A Multi-attribute Decision-Making Approach Based on Spherical Fuzzy Sets for Yunnan Baiyao’s R&D Project Selection Problem

As we all know, the research and development (R&D) is crucial for enterprises. How to choose the R&D project is an important research topic, and at the same time, it is also a typical multi-attribute decision-making (MADM) problem. In this paper, we propose a novel MADM method for selecting the Yunnan Baiyao’s R&D project of toothpastes. Firstly, we use T-spherical fuzzy sets (T-SFSs) to express the evaluation information of the toothpastes from decision makers to overcome the shortcomings existing that the traditional information form may cause information distortion. Secondly, we extend the generalized Maclaurin symmetric mean (GMSM) operator to T-spherical fuzzy environment and propose the T-spherical fuzzy GMSM operator (T-SFGMSM) and the T-spherical fuzzy weighted GMSM operator (T-SFWGMSM), where these operators are suitable for the situation which the input information is interrelated. Further, we put forward a novel MADM approach for selecting toothpaste product based upon the T-SFWGMSM operator, which can not only handle a more extensive scope of fuzzy information, but also process the interrelationships between multiple attributes. Lastly, we solve a R&D project selection problem for Yunnan Baiyao Co., Ltd. by the proposed method and make a detailed comparative analysis with other MADM methods to demonstrate the effectiveness and superiority of the novel approach.

This paper is supported by the National Natural Science Foundation of China (Nos. 71771140, 71471172), the Special Funds of Taishan Scholars Project of Shandong Province (No. ts201511045).

Authors and Affiliations

1. School of Management Science and Engineering, Shandong University of Finance and Economics, Jinan, 250014, Shandong, China

   Peide Liu, Baoying Zhu & Peng Wang

Corresponding author

Correspondence to Peide Liu.

Appendix 1

Suppose \(T = \left\langle {s,i,d} \right\rangle\), \(T_{A} = \left\langle {s_{A} ,i_{A} ,d_{A} } \right\rangle\), \(T_{B} = \left\langle {s_{B} ,i_{B} ,d_{B} } \right\rangle\)are any three T-SFNs, and \(\lambda ,\lambda_{1,} \lambda_{2} > 0\), Then they have the following operational properties.

Proof

1. 1.
   $$\begin{array}{*{20}l} {T_{A} \oplus T_{B} = \left\langle {\left( {1 - \left( {1 - s_{A}^{n} } \right)\left( {1{ - }s_{B}^{n} } \right)} \right)^{1/n} ,i_{A} i_{B} ,d_{A} d_{B} } \right\rangle } \hfill \\ { = \left\langle {\left( {1 - \left( {1 - s_{B}^{n} } \right)\left( {1{ - }s_{A}^{n} } \right)} \right)^{1/n} ,i_{B} i_{A} ,d_{B} d_{A} } \right\rangle = T_{B} \oplus T_{A} } \hfill \\ \end{array}$$
2. 2.
   $$\begin{array}{*{20}l} {T_{A} \otimes T_{B} = \left\langle {s_{A} s_{B} ,\left( {1 - \left( {1 - i_{A}^{n} } \right)\left( {1{ - }i_{B}^{n} } \right)} \right)^{1/n} ,\left( {1 - \left( {1 - d_{A}^{n} } \right)\left( {1{ - }d_{B}^{n} } \right)} \right)^{1/n} } \right\rangle } \hfill \\ { = \left\langle {s_{B} s_{A} ,\left( {1 - \left( {1 - i_{B}^{n} } \right)\left( {1{ - }i_{A}^{n} } \right)} \right)^{1/n} ,\left( {1 - \left( {1 - d_{B}^{n} } \right)\left( {1{ - }d_{A}^{n} } \right)} \right)^{1/n} } \right\rangle } \hfill \\ { = T_{B} \otimes T_{A} .} \hfill \\ \end{array}$$
3. 3.
   $$\begin{array}{*{20}l} {\lambda \left( {T_{A} \oplus T_{B} } \right) = \lambda \left\langle {\left( {1 - \left( {1 - s_{A}^{n} } \right)\left( {1{ - }s_{B}^{n} } \right)} \right)^{1/n} ,i_{A} i_{B} ,d_{A} d_{B} } \right\rangle } \hfill \\ { = \left\langle {\left( {1 - \left( {1 - \left( {\left( {1 - \left( {1 - s_{A}^{n} } \right)\left( {1{ - }s_{B}^{n} } \right)} \right)^{1/n} } \right)^{n} } \right)^{\lambda } } \right)^{1/n} ,i_{A}^{\lambda } ,d_{A}^{\lambda } } \right\rangle } \hfill \\ { = \left\langle {\left( {1 - \left( {1 - \left( {1 - \left( {1 - s_{A}^{n} } \right)\left( {1{ - }s_{B}^{n} } \right)} \right)} \right)^{\lambda } } \right)^{1/n} ,\left( {i_{A} i_{B} } \right)^{\lambda } ,\left( {d_{A} d_{B} } \right)^{\lambda } } \right\rangle } \hfill \\ { = \left\langle {\left( {1 - \left( {\left( {1 - s_{A}^{n} } \right)\left( {1{ - }s_{B}^{n} } \right)} \right)^{\lambda } } \right)^{1/n} ,\left( {i_{A} i_{B} } \right)^{\lambda } ,\left( {d_{A} d_{B} } \right)^{\lambda } } \right\rangle } \hfill \\ \end{array}$$

Because \(\lambda T_{A} = \left\langle {\left( {1 - \left( {1 - s_{A}^{n} } \right)^{\lambda } } \right)^{1/n} ,i_{A}^{\lambda } ,d_{A}^{\lambda } } \right\rangle ,\)\(\lambda T_{B} = \left\langle {\left( {1 - \left( {1 - s_{B}^{n} } \right)^{\lambda } } \right)^{1/n} ,i_{B}^{\lambda } ,d_{B}^{\lambda } } \right\rangle\)

So, \(\lambda \left( {T_{A} \oplus T_{B} } \right) = \lambda T_{A} \oplus \lambda T_{B}\).

1. 4.
   $$\begin{array}{*{20}l} {\lambda_{1} T \oplus \lambda_{2} T{ = }\left\langle {\left( {1 - \left( {1 - s^{n} } \right)^{{\lambda_{1} }} } \right)^{1/n} ,i^{{\lambda_{1} }} ,d^{{\lambda_{1} }} } \right\rangle \oplus \left\langle {\left( {1 - \left( {1 - s^{n} } \right)^{{\lambda_{2} }} } \right)^{1/n} ,i^{{\lambda_{2} }} ,d^{{\lambda_{2} }} } \right\rangle } \hfill \\ { = \left\langle {\left( {1 - \left( {1 - \left( {\left( {1 - \left( {1 - s^{n} } \right)^{{\lambda_{1} }} } \right)^{1/n} } \right)^{n} } \right)\left( {1{ - }\left( {\left( {1 - \left( {1 - s^{n} } \right)^{{\lambda_{2} }} } \right)^{1/n} } \right)^{n} } \right)} \right)^{1/n} ,i^{{\lambda_{1} }} i^{{\lambda_{2} }} ,d^{{\lambda_{1} }} d^{{\lambda_{2} }} } \right\rangle } \hfill \\ { = \left\langle {\left( {1 - \left( {1 - \left( {1 - \left( {1 - s^{n} } \right)^{{\lambda_{1} }} } \right)} \right)\left( {1{ - }\left( {1 - \left( {1 - s^{n} } \right)^{{\lambda_{2} }} } \right)} \right)} \right)^{1/n} ,i^{{\lambda_{1} }} i^{{\lambda_{2} }} ,d^{{\lambda_{1} }} d^{{\lambda_{2} }} } \right\rangle } \hfill \\ { = \left\langle {\left( {1 - \left( {1 - s^{n} } \right)^{{\lambda_{1} }} \left( {1 - s^{n} } \right)^{{\lambda_{2} }} } \right)^{1/n} ,i^{{\lambda_{1} }} i^{{\lambda_{2} }} ,d^{{\lambda_{1} }} d^{{\lambda_{2} }} } \right\rangle } \hfill \\ { = \left\langle {\left( {1 - \left( {1 - s^{n} } \right)^{{\lambda_{1} { + }\lambda_{2} }} } \right)^{1/n} ,i^{{\lambda_{1} { + }\lambda_{2} }} ,d^{{\lambda_{1} { + }\lambda_{2} }} } \right\rangle } \hfill \\ \end{array}$$

and \(\left( {\lambda_{1} \oplus \lambda_{2} } \right)T = \left\langle {\left( {1 - \left( {1 - s^{n} } \right)^{{\lambda_{1} { + }\lambda_{2} }} } \right)^{1/n} ,i^{{\lambda_{1} { + }\lambda_{2} }} ,d^{{\lambda_{1} { + }\lambda_{2} }} } \right\rangle\)So \(\lambda_{1} T \oplus \lambda_{2} T = \left( {\lambda_{1} \oplus \lambda_{2} } \right)T\)

1. 5.

   \(\begin{array}{*{20}l} {\left( {T_{A} \otimes T_{B} } \right)^{\lambda } { = }\left\langle {s_{A} s_{B} ,\left( {1 - \left( {1 - i_{A}^{n} } \right)\left( {1{ - }i_{B}^{n} } \right)} \right)^{1/n} ,\left( {1 - \left( {1 - d_{A}^{n} } \right)\left( {1{ - }d_{B}^{n} } \right)} \right)^{1/n} } \right\rangle^{\lambda } } \hfill \\ { = \left\langle {\left( {s_{A} s_{B} } \right)^{\lambda } ,\left( {1 - \left( {1 - \left( {\left( {1 - \left( {1 - i_{A}^{n} } \right)\left( {1{ - }i_{B}^{n} } \right)} \right)^{1/n} } \right)^{n} } \right)^{\lambda } } \right)^{1/n} ,\left( {1 - \left( {1 - \left( {\left( {1 - \left( {1 - d_{A}^{n} } \right)\left( {1{ - }d_{B}^{n} } \right)} \right)^{1/n} } \right)^{n} } \right)^{\lambda } } \right)^{1/n} } \right\rangle } \hfill \\ { = \left\langle {\left( {s_{A} s_{B} } \right)^{\lambda } ,\left( {1 - \left( {1 - \left( {1 - \left( {1 - i_{A}^{n} } \right)\left( {1{ - }i_{B}^{n} } \right)} \right)} \right)^{\lambda } } \right)^{1/n} ,\left( {1 - \left( {1 - \left( {1 - \left( {1 - d_{A}^{n} } \right)\left( {1{ - }d_{B}^{n} } \right)} \right)} \right)^{\lambda } } \right)^{1/n} } \right\rangle } \hfill \\ { = \left\langle {\left( {s_{A} s_{B} } \right)^{\lambda } ,\left( {1 - \left( {\left( {1 - i_{A}^{n} } \right)\left( {1{ - }i_{B}^{n} } \right)} \right)^{\lambda } } \right)^{1/n} ,\left( {1 - \left( {\left( {1 - d_{A}^{n} } \right)\left( {1{ - }d_{B}^{n} } \right)} \right)^{\lambda } } \right)^{1/n} } \right\rangle } \hfill \\ \end{array}\)

and

So \(\left( {T_{A} \otimes T_{B} } \right)^{\lambda } = T_{A}^{\lambda } \otimes T_{B}^{\lambda }\)

1. 6.
   $$\begin{array}{*{20}l} {T^{{\lambda_{1} }} \otimes T^{{\lambda_{2} }} = \left\langle {s^{{\lambda_{1} }} ,\left( {1 - \left( {1 - i^{n} } \right)^{{\lambda_{1} }} } \right)^{1/n} ,\left( {1 - \left( {1 - d^{n} } \right)^{{\lambda_{1} }} } \right)^{1/n} } \right\rangle \otimes \left\langle {s^{{\lambda_{2} }} ,\left( {1 - \left( {1 - i^{n} } \right)^{{\lambda_{2} }} } \right)^{1/n} ,\left( {1 - \left( {1 - d^{n} } \right)^{{\lambda_{2} }} } \right)^{1/n} } \right\rangle } \hfill \\ { = \left\langle \begin{aligned} s^{{\lambda_{1} }} s^{{\lambda_{2} }} ,\left( {1 - \left( {1 - \left( {\left( {1 - \left( {1 - i^{n} } \right)^{{\lambda_{1} }} } \right)^{1/n} } \right)^{n} } \right)\left( {1{ - }\left( {\left( {1 - \left( {1 - i^{n} } \right)^{{\lambda_{2} }} } \right)^{1/n} } \right)^{n} } \right)} \right)^{1/n} , \hfill \\ \left( {1 - \left( {1 - \left( {\left( {1 - \left( {1 - d^{n} } \right)^{{\lambda_{1} }} } \right)^{1/n} } \right)^{n} } \right)\left( {1{ - }\left( {\left( {1 - \left( {1 - d^{n} } \right)^{{\lambda_{2} }} } \right)^{1/n} } \right)^{n} } \right)} \right)^{1/n} \hfill \\ \end{aligned} \right\rangle } \hfill \\ { = \left\langle {s^{{\lambda_{1} }} s^{{\lambda_{2} }} ,\left( {1 - \left( {1 - \left( {1 - \left( {1 - i^{n} } \right)^{{\lambda_{1} }} } \right)} \right)\left( {1{ - }\left( {1 - \left( {1 - i^{n} } \right)^{{\lambda_{2} }} } \right)} \right)} \right)^{1/n} ,\left( {1 - \left( {1 - \left( {1 - \left( {1 - d^{n} } \right)^{{\lambda_{1} }} } \right)} \right)\left( {1{ - }\left( {1 - \left( {1 - d^{n} } \right)^{{\lambda_{2} }} } \right)} \right)} \right)^{1/n} } \right\rangle } \hfill \\ { = \left\langle {s^{{\lambda_{1} }} s^{{\lambda_{2} }} ,\left( {1 - \left( {\left( {1 - i^{n} } \right)^{{\lambda_{1} }} } \right)\left( {\left( {1 - i^{n} } \right)^{{\lambda_{2} }} } \right)} \right)^{1/n} ,\left( {1 - \left( {\left( {1 - d^{n} } \right)^{{\lambda_{1} }} } \right)\left( {\left( {1 - d^{n} } \right)^{{\lambda_{2} }} } \right)} \right)^{1/n} } \right\rangle } \hfill \\ { = \left\langle {s^{{\left( {\lambda_{1} + \lambda_{2} } \right)}} ,\left( {1 - \left( {1 - i^{n} } \right)^{{\left( {\lambda_{1} + \lambda_{2} } \right)}} } \right)^{1/n} ,\left( {1 - \left( {1 - d^{n} } \right)^{{\left( {\lambda_{1} + \lambda_{2} } \right)}} } \right)^{1/n} } \right\rangle } \hfill \\ \end{array}$$

and \(T^{{\left( {\lambda_{1} + \lambda_{2} } \right)}} = \left\langle {s,i,d} \right\rangle^{{\left( {\lambda_{1} + \lambda_{2} } \right)}}\)\(= \left\langle {s^{{\left( {\lambda_{1} + \lambda_{2} } \right)}} ,\left( {1 - \left( {1 - i^{n} } \right)^{{\left( {\lambda_{1} + \lambda_{2} } \right)}} } \right)^{1/n} ,\left( {1 - \left( {1 - d^{n} } \right)^{{\left( {\lambda_{1} + \lambda_{2} } \right)}} } \right)^{1/n} } \right\rangle\)

So \(T^{{\lambda_{1} }} \otimes T^{{\lambda_{2} }} = T^{{\left( {\lambda_{1} + \lambda_{2} } \right)}}\).□

Appendix 2

Theorem 1

Suppose\(T_{1} ,T_{2} , \ldots\)and\(T_{k}\)are T-SFNs, where\(T_{m} = \left\langle {s_{m} ,i_{m} ,d_{m} } \right\rangle\)\((m = 1,2, \ldots ,k)\). The synthesized result of the T-SFGMSM operator of the T-SFNs\(T_{1} ,T_{2} , \ldots\)and\(T_{k}\)is still a T-SFN, shown as follows:

Proof

According to the operational rules of T-SFNs, we can obtain

and

Then, we can calculate

and

So the comprehensive result of the T-SFGMSM operator is

That proves Eq. (16) is right.□

Appendix 3

(1) Idempotency: If \(T_{1} ,T_{2} , \ldots ,\) and \(T_{k}\) are T-SFNs meeting the condition with\(T_{m} = \left\{ {s_{m} ,i_{m} ,d_{m} } \right\} = T = \left\{ {s,i,d} \right\}\)\((m = 1,2, \ldots ,k)\). Then, \(\hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) = T\).

Proof.

According to the above Theorem 1, we can obtain that \(T_{m} = \left\langle {s_{m} ,i_{m} ,d_{m} } \right\rangle = T = \left\langle {s,i,d} \right\rangle\) for any \(m = 1,2, \ldots ,k\).

□

Appendix 4

(2) Monotonicity: If \(T_{m} = \left\langle {s_{m} ,i_{m} ,d_{m} } \right\rangle\)\(\left( {m = 1,2, \ldots ,k} \right)\) and \(\hat{T}_{m} = \left\langle {\hat{s}_{m} ,\hat{i}_{m} ,\hat{d}_{m} } \right\rangle\)\(\left( {m = 1,2, \ldots ,k} \right)\) are any two sets of T-SFNs. They accord the prerequisite “\(\hat{s}_{m} \le s_{m}\), \(\hat{i}_{m} \le i_{m}\) and \(\hat{d}_{m} \ge d_{m}\),” then\(\hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {\hat{T}_{1} ,\hat{T}_{2} , \ldots ,\hat{T}_{k} } \right) \le \hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right).\)

Proof

Suppose that \(\hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {\hat{T}_{1} ,\hat{T}_{2} , \ldots ,\hat{T}_{k} } \right) = \hat{T} = \left\langle {\hat{s},\hat{i},\hat{d}} \right\rangle\) and \(\hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) = T = \left\langle {s,i,d} \right\rangle\), then

In order to prove this property, we need to compute their score function values \(S(\hat{T})\) and \(S(T)\), and their accuracy function values \(A(\hat{T})\) and \(A(T)\) to compare their synthesized results, i.e., \(\hat{T} \le T\). Firstly, based on the conditions \(\hat{s}_{m} \le s_{m}\), \(\hat{i}_{m} \le i_{m}\) and \(\hat{d}_{m} \ge d_{m}\), we can get the compared results of their MD, AMD and NMD.

1. 1.

   For the membership degree

Based on \(\hat{s}_{m} \le s_{m}\), we can get

That is \(\hat{s}_{m} \le s_{m}\)

1. 2.

   For the abstinence degree

Based on \(\hat{i}_{m} \le i_{m}\) , we can obtain

That is \(\hat{i}_{m} \le i_{m}\)

1. 3.

   For the non-membership degree

Based on \(\hat{d}_{m} \ge d_{m}\) , we can obtain

That is \(\hat{d}_{m} \ge d_{m}\)Thus, it can be obtained that \(S(\hat{T}) = \hat{s}^{n} - \hat{d}^{n} \le S(T) = s^{n} - d^{n}\). Next, we explore two cases.

1. 1.

   If \(S(\hat{T}) < S(T)\), then \(\hat{T} < T\) according to Definition 4.

2. 2.

   If \(S(\hat{T}) = S(T)\), then \(\hat{s} = s\), \(\hat{d} = d\) Because \(\hat{i} \le i\), so \(A(\hat{T}) = \hat{s}^{n} + \hat{i}^{n} { + }\hat{d}^{n} = A(T) = s^{n} + i^{n} { + }d^{n}\), which testifies \(\hat{T} \le T\).

In conclusion, the synthesized result \(\hat{T} \le T\), which explains\(\hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {\hat{T}_{1} ,\hat{T}_{2} , \ldots ,\hat{T}_{k} } \right) \le \hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right)\)□

Appendix 5

(3) Boundedness: Suppose \(T_{m} = \left\langle {s_{m} ,i_{m} ,d_{m} } \right\rangle\)\(\left( {m = 1,2, \ldots ,k} \right)\) is any a set of T-SFNs, \(T^{ - } = \mathop { \hbox{min} }\limits_{m} T_{m}\) and \(T^{ + } = \mathop { \hbox{max} }\limits_{m} T_{m}\), then \(T^{ - } \le \hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \cdots ,T_{k} } \right) \le T^{ + }\).

Proof

Owing to \(T_{m} \ge T^{ - } = \mathop { \hbox{min} }\limits_{m} T_{m}\), based on the Monotonicity and Idempotency of the novel \(\hbox{T-SFGMSM}\) operator, the following results can be obtained:

comparably, the corresponding result for \(T^{ + } = \mathop { \hbox{max} }\limits_{m} T_{m}\) can be obtained:

Therefore, \(T^{ - } \le \hbox{T-SFGMSM}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) \le T^{ + }\).□

Appendix 6

Theorem 3

Let\(T_{1} ,T_{2} , \cdots\)and\(T_{k}\)be T-SFNs, where\(T_{m} = \left\langle {s_{m} ,i_{m} ,d_{m} } \right\rangle\)\(m = \left( {1,2, \ldots ,k} \right)\), and let the weight of input argument\(T_{m} = \left\langle {s_{m} ,i_{m} ,d_{m} } \right\rangle\)\(m = \left( {1,2, \ldots ,k} \right)\)be\(\omega_{m}\), where\(\omega_{m} \in \left[ {0,1} \right]\)and\(\sum\nolimits_{m = 1}^{k} {\omega_{m} } = 1\). Then, the synthesized result of the\(\hbox{T-SFGMSM}\)operator of the T-SFNs\(T_{1} ,T_{2} , \ldots\)and\(T_{k}\)is still a T-SFN, shown as follows:

Proof

Obtained by the operational laws of T-SFNs defined in Definition 5, Theorem 3 can be easily proved.

1. 1.

   For \(1 \le r < k\), we can have \(\mathop \otimes \limits_{j = 1}^{r} T_{{m_{j} }}^{{\lambda_{j} }} = \left\langle {\prod\limits_{j = 1}^{r} {s_{{m_{j} }}^{{\lambda_{j} }} } ,\left( {1 - \prod\limits_{j = 1}^{r} {\left( {1 - i_{{m_{j} }}^{n} } \right)^{{\lambda_{j} }} } } \right)^{1/n} ,\left( {1 - \prod\limits_{j = 1}^{r} {\left( {1 - d_{{m_{j} }}^{n} } \right)^{{\lambda_{j} }} } } \right)^{1/n} } \right\rangle\)

Obtained by the operational laws of T-SFNs, we can derive the below equations:

and

then

So formula (18) is right when \(1 \le r < n\).

1. 2.

   For \(r = k\), we have known that

   $$\mathop \otimes \limits_{m = 1}^{r} T_{m}^{{\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}}} = \left\langle {\prod\limits_{m = 1}^{r} {s_{m}^{{\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}}} } ,\left( {1 - \prod\limits_{m = 1}^{r} {(1 - i_{m}^{n} )^{{\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}}} } } \right)^{1/n} ,\left( {1 - \prod\limits_{m = 1}^{r} {(1 - d_{m}^{n} )^{{\lambda_{m} + \frac{{1 - k\omega_{m} }}{k - 1}}} } } \right)^{1/n} } \right\rangle$$

Based on the proof of the first case when \(1 \le r < n\), we can derive the below equations:

□

Appendix 7

(1) Idempotency: If \(T_{1} ,T_{2} , \ldots ,\) and \(T_{k}\) are T-SFNs and meet \(T_{m} = \left\langle {s_{m} , i_{m} ,d_{m} } \right\rangle = T = \left\langle {s, i,d} \right\rangle\)\((m = 1,2, \ldots ,k)\). Then, \({\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) = T\).

Proof

According to the above Theorem 3, we can obtain that

1. 1.

   For \(1 \le r < k\), we can get

1. 2.

   For \(r = k\), we can get

Because \(r = k\) and \(\sum\nolimits_{m = 1}^{k} {\omega_{m} } = 1\), we have \(\sum\nolimits_{m = 1}^{r} {\left( {1 - k\omega_{m} } \right)} = k - k\sum\nolimits_{m = 1}^{k} {\omega_{m} } = 0\) and the upper equality becomes the following form:

Therefore, the \({\hbox{T-SFWGMSM}}\) operator has the Idempotency property.□

Appendix 8

(2) Monotonicity: If \(T_{m} = \left\langle {s_{m} ,i_{m} ,d_{m} } \right\rangle\)\(\left( {m = 1,2, \ldots ,k} \right)\) and \(\hat{T}_{m} = \left\langle {\hat{s}_{m} ,\hat{i}_{m} ,\hat{d}_{m} } \right\rangle\)\(\left( {m = 1,2, \ldots ,k} \right)\) are any two sets about T-SFNs. They accord the prerequisite \(\hat{s}_{m} \le s_{m}\), \(\hat{i}_{m} \le i_{m}\) and \(\hat{d}_{m} \ge d_{m}\), then \({\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {\hat{T}_{1} ,\hat{T}_{2} , \ldots ,\hat{T}_{k} } \right) \le {\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right).\)

Proof

Similar to the monotonicity property of the \({\hbox{T-SFWGMSM}}\) operator, the following formulas can be assumed:\({\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {\hat{T}_{1} ,\hat{T}_{2} , \ldots ,\hat{T}_{k} } \right) = \hat{T} = \left\langle {\hat{s},\hat{i},\hat{d}} \right\rangle\) and \({\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) = T = \left\langle {s,i,d} \right\rangle\). Since formula (18) is divided into two cases, we discuss each case separately.

1. 1.

   For \(1 \le r < k\), we can get

Because \(\hat{s}_{m} \le s_{m}\), \(\hat{i}_{m} \le i_{m}\) and \(\hat{d}_{m} \ge d_{m}\), it is easy to obtain the following inequalities:

and

i.e., \(\hat{s} \le s\) and \(\hat{i} \le i\).Thus, it can be obtained that \(S(\hat{T}) = \hat{s}^{n} - \hat{d}^{n} \le S(T) = s^{n} - d^{n}\). Next, we discuss two cases.

• (a) If \(S(\hat{T}) < S(T)\), then \(\hat{T} < T\) according to Definition 4.

• (b) If \(S(\hat{T}) = S(T)\), then \(\hat{s} = s\), \(\hat{d} = d\) Because \(\hat{i} \le i\), so \(A(\hat{T}) = \hat{s}^{n} + \hat{i}^{n} { + }\hat{d}^{n} = A(T) = s^{n} + i^{n} { + }d^{n}\), which testifies \(\hat{T} < T\).

To sum up, the synthesized result is \(\hat{T} < T\).

1. 2.

   For \(r = k\), its proof process is similar to the one of the first case where \(1 \le r < k\), so we omit it here.

According to (i) and (ii), we can get \(\hat{T} \le T\), which explains \({\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {\hat{T}_{1} ,\hat{T}_{2} , \ldots ,\hat{T}_{k} } \right) \le {\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right).\)□

Appendix 9

(3) Boundedness: Let \(T_{m} = \left\langle {s_{m} ,i_{m} ,d_{m} } \right\rangle\)\(\left( {m = 1,2, \ldots ,k} \right)\) be any a set of T-SFNs, \(T^{ - } = \mathop { \hbox{min} }\limits_{m} T_{m}\) and\(T^{ + } = \mathop { \hbox{max} }\limits_{m} T_{m}\), then \(T^{ - } \le {\hbox{T-SFWGMSM}}^{{\left( {r, \lambda_{1} , \lambda_{2} , \ldots , \lambda_{r} } \right)}} \left( {T_{1} ,T_{2} , \ldots ,T_{k} } \right) \le T^{ + }\).

Because the certification process is same with the one of corresponding Boundedness property of the \(\hbox{T-SFGMSM}\) operator, we omit it here.

Similarly, we also observe many particular instances about the T-SFWGMSM operator with different parameters. Here, we just introduce a particular instance of this operator.

When \(\omega_{m} = \frac{1}{k}\)\(\left( {m = 1,2, \ldots ,k} \right)\), the proposed \({\hbox{T-SFWGMSM}}\) operator is simplified into the \(\hbox{T-SFGMSM}\) operator.

Proof

According to Eq. (18), we should take two cases into account.

1. 1.

   For \(1 \le r < k\), we know\(\omega_{m} = \frac{1}{k}\)\(\left( {m = 1,2, \ldots ,k} \right)\), so we can get

1. 2.

   For \(r = k\), based on\(\omega_{m} = \frac{1}{k}\)\(\left( {m = 1,2, \ldots ,k} \right)\), we have

By Theorem 1, we can see

Therefore, when the weights of input arguments are assigned to equal values, the proposed T-SFWGMSM operator is simplified as the T-SFGMSM operator.□

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