Abstract
Type-2 fuzzy sets/numbers (T2FSs/T2FNs) attract more and more attention in fuzzy decision field. The existing studies mostly focus on the general properties of T2FS, or interval type-2 fuzzy number whose membership degrees are denoted by intervals. A new form of T2FN named triangular type-2 fuzzy number is proposed, whose primary and secondary memberships both have the continuous triangular feature. For aggregating the triangular type-2 fuzzy information, two operators are also defined. Based on them, a method is developed to handle the duplex linguistic group multi-criteria decision making problems and rank the alternatives. Finally, an example is provided to show the feasibility of the method.
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Acknowledgments
We are grateful to the anonymous referees for their valuable comments that helped us considerably improve the paper. This work was supported by the National Natural Science Foundation of China (Nos. 71271218 and 71401185).
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Appendix
Appendix
The Proof of Property 1
Since the properties (1), (3), (4) are easy to proof, here the property (2) is proofed. As \(a_{i} \ge 0\) \(\left( {i = 1,2,3} \right)\), from definition 3:
Thus,
Similarly,
So \((\tilde{a}_{1} + \tilde{a}_{2} ) + \tilde{a}_{3} = \tilde{a}_{1} + (\tilde{a}_{2} + \tilde{a}_{3} )\).□
The Proof of Property 3
\(0 \le P(\tilde{a}_{1} \ge \tilde{a}_{2} ) \le 1\) is easy to proof.
□
The Proof of Property 4
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(1)
If \((b_{1} + c_{1} )\mu_{1} \le (a_{2} + b_{2} )\mu_{2}\), So, \((a_{1} + b_{1} )\mu_{1} \le (b_{1} + c_{1} )\mu_{1} \le (a_{2} + b_{2} )\mu_{2} \le (b_{2} + c_{2} )\mu_{2}\),
$$P(\tilde{a}_{1} \ge \tilde{a}_{2} ) = 0,$$$$\begin{aligned} P(\tilde{a}_{2} \ge \tilde{a}_{1} ) = \frac{{\hbox{min} \left\{ {l_{1} \mu_{1} + l_{2} \mu_{2} ,(b_{2} + c_{2} )\mu_{2} - (a_{1} + b_{1} )\mu_{1} } \right\}}}{{l_{1} \mu_{1} + l_{2} \mu_{2} }} \hfill \\ \quad \quad \quad \quad \,\, = \frac{{\hbox{min} \left\{ {(c_{1} - a_{1} )\mu_{1} + (c_{2} - a_{2} )\mu_{2} ,(b_{2} + c_{2} )\mu_{2} - (a_{1} + b_{1} )\mu_{1} } \right\}}}{{(c_{1} - a_{1} )\mu_{1} + (c_{2} - a_{2} )\mu_{2} }}. \hfill \\ \end{aligned}$$Because \([(c_{1} - a_{1} )\mu_{1} + (c_{2} - a_{2} )\mu_{2} ] - [(b_{2} + c_{2} )\mu_{2} - (a_{1} + b_{1} )\mu_{1} ] = (\mu_{1} c_{1} - \mu_{2} a_{2} ) - (\mu_{2} b_{2} - \mu_{1} b_{1} )\), and \((\mu_{1} c_{1} - \mu_{2} a_{2} ) - (\mu_{2} b_{2} - \mu_{1} b_{1} ) = (b_{1} + c_{1} )\mu_{1} - (a_{2} + b_{2} )\mu_{2} \le 0\), so \(\mu_{1} c_{1} - \mu_{2} a_{2} \le \mu_{2} b_{2} - \mu_{1} b_{1}\),\((c_{1} - a_{1} )\mu_{1} + (c_{2} - a_{2} )\mu_{2} \le (b_{2} + c_{2} )\mu_{2} - (a_{1} + b_{1} )\mu_{1}\). That is \(P(\tilde{a}_{2} \ge \tilde{a}_{1} ) = 1\). So \(P(\tilde{a}_{1} \ge \tilde{a}_{2} ) + P(\tilde{a}_{2} \ge \tilde{a}_{1} ) = 1.\)
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(2)
If \((b_{1} + c_{1} )\mu_{1} > (a_{2} + b_{2} )\mu_{2}\), \((b_{2} + c_{2} )\mu_{2} \le (a_{1} + b_{1} )\mu_{1}\), it is the same as 1)
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(3)
If \((b_{1} + c_{1} )\mu_{1} > (a_{2} + b_{2} )\mu_{2}\), \((b_{2} + c_{2} )\mu_{2} > (a_{1} + b_{1} )\mu_{1}\), then
$$\begin{aligned} P(\tilde{a}_{1} \ge \tilde{a}_{2} ) & = \frac{{\hbox{min} \left\{ {(c_{1} - a_{1} )\mu_{1} + (c_{2} - a_{2} )\mu_{2} ,(b_{1} + c_{1} )\mu_{1} - (a_{2} + b_{2} )\mu_{2} } \right\}}}{{(c_{1} - a_{1} )\mu_{1} + (c_{2} - a_{2} )\mu_{2} }} \\ & = \frac{{\hbox{min} \left\{ {c_{1} \mu_{1} - a_{1} \mu_{1} + c_{2} \mu_{2} - a_{2} \mu_{2} ,b_{1} \mu_{1} + c_{1} \mu_{1} - a_{2} \mu_{2} - b_{2} \mu_{2} } \right\}}}{{(c_{1} - a_{1} )\mu_{1} + (c_{2} - a_{2} )\mu_{2} }}. \\ \end{aligned}$$Because \((c_{1} \mu_{1} - a_{1} \mu_{1} + c_{2} \mu_{2} - a_{2} \mu_{2} ) - (b_{1} \mu_{1} + c_{1} \mu_{1} - a_{2} \mu_{2} - b_{2} \mu_{2} ) = c_{2} \mu_{2} - a_{1} \mu_{1} - b_{1} \mu_{1} + b_{2} \mu_{2}\), and \(c_{2} \mu_{2} - a_{1} \mu_{1} - b_{1} \mu_{1} + b_{2} \mu_{2} = (b_{2} + c_{2} )\mu_{2} - (a_{1} + b_{1} )\mu_{1} > 0\), so \(c_{1} \mu_{1} - a_{1} \mu_{1} + c_{2} \mu_{2} - a_{2} \mu_{2} > b_{1} \mu_{1} + c_{1} \mu_{1} - a_{2} \mu_{2} - b_{2} \mu_{2}\), that is \(P(\tilde{a}_{1} \ge \tilde{a}_{2} ) = \frac{{b_{1} \mu_{1} + c_{1} \mu_{1} - a_{2} \mu_{2} - b_{2} \mu_{2} }}{{(c_{1} - a_{1} )\mu_{1} + (c_{2} - a_{2} )\mu_{2} }}\). There also have \(P(\tilde{a}_{2} \ge \tilde{a}_{1} ) = \frac{{(b_{2} + c_{2} )\mu_{2} - (a_{1} + b_{1} )\mu_{1} }}{{l_{1} \mu_{1} + l_{2} \mu_{2} }} = \frac{{b_{2} \mu_{2} + c_{2} \mu_{2} - a_{1} \mu_{1} - b_{1} \mu_{1} }}{{(c_{1} - a_{1} )\mu_{1} + (c_{2} - a_{2} )\mu_{2} }}\), and \(P(\tilde{a}_{1} \ge \tilde{a}_{2} ) + P(\tilde{a}_{2} \ge \tilde{a}_{1} )\) \(= \frac{{b_{1} \mu_{1} + c_{1} \mu_{1} - a_{2} \mu_{2} - b_{2} \mu_{2} }}{{(c_{1} - a_{1} )\mu_{1} + (c_{2} - a_{2} )\mu_{2} }} + \frac{{b_{2} \mu_{2} + c_{2} \mu_{2} - a_{1} \mu_{1} - b_{1} \mu_{1} }}{{(c_{1} - a_{1} )\mu_{1} + (c_{2} - a_{2} )\mu_{2} }} = \frac{{c_{1} \mu_{1} - a_{2} \mu_{2} + c_{2} \mu_{2} - a_{1} \mu_{1} }}{{(c_{1} - a_{1} )\mu_{1} + (c_{2} - a_{2} )\mu_{2} }}\) \(= 1\). So \(P(\tilde{a}_{1} \ge \tilde{a}_{2} ) + P(\tilde{a}_{2} \ge \tilde{a}_{1} ) = 1\). □
The Proof of Theorem 2
Obviously, from definition 3, the sum of PTT2FNs is also a PTT2FN. In the following, equation (2) is proved by using mathematical induction on \(n\).
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(1)
For \(n = 2\), since
$$\begin{aligned} &\sum\limits_{i = 1}^{2} {\omega_{i} \tilde{a}_{i} } = \langle \left[ {\omega_{1} a_{1} + \omega_{2} a_{2} ,\omega_{1} b_{1} + \omega_{2} b_{2} ,\omega_{1} c_{1} + \omega_{2} c_{2} } \right]; \\ &\left[ {\frac{{\left\| {\tilde{a}_{1} } \right\|\omega_{1} \mu_{1}^{L} + \left\| {\tilde{a}_{2} } \right\|\omega_{2} \mu_{2}^{L} }}{{\omega_{1} \left\| {\tilde{a}_{1} } \right\| + \omega_{2} \left\| {\tilde{a}_{2} } \right\|}},\frac{{\left\| {\tilde{a}_{1} } \right\|\omega_{1} \mu_{1}^{M} + \left\| {\tilde{a}_{2} } \right\|\omega_{2} \mu_{2}^{M} }}{{\omega_{1} \left\| {\tilde{a}_{1} } \right\| + \omega_{2} \left\| {\tilde{a}_{2} } \right\|}},\frac{{\left\| {\tilde{a}_{1} } \right\|\omega_{1} \mu_{1}^{R} + \left\| {\tilde{a}_{2} } \right\|\omega_{2} \mu_{2}^{R} }}{{\omega_{1} \left\| {\tilde{a}_{1} } \right\| + \omega_{2} \left\| {\tilde{a}_{2} } \right\|}}} \right]\rangle \\ \end{aligned}$$then the Eq. (2) is clearly true.
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(2)
If Eq. (2) holds for \(n = k\), that is
$$\sum\limits_{i = 1}^{k} {\omega_{i} \tilde{a}_{i} = \left\langle {\left[ {\sum\limits_{i = 1}^{k} {\omega_{i} a_{i}^{{}} } ,\sum\limits_{i = 1}^{k} {\omega_{i} b_{i} } ,\sum\limits_{i = 1}^{k} {\omega_{i} c_{i} } } \right];\left[ {\frac{{\sum\nolimits_{i = 1}^{k} {\left\| {\tilde{a}_{i} } \right\|\omega_{i} \mu_{i}^{L} } }}{{\sum\nolimits_{i = 1}^{k} {\left\| {\tilde{a}_{i} } \right\|\omega_{i} } }},\frac{{\sum\nolimits_{i = 1}^{k} {\left\| {\tilde{a}_{i} } \right\|\omega_{i} \mu_{i}^{M} } }}{{\sum\nolimits_{i = 1}^{k} {\left\| {\tilde{a}_{i} } \right\|\omega_{i} } }},\frac{{\sum\nolimits_{i = 1}^{k} {\left\| {\tilde{a}_{i} } \right\|\omega_{i} \mu_{i}^{R} } }}{{\sum\nolimits_{i = 1}^{k} {\left\| {\tilde{a}_{i} } \right\|\omega_{i} } }}} \right]} \right\rangle }$$
Then, when \(n = k + 1\), by the operational laws in Definition 3, there is:
i.e. equation (2) holds for \(n = k + 1\).
Therefore, based on (1) and (2), Eq. (2) holds for all \(n \in N\), which completes the proof.□
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Han, Zq., Wang, Jq., Zhang, Hy. et al. Group Multi-criteria Decision Making Method with Triangular Type-2 Fuzzy Numbers. Int. J. Fuzzy Syst. 18, 673–684 (2016). https://doi.org/10.1007/s40815-015-0110-8
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DOI: https://doi.org/10.1007/s40815-015-0110-8