A probabilistic linguistic TODIM method considering cumulative probability-based Hellinger distance and its application in waste mobile phone recycling

Abstract

The prosperity of online mobile phone recycling platforms has brought convenience to waste mobile phone recycling. This study aims to propose a novel decision-making method to evaluate the recycling channels of mobile phones in an uncertain environment using the probabilistic linguistic term set (PLTS) to represent uncertain information. Given that each PLTS is associated with a probability distribution, the differences of different PLTSs can be represented by comparing different probability distributions. The Hellinger distance is a widely-used index to measure the similarity of two probability distributions, but it cannot be directly used to express the semantic differences of PLTSs. In this regard, we introduce a cumulative probability-based Hellinger distance measure of PLTSs. Then, we propose an extension of TODIM (an acronym in Portuguese of Interactive and Multicriteria Decision Making) method considering the bounded rationality and psychology of decision-makers based on the proposed distance measure of PLTSs and a criterion weight determination method, the Simos-Roy-Figueira method. Finally, an illustrative example of evaluating the recycling channels of mobile phones is given with sensitive and comparative analyses, showing the efficiency of the proposed method. This method can also be applied to other various fields.

Appendix

1.1 The Proof of Theorem 1

Proof: (1) $CHd\left({l}_2,{l}_1\right)=\sqrt{\sum \limits_{t=-\tau}^{\tau }{\left(\sqrt{f_t^{(2)}}-\sqrt{f_t^{(1)}}\right)}^2}=\sqrt{\sum \limits_{t=-\tau}^{\tau }{\left(\sqrt{f_t^{(1)}}-\sqrt{f_t^{(2)}}\right)}^2}= CHd\left({l}_1,{l}_2\right)$.

(2) $CHd\left({l}_1,{l}_1\right)=\sqrt{\sum \limits_{t=-\tau}^{\tau }{\left(\sqrt{f_t^{(1)}}-\sqrt{f_t^{(1)}}\right)}^2}=0$.

(3) Let $CHd\left({l}_1,{l}_2\right)=\sqrt{\sum \limits_{t=-\tau}^{\tau }{\left(\sqrt{f_t^{(1)}}-\sqrt{f_t^{(2)}}\right)}^2}$, $CHd\left({l}_2,{l}_3\right)=\sqrt{\sum \limits_{t=-\tau}^{\tau }{\left(\sqrt{f_t^{(2)}}-\sqrt{f_t^{(3)}}\right)}^2}$, and $CHd\left({l}_1,{l}_3\right)=\sqrt{\sum \limits_{t=-\tau}^{\tau }{\left(\sqrt{f_t^{(1)}}-\sqrt{f_t^{(3)}}\right)}^2}$. Because CHd(l₁, l₂) ≥ 0, CHd(l₁, l₃) + CHd(l₃, l₂) ≥ CHd(l₁, l₂)⇔ (CHd(l₁, l₃) + CHd(l₃, l₂))² ≥ CHd²(l₁, l₂). That is

$$ {\left(\sqrt{\sum \limits_{t=-\tau}^{\tau }{\left(\sqrt{f_t^{(1)}}-\sqrt{f_t^{(3)}}\right)}^2}+\sqrt{\sum \limits_{t=-\tau}^{\tau }{\left(\sqrt{f_t^{(3)}}-\sqrt{f_t^{(2)}}\right)}^2}\right)}^2\ge {\left(\sqrt{\sum \limits_{t=-\tau}^{\tau }{\left(\sqrt{f_t^{(1)}}-\sqrt{f_t^{(2)}}\right)}^2}\right)}^2 $$

(13)

Thus, we need to prove

$$ \sum \limits_{t=-\tau}^{\tau }{\left(\sqrt{f_t^{(1)}}-\sqrt{f_t^{(3)}}\right)}^2+\sum \limits_{t=-\tau}^{\tau }{\left(\sqrt{f_t^{(3)}}-\sqrt{f_t^{(2)}}\right)}^2\ge \sum \limits_{t=-\tau}^{\tau }{\left(\sqrt{f_t^{(1)}}-\sqrt{f_t^{(2)}}\right)}^2 $$

(14)

If Inequality (14) is held, Inequality (13) is necessary to be held. Inequality (14) can be simplified as:

$$ {\displaystyle \begin{array}{c}\sum \limits_{t=-\tau}^{\tau }{\left(\sqrt{f_t^{(1)}}-\sqrt{f_t^{(3)}}\right)}^2+\sum \limits_{t=-\tau}^{\tau }{\left(\sqrt{f_t^{(3)}}-\sqrt{f_t^{(2)}}\right)}^2-\sum \limits_{t=-\tau}^{\tau }{\left(\sqrt{f_t^{(1)}}-\sqrt{f_t^{(2)}}\right)}^2=2\sum \limits_{t=-\tau}^{\tau}\left(\sqrt{f_t^{(3)}}-\sqrt{f_t^{(1)}}\right)\left(\sqrt{f_t^{(3)}}-\sqrt{f_t^{(2)}}\right)\\ {}=2\sum \limits_{t=-\tau}^{\tau}\left(\ln \left(\frac{e^{\sqrt{f_t^{(3)}}}}{e^{\sqrt{f_t^{(1)}}}}\right)+\ln \left(\frac{e^{\sqrt{f_t^{(3)}}}}{e^{\sqrt{f_t^{(2)}}}}\right)\right)\ge 0\end{array}} $$

According to the information entropy inequality$\sum \limits_{t=-\tau}^{\tau }{w}_i\ln {w}_i\ge \sum \limits_{t=-\tau}^{\tau }{w}_i\ln {p}_i\ln \left(\frac{e^{\sqrt{f_t^{(2)}}}}{e^{\sqrt{f_t^{(3)}}}}\right)$, we can obtain that $\sum \limits_{t=-\tau}^{\tau }{e}^{\sqrt{f_t^{(1)}}}\ln \left(\frac{e^{\sqrt{f_t^{(3)}}}}{e^{\sqrt{f_t^{(1)}}}}\right)\ge 0$, $\sum \limits_{t=-\tau}^{\tau }{e}^{\sqrt{f_t^{(3)}}}\ln \left(\frac{e^{\sqrt{f_t^{(3)}}}}{e^{\sqrt{f_t^{(2)}}}}\right)\ge 0$. Because ${e}^{\sqrt{f_t^{(1)}}}$, $\sum \limits_{t=-\tau}^{\tau}\ln \left(\frac{e^{\sqrt{f_t^{(3)}}}}{e^{\sqrt{f_t^{(1)}}}}\right)\ge 0$, $\sum \limits_{t=-\tau}^{\tau}\ln \left(\frac{e^{\sqrt{f_t^{(3)}}}}{e^{\sqrt{f_t^{(2)}}}}\right)\ge 0$, it follows $\sum \limits_{t=-\tau}^{\tau}\ln \left(\left(\frac{e^{\sqrt{f_t^{(3)}}}}{e^{\sqrt{f_t^{(1)}}}}\right)+\left(\frac{e^{\sqrt{f_t^{(3)}}}}{e^{\sqrt{f_t^{(2)}}}}\right)\right)\ge 0$. That is $\sum \limits_{t=-\tau}^{\tau }{\left(\sqrt{f_t^{(1)}}-\sqrt{f_t^{(3)}}\right)}^2+\sum \limits_{t=-\tau}^{\tau }{\left(\sqrt{f_t^{(3)}}-\sqrt{f_t^{(2)}}\right)}^2-\sum \limits_{t=-\tau}^{\tau }{\left(\sqrt{f_t^{(1)}}-\sqrt{f_t^{(2)}}\right)}^2\ge 0$. Therefore, the triangle inequality is held.