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Multi-criteria sorting decision making based on dominance and opposition relations with probabilistic linguistic information

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Abstract

The probabilistic linguistic term set (PLTS) is a powerful tool for describing linguistic evaluations derived from expert teams and has adequate capability to identify preferences among different evaluations. Due to the practicability of PLTSs, probabilistic linguistic decision making problems have been widely investigated in recent years. However, no study on probabilistic linguistic outranking relations has been conducted. This study aims to explore effective processing for the complex two-dimension structure of PLTSs and formulate probabilistic linguistic dominance and opposition relations for multi-criteria sorting decision making. Linguistic scale functions, which can generate different semantics for linguistic variables under different decision making environments, are introduced to deal with the linguistic terms in PLTSs. In this way, the probabilistic linguistic dominance degree, concordance and discordance indices are defined by systematically comparing the probabilities of PLTSs. Then, two kinds of outranking relations with dominance and opposition for PLTSs are formulated based on the defined outranking indices. Subsequently, an innovative sorting decision making framework is constructed by exploring the outranking relations between alternatives and characteristic actions under multiple criteria and implementing the outranking aggregation and exploitation. Finally, this framework is demonstrated using an illustrative example with result analyses and comparison discussions.

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Acknowledgements

The authors are very grateful to the anonymous reviewers for their valuable comments and suggestions to help improve the overall quality of this paper. This work was supported by the National Natural Science Foundation of China (No. 71871228).

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  1. School of Business, Central South University, Changsha, 410083, People’s Republic of China

    Hong-gang Peng & Jian-qiang Wang

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  1. Hong-gang Peng
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Appendix

Appendix

Proof of Property 2

(1) Necessity If \( X_{1} (p) >_{sd} X_{2} (p) \), then \( C(X_{1} (p),X_{2} (p)) - C(X_{2} (p),X_{1} (p)) = 1 \) can be obtained from Definition 7. In this case, we have \( C(X_{1} (p),X_{2} (p)) = 1 \) and \( C(X_{2} (p),X_{1} (p)) = 0 \) in light of Property 1. Then, according to Definition 6, the following inequalities can be deduced:

$$ C(X_{1} (p),X_{2} (p)) = 1 \Rightarrow C(p_{1} ,p_{2} ) = 1 \Rightarrow x_{2} \le x_{1} ,p_{1} - p_{2} \ge - q_{j} , $$
$$ {\text{and}}\;C(X_{2} (p),X_{1} (p)) = 0 \Rightarrow C(p_{1} ,p_{2} ) = 0 \Rightarrow x_{2} < x_{1} ,\;{\text{or}}\;x_{1} \le x_{2} ,p_{1} - p_{2} \ge r_{j} , $$

for all \( x_{1} \in X_{1} (p) \) and \( x_{2} \in X_{2} (p) \).

Therefore, based on the intersection operation, \( x_{1} > x_{2} \) and \( p_{1} - p_{2} \ge - q_{j} \), and \( x_{1} = x_{2} \) and \( p_{1} - p_{2} \ge r_{j} \) are true for all \( x_{1} \in X_{1} (p) \) and \( x_{2} \in X_{2} (p) \).

(2) Sufficiency Since \( x_{1} > x_{2} \) and \( p_{1} - p_{2} \ge - q_{j} \), and \( x_{1} = x_{2} \) and \( p_{1} - p_{2} \ge r_{j} \) are valid for all \( x_{1} \in X_{1} (p) \) and \( x_{2} \in X_{2} (p) \), according to Definition 6, the following equalities can be deduced:

$$ x_{1} > x_{2} ,p_{1} - p_{2} \ge - q_{j} \Rightarrow C(p_{1} ,p_{2} ) = 1 \Rightarrow C(X_{1} (p),X_{2} (p)) = 1,\;{\text{and}}\;x_{1} > x_{2} \Rightarrow C(X_{2} (p),X_{1} (p)){ = 0;} $$
$$ x_{1} = x_{2} ,p_{1} - p_{2} \ge r_{j} \Rightarrow C(p_{1} ,p_{2} ) = 1 \Rightarrow C(X_{1} (p),X_{2} (p)) = 1,\;{\text{and}}\;x_{1} = x_{2} ,p_{1} - p_{2} \ge r_{j} \Rightarrow C(X_{2} (p),X_{1} (p)) = 0. $$

Then, \( C(X_{1} (p),X_{2} (p)) - C(X_{2} (p),X_{1} (p)) = 1 \) can be obtained. Therefore, \( X_{1} (p) >_{sd} X_{2} (p) \) is verified.□

Proof of Property 3

  1. (1.1)

    Since \( C(X_{1} (p),X_{1} (p)) - C(X_{1} (p),X_{1} (p)) = 0 \ne 1 \), \( X_{1} (p)\not >_{sd} X_{1} (p) \).

  2. (1.2)

    When \( X_{1} (p) >_{sd} X_{2} (p) \), \( C(X_{1} (p),X_{2} (p)) = 1 \) and \( C(X_{2} (p),X_{1} (p)) = 0 \) can be obtained according to Definition 7 and Property 1. Then, \( C(X_{2} (p),X_{1} (p)) - C(X_{1} (p),X_{2} (p)) = - 1 \ne 1 \) can be deduced. Therefore, \( X_{1} (p) >_{sd} X_{2} (p) \)\( \not \Rightarrow X_{2} (p) >_{sd} X_{1} (p) \).

  3. (1.3)

    When \( X_{1} (p) >_{sd} X_{2} (p) \) and \( X_{2} (p) >_{sd} X_{3} (p) \), \( x_{1} > x_{2} \) and \( p_{1} - p_{2} \ge - q_{j} \), \( x_{2} > x_{3} \) and \( p_{2} - p_{3} \ge - q_{j} \), \( x_{1} = x_{2} \) and \( p_{1} - p_{2} \ge r_{j} \), and \( x_{2} = x_{3} \) and \( p_{2} - p_{3} \ge r_{j} \) for all \( x_{1} \in X_{1} (p) \), \( x_{2} \in X_{2} (p) \) and \( x_{3} \in X_{3} (p) \) can be obtained according to Property 2. Then, \( x_{1} > x_{3} \) and \( p_{1} - p_{3} \ge - 2q_{j} \) can be deduced. However, \( p_{1} - p_{3} \ge - 2q_{j} \)\( \not \Rightarrow p_{1} - p_{3} \ge - q_{j} \). Therefore, \( X_{1} (p) >_{sd} X_{2} (p) \) and \( X_{2} (p) >_{sd} X_{3} (p) \)\( \not \Rightarrow X_{1} (p) >_{sd} X_{3} (p) \).

Similar to (1.1) and (1.2), (2.1), (2.2), (3.1), (3.2), (4.1) and (4.2) can be proved easily.□

Property 2 presents the conditions of the strong dominance relation. However, the conditions of the moderate, weak and indifferent dominance relations cannot be obtained because \( C(X_{1} (p),X_{2} (p)) \) and \( C(X_{2} (p),X_{1} (p)) \) in cases (2), (3) and (4) in Definition 7 cannot be determined explicitly. In this case, it is difficult to prove the non-transitivity of the moderate, weak and indifferent dominance relations. Motivated by the idea in Peng et al. (2015), where the non-transitivity of outranking relations is exemplified, several examples are provided to demonstrate the non-transitivity of the moderate, weak and indifferent dominance relations in the following.

Example 6

Let \( q_{j} = 0.05 \) and \( r_{j} = 0.15 \), then the non-transitivity can be exemplified as follows:

  1. (1)

    Let \( X_{1} (p) = \{ s_{3} (0.3),s_{4} (0.3),s_{5} (0.2)\} \), \( X_{2} (p) = \{ s_{2} (0.1),s_{3} (0.3),s_{4} (0.3),s_{5} (0.1)\} \), and \( X_{3} (p) = \{ s_{5} (0.3),s_{6} (0.3),s_{7} (0.3)\} \) be three PLTSs, then \( C(X_{1} (p),X_{2} (p)) - C(X_{2} (p),X_{1} (p)) = 0.3 \), \( C(X_{3} (p),X_{1} (p)) - C(X_{1} (p),X_{3} (p)) = 0.9407 \), and \( C(X_{3} (p),X_{2} (p)) - C(X_{2} (p),X_{3} (p)) = 1 \). Therefore, \( X_{3} (p) >_{md} X_{1} (p) \), \( X_{1} (p) >_{md} X_{2} (p) \), but \( X_{4} (p) >_{sd} X_{2} (p) \). This indicates that the non-transitivity of the moderate dominance relation is true.

  2. (2)

    Let \( X_{1} (p) = \{ s_{2} (0.2),s_{3} (0.2),s_{4} (0.2),s_{5} (0.2)\} \), \( X_{2} (p) = \{ s_{2} (0.3),s_{3} (0.1),s_{4} (0.3),s_{5} (0.2)\} \), and \( X_{3} (p) = \{ s_{2} (0.4),s_{3} (0.3),s_{4} (0.2)\} \) be three PLTSs, then \( C(X_{2} (p),X_{1} (p)) - C(X_{1} (p),X_{2} (p)) = 0.1250 \), \( C(X_{1} (p),X_{3} (p)) - C(X_{3} (p),X_{1} (p)) = 0.0833 \), and \( C(X_{2} (p),X_{3} (p)) - C(X_{3} (p),X_{2} (p)) = 0.2333 \). Therefore, \( X_{2} (p) >_{wd} X_{1} (p) \), \( X_{1} (p) >_{wd} X_{3} (p) \), but \( X_{2} (p) >_{md} X_{3} (p) \). This indicates that the non-transitivity of the weak dominance relation is true.

  3. (3)

    Let \( X_{1} (p) = \{ s_{2} (0.1),s_{3} (0.3),s_{4} (0.4),s_{5} (0.2)\} \), \( X_{2} (p) = \{ s_{2} (0.2),s_{3} (0.3),s_{4} (0.2),s_{5} (0.3)\} \), and \( X_{3} (p) = \{ s_{2} (0.1),s_{3} (0.4),s_{4} (0.3),s_{5} (0.2)\} \) be three PLTSs, then \( C(X_{1} (p),X_{2} (p)) - C(X_{2} (p),X_{1} (p)){ = }0. 055 \), \( C(X_{1} (p),X_{3} (p)) - C(X_{3} (p),X_{1} (p)){ = }0.045 \), and \( C(X_{3} (p),X_{2} (p)){ - }C(X_{2} (p),X_{3} (p)){ = 0} . 0 4 \). Therefore, \( X_{1} (p) \sim_{id} X_{3} (p) \), \( X_{3} (p) \sim_{id} X_{2} (p) \), but \( X_{1} (p) >_{wd} X_{2} (p) \). This indicates that the non-transitivity of the indifferent dominance relation is true.

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Peng, Hg., Wang, Jq. Multi-criteria sorting decision making based on dominance and opposition relations with probabilistic linguistic information. Fuzzy Optim Decis Making 19, 435–470 (2020). https://doi.org/10.1007/s10700-020-09330-z

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