An Integrated Group Decision-Making Method with Hesitant Qualitative Information Based on DEA Cross-Efficiency and Priority Aggregation for Evaluating Factors Affecting a Resilient City

Abstract

A city is a carrier of the development of human society; the higher the resilience of a city is, the better it can resist invasion from the outside. Therefore, the evaluation factors affecting resilient cities are of great practical significance in the study of resilient cities. The assessment method with hesitant fuzzy linguistic preference relation (HFLPR) has recently widely used in evaluation problems. However, some existing priority vector solving methods in the assessment method often ignore much decision-making information. Therefore, the aim of this paper is to present a new priority vector solving method for HFLPR by integrating data envelopment analysis (DEA) cross-efficiency and a distance-based priority aggregation (DPA) model. The innovation of this method contains: (1) DEA model is introduced to solve the priority vector of linguistic preference relation (LPR); (2) a DPA model is constructed to obtain the HFLPR’s priority vector, which can avoid the loss of decision-making information. For completely additive consistent LPR and incompletely additive consistent LPR, an output-oriented DEA model and DEA cross-efficiency model are introduced to derive its priority vector, respectively. An HFLPR is viewed as being composed of many LPRs, a DPA model is constructed based on the L2 norm to find the priority vector of HFLPR that minimizes the distance among the LPRs’ priority vectors. Based on the above, an integrated group decision-making method is proposed and then applied to an illustrative example of the evaluation factors affecting resilient cities to show its performance and advantages by comparing with the existing methods.

Appendix A

Proof of Theorem 1:

Based on Eqs. (1) and (4), for $\forall i,j = 1,2, \cdots ,n$, we have.

$$ e_{ij} + e_{ji} = \frac{{4\Delta^{ - 1} (l_{ij} )}}{5g} + \frac{{4\Delta^{ - 1} (l_{ji} )}}{5g} + 0.1 + 0.1 $$

$$ = \frac{{4\Delta^{ - 1} (l_{ij} ) + 4\Delta^{ - 1} (l_{ji} )}}{5g} + 0.2 $$

$$ = 1 $$

On the other hand, when $i = j$, we have

$$ e_{ii} = \frac{{4\Delta^{ - 1} (l_{ii} )}}{5g} + 0.1 = 0.5. $$

According to Eqs. (2) and (4), it follows that

$$ e_{ij} + e_{jk} = \frac{{4\Delta^{ - 1} (l_{ij} )}}{5g} + \frac{{4\Delta^{ - 1} (l_{jk} )}}{5g} + 0.1 + 0.1 $$

$$ = \frac{{4\Delta^{ - 1} (l_{ik} ) + 2g}}{5g} + 0.1 + 0.1 $$

$$ = \frac{{4\Delta^{ - 1} (l_{ik} )}}{5g} + 0.1 + 0.5 $$

$$ = e_{ik} + 0.5. $$

Based on the above analyses, the proof of the theorem is complete.

Proof of Theorem 2:

According to Eq. (5), model (M-1) is equivalent to following model:

$$ \begin{array}{*{20}l} {(M - 2)\max \theta_{m} } \hfill \\ {s.t.\left\{ {\begin{array}{*{20}l} {\sum\nolimits_{i = 1}^{n} {u_{i} (\omega_{i} - \omega_{j} + 1)} \ge \theta_{m} (\omega_{m} - \omega_{j} + 1),} \hfill \\ {\sum\nolimits_{i = 1}^{n} {u_{i} } \le 1,} \hfill \\ {u_{i} \ge 0,} \hfill \\ {i = 1,2, \cdots ,n.} \hfill \\ \end{array} } \right.} \hfill \\ \end{array} $$

Without loss of generality, assuming that the priority vector satisfies $\omega_{1} \le \omega_{2} \le \cdots \le \omega_{n}$, one can obtain $\omega_{i} - \omega_{j} + 1 \ge 0$ and $\omega_{m} - \omega_{j} + 1 \ge 0$. According to the second constraint condition of model (M-2), when it holds, the objective function $\theta_{m}$ reaches the maximum value.

$\forall j = 1, \cdots ,n$, $0 \le \omega_{1} - \omega_{j} + 1 \le \omega_{2} - \omega_{j} + 1 \le \cdots \le \omega_{n} - \omega_{j} + 1$ holds. Based on the first constraint condition of model (M-2), the optimal solutions are $u_{1}^{*} = u_{2}^{*} = \cdots u_{n - 1}^{*} = 0$, $u_{n}^{*} = 1$. Additionally, the first constraint condition is equivalent to

$$ \frac{{\omega_{n} - \omega_{j} + 1}}{{\omega_{m} - \omega_{j} + 1}} \ge \theta_{m} . $$

Then, the optimal solution of model (M-2) can be obtained as follows:

$$ \theta_{m}^{*} = \mathop {\min }\limits_{{j \in \{ 1, \cdots ,n\} }} \left\{ {\frac{{\omega_{n} - \omega_{j} + 1}}{{\omega_{m} - \omega_{j} + 1}}} \right\} = \frac{1}{{\omega_{m} - \omega_{n} + 1}}. $$

Then, the efficiency of alternative $x_{m}$ is

$$ \frac{1}{{\theta_{m}^{*} }} = \omega_{m} - \omega_{n} + 1 $$

Furthermore, since $\sum\nolimits_{i = 1}^{n} {\omega_{i} } = 1$, it follows that

$$ \sum\limits_{m = 1}^{n} {\frac{1}{{\theta_{m}^{*} }}} = \sum\limits_{m = 1}^{n} {(\omega_{m} - \omega_{n} + 1)} = n + 1 - n\omega_{n} $$

In line with the above equation, we have

$$ \omega_{n} = \frac{n + 1}{n} - \frac{1}{n}\sum\limits_{m = 1}^{n} {\frac{1}{{\theta_{m}^{*} }}} $$

Based on $\frac{1}{{\theta_{m}^{*} }}$ and $\omega_{n}$, we also have

$$ \omega_{m} = \frac{1}{{\theta_{m}^{*} }} + \frac{1}{n} - \frac{1}{n}\sum\limits_{m = 1}^{n} {\frac{1}{{\theta_{m}^{*} }}} ,m = 1,2, \cdots ,n - 1 $$

This theorem is proven.

Proof of Theorem 3:

Since $\gamma_{i}$, $\omega_{i}^{k} \in [0,1]$, we have $0 \le \left( {\gamma_{i} - \omega_{i}^{k} } \right)^{2} \le 1$. It follows that.

$$ 0 \le \left( {\sum\limits_{i = 1}^{n} {(\gamma_{i} - \omega_{i}^{k} )^{2} } } \right) \le 1 $$

Additionally, we have $\sum\nolimits_{k = 1}^{\aleph } {\lambda_{k} } = 1$. Then, we have

$$ 0 \le \sum\limits_{k = 1}^{\aleph } {\lambda_{k} \left( {\sum\limits_{i = 1}^{n} {(\gamma_{i} - \omega_{i}^{k} )^{2} } } \right)} \le 1 $$

This completes the proof of the theorem.