New ranking model with evidence theory under probabilistic hesitant fuzzy context and unknown weights

Abstract

This paper proposes a novel ranking model under probabilistic hesitant fuzzy set (PHFS) by extending the idea of evidence theory (ET). PHFS is a strong variant of hesitant fuzzy set that associates occurrence probabilities to multiple membership grades. This offers flexibility to experts during preference elicitation and aids proper handling of uncertainty. Evidence theory is also a powerful concept for managing uncertainty/hesitation. By integrating Bayesian approximation with ET, a flexible ranking model is developed that complements ET. Due to the partial availability of evidences for decision-making under uncertainty, an approximation strategy is combined. Previous studies on PHFS have not handled hesitation of experts better and to alleviate the issue, a new regret-rejoice approach is put forward that calculates weights of criteria by handling hesitation efficiently. Ranking values from each expert are obtained that are further fused by the Maclaurin operator to get the final ranking order. These approaches are integrated into a framework and its usefulness is exemplified using renewable energy technology selection for Tamil Nadu. Finally, comparative analysis with other approaches reveals the strengths and weaknesses of the proposed work.

Appendix

Proof of Theorem 1: (Idempotent)

$$ \begin{aligned} RV_{i}^{{\left( {g,\lambda _{1} ,\lambda _{2} , \ldots ,\lambda _{g} } \right)}} = & \left( {\left( {1 - \left( {\left( {1 - \prod\limits_{{ll = 1}}^{g} {BAX_{{lj}}^{{\lambda _{{ll}} }} } } \right)^{{ewt^{l} }} } \right)} \right)} \right)^{{\frac{1}{{\sum\limits_{{ll}} {\lambda _{{ll}} } }}}} \\ = & \left( {\prod\limits_{{ll = 1}}^{g} {BAX_{{lj}}^{{\lambda _{{ll}} }} } } \right)^{{\frac{1}{{\sum\limits_{{ll}} {\lambda _{{ll}} } }}}} = BAXr \\ \end{aligned} $$

By expanding, we get

(Commutative)

$$ \begin{aligned} RV_{i}^{{\left( {g,\lambda _{1} ,\lambda _{2} , \ldots ,\lambda _{g} } \right)}} \left( {BAX_{l}^{'} } \right) = & \left( {\left( {1 - \left( {\left( {1 - \mathop \prod \limits_{{ll = 1}}^{g} BAX_{{lj}}^{{'\lambda _{{ll}} }} } \right)^{{ewt^{l} }} } \right)} \right)} \right)^{{\frac{1}{{\mathop \sum \nolimits_{{ll}} \lambda _{{ll}} }}}} \\ = & \left( {\left( {1 - \left( {\left( {1 - \prod\limits_{{ll = 1}}^{g} {BAX_{{lj}}^{{\lambda _{{ll}} }} } } \right)^{{ewt^{l} }} } \right)} \right)} \right)^{{\frac{1}{{\sum\limits_{{ll}} {\lambda _{{ll}} } }}}} = RV_{i}^{{\left( {g,\lambda _{1} ,\lambda _{2} , \ldots ,\lambda _{g} } \right)}} (BAX_{l} ) \\ \end{aligned} $$

Since ${BAX}_{l}^{^{\prime}}$ is any permutation of ${BAX}_{l}$, we get

(Monotonicity).

It is given that $BAX_{'}^{{''}}$ is greater than or equal ${BAX}_{l}$ and hence, score of $BAX_{'}^{{''}}$ is greater than or equal to the score of ${BAX}_{l}$ for all $l=\mathrm{1,2},\dots ,s$. In case of equal scores, deviation is determined and it can be seen that deviation of $BAX_{'}^{{''}}$ is less than the deviation of ${BAX}_{l}$ for all $l=\mathrm{1,2},\dots s$. The formula for score $s$ and deviation $d$ are adapted from [61]. Without loss of generality, it is also true that $s(BAX^{{''}} ) \ge s(BAX)$ and when scores are equal, deviation is calculated and $d\left( {BAX^{{''}} } \right) < d\left( {BAX} \right)$. Thus,

$$ \begin{aligned} RV_{i}^{{\left( {g,\lambda _{1} ,\lambda _{2} , \ldots ,\lambda _{g} } \right)}} \left( {BAX_{l} } \right) = & \left( {\left( {1 - \left( {\left( {1 - \mathop \prod \limits_{{ll = 1}}^{g} BAX_{{lj}}^{{\lambda _{{ll}} }} } \right)^{{ewt^{l} }} } \right)} \right)} \right)^{{\frac{1}{{\mathop \sum \nolimits_{{ll}} \lambda _{{ll}} }}}} \le RV_{i}^{{\left( {g,\lambda _{1} ,\lambda _{2} , \ldots ,\lambda _{g} } \right)}} \left( {BAX_{l}^{{''}} } \right)\_ \\ = & \left( {\left( {1 - \left( {\left( {1 - \mathop \prod \limits_{{ll = 1}}^{g} BAX_{{lj}}^{{''\lambda _{{ll}} }} } \right)^{{ewt^{l} }} } \right)} \right)} \right)^{{\frac{1}{{\mathop \sum \nolimits_{{ll}} \lambda _{{ll}} }}}} \\ \end{aligned} $$

(Bounded).

Based on the monotonicity and idempotent properties, it is clear that ${RV}_{i}^{\left(g,{\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{g}\right)}({BAX}_{l})>{RV}_{i}^{\left(g,{\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{g}\right)}\left({BAX}^{-},{BAX}^{-},\dots .,{BAX}^{-}\right)$ and

$${RV}_{i}^{\left(g,{\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{g}\right)}({BAX}_{l})\le {RV}_{i}^{\left(g,{\lambda }_{1},{\lambda }_{2},\dots ,{\lambda }_{g}\right)}\left({BAX}^{+},{BAX}^{+},\dots .,{BAX}^{+}\right)$$

Proof of Theorem 2:

Let us consider ${v}_{lj}\left(hf\right)={v}_{lj}=\frac{{\sigma }_{j}^{l}}{\sum_{j}{\sigma }_{j}^{l}}$ for proof purpose.

Given that ${v}_{lj}(0)=0$ and ${v}_{lj}(1)=1$. Based on Eq. (17), we get values in the unit interval. Also, ${SE}_{l}\left(0\right)={SE}_{l}\left(1\right)=0$. Thus, ${SE}_{l}\left(0(1)\right)={SE}_{l}\left(1(1)\right)={SE}_{l}\left(0\left(p\right),1(1-p)\right)=0$.

Given that ${v}_{lj}(0.5)=0.5$ and according to the binary entropic, ${ASE}_{l}\left(1(1)\right)=\mathrm{max}({SE}_{l})$.

If ${\gamma }_{lj}^{(1)}\le {\gamma }_{lj}^{\left(2\right)}\le 0.5$, ${v}_{i}^{(1)}\le {v}_{i}^{\left(2\right)}\le 0.5$. From the formulation, it is evident that ${v}_{lj}^{(1)}ln{v}_{lj}^{(1)}\le {v}_{lj}^{(2)}ln{v}_{lj}^{(3)}$ and hence, ${SE}_{l}\left({hf}^{(1)}\right)\le {SE}_{l}\left({hf}^{(2)}\right)$. Similarly, we can also be prove in a similar fashion.

By binary entropic measure, it is observe that

$$ \begin{aligned} SE_{l} \left( {hf\left( {p,1 - p} \right)} \right) = & - \sum {v_{{lj}} } ln(v_{{lj}} ) + ( - \sum (1 - v_{{lj}} )ln(1 - v_{{lj}} )) - \sum {(1 - v_{{lj}} )} ln(1 - v_{{lj}} ) \\ + ( - \sum 1 - (1 - v_{{lj}} )ln1 - (1 - v_{{lj}} ) = ASE_{l} \left( {hf(p)^{{(c)}} } \right). \\ \end{aligned} $$

By taking complement, we get . Thus, ${ASE}_{l}\left(hf\right)={ASE}_{l}\left({hf}^{c}\right)$. All four axioms are satisfied. Axioms (P2) and (P4) adopt binary entropic measure formulation for deriving efficient proofs