L_{A} \left( p \right) \ne 1$,

$\partial F\left( {L_{A} \left( p \right)} \right)/\partial \# L_{A} \left( p \right) = \partial \left( {\frac{1}{{\left( {2\left( {1 + \log_{2} \# L_{A} \left( p \right)} \right)} \right)}}} \right)/\partial \# L_{A} \left( p \right) = - \frac{1}{{\left( {2\left( {1 + \log_{2} \# L_{A} \left( p \right)} \right)^{2} \cdot \# L_{A} \left( p \right) \cdot \ln 2} \right)}}$.

As $\# L_{A} \left( p \right) \in N^{ + }$, then $\# L_{A} \left( p \right) \cdot \ln 2 > 0$.

As $\# L_{A} \left( p \right) \ne 1$, then $\left( {1 + \log_{2} \# L_{A} \left( p \right)} \right)^{2} > 0$.

Thus, $\partial F\left( {L_{A} \left( p \right)} \right)/\partial \# L_{A} \left( p \right) = - \frac{1}{{\left( {2\left( {1 + \log_{2} \# L_{A} \left( p \right)} \right)^{2} \cdot \# L_{A} \left( p \right) \cdot \ln 2} \right)}} < 0$ holds for $\# L_{A} \left( p \right) \ne 1$.

Then, $\partial F\left( {L_{A} \left( p \right)} \right)/\partial \# L_{A} \left( p \right) \le 0$ holds for $\# L_{A} \left( p \right) \in N^{ + }$. Thus, it can be derived that the novel score function $F\left( {L_{A} \left( p \right)} \right)$ for $L_{A} \left( p \right)$ is a decreasing function of parameter $\# L_{A} \left( p \right)$.

The proof of Theorem 1 (1) is completed. □

1.2 A.2 The proof of Theorem 1 (2)

Proof

$F\left( {L_{A} \left( p \right)} \right)$ strives for the partial derivative with respect to $\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right).}$

$\begin{aligned} \frac{{\partial F\left( {{L_A}\left( p \right)} \right)}}{{\partial \sum\limits_{k = 1}^{\# {L_A}\left( p \right)} {{p^{\left( k\right)}}} }} = \frac{1}{{\left( {1 + \sum\limits_{k = 1}^{\# {L_A}\left( p \right)} {{p^{\left( k \right)}}} } \right) \cdot \ln 2}} - \frac{{\sum\limits_{k = 1}^{\# {L_A}\left( p \right)} {{r_A}^{\left( k \right)}{p^{\left( k \right)}}} }}{{\tau {{\left({\sum\limits_{k = 1}^{\# {L_A}\left( p \right)} {{p^{\left( k \right)}}} } \right)}^2}}} + \frac{{\sum\limits_{k = 1}^{\#{L_A}\left( p \right)} {\left( {{p^{\left( k \right)}}{{\left( {{r_A}^{\left( k \right)}} \right)}^2}} \right)} }}{{8{\tau ^2}{{\left({\sum\limits_{k = 1}^{\# {L_A}\left( p \right)} {{p^{\left( k \right)}}} } \right)}^2}}} - \frac{{{{\left( {\sum\limits_{k = 1}^{\#{L_A}\left( p \right)} {{r_A}^{\left( k \right)}{p^{\left( k \right)}}} } \right)}^2}}}{{4{\tau ^2}{{\left( {\sum\limits_{k = 1}^{\#{L_A}\left( p \right)} {{p^{\left( k \right)}}} } \right)}^3}}} = {1 \mathord{\left/{\vphantom {1 {\left( {\left( {1 + \sum\limits_{k = 1}^{\# {L_A}\left( p \right)} {{p^{\left( k \right)}}} } \right) \cdot \ln 2} \right)}}} \right. \kern-\nulldelimiterspace} {\left( {\left( {1 + \sum\limits_{k = 1}^{\# {L_A}\left( p \right)} {{p^{\left( k \right)}}} } \right)\cdot \ln 2} \right)}} + {{\left( {\sum\limits_{k = 1}^{\# {L_A}\left( p \right)} {{p^{\left( k \right)}}} \sum\limits_{k = 1}^{\#{L_A}\left( p \right)} {\left( {{p^{\left( k \right)}}{{\left( {{r_A}^{\left( k \right)}} \right)}^2}} \right)} - 2{{\left({\sum\limits_{k = 1}^{\# {L_A}\left( p \right)} {{r_A}^{\left( k \right)}{p^{\left( k \right)}}} } \right)}^2} - 8\tau\sum\limits_{k = 1}^{\# {L_A}\left( p \right)} {{p^{\left( k \right)}}} \sum\limits_{k = 1}^{\# {L_A}\left( p \right)} {{r_A}^{\left( k \right)}{p^{\left( k \right)}}} } \right)} \mathord{\left/{\vphantom {{\left( {\sum\limits_{k = 1}^{\# {L_A}\left( p \right)} {{p^{\left( k \right)}}} \sum\limits_{k = 1}^{\#{L_A}\left( p \right)} {\left( {{p^{\left( k \right)}}{{\left( {{r_A}^{\left( k \right)}} \right)}^2}} \right)} - 2{{\left({\sum\limits_{k = 1}^{\# {L_A}\left( p \right)} {{r_A}^{\left( k \right)}{p^{\left( k \right)}}} } \right)}^2} - 8\tau\sum\limits_{k = 1}^{\# {L_A}\left( p \right)} {{p^{\left( k \right)}}} \sum\limits_{k = 1}^{\# {L_A}\left( p \right)}{{r_A}^{\left( k \right)}{p^{\left( k \right)}}} } \right)} {\left( {8{\tau ^2}{{\left( {\sum\limits_{k = 1}^{\# {L_A}\left( p \right)} {{p^{\left( k \right)}}} } \right)}^3}} \right)}}} \right.\kern-\nulldelimiterspace} {\left( {8{\tau ^2}{{\left( {\sum\limits_{k = 1}^{\# {L_A}\left( p \right)} {{p^{\left( k \right)}}} }\right)}^3}} \right)}} \end{aligned}$

Let $R = \mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} \mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} \left( {p^{\left( k \right)} \left( {r_{A}^{\left( k \right)} } \right)^{2} } \right) - 2\left( {\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} r_{A}^{\left( k \right)} p^{\left( k \right)} } \right)^{2} - 8\tau \mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} \mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} r_{A}^{\left( k \right)} p^{\left( k \right)}$.

Then, $\partial R/\partial \mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} = \mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} \left( {p^{\left( k \right)} \left( {r_{A}^{\left( k \right)} } \right)^{2} } \right) - 8\tau \mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} r_{A}^{\left( k \right)} p^{\left( k \right)} = Q$.

As $\partial Q/\partial \mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} \left( {p^{\left( k \right)} \left( {r_{A}^{\left( k \right)} } \right)^{2} } \right) = 1 > 0$,

$Q$ is a monotonically increasing function for $\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} \left( {p^{\left( k \right)} \left( {r_{A}^{\left( k \right)} } \right)^{2} } \right)$.

As $p^{\left( k \right)} > 0$ and $\left( {r_{A}^{\left( k \right)} } \right)^{2} \ge 0$, then $\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} \left( {p^{\left( k \right)} \left( {r_{A}^{\left( k \right)} } \right)^{2} } \right) \ge 0$.

For $\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} \left( {p^{\left( k \right)} \left( {r_{A}^{\left( k \right)} } \right)^{2} } \right) = 0$, $r_{A}^{\left( k \right)} = 0$ holds for $\forall k$.

Therefore, $\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} r_{A}^{\left( k \right)} p^{\left( k \right)} = 0$ when $\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} \left( {p^{\left( k \right)} \left( {r_{A}^{\left( k \right)} } \right)^{2} } \right) = 0$.

Then, $Q = 0$ when $\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} \left( {p^{\left( k \right)} \left( {r_{A}^{\left( k \right)} } \right)^{2} } \right) = 0$. Hence, $Q \ge 0$.

Thus, $R$ is an increasing function for $\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)}$.

For $\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} = 0$, $p^{\left( k \right)} = 0$ holds for $\forall k$.

Therefore, $R = 0$ when $\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} = 0$.

As $\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} \in \left( {0,} \right.\left. 1 \right]$, then $R > 0$.

As $1/\left( {\ln 2 \cdot \left( {1 + \mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} } \right)} \right) > 0$ and $8\tau^{2} \left( {\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} } \right)^{3} > 0$.

Hence, $\partial F\left( {L_{A} \left( p \right)} \right)/\partial \mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} > 0$, which can be derived that the novel score function $F\left( {L_{A} \left( p \right)} \right)$ for $L_{A} \left( p \right)$ is an increasing function of parameter $\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)}$.

The proof of Theorem 1 (2) is completed.□

1.3 A.3 The proof of Theorem 1 (3)

Proof

According to Theorems 1 (1) and 1 (2),when $F\left( {L_{A} \left( p \right)} \right)$ reaches the maximum value, $\# L_{A} \left( p \right) = 1$ and $\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} = 1$.

Hence, it can be obtained that $(r_{A}^{\left( k \right)} )_{{{\text{max}}}} = \tau$, $\left( {E\left( {L_{A} \left( p \right)} \right)} \right)_{{{\text{max}}}} = 1$, $\sigma \left( {L_{A} \left( p \right)} \right) = 0$,$\log_{2} \left( {1 + \mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} } \right) = 1$, and $1 - 1/\left( {1 + \log_{2} \# L_{A} \left( p \right)} \right) = 0$.

Then, $F\left( {L_{A} \left( p \right)} \right)_{{{\text{max}}}} = 2$.

According to Theorems 1 (1) and 1 (2),when $F\left( {L_{A} \left( p \right)} \right)$ reaches the minimum value, $\# L_{A} \left( p \right)$ is infinitely close to $+ \infty$, and $\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)}$ is infinitely close to 0.

Hence, $\log_{2} \left( {1 + \mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} } \right)$ is infinitely close to 0, and $1 - 1/\left( {1 + \log_{2} \# L_{A} \left( p \right)} \right)$ is infinitely close to 1.

$$ E\left( {L_{A} \left( p \right)} \right) $$

$= \left( {\frac{1}{2\tau }\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} r_{A}^{\left( k \right)} p^{\left( k \right)} + \frac{1}{2}\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} } \right)/\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} = 1/2 + \left( {\frac{1}{2\tau }\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} r_{A}^{\left( k \right)} p^{\left( k \right)} } \right)/\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)}$.

When $\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)}$ is infinitely close to 0,

$\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)}$ and $\frac{1}{2\tau }\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} r_{A}^{\left( k \right)} p^{\left( k \right)}$ are both infinitely close to 0.

Then, expanding $E\left( {L_{A} \left( p \right)} \right)$ to a continuous function, according to L’Hospital’s rule,

$\left( {\frac{1}{2\tau }\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} r_{A}^{\left( k \right)} p^{\left( k \right)} } \right)/\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} = \left( {\frac{{\partial \left( {\frac{1}{2\tau }\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} r_{A}^{\left( k \right)} p^{\left( k \right)} } \right)}}{{\partial \left( {\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} } \right)}}} \right)/\left( {\frac{{\partial \left( {\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} } \right)}}{{\partial \left( {\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} } \right)}}} \right) = 0$.

Hence, when $\# L_{A} \left( p \right)$ is infinitely close to $+ \infty$ and $\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)}$ is infinitely close to 0, $E\left( {L_{A} \left( p \right)} \right)$ is infinitely close to $1/2$.

$$ \sigma \left( {L_{A} \left( p \right)} \right) $$

$= \mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} \left( {r_{A}^{\left( k \right)} } \right)^{2} /\left( {4\tau^{2} \mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} } \right) + \left( {\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} r_{A}^{\left( k \right)} p^{\left( k \right)} } \right)^{2} /\left( {4\tau^{2} \left( {\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} } \right)^{2} } \right) - \left( {\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} r_{A}^{\left( k \right)} p^{\left( k \right)} } \right)^{2} /\left( {2\tau^{2} } \right)$.

When $\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)}$ is infinitely close to 0,

$\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)}$, $\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} \left( {r_{A}^{\left( k \right)} } \right)^{2}$, and $\left( {\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} r_{A}^{\left( k \right)} } \right)^{2}$ are all infinitely close to 0.

Then, expanding $\sigma \left( {L_{A} \left( p \right)} \right)$ to a continuous function, according to L’Hospital’s rule, $\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} \left( {r_{A}^{\left( k \right)} } \right)^{2} /\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} = \left( {\frac{{\partial \left( {\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} \left( {r_{A}^{\left( k \right)} } \right)^{2} } \right)}}{{\partial \left( {\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} } \right)}}} \right)/\left( {\frac{{\partial \left( {\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} } \right)}}{{\partial \left( {\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} } \right)}}} \right) = 0$and $\left( {\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} r_{A}^{\left( k \right)} p^{\left( k \right)} } \right)^{2} /\left( {\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} } \right)^{2} = \left( {\frac{{\partial \left( {\left( {\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} r_{A}^{\left( k \right)} p^{\left( k \right)} } \right)^{2} } \right)}}{{\partial \left( {\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} } \right)}}} \right)/\left( {\frac{{\partial \left( {\left( {\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} } \right)^{2} } \right)}}{{\partial \left( {\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)} } \right)}}} \right) = 0$.

Hence, when $\# L_{A} \left( p \right)$ is infinitely close to $+ \infty$ and $\mathop \sum \nolimits_{k = 1}^{{\# L_{A} \left( p \right)}} p^{\left( k \right)}$ is infinitely close to 0, $\sigma \left( {L_{A} \left( p \right)} \right)$ is infinitely close to $0$.

Then, when $F\left( {L_{A} \left( p \right)} \right)$ reaches the minimum value, $F\left( {L_{A} \left( p \right)} \right)$ is infinitely close to $0$.

Thus, $F\left( {L_{A} \left( p \right)} \right) \in \left( {0,\left. 2 \right]} \right.$ holds.

The proof of Theorem 1 (3) is completed. □