Distributed linguistic GDM model in social network
Stage 1 Calculate the weight of each expert under SNA
Step 1: collect the trust relationship among experts to construct DLTRM \(TD{ = (}TD_{pq} )_{l \times l}\). Simultaneously, every expert expresses their evaluation information with decision matrices \(D^{p} = (d_{kj}^{p} )_{m \times n} (p = 1,2, \ldots ,l)\).
Step 2: construct the complete trust networks matrix
\(\overline{TD} { = (}\overline{TD}_{pq} )_{l \times l}\) by propagating and aggregating trust relationships in Eqs. (
13) and (
16).
Step 3: based on the complete trust networks matrix
\(\overline{TD} { = (}\overline{TD}_{pq} )_{l \times l}\), we determine the weights
\(w_{q} (q = 1,2, \ldots ,l)\) of experts
\(\left\{ {e_{1} ,e_{2} , \ldots ,e_{l} } \right\}\) by utilizing Eq. (
18).
Stage 2 Aggregation of DLTDMMs.
Step 4: based on the weights of experts \(w_{q} (q = 1,2, \ldots ,l)\), the DLTDMMs \(D^{p} = (d_{kj}^{p} )_{m \times n} (p = 1,2, \ldots ,l)\) is aggregated into a collective DLTDMM \(\overline{D} { = (}\overline{D}_{kj} )_{m \times n}\) with DLTWA operator.
Stage 3 Consensus test and adjustment
Step 5: calculate three-level consensus index at evaluation information levels, expert level and group level by using Eqs. (
21)–(
23). Then, if the value of
\({\text{GCD}}\) reaches the predefined threshold
\(\lambda\), turn to Step 7. Otherwise, turn to step 6.
Step 6: checking the most inconsistent evaluation information and applying the feedback mechanism to adjust the evaluation information with Eqs. (
24)–(
26), then we obtain the adjusted individual DLTDMM
\(D^{\prime {p}} = (d^{\prime {p}}_{kj})_{m \times n} (p = 1,2, \ldots ,l)\).
Stage 4 Determination of the ranking among
\(m\)
alternatives.
Step 7: Renew the collective DLTDMM. Based on the weight vector of experts
\(W = \left( {w_{1} ,w_{2} , \ldots ,w_{l} } \right)^{T}\) and the adjusted individual DLTDMM
\(D^{\prime {p}} = (d^{\prime {p}}_{kj} )_{m \times n} (p = 1,2, \ldots ,l)\), the collective adjusted DLTDMM
\(D^{\prime} = \left( {d^{\prime}_{kj} } \right)_{m \times n}\) is derived, where
$$ \begin{aligned}& d^{\prime}_{kj} = {\text{DLTWA}}(d^{\prime {1}}_{kj} ,d^{\prime {2}}_{kj} , \ldots ,d^{\prime {l}}_{kj} ) \\ &\quad = \left\{ {(s_{i} ,\sum\nolimits_{p = 1}^{l} {w_{p} T^{\prime {p(i)}}_{kj} } )|i = 0,1, \ldots ,2r} \right\} \\ &\quad \triangleq \left\{ {(s_{i} ,T^{\prime {(i)}}_{kj} )|i = 0,1, \ldots ,2r} \right\}. \end{aligned}$$
Step 8: Calculate the expectation
\(E(a_{k} )\) of each alternative
\(\left\{ {a_{1} ,a_{2} , \ldots ,a_{m} } \right\}\) as follows:
$$ E(a_{k} ) = \sum\limits_{j = 1}^{n} {\sum\limits_{i = 0}^{2r} {T^{\prime {(i)}}_{kj} \cdot I(s_{i} )} } , $$
and the complete ranking of the alternatives is determined in accordance with the decreasing
\(E(a_{k} )(k = 1,2, \ldots ,m)\).
Numerical experiment
In 2019, the Chinese government issued the “Outline of the Yangtze River Delta Regional Integration Development Plan” which clearly pointed out that the Jiangsu, Zhejiang and Anhui provinces should promote their respective strengths and strengthen cross-regional coordination and interaction. The construction of inter-provincial cooperative industrial parks is not only an important way to achieve regional integration, but also is of great significance to improve the market operation level of Anhui's economy.
Recently, Anhui Province and Jiangsu Province have cooperated to build an industrial park. In order to enhance the pertinence of investment, these two local governments need to strictly control the conditions of the companies that want to settle. The three criteria are the company’s development prospects (\(x_{1}\)), pollution control capabilities (\(x_{2}\)), and the company’s profitability (\(x_{3}\)). Now Anhui Provincial Government invites four experts \({\text{\{ }}e_{1} ,e_{2} ,e_{3} ,e_{4} {\text{\} }}\) to evaluate the three alternative companies \({\text{\{ }}a_{1} ,a_{2} ,a_{3} {\text{\} }}\) to select the most appropriate one from the three aspects. The above three alternative companies compete with each other. Let \(S = \left\{ {s_{0} :{\text{poor}},s_{1} :{\text{medium}},s_{2} :{\text{good}}} \right\}\) be a LST. In order to make the four experts express their evaluation information reasonably, we explain the related concepts of distributed linguistic in detail. Then, they are required to provide their preference using distributed linguistic. For example, after expert \(e_{1}\) compares the alternative company \(x_{1}\) and alternative company \(x_{3}\), he/she think that the linguistic preference degree of alternative company \(x_{1}\) over alternative company \(x_{3}\) may be \(``poor^{\prime\prime},``medium^{\prime\prime}\) or \(``good^{\prime\prime}\), and the their corresponding probabilities are 20%, 10% and 70%, respectively. Thus, the evaluation information of alternative company \(x_{1}\) over alternative company \(x_{3}\) from expert \(e_{1}\) can be depicted by DLTS \(d_{13}^{1} {\text{ = \{ }}(s_{0} ,0.2),(s_{1} ,0.1),(s_{2} ,0.7)\}\). Therefore, after interviewing 4 experts and selecting the evaluation information in a similar way, four DLTDMMs \(D^{p} = (d_{kj}^{p} )_{3 \times 3} (p = 1,2,3,4)\). In the meanwhile, they are required to give the DLTRM \(D^{p} = (d_{kj}^{p} )_{3 \times 3} (p = 1,2,3,4)\) and a DLTRM \(TD{ = (}TD_{pq} )_{4 \times 4}\) are obtained.
Stage 1 Calculate the weight of each expert under SNA
Step 1: Collect expert’s individual DLTDMMs
\(D^{p} = (d_{kj}^{p} )_{3 \times 3} (p = 1,2,3,4)\) and the DLTRM
\(TD{ = (}TD_{pq} )_{4 \times 4}\) under social network:
$$ D^{1} = \left( {\begin{array}{*{20}c} {\left\{ \begin{array}{l} (s_{0} ,0.0) \\ (s_{1} ,0.3) \\ (s_{2} ,0.7) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.3) \\ (s_{1} ,0.7) \\ (s_{2} ,0.0) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.2) \\ (s_{1} ,0.1) \\ (s_{2} ,0.7) \\ \end{array} \right\}} \\ {\left\{ \begin{array}{l} (s_{0} ,0.1) \\ (s_{1} ,0.2) \\ (s_{2} ,0.7) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.4) \\ (s_{1} ,0.6) \\ (s_{2} ,0.0) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.8) \\ (s_{1} ,0.0) \\ (s_{2} ,0.2) \\ \end{array} \right\}} \\ {\left\{ \begin{array}{l} (s_{0} ,0.0) \\ (s_{1} ,0.2) \\ (s_{2} ,0.8) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.6) \\ (s_{1} ,0.0) \\ (s_{2} ,0.4) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.1) \\ (s_{1} ,0.2) \\ (s_{2} ,0.7) \\ \end{array} \right\}} \\ \end{array} } \right), D^{2} = \left( {\begin{array}{*{20}c} {\left\{ \begin{array}{l} (s_{0} ,0.2) \\ (s_{1} ,0.0) \\ (s_{2} ,0.8) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.2) \\ (s_{1} ,0.5) \\ (s_{2} ,0.3) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.1) \\ (s_{1} ,0.4) \\ (s_{2} ,0.5) \\ \end{array} \right\}} \\ {\left\{ \begin{array}{l} (s_{0} ,0.0) \\ (s_{1} ,0.1) \\ (s_{2} ,0.9) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.2) \\ (s_{1} ,0.4) \\ (s_{2} ,0.4) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.3) \\ (s_{1} ,0.3) \\ (s_{2} ,0.4) \\ \end{array} \right\}} \\ {\left\{ \begin{array}{l} (s_{0} ,0.1) \\ (s_{1} ,0.2) \\ (s_{2} ,0.7) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.2) \\ (s_{1} ,0.0) \\ (s_{2} ,0.8) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.1) \\ (s_{1} ,0.9) \\ (s_{2} ,0.0) \\ \end{array} \right\}} \\ \end{array} } \right), $$
$$ D^{3} = \left( {\begin{array}{*{20}c} {\left\{ \begin{array}{l} (s_{0} ,0.1) \\ (s_{1} ,0.9) \\ (s_{2} ,0.0) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.3) \\ (s_{1} ,0.5) \\ (s_{2} ,0.2) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.5) \\ (s_{1} ,0.5) \\ (s_{2} ,0.0) \\ \end{array} \right\}} \\ {\left\{ \begin{array}{l} (s_{0} ,0.1) \\ (s_{1} ,0.1) \\ (s_{2} ,0.8) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.2) \\ (s_{1} ,0.7) \\ (s_{2} ,0.1) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.6) \\ (s_{1} ,0.2) \\ (s_{2} ,0.2) \\ \end{array} \right\}} \\ {\left\{ \begin{array}{l} (s_{0} ,0.2) \\ (s_{1} ,0.5) \\ (s_{2} ,0.3) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.7) \\ (s_{1} ,0.2) \\ (s_{2} ,0.1) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.2) \\ (s_{1} ,0.0) \\ (s_{2} ,0.8) \\ \end{array} \right\}} \\ \end{array} } \right), D^{4} = \left( {\begin{array}{*{20}c} {\left\{ \begin{array}{l} (s_{0} ,0.1) \\ (s_{1} ,0.2) \\ (s_{2} ,0.7) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.0) \\ (s_{1} ,0.6) \\ (s_{2} ,0.4) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.3) \\ (s_{1} ,0.2) \\ (s_{2} ,0.5) \\ \end{array} \right\}} \\ {\left\{ \begin{array}{l} (s_{0} ,0.2) \\ (s_{1} ,0.1) \\ (s_{2} ,0.7) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.3) \\ (s_{1} ,0.7) \\ (s_{2} ,0.0) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.5) \\ (s_{1} ,0.1) \\ (s_{2} ,0.4) \\ \end{array} \right\}} \\ {\left\{ \begin{array}{l} (s_{0} ,0.1) \\ (s_{1} ,0.2) \\ (s_{2} ,0.7) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.5) \\ (s_{1} ,0.0) \\ (s_{2} ,0.5) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.1) \\ (s_{1} ,0.2) \\ (s_{2} ,0.7) \\ \end{array} \right\}} \\ \end{array} } \right), $$
$$ TD = \left( {\begin{array}{*{20}c} - & - & {\left\{ \begin{array}{l} (s_{0} ,0.2) \\ (s_{1} ,0.3) \\ (s_{2} ,0.5) \\ \end{array} \right\}} & - \\ {\left\{ \begin{array}{l} (s_{0} ,0.3) \\ (s_{1} ,0.5) \\ (s_{2} ,0.2) \\ \end{array} \right\}} & - & - & - \\ {\left\{ \begin{array}{l} (s_{0} ,0.1) \\ (s_{1} ,0.3) \\ (s_{2} ,0.6) \\ \end{array} \right\}} & {\left\{ \begin{array}{l} (s_{0} ,0.2) \\ (s_{1} ,0.2) \\ (s_{2} ,0.6) \\ \end{array} \right\}} & - & {\left\{ \begin{array}{l} (s_{0} ,0.1) \\ (s_{1} ,0.1) \\ (s_{2} ,0.8) \\ \end{array} \right\}} \\ - & {\left\{ \begin{array}{l} (s_{0} ,0.3) \\ (s_{1} ,0.1) \\ (s_{2} ,0.6) \\ \end{array} \right\}} & - & - \\ \end{array} } \right). $$
Step 2: complement the uncomplete social networks matrix \(TD{ = (}TD_{pq} )_{4 \times 4}\) by propagating and aggregating trust relationship.
According to
\(TD{ = (}TD_{pq} )_{4 \times 4}\), the expert
\(e_{1}\) does not directly express his trust relationship towards expert
\(e_{2}\), so taking the process of the trust relationship’s propagation and aggregation between expert
\(e_{1}\) and expert
\(e_{2}\) as an example. There are two indirect paths connecting
\(e_{1}\) and
\(e_{2}\). Path 1:
\(e_{1} \to e_{3} \to e_{4} \to e_{2}\) and Path 2:
\(e_{1} \to e_{3} \to e_{2}\). The weights of each path are as follows:
$$ pe_{1} = \frac{1/2}{{1 + 1/2}} \approx 0.3333;\quad pe_{2} = \frac{1}{1 + 1/2} \approx 0.6667. $$
Then calculate the trust relationship conveyed by each path:
$$ TD_{12}^{1} = P_{DL} (P_{DL} (TD_{13} ,TD_{34} ),TD_{42} ) = \left\{ {\begin{array}{*{20}c} {s_{2} ,E_{ \otimes } (T_{13}^{(2)} ,T_{34}^{(2)} ,T_{42}^{(2)} )} \\ {s_{1} ,E_{ \otimes } (T_{13}^{(1)} ,T_{34}^{(1)} ,T_{42}^{(1)} )} \\ {s_{0} ,1 - E_{ \otimes } (T_{13}^{(0)} ,T_{34}^{(0)} ,T_{42}^{(0)} )} \\ \end{array} } \right\} = \left\{ {\begin{array}{*{20}c} {s_{2} ,0.1743} \\ {s_{1} ,0.0012} \\ {s_{0} ,0.8245} \\ \end{array} } \right\}; $$
$$ TD_{12}^{2} = P_{DL} (TD_{13} ,TD_{32} ) = \left\{ {\begin{array}{*{20}c} {s_{2} ,E_{ \otimes } (T_{13}^{(2)} ,T_{32}^{(2)} )} \\ {s_{1} ,E_{ \otimes } (T_{13}^{(1)} ,T_{32}^{(1)} )} \\ {s_{0} ,1 - E_{ \otimes } (T_{13}^{(0)} ,T_{32}^{(0)} )} \\ \end{array} } \right\} = \left\{ {\begin{array}{*{20}c} {s_{2} ,0.2501} \\ {s_{1} ,0.0373} \\ {s_{0} ,0.7126)} \\ \end{array} } \right\}. $$
The trust relationship between expert
\(e_{1}\) and expert
\(e_{2}\) can be calculated by Eq. (
16):
$$ TD_{13} = {\text{HOWA}}_{w} \left( {Q_{1} ,Q_{2} } \right) = \sum\limits_{i = 1}^{n} {pe_{i} Q_{\sigma } (i)} = 0.3333{*}\left\{ {\begin{array}{*{20}c} {s_{2} ,0.1743} \\ {s_{1} ,0.0012} \\ {s_{0} ,0.8245)} \\ \end{array} } \right\} + 0.6667*\left\{ {\begin{array}{*{20}c} {s_{2} ,0.2501} \\ {s_{1} ,0.0373} \\ {s_{0} ,0.7126)} \\ \end{array} } \right\} = \left\{ {\begin{array}{*{20}c} {s_{2} ,0.2248} \\ {s_{1} ,0.0253} \\ {s_{0} ,0.7499} \\ \end{array} } \right\} $$
The complete DLTRM is shown as follows:
$$ \overline{TD} = \left( {\begin{array}{*{20}c} - & {\left\{ \begin{gathered} (s_{0} ,0.7499) \\ (s_{1} ,0.0253) \\ (s_{2} ,0.2248) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.2000) \\ (s_{1} ,0.3000) \\ (s_{2} ,0.5000) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.6182) \\ (s_{1} ,0.0175) \\ (s_{2} ,0.3643) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.3000) \\ (s_{1} ,0.5000) \\ (s_{2} ,0.2000) \\ \end{gathered} \right\}} & - & {\left\{ \begin{gathered} (s_{0} ,0.8176) \\ (s_{1} ,0.1111) \\ (s_{2} ,0.0713) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.9456) \\ (s_{1} ,0.0057) \\ (s_{2} ,0.0487) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.1000) \\ (s_{1} ,0.3000) \\ (s_{2} ,0.6000) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.2000) \\ (s_{1} ,0.2000) \\ (s_{2} ,0.6000) \\ \end{gathered} \right\}} & - & {\left\{ \begin{gathered} (s_{0} ,0.1000) \\ (s_{1} ,0.1000) \\ (s_{2} ,0.8000) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.7253) \\ (s_{1} ,0.1834) \\ (s_{2} ,0.0913) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.3000) \\ (s_{1} ,0.1000) \\ (s_{2} ,0.6000) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.9634) \\ (s_{1} ,0.0054) \\ (s_{2} ,0.0312) \\ \end{gathered} \right\}} & - \\ \end{array} } \right). $$
Step 3: compute the weight
\(w_{q} (q = 1,2,3,4)\) of each expert
\(e = \left\{ {e_{1} ,e_{2} ,e_{3} ,e_{4} } \right\}\) by Eqs. (
17) and (
18):
$$ w_{1} = 0.2495;\quad w_{1} = 0.2835;\quad w_{1} = 0.2123;\quad w_{1} = 0.2547. $$
Stage 2 Aggregation of DLTDM.
Step 4: connecting the weight of each expert and the individual distributed linguistic trust decision matrices
\(D^{p} = (d_{kj}^{p} )_{3 \times 3} (p = 1,2,3,4)\), a collective DLTDMM
\(\overline{D} { = (}\overline{D}_{kj} )_{3 \times 3}\) can be obtained:
$$ \overline{D} = \left( {\begin{array}{*{20}c} {\left\{ \begin{gathered} (s_{0} ,0.1034) \\ (s_{1} ,0.3167) \\ (s_{2} ,0.5797) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.1952) \\ (s_{1} ,0.5754) \\ (s_{2} ,0.2294) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.2608) \\ (s_{1} ,0.2954) \\ (s_{2} ,0.4438) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.0971) \\ (s_{1} ,0.1250) \\ (s_{2} ,0.7779) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.2758) \\ (s_{1} ,0.5900) \\ (s_{2} ,0.1342) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.5394) \\ (s_{1} ,0.1530) \\ (s_{2} ,0.3076) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.0963) \\ (s_{1} ,0.2637) \\ (s_{2} ,0.6400) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.4824) \\ (s_{1} ,0.0424) \\ (s_{2} ,0.4752) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.1212) \\ (s_{1} ,0.3560) \\ (s_{2} ,0.5228) \\ \end{gathered} \right\}} \\ \end{array} } \right). $$
Stage 3 Consensus test and adjustment
Step 5: by using Eqs. (
21)–(
23), calculate the three-level consensus degree at the evaluation information level
, expert level and group level:
(1) The consensus degree at evaluation information levels can be calculated via Eq. (
21):
$$\begin{aligned}& DS^{12} = \left( {\begin{array}{*{20}c} {0.80} & {0.80} & {0.80} \\ {0.87} & {0.73} & {0.67} \\ {0.93} & {0.73} & {0.53} \\ \end{array} } \right);\quad DS^{13} = \left( {\begin{array}{*{20}c} {0.50} & {0.87} & {0.53} \\ {0.93} & {0.87} & {0.87} \\ {0.67} & {0.80} & {0.93} \\ \end{array} } \right);\\ &\quad DS^{14} = \left( {\begin{array}{*{20}c} {0.93} & {0.73} & {0.87} \\ {0.93} & {0.93} & {0.80} \\ {0.93} & {0.93} & {1.00} \\ \end{array} } \right);\end{aligned}$$
$$\begin{aligned}& DS^{23} = \left( {\begin{array}{*{20}c} {0.40} & {0.93} & {0.67} \\ {0.93} & {0.70} & {0.70} \\ {0.73} & {0.80} & {0.87} \\ \end{array} } \right);\quad DS^{24} = \left( {\begin{array}{*{20}c} {0.87} & {0.87} & {0.87} \\ {0.67} & {0.73} & {0.87} \\ {1.00} & {0.80} & {0.53} \\ \end{array} } \right);\\ &\quad DS^{34} = \left( {\begin{array}{*{20}c} {0.53} & {0.80} & {0.67} \\ {0.93} & {0.93} & {0.87} \\ {0.73} & {0.73} & {0.87} \\ \end{array} } \right).\end{aligned}$$
(2) The consensus degree at experts’ level is:
$$ DB = \left( {\begin{array}{*{20}c} - & \quad {0.7622} &\quad {0.7644} &\quad {0.8944} \\ {0.7622} &\quad - &\quad {0.7477} &\quad {0.8011} \\ {0.7644} &\quad {0.7477} &\quad - &\quad {0.7800} \\ {0.8944} &\quad {0.8011} &\quad {0.7800} &\quad - \\ \end{array} } \right). $$
\(db^{12} = 0.7622\);\(db^{13} = 0.7644\);\(db^{14} = 0.8944\);\(db^{23} = 0.7477\);\(db^{24} = 0.8011\);\(db^{34} = 0.7800\).
(3) The value of GCD is:
\({\text{GCD}} = 0.7916 < 0.8000\).
The GCD is lower than the threshold value \(\lambda = 0.8000\), the feedback mechanism is activated to adjust the most inconsistent evaluation information.
Step 6: checking the most inconsistent distribute information with the lowest adjustment cost through Eqs. (
24) and (
25).
The order of the consensus index at experts’ level:
0.7477 < 0.7622 < 0.7644 < 0.7800 < 0.8011 < 0.8944.
The 0.7477 is derived from
\(DS^{23}\), then the smallest number 0.4 in
\(DS^{23}\) can be pinpointed. The 0.4 is the consensus degree at evaluation information levels between
\(d_{11}^{2}\) and
\(d_{11}^{3}\):
$$ d_{11}^{2} = \left\{ \begin{gathered} (s_{0} ,0.2) \\ (s_{1} ,0.0) \\ (s_{2} ,0.8) \\ \end{gathered} \right\};\quad d_{11}^{3} = \left\{ \begin{gathered} (s_{0} ,0.1) \\ (s_{1} ,0.9) \\ (s_{2} ,0.0) \\ \end{gathered} \right\}. $$
Then, calculate the distance between
\(d_{11}^{2}\),
\(d_{11}^{3}\) and
\(\overline{{d_{11} }}\) separately to decide which evaluation information needs to be adjusted:
$$ H(d_{11}^{2} ,\overline{{d_{11} }} ) = \frac{{\sum\nolimits_{i = 0}^{2} {|T_{11}^{2(i)} - \overline{{T_{11}^{i} }} |} }}{3} = 0.2112. $$
$$ H(d_{11}^{3} ,\overline{{d_{11} }} ) = \frac{{\sum\nolimits_{i = 0}^{2} {|T_{11}^{3(i)} - \overline{{T_{11}^{i} }} |} }}{3} = 0.3888. $$
Owing to 0.2112 < 0.3888, replace the evaluation information
\(d_{11}^{3}\) with
\(d_{11}^{2}\).Then recalculate the value of
\({\text{GCD}}^{*}\):
$$ DB_{1} = \left( {\begin{array}{*{20}c} - & \quad {0.7622} & \quad {0.8071} & \quad {0.8944} \\ {0.7616} & \quad - & \quad {0.8242} & \quad {0.8011} \\ {0.8071} & \quad {0.8242} & \quad - & \quad {0.8023} \\ {0.8944} & \quad {0.8011} & \quad {0.8023} & \quad - \\ \end{array} } \right). $$
$$ {\text{GCD}}^{ * } = 0.8152 > 0.8000. $$
The value of \({\text{GCD}}^{*}\) is higher than the threshold value \(\lambda = 0.800\), so end the feedback mechanism and turn to step 7.
Stage 4 Determining the order relationship among
\(m\)
alternatives.
Step 7: renew the collective distributed linguistic trust decision matrices. Combining the weight of each expert
\(\left( {w_{1} ,w_{2} ,w_{3} ,w_{4} } \right)^{T}\) and the adjusted individual distributed linguistic trust decision matrices
\(D^{\prime {p}} = (d^{\prime {p}}_{kj} )_{3 \times 3} (p = 1,2,3,4)\), the new collective DLTDMM can be obtained:
$$ \overline{{D_{1} }} = \left( {\begin{array}{*{20}c} {\left\{ \begin{gathered} (s_{0} ,0.1246) \\ (s_{1} ,0.1258) \\ (s_{2} ,0.7496) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.1952) \\ (s_{1} ,0.5754) \\ (s_{2} ,0.2294) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.2608) \\ (s_{1} ,0.2954) \\ (s_{2} ,0.4438) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.0971) \\ (s_{1} ,0.1250) \\ (s_{2} ,0.7779) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.2754) \\ (s_{1} ,0.5900) \\ (s_{2} ,0.1346) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.5394) \\ (s_{1} ,0.1530) \\ (s_{2} ,0.3076) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.0963) \\ (s_{1} ,0.2637) \\ (s_{2} ,0.6400) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.4824) \\ (s_{1} ,0.0424) \\ (s_{2} ,0.4752) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.1212) \\ (s_{1} ,0.3560) \\ (s_{2} ,0.5228) \\ \end{gathered} \right\}} \\ \end{array} } \right). $$
Step 8: calculate the expectation of each alternative
\(\left\{ {a_{1} ,a_{2} ,a_{3} } \right\}\) as follows:
$$ E(a_{1} ) = 4.6444;\quad E(a_{2} ) = 3.6606;\quad E(a_{3} ) = 4.9140. $$
As \(E(a_{2} ) < E(a_{1} ) < E(a_{3} )\), thus the company \(a_{3}\) performs best in the three aspects of the company’s development prospects \((x_{1} )\), pollution control capabilities \((x_{2} )\) and the company’s profitability \((x_{3} )\). The government can choose \(a_{3}\) as a resident enterprise in the industrial park.
Comparative analysis
To prove our model’s validity and applicability, this subsection applies three models proposed in [
12,
14,
31] to solve the problem mentioned in Sect. "
Numerical experiment".
To solve the unreasonable assumption that the decision maker knows the weight in advance, Wu et al. [
12] developed the DLTDMS composed of related properties of DLTFs. Then the weight of expert can be obtained by calculating in-degree of centrality. Finally, a novel feedback mechanism based on the minimum adjustment cost which can produce the boundary feedback parameter was constructed to recommend personalized advice for inconsistent experts.
Step 1: calculate the trust in-degree centrality by Eq. (
6) and weight of each expert by Eq. (
7) in [
12]:
$$ C_{D}^{L} (e_{1} ) = \left\{ {\begin{array}{*{20}c} {(s_{0} ,0.20)} & {(s_{1} ,0.40)} & {(s_{2} ,0.20)} \\ \end{array} } \right\}; $$
$$ C_{D}^{L} (e_{2} ) = \left\{ {\begin{array}{*{20}c} {(s_{0} ,0.25)} & {(s_{1} ,0.15)} & {(s_{2} ,0.60)} \\ \end{array} } \right\}; $$
$$ C_{D}^{L} (e_{3} ) = \left\{ {\begin{array}{*{20}c} {(s_{0} ,0.20)} & {(s_{1} ,0.30)} & {(s_{2} ,0.50)} \\ \end{array} } \right\}; $$
$$ C_{D}^{L} (e_{4} ) = \left\{ {\begin{array}{*{20}c} {(s_{0} ,0.10)} & {(s_{1} ,0.10)} & {(s_{2} ,0.80)} \\ \end{array} } \right\}. $$
$$ w_{1} = 0.1787;\quad w_{2} = 0.2625;\quad w_{3} = 0.2569;\quad w_{4} = 0.3019. $$
Step 2: get a collective DLTDMM
\(\overline{D} { = (}\overline{D}_{pq} )_{3 \times 3}\) by Eq. (
8) in [
12]:
$$ \overline{D} = \left( {\begin{array}{*{20}c} {\left\{ \begin{gathered} (s_{0} ,0.1340) \\ (s_{1} ,0.1141) \\ (s_{2} ,0.7519) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.1832) \\ (s_{1} ,0.5660) \\ (s_{2} ,0.2508) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.2766) \\ (s_{1} ,0.3077) \\ (s_{2} ,0.4157) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.1040) \\ (s_{1} ,0.1179) \\ (s_{2} ,0.7781) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.2674) \\ (s_{1} ,0.6078) \\ (s_{2} ,0.1249) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.5241) \\ (s_{1} ,0.1650) \\ (s_{2} ,0.3110) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.1059) \\ (s_{1} ,0.2725) \\ (s_{2} ,0.6216) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.4925) \\ (s_{1} ,0.0494) \\ (s_{2} ,0.4581) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.1257) \\ (s_{1} ,0.3320) \\ (s_{2} ,0.5423) \\ \end{gathered} \right\}} \\ \end{array} } \right). $$
Step 3: calculate the consensus levels with the group by Eqs. (
9)–(
11) in [
12]:
(1) Consensus degree at the level of evaluation information:
$$ CE^{1} = \left( {\begin{array}{*{20}c} {0.8761} & {0.8328} & {0.8049} \\ {0.9453} & {0.9107} & {0.8180} \\ {0.8767} & {0.9271} & {0.8949} \\ \end{array} } \right);\quad CE^{2} = \left( {\begin{array}{*{20}c} {0.9239} & {0.9560} & {0.8793} \\ {0.9187} & {0.8203} & {0.8487} \\ {0.9434} & {0.7719} & {0.6213} \\ \end{array} } \right); $$
$$ CE^{3} = \left( {\begin{array}{*{20}c} {0.9239} & {0.9221} & {0.7285} \\ {0.9854} & {0.9357} & {0.9248} \\ {0.7899} & {0.7614} & {0.7787} \\ \end{array} } \right);\quad CE^{4} = \left( {\begin{array}{*{20}c} {0.9427} & {0.8779} & {0.9256} \\ {0.9360} & {0.9130} & {0.9419} \\ {0.9434} & {0.9657} & {0.8949} \\ \end{array} } \right). $$
(2) Consensus degree at the level of alternatives:
$$ CA^{1} { = (}\begin{array}{*{20}c} {0.8379} & {0.8913} & {0.8996} \\ \end{array} {);} $$
$$ CA^{2} { = (}\begin{array}{*{20}c} {0.9197} & {0.8626} & {0.7789} \\ \end{array} {);} $$
$$ CA^{3} { = (}\begin{array}{*{20}c} {0.8582} & {0.9486} & {0.7767} \\ \end{array} {);} $$
$$ CA^{{4}} { = (}\begin{array}{*{20}c} {0.9154} & {0.9303} & {0.{9347) }} \\ \end{array} . $$
(3) Consensus degree at the level of group decision matrix level:
$$ {\text{CI}}^{{1}} { = 0}{\text{.8763;}}\quad {\text{CI}}^{{2}} { = 0}{\text{.8612}};\quad {\text{CI}}^{{3}} { = 0}{\text{.8537}};\quad {\text{CI}}^{{4}} { = 0}{\text{.9268}}. $$
Under the condition of \(\lambda { = 0}{\text{.87}}\),then the \({\text{CI}}^{2} < \lambda\) and \({\text{CI}}^{{3}} < \lambda\).
Step 4: identification of the inconsistent evaluation elements:
$$\begin{aligned} APS& = \left\{ (2,2,2),(2,2,3),(2,3,2),(2,3,3),\right.\\ & \left.\quad (3,1,3),(3,3,1),(3,3,2),(3,3,3) \right\}.\end{aligned}$$
Step 5: calculate the boundary feedback parameter by solving the optimization model to obtain the recommendation advice:
$$ \begin{gathered} {\text{Min}}\sum\limits_{h,i,j \in APS} {\delta |d^{h} - \overline{d}_{ij}^{h} |} \\ {\text{s.t.}} \left\{ \begin{gathered} {\text{CI}}^{h} ({\text{RD}}^{h} ,\overline{{{\text{RD}}}} ) \ge \gamma \\ {\text{CI}}^{s} ({\text{RD}}^{s} ,\overline{{{\text{RD}}}} ) \ge \gamma \\ \overline{{{\text{RD}}}} = {\text{DTWA}}({\text{RD}}^{h} ,{\text{RD}}^{1} \ldots ,{\text{RD}}^{s} , \ldots ,{\text{RD}}^{k} ) \\ \end{gathered} \right.. \\ \end{gathered} $$
$$ \delta_{\min } = 0.08 $$
The recommendations advice for expert
\(e_{2}\) and
\(e_{3}\) are:
$$ (2,2,2) \to \left\{ {(s_{0} ,0.2053),(s_{1} ,0.4164),(s_{2} ,0.3773)} \right\}; $$
$$ (2,2,3) \to \left\{ {(s_{0} ,0.2891),(s_{1} ,0.3181),(s_{2} ,0.3928)} \right\}; $$
$$ (2,3,2) \to \left\{ {(s_{0} ,0.2235),(s_{1} ,0.0045),(s_{2} ,0.7720)} \right\}; $$
$$ (2,3,3) \to \left\{ {(s_{0} ,0.1017),(s_{1} ,0.8551),(s_{2} ,0.0432)} \right\}; $$
$$ (3,1,3) \to \left\{ {(s_{0} ,0.4819),(s_{1} ,0.4850),(s_{2} ,0.0331)} \right\}; $$
$$ (3,3,2) \to \left\{ {(s_{0} ,0.6835),(s_{1} ,0.1885),(s_{2} ,0.1280)} \right\}; $$
$$ (3,3,3) \to \left\{ {(s_{0} ,0.1946),(s_{1} ,0.0261),(s_{2} ,0.7793)} \right\} $$
Step 6: after modifying the inconsistent evaluation information, the new collective DLTDMM would be:
$$ \overline{{D^{^{\prime}} }} = \left( {\begin{array}{*{20}c} {\left\{ \begin{gathered} (s_{0} ,0.1340) \\ (s_{1} ,0.1141) \\ (s_{2} ,0.7519) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.1832) \\ (s_{1} ,0.5660) \\ (s_{2} ,0.2508) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.2766) \\ (s_{1} ,0.3077) \\ (s_{2} ,0.4157) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.1040) \\ (s_{1} ,0.1179) \\ (s_{2} ,0.7781) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.2674) \\ (s_{1} ,0.6078) \\ (s_{2} ,0.1248) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.5240) \\ (s_{1} ,0.1650) \\ (s_{2} ,0.3110) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.1059) \\ (s_{1} ,0.2725) \\ (s_{2} ,0.6216) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.4925) \\ (s_{1} ,0.0494) \\ (s_{2} ,0.4581) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.1247) \\ (s_{1} ,0.3269) \\ (s_{2} ,0.5484) \\ \end{gathered} \right\}} \\ \end{array} } \right). $$
Step 7: their corresponding expected trust scores are by Eq. (
3):
$$ E(a_{1} ) = 4.6173;\quad E(a_{2} ) = 3.6418;\quad E(a_{3} ) = 4.8841. $$
$$ E(a_{2} ) < E(a_{1} ) < E(a_{3} ). $$
Therefore, the alternative \(a_{3}\) is the best choose.
Wu et al. [
14] first developed a propagation operator on the basis of
t-norms to get a complete DLTRM. Then to complement the incomplete individual DLTDMM, a trust estimation mechanism in which the evaluation information of unknown experts was estimated from other experts’ evaluation information was set up. Finally, an optimization model with the maximum retention of self-esteem degree was designed to obtain optimal feedback parameters in the process of reaching the GCD.
Step 1: complete the DLTRM with the support of the trust propagation operator
\(P_{LD}\) in [
14]:
$$ \overline{TD} = \left( {\begin{array}{*{20}c} - & {\left\{ \begin{gathered} (s_{0} ,0.7121) \\ (s_{1} ,0.0378) \\ (s_{2} ,0.2501) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.2000) \\ (s_{1} ,0.3000) \\ (s_{2} ,0.5000) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.6175) \\ (s_{1} ,0.0184) \\ (s_{2} ,0.3641) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.3000) \\ (s_{1} ,0.5000) \\ (s_{2} ,0.2000) \\ \end{gathered} \right\}} & - & {\left\{ \begin{gathered} (s_{0} ,0.8200) \\ (s_{1} ,0.1100) \\ (s_{2} ,0.0700) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.9462) \\ (s_{1} ,0.0064) \\ (s_{2} ,0.0474) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.1000) \\ (s_{1} ,0.3000) \\ (s_{2} ,0.6000) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.2000) \\ (s_{1} ,0.2000) \\ (s_{2} ,0.6000) \\ \end{gathered} \right\}} & - & {\left\{ \begin{gathered} (s_{0} ,0.1000) \\ (s_{1} ,0.1000) \\ (s_{2} ,0.8000) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.8084) \\ (s_{1} ,0.1535) \\ (s_{2} ,0.0381) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.3000) \\ (s_{1} ,0.4000) \\ (s_{2} ,0.300) \\ \end{gathered} \right\}0} & {\left\{ \begin{gathered} (s_{0} ,0.9623) \\ (s_{1} ,0.0065) \\ (s_{2} ,0.0312) \\ \end{gathered} \right\}} & - \\ \end{array} } \right). $$
Step 2: calculate the relative node in-degree centrality index by Eq. (
10) and the weight of each expert by Eq. (
11) in [
14]:
$$ C_{D}^{L} (e_{1} ) = \left\{ {(s_{1} ,0.4032),(s_{1} ,0.3176),(s_{2} ,0.2792)} \right\}; $$
$$ C_{D}^{L} (e_{2} ) = \left\{ {(s_{1} ,0.4041),(s_{1} ,0.2124),(s_{2} ,0.3835)} \right\}; $$
$$ C_{D}^{L} (e_{3} ) = \left\{ {(s_{1} ,0.6614),(s_{1} ,0.1045),(s_{2} ,0.2341)} \right\}; $$
$$ C_{D}^{L} (e_{4} ) = \left\{ {(s_{1} ,0.5553),(s_{1} ,0.0405),(s_{2} ,0.4042)} \right\}; $$
$$ w_{1} = 0.2581;\quad w_{2} = 0.2722;\quad w_{3} = 0.2153;\quad w_{4} = 0.2544. $$
Step 3: get a collective DLTDMM
\(\overline{D} { = (}\overline{D}_{kj} )_{3 \times 3}\) by Eq. (
18) in [
14]:
$$ \overline{D} = \left( {\begin{array}{*{20}c} {\left\{ \begin{gathered} (s_{0} ,0.1014) \\ (s_{1} ,0.3226) \\ (s_{2} ,0.5760) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.1966) \\ (s_{1} ,0.5770) \\ (s_{2} ,0.2264) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.2630) \\ (s_{1} ,0.2934) \\ (s_{2} ,0.4436) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.0982) \\ (s_{1} ,0.1258) \\ (s_{2} ,0.7760) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.2770) \\ (s_{1} ,0.5926) \\ (s_{2} ,0.1304) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.5446) \\ (s_{1} ,0.1502) \\ (s_{2} ,0.3052) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.0958) \\ (s_{1} ,0.2648) \\ (s_{2} ,0.6394) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.4874) \\ (s_{1} ,0.0432) \\ (s_{2} ,0.4694) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.1216) \\ (s_{1} ,0.3472) \\ (s_{2} ,0.5312) \\ \end{gathered} \right\}} \\ \end{array} } \right). $$
Step 4: consensus test.
(1) Deviation indexes at the level of evaluation information:
$$ DE^{1} = \left( {\begin{array}{*{20}c} {0.0087} & {0.0257} & {0.0357} \\ {0.0038} & {0.0107} & {0.0330} \\ {0.0131} & {0.0065} & {0.0169} \\ \end{array} } \right);\quad DE^{2} = \left( {\begin{array}{*{20}c} {0.0547} & {0.0038} & {0.0137} \\ {0.0086} & {0.0386} & {0.0304} \\ {0.0026} & {0.0646} & {0.1961} \\ \end{array} } \right); $$
$$ DE^{3} = \left( {\begin{array}{*{20}c} {0.2217} & {0.0058} & {0.0985} \\ {0.0004} & {0.0061} & {0.0055} \\ {0.0605} & {0.0687} & {0.0663} \\ \end{array} } \right);\quad DE^{4} = \left( {\begin{array}{*{20}c} {0.0101} & {0.0231} & {0.0044} \\ {0.0056} & {0.0097} & {0.0045} \\ {0.0026} & {0.0010} & {0.0169} \\ \end{array} } \right). $$
(2) Deviation indexes at the level of alternatives:
$$ DA^{1} = (\begin{array}{*{20}c} {0.0234} & {0.0158} & {0.0121} \\ \end{array} );\quad DA^{2} = (\begin{array}{*{20}c} {0.0240} & {0.0258} & {0.0878} \\ \end{array} ); $$
$$ DA^{3} = (\begin{array}{*{20}c} {0.1087} & {0.0040} & {0.0652} \\ \end{array} );\quad DA^{4} = (\begin{array}{*{20}c} {0.0126} & {0.0066} & {0.0068} \\ \end{array} ). $$
(3) Deviation indexes at the level of group decision matrix:
$$ {\text{DI}}^{1} = 0.0171;\quad {\text{DI}}^{2} = 0.0459;\quad {\text{DI}}^{3} = 0.0593;\quad {\text{DI}}^{4} = 0.0086. $$
The evaluation information that does not meet the threshold
\(\lambda = 0.04\) are classified into the set APS by Eqs. (
23)–(
25):
$$ APS = \left( {\begin{array}{*{20}c} {(2,3,2)} &\quad {(3,1,2)} &\quad {(3,3,3)} \\ {(2,3,3)} &\quad {(3,3,1)} &\quad {} \\ {(3,1,1)} &\quad {(3,3,2)} \quad & {} \\ \end{array} } \right). $$
Step 5: determine the optimal boundary feedback parameter:
$$ \left\{ \begin{gathered} {\text{Max}}:2 - \frac{{\delta_{2}^{2} }}{3*2} \cdot \sum\limits_{i.j \in APS} {\sum\limits_{a = 0}^{\xi } {\left[ {(s_{\alpha } )_{ij}^{2} - (\overline{s}_{\alpha } )_{ij} } \right]^{2} - \frac{{\delta_{3}^{2} }}{3*5} \cdot \sum\limits_{i.j \in APS} {\sum\limits_{a = 0}^{\xi } {\left[ {(s_{\alpha } )_{ij}^{3} - (\overline{s}_{\alpha } )_{ij} } \right]^{2} } } } } \\ {\text{s.t.}} \left\{ \begin{gathered} {\text{DI}}^{1} \le 0.04 \\ {\text{DI}}^{2} \le 0.04 \\ {\text{DI}}^{3} \le 0.04 \\ {\text{DI}}^{4} \le 0.04 \\ r(s_{\alpha } )_{ij}^{2} = (1 - \delta_{2} ) \cdot (s_{\alpha } )_{ij}^{1} + \delta_{2} \cdot (\overline{s}_{\alpha } )_{ij} ,\alpha \in \left\{ {0,1,2} \right\} \\ r(s_{\alpha } )_{ij}^{2} = (1 - \delta_{3} ) \cdot (s_{\alpha } )_{ij}^{1} + \delta_{3} \cdot (\overline{s}_{\alpha } )_{ij} ,\alpha \in \left\{ {0,1,2} \right\} \\ 0 \le \delta_{2} \le 1 \\ 0 \le \delta_{3} \le 1 \\ \end{gathered} \right. \\ \end{gathered} \right.. $$
The value of
\(\delta_{2} { = 0}{\text{.1271 }}\delta_{3} { = 0}{\text{.2352}}\) can be obtained by solving the nonlinear model. The recommendation advice for
\(e_{2}\) are:
$$ (2,3,2) \to \left\{ {(s_{0} ,0.2362),(s_{1} ,0.0048),(s_{2} ,0.7590)} \right\};\quad (2,3,3) \to \left\{ {(s_{0} ,0.1031),(s_{1} ,0.8303),(s_{2} ,0.0666)} \right\}. $$
The recommendation advice for
\(e_{3}\) are:
$$ (3,1,2) \to \left\{ {(s_{0} ,0.2754),(s_{1} ,0.5183),(s_{2} ,0.2063)} \right\}; $$
$$ (3,1,3) \to \left\{ {(s_{0} ,0.4824),(s_{1} ,0.4845),(s_{2} ,0.0331)} \right\} $$
$$ (3,3,1) \to \left\{ {(s_{0} ,0.1743),(s_{1} ,0.4454),(s_{2} ,0.3803)} \right\} $$
$$ (3,3,2) \to \left\{ {(s_{0} ,0.6526),(s_{1} ,0.1624),(s_{2} ,0.1850)} \right\}; $$
$$ (3,3,3) \to \left\{ {(s_{0} ,0.1827),(s_{1} ,0.0761),(s_{2} ,0.7412)} \right\} $$
After modifying the inconsistent evaluation information, the new collective DLTDMM will be:
$$ \overline{{D^{^{\prime}} }} = \left( {\begin{array}{*{20}c} {\left\{ \begin{gathered} (s_{0} ,0.1015) \\ (s_{1} ,0.2933) \\ (s_{2} ,0.6052) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.1914) \\ (s_{1} ,0.5809) \\ (s_{2} ,0.2277) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.2630) \\ (s_{1} ,0.2934) \\ (s_{2} ,0.4436) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.0982) \\ (s_{1} ,0.1258) \\ (s_{2} ,0.776) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.2770) \\ (s_{1} ,0.5926) \\ (s_{2} ,0.1304) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.5446) \\ (s_{1} ,0.1502) \\ (s_{2} ,0.3052) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.0905) \\ (s_{1} ,0.2529) \\ (s_{2} ,0.6566) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.4870) \\ (s_{1} ,0.0368) \\ (s_{2} ,0.4762) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.1184) \\ (s_{1} ,0.3448) \\ (s_{2} ,0.5368) \\ \end{gathered} \right\}} \\ \end{array} } \right). $$
The new deviation indexes at the level of decision matrix are as follows:
$$ {\text{DI}}^{1} = 0.0153;\quad {\text{DI}}^{2} = 0.0400;\quad {\text{DI}}^{3} = 0.0400;\quad {\text{DI}}^{4} = 0.0075. $$
Step 6: calculate the expectation degree of each alternative and make a rank of them:
$$ E(a_{1} ) = 6.7208;\quad E(a_{2} ) = 6.2918;\quad E(a_{3} ) = 6.9735. $$
$$ E(a_{2} ) < E(a_{1} ) < E(a_{3} ). $$
Therefore, the best alternative is the company \(a_{3}\).
In Ref. [
31], Zhang et al. defined the consistency of PLPR based on graph theory's preference graph. As for the unacceptable consistent probabilistic linguistic preference relation, an automatic optimization method was designed to improve GCD. Finally, Zhang et al. used the aggregation operator to calculate the collective preference value of all the alternatives and make an order of them.
Step 1: see Step 1 in the above method.
Step 2: calculate the consistency indices (CI) by Eq. (
13) in [
31]:
$$ {\text{CI}}^{1} = 0.0826 > \overline{{{\text{CI}}}} ;\quad {\text{CI}}^{2} = 0.0793 < \overline{{{\text{CI}}}} ;\quad {\text{CI}}^{3} = 0.0793 < \overline{{{\text{CI}}}} ;\quad {\text{CI}}^{4} = 0.0779 < \overline{{{\text{CI}}}} {.} $$
Step 3: let
\(\theta = 0.1\). The modified NPLPR is obtained by Eq. (
10) in [
31]:
$$ D^{(1)^{\prime}} = \left( {\begin{array}{*{20}c} {\left\{ \begin{gathered} (s_{0} ,0.0131) \\ (s_{1} ,0.2819) \\ (s_{2} ,0.7050) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.2886) \\ (s_{1} ,0.6864) \\ (s_{2} ,0.0250) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.2078) \\ (s_{1} ,0.1216) \\ (s_{2} ,0.6706) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.1003) \\ (s_{1} .0.1919) \\ (s_{2} ,0.7078) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.3868) \\ (s_{1} ,0.6001) \\ (s_{2} ,0.0131) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.7734) \\ (s_{1} ,0.0156) \\ (s_{2} ,0.2110) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.0116) \\ (s_{1} ,0.2073) \\ (s_{2} ,0.7821) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.5888) \\ (s_{1} ,0.0049) \\ (s_{2} ,0.4063) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.1035) \\ (s_{1} ,0.2133) \\ (s_{2} ,0.6832) \\ \end{gathered} \right\}} \\ \end{array} } \right). $$
Then we can get \({\text{CI}}^{\prime{1}} = 0.0818 < \overline{{{\text{CI}}}}\).
Step 4: get a collective DLTDMM by applying the model in Stages II and III:
$$ \overline{D} = \left( {\begin{array}{*{20}c} {\left\{ \begin{gathered} (s_{0} ,0.1365) \\ (s_{1} ,0.1107) \\ (s_{2} ,0.7528) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.1811) \\ (s_{1} ,0.5636) \\ (s_{2} ,0.2553) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.2825) \\ (s_{1} ,0.3154) \\ (s_{2} ,0.4021) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.1041) \\ (s_{1} ,0.1164) \\ (s_{2} ,0.7795) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.2636) \\ (s_{1} ,0.6036) \\ (s_{2} ,0.1328) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.5221) \\ (s_{1} ,0.1631) \\ (s_{2} ,0.3148) \\ \end{gathered} \right\}} \\ {\left\{ \begin{gathered} (s_{0} ,0.1097) \\ (s_{1} ,0.2785) \\ (s_{2} ,0.6118) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.4887) \\ (s_{1} ,0.0524) \\ (s_{2} ,0.4589) \\ \end{gathered} \right\}} & {\left\{ \begin{gathered} (s_{0} ,0.1262) \\ (s_{1} ,0.3344) \\ (s_{2} ,0.5394) \\ \end{gathered} \right\}} \\ \end{array} } \right). $$
Step 5: calculate the comprehensive preference values (PVs):
$$ {\text{PV}}_{1} = \left\{ {s_{0.6003} ,s_{1.9800} ,s_{4.2310} } \right\}; $$
$$ {\text{PV}}_{2} = \left\{ {s_{0.8903} ,s_{1.7661} ,s_{3.6815} } \right\}; $$
$$ {\text{PV}}_{3} = \left\{ {s_{0.7254} ,s_{1.3305} ,s_{4.8318} } \right\}. $$
Step 6: calculate the expectation of PV by definition 4 in [
31]:
$$ E({\text{PV}}_{1} ) = s_{2.2704} ; $$
$$ E({\text{PV}}_{2} ) = s_{2.1126} ; $$
$$ E({\text{PV}}_{3} ) = s_{2.2959} . $$
Then the order of \({\text{PV}}_{i}\) is as follows:\({\text{PV}}_{3} > {\text{PV}}_{1} > {\text{PV}}_{2}\), and the best option is the company \(a_{3}\).
The GDM results with different methods are displayed in Table
2.
Table 2
GDM results with different methods
Our GDM method | \(a_{2} \prec a_{1} \prec a_{3}\) | \(a_{3}\) |
| \(a_{2} \prec a_{1} \prec a_{3}\) | \(a_{3}\) |
| \(a_{2} \prec a_{1} \prec a_{3}\) | \(a_{3}\) |
Zhang et al. [ 31]’s method | \(a_{2} \prec a_{1} \prec a_{3}\) | \(a_{3}\) |
Compared with the models in Refs. [
12,
14,
31], the advantages of our model are summarized as follows:
1.
Using the distributed linguistic group decision making model with SNA, we can find that the final selection of the settled company is entirely consistent with the result in [
12,
14,
31], which verifies the effectiveness of our model.
2.
SNA is an important method to determine the weight of DMs, which requires that the designed models are supposed to have the ability to dig deeply into the available information in the DLTRM. The methods proposed in Wu et al. [
12] and Zhang et al. [
31] did not research the trust’s propagation and aggregation operators in social networks but directly calculated the weight of each expert based on the incomplete DLTRM. In Ref. [
14], Wu investigated the propagation operator for experts who are not directly connected and used the shortest indirect path (use the average value of them on the assumption that there is more than one shortest path) as the path of trust transfer. However, it is more common that a series of trust paths of different lengths that transfer trust relationships between indirectly connected experts. This method does not consider the influence of trust paths of different length on the final result. To solve this problem, the model in this article designs a POWA operator, which not only considers the trust relationship from all trust paths between experts, but also takes the weights of each path into consideration. In general, compared with [
12,
14,
31], the trust model based on SNA in this paper more comprehensively mines trust relationships in social networks to obtain more accurate weights.
3.
Owing to DMs’ different background and knowledge, they may be inconsistent with each other when making decisions. Therefore, CRP is important for selecting a reliable decision. In Ref. [
14], Wu designed an optimization model with the maximum retention of self-esteem degree to get optimal feedback parameters in CRP. However, the proposed objective function and solution process are complicated. At the same time, we can find
\(\delta_{2} { = 0}{\text{.127}}\),
\(\delta_{3} { = 0}{\text{.235}}\), which means that 12.7% and 23.5% of the initial inconsistent information of
\(e_{2}\) and
\(e_{3}\) need to be adjusted which greatly damages the integrity of the original data. In Ref. [
12], an optimization model based on the minimum adjustment cost was established to maintain individual independence while ensuring that the group consensus reaches the threshold. However, it only takes the model to minimize the adjustment cost into account, ignoring the deviation degree between experts and the group. In view of the advantages and disadvantages of the above two models in [
12,
14], our model finds the most inconsistent evaluation information in each cycle of CRP, which greatly maintains the integrity of the original information. At the same time, to simplify the complexity of the model, we choose the evaluation information that needs to be adjusted based on the principle of minimum adjustment costs.