Introduction

Fuzzy set (FS) theory, firstly proposed by Zadeh [1], has been a hot research topic. Further, in order to express some types of fuzzy information, Atanassov [2, 3] presented the intuitionistic fuzzy set (IFS) by adding a non-membership function based on FS. Furthermore, Atanassov [4] and Atanassov and Gargov [5] extended the IFS to interval-valued IFS (IVIFS) in which the membership and non-membership degrees are described by interval numbers. Then, some operational laws and relations of IVIFS were defined. Liu [6] and Zhang [7] presented some information entropy for IVIFS. Based on the prospect theory, Wang [8] proposed a new score function to overcome the weakness of not comparing two interval-valued intuitionistic fuzzy numbers (IVIFNs). Many researchers also developed some similarity measurements of IVIFS [9–11] to compare two IVIFNs. In addition, Tan and Zhang [12] developed an extended TOPSIS method on the basis of IVIFNs to solve the MADM problems. Hashemi et al. [13] proposed the extended ELECTRE III method for IVIFNs. Wang and Xu [14] provided a fractional programming method to solve the IVIF-MADM problems.

The aggregation operators are a powerful method for the MAGDM problems [15–23]. Particularly, the information aggregation operators on the basis of IVIFS have attracted more and more attentions [19, 24–32]. Yager [33] firstly proposed the power average (PA) operator, which could eliminate the influence of unreasonable data from the biased decision makers. Further, Xu and Yager [34] developed power geometric operator. Bonferroni [35] introduced Bonferroni mean (BM) operator, which could capture the interrelationships of two arguments. Zhu [36] proposed the geometric Bonferroni mean by combining BM and geometric mean operators. He [37, 38] introduced the interaction of BM operator for intuitionistic fuzzy information. To consider the advantages of PA and BM operators together, He et al. [39–41] proposed some power Bonferroni mean (PBM) operators by combining the PA operator and BM operator.

The PBM operator can take the advantages of PA and BM operators. However, up to now, there is no research on how to use PBM operator to aggregate the IVIFNs, so the goal and motivation of this study are to extend the PBM operator to IVIFNs and to propose some MAGDM methods for IVIFNs.

For that, the structure of this paper is shown as follows. In the “Preliminaries” section, we introduce the definition of the IVIFNs, the PBM, and PGBM operators in brief. In the “Some interval-valued intuitionistic fuzzy PBM operators” section, we combine the IVIFNs with PBM aggregation operators and develop some new operators to aggregate the IVIFNs. In the “The MAGDM approach based on IVIFWPBM and IVIFWPGBM operators” section, on the basis of these operators, an effective method is developed for MAGDM problems with the IVIFNs. The “An application example” section presented an application example to verify the feasibility of the novel developed method. In the “Conclusion” section, some concluding remarks are given.

Preliminaries

Interval-Valued Intuitionistic Fuzzy Set

Definition 1 [2]. Let Z = {z 1, z 2,  ⋯ , z n } be a fixed set, then an IFS named A in Z is expressed as

$$ A=\left\{< z,{u}_A(z),{v}_A(z)>| z\in Z\right\} $$
(1)

where 0 ≤ u A (z) ≤ 1, 0 ≤ v A (z) ≤ 1 and 0 ≤ u A (z) + v A (z) ≤ 1. u A (z) and v A (z) represent membership and non-membership degrees of the element z to A, respectively.

In addition, suppose π(z) = 1 − u A (z) − v A (z), then π(z) is named the hesitancy degree of z to A [2, 3]. It is apparent that 0 ≤ π(z) ≤ 1 for ∀z ∈ Z.

To element z ∈ Z from IFS A, the pair (u A (z), v A (z)) represents an intuitionistic fuzzy value (IFV). For convenience, it can be denoted as \( \tilde{a}=\left({u}_{\tilde{a}},{v}_{\tilde{a}}\right) \), satisfying that \( 0\le {u}_{\tilde{a}}\le 1 \), \( 0\le {v}_{\tilde{a}}\le 1 \) and \( 0\le {u}_{\tilde{a}}+{v}_{\tilde{a}}\le 1 \).

Definition 2 [3, 4]. Let Z = {z 1, z 2,  ⋯ , z n } be a fixed set, and then an IVIFS AL is expressed as

$$ AL=\left\{< z,{u}_{AL}(z),{v}_{AL}(z) > \left| z\in Z\right.\right\} $$
(2)

where the interval numbers u AL (z) ⊆ [0, 1] and v AL (z) ⊆ [0, 1] satisfies 0 ≤  sup (u AL (z)) +  sup (v AL (z)) ≤ 1. u AL (z) and v AL (z) represent the membership and non-membership degrees of the element z to AL respectively. For simplicity, al = ([μa, μb], [vc, vd]) is called an IVIFN.

Definition 3 [42]. Suppose al 1 = ([μa 1, μb 1], [vc 1, vd 1]) and al 2 = ([μa 2, μb 2], [vc 2, vd 2]) are two IVIFNs, then the Euclidean distance between them is defined as follows:

$$ d\left({ a l}_1,{ a l}_2\right)=\sqrt{\frac{1}{4}\left({\left(\mu {a}_1-\mu {a}_2\right)}^2+{\left(\mu {b}_1-\mu {b}_2\right)}^2+{\left({vc}_1-{vc}_2\right)}^2+{\left({vd}_1-{vd}_2\right)}^2\right)} $$
(3)

Definition 4 [43]. Suppose al 1 = ([μa 1, μb 1], [vc 1, vd 1]) and al 2 = ([μa 2, μb 2], [vc 2, vd 2]) are two IVIFNs, then the operational laws can be expressed as follows:

$$ { a l}_1\otimes { a l}_2=\left(\left[\mu {a}_1\mu {a}_2,\mu {b}_1\mu {b}_2\right],\left[{vc}_1+{vc}_2-{vc}_1{vc}_2,{vd}_1+{vd}_2-{vd}_1{vd}_2\right]\right), $$
(4)
$$ { a l}_1\oplus { a l}_2=\left(\left[\mu {a}_1+\mu {a}_2-\mu {a}_1\mu {a}_2,\mu {b}_1+\mu {b}_2-\mu {b}_1\mu {b}_2\right],\left[{vc}_1{vc}_2,{vd}_1{vd}_2\right]\right), $$
(5)
$$ n\cdot { a l}_1=\left(\left[1-{\left(1-\mu {a}_1\right)}^n,1-{\left(1-\mu {b}_1\right)}^n\right],\left[{vc}_1^n,{vd}_1^n\right]\right)\kern0.75em n>0, $$
(6)
$$ { a l}_1^n=\left(\left[\mu {a}_1^n,\mu {b}_1^n\right],\left[1-{\left(1-{vc}_1\right)}^n,1-{\left(1-{vd}_1\right)}^n\right]\right)\kern0.75em n>0. $$
(7)

Example 1. Suppose al 1 = ([0.1, 0.3], [0.4, 0.5]) and al 2 = ([0.2, 0.4], [0.3, 0.5]) are two IVIFNs, and n = 2, then on the basis of Definition 4, we can get

\( \begin{array}{l}{al}_1\oplus {al}_2=\left(\left[0.1+0.2-0.1\times 0.2,0.3+0.4-0.3\times 0.4\right],\left[0.4\times 0.3,0.5\times 0.5\right]\right)=\left(\left[0.28,0.58\right],\left[0.12,0.25\right]\right),\hfill \\ {}{al}_1\otimes {al}_2=\left(\left[0.1\times 0.2,0.3\times 0.4\right],\left[0.4+0.3\hbox{-} 0.4\times 0.3,0.5+0.5\hbox{-} 0.5\times 0.5\right]\right)=\left(\left[0.02,0.12\right],\left[0.58,0.75\right]\right),\hfill \\ {} n\cdot {al}_1=2{al}_1=\left(\left[1-{\left(1-0.1\right)}^2,1-{\left(1\hbox{-} 0.3\right)}^2\right],\left[0{.4}^2,0{.5}^2\right]\right)=\left(\left[0.19,0.51\right],\left[0.16,0.25\right]\right),\hfill \\ {}{al}_1^n={al}_1^2=\left(\left[{0.1}^2,0{.3}^2\right],\left[1-{\left(1\hbox{-} 0.4\right)}^2,1-{\left(1\hbox{-} 0.5\right)}^2\right]\right)=\left(\left[0.01,0.09\right],\left[0.64,0.75\right]\right).\hfill \end{array} \) Theorem 1 [43]. Suppose al 1 = ([μa 1, μb 1], [vc 1, vd 1]) and al 2 = ([μa 2, μb 2], [vc 2, vd 2]) are two IVIFNs, then

$$ (1)\kern0.5em {al}_1\oplus {al}_2={al}_2\oplus {al}_1 $$
(8)
$$ (2)\kern0.5em {al}_1\otimes {al}_2={al}_2\otimes {al}_1 $$
(9)
$$ (3)\eta \left({al}_1\oplus {al}_2\right)=\eta \cdot {al}_1\oplus \eta \cdot {al}_2,\eta \ge 0 $$
(10)
$$ (4)\kern0.5em \eta \cdot {al}_1\oplus {\eta}_2\cdot {al}_1=\left({\eta}_1+{\eta}_2\right){al}_1,{\eta}_1,{\eta}_2\ge 0 $$
(11)
$$ (5)\kern0.5em {al}_1^{\eta_1}\otimes {al}_1^{\eta_2}={\left({al}_1\right)}^{\eta_1+{\eta}_2},{\eta}_1,{\eta}_2\ge 0 $$
(12)
$$ (6)\kern0.5em {al}_1^{\eta}\otimes {al}_2^{\eta}={\left({al}_1\otimes {al}_2\right)}^{\eta} $$
(13)

Definition 5 [44]. Supposing al i  = ([μa i , μb i ], [vc i , vd i ]) is an IVIFN, we can define the score function sf of al i as follows:

$$ sf\left({ a l}_i\right)=\frac{\mu {a}_i+\mu {b}_i-{vc}_i-{vd}_i}{2} $$
(14)

Definition 6 [44]. Supposing al i  = ([μa i , μb i ], [vc i , vd i ]) is an IVIFN, we can define the accuracy function af of the IVIFN al i as follows:

$$ af\left({ a l}_i\right)=\frac{\mu {a}_i+\mu {b}_i+{vc}_i+{vd}_i}{2} $$
(15)

Definition 7 [44]. If al 1 = ([μa 1, μb 1], [vc 1, vd 1]) and al 2 = ([μa 2, μb 2], [vc 2, vd 2]) are two IVIFNs, we can get

(1) If sf(al 1) > sf(al 2), then al 1 > al 2;

(2) If sf(al 1) = sf(al 2), then.

If af(al 1) > af(al 2), then al 1 > al 2;

If af(al 1) = af(al 2), then al 1 = al 2.

Example 2. Supposing al 1 = ([0.4, 0.5], [0.2, 0.3]) and al 2 = ([0.2, 0.5], [0.1, 0.3]) are two IVIFNs, then based on the Definition 7, we can get the following results:

$$ s\left({al}_1\right)=\frac{0.4+0.5-0.2-0.3}{2}=0.2,\kern0.5em s\left({al}_2\right)=\frac{0.2+0.5-0.1-0.3}{2}=0.15. $$

Because sf(al 1) > sf(al 2), we can get al 1 > al 2.

If al 1 = ([0.4, 0.5], [0.2, 0.3]) and al 2 = ([0.2, 0.5], [0.1, 0.2]), then we can get

$$ \begin{array}{l}\begin{array}{ll} sf\left({al}_1\right)=\frac{0.4+0.5-0.2-0.3}{2}=0.2,\hfill & sf\left({al}_2\right)=\frac{0.2+0.5-0.1-0.2}{2}=0.2;\hfill \end{array}\\ {}\begin{array}{ll} af\left({al}_1\right)=\frac{0.4+0.5+0.2+0.3}{2}=0.7,\hfill & af\left({al}_2\right)=\frac{0.2+0.5+0.1+0.2}{2}=0.5.\hfill \end{array}\end{array} $$

Because sf(al 1) = sf(al 2) and af(al 1) > af(al 2), we can get al 1 > al 2.

The Power Bonferroni Mean Operator and Power Geometric Bonferroni Mean Operator

Definition 8 [41]. Let ra k (k = 1, 2,  ⋯ , n) be a set of positive real numbers and x , y ≥ 0 the aggregation function

$$ {\mathrm{PBM}}^{x, y}\left({ra}_1,{ra}_2,\cdot \cdot \cdot, {ra}_n\right)={\left(\frac{1}{n^2- n}\sum_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em g\ne h\end{array}}^n\left({\left(\frac{n\left( T\left({ra}_g\right)+1\right)}{\sum_{t=1}^n\left( T\left({ra}_t\right)+1\right)}{ra}_g\right)}^x\otimes {\left(\frac{n\left( T\left({ra}_h\right)+1\right)}{\sum_{t=1}^n\left( T\left({ra}_t\right)+1\right)}{ra}_h\right)}^y\right)\right)}^{\frac{1}{x+ y}} $$
(16)

is called power Bonferroni mean (PBM) operator.

Definition 9 [41]. Let ra k (k = 1, 2,  ⋯ , n) be a set of positive real numbers and x , y > 0 the aggregation function

$$ {\mathrm{PGBM}}^{x, y}\left({ra}_1,{ra}_2,\dots, {ra}_n\right)=\frac{1}{x+ y}{\left(\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em g\ne h\end{array}}^n\left({{ x ra}_g}^{\frac{n\left( T\left({ra}_g\right)+1\right)}{\sum_{t=1}^n\left( T\left({ra}_t\right)+1\right)}}+{yra_h}^{\frac{n\left( T\left({ra}_h\right)+1\right)}{\sum_{t=1}^n\left( T\left({ra}_t\right)+1\right)}}\right)\right)}^{\frac{1}{n^2- n}} $$
(17)

is called power geometric Bonferroni mean (PGBM) operator.

In definitions 8 and 9, \( T\left({ra}_g\right)=\sum_{\begin{array}{l} h=1\\ {} h\ne g\end{array}}^n Sup\left({ra}_g,{ra}_h\right) \), and Sup(ra g , ra h ) is the support degree for ra g from ra h satisfying the properties as

  1. 1.

    Sup(ra g , ra h ) = 1 − d(ra g , ra h ), so Sup(ra g , ra h ) ∈ [0, 1];

  2. 2.

    Sup(ra g , ra h ) = Sup(ra h , ra g );

  3. 3.

    If |ra g  − ra h | < |ra l  − ra r |, then Sup(ra g , ra h ) > Sup(ra l , ra r ).

where d is Euclidean distance from Definition 3. T(ra g ) can represent the support of ra g by all the other numbers, and the closer two values are, the bigger the support degree is.

Some Interval-Valued Intuitionistic Fuzzy PBM Operators

On the basis of IVIFNs, the PBM and PGBM operators, we shall propose the weighted PBM (IVIFWPBM) operator of the IVIFNs and the weighted PGBM (IVIFWPGBM) operator of the IVIFNs.

The Interval-Valued Intuitionistic Fuzzy Power Bonferroni Mean Operator

Definition 10 [41]. Suppose al j  = ([μa j , μb j ], [vc j , vd j ]) is a set of the IVIFNs (j = 1, 2,  ⋯ , n), then the IVIFPBM operator is defined as

$$ {IVIFPBM}^{x, y}\left({al}_1,{al}_2,\cdot \cdot \cdot, {al}_n\right)={\left(\frac{1}{n^2- n}\sum_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em g\ne h\end{array}}^n\left({\left(\frac{n\left( T\left({al}_g\right)+1\right)}{\sum_{t=1}^n\left( T\left({al}_t\right)+1\right)}{al}_g\right)}^x\otimes {\left(\frac{n\left( T\left({al}_h\right)+1\right)}{\sum_{t=1}^n\left( T\left({al}_t\right)+1\right)}{al}_h\right)}^y\right)\right)}^{\frac{1}{x+ y}} $$
(18)

where \( T\left({al}_g\right)=\sum_{h=1, h\ne g}^n Sup\left({al}_g,{al}_h\right) \), x , y > 0.

Theorem 2. Based on the IVIFNs al j  = ([μa j , μb j ], [vc j , vd j ]) (j = 1, 2, 3,  ⋯ , n), the aggregated result from Definition 10 is expressed by

$$ {IVIFPBM}^{x, y}\left({a l}_1,{ a l}_2,\cdot \cdot \cdot, { a l}_n\right)=\left(\left[{\left(1-{\left(\prod_g^{=}1, h=1\kern0.5em g\ne hn\left(1-{\left(1-{\left(1-\mu {a}_g\right)}^{\frac{n\left( T\left({a l}_g\right)+1\right)}{\sum_{t=1}^n\left( T\left({a}_{lt}\right)+1\right)}}\right)}^x\times {\left(1-{\left(1-\mu {a}_h\right)}^{\frac{n\left( T\left({a l}_h\right)+1\right)}{\sum_{t=1}^n\left( T\left({a l}_t\right)+1\right)}}\right)}^y\right)\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}},\right.\right.\left.{\left(1-{\left(\prod_g^{=}1, h=1\kern0.5em \mathrm{g}\ne hn\left(1-{\left(1-{\left(1-\mu {b}_g\right)}^{\frac{n\left( T\left({a l}_g\right)+1\right)}{\sum_{t=1}^n\left( T\left({a l}_t\right)+1\right)}}\right)}^x\times {\left(1-{\left(1-\mu {b}_h\right)}^{\frac{n\left( T\left({a l}_h\right)+1\right)}{\sum_{t=1}^n\left( T\left({a l}_t\right)+1\right)}}\right)}^y\right)\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}}\right],\left[1-{\left(1-{\left(\prod_g^{=}1, h=1\kern0.5em \mathrm{g}\ne hn\left(1-{\left(1-{vc_g}^{\frac{n\left( T\left({a l}_g\right)+1\right)}{\sum_{t=1}^n\left( T\left({a l}_t\right)+1\right)}}\right)}^x{\left(1-{vc_h}^{\frac{n\left( T\left({a l}_h\right)+1\right)}{\sum_{t=1}^n\left( T\left({a l}_t\right)+1\right)}}\right)}^y\right)\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}},\right.\left.\left.1-{\left(1-{\left(\prod_g^{=}1, h=1\kern0.5em \mathrm{g}\ne hn\left(1-{\left(1-{vd_g}^{\frac{n\left( T\left({a l}_g\right)+1\right)}{\sum_{t=1}^n\left( T\left({a l}_t\right)+1\right)}}\right)}^x{\left(1-{vd_h}^{\frac{n\left( T\left({a l}_h\right)+1\right)}{\sum_{t=1}^n\left( T\left({a l}_t\right)+1\right)}}\right)}^y\right)\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}}\right]\right) $$
(19)

Proof.

Let \( {\tau}_k=\frac{n\left( T\left({al}_k\right)+1\right)}{\sum_{t=1}^n\left( T\left({al}_t\right)+1\right)}\left( k=1,2,\cdots, n\right) \), we can get

$$ {IVIFPBM}^{x, y}\left({al}_1,{al}_2,\cdot \cdot \cdot, {al}_n\right)={\left(\frac{1}{n^2- n}\sum_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left({\left({\tau}_g\cdot {al}_g\right)}^x\otimes {\left({\tau}_h\cdot {al}_h\right)}^y\right)\right)}^{\frac{1}{x+ y}}. $$

Calculate τ g  ⋅ al g and τ h  ⋅ al h , and we can get

$$ \begin{array}{l}{\tau}_g\cdot { a l}_g=\left(\left[1-{\left(1-\mu {a}_g\right)}^{\tau_g},1-{\left(1-\mu {b}_g\right)}^{\tau_g}\right],\left[{vc}_g^{\tau_g},{vd}_g^{\tau_g}\right]\right),\hfill \\ {}{\tau}_h\cdot { a l}_h=\left(\left[1-{\left(1-\mu {a}_h\right)}^{\tau_h},1-{\left(1-\mu {b}_h\right)}^{\tau_h}\right],\left[{vc}_h^{\tau_h},{vd}_h^{\tau_h}\right]\right).\hfill \end{array} $$
  1. 1.

    Calculate (τ g  ⋅ al g )x and (τ h  ⋅ al h )y, and we can get

$$ \begin{array}{l}{\left({\tau}_g\cdot { a l}_g\right)}^x=\left(\left[{\left(1-{\left(1-\mu {a}_g\right)}^{\tau_g}\right)}^x,{\left(1-{\left(1-\mu {b}_g\right)}^{\tau_g}\right)}^x\right],\left[1-{\left(1-{vc}_g^{\tau_g}\right)}^x,1-{\left(1-{vd}_g^{\tau_g}\right)}^x\right]\right),\hfill \\ {}{\left({\tau}_h\cdot { a l}_h\right)}^y=\left(\left[{\left(1-{\left(1-\mu {a}_h\right)}^{\tau_h}\right)}^y,{\left(1-{\left(1-\mu {b}_h\right)}^{\tau_h}\right)}^y\right],\left[1-{\left(1-{vc}_h^{\tau_h}\right)}^y,1-{\left(1-{vd}_h^{\tau_h}\right)}^y\right]\right).\hfill \end{array} $$
  1. 2.

    Calculate (τ g  ⋅ al g )x ⊗ (τ h  ⋅ al h )y, and we can get

\( {\left({\tau}_g\cdot { a l}_g\right)}^x\otimes {\left({\tau}_h\cdot { a l}_h\right)}^y=\left(\left[{\left(1-{\left(1-\mu {a}_g\right)}^{\tau_g}\right)}^x\times {\left(1-{\left(1-\mu {a}_h\right)}^{\tau_h}\right)}^y,{\left(1-{\left(1-\mu {b}_g\right)}^{\tau_g}\right)}^x\times {\left(1-{\left(1-\mu {b}_h\right)}^{\tau_h}\right)}^y\right]\right.,\left.\left[1-{\left(1-{vc}_g^{\tau_g}\right)}^x\times {\left(1-{vc}_h^{\tau_h}\right)}^y,1-{\left(1-{vd}_g^{\tau_g}\right)}^x\times {\left(1-{vd}_h^{\tau_h}\right)}^y\right]\right) \)Calculate \( \sum_{\begin{array}{l} g=1, h=1\\ {} g\ne h\end{array}}^n\left({\left({\tau}_g\cdot {al}_g\right)}^x\otimes {\left({\tau}_h\cdot {al}_h\right)}^y\right) \), and we get

$$ \sum_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left({\left({\tau}_g\cdot { a l}_g\right)}^x\otimes {\left({\tau}_h\cdot { a l}_h\right)}^y\right)=\left(\left[1-\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left(1-{\left(1-{\left(1-\mu {a}_g\right)}^{\tau_g}\right)}^x\times {\left(1-{\left(1-\mu {a}_h\right)}^{\tau_h}\right)}^y\right),1-\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left(1-{\left(1-{\left(1-\mu {b}_g\right)}^{\tau_g}\right)}^x\times {\left(1-{\left(1-\mu {b}_h\right)}^{\tau_h}\right)}^y\right)\right]\right.,\left.\left[\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left(1-{\left(1-{vc}_g^{\tau_g}\right)}^x\times {\left(1-{vc}_h^{\tau_h}\right)}^y\right),\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left(1-{\left(1-{vd}_g^{\tau_g}\right)}^x\times {\left(1-{vd}_h^{\tau_h}\right)}^y\right)\right]\right). $$
  1. 3.

    Calculate \( \frac{1}{n\left( n-1\right)}\sum_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left({\left({\tau}_g\cdot {al}_g\right)}^x\otimes {\left({\tau}_h\cdot {al}_h\right)}^y\right) \), and we get

$$ \frac{1}{n\left( n-1\right)}\sum_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left({\left({\tau}_g\cdot { a l}_h\right)}^x\otimes {\left({\tau}_h\cdot { a l}_h\right)}^y\right)=\left(\left[1-\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n{\left(1-{\left(1-{\left(1-\mu {a}_g\right)}^{\tau_g}\right)}^x\times {\left(1-{\left(1-\mu {a}_h\right)}^{\tau_h}\right)}^y\right)}^{\frac{1}{n^2- n}},1-\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n{\left(1-{\left(1-{\left(1-\mu {b}_g\right)}^{\tau_g}\right)}^x\times {\left(1-{\left(1-\mu {b}_h\right)}^{\tau_h}\right)}^y\right)}^{\frac{1}{n^2- n}}\right]\right.,\left.\left[\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n{\left(1-{\left(1-{vc}_g^{\tau_g}\right)}^x\times {\left(1-{vc}_h^{\tau_h}\right)}^y\right)}^{\frac{1}{n^2- n}},\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n{\left(1-{\left(1-{vd}_g^{\tau_g}\right)}^x\times {\left(1-{vd}_h^{\tau_h}\right)}^y\right)}^{\frac{1}{n^2- n}}\right]\right). $$
  1. 4.

    Calculate \( {\left(\frac{1}{n\left( n-1\right)}\sum_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left({\left({\tau}_g\cdot {al}_g\right)}^x\otimes {\left({\tau}_h\cdot {al}_h\right)}^y\right)\right)}^{\frac{1}{x+ y}} \), and we get

$$ {\left(\frac{1}{n\left( n-1\right)}\sum_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left({\left({\tau}_g\cdot { a l}_g\right)}^x\otimes {\left({\tau}_h\cdot { a l}_h\right)}^y\right)\right)}^{\frac{1}{x+ y}}=\left(\left[{\left(1-\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n{\left(1-{\left(1-{\left(1-\mu {a}_g\right)}^{\tau_g}\right)}^x\times {\left(1-{\left(1-\mu {a}_h\right)}^{\tau_h}\right)}^y\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}},\right.\right.\left.{\left(1-\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n{\left(1-{\left(1-{\left(1-\mu {b}_g\right)}^{\tau_g}\right)}^x\times {\left(1-{\left(1-\mu {b}_h\right)}^{\tau_h}\right)}^y\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}}\right],\left.\left[1-{\left(1-\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n{\left(1-{\left(1-{vc}_g^{\tau_g}\right)}^x\times {\left(1-{vc}_h^{\tau_h}\right)}^y\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}},1-{\left(1-\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n{\left(1-{\left(1-{vd}_g^{\tau_g}\right)}^x\times {\left(1-{vd}_h^{\tau_h}\right)}^y\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}}\right]\right). $$
  1. 5.

    Replace \( {\tau}_k=\frac{n\left( T\left({al}_k\right)+1\right)}{\sum_{t=1}^n\left( T\left({al}_t\right)+1\right)} \), and we get

$$ \begin{array}{l}{\left(\frac{1}{n\left( n-1\right)}\sum_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left({\left(\frac{n\left( T\left({ a l}_g\right)+1\right)}{\sum_{t=1}^n\left( T\left({ a l}_t\right)+1\right)}{a l}_g\right)}^x\otimes {\left(\frac{n\left( T\left({ a l}_h\right)+1\right)}{\sum_{t=1}^n\left( T\left({ a l}_t\right)+1\right)}{a l}_h\right)}^y\right)\right)}^{\frac{1}{x+ y}}=\\ {}\left(\left[{\left(1-{\left(\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left(1-{\left(1-{\left(1-\mu {a}_g\right)}^{\frac{n\left( T\left({ a l}_g\right)+1\right)}{\sum_{t=1}^n\left( T\left({ a l}_t\right)+1\right)}}\right)}^x\times {\left(1-{\left(1-\mu {a}_h\right)}^{\frac{n\left( T\left({ a l}_h\right)+1\right)}{\sum_{t=1}^n\left( T\left({ a l}_t\right)+1\right)}}\right)}^y\right)\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}},\right.\right.\\ {}\left.{\left(1-{\left(\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left(1-{\left(1-{\left(1-\mu {b}_g\right)}^{\frac{n\left( T\left({ a l}_g\right)+1\right)}{\sum_{t=1}^n\left( T\left({ a l}_t\right)+1\right)}}\right)}^x\times {\left(1-{\left(1-\mu {b}_h\right)}^{\frac{n\left( T\left({ a l}_h\right)+1\right)}{\sum_{t=1}^n\left( T\left({ a l}_t\right)+1\right)}}\right)}^y\right)\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}}\right],\\ {}\left[1-{\left(1-{\left(\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left(1-{\left(1-{vc_g}^{\frac{n\left( T\left({ a l}_{\mathrm{g}}\right)+1\right)}{\sum_{t=1}^n\left( T\left({ a l}_t\right)+1\right)}}\right)}^x{\left(1-{vc_h}^{\frac{n\left( T\left({ a l}_h\right)+1\right)}{\sum_{t=1}^n\left( T\left({ a l}_t\right)+1\right)}}\right)}^y\right)\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}},\right.\\ {}\left.\left.1-{\left(1-{\left(\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left(1-{\left(1-{vd_g}^{\frac{n\left( T\left({ a l}_g\right)+1\right)}{\sum_{t=1}^n\left( T\left({ a l}_t\right)+1\right)}}\right)}^x{\left(1-{vd_h}^{\frac{n\left( T\left({ a l}_h\right)+1\right)}{\sum_{t=1}^n\left( T\left({ a l}_t\right)+1\right)}}\right)}^y\right)\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}}\right]\right).\end{array} $$
(20)

The proof ends.

Now, we will give an example to demonstrate the aggregation process.

Example 3. Suppose that there are two IVIFNs al 1 = ([0.1, 0.3], [0.4, 0.5]) and al 2 = ([0.2, 0.4], [0.3, 0.5]), and let x = 1 , y = 2, then we can derive the following results:

Calculate \( {\tau}_k=\frac{n\left( T\left({al}_k\right)+1\right)}{\sum_{t=1}^n\left( T\left({al}_t\right)+1\right)} \), we can get \( {\tau}_1=\frac{2\left( T\left({al}_1\right)+1\right)}{\sum_{t=1}^2\left( T\left({al}_t\right)+1\right)}=1 \) and \( {\tau}_2=\frac{2\left( T\left({al}_2\right)+1\right)}{\sum_{t=1}^2\left( T\left({al}_t\right)+1\right)}=1 \). So, \( {IVIFPBM}^{1,2}\left({al}_1,{al}_2\right)=\left(\left[{\left(1-{\left(\left(1-{\left(1-{\left(1-0.1\right)}^1\right)}^1\otimes {\left(1-{\left(1-0.2\right)}^1\right)}^2\right)\otimes \left(1-{\left(1-{\left(1-0.2\right)}^1\right)}^1\otimes {\left(1-{\left(1-0.1\right)}^1\right)}^2\right)\right)}^{\frac{1}{2}}\right)}^{\frac{1}{3}},\right.\right.\left.{\left(1-{\left(\left(1-{\left(1-{\left(1-0.3\right)}^1\right)}^1\otimes {\left(1-{\left(1-0.4\right)}^1\right)}^2\right)\otimes \left(1-{\left(1-{\left(1-0.4\right)}^1\right)}^1\otimes {\left(1-{\left(1-0.3\right)}^1\right)}^2\right)\right)}^{\frac{1}{2}}\right)}^{\frac{1}{3}}\right],\left[1-{\left(1-{\left(\left(1-{\left(1-{0.4}^1\right)}^1{\left(1-{0.3}^1\right)}^2\right)\otimes \left(1-{\left(1-{0.3}^1\right)}^1{\left(1-{0.4}^1\right)}^2\right)\right)}^{\frac{1}{2}}\right)}^{\frac{1}{3}},\right.1-\left.\left.{\left(1-{\left(\left(1-{\left(1-{0.5}^1\right)}^1{\left(1-{0.5}^1\right)}^2\right)\otimes \left(1-{\left(1-{0.5}^1\right)}^1{\left(1-{0.5}^1\right)}^2\right)\right)}^{\frac{1}{2}}\right)}^{\frac{1}{3}}\right]\right)=\left(\left[0.1442,0.6490\right],\left[0.3510,0.5\right]\right). \)

By the operations of IVIFNs, several properties of the IVIFPBM operator shall be proved.

Theorem 3 (idempotency). Suppose al k  = al = ([ua, ub], [vc, vd])(k = 1, 2,  ... , n), then

$$ IVIFPBM\left({al}_1,{al}_2,\dots, {al}_n\right)= al. $$

Proof.

Since al k  = al(k = 1, 2,  ... , n), then according to Definition 10,

$$ \begin{array}{l}{IVIFPBM}^{x, y}\left({al}_1,{al}_2,\dots, {al}_n\right)={\left(\frac{1}{n^2- n}\sum_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left({\left(\frac{n\left( T\left({al}_g\right)+1\right)}{\sum_{t=1}^n\left( T\left({al}_t\right)+1\right)}{al}_g\right)}^x\otimes {\left(\frac{n\left( T\left({al}_h\right)+1\right)}{\sum_{t=1}^n\left( T\left({al}_t\right)+1\right)}{al}_h\right)}^y\right)\right)}^{\frac{1}{x+ y}}=\\ {}{\left(\frac{1}{n^2- n}\sum_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left({\left(\frac{n\left( T(al)+1\right)}{\sum_{t=1}^n\left( T(al)+1\right)} al\right)}^x\otimes {\left(\frac{n\left( T(al)+1\right)}{\sum_{t=1}^n\left( T(al)+1\right)} al\right)}^y\right)\right)}^{\frac{1}{x+ y}}={\left(\frac{1}{n^2- n}\sum_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n{al}^{\left( x+ y\right)}\right)}^{\frac{1}{x+ y}}= al.\end{array} $$

Theorem 4 (commutativity). Suppose \( {al}_k^{\prime } \) is any permutation of al k (k = 1, 2,  ... , n), then

$$ \mathrm{IVIFPBM}\left({al}_1^{\prime },{al}_2^{\prime },\dots, {al}_n^{\prime}\right)=\mathrm{IVIFPBM}\left({al}_1,{al}_2,\dots, {al}_n\right). $$

Proof.

Based on Definition 10, we get

$$ {\mathrm{IVIFPBM}}^{x, y}\left({al}_1^{\prime },{al}_2^{\prime },\dots, {al}_n^{\prime}\right)={\left(\frac{1}{n^2- n}\sum_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left({\left(\frac{n\left( T\left({al}_g^{\prime}\right)+1\right)}{\sum_{t=1}^n\left( T\left({al}_t^{\prime}\right)+1\right)}{al}_i^{\prime}\right)}^x\otimes {\left(\frac{n\left( T\left({al}_h^{\prime}\right)+1\right)}{\sum_{t=1}^n\left( T\left({al}_t^{\prime}\right)+1\right)}{al}_j^{\prime}\right)}^y\right)\right)}^{\frac{1}{x+ y}}, $$
(and)
$$ {\mathrm{IVIFPBM}}^{x, y}\left({al}_1,{al}_2,\dots, {al}_n\right)={\left(\frac{1}{n^2- n}\sum_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left({\left(\frac{n\left( T\left({al}_g\right)+1\right)}{\sum_{t=1}^n\left( T\left({al}_t\right)+1\right)}{al}_g\right)}^x\otimes {\left(\frac{n\left( T\left({al}_h\right)+1\right)}{\sum_{t=1}^n\left( T\left({al}_t\right)+1\right)}{al}_h\right)}^y\right)\right)}^{\frac{1}{x+ y}} $$

Because

$$ \sum_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left({\left(\frac{n\left( T\left({al}_g^{\prime}\right)+1\right)}{\sum_{t=1}^n\left( T\left({al}_t^{\prime}\right)+1\right)}{al}_g^{\prime}\right)}^x\otimes {\left(\frac{n\left( T\left({al}_h^{\prime}\right)+1\right)}{\sum_{t=1}^n\left( T\left({al}_t^{\prime}\right)+1\right)}{al}_h^{\prime}\right)}^y\right)=\sum_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left({\left(\frac{n\left( T\left({al}_g\right)+1\right)}{\sum_{t=1}^n\left( T\left({al}_t\right)+1\right)}{al}_g\right)}^x\otimes {\left(\frac{n\left( T\left({al}_h\right)+1\right)}{\sum_{t=1}^n\left( T\left({al}_t\right)+1\right)}{al}_h\right)}^y\right), $$

Thus, \( \mathrm{IVIFPBM}\left({al}_1^{\prime },{al}_2^{\prime },\dots, {al}_n^{\prime}\right)=\mathrm{IVIFPBM}\left({al}_1,{al}_2,\dots, {al}_n\right) \).

In the IVIFPBM operator, it is noted that we only consider the power weight vector and the interrelationship among input arguments and do not take the importance of the input arguments into account. In what follows, the IVIFWPBM operator shall be proposed to overcome the shortcoming.

Definition 11. Suppose al j  = ([μa j , μb j ], [vc j , vd j ]) is a set of the IVIFNs (j = 1, 2,  ⋯ , n), then the IVIFWPBM operator is defined as

$$ {\mathrm{IVIFWPBM}}^{x, y}\left({al}_1,{al}_2,\cdot \cdot \cdot, {al}_n\right)={\left(\frac{1}{n^2- n}\sum_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left({\left(\frac{n{\omega}_g\left( T\left({al}_g\right)+1\right){al}_g}{\sum_{t=1}^n{\omega}_t\left( T\left({al}_t\right)+1\right)}\right)}^x{\left(\frac{n{\omega}_h\left( T\left({al}_h\right)+1\right){al}_h}{\sum_{t=1}^n{\omega}_t\left( T\left({al}_t\right)+1\right)}\right)}^y\right)\right)}^{\frac{1}{x+ y}} $$
(21)

where \( T\left({al}_g\right)=\sum_{h=1, h\ne g}^n Sup\left({al}_g,{al}_h\right) \), x , y > 0. ω = (ω 1, ω 2,  ⋯ , ω n )T is the weight vector of the IVIFNs, 0 ≤ ω k  ≤ 1 (k = 1, 2,  ... , n) and \( \sum_{k=1}^n{\omega}_k=1 \).

Theorem 5. Suppose al j  = ([μa j , μb j ], [vc j , vd j ]) is a set of the IVIFNs (j = 1, 2⋯, n) and x , y > 0, then the aggregated result from Definition 11 is expressed by

$$ \begin{array}{l}{\mathrm{IVIFWPBM}}^{x, y}\left({ a l}_1,{ a l}_2,\cdot \cdot \cdot, { a l}_n\right)=\\ {}\left(\left[{\left(1-{\left(\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left(1-{\left(1-{\left(1-\mu {a}_g\right)}^{\frac{n{\omega}_g\left( T\left({ a l}_g\right)+1\right)}{\sum_{t=1}^n{\omega}_t\left( T\left({ a l}_t\right)+1\right)}}\right)}^x\times {\left(1-{\left(1-\mu {a}_h\right)}^{\frac{n{\omega}_h\left( T\left({ a l}_h\right)+1\right)}{\sum_{t=1}^n{\omega}_t\left( T\left({ a l}_t\right)+1\right)}}\right)}^y\right)\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}},\right.\right.\\ {}\left.{\left(1-{\left(\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left(1-{\left(1-{\left(1-\mu {b}_g\right)}^{\frac{n{\omega}_g\left( T\left({ a l}_g\right)+1\right)}{\sum_{t=1}^n{\omega}_t\left( T\left({ a l}_t\right)+1\right)}}\right)}^x\times {\left(1-{\left(1-\mu {b}_h\right)}^{\frac{n{\omega}_h\left( T\left({ a l}_h\right)+1\right)}{\sum_{t=1}^n{\omega}_t\left( T\left({ a l}_t\right)+1\right)}}\right)}^y\right)\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}}\right],\\ {}\begin{array}{l}\left[1-{\left(1-{\left(\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left(1-{\left(1-{vc_g}^{\frac{n{\omega}_g\left( T\left({ a l}_g\right)+1\right)}{\sum_{t=1}^n{\omega}_t\left( T\left({ a l}_t\right)+1\right)}}\right)}^x{\left(1-{vc_h}^{\frac{n{\omega}_h\left( T\left({ a l}_h\right)+1\right)}{\sum_{t=1}^n{\omega}_t\left( T\left({ a l}_t\right)+1\right)}}\right)}^y\right)\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}},\right.\hfill \\ {}\left.\left.1-{\left(1-{\left(\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left(1-{\left(1-{vd_g}^{\frac{n{\omega}_g\left( T\left({ a l}_g\right)+1\right)}{\sum_{t=1}^n{\omega}_t\left( T\left({ a l}_t\right)+1\right)}}\right)}^x{\left(1-{vd_h}^{\frac{n{\omega}_h\left( T\left({ a l}_h\right)+1\right)}{\sum_{t=1}^n{\omega}_t\left( T\left({ a l}_t\right)+1\right)}}\right)}^y\right)\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}}\right]\right).\hfill \end{array}\end{array} $$
(22)

The Interval-Valued Intuitionistic Fuzzy Power Geometric Bonferroni Mean Operator

Definition 12 [41]. Suppose al j  = ([μa j , μb j ], [vc j , vd j ]) is a set of the IVIFNs (j = 1, 2⋯, n), then the IVIFPGBM operator is defined as

$$ {\mathrm{IVIFPGBM}}^{x, y}\left({al}_1,{al}_2,\cdot \cdot \cdot, {al}_n\right)=\frac{1}{x+ y}{\left(\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left( x{al_g}^{\frac{n\left( T\left({al}_g\right)+1\right)}{\sum_{t=1}^n\left( T\left({al}_t\right)+1\right)}}+ y{al_h}^{\frac{n\left( T\left({al}_h\right)+1\right)}{\sum_{t=1}^n\left( T\left({al}_t\right)+1\right)}}\right)\right)}^{\frac{1}{n\left( n-1\right)}} $$
(23)

where \( T\left({al}_g\right)=\sum_{h=1, h\ne g}^n Sup\left({al}_g,{al}_h\right) \), x , y > 0.

Theorem 8. Suppose al j  = ([μa j , μb j ], [vc j , vd j ]) is a set of IVIFNs (j = 1, 2⋯, n), then the aggregated result according to Definition 12 is expressed by

$$ \begin{array}{l}{\mathrm{IVIFPGBM}}^{x, y}\left({ a l}_1,{ a l}_2,\cdot \cdot \cdot, { a l}_n\right)=\\ {}\left(\left[\left(1-{\left(1-\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n{\left(1-{\left(1-\mu {a_g}^{\frac{n\left( T\left({ a l}_g\right)+1\right)}{\sum_{t=1}^n\left( T\left({ a l}_t\right)+1\right)}}\right)}^x\times {\left(1-\mu {a_h}^{\frac{n\left( T\left({ a l}_h\right)+1\right)}{\sum_{t=1}^n\left( T\left({ a l}_t\right)+1\right)}}\right)}^y\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}},\right.\right.\right.\\ {}\left.1-{\left(1-\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n{\left(1-{\left(1-\mu {b_g}^{\frac{n\left( T\left({ a l}_g\right)+1\right)}{\sum_{t=1}^n\left( T\left({ a l}_t\right)+1\right)}}\right)}^x\times {\left(1-\mu {b_h}^{\frac{n\left( T\left({ a l}_h\right)+1\right)}{\sum_{t=1}^n\left( T\left({ a l}_t\right)+1\right)}}\right)}^y\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}}\right],\\ {}\begin{array}{l}\left[{\left(1-\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n{\left(1-{\left(1-{\left(1-{vc}_g\right)}^{\frac{n\left( T\left({ a l}_g\right)+1\right)}{\sum_{t=1}^n\left( T\left({ a l}_t\right)+1\right)}}\right)}^x\times {\left(1-{\left(1-{vc}_h\right)}^{\frac{n\left( T\left({ a l}_h\right)+1\right)}{\sum_{t=1}^n\left( T\left({ a l}_t\right)+1\right)}}\right)}^y\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}},\right.\hfill \\ {}\left.\left.{\left(1-\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n{\left(1-{\left(1-{\left(1-{vd}_g\right)}^{\frac{n\left( T\left({ a l}_g\right)+1\right)}{\sum_{t=1}^n\left( T\left({ a l}_t\right)+1\right)}}\right)}^x\times {\left(1-{\left(1-{vd}_h\right)}^{\frac{n\left( T\left({ a l}_h\right)+1\right)}{\sum_{t=1}^n\left( T\left({ a l}_t\right)+1\right)}}\right)}^y\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}},\right]\right).\hfill \end{array}\end{array} $$
(24)

Similar to Theorem 2, the proof of Theorem 8 is omitted.

Now, we will give an example to demonstrate the aggregation process.

Example 4. Suppose that al 1 = ([0.1, 0.3], [0.4, 0.5]) and al 2 = ([0.2, 0.4], [0.3, 0.5]) are two IVIFNs, and let x = 1 , y = 2, then we can derive the following results.

Calculate \( {\tau}_k=\frac{n\left( T\left({al}_k\right)+1\right)}{\sum_{t=1}^n\left( T\left({al}_t\right)+1\right)} \), we can get \( {\tau}_1=\frac{2\left( T\left({al}_1\right)+1\right)}{\sum_{t=1}^2\left( T\left({al}_t\right)+1\right)}=1 \), \( {\tau}_2=\frac{2\left( T\left({al}_2\right)+1\right)}{\sum_{t=1}^2\left( T\left({al}_t\right)+1\right)}=1 \). So, \( \begin{array}{l}{\mathrm{IVIFPGBM}}^{1,2}\left({al}_1,{al}_2\right)=\left(\left[\left(1-{\left(1-{\left(1-{\left(1-{0.1}^1\right)}^1\times {\left(1-{0.2}^1\right)}^2\right)}^{\frac{1}{2}}\times {\left(1-{\left(1-{0.2}^1\right)}^1\times {\left(1-{0.1}^1\right)}^2\right)}^{\frac{1}{2}}\right)}^{\frac{1}{3}},\right.\right.\right.\hfill \\ {}\left(1-{\left(1-{\left(1-{\left(1-{0.3}^1\right)}^1\times {\left(1-{0.4}^1\right)}^2\right)}^{\frac{1}{2}}\times {\left(1-{\left(1-{0.4}^1\right)}^1\times {\left(1-{0.3}^1\right)}^2\right)}^{\frac{1}{2}}\right)}^{\frac{1}{3}}\right],\hfill \\ {}\left[{\left(1-{\left(1-{\left(1-{\left(1-0.4\right)}^1\right)}^1\times {\left(1-{\left(1-0.3\right)}^1\right)}^2\right)}^{\frac{1}{2}}\times {\left(1-{\left(1-{\left(1-0.3\right)}^1\right)}^1\times {\left(1-{\left(1-0.4\right)}^1\right)}^2\right)}^{\frac{1}{2}}\right)}^{\frac{1}{3}},\right.\hfill \\ {}\left.\left[{\left(1-{\left(1-{\left(1-{\left(1-0.4\right)}^1\right)}^1\times {\left(1-{\left(1-0.3\right)}^1\right)}^2\right)}^{\frac{1}{2}}\times {\left(1-{\left(1-{\left(1-0.3\right)}^1\right)}^1\times {\left(1-{\left(1-0.4\right)}^1\right)}^2\right)}^{\frac{1}{2}}\right)}^{\frac{1}{3}}\right]\right)\hfill \\ {}=\left(\left[0.1502,0.3510\right],\left[0.3477,0.5\right]\right).\hfill \end{array} \)

By the operations of IVIFNs, several properties of the IVIFPGBM operator shall be proved.

Theorem 9 (idempotency). Suppose al k  = al = ([ua, ub], [vc, vd])(k = 1, 2,  ... , n), then

$$ \mathrm{IVIFPGBM}\left({al}_1,{al}_2,\dots, {al}_n\right)= al. $$

Similar to Theorem 3, the proof of Theorem 9 is omitted.

Theorem 10 (commutativity). Let \( {al}_k^{\prime } \) be any permutation of al k (k = 1, 2,  ... , n), then

$$ \mathrm{IVIFPGBM}\left({al}_1,{al}_2,\dots, {al}_n\right)=\mathrm{IVIFPGBM}\left({al}_1^{\prime },{al}_2^{\prime },\dots, {al}_n^{\prime}\right). $$

Similar to Theorem 4, the proof of Theorem 10 is omitted.

Similar to the IVIFWPBM operator, the IVIFWPGBM operator shall be given to overcome the shortcoming of the IVIFPGBM operator.

Definition 13. Suppose al j  = ([μa j , μb j ], [vc j , vd j ]) is a set of the IVIFNs (j = 1, 2⋯, n), IVIFWPGBM: Ωn → Ω, then

$$ {\mathrm{IVIFWPGBM}}^{x, y}\left({al}_1,{al}_2,\cdot \cdot \cdot, {al}_n\right)=\frac{1}{x+ y}{\left(\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n\left( x{al_g}^{\frac{n{\omega}_g\left( T\left({al}_g\right)+1\right)}{\sum_{t=1}^n{\omega}_t\left( T\left({al}_t\right)+1\right)}}+ y{al_h}^{\frac{n{\omega}_h\left( T\left({al}_h\right)+1\right)}{\sum_{t=1}^n{\omega}_t\left( T\left({al}_t\right)+1\right)}}\right)\right)}^{\frac{1}{n^2- n}} $$
(25)

where Ω is the set of all IVIFNs, and \( T\left({al}_g\right)=\sum_{h=1, h\ne g}^n Sup\left({al}_g,{al}_h\right) \), x , y > 0. ω = (ω 1, ω 2,  ⋯ , ω n )T is the weight vector of the IVIFNs, 0 ≤ ω k  ≤ 1 (k = 1, 2,  ... , n) and \( \sum_{k=1}^n{\omega}_k=1 \).

Theorem 11. Let al j  = ([μa j , μb j ], [vc j , vd j ]) be a set of the IVIFNs (j = 1, 2⋯, n), x , y > 0, the result aggregated based on Definition 13 is expressed by

$$ \begin{array}{l}{\mathrm{IVIFWPGBM}}^{x, y}\left({ a l}_1,{ a l}_2,\cdot \cdot \cdot, { a l}_n\right)=\\ {}\left(\left[\left(1-{\left(1-\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n{\left(1-{\left(1-\mu {a_g}^{\frac{n{\omega}_g\left( T\left({ a l}_g\right)+1\right)}{\sum_{t=1}^n{\omega}_t\left( T\left({ a l}_t\right)+1\right)}}\right)}^x\times {\left(1-\mu {a_h}^{\frac{n{\omega}_h\left( T\left({ a l}_h\right)+1\right)}{\sum_{t=1}^n{\omega}_t\left( T\left({ a l}_t\right)+1\right)}}\right)}^y\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}},\right.\right.\right.\\ {}\left.1-{\left(1-\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n{\left(1-{\left(1-\mu {b_g}^{\frac{n{\omega}_g\left( T\left({ a l}_g\right)+1\right)}{\sum_{t=1}^n{\omega}_t\left( T\left({ a l}_t\right)+1\right)}}\right)}^x\times {\left(1-\mu {b_h}^{\frac{n{\omega}_h\left( T\left({ a l}_h\right)+1\right)}{\sum_{t=1}^n{\omega}_t\left( T\left({ a l}_t\right)+1\right)}}\right)}^y\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}}\right],\\ {}\begin{array}{l}\left[{\left(1-\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n{\left(1-{\left(1-{\left(1-{vc}_g\right)}^{\frac{n{\omega}_g\left( T\left({ a l}_g\right)+1\right)}{\sum_{t=1}^n{\omega}_t\left( T\left({ a l}_t\right)+1\right)}}\right)}^x\times {\left(1-{\left(1-{vc}_h\right)}^{\frac{n{\omega}_h\left( T\left({ a l}_h\right)+1\right)}{\sum_{t=1}^n{\omega}_t\left( T\left({ a l}_t\right)+1\right)}}\right)}^y\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}},\right.\hfill \\ {}\left.\left.{\left(1-\prod_{\begin{array}{l} g=1, h=1\\ {}\kern0.5em \mathrm{g}\ne h\end{array}}^n{\left(1-{\left(1-{\left(1-{vd}_g\right)}^{\frac{n{\omega}_g\left( T\left({ a l}_g\right)+1\right)}{\sum_{t=1}^n{\omega}_t\left( T\left({ a l}_t\right)+1\right)}}\right)}^x\times {\left(1-{\left(1-{vd}_h\right)}^{\frac{n{\omega}_h\left( T\left({ a l}_h\right)+1\right)}{\sum_{t=1}^n{\omega}_t\left( T\left({ a l}_t\right)+1\right)}}\right)}^y\right)}^{\frac{1}{n^2- n}}\right)}^{\frac{1}{x+ y}},\right]\right).\hfill \end{array}\end{array} $$
(26)

The MAGDM Approach Based on Interval-Valued Intuitionistic Fuzzy Power Bonferroni Mean and Interval-Valued Intuitionistic Fuzzy Power Geometric Bonferroni Mean Operators

For a MAGDM problem with IVIFNs, in which the attributes’ and experts’ weights are known, let Z = {z 1, z 2,  ⋯ , z m } be the set of all alternatives, A = {a 1, a 2,  ⋯ , a n } be the set of attributes, and E = {e 1, e 2,  ⋯ , e t } be the set of all experts. Assume that \( {\tilde{a}}_{gh}^k=\left(\left[{a}_{gh}^k,{b}_{gh}^k\right],\left[{c}_{gh}^k,{d}_{gh}^k\right]\right) \) is the attribute evaluation value given by the expert e k for the alternative z g about the attribute a h . ω = (ω 1, ω 2,  ⋯ , ω n ) is the weight vector of {a 1, a 2,  ⋯ , a n } satisfying with \( {\omega}_h\in \left[0,1\right],\sum_{h=1}^n{\omega}_h=1 \). γ = (γ 1, γ 2,  ⋯ , γ t ) is the weight vector of {e 1, e 2,  ⋯ , e t }, and \( {\gamma}_k\in \left[0,1\right],\sum_{k=1}^t{\gamma}_k=1\left( k=1,2,\cdots, t\right) \), then the goal of this MAGDM problem is to rank the alternatives.

The Decision-Making Steps Based on Interval-Valued Intuitionistic Fuzzy Weighted Power Bonferroni Mean and Interval-Valued Intuitionistic Fuzzy Weighted Power Geometric Bonferroni Mean Operators

  1. Step 1.

    Normalize the decision matrix.

Generally, if there are the different types in attributes, we need to convert them to the same type. For convenience, we need to convert the cost type to the benefit type by the following method:

$$ {\tilde{r}}_{gh}^k=\left(\left[{\underset{\bar{\mkern6mu}}{u}}_{gh}^k,{\overline{u}}_{gh}^k\right],\left[{\underset{\bar{\mkern6mu}}{f}}_{gh}^k,{\overline{f}}_{gh}^k\right]\right)=\left\{\begin{array}{l}\left(\left[{a}_{gh}^k,{b}_{gh}^k\right],\left[{c}_{gh}^k,{d}_{gh}^k\right]\right)\;\mathrm{for}\;\mathrm{benefit}\;\mathrm{attribute}\;{a}_h\hfill \\ {}\left(\left[{c}_{gh}^k,{d}_{gh}^k\right],\left[{a}_{gh}^k,{b}_{gh}^k\right]\right)\;\mathrm{for}\;\mathrm{cost}\;\mathrm{attribute}\;{a}_h\hfill \end{array}\right. $$
(27)

So, the decision matrices \( \tilde{A}={\left[{\tilde{a}}_{gh}^k\right]}_{m\times n} \) can be converted to matrices \( \tilde{R}={\left[{\tilde{r}}_{gh}^k\right]}_{m\times n} \).

  1. Step 2.

    Calculate the supports \( Sup\left({\tilde{r}}_{gh}^k,{\tilde{r}}_{gl}^k\right)\left( g=1,2,\cdots, m; k=1,2\cdots, t; h, l=1,2,\cdots, n\right) \) by

$$ Sup\left({\tilde{r}}_{gh}^k,{\tilde{r}}_{gl}^k\right)=1- d\left({\tilde{r}}_{gh}^k,{\tilde{r}}_{gl}^k\right) $$
(28)

where \( d\left({\tilde{r}}_{gh}^k,{\tilde{r}}_{gl}^k\right) \) is the Euclidean distance between two IVIFNs \( {\tilde{r}}_{gh}^k \) and \( {\tilde{r}}_{gl}^k \), which is from Definition 3.

  1. Step 3.

    Calculate \( T\left({\tilde{r}}_{gh}^k\right) \) by

$$ T\left({\tilde{r}}_{gh}^k\right)=\sum_{\begin{array}{l} l=1\\ {} l\ne h\end{array}}^n Sup\left({\tilde{r}}_{gh}^k,{\tilde{r}}_{gl}^k\right)\kern0.5em \left( g=1,2,\cdots, m; k=1,2\cdots, t; h=1,2,\cdots, n\right) $$
(29)
  1. Step 4.

    Calculate\( {\tau}_{gh}^k=\frac{n{\omega}_h\left(1+ T\left({\tilde{r}}_{gh}^k\right)\right)}{\sum_{t=1}^n{\omega}_t\left(1+ T\left({\tilde{r}}_{gt}^k\right)\right)}\ \left( g=1,2,\cdots, m; k=1,2\cdots, t; h=1,2,\cdots, n\right) \) .

  2. Step 5.

    Utilize the IVIFWPBM or IVIFWPGBM operator.

$$ {\tilde{r}}_g^k=\left(\left[{\underset{\bar{\mkern6mu}}{u}}_g^k,{\overline{u}}_g^k\right],\Big[{\underset{\bar{\mkern6mu}}{f}}_g^k,{\overline{f}}_g^k\Big]\right)=\mathrm{IVIFWPBM}\left({\tilde{r}}_{g1}^k,{\tilde{r}}_{g2}^k,\cdots, {\tilde{r}}_{g n}^k\right)\ \mathrm{or}\ \mathrm{IVIFWPGBM}\left({\tilde{r}}_{g1}^k,{\tilde{r}}_{g2}^k,\cdots, {\tilde{r}}_{g n}^k\right) $$
(30)

to determine the overall IVIFNs \( {{\tilde{r}}^k}_g\ \left( g=1,2,\cdots, m; k=1,2\cdots, t\right) \).

  1. Step 6.

    Calculate the supports \( Sup\left({{\tilde{r}}^k}_g,{{\tilde{r}}^l}_g\right)\kern0.5em \left( g=1,2,\cdots, m; k, l=1,2,\cdots, t\right) \) by

$$ Sup\left({{\tilde{r}}^k}_g,{{\tilde{r}}^l}_g\right)=1- d\left({{\tilde{r}}^k}_g,{{\tilde{r}}^l}_g\right),\kern0.5em $$
(31)

where \( d\left({{\tilde{r}}^k}_g,{{\tilde{r}}^l}_g\right) \)is the Euclidean distance between two IVIFNs \( {{\tilde{r}}^k}_g \) and\( {{\tilde{r}}^l}_g \), which is from Definition 3.

  1. Step 7.

    Calculate\( T\left({\tilde{r}}_g^k\right) \)by

$$ T\left({\tilde{r}}_g^k\right)=\sum_{\begin{array}{l} l=1\\ {} l\ne g\end{array}}^t Sup\left({\tilde{r}}_g^k,{\tilde{r}}_g^l\right)\kern0.5em \left( g=1,2,\cdots, m; k=1,2,\cdots, t\right) $$
(32)
  1. Step 8.

    Calculate\( {\tau}_g^k=\frac{t{\gamma}_k\left(1+ T\left({\tilde{r}}_g^k\right)\right)}{\sum_{k=1}^t{\gamma}_k\left(1+ T\left({\tilde{r}}_g^k\right)\right)}\kern1em \left( g=1,2,\cdots, m; k=1,2\cdots, t\right) \).

Step 9: Use IVIFWPBM or IVIFWPGBM operators to get the collective IVIFNs \( {\tilde{r}}_g\left( g=1,2,\dots, m\right) \).

$$ {\tilde{r}}_g=\left(\left[{\underset{\bar{\mkern6mu}}{u}}_g,{\overline{u}}_g\right],\Big[{\underset{\bar{\mkern6mu}}{f}}_g,{\overline{f}}_g\Big]\right)=\mathrm{IVIFWPBM}\left({\tilde{r}}_g^1,{\tilde{r}}_g^2,\cdots, {\tilde{r}}_g^t\right)\ \mathrm{or}\ \mathrm{IVIFWPGBM}\left({\tilde{r}}_g^1,{\tilde{r}}_g^2,\cdots, {\tilde{r}}_g^t\right) $$
(33)

Step 10: Calculate the score function \( sf\left({\tilde{r}}_g\right) \) and accuracy function \( af\left({\tilde{r}}_g\right) \) of the collective IVIFNs\( {\tilde{r}}_g\left( g=1,2,\dots, m\right) \).

Step 11: Rank all the alternatives {z 1, z 2,  ⋯ , z m } by comparison method of IVIFNs, and opt for the most eligible alternative(s).

Step 12: End.

In order to easily perform the steps, we can give some pseudo codes as follows:

figure a

An application example

This example is adapted from Liu [19]. Suppose that four alternatives (z 1, z 2, z 3, z 4) representing the air quality of 2006, 2007, 2008, and 2009 are evaluated (the air quality of Guangzhou). Three attributes are taken into consideration, including the SO2 (a 1), the NO2 (a 2), and the PM10 (a 3). The weight vector about criteria is provided by (0.40, 0.20, 0.40)T. The possible alternatives z g (g = 1, 2, 3, 4) are assessed by three air-quality monitoring stations regarded as experts (e 1, e 2, e 3). The weight vector about experts is provided by(0.314, 0.355, 0.331)T. The assessment values are represented by the IVIFNs, which are listed in Tables 1, 2, and 3.

Table 1 Air quality data from station e 1
Full size table
Table 2 Air quality data from station e 2
Full size table
Table 3 Air quality data from station e 3
Full size table

Rank the Alternatives by the Proposed Method Based on the Interval-Valued Intuitionistic Fuzzy Power Bonferroni Mean Operator

Step 1: Transform the decision matrix \( {\tilde{A}}^k={\left[{{\tilde{a}}^k}_{gh}\right]}_{m\times n} \) into the normalized matrix \( {\tilde{R}}^k={\left[{{\tilde{r}}^k}_{gh}\right]}_{m\times n} \).

Because all the attributes are the same type, they do not need to be normalized.

Step 2: Calculate the supports\( Sup\left({\tilde{r}}_{gh}^k,{\tilde{r}}_{gl}^k\right) \).

By formula (28), calculate the supports \( Sup\left({\tilde{r}}_{gh}^k,{\tilde{r}}_{gl}^k\right) \) (for simplicity, we denote\( Sup\left({\tilde{r}}_{gh}^k,{\tilde{r}}_{gl}^k\right) \) with\( {S}_{gh, gl}^k\left( h, l=1,2,3;\mathrm{g}=1,2,3,4; k=1,2,3.\right) \)). We can get

$$ \begin{array}{l}{S^1}_{11,12}={S^1}_{12,11}=0.8502,{S^1}_{12,13}={S^1}_{13,12}=0.8374,{S^1}_{11,13}={S^1}_{13,11}=0.8964\\ {}{S^1}_{21,22}={S^1}_{22,21}=0.8874,{S^1}_{22,23}={S^1}_{23,22}=0.8503,{S^1}_{21,23}={S^1}_{23,21}=0.9149\\ {}{S^1}_{31,32}={S^1}_{32,31}=0.8701,{S^1}_{32,33}={S^1}_{33,32}=0.8742,{S^1}_{31,33}={S^1}_{33,31}=0.9388\\ {}{S^1}_{41,42}={S^1}_{42,41}=0.9423,{S^1}_{42,43}={S^1}_{43,42}=0.8569,{S^1}_{41,43}={S^1}_{43,41}=0.8873\\ {}{S^2}_{11,12}={S^2}_{12,11}=0.9178,{S^2}_{12,13}={S^2}_{13,12}=0.8188,{S^2}_{11,13}={S^2}_{13,11}=0.7655\\ {}{S^2}_{21,22}={S^2}_{22,21}=0.8280,{S^2}_{22,23}={S^2}_{23,22}=0.7916,{S^2}_{21,23}={S^2}_{23,21}=0.8402\\ {}{S^2}_{31,32}={S^2}_{32,31}=0.8504,{S^2}_{32,33}={S^2}_{33,32}=0.9150,{S^2}_{31,33}={S^2}_{33,31}=0.9209\\ {}{S^2}_{41,42}={S^2}_{42,41}=0.9390,{S^2}_{42,43}={S^2}_{43,42}=0.8184,{S^2}_{41,43}={S^2}_{43,41}=0.8642\\ {}{S^3}_{11,12}={S^3}_{12,11}=0.9573,{S^3}_{12,13}={S^3}_{13,12}=0.9348,{S^3}_{11,13}={S^3}_{13,11}=0.9441\\ {}{S^3}_{21,22}={S^3}_{22,21}=0.8818,{S^3}_{22,23}={S^3}_{23,22}=0.7967,{S^3}_{21,23}={S^3}_{23,21}=0.8725\\ {}{S^3}_{31,32}={S^3}_{32,31}=0.8972,{S^3}_{32,33}={S^3}_{33,32}=0.9333,{S^3}_{31,33}={S^3}_{33,31}=0.8948\\ {}{S^3}_{41,42}={S^3}_{42,41}=0.8364,{S^3}_{42,43}={S^3}_{43,42}=0.8249,{S^3}_{41,43}={S^3}_{43,41}=0.8042\end{array} $$

Step3: Calculate\( T\left({\tilde{r}}_{gh}^k\right)\ \left( h=1,2,3; g=1,2,3,4; k=1,2,3\right) \) by formula (29) (for simplicity, we denote \( T\left({{\tilde{r}}^k}_{gh}\right) \)withT gh k).

$$ \begin{array}{l}{T^1}_{11}=1.7466,\kern0.5em {T^1}_{12}=1.6876,\kern0.5em {T^1}_{13}=1.7339,{T^1}_{21}=1.8023,\kern0.5em {T^1}_{22}=1.7377,\kern0.5em {T^1}_{23}=1.7651\\ {}{T^1}_{31}=1.8089,\kern0.5em {T^1}_{32}=1.7443,\kern0.5em {T^1}_{33}=1.8130,{T^1}_{41}=1.8296,\kern0.5em {T^1}_{42}=1.7992,\kern0.5em {T^1}_{43}=1.7442\\ {}{T^2}_{11}=1.6833,\kern0.5em {T^2}_{12}=1.7366,\kern0.5em {T^2}_{13}=1.5842,{T^2}_{21}=1.6683,\kern0.5em {T^2}_{22}=1.6196,\kern0.5em {T^2}_{23}=1.6318\\ {}{T^2}_{31}=1.7714,\kern0.5em {T^2}_{32}=1.7654,\kern0.5em {T^2}_{33}=1.8359,{T^2}_{41}=1.8031,\kern0.5em {T^2}_{42}=1.7574,\kern0.5em {T^2}_{43}=1.6826\\ {}{T^3}_{11}=1.9014,\kern0.5em {T^3}_{12}=1.8921,\kern0.5em {T^3}_{13}=1.8789,{T^3}_{21}=1.7543,\kern0.5em {T^3}_{22}=1.6785,\kern0.5em {T^3}_{23}=1.6692\\ {}{T^3}_{31}=1.7919,\kern0.5em {T^3}_{32}=1.8305,\kern0.5em {T^3}_{33}=1.8281,{T^3}_{41}=1.6406,\kern0.5em {T^3}_{42}=1.6613,\kern0.5em {T^3}_{43}=1.6292\end{array} $$

Step 4: Calculateτ k gh (g = 1, 2, 3, 4; h = 1, 2, 3; k = 1, 2, 3.), we get

$$ \begin{array}{l}{\tau^1}_{11}=1.2074,\kern0.5em {\tau^1}_{12}=0.5907,\kern0.5em {\tau^1}_{13}=1.2018,{\tau^1}_{21}=1.2120,\kern0.5em {\tau^1}_{22}=0.5920,\kern0.5em {\tau^1}_{23}=1.1959\\ {}{\tau^1}_{31}=1.2048,\kern0.5em {\tau^1}_{32}=0.5886,\kern0.5em {\tau^1}_{33}=1.2066,{\tau^1}_{41}=1.2173,\kern0.5em {\tau^1}_{42}=0.6021,\kern0.5em {\tau^1}_{43}=1.1806\\ {}{\tau^2}_{11}=1.2131,\kern0.5em {\tau^2}_{12}=0.6186,\kern0.5em {\tau^2}_{13}=1.1683,{\tau^2}_{21}=1.2110,\kern0.5em {\tau^2}_{22}=0.5945,\kern0.5em {\tau^2}_{23}=1.1945\\ {}{\tau^2}_{31}=1.1894,\kern0.5em {\tau^2}_{32}=0.5934,\kern0.5em {\tau^2}_{33}=1.2171,{\tau^2}_{41}=1.2251,\kern0.5em {\tau^2}_{42}=0.6025,\kern0.5em {\tau^2}_{43}=1.1724\\ {}{\tau^3}_{11}=1.2045,\kern0.5em {\tau^3}_{12}=0.6003,\kern0.5em {\tau^3}_{13}=1.1952,{\tau^3}_{21}=1.2218,\kern0.5em {\tau^3}_{22}=0.5941,\kern0.5em {\tau^3}_{23}=1.1841\\ {}{\tau^3}_{31}=1.1906,\kern0.5em {\tau^3}_{32}=0.6035,\kern0.5em {\tau^3}_{33}=1.2060,{\tau^1}_{41}=1.2002,\kern0.5em {\tau^1}_{42}=0.6048,\kern0.5em {\tau^1}_{43}=1.1950\end{array} $$

Step 5: Utilize the IVIFWPBM operator to determine the overall IVIFNs \( {{\tilde{r}}^k}_g \), which is listed in Table 4 (suppose x , y = 1).

Table 4 the overall IVIFNs \( {{\tilde{r}}^k}_g \) from three monitoring stations (e 1, e 2, e 3)
Full size table

Step 6: Calculate the supports \( Sup\left({\tilde{r}}_g^k,{\tilde{r}}_g^l\right) \) based on formula (31) (for simplicity, we denote \( Sup\left({\tilde{r}}_g^k,{\tilde{r}}_g^l\right) \)with\( {S}_g^{k, l}\left( g=1,2,3,4; k, l=1,2,3\right) \)). We can get

$$ \begin{array}{l}{S_1}^{1,2}={S_1}^{2,1}=0.9192,{S_1}^{2,3}={S_1}^{3,2}=0.9489,{S_1}^{1,3}={S_1}^{3,1}=0.9027\\ {}{S_2}^{1,2}={S_2}^{2,1}=0.9165,{S_2}^{2,3}={S_2}^{3,2}=0.9281,{S_2}^{1,3}={S_2}^{3,1}=0.9128\\ {}{S_3}^{1,2}={S_3}^{2,1}=0.9190,{S_3}^{2,3}={S_3}^{3,2}=0.9222,{S_3}^{1,3}={S_3}^{3,1}=0.8527\\ {}{S_4}^{1,2}={S_4}^{2,1}=0.9809,{S_4}^{2,3}={S_4}^{3,2}=0.9168,{S_4}^{1,3}={S_4}^{3,1}=0.9022\end{array} $$

Step 7: Calculate \( T\left({\tilde{r}}_g^k\right)\ \left( g=1,2,3,4; k=1,2,3.\right) \) based on formula (32) (for simplicity, we denote \( T\left({{\tilde{r}}^k}_g\right) \)withT g k).

$$ \begin{array}{l}{T^1}_1=1.8219,\kern0.5em {T^2}_1=1.8681,\kern0.5em {T^3}_1=1.8516,{T^1}_2=1.8293,\kern0.5em {T^2}_2=1.8446,\kern0.5em {T^3}_2=1.8409\\ {}{T^1}_3=1.7717,\kern0.5em {T^2}_3=1.8413,\kern0.5em {T^3}_3=1.7750,{T^1}_4=1.8831,\kern0.5em {T^2}_4=1.8977,\kern0.5em {T^3}_4=1.8190\end{array} $$

Step 8: Calculateτ k g (g = 1, 2, 3, 4; k = 1, 2, 3.), we get

$$ {\tau^1}_1=0.9333,\kern0.5em {\tau^2}_1=1.0725,\kern0.5em {\tau^3}_1=0.9942,{\tau^1}_2=0.9389,\kern0.5em {\tau^2}_2=1.0673,\kern0.5em {\tau^3}_2=0.9938 $$
$$ {\tau^1}_3=0.9333,\kern0.5em {\tau^2}_3=1.0817,\kern0.5em {\tau^3}_3=0.9850,{\tau^1}_4=0.9473,\kern0.5em {\tau^2}_4=1.0764,\kern0.5em {\tau^3}_4=0.9764 $$

Step 9: Utilize the IVIFWPBM operator to determine the collective IVIFNs\( {\tilde{r}}_g \)which is listed in Table 5 (suppose x , y = 1).

Table 5 The collective IVIFNs \( {\tilde{r}}_g \) for four alternatives
Full size table

Step 10: Calculate the score function\( sf\left({\tilde{r}}_g\right) \), we get

$$ sf\left({\tilde{r}}_1\right)=-0.1143, sf\left({\tilde{r}}_2\right)=-0.0809, sf\left({\tilde{r}}_3\right)=0.0433, sf\left({\tilde{r}}_4\right)=0.1410 $$

Step 11: Rank all the alternatives.

According to \( sf\left({\tilde{r}}_g\right) \), we rank the alternatives {z 1, z 2, z 3, z 4} shown as follows:

z 4 ≻ z 3 ≻ z 2 ≻ z 1.

Rank the Alternatives by the Proposed Method Based on the Interval-Valued Intuitionistic Fuzzy Weighted Power Geometric Bonferroni Mean Operator

Step 1 to Step 4 is the same as those in the “Rank the alternatives by the proposed method based on the IVIFWPBM operator” section.

Step 5: Utilize the IVIFWPGBM operator to determine the overall IVIFNs \( {{\tilde{r}}^k}_g \), which is listed in Table 6 (supposex , y = 1).

Table 6 The overall IVIFNs \( {{\tilde{r}}^k}_g \) from three monitoring stations (e 1, e 2, e 3) by IVIFWPGBM operator
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Step 6: Calculate the supports \( Sup\left({\tilde{r}}_g^k,{\tilde{r}}_g^l\right) \) based on formula (31) (for simplicity, we denote \( Sup\left({\tilde{r}}_g^k,{\tilde{r}}_g^l\right) \)with \( {S}_g^{k, l}\left( k, l=1,2,3;\mathrm{g}=1,2,3,4.\right) \)). We can get

$$ \begin{array}{l}{S_1}^{1,2}={S_1}^{2,1}=0.9232,{S_1}^{2,3}={S_1}^{3,2}=0.9526,{S_1}^{1,3}={S_1}^{3,1}=0.9106\\ {}{S_2}^{1,2}={S_2}^{2,1}=0.9280,{S_2}^{2,3}={S_2}^{3,2}=0.9312,{S_2}^{1,3}={S_2}^{3,1}=0.9211\\ {}{S_3}^{1,2}={S_3}^{2,1}=0.9209,{S_3}^{2,3}={S_3}^{3,2}=0.9263,{S_3}^{1,3}={S_3}^{3,1}=0.8698\\ {}{S_4}^{1,2}={S_4}^{2,1}=0.9831,{S_4}^{2,3}={S_4}^{3,2}=0.9275,{S_4}^{1,3}={S_4}^{3,1}=0.9143\end{array} $$

Step 7: Calculate \( T\left({\tilde{r}}_g^k\right)\ \left( g=1,2,3,4; k=1,2,3\right) \) based on formula (32) (for simplicity, we denote \( T\left({{\tilde{r}}^k}_g\right) \)withT g k).

$$ \begin{array}{l}{T^1}_1=1.8338,\kern0.5em {T^2}_1=1.8758,\kern0.5em {T^3}_1=1.8632,{T^1}_2=1.8490,\kern0.5em {T^2}_2=1.8592,\kern0.5em {T^3}_2=1.8523\\ {}{T^1}_3=1.7907,\kern0.5em {T^2}_3=1.8472,\kern0.5em {T^3}_3=1.7961,{T^1}_4=1.8975,\kern0.5em {T^2}_4=1.9106,\kern0.5em {T^3}_4=1.8418,,\end{array} $$

Step 8: Calculateτ k g (g = 1, 2, 3, 4; k = 1, 2, 3.), we get

$$ \begin{array}{l}{\tau^1}_1=0.9339,\kern0.5em {\tau^2}_1=1.0715,\kern0.5em {\tau^3}_1=0.9947,{\tau^1}_2=0.9405,\kern0.5em {\tau^2}_2=1.0670,\kern0.5em {\tau^3}_2=0.9925\\ {}{\tau^1}_3=0.9347,\kern0.5em {\tau^2}_3=1.0781,\kern0.5em {\tau^3}_3=0.9872,{\tau^1}_4=0.9465,\kern0.5em {\tau^2}_4=1.0749,\kern0.5em {\tau^3}_4=0.9786\end{array} $$

Step 9: Utilize the IVIFWPGBM operator to determine the IVIFNs \( {\tilde{r}}_g\left( g=1,2,3,4\right) \), which is listed in Table 7 (supposex , y = 1).

Table 7 The collective IVIFNs\( {\tilde{r}}_g \) for four alternatives by IVIFWPGBM operator
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Step 10: Calculate the score function\( sf\left({\tilde{r}}_g\right) \), we get

$$ sf\left({\tilde{r}}_1\right)=-0.1087, sf\left({\tilde{r}}_2\right)=-0.0756, sf\left({\tilde{r}}_3\right)=0.0514, sf\left({\tilde{r}}_4\right)=0.1448 $$

Step 11: Rank the alternatives.

According to \( sf\left({\tilde{r}}_g\right) \), we rank the alternatives {z 1, z 2, z 3, z 4} shown as follows:

z 4 ≻ z 3 ≻ z 2 ≻ z 1.

The Influence of the Parameters x , y on the Decision-Making Result

To observe the influence of parameters x , y on decision making, we set the different values x , y in Step 5 and Step 9, then to rank {z 1, z 2, z 3, z 4}. The results are listed in Tables 8 and 9.

Table 8 Ordering of the alternatives based on IVIFWPBM by using the different x , y
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Table 9 Ordering of the alternatives based on IVIFWPGBM by using the different x , y
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As we can see from Tables 8 and 9, the aggregation results based on IVIFWPBM operator or IVIFWPGBM operator are different, but the orderings are the same. Furthermore, orderings produced by the different parameters x , y are the same. So, the proposed method is practical and effective. In general, we set the parameter x = y = 1.

Comparison with Other Methods

To further demonstrate the validity of the proposed methods in this paper, we solve the same illustrative example [19] by using the three existing MAGDM methods, which are the IVIFWA operator-based approach proposed by Xu [27], the IVIFWPA operator-based approach proposed by He [45], and the IVIFWBM operator-based approach proposed by Xu [46]. The final orders of the alternatives obtained by the above three methods are listed in Table 10.

Table 10 Comparisons of ranking results for different methods
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From Table 10, the methods proposed in [27, 45, 46] have the same ranking results with the proposed method. This can verify the proposed method. In the following, we give some characteristic comparisons of our proposed method and the aforementioned three methods, which are listed in Table 11.

Table 11 Characteristic comparisons of different operators
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Conclusion

In this paper, we propose several PBM aggregation operators for IVIFNs, such as IVIFPBM operator, IVIFWPBM operator, IVIFPGBM operator, and IVIFWPGBM operator, and then we discussed several properties and special cases of the proposed operators. Obviously, these operators can take the advantages of power operator and Bonferroni mean operator, i.e., they can overcome the influence of the unreasonable attribute values and can also consider the interaction between two attributes. In addition, we utilized these operators to solve the MAGDM problem with IVIFNs, and an example is provided to illustrate the validity and advantages of the proposed methods by comparing with three existing methods.

In further researches, we will develop some real applications of these proposed operators in other areas, such as supplier selection evaluation, product scheme selection evaluation, fuzzy cluster analysis, and so on. In addition, we can also extend the PBM operators to some new fuzzy information, such as Pythagorean fuzzy set, linguistic interval hesitant fuzzy set, neutrosophic set, and so on.