A decision support model for group decision making with intuitionistic fuzzy linguistic preferences relations

Abstract

As a new preference structure, the intuitionistic fuzzy linguistic preference relation (IFLPR) was introduced to efficiently cope with situations in which the membership degree and non-membership degree are represented as linguistic terms. For group decision making (GDM) problems with IFLPRs, two significant and challenging issues are individual consistency and group consensus before deriving the reliable priority weights of alternatives. In this paper, a novel decision support model is investigated to simultaneously deal with the individual consistency and group consensus for GDM with IFLPRs. First, the concepts of multiplicative consistency and weak transitivity for IFLPRs are introduced and followed by a discussion of their desirable properties. Then, a transformation approach is developed to convert the normalized intuitionistic fuzzy priority weights into multiplicative consistent IFLPR. Based on the distance of IFLPRs, the consistency index, individual consensus degree and group consensus degree for IFLPRs are further defined. In addition, two convergent automatic iterative algorithms are proposed in the investigated decision support model. The first algorithm is utilized to convert an unacceptable multiplicative consistent IFLPR to an acceptable one. The second algorithm can assist the group decision makers to achieve a predefined consensus level. The main characteristic of the investigated decision support model is that it guarantees each IFLPR is still acceptable multiplicative consistent when the predefined consensus level is achieved. Finally, several numerical examples are provided, and comparative analyses with existing approaches are performed to demonstrate the effectiveness and practicality of the investigated model.

Appendix

Proof of Theorem 1

First, we prove the necessary part. If IFLPR $B = (b_{ij} )_{n \times n}$ is multiplicative consistent, by Definition 3, we have

$$ I(b_{ij\mu } )I(b_{jk\mu } )I(b_{ki\mu } ) = I(b_{ji\mu } )I(b_{ik\mu } )I(b_{kj\mu } ),\quad i,j,k \in N. $$

As $b_{ij\mu } = b_{ji\nu } ,b_{ij\nu } = b_{ji\mu } ,i,j \in N$, one can get

$$ I(b_{ij\mu } )I(b_{kj\nu } )I(b_{ik\nu } ) = I(b_{ij\nu } )I(b_{kj\mu } )I(b_{ik\mu } ),\quad i,j,k \in N, $$

(33)

then

$$ \frac{{I(b_{ij\mu } )}}{{I(b_{ij\nu } )}} = \frac{{I(b_{ik\mu } )}}{{I(b_{ik\nu } )}} \cdot \frac{{I(b_{kj\mu } )}}{{I(b_{kj\nu } )}},\quad i,j,k \in N. $$

(34)

Therefore, it follows from Eq. (34) that

$$ \varPhi (b_{ij} ) = \varPhi (b_{ik} ) \cdot \varPhi (b_{kj} ),\quad i,j,k \in N. $$

Next, we prove the sufficient part. Since $\varPhi (b_{ij} ) = \varPhi (b_{ik} ) \cdot \varPhi (b_{kj} ),i,j,k \in N$, and $\varPhi (b_{ij} ) = \frac{{I(b_{ij\mu } )}}{{I(b_{ij\nu } )}}$, we have

$$ \frac{{I(b_{ij\mu } )}}{{I(b_{ij\nu } )}} = \frac{{I(b_{ik\mu } )}}{{I(b_{ik\nu } )}} \cdot \frac{{I(b_{kj\mu } )}}{{I(b_{kj\nu } )}},\quad i,j,k \in N, $$

then by reversing the aforesaid proof of the necessary condition, one can obtain that

$$ I(b_{ij\mu } )I(b_{jk\mu } )I(b_{ki\mu } ) = I(b_{ji\mu } )I(b_{ik\mu } )I(b_{kj\mu } ),\quad i,j,k \in N, $$

which means that $B$ is multiplicative consistent. □

Proof of Theorem 2

Since $B = (b_{ij} )_{n \times n}$ is a multiplicative consistent IFLPR, then we have $\varPhi (b_{ij} ) = \varPhi (b_{ik} ) \cdot \varPhi (b_{kj} ),i,j,k \in N$. If $\varPhi (b_{ik} ) \ge 1$ and $\varPhi (b_{kj} ) \ge 1$, thus $\varPhi (b_{ij} ) = \varPhi (b_{ik} ) \cdot \varPhi (b_{kj} ) \ge 1$, for any $i,j,k \in N$. By Definition 4, the proof of Theorem 2 is completed. □

Proof of Theorem 3

First, we prove that the matrix $F = (f_{ij} )_{n \times n}$ is an IFLPR.

For any $i,j \in N$, we have $f_{ij\mu } = I^{ - 1} \left( {2\tau \sqrt {\omega_{i\mu } \omega_{j\nu } } } \right) = I^{ - 1} \left( {2\tau \sqrt {\omega_{j\nu } \omega_{i\mu } } } \right) = f_{ji\nu }$.

As $\omega_{i\mu } ,\omega_{i\nu } \in [0,1],\omega_{i\mu } + \omega_{i\nu } \le 1,i \in N$, we have

$$ 0 \le \sqrt {\omega_{i\mu } \omega_{j\nu } } \le \frac{{\omega_{i\mu } + \omega_{j\nu } }}{2} \le \frac{1 + 1}{2} \le 1,\quad 0 \le \sqrt {\omega_{i\nu } \omega_{j\mu } } \le \frac{{\omega_{i\nu } + \omega_{j\mu } }}{2} \le \frac{1 + 1}{2} \le 1,\quad i,j \in N, $$

then

$$ 0 \le 2\tau \sqrt {\omega_{i\mu } \omega_{j\nu } } \le 2\tau \,{\text{and}}\,0 \le 2\tau \sqrt {\omega_{i\nu } \omega_{j\mu } } \le 2\tau ,\quad i,j \in N, $$

it follows that

$$ s_{0} \le I^{ - 1} \left( {2\tau \sqrt {\omega_{i\mu } \omega_{j\nu } } } \right) \le s_{2\tau } ,\quad s_{0} \le I^{ - 1} \left( {2\tau \sqrt {\omega_{i\nu } \omega_{j\mu } } } \right) \le s_{2\tau } ,\quad i,j \in N, $$

i.e.,

$$ s_{0} \le f_{ij\mu } \le s_{2\tau } \,{\text{and}}\,s_{0} \le f_{ij\nu } \le s_{2\tau } ,i,j \in N. $$

(35)

Moreover,

$$ \begin{aligned} I(f_{ij\mu } ) + I(f_{ij\nu } ) & = 2\tau \sqrt {\omega_{i\mu } \omega_{j\nu } } + 2\tau \sqrt {\omega_{i\nu } \omega_{j\mu } } \\ & \le 2\tau \times \frac{{\omega_{i\mu } + \omega_{j\nu } }}{2} + 2\tau \times \frac{{\omega_{i\nu } + \omega_{j\mu } }}{2} \\ & = \tau \times \left( {(\omega_{i\mu } + \omega_{i\nu } ) + (\omega_{j\mu } + \omega_{j\nu } )} \right) = 2\tau ,\quad i,j \in N, \\ \end{aligned} $$

that is

$$ f_{ij\mu } + f_{ij\nu } \le s_{2\tau } ,i,j \in N. $$

(36)

According to Definition 1, $F = (f_{ij} )_{n \times n}$ is an IFLPR.

Next, we prove that the IFLPR $F = (f_{ij} )_{n \times n}$ is multiplicative consistent.

For any $i,j,k \in N$, one can get that

$$ \begin{aligned} \varPhi (f_{ik} ) \cdot \varPhi (f_{kj} ) & = \frac{{I(f_{ik\mu } )}}{{I(f_{ik\nu } )}} \cdot \frac{{I(f_{kj\mu } )}}{{I(f_{kj\nu } )}} = \frac{{2\tau \sqrt {\omega_{i\mu } \omega_{k\nu } } }}{{2\tau \sqrt {\omega_{i\nu } \omega_{k\mu } } }} \cdot \frac{{2\tau \sqrt {\omega_{k\mu } \omega_{j\nu } } }}{{2\tau \sqrt {\omega_{k\nu } \omega_{j\mu } } }} \\ & = \frac{{2\tau \sqrt {\omega_{i\mu } \omega_{j\nu } } }}{{2\tau \sqrt {\omega_{i\nu } \omega_{j\mu } } }} = \frac{{I(f_{ij\mu } )}}{{I(f_{ij\nu } )}} = \varPhi (f_{ij} ). \\ \end{aligned} $$

(37)

By Theorem 1, it is certified that IFLPR $F = (f_{ij} )_{n \times n}$ is multiplicative consistent, which competes the proof of Theorem 3. □

Proof of Theorem 4

Since $\tilde{d}_{ij}^{ - } ,\tilde{d}_{ij}^{ + } ,\tilde{e}_{ij}^{ - } ,\tilde{e}_{ij}^{ + } ,i,j \in N$ are the optimal deviation values and $\tilde{\omega } = (\tilde{\omega }_{1} ,\tilde{\omega }_{2} , \ldots ,\tilde{\omega }_{n} )^{\rm T}$ is the optimal normalized intuitionistic fuzzy weight vector in Model 2, then for any $i,j \in N$, we have

$$ \begin{array}{*{20}c} {\tilde{d}_{ij}^{ - } - \tilde{d}_{ij}^{ + } = 0.5(\ln \tilde{\omega }_{i\mu } + \ln \tilde{\omega }_{j\nu } ) + \ln 2\tau - \ln I(b_{ij\mu } ),\;} \\ {\tilde{e}_{ij}^{ - } - \tilde{e}_{ij}^{ + } = 0.5(\ln \tilde{\omega }_{i\nu } + \ln \tilde{\omega }_{j\mu } ) + \ln 2\tau - \ln I(b_{ij\nu } ),} \\ \end{array} $$

that is

$$ \begin{aligned} \ln I(b_{ij\mu } ) + \tilde{d}_{ij}^{ - } - \tilde{d}_{ij}^{ + } = 0.5(\ln \tilde{\omega }_{i\mu } + \ln \tilde{\omega }_{j\nu } ) + \ln 2\tau , \hfill \\ \ln I(b_{ij\nu } ) + \tilde{e}_{ij}^{ - } - \tilde{e}_{ij}^{ + } = 0.5(\ln \tilde{\omega }_{i\nu } + \ln \tilde{\omega }_{j\mu } ) + \ln 2\tau , \hfill \\ \end{aligned} $$

it follows that

$$ I(b_{ij\mu } ) \cdot \exp (\tilde{d}_{ij}^{ - } - \tilde{d}_{ij}^{ + } ) = 2\tau \sqrt {\tilde{\omega }_{i\mu } \tilde{\omega }_{j\nu } } ,\quad I(b_{ij\nu } ) \cdot \exp (\tilde{e}_{ij}^{ - } - \tilde{e}_{ij}^{ + } ) = 2\tau \sqrt {\tilde{\omega }_{i\nu } \tilde{\omega }_{j\mu } } ,\quad i \ne j, $$

i.e.,

$$ \begin{aligned} I^{ - 1} \left( {I(b_{ij\mu } ) \cdot \exp (\tilde{d}_{ij}^{ - } - \tilde{d}_{ij}^{ + } )} \right) = I^{ - 1} \left( {2\tau \sqrt {\tilde{\omega }_{i\mu } \tilde{\omega }_{j\nu } } } \right),\quad i \ne j, \hfill \\ I^{ - 1} \left( {I(b_{ij\nu } ) \cdot \exp (\tilde{e}_{ij}^{ - } - \tilde{e}_{ij}^{ + } )} \right) = I^{ - 1} \left( {2\tau \sqrt {\tilde{\omega }_{i\nu } \tilde{\omega }_{j\mu } } } \right),\quad i \ne j, \hfill \\ \end{aligned} $$

thus, from Eq. (17), one can get that

$$ \tilde{b}_{ij\mu } { = }I^{ - 1} \left( {2\tau \sqrt {\tilde{\omega }_{i\mu } \tilde{\omega }_{j\nu } } } \right),\quad \tilde{b}_{ij\nu } { = }I^{ - 1} \left( {2\tau \sqrt {\tilde{\omega }_{i\nu } \tilde{\omega }_{j\mu } } } \right),\quad i \ne j. $$

According to Corollary 1, $\tilde{B} = (\tilde{b}_{ij} )_{n \times n}$ is a multiplicative consistent IFLPR, which completes the proof of Theorem 4. □

Proof of Theorem 6

(i) and (iii) are obvious. In the following, we prove that (ii) and (iv) are valid.

(ii) Since $a_{ij\mu } = a_{ji\nu } ,a_{ij\nu } = a_{ji\mu } ,b_{ij\mu } = b_{ji\nu } ,b_{ij\nu } = b_{ji\mu } ,i,j \in N$, then

$$ \begin{aligned} d(A,B) = 0 & \Leftrightarrow \left| {\ln I(a_{ij\mu } ) - \ln I(b_{ij\mu } )} \right| + \left| {\ln I(a_{ij\nu } ) - \ln I(b_{ij\nu } )} \right| = 0,\quad i < j, \\ & \Leftrightarrow I(a_{ij\mu } ) = I(b_{ij\mu } ),I(a_{ij\nu } ) = I(b_{ij\nu } ),\quad i < j \\ & \Leftrightarrow I(a_{ij\mu } ) = I(b_{ij\mu } ),I(a_{ij\nu } ) = I(b_{ij\nu } ),\quad i,j \in N \\ & \Leftrightarrow a_{ij\mu } = b_{ij\mu } ,a_{ij\nu } = b_{ij\nu } ,\quad i,j \in N \\ & \Leftrightarrow A = B. \\ \end{aligned} $$

(iv) According to Eq. (18), we have

$$ \begin{aligned} d(A,B) & = \frac{1}{n(n - 1)\ln 2\tau }\sum\limits_{i < j} {\left( {\left| {\ln I(a_{ij\mu } ) - \ln I(b_{ij\mu } )} \right| + \left| {\ln I(a_{ij\nu } ) - \ln I(b_{ij\nu } )} \right|} \right)} \\ & = \frac{1}{n(n - 1)\ln 2\tau }\sum\limits_{i < j} {\left( {\left| {\left( {\ln I(a_{ij\mu } ) - \ln I(c_{ij\mu } )} \right) + \left( {\ln I(c_{ij\mu } ) - \ln I(b_{ij\mu } )} \right)} \right|} \right.} \\ & \left. {\quad + \left| {\left( {\ln I(a_{ij\nu } ) - \ln I(c_{ij\nu } )} \right) + \left( {\ln I(c_{ij\nu } ) - \ln I(b_{ij\nu } )} \right)} \right|} \right) \\ & \le \frac{1}{n(n - 1)\ln 2\tau }\sum\limits_{i < j} {\left( {\left| {\ln I(a_{ij\mu } ) - \ln I(c_{ij\mu } )} \right| + \left| {\ln I(c_{ij\mu } ) - \ln I(b_{ij\mu } )} \right|} \right.} \\ & \left. {\quad + \left| {\ln I(a_{ij\nu } ) - \ln I(c_{ij\nu } )} \right| + \left| {\ln I(c_{ij\nu } ) - \ln I(b_{ij\nu } )} \right|} \right) \\ & = \frac{1}{n(n - 1)\ln 2\tau }\sum\limits_{i < j} {\left( {\left| {\ln I(a_{ij\mu } ) - \ln I(c_{ij\mu } )} \right| + \left| {\ln I(c_{ij\mu } ) - \ln I(b_{ij\mu } )} \right|} \right)} \\ & \quad + \frac{1}{n(n - 1)\ln 2\tau }\sum\limits_{i < j} {\left( {\left| {\ln I(a_{ij\nu } ) - \ln I(c_{ij\nu } )} \right| + \left| {\ln I(c_{ij\nu } ) - \ln I(b_{ij\nu } )} \right|} \right)} \\ & = d(A,C) + d(C,B). \\ \end{aligned} $$

This completes the proof of Theorem 6. □

Proof of Theorem 7

Let $d_{ij}^{(t) - } ,d_{ij}^{(t) + } ,e_{ij}^{(t) - }$ and $e_{ij}^{(t) + } ,i,j \in N$ be the deviation variables and $\omega = (\omega_{1}^{(t)} ,\omega_{2}^{(t)} , \ldots ,\omega_{n}^{(t)} )^{\rm T}$ be the normalized intuitionistic fuzzy weight vector in Model 2 for $B^{(t)}$, then for each $t$, we have

$$ d_{ij}^{(t) - } - d_{ij}^{(t) + } = 0.5(\ln \omega_{i\mu }^{(t)} + \ln \omega_{j\nu }^{(t)} ) + \ln 2\tau - \ln I(b_{ij\mu }^{(t)} ),\quad i < j, $$

(38)

$$ e_{ij}^{(t) - } - e_{ij}^{(t) + } = 0.5(\ln \omega_{i\nu }^{(t)} + \ln \omega_{j\mu }^{(t)} ) + \ln 2\tau - \ln I(b_{ij\nu }^{(t)} ),\quad i < j, $$

(39)

Since $d_{ij}^{(t) - } \ge 0,d_{ij}^{(t) + } \ge 0,e_{ij}^{(t) - } \ge 0,e_{ij}^{(t) + } \ge 0$ and $d_{ij}^{(t) - } \cdot d_{ij}^{(t) + } = 0,e_{ij}^{(t) - } \cdot e_{ij}^{(t) + } = 0,i,j \in N$, for each $t$, then

$$ d_{ij}^{(t) - } + d_{ij}^{(t) + } = \left| {d_{ij}^{(t) + } - d_{ij}^{(t) - } } \right|,e_{ij}^{(t) - } + e_{ij}^{(t) + } = \left| {e_{ij}^{(t) + } - e_{ij}^{(t) - } } \right|,\quad i,j \in N. $$

(40)

Assume that $\tilde{\omega }^{(t)} = (\tilde{\omega }_{1}^{(t)} ,\tilde{\omega }_{2}^{(t)} , \ldots ,\tilde{\omega }_{n}^{(t)} )^{\rm T}$ is the optimal normalized intuitionistic fuzzy weight vector, $\tilde{d}_{ij}^{(t) - } ,\tilde{d}_{ij}^{(t) + } ,\tilde{e}_{ij}^{(t) - } ,\tilde{e}_{ij}^{(t) + } ,i,j \in N$ are the optimal deviation values in Model 2 $B^{(t)}$. Since $\tilde{B}^{(t)}$ is a multiplicative consistent IFLPR, based on Theorem 4, we have

$$ I(\tilde{b}_{ij\mu }^{(t)} ) = 2\tau \sqrt {\tilde{\omega }_{i\mu }^{(t)} \tilde{\omega }_{j\nu }^{(t)} } ,\;I(\tilde{b}_{ij\nu }^{(t)} ) = 2\tau \sqrt {\tilde{\omega }_{i\nu }^{(t)} \tilde{\omega }_{j\mu }^{(t)} } ,\quad i \ne j,\quad {\text{for}}\,{\text{each}}\,t, $$

i.e.,

$$ \ln I(\tilde{b}_{ij\mu }^{(t)} ) = 0.5(\ln \tilde{\omega }_{i\mu }^{(t)} + \ln \tilde{\omega }_{j\nu }^{(t)} ) + \ln 2\tau ,\quad \ln I(\tilde{b}_{ij\nu }^{(t)} ) = 0.5(\ln \tilde{\omega }_{i\nu }^{(t)} + \ln \tilde{\omega }_{j\mu }^{(t)} ) + \ln 2\tau ,\quad i \ne j. $$

(41)

According to Eq. (20), for each $t$, we have

$$ \ln I(\tilde{b}_{ij\mu }^{(t)} ) - \ln I(b_{ij\mu }^{(t)} ) = \tilde{d}_{ij}^{(t) - } - \tilde{d}_{ij}^{(t) + } ,\quad \ln I(\tilde{b}_{ij\nu }^{(t)} ) - \ln I(b_{ij\nu }^{(t)} ) = \tilde{e}_{ij}^{(t) - } - \tilde{e}_{ij}^{(t) + } ,\quad i \ne j. $$

(42)

As $\tilde{d}_{ij}^{(t + 1) - } ,\tilde{d}_{ij}^{(t + 1) + } ,\tilde{e}_{ij}^{(t + 1) - } ,\tilde{e}_{ij}^{(t + 1) + } ,i,j \in N$ are the smallest deviation variables respect to IFLPR $B^{(t + 1)}$ in Model 2, by using Eqs. (38)–(40), one can obtain that

$$ \begin{aligned} & \sum\limits_{i < j} {\left( {\tilde{d}_{ij}^{(t + 1) - } + \tilde{d}_{ij}^{(t + 1) + } + \tilde{e}_{ij}^{(t + 1) - } + \tilde{e}_{ij}^{(t + 1) + } } \right)} \\ & \quad \le \sum\limits_{i < j} {\left( {d_{ij}^{(t + 1) - } + d_{ij}^{(t + 1) + } + e_{ij}^{(t + 1) - } + e_{ij}^{(t + 1) + } } \right)} \\ & \quad = \sum\limits_{i < j} {\left( {\left| {d_{ij}^{(t + 1) - } - d_{ij}^{(t + 1) + } } \right| + \left| {e_{ij}^{(t + 1) - } - e_{ij}^{(t + 1) + } } \right|} \right)} \\ & \quad = \sum\limits_{i < j} {\left( {\left| {0.5(\ln \omega_{i\mu }^{(t + 1)} + \ln \omega_{j\nu }^{(t + 1)} ) + \ln 2\tau - \ln I(b_{ij\mu }^{(t + 1)} )} \right|} \right.} \\ & \quad \quad \left. { + \left| {0.5(\ln \omega_{i\nu }^{(t + 1)} + \ln \omega_{j\mu }^{(t + 1)} ) + \ln 2\tau - \ln I(b_{ij\nu }^{(t + 1)} )} \right|} \right). \\ \end{aligned} $$

Let $\omega_{i}^{(t + 1)} = \tilde{\omega }_{i}^{(t)} ,i \in N$, it follows that

$$ \begin{aligned} & \sum\limits_{i < j} {\left( {\tilde{d}_{ij}^{(t + 1) - } + \tilde{d}_{ij}^{(t + 1) + } + \tilde{e}_{ij}^{(t + 1) - } + \tilde{e}_{ij}^{(t + 1) + } } \right)} \\ & \quad \le \sum\limits_{i < j} {\left( {\left| {0.5(\ln \tilde{\omega }_{i\mu }^{(t)} + \ln \tilde{\omega }_{j\nu }^{(t)} ) + \ln 2\tau - \ln I(b_{ij\mu }^{(t + 1)} )} \right|} \right.} \\ & \quad \quad \left. { + \left| {0.5(\ln \tilde{\omega }_{i\nu }^{(t)} + \ln \tilde{\omega }_{j\mu }^{(t)} ) + \ln 2\tau - \ln I(b_{ij\nu }^{(t + 1)} )} \right|} \right) \\ & \quad = \sum\limits_{i < j} {\left( {\left| {\ln I(\tilde{b}_{ij\mu }^{(t)} ) - \ln I(b_{ij\mu }^{(t + 1)} )} \right| + \left| {\ln I(\tilde{b}_{ij\nu }^{(t)} ) - \ln I(b_{ij\nu }^{(t + 1)} )} \right|} \right)} . \\ \end{aligned} $$

(43)

In addition, from Eq. (22), we have

$$ \ln I(b_{ij\mu }^{(t + 1)} ) = (1 - \theta )\ln I(b_{ij\mu }^{(t)} ) + \theta \ln I(\tilde{b}_{ij\mu }^{(t)} ),\quad \ln I(b_{ij\nu }^{(t + 1)} ) = (1 - \theta )\ln I(b_{ij\nu }^{(t)} ) + \theta \ln I(\tilde{b}_{ij\nu }^{(t)} ),\quad i \ne j. $$

(44)

Therefore, by using Eqs. (40)–(44), for each t, we can get

$$ \begin{aligned} {\text{CI}}(B^{(t + 1)} ) & = \frac{1}{\tau n(n - 1)\ln 2}\sum\limits_{i < j} {\left( {\left| {\ln I(\tilde{b}_{ij\mu }^{(t + 1)} ) - \ln I(b_{ij\mu }^{(t + 1)} )} \right| + \left| {\ln I(\tilde{b}_{ij\nu }^{(t + 1)} ) - \ln I(b_{ij\nu }^{(t + 1)} )} \right|} \right)} \\ & = \frac{1}{\tau n(n - 1)\ln 2}\sum\limits_{i < j} {\left( {\left| {\tilde{d}_{ij}^{(t + 1) - } - \tilde{d}_{ij}^{(t + 1) + } } \right| + \left| {\tilde{e}_{ij}^{(t + 1) - } - \tilde{e}_{ij}^{(t + 1) + } } \right|} \right)} \\ & = \frac{1}{\tau n(n - 1)\ln 2}\sum\limits_{i < j} {\left( {\tilde{d}_{ij}^{(t + 1) - } + \tilde{d}_{ij}^{(t + 1) + } + \tilde{e}_{ij}^{(t + 1) - } + \tilde{e}_{ij}^{(t + 1) + } } \right)} \\ & \le \frac{1}{\tau n(n - 1)\ln 2}\sum\limits_{i < j} {\left( {\left| {\ln I(\tilde{b}_{ij\mu }^{(t)} - \ln I(b_{ij\mu }^{(t + 1)} )} \right| + \left| {\ln I(\tilde{b}_{ij\nu }^{(t)} ) - \ln I(b_{ij\nu }^{(t + 1)} )} \right|} \right)} \\ & = \frac{1}{\tau n(n - 1)\ln 2}\sum\limits_{i < j} {\left( {\left| {\ln I(\tilde{b}_{ij\mu }^{(t)} - (1 - \theta )\ln I(b_{ij\mu }^{(t)} ) - \theta \ln I(\tilde{b}_{ij\mu }^{(t)} )} \right|} \right.} \\ & \quad + \left. {\left| {\ln I(\tilde{b}_{ij\nu }^{(t)} ) - (1 - \theta )\ln I(b_{ij\nu }^{(t)} ) - \theta \ln I(\tilde{b}_{ij\nu }^{(t)} )} \right|} \right) \\ & = (1 - \theta ) \cdot \frac{1}{\tau n(n - 1)\ln 2}\sum\limits_{i < j} {\left( {\left| {\ln I(\tilde{b}_{ij\mu }^{(t)} - \ln I(b_{ij\mu }^{(t)} )} \right| + \left| {\ln I(\tilde{b}_{ij\nu }^{(t)} ) - \ln I(b_{ij\nu }^{(t)} )} \right|} \right)} \\ & = (1 - \theta ) \cdot {\text{CI}}(B^{(t)} ) < {\text{CI}}(B^{(t)} ). \\ \end{aligned} $$

(45)

Furthermore, by Eq. (45), we have

$$ \mathop {\lim }\limits_{t \to + \infty } {\text{CI}}(B^{(t)} ) \le \mathop {\lim }\limits_{t \to + \infty } (1 - \theta ){\text{CI}}(B^{(t - 1)} ) \le \mathop {\lim }\limits_{t \to + \infty } (1 - \theta )^{2} {\text{CI}}(B^{(t - 2)} ) \le \ldots \le \mathop {\lim }\limits_{t \to + \infty } (1 - \theta )^{t} {\text{CI}}(B^{(0)} ) = 0. $$

Since ${\text{CI}}(B^{(t)} ) \ge 0$ for each $t$, then $\mathop {\lim }\limits_{t \to + \infty } {\text{CI}}(B^{(t)} ) = 0$.

This statement completes the proof of Theorem 7. □

Proof of Theorem 8

Since $B_{k} = \left( {b_{ij,k} } \right)_{n \times n} (k \in M)$ be a collection of IFLPRs, according to Definition 1, for any $i,j \in N,k \in M$, we have $b_{ij\mu ,k} = b_{ji\nu ,k} ,b_{ij\nu ,k} = b_{ji\mu ,k} ,b_{ij\mu ,k} + b_{ij\nu ,k} \le s_{2\tau }$, and then

$$ b_{ij\mu ,c} = I^{ - 1} \left( {\prod\limits_{k = 1}^{m} {\left( {I(b_{ij\mu ,k} )} \right)^{{\lambda_{k} }} } } \right) = I^{ - 1} \left( {\prod\limits_{k = 1}^{m} {\left( {I(b_{ji\nu ,k} )} \right)^{{\lambda_{k} }} } } \right) = b_{ji\nu ,c} . $$

From Eq. (26), we have

$$ I(b_{ij\mu ,c} ) = \prod\limits_{k = 1}^{m} {\left( {I(b_{ij\mu ,k} )} \right)^{{\lambda_{k} }} } ,\quad I(b_{ij\nu ,c} ) = \prod\limits_{k = 1}^{m} {\left( {I(b_{ij\nu ,k} )} \right)^{{\lambda_{k} }} } ,\quad i,j \in N. $$

(46)

As $I(b_{ij\mu ,k} ),I(b_{ij\nu ,k} ) \in [0,2\tau ],i,j \in N,k \in M$, using the Lemma 1, we have

$$ \begin{aligned} 0 \le I(b_{ij\mu ,c} ) & = \prod\limits_{k = 1}^{m} {\left( {I(b_{ij\mu ,k} )} \right)^{{\lambda_{k} }} } \le \sum\limits_{k = 1}^{m} {\lambda_{k} I(b_{ij\mu ,k} )} \le 2\tau \cdot \sum\limits_{k = 1}^{m} {\lambda_{k} } = 2\tau , \\ 0 \le I(b_{ij\nu ,c} ) & = \prod\limits_{k = 1}^{m} {\left( {I(b_{ij\nu ,k} )} \right)^{{\lambda_{k} }} } \le \sum\limits_{k = 1}^{m} {\lambda_{k} I(b_{ij\nu ,k} )} \le 2\tau \cdot \sum\limits_{k = 1}^{m} {\lambda_{k} } = 2\tau , \\ \end{aligned} $$

it follows that

$$ s_{0} \le b_{ij\mu ,c} \le s_{2\tau } \,{\text{and}}\,s_{0} \le b_{ij\nu ,c} \le s_{2\tau } ,\quad i,j \in N. $$

(47)

Furthermore, since $I(b_{ij\mu ,k} ) + I(b_{ij\nu ,k} ) \in [0,2\tau ],i,j \in N,k \in M$, by Lemma 1, we have

$$ \begin{aligned} I(b_{ij\mu ,c} ) + I(b_{ij\nu ,c} ) & = \prod\limits_{k = 1}^{m} {\left( {I(b_{ij\mu ,k} )} \right)^{{\lambda_{k} }} } + \prod\limits_{k = 1}^{m} {\left( {I(b_{ij\nu ,k} )} \right)^{{\lambda_{k} }} } \\ & \le \sum\limits_{k = 1}^{m} {\lambda_{k} I(b_{ij\mu ,k} )} + \sum\limits_{k = 1}^{m} {\lambda_{k} I(b_{ij\nu ,k} )} \\ & = \sum\limits_{k = 1}^{m} {\lambda_{k} \left( {I(b_{ij\mu ,k} ) + I(b_{ij\nu ,k} )} \right)} \\ & = \sum\limits_{k = 1}^{m} {(\lambda_{k} \cdot 2\tau )} = 2\tau , \\ \end{aligned} $$

i.e.,

$$ b_{ij\mu ,c} + b_{ij\nu ,c} \le s_{2\tau } ,\quad i,j \in N. $$

(48)

According to Definition 1, it is certified that $B_{c} = \left( {b_{ij,c} } \right)_{n \times n}$ is an IFLPR, which competes the proof of Theorem 8. □

Proof of Theorem 9

Suppose that $\tilde{d}_{ij,k}^{ - } ,\tilde{d}_{ij,k}^{ + } ,\tilde{e}_{ij,k}^{ - } ,\tilde{e}_{ij,k}^{ + } ,i,j \in N,k \in M$ and $\tilde{d}_{ij,c}^{ - } ,\tilde{d}_{ij,c}^{ + } ,\tilde{e}_{ij,c}^{ - } ,$$\tilde{e}_{ij,c}^{ + } ,i,j \in N$ be the optimal deviation values of Model 3 and Model 4, respectively, and then

$$ \tilde{d}_{ij,c}^{ - } = \sum\limits_{k = 1}^{m} {\lambda_{k} \tilde{d}_{ij,k}^{ - } } ,\tilde{d}_{ij,c}^{ + } = \sum\limits_{k = 1}^{m} {\lambda_{k} \tilde{d}_{ij,k}^{ + } } ,\tilde{e}_{ij,c}^{ - } = \sum\limits_{k = 1}^{m} {\lambda_{k} \tilde{e}_{ij,k}^{ - } } \,{\text{and}}\,\tilde{e}_{ij,c}^{ + } = \sum\limits_{k = 1}^{m} {\lambda_{k} \tilde{e}_{ij,k}^{ + } } . $$

(49)

Let $\tilde{B}_{c} = (\tilde{b}_{ij,c} )_{n \times n} = (\langle \tilde{b}_{ij\mu ,c} ,\tilde{b}_{ij\nu ,c} \rangle )_{n \times n}$ and $\tilde{B}_{k} = (\tilde{b}_{ij,k} )_{n \times n} = (\langle \tilde{b}_{ij\mu ,k} ,\tilde{b}_{ij\nu ,k} \rangle )_{n \times n} ,k \in M$, where

$$ \tilde{b}_{ij\mu ,k} = I^{ - 1} \left( {I(b_{ij\mu ,k} ) \cdot \exp (\tilde{d}_{ij,k}^{ - } - \tilde{d}_{ij,k}^{ + } )} \right),\quad \tilde{b}_{ij\nu ,k} = I^{ - 1} \left( {I(b_{ij\nu ,k} ) \cdot \exp (\tilde{e}_{ij,k}^{ - } - \tilde{e}_{ij,k}^{ + } )} \right),\quad i,j \in N,k \in M, $$

(50)

$$ \tilde{b}_{ij\mu ,c} = I^{ - 1} \left( {I(b_{ij\mu ,c} ) \cdot \exp (\tilde{d}_{ij,c}^{ - } - \tilde{d}_{ij,c}^{ + } )} \right),\quad \tilde{b}_{ij\nu ,c} = I^{ - 1} \left( {I(b_{ij\nu ,c} ) \cdot \exp (\tilde{e}_{ij,c}^{ - } - \tilde{e}_{ij,c}^{ + } )} \right),\quad i,j \in N. $$

(51)

According to Theorem 4, one can obtain that $\tilde{B}_{k} ,k \in M$ and $\tilde{B}_{c}$ are multiplicative consistent IFLPRs.

From Eqs. (50) and (51), we have

$$ \begin{aligned} I(\tilde{b}_{ij\mu ,k} ) & = I(b_{ij\mu ,k} ) \cdot \exp (\tilde{d}_{ij,k}^{ - } - \tilde{d}_{ij,k}^{ + } ),\quad I(\tilde{b}_{ij\nu ,k} ) = I(b_{ij\nu ,k} ) \cdot \exp (\tilde{e}_{ij,k}^{ - } - \tilde{e}_{ij,k}^{ + } ),\quad i,j \in N,k \in M, \\ I(\tilde{b}_{ij\mu ,c} ) & = I(b_{ij\mu ,c} ) \cdot \exp (\tilde{d}_{ij,c}^{ - } - \tilde{d}_{ij,c}^{ + } ),\quad I(\tilde{b}_{ij\nu ,c} ) = I(b_{ij\nu ,c} ) \cdot \exp (\tilde{e}_{ij,c}^{ - } - \tilde{e}_{ij,c}^{ + } ),\quad i,j \in N, \\ \end{aligned} $$

it follows that

$$ \ln I(\tilde{b}_{ij\mu ,k} ) - \ln I(b_{ij\mu ,k} ) = \tilde{d}_{ij,k}^{ - } - \tilde{d}_{ij,k}^{ + } ,\quad \ln I(\tilde{b}_{ij\nu ,k} ) - \ln I(b_{ij\nu ,k} ) = \tilde{e}_{ij,k}^{ - } - \tilde{e}_{ij,k}^{ + } ,\quad i,j \in N,\quad k \in M, $$

(52)

$$ \ln I(\tilde{b}_{ij\mu ,c} ) - \ln I(b_{ij\mu ,c} ) = \tilde{d}_{ij,c}^{ - } - \tilde{d}_{ij,c}^{ + } ,\quad \ln I(\tilde{b}_{ij\nu ,c} ) - \ln I(b_{ij\nu ,c} ) = \tilde{e}_{ij,c}^{ - } - \tilde{e}_{ij,c}^{ + } ,\quad i,j \in N. $$

(53)

Since ${\text{CI}}(B_{k} ) \le \overline{\text{CI}} ,k \in M$, by Eqs. (49), (52) and (53), we have

$$ \begin{aligned} {\text{CI}}(B_{c} ) & = \frac{1}{n(n - 1)\ln 2\tau }\sum\limits_{i < j} {\left( {\left| {\ln I(\tilde{b}_{ij\mu ,c} ) - \ln I(b_{ij\mu ,c} )} \right| + \left| {\ln I(\tilde{b}_{ij\nu ,c} ) - \ln I(b_{ij\nu ,c} )} \right|} \right)} \\ & = \frac{1}{n(n - 1)\ln 2\tau }\sum\limits_{i < j} {\left( {\left| {\tilde{d}_{ij,c}^{ - } - \tilde{d}_{ij,c}^{ + } } \right| + \left| {\tilde{e}_{ij,c}^{ - } - \tilde{e}_{ij,c}^{ + } } \right|} \right)} \\ & = \frac{1}{n(n - 1)\ln 2\tau }\sum\limits_{i < j} {\left( {\left| {\sum\limits_{k = 1}^{m} {\lambda_{k} \tilde{d}_{ij,k}^{ - } } - \sum\limits_{k = 1}^{m} {\lambda_{k} \tilde{d}_{ij,k}^{ + } } } \right| + \left| {\sum\limits_{k = 1}^{m} {\lambda_{k} \tilde{e}_{ij,k}^{ - } } - \sum\limits_{k = 1}^{m} {\lambda_{k} \tilde{e}_{ij,k}^{ + } } } \right|} \right)} \\ & = \frac{1}{n(n - 1)\ln 2\tau }\sum\limits_{i < j} {\left( {\left| {\sum\limits_{k = 1}^{m} {\lambda_{k} (\tilde{d}_{ij,k}^{ - } - \tilde{d}_{ij,k}^{ + } )} } \right| + \left| {\sum\limits_{k = 1}^{m} {\lambda_{k} (\tilde{e}_{ij,k}^{ - } - \tilde{e}_{ij,k}^{ + } )} } \right|} \right)} \\ & \le \sum\limits_{k = 1}^{m} {\lambda_{k} \left( {\frac{1}{n(n - 1)\ln 2\tau }\sum\limits_{i < j} {\left( {\left| {\tilde{d}_{ij,k}^{ - } - \tilde{d}_{ij,k}^{ + } } \right| + \left| {\tilde{e}_{ij,k}^{ - } - \tilde{e}_{ij,k}^{ + } } \right|} \right)} } \right)} \\ & = \sum\limits_{k = 1}^{m} {\lambda_{k} \left( {\frac{1}{n(n - 1)\ln 2\tau }\sum\limits_{i < j} {\left( {\left| {\ln I(\tilde{b}_{ij\mu ,k} ) - \ln I(b_{ij\mu ,k} )} \right| + \left| {\ln I(\tilde{b}_{ij\nu ,k} ) - \ln I(b_{ij\nu ,k} )} \right|} \right)} } \right)} \\ & = \sum\limits_{k = 1}^{m} {\lambda_{k} \cdot {\text{CI}}(B_{k} )} \\ & \le \sum\limits_{k = 1}^{m} {\lambda_{k} \cdot \mathop {\hbox{max} }\limits_{1 \le k \le m} \{ {\text{CI}}(B_{k} )\} } = \mathop {\hbox{max} }\limits_{1 \le k \le m} \{ {\text{CI}}(B_{k} )\} . \\ \end{aligned} $$

(54)

Therefore, we complete the proof of Theorem 9. □

Proof of Theorem 10

According to Eq. (29), we have

$$ I(b_{{ij\mu ,\delta_{1} }} ) = \left( {I(b_{ij\mu ,\psi }^{(t)} )} \right)^{{\delta_{1} }} \left( {\prod\limits_{k \ne \psi } {I(b_{ij\mu ,k}^{(t)} )} } \right)^{{\frac{{1 - \delta_{1} }}{m - 1}}} ,\quad I(b_{{ij\nu ,\delta_{1} }} ) = \left( {I(b_{ij\nu ,\psi }^{(t)} )} \right)^{{\delta_{1} }} \left( {\prod\limits_{k \ne \psi } {I(b_{ij\nu ,k}^{(t)} )} } \right)^{{\frac{{1 - \delta_{1} }}{m - 1}}} , $$

and

$$ I(b_{{ij\mu ,\delta_{2} }} ) = \left( {I(b_{ij\mu ,\psi }^{(t)} )} \right)^{{\delta_{2} }} \left( {\prod\limits_{k \ne \psi } {I(b_{ij\mu ,k}^{(t)} )} } \right)^{{\frac{{1 - \delta_{2} }}{m - 1}}} ,\quad I(b_{{ij\nu ,\delta_{2} }} ) = \left( {I(b_{ij\nu ,\psi }^{(t)} )} \right)^{{\delta_{2} }} \left( {\prod\limits_{k \ne \psi } {I(b_{ij\nu ,k}^{(t)} )} } \right)^{{\frac{{1 - \delta_{2} }}{m - 1}}} , $$

then, by Definition 6, we have

$$ \begin{aligned} d(B_{{\delta_{1} }} ,B_{\psi }^{(t)} ) & = \frac{1}{n(n - 1)\ln 2\tau }\sum\limits_{i < j} {\left( {\left| {\ln I(b_{{ij\mu ,\delta_{1} }} ) - \ln I(b_{ij\mu ,\psi }^{(t)} )} \right| + \left| {\ln I(b_{{ij\nu ,\delta_{1} }} ) - \ln I(b_{ij\nu ,\psi }^{(t)} )} \right|} \right)} \\ & = \frac{1}{n(n - 1)\ln 2\tau }\sum\limits_{i < j} {\left( {\left| {\delta_{1} \ln I(b_{ij\mu ,\psi } ) + \frac{{1 - \delta_{1} }}{m - 1}\sum\limits_{k \ne \psi } {I(b_{ij\mu ,k}^{(t)} )} - \ln I(b_{ij\mu ,\psi }^{(t)} )} \right|} \right.} \\ & \left. {\quad + \left| {\delta_{1} \ln I(b_{ij\nu ,\psi } ) + \frac{{1 - \delta_{1} }}{m - 1}\sum\limits_{k \ne \psi } {I(b_{ij\nu ,k}^{(t)} )} - \ln I(b_{ij\nu ,\psi }^{(t)} )} \right|} \right) \\ & = \frac{{1 - \delta_{1} }}{n(n - 1)\ln 2\tau }\sum\limits_{i < j} {\left( {\left| {\frac{1}{m - 1}\sum\limits_{k \ne \psi } {I(b_{ij\mu ,k}^{(t)} )} - \ln I(b_{ij\mu ,\psi }^{(t)} )} \right| + \left| {\frac{1}{m - 1}\sum\limits_{k \ne \psi } {I(b_{ij\nu ,k}^{(t)} )} - \ln I(b_{ij\nu ,\psi }^{(t)} )} \right|} \right)} . \\ \end{aligned} $$

Similarly, we have

$$ d(B_{{\delta_{2} }} ,B_{\psi }^{(t)} ) = \frac{{1 - \delta_{2} }}{n(n - 1)\ln 2\tau }\sum\limits_{i < j} {\left( {\left| {\frac{1}{m - 1}\sum\limits_{k \ne \psi } {I(b_{ij\mu ,k}^{(t)} )} - \ln I(b_{ij\mu ,\psi }^{(t)} )} \right| + \left| {\frac{1}{m - 1}\sum\limits_{k \ne \psi } {I(b_{ij\nu ,k}^{(t)} )} - \ln I(b_{ij\nu ,\psi }^{(t)} )} \right|} \right)} . $$

Since $\delta_{1} \ge \delta_{2}$, then $d(B_{{\delta_{1} }} ,B_{\psi }^{(t)} ) \le d(B_{{\delta_{2} }} ,B_{\psi }^{(t)} )$, which completes the proof of Theorem 10. □

Proof of Theorem 11

According to Eq. (51), we have $B_{\psi }^{(t + 1)} = \left( {b_{ij,\psi }^{(t + 1)} } \right)_{n \times n} = \left( {\langle b_{ij\mu ,\psi }^{(t + 1)} ,} \right.\left. {b_{ij\nu ,\psi }^{(t + 1)} \rangle } \right)_{n \times n},$ where

$$ I(b_{ij\mu ,\psi }^{(t + 1)} ) = \left( {I(b_{ij\mu ,\psi }^{(t)} )} \right)^{\delta } \left( {\prod\limits_{k \ne \psi } {I(b_{ij\mu ,k}^{(t)} )} } \right)^{{\frac{1 - \delta }{m - 1}}} ,\quad I(b_{ij\nu ,\psi }^{(t + 1)} ) = \left( {I(b_{ij\nu ,\psi }^{(t)} )} \right)^{\delta } \left( {\prod\limits_{k \ne \psi } {I(b_{ij\nu ,k}^{(t)} )} } \right)^{{\frac{1 - \delta }{m - 1}}} . $$

(55)

For all $l \ne \psi$, we have $B_{l}^{(t + 1)} = B_{l}^{(t)}$, and ${\text{ICD}}(B_{l}^{(t)} ) < {\text{ICD}}(B_{\psi }^{(t)} ),l \ne \psi$.

From Definition 6 and Eq. (55), for all $l \ne \psi$, one can obtain that

$$ \begin{aligned} & d(B_{\psi }^{(t + 1)} ,B_{l}^{(t + 1)} ) \\ & \quad = \frac{1}{n(n - 1)\ln 2\tau }\sum\limits_{i < j} {\left( {\left| {\ln I(b_{ij\mu ,\psi }^{(t + 1)} ) - \ln I(b_{ij\mu ,l}^{(t + 1)} )} \right| + \left| {\ln I(b_{ij\nu ,\psi }^{(t + 1)} ) - \ln I(b_{ij\nu ,l}^{(t + 1)} )} \right|} \right)} \\ & \quad = \frac{1}{n(n - 1)\ln 2\tau }\sum\limits_{i < j} {\left( {\left| {\ln I(b_{ij\mu ,\psi }^{(t + 1)} ) - \ln I(b_{ij\mu ,l}^{(t)} )} \right| + \left| {\ln I(b_{ij\nu ,\psi }^{(t + 1)} ) - \ln I(b_{ij\nu ,l}^{(t)} )} \right|} \right)} \\ & \quad = \frac{1}{n(n - 1)\ln 2\tau }\sum\limits_{i < j} {\left( {\left| {\delta \ln I(b_{ij\mu ,\psi }^{(t)} ) + \frac{1 - \delta }{m - 1}\sum\limits_{k \ne \psi } {I(b_{ij\mu ,k}^{(t)} )} - \delta \ln I(b_{ij\mu ,l}^{(t)} ) - (1 - \delta )\ln I(b_{ij\mu ,l}^{(t)} )} \right|} \right.} \\ & \left. {\quad \quad + \left| {\delta \ln I(b_{ij\nu ,\psi }^{(t)} ) + \frac{1 - \delta }{m - 1}\sum\limits_{k \ne \psi } {I(b_{ij\nu ,k}^{(t)} )} - \delta \ln I(b_{ij\nu ,l}^{(t)} ) - (1 - \delta )\ln I(b_{ij\nu ,l}^{(t)} )} \right|} \right) \\ & \quad \le \frac{\delta }{n(n - 1)\ln 2\tau }\sum\limits_{i < j} {\left( {\left| {\ln I(b_{ij\mu ,\psi }^{(t)} ) - \ln I(b_{ij\mu ,l}^{(t)} )} \right| + \left| {\ln I(b_{ij\nu ,\psi }^{(t)} ) - \ln I(b_{ij\nu ,l}^{(t)} )} \right|} \right)} \\ & \quad \quad + \frac{1 - \delta }{m - 1}\sum\limits_{k \ne \psi } {\frac{1}{n(n - 1)\ln 2\tau }\sum\limits_{i < j} {\left( {\left| {\ln I(b_{ij\mu ,k}^{(t)} ) - \ln I(b_{ij\mu ,l}^{(t)} )} \right| + \left| {\ln I(b_{ij\nu ,k}^{(t)} ) - \ln I(b_{ij\nu ,l}^{(t)} )} \right|} \right)} } \\ & \quad = \delta d(B_{\psi }^{(t)} ,B_{l}^{(t)} ) + \frac{1 - \delta }{m - 1}\sum\limits_{k \ne \psi } {d(B_{k}^{(t)} ,B_{l}^{(t)} )} \\ & \quad < \delta d(B_{\psi }^{(t)} ,B_{l}^{(t)} ) + \frac{1 - \delta }{m - 1}\sum\limits_{k \ne l} {d(B_{k}^{(t)} ,B_{l}^{(t)} )} \\ & \quad = \delta d(B_{\psi }^{(t)} ,B_{l}^{(t)} ) + (1 - \delta ){\text{ICD}}(B_{l}^{(t)} ), \\ \end{aligned} $$

it follows that

$$ \begin{aligned} {\text{ICD}}(B_{\psi }^{(t + 1)} ) & = \frac{1}{m - 1}\sum\limits_{l \ne \psi } {d(B_{\psi }^{(t + 1)} ,B_{l}^{(t + 1)} )} \\ & < \frac{1}{m - 1}\sum\limits_{l \ne \psi } {\left( {\delta d(B_{\psi }^{(t)} ,B_{l}^{(t)} ) + (1 - \delta ){\text{ICD}}(B_{l}^{(t)} )} \right)} \\ & = \delta \frac{1}{m - 1}\sum\limits_{l \ne \psi } {d(B_{\psi }^{(t)} ,B_{l}^{(t)} )} + (1 - \delta )\frac{1}{m - 1}\sum\limits_{l \ne k} {{\text{ICD}}(B_{l}^{(t)} )} \\ & < \delta {\text{ICD}}(B_{\psi }^{(t)} ) + (1 - \delta )\frac{1}{m - 1}\sum\limits_{l \ne k} {{\text{ICD}}(B_{\psi }^{(t)} )} \\ & = {\text{ICD}}(B_{\psi }^{(t)} ). \\ \end{aligned} $$

Therefore, we complete the proof of Theorem 11. □

Proof of Theorem 12

Suppose that $B_{\psi }^{(t)}$ is the IFLPR which needs to be adjusted in the tth iteration, then we have $B_{l}^{(t + 1)} = B_{l}^{(t)}$ for all $l \ne \psi$.

According to Theorem 11, we have ${\text{ICD}}(B_{\psi }^{(t + 1)} ) < {\text{ICD}}(B_{\psi }^{(t)} )$, that is

$$ {\text{ICD}}(B_{\psi }^{(t + 1)} ) - {\text{ICD}}(B_{\psi }^{(t)} ) < 0. $$

(56)

Therefore, by Definition 10 and Eq. (56), for each $t$, we have

$$ \begin{aligned} {\text{GCD}}^{(t + 1)} - {\text{GCD}}^{(t)} & = \frac{1}{m}\sum\limits_{k = 1}^{m} {\frac{1}{m - 1}\sum\limits_{l \ne k} {d(B_{k}^{(t + 1)} ,B_{l}^{(t + 1)} )} } - \frac{1}{m}\sum\limits_{k = 1}^{m} {\frac{1}{m - 1}\sum\limits_{l \ne k} {d(B_{k}^{(t)} ,B_{l}^{(t)} )} } \\ & = \frac{2}{m}\left( {\frac{1}{m - 1}\sum\limits_{l \ne \psi } {d(B_{\psi }^{(t + 1)} ,B_{l}^{(t + 1)} )} - \frac{1}{m - 1}\sum\limits_{l \ne \psi } {d(B_{\psi }^{(t)} ,B_{l}^{(t)} )} } \right) \\ & = \frac{2}{m}\left( {{\text{ICD}}(B_{\psi }^{(t + 1)} ) - {\text{ICD}}(B_{\psi }^{(t)} )} \right) < 0, \\ \end{aligned} $$

i.e.,

$$ {\text{GCD}}^{(t + 1)} < {\text{GCD}}^{(t)} \,{\text{for}}\,{\text{each}}\,t. $$

This completes the Proof of Theorem 12. □