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Some new intuitionistic linguistic aggregation operators based on Maclaurin symmetric mean and their applications to multiple attribute group decision making

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Abstract

With respect to multiple attribute group decision making (MAGDM) problems in which the attributes are dependent and the attribute values take the forms of intuitionistic linguistic numbers and intuitionistic uncertain linguistic numbers, this paper investigates two novel MAGDM methods based on Maclaurin symmetric mean (MSM) aggregation operators. First, the Maclaurin symmetric mean is extended to intuitionistic linguistic environment and two new aggregation operators are developed for aggregating the intuitionistic linguistic information, such as the intuitionistic linguistic Maclaurin symmetric mean (ILMSM) operator and the weighted intuitionistic linguistic Maclaurin symmetric mean (WILMSM) operator. Then, some desirable properties and special cases of these operators are discussed in detail. Furthermore, this paper also develops two new Maclaurin symmetric mean operators for aggregating the intuitionistic uncertain linguistic information, including the intuitionistic uncertain linguistic Maclaurin symmetric mean (IULMSM) operator and the weighted intuitionistic uncertain linguistic Maclaurin symmetric mean (WIULMSM) operator. Based on the WILMSM and WIULMSM operators, two approaches to MAGDM are proposed under intuitionistic linguistic environment and intuitionistic uncertain linguistic environment, respectively. Finally, two practical examples of investment alternative evaluation are given to illustrate the applications of the proposed methods.

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Acknowledgments

This research is supported by Program for New Century Excellent Talents in University (NCET-13-0037), Humanities and Social Sciences Foundation of Ministry of Education of China (14YJA630019).

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Corresponding author

Correspondence to Yanbing Ju.

Additional information

Communicated by V. Loia.

Appendices

Appendix A: Proof of Theorem 2

According to the operational laws in Definition 2, we have

$$\begin{aligned} \mathop \otimes \limits _{j=1}^k a_{i_j } =\left\langle {s_{\prod \limits _{j=1}^k {\theta (a_{i_j } )} } ,\left( {\prod \limits _{j=1}^k {\mu (a_{i_j } )} ,1-\prod \limits _{j=1}^k {(1-v(a_{i_j } ))} }\right) }\right\rangle , \end{aligned}$$

and

$$\begin{aligned} \begin{array}{l} \displaystyle \mathop \oplus \limits _{1\le i_1 <\cdots <i_k \le n} \left( {\mathop \otimes \limits _{j=1}^k a_{i_j } }\right) \\ \displaystyle \quad =\left\langle s_{\sum \limits _{1\le i_1 <\cdots <i_k \le n} {\prod \limits _{j=1}^k {\theta (a_{i_j } )} } } ,\left( 1-\prod \limits _{1\le i_1 <\cdots <i_k \le n} \left( {1-\prod \limits _{j=1}^k {\mu (a_{i_j } )} }\right) ,\right. \right. \\ \displaystyle \quad \left. \left. \displaystyle \prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {(1-v(a_{i_j } ))} }\right) } \right) \right\rangle ; \\ \end{array} \end{aligned}$$

then we obtain

Therefore, we have

In addition, since

$$\begin{aligned}&0\le 1-\prod \limits _{j=1}^k {\mu (a_{i_j } )} \le 1\quad \;\mathrm{and}\\&\quad 0\le \prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\mu (a_{i_j } )} }\right) } \le 1, \end{aligned}$$

it follows that

$$\begin{aligned}&0\le \left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\mu (a_{i_j } )} }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k}\\&\quad \le 1. \end{aligned}$$

Similarly, we have

$$\begin{aligned} 0\le & {} \left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {(1-v(a_{i_j } ))} }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k}\\&\quad \le 1, \end{aligned}$$

which completes the proof of Theorem 2.

Appendix B: Proof of Theorem 3

  1. (1)

    Since \(a_{i}(i\)=1,2,...,n) are equal, i.e., \(a_i =\langle s_{\theta (a)} ,(\mu (a), v(a))\rangle \) for all i=1,2,...,n, according to Theorem 2, we have

  2. (2)

    Since \(s_{\theta (a_i )} \le s_{\theta (a_i^{'})} \) for any i, we have \(\theta (a_{i_j } )\le \theta (a_{i_j }^{'} )\). Based on the assumpation condition, for all j=1,2,...,k, we have

    $$\begin{aligned} \begin{array}{l} \displaystyle s_{\prod \limits _{j=1}^k {\theta (a_{i_j } )} } \le s_{\prod \limits _{j=1}^k {\theta (a_{i_j }^{'} )} } \\ \displaystyle \Rightarrow s_{\sum \limits _{1\le i_1 <\cdots <i_k \le n} {\prod \limits _{j=1}^k {\theta (a_{i_j } )} } } \le s_{\sum \limits _{1\le i_1 <\cdots <i_k \le n} {\prod \limits _{j=1}^k {\theta (a_{i_j }^{'} )} } } \\ \displaystyle \Rightarrow s_{\frac{\sum \limits _{1\le i_1 <\cdots <i_k \le n} {\prod \limits _{j=1}^k {\theta (a_{i_j } )} } }{C_n^k }} \le s_{\frac{\sum \limits _{1\le i_1 <\cdots <i_k \le n} {\prod \limits _{j=1}^k {\theta (a_{i_j }^{'} )} } }{C_n^k }.} \\ \end{array} \end{aligned}$$

    Therefore, we have

    $$\begin{aligned} s_{\left( {\frac{\sum \limits _{1\le i_1 <\cdots <i_k \le n} {\prod \limits _{j=1}^k {\theta (a_{i_j } )} } }{C_n^k }}\right) ^{\frac{1}{k}}} \le s_{\left( {\frac{\sum \limits _{1\le i_1 <\cdots <i_k \le n} {\prod \limits _{j=1}^k {\theta (a_{i_j }^{'} )} } }{C_n^k }}\right) ^{\frac{1}{k}}}. \end{aligned}$$
    (47)

    In addition, since \(\mu (a_i )\le \mu (a_i^{'} )\) and \(v(a_i )\ge v(a_i^{'} )\) for all i, we have

    $$\begin{aligned} \begin{array}{l} \prod \limits _{j=1}^k {\mu (a_{i_j } )} \le \prod \limits _{j=1}^k {\mu (a_{i_j }^{'} )} \\ \Rightarrow 1-\prod \limits _{j=1}^k {\mu (a_{i_j } )} \ge 1-\prod \limits _{j=1}^k {\mu (a_{i_j }^{'} )} \\ \Rightarrow \prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\mu (a_{i_j } )} }\right) } \\ \quad \ge \prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\mu (a_{i_j }^{'} )} }\right) }. \end{array} \end{aligned}$$

    Therefore, we can obtain

    $$\begin{aligned}&\left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\mu \left( a_{i_j } \right) } }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k} \nonumber \\&\quad \le \left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\mu \left( a_{i_j }^{'} \right) } }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k}.\nonumber \\ \end{aligned}$$
    (48)

    Similarly, we have

    $$\begin{aligned}&\displaystyle 1-\left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\left( 1-v\left( a_{i_j } \right) \right) } }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k} \nonumber \\&\quad \displaystyle \quad \ge 1-\left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\left( 1-v\left( a_{i_j }^{'} \right) \right) } }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k}. \nonumber \\ \end{aligned}$$
    (49)

    Therefore, based on Eqs.(48) and (49), we can obtain

    $$\begin{aligned} \begin{array}{l} \displaystyle \left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\mu \left( a_{i_j } \right) } }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k}+ \\ \displaystyle 1-\left( {1-\left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\left( 1-v\left( a_{i_j } \right) \right) } }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k}}\right) \le \\ \displaystyle \left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\mu \left( a_{i_j }^{'} \right) } }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k}+ \\ \displaystyle 1-\left( {1-\left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\left( 1-v\left( a_{i_j }^{'} \right) \right) } }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k}}\right) . \end{array} \end{aligned}$$
    (50)

    Let \(a\mathrm{=ILMSM(}a_\mathrm{1}, a_\mathrm{2} \mathrm{,}\ldots \mathrm{,}a_n )\) and \(a^{'}=\mathrm{ILMSM(}a_1^{'} \mathrm{,}a_2^{'} \mathrm{,}\ldots \mathrm{,}a_n^{'} )\); based on the Eqs. (47) and (50), using the score function of ILN in Eq. (6), we can easily obtain that \(S(a)\le S(a^{'})\). Therefore, we should discuss the following two cases:

  1. (1)

    If \(S(a)<S(a^{'})\), according to Theorem 1, we can obtain

    $$\begin{aligned} \mathrm{ILMSM}^{(k)}(a_1 ,a_2 ,\ldots ,a_n )<\mathrm{ILMSM}^{(k)}(a_1^{'} ,a_2^{'} ,\ldots ,a_n^{'} ). \end{aligned}$$
  2. (2)

    If \(S(a)=S(a^{'})\), then

    $$\begin{aligned} \begin{array}{l} \displaystyle \left( \left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\mu \left( a_{i_j } \right) } }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k}+\right. \displaystyle \left. 1-\left( {1-\left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\left( 1-v\left( a_{i_j } \right) \right) } }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k}}\right) \right) \,\\ \displaystyle \quad \quad \times \left( {\frac{\sum \limits _{1\le i_1 <\cdots <i_k \le n} {\prod \limits _{j=1}^k {\theta \left( a_{i_j } \right) } } }{C_n^k }}\right) ^{\frac{1}{k}} \\ \displaystyle \quad =\left( \left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\mu \left( a_{i_j }^{'} \right) } }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k}\right. \\ \displaystyle +\quad \left. 1-\left( {1-\left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\left( 1-v\left( a_{i_j }^{'} \right) \right) } }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k}}\right) \right) \displaystyle \times \left( {\frac{\sum \limits _{1\le i_1 <\cdots <i_k \le n} {\prod \limits _{j=1}^k {\theta \left( a_{i_j }^{'} \right) } } }{C_n^k }}\right) ^{\frac{1}{k}}. \end{array}\nonumber \\ \end{aligned}$$
    (51)

    Based on the Eqs. (47) and (48), we can get the following inequality:

    $$\begin{aligned}&\left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\mu \left( a_{i_j } \right) } }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k}\\&\quad =\left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\mu \left( a_{i_j }^{'} \right) } }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k} \end{aligned}$$

    and

    $$\begin{aligned}&\left( {\frac{\sum \limits _{1\le i_1 <\cdots <i_k \le n} {\prod \limits _{j=1}^k {\theta \left( a_{i_j } \right) } } }{C_n^k }}\right) ^{\frac{1}{k}}\\&\quad =\left( {\frac{\sum \limits _{1\le i_1 <\cdots <i_k \le n} {\prod \limits _{j=1}^k {\theta \left( a_{i_j }^{'} \right) } } }{C_n^k }}\right) ^{\frac{1}{k}}, \end{aligned}$$

    i.e.,

    $$\begin{aligned}&\left( {2\times \left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\mu \left( a_{i_j } \right) } }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k}+1}\right) \nonumber \\&\quad \quad \times \left( {\frac{\sum \limits _{1\le i_1 <\cdots <i_k \le n} {\prod \limits _{j=1}^k {\theta \left( a_{i_j } \right) } } }{C_n^k }}\right) ^{\frac{1}{k}} \nonumber \\&\quad =\left( {2\times \left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\mu \left( a_{i_j }^{'} \right) } }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k}+1}\right) \nonumber \\&\quad \quad \times \left( {\frac{\sum \limits _{1\le i_1 <\cdots <i_k \le n} {\prod \limits _{j=1}^k {\theta \left( a_{i_j }^{'} \right) } } }{C_n^k }}\right) ^{\frac{1}{k}}, \end{aligned}$$
    (52)

    and then subtracting Eq. (51) from Eq. (52), we can obtain that

    $$\begin{aligned}&\left( \left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\mu \left( a_{i_j } \right) } }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k}\right. \\&\quad \left. +\left( {1-\left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\left( 1-v\left( a_{i_j } \right) \right) } }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k}}\right) \right) \, \\&\quad \times \left( {\frac{\sum \limits _{1\le i_1 <\cdots <i_k \le n} {\prod \limits _{j=1}^k {\theta \left( a_{i_j } \right) } } }{C_n^k }}\right) ^{\frac{1}{k}} \\&\quad =\left( \left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\mu \left( a_{i_j }^{'} \right) } }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k}\right. \\&\quad \left. +\left( {1-\left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\left( 1-v\left( a_{i_j }^{'} \right) \right) } }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k}}\right) \right) \, \\&\quad \times \left( {\frac{\sum \limits _{1\le i_1 <\cdots <i_k \le n} {\prod \limits _{j=1}^k {\theta \left( a_{i_j }^{'} \right) } } }{C_n^k }}\right) ^{\frac{1}{k}}, \end{aligned}$$

    that is,

    $$\begin{aligned} H(a)=H(a^{'}). \end{aligned}$$

    Therefore, according to Theorem 1, we can obtain

    $$\begin{aligned}&\mathrm{ILMSM}^{(k)}(a_1 ,a_2 ,\ldots ,a_n )\\&\quad \le \mathrm{ILMSM}^{(k)}(a_1^{'} ,a_2^{'} ,\ldots ,a_n^{'} ). \end{aligned}$$
  3. (3)

    Since \(s_{\theta ^-} =\mathop {\min }\limits _{1\le i\le n} \{s_{\theta (a_i )} \}\),\(s_{\theta ^+} =\mathop {\max }\limits _{1\le i\le n} \{s_{\theta (a_i )} \}\), \(\mu ^-=\mathop {\min }\limits _{1\le i\le n} \{\mu (a_i )\}\),\(\mu ^+=\mathop {\max }\limits _{1\le i\le n} \{\mu (a_i )\}\), \(v^-=\mathop {\min }\limits _{1\le i\le n} \{v(a_i )\}\) and \(v^+=\mathop {\max }\limits _{1\le i\le n} \{v(a_i )\}\), then we have \(s_{\theta ^-} \le s_{\theta (a_i )} \le s_{\theta ^+} , \quad \mu ^-\le \mu (a_i )\le \mu ^+,\) and \(v^-\le v(a_i )\le v^+\) for all i=1,2,...,n. Therefore, according to the monotonicity and idempotency of Theorem 3, we can obtain

    $$\begin{aligned}&\langle s_{\theta ^-} ,(\mu ^-,v^+)\rangle \le \mathrm{ILMSM}^{(k)}(a_1 ,a_2 ,...,a_n )\\&\quad \le \langle s_{\theta ^+} ,(\mu ^+,v^-)\rangle . \end{aligned}$$
  4. (4)

    Since \(a_i^{'} =\langle s_{\theta (a_i^{'} )} ,(\mu (a_i^{'} ),v(a_i^{'} ))\rangle \) is any permutation of \(a_i =\langle s_{\theta (a_i )} ,(\mu (a_i ),v(a_i ))\rangle \) (i=1,2,...,n), then

    $$\begin{aligned}&\left( {\frac{\mathop \oplus \limits _{1\le i_1 <...<i_k \le n} (\mathop \otimes \limits _{j=1}^n a_{i_j } )}{C_n^k }}\right) ^{1/k}\\&\quad =\left( {\frac{\mathop \oplus \limits _{1\le i_1 <...<i_k \le n} (\mathop \otimes \limits _{j=1}^n a_{i_j }^{'} )}{C_n^k }}\right) ^{1/k}. \end{aligned}$$

    Therefore, according to Theorem 2, we can obtain

    $$\begin{aligned} \mathrm{ILMSM}^{(k)}(a_1 ,a_2 ,\ldots ,a_n )=\mathrm{ILMSM}^{(k)}(a_1^{'} ,a_2^{'} ,\ldots ,a_n^{'} ). \end{aligned}$$

Appendix C: Proof of Theorem 4

For \(a_i =\langle s_{\theta (a_i )} ,(\mu (a_i ),v(a_i ))\rangle (i\)=1,2,...,n), based on Theorem 2, we have

Let

$$\begin{aligned}&f(k)=\left( {\frac{\sum \limits _{1\le i_1 <\cdots <i_k \le n} {\prod \limits _{j=1}^k {\theta (a_{i_j } )} } }{C_n^k }}\right) ^{1/k}, \\&p(k)=\left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {\mu (a_{i_j } )} }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k} \mathrm{and}\\&q(k)=1-\left( {1-\left( {\prod \limits _{1\le i_1 <\cdots <i_k \le n} {\left( {1-\prod \limits _{j=1}^k {(1-v(a_{i_j } ))} }\right) } }\right) ^{\frac{1}{C_n^k }}}\right) ^{1/k} \end{aligned}$$

Based on Lemma 1, we can see that the function \(f(k)_{ }\)is monotonically decreasing with respect to the parameter k. Moreover, based on Lemma 2, we can see that the functions p(k) and q(k) are monotonically decreasing and increasing with respect to the parameter k, respectively.

According to the score function in Eq. (6), we have

$$\begin{aligned}&S(\mathrm{ILMSM}^{(k)}(a_1 ,a_2 ,\ldots ,a_n ))=( {p(k)+1-q(k)})\cdot f(k),\\&S(\mathrm{ILMSM}^{(k+1)}(a_1 ,a_2 ,\ldots ,a_n ))=( {p(k+1)+}\\&\quad {1-q(k+1)})\cdot f(k+1). \end{aligned}$$

For any positive integer \(k\in [1,n-1]\), we can obtain

$$\begin{aligned} \begin{array}{l} S(\mathrm{ILMSM}^{(k+1)}(a_1 ,a_2 ,\ldots ,a_n ))-S(\mathrm{ILMSM}^{(k)}(a_1 ,a_2 ,\ldots ,a_n )) \\ =( {p(k+1)+1-q(k+1)})\cdot f(k+1)-( {p(k)+1-q(k)})\cdot f(k) \\ \le ( {p(k+1)+1-q(k+1)})\cdot f(k)-( {p(k)+1-q(k)})\cdot f(k) \\ =( {(p(k+1)-p(k))-(q(k+1)-q(k))})\cdot f(k)\le 0, \\ \end{array} \end{aligned}$$

i.e., \(S(\mathrm{ILMSM}^{(k+1)}(a_1 ,a_2 ,\ldots ,a_n ))\le S(\mathrm{ILMSM}^{(k)} (a_1 ,a_2 ,\ldots ,a_n ))\) for any integer \(k\in [1,n-1]\), which completes the proof of Theorem 4.

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Ju, Y., Liu, X. & Ju, D. Some new intuitionistic linguistic aggregation operators based on Maclaurin symmetric mean and their applications to multiple attribute group decision making. Soft Comput 20, 4521–4548 (2016). https://doi.org/10.1007/s00500-015-1761-y

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