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An approach for solving fuzzy multi-criteria decision problem under linguistic information

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Abstract

Linguistic information processing exists in multi-criteria decision making, and linguistic truth-valued lattice implication algebra (LTV-LIA) has definite advantages in handling comparable and incomparable linguistic values. To deal with the preference relations with linguistic evaluation information, we establish a novel approach for solving fuzzy multi-criteria decision problem under linguistic information based on LTV-LIA. In this paper, we propose linguistic lattice-valued preference relation (LLVPR). LLVPR positive and negative matrixes are introduced to evaluate the advantages and disadvantages of alternatives respectively. In order to get a reasonable result, we introduce a new algorithm to check and repair the consistency of a LLVPR. A linguistic lattice-valued 2-tuple representation model (LLV2-tuple) and some new aggregation operations are presented to get the comprehensive linguistic information without information loss. Considering different decision makers have different preferences, a multiple preferences implication operation of LLV2-tuple is introduced. Finally, we propose a novel linguistic analytic hierarchy process embedded in aggregation layer and implication layer, introducing algorithm and numerical examples. A comparative analysis is adopted to illustrate the rationality.

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Acknowledgements

This work is partially supported by the National Natural Science Foundation of P.R. China (Nos. 61502068, 61772250) and the Scientific Research Project of Department of Education of Liaoning Province (No. LJ2020007).

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Authors and Affiliations

  1. Information Science and Technology College, Dalian Maritime University, Dalian, China

    Hongyue Diao, Ansheng Deng & Hui Cui

  2. School of Mathematics, Liaoning Normal University, Dalian, China

    Xin Liu

  3. School of Computer Science and Technology, Shandong Jianzhu University, Jinan, China

    Li Zou

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  1. Hongyue Diao
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  2. Ansheng Deng
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  3. Hui Cui
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  4. Xin Liu
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  5. Li Zou
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Correspondence to Ansheng Deng or Xin Liu.

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Appendix

Appendix

In this appendix, detailed calculation process of numerical examples is introduced.

Determine the LLVPR according to the linguistic fundamental scale by pair-wise comparisons respectively.

$$\begin{aligned} C= & {} \left[ {\begin{array}{*{20}{c}} - &{}{\left( {{h_1},t} \right) }&{}{\left( {{h_1},t} \right) }&{}{\left( {{h_2},t} \right) }&{}{\left( {{h_2},t} \right) }\\ {\left( {{h_1},f} \right) }&{} - &{}{\left( {{h_1},t} \right) }&{}{\left( {{h_2},t} \right) }&{}{\left( {{h_2},t} \right) }\\ {\left( {{h_1},f} \right) }&{}{\left( {{h_1},f} \right) }&{} - &{}{\left( {{h_1},t} \right) }&{}{\left( {{h_1},t} \right) }\\ {\left( {{h_2},f} \right) }&{}{\left( {{h_2},f} \right) }&{}{\left( {{h_1},f} \right) }&{} - &{}{\left( {{h_1},t} \right) }\\ {\left( {{h_2},f} \right) }&{}{\left( {{h_2},f} \right) }&{}{\left( {{h_1},f} \right) }&{}{\left( {{h_1},f} \right) }&{} - \end{array}} \right] ;\\ {A^{\left( 1 \right) }}= & {} \left[ {\begin{array}{*{20}{c}} - &{}{\left( {{h_1},t} \right) }&{}{\left( {{h_0},t} \right) }\\ {\left( {{h_1},f} \right) }&{} - &{}{\left( {{h_1},f} \right) }\\ {\left( {{h_0},f} \right) }&{}{\left( {{h_1},t} \right) }&{} - \end{array}} \right] ; {A^{\left( 2 \right) }} = \left[ {\begin{array}{*{20}{c}} - &{}{\left( {{h_1},t} \right) }&{}{\left( {{h_0},t} \right) }\\ {\left( {{h_1},f} \right) }&{} - &{}{\left( {{h_1},f} \right) }\\ {\left( {{h_0},f} \right) }&{}{\left( {{h_1},t} \right) }&{} - \end{array}} \right] ;\\ {A^{\left( 3 \right) }}= & {} \left[ {\begin{array}{*{20}{c}} - &{}{\left( {{h_1},t} \right) }&{}{\left( {{h_2},t} \right) }\\ {\left( {{h_1},f} \right) }&{} - &{}{\left( {{h_1},t} \right) }\\ {\left( {{h_2},f} \right) }&{}{\left( {{h_1},f} \right) }&{} - \end{array}} \right] ; {A^{\left( 4 \right) }} = \left[ {\begin{array}{*{20}{c}} - &{}{\left( {{h_3},t} \right) }&{}{\left( {{h_2},t} \right) }\\ {\left( {{h_3},f} \right) }&{} - &{}{\left( {{h_1},t} \right) }\\ {\left( {{h_2},f} \right) }&{}{\left( {{h_1},f} \right) }&{} - \end{array}} \right] ;\\ {A^{\left( 5 \right) }}= & {} \left[ {\begin{array}{*{20}{c}} - &{}{\left( {{h_2},t} \right) }&{}{\left( {{h_2},t} \right) }\\ {\left( {{h_2},f} \right) }&{} - &{}{\left( {{h_1},t} \right) }\\ {\left( {{h_2},f} \right) }&{}{\left( {{h_1},f} \right) }&{} - \end{array}} \right] . \end{aligned}$$

Compute the LLVPR positive matrix and negative matrix of every preference relation matrix.

$$\begin{aligned} {C^ + }= & {} \left[ {\begin{array}{*{20}{c}} *&{}{\left( {{h_1},t} \right) }&{}{\left( {{h_1},t} \right) }&{}{\left( {{h_2},t} \right) }&{}{\left( {{h_2},t} \right) }\\ *&{}*&{}{\left( {{h_1},t} \right) }&{}{\left( {{h_2},t} \right) }&{}{\left( {{h_2},t} \right) }\\ *&{}*&{}*&{}{\left( {{h_1},t} \right) }&{}{\left( {{h_1},t} \right) } \\ *&{}*&{}*&{}*&{}{\left( {{h_1},t} \right) }\\ *&{}*&{}*&{}*&{}* \end{array}} \right] , {C^ - } = \left[ {\begin{array}{*{20}{c}} *&{}*&{}*&{}*&{}* \\ {\left( {{h_1},f} \right) }&{}*&{}*&{}*&{}*\! \\ {\left( {{h_1},f} \right) }&{}{\left( {{h_1},f} \right) }&{}*&{}*&{}*\\ {\left( {{h_2},f} \right) }&{}{\left( {{h_2},f} \right) }&{}{\left( {{h_1},f} \right) }&{}*&{}*\\ {\left( {{h_2},f} \right) }&{}{\left( {{h_2},f} \right) }&{}{\left( {{h_1},f} \right) }&{}{\left( {{h_1},f} \right) }&{}* \end{array}} \right] ;\\ {A^{\left( 1 \right) , + }}= & {} \left[ {\begin{array}{*{20}{c}} *&{}{\left( {{h_1},t} \right) }&{}{\left( {{h_0},t} \right) }\\ *&{}*&{}*\\ *&{}{\left( {{h_1},t} \right) }&{}* \end{array}} \right] , {A^{\left( 1 \right) , - }} = \left[ {\begin{array}{*{20}{c}} *&{}*&{}*\\ {\left( {{h_1},f} \right) }&{}*&{}{\left( {{h_1},f} \right) }\\ {\left( {{h_0},f} \right) }&{}*&{}* \end{array}} \right] ;\\ {A^{\left( 2 \right) , + }}= & {} \left[ {\begin{array}{*{20}{c}} *&{}{\left( {{h_1},t} \right) }&{}{\left( {{h_0},t} \right) }\\ *&{}*&{}*\\ *&{}{\left( {{h_1},t} \right) }&{}* \end{array}} \right] , {A^{\left( 2 \right) , - }} = \left[ {\begin{array}{*{20}{c}} *&{}*&{}*\\ {\left( {{h_1},f} \right) }&{}*&{}{\left( {{h_1},f} \right) }\\ {\left( {{h_0},f} \right) }&{}*&{}* \end{array}} \right] ;\\ {A^{\left( 3 \right) , + }}= & {} \left[ {\begin{array}{*{20}{c}} *&{}{\left( {{h_1},t} \right) }&{}{\left( {{h_2},t} \right) }\\ *&{}*&{}{\left( {{h_1},t} \right) }\\ *&{}*&{}* \end{array}} \right] , {A^{\left( 3 \right) , - }} = \left[ {\begin{array}{*{20}{c}} *&{}*&{}*\\ {\left( {{h_1},f} \right) }&{}*&{}*\\ {\left( {{h_2},f} \right) }&{}{\left( {{h_1},f} \right) }&{}* \end{array}} \right] ;\\ {A^{\left( 4 \right) , + }}= & {} \left[ {\begin{array}{*{20}{c}} *&{}{\left( {{h_3},t} \right) }&{}{\left( {{h_2},t} \right) }\\ *&{}*&{}{\left( {{h_1},t} \right) }\\ *&{}*&{}* \end{array}} \right] , {A^{\left( 4 \right) , - }} = \left[ {\begin{array}{*{20}{c}} *&{}*&{}*\\ {\left( {{h_3},f} \right) }&{}*&{}*\\ {\left( {{h_2},f} \right) }&{}{\left( {{h_1},f} \right) }&{}* \end{array}} \right] ;\\ {A^{\left( 5 \right) , + }}= & {} \left[ {\begin{array}{*{20}{c}} *&{}{\left( {{h_2},t} \right) }&{}{\left( {{h_2},t} \right) }\\ *&{}*&{}{\left( {{h_1},t} \right) }\\ *&{}*&{}* \end{array}} \right] , {A^{\left( 5 \right) , - }} = \left[ {\begin{array}{*{20}{c}} *&{}*&{}*\\ {\left( {{h_2},f} \right) }&{}*&{}*\\ {\left( {{h_2},f} \right) }&{}{\left( {{h_1},f} \right) }&{}* \end{array}} \right] . \end{aligned}$$

Aggregate every row of every matrix and calculate the comprehensive aggregate value. We take the \({G_{LAAO}}\) as an example.

$$\begin{aligned} {C^ + }= & {} \left( {\begin{array}{*{20}{c}} {\left( {\left( {{h_1},t} \right) ,0.2} \right) }\\ {\left( {\left( {{h_1},t} \right) ,0} \right) }\\ {\left( {\left( {{h_0},t} \right) ,0.4} \right) }\\ {\left( {\left( {{h_0},t} \right) ,0.2} \right) }\\ * \end{array}} \right) , {C^ - } = \left( {\begin{array}{*{20}{c}} * \\ {\left( {\left( {{h_0},f} \right) ,0.2} \right) }\\ {\left( {\left( {{h_0},f} \right) ,0.4} \right) }\\ {\left( {\left( {{h_1},f} \right) ,0} \right) }\\ {\left( {\left( {{h_1},f} \right) ,0.2} \right) } \end{array}} \right) , {\tilde{C}} = \left( {\begin{array}{*{20}{c}} {\left( {\left( {{h_1},t} \right) ,0.2} \right) }\\ {\left( {\left( {{h_0},t} \right) ,0.8} \right) }\\ - \\ {\left( {\left( {{h_0},f} \right) ,0.8} \right) }\\ {\left( {\left( {{h_1},f} \right) ,0.2} \right) } \end{array}} \right) ;\\ {A^{\left( 1 \right) , + }}= & {} \left( {\begin{array}{*{20}{c}} {\left( {\left( {{h_0},t} \right) ,0.33} \right) }\\ * \\ {\left( {\left( {{h_0},t} \right) ,0.33} \right) } \end{array}} \right) , {A^{\left( 1 \right) , - }} = \left( {\begin{array}{*{20}{c}} * \\ {\left( {\left( {{h_0},f} \right) ,0.67} \right) }\\ {\left( {\left( {{h_0},f} \right) ,0} \right) } \end{array}} \right) , {{\tilde{A}}^{\left( 1 \right) }} = \left( {\begin{array}{*{20}{c}} {\left( {\left( {{h_0},t} \right) ,0.33} \right) }\\ {\left( {\left( {{h_0},f} \right) ,0.67} \right) }\\ {\left( {\left( {{h_0},t} \right) ,0.17} \right) } \end{array}} \right) ;\\ {A^{\left( 2 \right) , + }}= & {} \left( {\begin{array}{*{20}{c}} {\left( {\left( {{h_0},t} \right) ,0.33} \right) }\\ * \\ {\left( {\left( {{h_0},t} \right) ,0.33} \right) } \end{array}} \right) , {A^{\left( 2 \right) , - }} = \left( {\begin{array}{*{20}{c}} * \\ {\left( {\left( {{h_0},f} \right) ,0.67} \right) }\\ {\left( {\left( {{h_0},f} \right) ,0} \right) } \end{array}} \right) , {{\tilde{A}}^{\left( 2 \right) }} = \left( {\begin{array}{*{20}{c}} {\left( {\left( {{h_0},t} \right) ,0.33} \right) }\\ {\left( {\left( {{h_0},f} \right) ,0.67} \right) }\\ {\left( {\left( {{h_0},t} \right) ,0.17} \right) } \end{array}} \right) ;\\ {A^{\left( 3 \right) , + }}= & {} \left( {\begin{array}{*{20}{c}} {\left( {\left( {{h_1},t} \right) ,0} \right) }\\ {\left( {\left( {{h_0},t} \right) ,0.33} \right) }\\ * \end{array}} \right) , {A^{\left( 3 \right) , - }} = \left( {\begin{array}{*{20}{c}} * \\ {\left( {\left( {{h_0},f} \right) ,0.33} \right) }\\ {\left( {\left( {{h_1},f} \right) ,0} \right) } \end{array}} \right) , {{\tilde{A}}^{\left( 3 \right) }} = \left( {\begin{array}{*{20}{c}} {\left( {\left( {{h_1},t} \right) ,0} \right) }\\ - \\ {\left( {\left( {{h_1},f} \right) ,0} \right) } \end{array}} \right) ;\\ {A^{\left( 4 \right) , + }}= & {} \left( {\begin{array}{*{20}{c}} {\left( {\left( {{h_1},t} \right) ,0.67} \right) }\\ {\left( {\left( {{h_0},t} \right) ,0.33} \right) }\\ * \end{array}} \right) , {A^{\left( 4 \right) , - }} = \left( {\begin{array}{*{20}{c}} * \\ {\left( {\left( {{h_1},f} \right) ,0} \right) }\\ {\left( {\left( {{h_1},f} \right) ,0} \right) } \end{array}} \right) , {{\tilde{A}}^{\left( 4 \right) }} = \left( {\begin{array}{*{20}{c}} {\left( {\left( {{h_1},t} \right) ,0.67} \right) }\\ {\left( {\left( {{h_0},f} \right) ,0.67} \right) }\\ {\left( {\left( {{h_1},f} \right) ,0} \right) } \end{array}} \right) ;\\ {A^{\left( 5 \right) , + }}= & {} \left( {\begin{array}{*{20}{c}} {\left( {\left( {{h_1},t} \right) ,0.33} \right) }\\ {\left( {\left( {{h_0},t} \right) ,0.33} \right) }\\ * \end{array}} \right) , {A^{\left( 5 \right) , - }} = \left( {\begin{array}{*{20}{c}} * \\ {\left( {\left( {{h_0},f} \right) ,0.67} \right) }\\ {\left( {\left( {{h_{\mathrm{1}}},f} \right) ,0} \right) } \end{array}} \right) , {{\tilde{A}}^{\left( 5 \right) }} = \left( {\begin{array}{*{20}{c}} {\left( {\left( {{h_1},t} \right) ,0.33} \right) }\\ {\left( {\left( {{h_0},f} \right) ,0.17} \right) }\\ {\left( {\left( {{h_{\mathrm{1}}},f} \right) ,0} \right) } \end{array}} \right) . \end{aligned}$$

According to the above calculation, the criterion aggregation value and alternative aggregation value can be obtained as follows

$$\begin{aligned} {\left( {{\tilde{C}}} \right) ^T}= & {} \left( {\begin{array}{*{20}{c}} {\left( {\left( {{h_1},t} \right) ,0.2} \right) }&{\left( {\left( {{h_0},t} \right) ,0.8} \right) }- & {} {\left( {\left( {{h_0},f} \right) ,0.8} \right) }&{\left( {\left( {{h_1},f} \right) ,0.2} \right) } \end{array}} \right) \\ {\tilde{A}}= & {} \left( {\begin{array}{*{20}{c}} {{{\left( {{{{\tilde{A}}}^{\left( 1 \right) }}} \right) }^T}}\\ {{{\left( {{{{\tilde{A}}}^{\left( 2 \right) }}} \right) }^T}}\\ {\begin{array}{*{20}{c}} {{{\left( {{{{\tilde{A}}}^{\left( 3 \right) }}} \right) }^T}}\\ {{{\left( {{{{\tilde{A}}}^{\left( 4 \right) }}} \right) }^T}} \end{array}}\\ {{{\left( {{{{\tilde{A}}}^{\left( 5 \right) }}} \right) }^T}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {\left( {\left( {{h_0},t} \right) ,0.33} \right) }&{}{\left( {\left( {{h_0},f} \right) ,0.67} \right) }&{}{\left( {\left( {{h_0},t} \right) ,0.17} \right) }\\ {\left( {\left( {{h_0},t} \right) ,0.33} \right) }&{}{\left( {\left( {{h_0},f} \right) ,0.67} \right) }&{}{\left( {\left( {{h_0},t} \right) ,0.17} \right) }\\ {\left( {\left( {{h_1},t} \right) ,0} \right) }&{} - &{}{\left( {\left( {{h_1},f} \right) ,0} \right) }\\ {\left( {\left( {{h_1},t} \right) ,0.67} \right) }&{}{\left( {\left( {{h_0},f} \right) ,0.67} \right) }&{}{\left( {\left( {{h_1},f} \right) ,0} \right) }\\ {\left( {\left( {{h_1},t} \right) ,0.33} \right) }&{}{\left( {\left( {{h_0},f} \right) ,0.17} \right) }&{}{\left( {\left( {{h_{\mathrm{1}}},f} \right) ,0} \right) } \end{array}} \right) \\ R= & {} {\left( {{\tilde{C}}} \right) ^T} \rightarrow {\tilde{A}} = \left( {\begin{array}{*{20}{c}} {\left( {\left( {{h_2},t}\right) ,0.28}\right) }&{\left( {\left( {{h_1},t}\right) ,0.942}\right) }&{\left( {\left( {{h_2},t}\right) ,0.868}\right) } \end{array}}\right) \end{aligned}$$

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Diao, H., Deng, A., Cui, H. et al. An approach for solving fuzzy multi-criteria decision problem under linguistic information. Fuzzy Optim Decis Making 21, 45–69 (2022). https://doi.org/10.1007/s10700-021-09356-x

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