Possibility Degree and Power Aggregation Operators of Single-Valued Trapezoidal Neutrosophic Numbers and Applications to Multi-Criteria Group Decision-Making

Abstract

Single-valued trapezoidal neutrosophic numbers (SVTNNs) are very useful tools to describe complex cognitive information because of their advantage in maintaining the completeness and accuracy of information. This paper develops a method based on the single-valued trapezoidal neutrosophic power-weighted aggregation operators and possibility degree of SVTNNs for dealing with multi-criteria group decision-making (MCGDM) problems. First, the limitations of the existing operations for SVTNNs are discussed, and then an improved operation is defined. Moreover, the possibility degree of two SVTNNs with consideration of the influence of risk attitudes is proposed, and the comparison rules for SVTNNs are thereby established. Based on the new operation and possibility degree of SVTNNs, the single-valued trapezoidal neutrosophic power average and single-valued trapezoidal neutrosophic power geometric operators are proposed to aggregate the single-valued trapezoidal neutrosophic information. Furthermore, a single-valued trapezoidal neutrosophic MCGDM method is developed. Finally, an example of a company selecting the most suitable green supplier is provided to present a comparative analysis between the proposed approach and other related methods. This example can demonstrate the effectiveness and flexibility of the proposed methodology.

Appendices

Appendix 1. Proof of Theorem 2

Proof. We first prove Property (2) and prepare for the proof of the other properties.

• (A) First,

  $$ {\displaystyle \begin{array}{c}x\left(\alpha, \beta \right)+x\left(\beta, \alpha \right)=\frac{\sum_{i=1}^4\max \left({b}_i-{a}_i,0\right)+\left({b}_4-{a}_1\right)+2\max \left(T\left(\beta \right)-T\left(\alpha \right),0\right)}{\sum_{i=1}^4\left|{b}_i-{a}_i\right|+\left({b}_4-{b}_1\right)+\left({a}_4-{a}_1\right)+2\left|T\left(\beta \right)-T\left(\alpha \right)\right|}\\ {}+\frac{\sum_{i=1}^4\max \left({a}_i-{b}_i,0\right)+\left({a}_4-{b}_1\right)+2\max \left(T\left(\alpha \right)-T\left(\beta \right),0\right)}{\sum_{i=1}^4\left|{a}_i-{b}_i\right|+\left({a}_4-{a}_1\right)+\left({b}_4-{b}_1\right)+2\left|T\left(\alpha \right)-T\left(\beta \right)\right|}\\ {}\begin{array}{c}=\frac{\sum_{i=1}^4\left|{a}_i-{b}_i\right|+\left({a}_4-{b}_1\right)+\left({b}_4-{a}_1\right)+2\left|T\left(\alpha \right)-T\left(\beta \right)\right|}{\sum_{i=1}^4\left|{a}_i-{b}_i\right|+\left({a}_4-{a}_1\right)+\left({b}_4-{b}_1\right)+2\left|T\left(\alpha \right)-T\left(\beta \right)\right|}=1;\\ {}y\left(a,\beta \right)+y\left(\beta, a\right)=1;z\left(a,\beta \right)+z\left(\beta, a\right)=1.\end{array}\end{array}} $$

Second, there are three cases being considered.

• (i) If x(α, β) < 0, then x(β, α) = 1 − x(α, β) > 1.

  $$ \max \left\{1-\max \left[x\left(\alpha, \beta \right),0\right],0\right\}+\max \left\{1-\max \left[x\left(\beta, \alpha \right),0\right],0\right\}=1+0=1. $$

• (ii) If 0 ≤ x(α, β) < 1, then 0 < x(β, α) = 1 − x(α, β) ≤ 1.

  $$ {\displaystyle \begin{array}{c}\max \left\{1-\max \left[x\left(\alpha, \beta \right),0\right],0\right\}+\max \left\{1-\max \left[x\left(\beta, \alpha \right),0\right],0\right\}\\ {}=\max \left\{1-x\left(\alpha, \beta \right),0\right\}+\max \left\{1-x\left(\beta, \alpha \right),0\right\}\\ {}=1-x\left(\alpha, \beta \right)+1-x\left(\beta, \alpha \right)=1.\end{array}}. $$

• (iii) If x(α, β) ≥ 1, then x(β, α) = 1 − x(α, β) ≤ 0.

$$ \max \left\{1-\max \left[x\left(\alpha, \beta \right),0\right],0\right\}+\max \left\{1-\max \left[x\left(\beta, \alpha \right),0\right],0\right\}=0+1=1. $$

Thus, max{1 − max[x(α, β), 0], 0} + max {1 − max[x(β, α), 0], 0} = 1.

Similarly, max{1 − max[y(α, β), 0], 0} + max {1 − max[y(β, α), 0], 0} = 1;

$$ 1-\max \left\{1-\max \left[z\left(\alpha, \beta \right),0\right],0\right\}+1-\max \left\{1-\max \left[z\left(\beta, \alpha \right),0\right],0\right\}=1. $$

Third,

$$ {\displaystyle \begin{array}{c}p\left(\alpha \succ \beta \right)+P\left(\beta \succ \alpha \right)\\ {}=\frac{1}{2+\lambda}\left(\max \left\{1-\max \left[x\left(\alpha, \beta \right),0\right],0\right\}+\lambda \max \left\{1-\max \left[y\left(\alpha, \beta \right),0\right],0\right\}\left.+1-\max \left\{1-\max \left[z\left(\alpha, \beta \right),0\right],0\right\}\right)\right.\\ {}+\frac{1}{2+\lambda}\left(\max \left\{1-\max \left[x\left(\beta, \alpha \right),0\right],0\right\}+\lambda \max \left\{1-\max \left[y\left(\beta, \alpha \right),0\right],0\right\}\right.\left.+1-\max \left\{1-\max \left[z\left(\beta, \alpha \right),0\right],0\right\}\right)\\ {}=\frac{1}{2+\lambda}\left(1+\lambda +1\right)=1.\end{array}} $$

Thus, Property (2) is true.

• (B) First, max{1 − max[x(α, β), 0], 0} ≥ 0, and max{1 − max[y(α, β), 0], 0} ≥ 0.

  • (i) If 1 − max[z(α, β), 0] > 0, then 1 − max {1 − max[z(α, β), 0], 0} = max[z(α, β), 0] ≥ 0.

  • (ii) If 1 − max[z(α, β), 0] ≤ 0, then 1 − max {1 − max[z(α, β), 0], 0} = 1 − 0 = 1_.

Thus, 1 − max {1 − max[z(α, β), 0], 0} ≥ 0

p(α ≻ β) is the sum of max{1 − max[x(α, β), 0], 0} ≥ 0, max{1 − max[y(α, β), 0], 0} ≥ 0, and 1 − max {1 − max[z(α, β), 0], 0} ≥ 0, and, therefore, p(α ≻ β) ≥ 0.

Second, since p(α ≻ β) + p(β ≻ α) = 1, p(α ≻ β) ≤ 1.

Finally, Property (1) is true.

• (C) We can easily infer p(α ≻ β) = p(β ≻ α) = 0.5 from p(α ≻ β) + p(β ≻ α) = 1, supposing that a_i = b_i, T(α) = T(β), I(α) = I(β), and F(α) = F(β). Therefore, Property (3) is true.

• (D) If a₁ ≥ b₄, T(α) ≥ T(β), I(α) ≥ I(β), and F(α) ≤ F(β), then

$x\left(\alpha, \beta \right)=\frac{b_4-{a}_1}{\sum_{i=1}^4\left|{b}_i-{a}_i\right|+\left({b}_4-{b}_1\right)+\left({a}_4-{a}_1\right)+2\left|T\left(\beta \right)-T\left(\alpha \right)\right|}\le 0$, and, similarly, y(α, β) ≤ 0; additionally,

$$ z\left(\alpha, \beta \right)=\frac{\sum_{i=1}^4\left({a}_i-{b}_i\right)+\left({a}_4-{b}_1\right)+2\left(F\left(\beta \right)-F\left(\alpha \right)\right)}{\sum_{i=1}^4\left({a}_i-{b}_i\right)+\left({b}_4-{b}_1\right)+\left({a}_4-{a}_1\right)+2\left(F\left(\beta \right)-F\left(\alpha \right)\right)}\ge 1. $$

Thus, $p\left(\alpha \succ \beta \right)=\frac{1}{2+\gamma}\left(1+\gamma +1\right)=1$, and Property (4) holds under the given conditions.

• (E) According to Minkowski’s inequality, Property (5) can be proven true. However, there are many cases of x(α, β), y(α, β), and z(α, β) being discussed; for instance, x(α, β) < 0, 0 ≤ x(α, β) < 1, and x(α, β) ≥ 1. Thus, the proof is omitted here because of the limited space for publication.

Appendix 2. Proof of Theorem 4

Proof. According to Definition 4, Theorem 4 can be easily proven by using a mathematical induction on n.

• (A) For n = 2, since

  $$ {\displaystyle \begin{array}{c}\frac{1+G{\left({\alpha}_1\right)}_1}{\sum_{i=1}^2\left(1+G\left({\alpha}_i\right)\right)}{\alpha}_1=\left\langle \left[\tau \left({\alpha}_1\right){a}_{11},\tau \left({\alpha}_1\right){a}_{12},\tau \left({\alpha}_1\right){a}_{13},\tau \left({\alpha}_1\right){a}_{14}\right],\left(T\left({\alpha}_1\right),I\left({\alpha}_1\right),F\left({\alpha}_1\right)\right)\right\rangle\ \mathrm{and}\\ {}\frac{1+G\left({\alpha}_2\right)}{\sum_{i=1}^2\left(1+G\left({\alpha}_i\right)\right)}{\alpha}_2=\left\langle \left[\tau \left({\alpha}_2\right){a}_{21},\tau \left({\alpha}_2\right){a}_{22},\tau \left({\alpha}_2\right){a}_{23},\tau \left({\alpha}_2\right){a}_{24}\right],\left(T\left({\alpha}_2\right),I\left({\alpha}_2\right),F\left({\alpha}_2\right)\right)\right\rangle, \end{array}} $$

then

$$ {\displaystyle \begin{array}{c} SVTNPA\left({\alpha}_1,{\alpha}_2\right)=\frac{1+G\left({\alpha}_1\right)}{\sum_{i=1}^2\left(1+G\left({\alpha}_i\right)\right)}{\alpha}_1\oplus \frac{1+G\left({\alpha}_2\right)}{\sum_{i=1}^2\left(1+G\left({\alpha}_i\right)\right)}{\alpha}_2\\ {}=\left\langle \left[\tau \left({\alpha}_1\right){a}_{11}+\tau \left({\alpha}_2\right){a}_{21},\tau \left({\alpha}_1\right){a}_{12}+\tau \left({\alpha}_2\right){a}_{22},\tau \left({\alpha}_1\right){a}_{13}+\tau \left({\alpha}_2\right){a}_{23},\tau \left({\alpha}_1\right){a}_{14}+\tau \left({\alpha}_2\right){a}_{24}\right]\right.,\\ {}\left.\left(\frac{\varphi \left({\alpha}_1\right)T\left({\alpha}_1\right)+\varphi \left({\alpha}_2\right)T\left({\alpha}_2\right)}{\varphi \left({\alpha}_1\right)+\varphi \left({\alpha}_2\right)},\frac{\varphi \left({\alpha}_1\right)I\left({\alpha}_1\right)+\varphi \left({\alpha}_2\right)I\left({\alpha}_2\right)}{\varphi \left({\alpha}_1\right)+\varphi \left({\alpha}_2\right)},\frac{\varphi \left({\alpha}_1\right)I\left({\alpha}_1\right)+\varphi \left({\alpha}_2\right)I\left({\alpha}_2\right)}{\varphi \left({\alpha}_1\right)+\varphi \left({\alpha}_2\right)}\right)\right\rangle .\end{array}} $$

Therefore, Theorem 4 is true for n = 2.

• (B) If Formula (18) holds for n = k, then

  $$ {\displaystyle \begin{array}{c} SVTNPA\left({\alpha}_1,{\alpha}_2,\cdots, {\alpha}_n\right)=\overset{n}{\underset{i=1}{\oplus }}\left(\frac{1+G\left({\alpha}_i\right)}{\sum_{i=1}^k\left(1+G\left({\alpha}_i\right)\right)}{\alpha}_i\right)\\ {}\kern10.66em =\left\langle \left[{\sum}_{i=1}^k\left(\tau \left({\alpha}_i\right){a}_{i1}\right),{\sum}_{i=1}^k\left(\tau \left({\alpha}_i\right){a}_{i2}\right),{\sum}_{i=1}^k\left(\tau \left({\alpha}_i\right){a}_{i3}\right),{\sum}_{i=1}^k\left(\tau \left({\alpha}_i\right){a}_{i4}\right)\right]\right.,\\ {}\left.\left(\frac{\sum_{i=1}^k\left(\varphi \left({\alpha}_i\right)T\left({\alpha}_i\right)\right)}{\sum_{i=1}^k\varphi \left({\alpha}_i\right)},\frac{\sum_{i=1}^k\left(\varphi \left({\alpha}_i\right)I\left({\alpha}_i\right)\right)}{\sum_{i=1}^k\varphi \left({\alpha}_i\right)},\frac{\sum_{i=1}^k\left(\varphi \left({\alpha}_i\right)F\left({\alpha}_i\right)\right)}{\sum_{i=1}^k\varphi \left({\alpha}_i\right)}\right)\right\rangle .\end{array}} $$

When n = k + 1, according to the operations described in Definition 4, we have

$$ {\displaystyle \begin{array}{c} SVTNPA\left({\alpha}_1,{\alpha}_2,\cdots, {\alpha}_n\right)=\overset{n}{\underset{i=1}{\oplus }}\left(\frac{1+G\left({\alpha}_i\right)}{\sum_{i=1}^n\left(1+G\left({\alpha}_i\right)\right)}{\alpha}_i\right)\oplus \frac{1+G\left({\alpha}_{k+1}\right)}{\sum_{i=1}^n\left(1+G\left({\alpha}_i\right)\right)}{\alpha}_{k+1}\\ {}\kern12.66em =\left\langle \left[{\sum}_{i=1}^{k+1}\left(\tau \left({\alpha}_i\right){a}_{i1}\right),{\sum}_{i=1}^{k+1}\left(\tau \left({\alpha}_i\right){a}_{i2}\right),{\sum}_{i=1}^{k+1}\left(\tau \left({\alpha}_i\right){a}_{i3}\right),{\sum}_{i=1}^{k+1}\left(\tau \left({\alpha}_i\right){a}_{i4}\right)\right]\right.,\\ {}\left.\left(\frac{\sum_{i=1}^{k+1}\left(\varphi \left({\alpha}_i\right)T\left({\alpha}_i\right)\right)}{\sum_{i=1}^{k+1}\varphi \left({\alpha}_i\right)},\frac{\sum_{i=1}^{k+1}\left(\varphi \left({\alpha}_i\right)I\left({\alpha}_i\right)\right)}{\sum_{i=1}^{k+1}\varphi \left({\alpha}_i\right)},\frac{\sum_{i=1}^{k+1}\left(\varphi \left({\alpha}_i\right)F\left({\alpha}_i\right)\right)}{\sum_{i=1}^{k+1}\varphi \left({\alpha}_i\right)}\right)\right\rangle .\end{array}} $$

Therefore, Theorem 4 is also true for n = k + 1.

Based on Part (A) and (B), Theorem 4 holds for any n.

Appendix 3. Proof of Theorem 5

Proof. Because $Supp\left({\alpha}_i,{\alpha}_j\right)=c\ \left(c\in \left[0,1\right],\kern0.5em i\ne j,j=1,2,\cdots, n\right),\mathrm{we}\ \mathrm{have}\ \mathrm{G}\left({\alpha}_i\right)={\sum}_{j=1,j\ne i}^n Sup\left(\left({\alpha}_i,{\alpha}_j\right)\right)=\left(n-1\right)c$.

Therefore,$SVTNPA\left({\alpha}_1,{\alpha}_2,\cdots, {\alpha}_n\right)=\overset{n}{\underset{i=1}{\oplus }}\left(\frac{1+G{\left({\alpha}_i\right)}_i}{\sum_{i=1}^n\left(1+G\left({\alpha}_i\right)\right)}{\alpha}_i\right)=\overset{n}{\underset{i=1}{\oplus }}\left(\frac{1+\left(n-1\right)c}{\sum_{i=1}^n\left(1+\left(n-1\right)c\right)}{\alpha}_i\right)=\frac{1}{n}\overset{n}{\underset{i=1}{\oplus }}\left({\alpha}_i\right)$.