A linguistic intuitionistic multi-criteria decision-making method based on the Frank Heronian mean operator and its application in evaluating coal mine safety

Abstract

Coal mine safety has been a pressing issue for many years, and it is a constant and non-negligible problem that must be addressed during any coal mining process. This paper focuses on developing an innovative multi-criteria decision-making (MCDM) method to address coal mine safety evaluation problems. Because lots of uncertain and fuzzy information exists in the process of evaluating coal mine safety, linguistic intuitionistic fuzzy numbers (LIFNs) are introduced to depict the evaluation information necessary to the process. Furthermore, the handling of qualitative information requires the effective support of quantitative tools, and the linguistic scale function (LSF) is therefore employed to deal with linguistic intuitionistic information. First, the distance, a valid ranking method, and Frank operations are proposed for LIFNs. Subsequently, the linguistic intuitionistic fuzzy Frank improved weighted Heronian mean (LIFFIWHM) operator is developed. Then, a linguistic intuitionistic MCDM method for coal mine safety evaluation is constructed based on the developed operator. Finally, an illustrative example is provided to demonstrate the proposed method, and its feasibility and validity are further verified by a sensitivity analysis and comparison with other existing methods.

Appendix

Proof of Theorem.

In the following, Theorem 1 will be proved utilizing the mathematical induction on $n$.

Proof

Firstly, the following equation needs to be proved.

$$\begin{gathered} \mathop { \oplus _{F} }\limits_{{i = 1,j = i}}^{n} w_{i} w_{j} \cdot _{F} ((\phi _{i} )^{{\wedge_{F}} p} \otimes _{F} (\phi _{j} )^{{\wedge_{F} q}} ) = \left( {f^{{* - 1}} \left( {1 - \log _{\lambda } \left( {1 + (\lambda - 1)\prod\limits_{{i = 1,j = i}}^{n} {\left( {\frac{{(\lambda - 1)^{{p + q}} - (\lambda ^{{f^{*} (s_{{u_{i} }} )}} - 1)^{p} (\lambda ^{{f^{*} (s_{{u_{j} }} )}} - 1)^{q} }}{{(\lambda - 1)^{{p + q}} + (\lambda - 1)(\lambda ^{{f^{*} (s_{{u_{i} }} )}} - 1)^{p} (\lambda ^{{f^{*} (s_{{u_{j} }} )}} - 1)^{q} }}} \right)^{{w_{i} w_{j} }} } } \right)} \right)} \right., \hfill \\ {\kern 1pt} \left. {f^{{* - 1}} \left( {\log _{\lambda } \left( {1 + (\lambda - 1)\prod\limits_{{i = 1,j = i}}^{n} {\left( {\frac{{(\lambda - 1)^{{p + q}} - (\lambda ^{{1 - f^{*} (s_{{v_{i} }} )}} - 1)^{p} (\lambda ^{{1 - f^{*} (s_{{v_{j} }} )}} - 1)^{q} }}{{(\lambda - 1)^{{p + q}} + (\lambda - 1)(\lambda ^{{1 - f^{*} (s_{{v_{i} }} )}} - 1)^{p} (\lambda ^{{1 - f^{*} (s_{{v_{j} }} )}} - 1)^{q} }}} \right)^{{w_{i} w_{j} }} } } \right)} \right)} \right). \hfill \\ \end{gathered}$$

(12)

• 1. For n = 2, the following equation can be calculated easily.

$$\begin{gathered} \mathop { \oplus _{F} }\limits_{{i = 1,j = i}}^{2} w_{i} w_{j} \cdot _{F} ((\phi _{i} )^{{\wedge_{F} p}} \otimes _{F} (\phi _{j} )^{{\wedge_{F} q}} ) = \left( {f^{{* - 1}} \left( {1 - \log _{\lambda } \left( {1 + (\lambda - 1)\prod\limits_{{i = 1,j = i}}^{2} {\left( {\frac{{(\lambda - 1)^{{p + q}} - (\lambda ^{{f^{*} (s_{{u_{i} }} )}} - 1)^{p} (\lambda ^{{f^{*} (s_{{u_{j} }} )}} - 1)^{q} }}{{(\lambda - 1)^{{p + q}} + (\lambda - 1)(\lambda ^{{f^{*} (s_{{u_{i} }} )}} - 1)^{p} (\lambda ^{{f^{*} (s_{{u_{j} }} )}} - 1)^{q} }}} \right)^{{w_{i} w_{j} }} } } \right)} \right)} \right., \hfill \\ {\kern 1pt} \left. {f^{{* - 1}} \left( {\log _{\lambda } \left( {1 + (\lambda - 1)\prod\limits_{{i = 1,j = i}}^{2} {\left( {\frac{{(\lambda - 1)^{{p + q}} - (\lambda ^{{1 - f^{*} (s_{{v_{i} }} )}} - 1)^{p} (\lambda ^{{1 - f^{*} (s_{{v_{j} }} )}} - 1)^{q} }}{{(\lambda - 1)^{{p + q}} + (\lambda - 1)(\lambda ^{{1 - f^{*} (s_{{v_{i} }} )}} - 1)^{p} (\lambda ^{{1 - f^{*} (s_{{v_{j} }} )}} - 1)^{q} }}} \right)^{{w_{i} w_{j} }} } } \right)} \right)} \right). \hfill \\ \end{gathered}$$

• 2. If Eq. (11) holds for $n=k$, there is

$$\begin{gathered} \mathop { \oplus _{F} }\limits_{{i = 1,j = i}}^{k} w_{i} w_{j} \cdot _{F} ((\phi _{i} )^{{\wedge_{F} p}} \otimes _{F} (\phi _{j} )^{{\wedge_{F} q}} ) = \left( {f^{{* - 1}} \left( {1 - \log _{\lambda } \left( {1 + (\lambda - 1)\prod\limits_{{i = 1,j = i}}^{k} {\left( {\frac{{(\lambda - 1)^{{p + q}} - (\lambda ^{{f^{*} (s_{{u_{i} }} )}} - 1)^{p} (\lambda ^{{f^{*} (s_{{u_{j} }} )}} - 1)^{q} }}{{(\lambda - 1)^{{p + q}} + (\lambda - 1)(\lambda ^{{f^{*} (s_{{u_{i} }} )}} - 1)^{p} (\lambda ^{{f^{*} (s_{{u_{j} }} )}} - 1)^{q} }}} \right)^{{w_{i} w_{j} }} } } \right)} \right)} \right., \hfill \\ {\kern 1pt} \left. {f^{{* - 1}} \left( {\log _{\lambda } \left( {1 + (\lambda - 1)\prod\limits_{{i = 1,j = i}}^{k} {\left( {\frac{{(\lambda - 1)^{{p + q}} - (\lambda ^{{1 - f^{*} (s_{{v_{i} }} )}} - 1)^{p} (\lambda ^{{1 - f^{*} (s_{{v_{j} }} )}} - 1)^{q} }}{{(\lambda - 1)^{{p + q}} + (\lambda - 1)(\lambda ^{{1 - f^{*} (s_{{v_{i} }} )}} - 1)^{p} (\lambda ^{{1 - f^{*} (s_{{v_{j} }} )}} - 1)^{q} }}} \right)^{{w_{i} w_{j} }} } } \right)} \right)} \right). \hfill \\ \end{gathered}$$

(13)

Then, when $n=k+1$, the following equation can be obtained

$$\underset{i=1,j=i}{\overset{k+1}{\mathop{{{\oplus }_{F}}}}}\,{{w}_{i}}{{w}_{j}}{{\cdot }_{F}}({{({{\phi }_{i}})}^{{{\hat{\ }}_{F}}p}}{{\otimes }_{F}}{{({{\phi }_{j}})}^{{{\hat{\ }}_{F}}q}})=\underset{i=1,j=i}{\overset{k}{\mathop{{{\oplus }_{F}}}}}\,{{w}_{i}}{{w}_{j}}{{\cdot }_{F}}({{({{\phi }_{i}})}^{{{\hat{\ }}_{F}}p}}{{\otimes }_{F}}{{({{\phi }_{j}})}^{{{\hat{\ }}_{F}}q}}){{\oplus }_{F}}\underset{i=1}{\overset{k+1}{\mathop{{{\oplus }_{F}}}}}\,{{w}_{i}}{{w}_{k+1}}{{\cdot }_{F}}({{({{\phi }_{i}})}^{{{\hat{\ }}_{F}}p}}{{\otimes }_{F}}{{({{\phi }_{k+1}})}^{{{\hat{\ }}_{F}}q}}).$$

(14)

According to the operations of LIFNs, the following result can be calculated.

$$\begin{gathered} \mathop { \oplus _{F} }\limits_{{i = 1}}^{{k + 1}} w_{i} w_{{k + 1}} \cdot _{F} ((\phi _{i} )^{{\wedge_{F} p}} \otimes _{F} (\phi _{{k + 1}} )^{{\wedge_{F} q}} ) = \left( {f^{{* - 1}} \left( {1 - \log _{\lambda } \left( {1 + (\lambda - 1)\prod\limits_{{i = 1}}^{{k + 1}} {\left( {\frac{{(\lambda - 1)^{{p + q}} - (\lambda ^{{f^{*} (s_{{u_{i} }} )}} - 1)^{p} (\lambda ^{{f^{*} (s_{{u_{{k + 1}} }} )}} - 1)^{q} }}{{(\lambda - 1)^{{p + q}} + (\lambda - 1)(\lambda ^{{f^{*} (s_{{u_{i} }} )}} - 1)^{p} (\lambda ^{{f^{*} (s_{{u_{{k + 1}} }} )}} - 1)^{q} }}} \right)^{{w_{i} w_{{k + 1}} }} } } \right)} \right)} \right., \hfill \\ {\kern 1pt} \left. {f^{{* - 1}} \left( {\log _{\lambda } \left( {1 + (\lambda - 1)\prod\limits_{{i = 1}}^{{k + 1}} {\left( {\frac{{(\lambda - 1)^{{p + q}} - (\lambda ^{{1 - f^{*} (s_{{v_{i} }} )}} - 1)^{p} (\lambda ^{{f^{*} 1 - (s_{{v_{{k + 1}} }} )}} - 1)^{q} }}{{(\lambda - 1)^{{p + q}} + (\lambda - 1)(\lambda ^{{1 - f^{*} (s_{{v_{i} }} )}} - 1)^{p} (\lambda ^{{f^{*} 1 - (s_{{v_{{k + 1}} }} )}} - 1)^{q} }}} \right)^{{w_{i} w_{{k + 1}} }} } } \right)} \right)} \right). \hfill \\ \end{gathered}$$

(15)

Equation (15) can be easily proved by utilizing the mathematical induction on $k+1$, and the proof is omitted here.

Thus, by utilizing Eqs. (13) and (15), Eq. (14) can be converted into

$$\begin{gathered} \mathop { \oplus _{F} }\limits_{{i = 1,j = i}}^{{k + 1}} w_{i} w_{j} \cdot _{F} ((\phi _{i} )^{{\wedge_{F} p}} \otimes _{F} (\phi _{j} )^{{\wedge_{F} q}} ) = \mathop { \oplus _{F} }\limits_{{i = 1,j = i}}^{k} w_{i} w_{j} \cdot _{F} ((\phi _{i} )^{{\wedge_{F} p}} \otimes _{F} (\phi _{j} )^{{\wedge_{F} q}} ) \oplus _{F} \mathop { \oplus _{F} }\limits_{{i = 1}}^{{k + 1}} w_{i} w_{{k + 1}} \cdot _{F} ((\phi _{i} )^{{\wedge_{F} p}} \otimes _{F} (\phi _{{k + 1}} )^{{\wedge_{F} q}} ) \hfill \\ = \left( {f^{{* - 1}} \left( {1 - \log _{\lambda } \left( {1 + (\lambda - 1)\prod\limits_{{i = 1,j = i}}^{{k + 1}} {\left( {\frac{{(\lambda - 1)^{{p + q}} - (\lambda ^{{f^{*} (s_{{u_{i} }} )}} - 1)^{p} (\lambda ^{{f^{*} (s_{{u_{j} }} )}} - 1)^{q} }}{{(\lambda - 1)^{{p + q}} + (\lambda - 1)(\lambda ^{{f^{*} (s_{{u_{i} }} )}} - 1)^{p} (\lambda ^{{f^{*} (s_{{u_{j} }} )}} - 1)^{q} }}} \right)^{{w_{i} w_{j} }} } } \right)} \right)} \right., \hfill \\ {\kern 1pt} \left. {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} f^{{* - 1}} \left( {\log _{\lambda } \left( {1 + (\lambda - 1)\prod\limits_{{i = 1,j = i}}^{{k + 1}} {\left( {\frac{{(\lambda - 1)^{{p + q}} - (\lambda ^{{1 - f^{*} (s_{{v_{i} }} )}} - 1)^{p} (\lambda ^{{1 - f^{*} (s_{{v_{j} }} )}} - 1)^{q} }}{{(\lambda - 1)^{{p + q}} + (\lambda - 1)(\lambda ^{{1 - f^{*} (s_{{v_{i} }} )}} - 1)^{p} (\lambda ^{{1 - f^{*} (s_{{v_{j} }} )}} - 1)^{q} }}} \right)^{{w_{i} w_{j} }} } } \right)} \right)} \right). \hfill \\ \end{gathered}$$

That is, Eq. (12) also holds for $n=k+1.$ Thus, Eq. (12) is true for all $n$.

Then, by using Eq. (12), Eq. (6) can be calculated easily based on the operations of LIFNs, and Eq. (7) can be eventually acquired. Therefore, the proof Theorem 1 is completed.

Proof of Theorem 3.

Proof

Let $LIFFIWH{{M}^{p,q}}({{a}_{1}},{{a}_{2}},...,{{a}_{n}})=({{s}_{u(a)}},{{s}_{v(a)}})$and $LIFFIWH{{M}^{p,q}}({{b}_{1}},{{b}_{2}},...,{{b}_{n}})=({{s}_{u(b)}},{{s}_{v(b)}})$. Since ${{f}^{*}}$, ${{f}^{*-1}}$ and ${{\log }_{\lambda }},(\lambda>1)$ is a strictly monotonously increasing and continuous function, and ${{s}_{u({{a}_{i}})}}\ge {{s}_{u({{b}_{i}})}}$ and ${{s}_{u({{a}_{j}})}}\ge {{s}_{u({{b}_{j}})}}$ for all $i,j=1,2,...,n$, then the following inequalities can be obtained.

$$\begin{aligned} {\kern 1pt} f^{*} (s_{{u(a_{i} )}} ) \ge f^{*} (s_{{u(b_{i} )}} ){\kern 1pt} ,f^{*} (s_{{u(a_{j} )}} ) \ge f^{*} (s_{{u(b_{j} )}} ) &\Rightarrow (\lambda ^{{f^{*} (s_{{u(a_{i} )}} )}} - 1)^{p} (\lambda ^{{f^{*} (s_{{u(a_{j} )}} )}} - 1)^{q} \ge (\lambda ^{{f^{*} (s_{{u(b_{i} )}} )}} - 1)^{p} (\lambda ^{{f^{*} (s_{{u(b_{j} )}} )}} - 1)^{q} \hfill \\ & \Rightarrow \left( {\prod\limits_{{i = 1,j = i}}^{n} {\left( {\frac{{(\lambda - 1)^{{p + q}} - (\lambda ^{{f^{*} (s_{{u(a_{i} )}} )}} - 1)^{p} (\lambda ^{{f^{*} (s_{{u(a_{j} )}} )}} - 1)^{q} }}{{(\lambda - 1)^{{p + q}} + (\lambda - 1)(\lambda ^{{f^{*} (s_{{u(a_{i} )}} )}} - 1)^{p} (\lambda ^{{f^{*} (s_{{u(a_{j} )}} )}} - 1)^{q} }}} \right)^{{w_{i} w_{j} }} } } \right)^{k} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \hfill \\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} & \le \left( {\prod\limits_{{i = 1,j = i}}^{n} {\left( {\frac{{(\lambda - 1)^{{p + q}} - (\lambda ^{{f^{*} (s_{{u(b_{i} )}} )}} - 1)^{p} (\lambda ^{{f^{*} (s_{{u(b_{j} )}} )}} - 1)^{q} }}{{(\lambda - 1)^{{p + q}} + (\lambda - 1)(\lambda ^{{f^{*} (s_{{u(b_{i} )}} )}} - 1)^{p} (\lambda ^{{f^{*} (s_{{u(b_{j} )}} )}} - 1)^{q} }}} \right)^{{w_{i} w_{j} }} } } \right)^{k} \hfill \\ & \Rightarrow s_{{u(a)}} \ge s_{{u(b)}} . \hfill \\ \end{aligned}$$

In the same way, the inequality ${{s}_{v(a)}}\le {{s}_{v(b)}}$ can also be obtained. Since ${{s}_{u(a)}}\ge {{s}_{u(b)}}$ and ${{s}_{v(a)}}\le {{s}_{v(b)}}$, then, $LIFFIWH{{M}^{p,q}}\left( {{a}_{1}},{{a}_{2}},...,{{a}_{n}} \right)\ge LIFFIWH{{M}^{p,q}}\left( {{b}_{1}},{{b}_{2}},...,{{b}_{n}} \right).$

Proof of Theorem 4.

Proof

Since ${{s}_{u(a)}}\ge {{s}_{{{u}_{i}}}}$ and ${{s}_{v(a)}}\le {{s}_{{{v}_{i}}}}$ for all $i=1,2,...,n,$ then according to Theorems 2 and 3, we can obtain$LIFFIWH{{M}^{p,q}}({{\phi }_{1}},{{\phi }_{2}},...,{{\phi }_{n}})\le LIFFIWH{{M}^{p,q}}(a,a,...,a)=a.$ Since ${{s}_{u(b)}}\le {{s}_{{{u}_{i}}}}$ and ${{s}_{v(b)}}\ge {{s}_{{{v}_{i}}}}$ for all $i=1,2,...,n$, then we can also obtain $b=LIFFIWH{{M}^{p,q}}(b,b,...,b)\le LIFFIWH{{M}^{p,q}}({{\phi }_{1}},{{\phi }_{2}},...,{{\phi }_{n}})$. Thus, $b\le LIFFIWH{{M}^{p,q}}({{\phi }_{1}},{{\phi }_{2}},...,{{\phi }_{n}})\le a$ is true.