A Likelihood-Based Qualitative Flexible Approach with Hesitant Fuzzy Linguistic Information

Abstract

The qualitative flexible multiple criteria method (QUALIFLEX) is a useful outranking method for multi-criteria decision analysis due to its flexibility in regard to cardinal and ordinal information. This paper puts forward an extended QUALIFLEX approach with a new likelihood-based comparison method to address multi-criteria decision-making problems in a hesitant fuzzy linguistic environment. The rankings produced by our new comparison method are more convincing than those obtained by existing methods, such as likelihood, distance measures, and the score function of hesitant fuzzy linguistic term sets or hesitant fuzzy linguistic elements. The proposed QUALIFLEX model, which is based on the likelihood-based comparison method, can measure the level of concordance or discordance of the complete preference order for tackling multi-criteria decision-making problems. Finally, two cases are presented as a comparative analysis between the proposed approach and other related methods. This example demonstrates the effectiveness and flexibility of the proposed methodology in the context of hesitant fuzzy linguistic information.

Appendix

Proof of Property 1.

Proof

Only (4) of Property 1 will be proven; the proofs of the other properties are omitted.

• a. If $H_{S}^{a + } < H_{S}^{b - }$ or $H_{S}^{b + } < H_{S}^{a - }$, then Definition 7 reveals that $L(H_{S}^{a} \ge H_{S}^{b} ) + L(H_{S}^{a} \le H_{S}^{b} ) = 1$.

• b. If $H_{S}^{a - } \le H_{S}^{b + } ,\;H_{S}^{b - } \le H_{S}^{a + } \;{\text{and}}\;H_{S}^{a - } = H_{S}^{b - } = s_{0}$, then Definition 7 reveals the following:

  $$ \begin{aligned} L(H_{S}^{a} \ge H_{S}^{b} ) & = \frac{1}{{l_{a} l_{b} }}\left( {\sum\limits_{i = 2}^{{l_{a} }} {\sum\limits_{j = 1}^{{l_{b} }} {\frac{{I(a_{i} )}}{{I(a_{i} ) + I(b_{j} )}} + \frac{1}{2}} } } \right) \\ & = \frac{1}{{l_{a} l_{b} }}\left( {\frac{{I(a_{2} )}}{{I(a_{2} ) + I(b_{1} )}} + \frac{{I(a_{2} )}}{{I(a_{2} ) + I(b_{2} )}} + \cdots + \frac{{I(a_{2} )}}{{I(a_{2} ) + I(b_{{l_{b} }} )}}} \right. \\ & \quad + \frac{{I(a_{3} )}}{{I(a_{3} ) + I(b_{1} )}} + \frac{{I(a_{3} )}}{{I(a_{3} ) + I(b_{2} )}} + \cdots + \frac{{I(a_{3} )}}{{I(a_{3} ) + I(b_{{l_{b} }} )}} + \cdots \\ & \quad + \frac{{I(a_{{l_{a} }} )}}{{I(a_{{l_{a} }} ) + I(b_{1} )}} + \frac{{I(a_{{l_{a} }} )}}{{I(a_{{l_{a} }} ) + I(b_{2} )}} + \cdots + \frac{{I(a_{{l_{a} }} )}}{{I(a_{{l_{a} }} ) + I(b_{{l_{b} }} )}} + \left. {\frac{1}{2}} \right). \\ \end{aligned} $$

  (14)
  $$ \begin{aligned} L(H_{S}^{a} \le H_{S}^{b} ) & = L(H_{S}^{b} \ge H_{S}^{a} ) = \frac{1}{{l_{a} l_{b} }}\left( {\sum\limits_{j = 2}^{{l_{b} }} {\sum\limits_{i = 1}^{{l_{a} }} {\frac{{I(b_{j} )}}{{I(b_{j} ) + I(a_{i} )}} + \frac{1}{2}} } } \right) \\ & = \frac{1}{{l_{a} l_{b} }}\left( {\frac{{I(b_{2} )}}{{I(b_{2} ) + I(a_{1} )}} + \frac{{I(b_{2} )}}{{I(b_{2} ) + I(a_{2} )}} + \cdots + \frac{{I(b_{2} )}}{{I(b_{2} ) + I(a_{{l_{a} }} )}}} \right. \\ & \quad + \frac{{I(b_{3} )}}{{I(b_{3} ) + I(a_{1} )}} + \frac{{I(b_{3} )}}{{I(b_{3} ) + I(a_{2} )}} + \cdots + \frac{{I(b_{3} )}}{{I(b_{3} ) + I(a_{{l_{a} }} )}} + \cdots \\ & \quad + \frac{{I(b_{{l_{b} }} )}}{{I(b_{{l_{b} }} ) + I(a_{1} )}} + \frac{{I(b_{{l_{b} }} )}}{{I(b_{{l_{b} }} ) + I(a_{2} )}} + \cdots + \frac{{I(b_{{l_{b} }} )}}{{I(b_{{l_{b} }} ) + I(a_{{l_{a} }} )}} + \left. {\frac{1}{2}} \right). \\ \end{aligned} $$

  (15)

By combining Eqs. (14) and (15), Eq. (16) can be obtained:

$$ \begin{aligned} & L(H_{S}^{a} \ge H_{S}^{b} ) + L(H_{S}^{a} \le H_{S}^{b} ) \\ & \quad = \frac{1}{{l_{a} l_{b} }}\left( {\frac{{I(a_{2} )}}{{I(a_{2} ) + I(b_{2} )}} + \frac{{I(b_{2} )}}{{I(b_{2} ) + I(a_{2} )}} + \frac{{I(a_{2} )}}{{I(a_{2} ) + I(b_{3} )}} + \frac{{I(b_{3} )}}{{I(b_{3} ) + I(a_{2} )}} + \cdots } \right. \\ & \quad \quad + \frac{{I(a_{2} )}}{{I(a_{2} ) + I(b_{{l_{b} }} )}} + \frac{{I(b_{{l_{b} }} )}}{{I(b_{{l_{b} }} ) + I(a_{2} )}} \\ & \quad \quad + \frac{{I(a_{3} )}}{{I(a_{3} ) + I(b_{2} )}} + \frac{{I(b_{2} )}}{{I(b_{2} ) + I(a_{3} )}} + \frac{{I(a_{3} )}}{{I(a_{3} ) + I(b_{3} )}} + \frac{{I(b_{3} )}}{{I(b_{3} ) + I(a_{3} )}} + \cdots \\ & \quad \quad + \frac{{I(a_{3} )}}{{I(a_{3} ) + I(b_{{l_{b} }} )}} + \frac{{I(b_{{l_{b} }} )}}{{I(b_{{l_{b} }} ) + I(a_{3} )}} + \cdots \\ & \quad \quad + \frac{{I(a_{{l_{a} }} )}}{{I(a_{{l_{a} }} ) + I(b_{2} )}} + \frac{{I(b_{2} )}}{{I(b_{2} ) + I(a_{{l_{a} }} )}} + \frac{{I(a_{{l_{a} }} )}}{{I(a_{{l_{a} }} ) + I(b_{3} )}} + \frac{{I(b_{3} )}}{{I(b_{3} ) + I(a_{{l_{a} }} )}} + \cdots \\ & \quad \quad + \frac{{I(a_{{l_{a} }} )}}{{I(a_{{l_{a} }} ) + I(b_{{l_{b} }} )}} + \frac{{I(b_{{l_{b} }} )}}{{I(b_{{l_{b} }} ) + I(a_{{l_{a} }} )}} \\ & \quad \quad + \frac{{I(a_{2} )}}{{I(a_{2} ) + I(b_{1} )}} + \frac{{I(a_{3} )}}{{I(a_{3} ) + I(b_{1} )}} + \cdots + \frac{{I(a_{{l_{a} }} )}}{{I(a_{{l_{a} }} ) + I(b_{1} )}} \\ & \quad \quad + \frac{{I(b_{2} )}}{{I(b_{2} ) + I(a_{1} )}} + \frac{{I(b_{3} )}}{{I(b_{3} ) + I(a_{1} )}} + \cdots + \frac{{I(b_{{l_{b} }} )}}{{I(b_{{l_{b} }} ) + I(a_{1} )}} + \left. {\frac{1}{2} + \frac{1}{2}} \right) & \\ & \quad = \frac{1}{{l_{a} l_{b} }}\left( {(l_{a} - 1)(l_{b} - 1) + (l_{a} - 1) + (l_{b} - 1) + 1} \right) = 1. \\ \end{aligned} $$

(16)

• c. If $H_{S}^{a - } \le H_{S}^{b + } ,\;H_{S}^{b - } \le H_{S}^{a + } \;{\text{and}}\;H_{S}^{a - } \ne s_{0} \;{\text{or}}\;H_{S}^{b - } \ne s_{0}$, then, similar to Eqs. (14) and (15):

  $$ L(H_{S}^{a} \ge H_{S}^{b} ) + L(H_{S}^{a} \le H_{S}^{b} ) = \frac{1}{{l_{a} l_{b} }}(l_{a} l_{b} ) = 1. $$

  (17)

Therefore, $L(H_{S}^{a} \ge H_{S}^{b} ) + L(H_{S}^{a} \le H_{S}^{b} ) = 1$.

The proof is now complete.