New MULTIMOORA and Pairwise Evaluation-Based MCDM Methods for Hotel Selection Based on the Projection Measure of Z-Numbers

Abstract

The problem of hotel selection involves lots of uncertain information and multiple factors and can be identified as a multi-criteria decision-making problem. Aiming at the information description, measure, and fusion of hotel selection problems, this study develops multi-criteria decision-making methods based on Z-numbers and their projection measure and fusion techniques. To manage the three-dimensional structure of Z-numbers effectively, an optimization model is introduced to determine the potential probability distributions involved in Z-numbers. Then, the module, inner product, and cosine of Z-numbers are defined, and the projection measure of Z-numbers is presented by dealing with the three-dimensional structure of Z-numbers directly. Moreover, some Z-number Choquet integral projection operators are proposed for fusing Z-number evaluation information. Subsequently, an improved Multi-Objective Optimization by Ratio Analysis plus the Full Multiplicative Form method and an innovative pairwise evaluation-based multi-criteria decision-making method are developed. Based on the ideas of content-based recommendation and collaborative filtering recommendation, the above two methods are applied to solve hotel selection problems. And the sensitivity analysis and comparison discussion are conducted to demonstrate the applicability and validity of the two methods.

Appendix

1.1 Proof of Theorem 1

Proof

(1) As described in Definition 3, $a_{{ij}} \in R$, $b_{{il}} \in [0,\;1]$, $\mu (a_{{ij}} ) \in [0,\;1]$ and $\mu (b_{{il}} ) \in [0,\;1]$. Moreover, $1 - F_{{il}} (a_{{ij}} ) \in [0,\;1]$ is obvious. Then, $0 \le Z_{1} \cdot Z_{2}$ and $0 \le \left| {Z_{1} } \right|\left| {Z_{2} } \right|$ can be easily determined.

Moreover, according to the Cauchy–Schwarz inequality:

$(x_{1} y_{1} + x_{2} y_{2} + \cdots + x_{n} y_{n} )^{2} \le (x_{1}^{2} + x_{2}^{2} + \cdots + x_{n}^{2} )(y_{1}^{2} + y_{2}^{2} + \cdots + y_{n}^{2} )$,

the following inequality can be deduced:

$$\begin{aligned} & \left( {\sum\limits_{{j = 1}}^{n} {(a_{{1j}} \mu (a_{{1j}} ))(a_{{2j}} \mu (a_{{2j}} ))} + \sum\limits_{{l = 1}}^{m} {(b_{{1l}} \mu (b_{{1l}} ))(b_{{2l}} \mu (b_{{2l}} ))} + \sum\limits_{{j = 1}}^{n} {(a_{{1j}} (1 - F_{{1l}} (a_{{1j}} )))(a_{{2j}} (1 - F_{{2l}} (a_{{2j}} )))} } \right)^{2} \\ & \quad \le \left( {\sum\limits_{{j = 1}}^{n} {(a_{{1j}} \mu (a_{{1j}} ))^{2} } + \sum\limits_{{l = 1}}^{m} {(b_{{1l}} \mu (b_{{1l}} ))^{2} } + \sum\limits_{{j = 1}}^{n} {(a_{{1j}} (1 - F_{{1l}} (a_{{1j}} )))^{2} } } \right)\left( {\sum\limits_{{j = 1}}^{n} {(a_{{2j}} \mu (a_{{2j}} ))^{2} } + \sum\limits_{{l = 1}}^{m} {(b_{{2l}} \mu (b_{{2l}} ))^{2} } + \sum\limits_{{j = 1}}^{n} {(a_{{2j}} (1 - F_{{2l}} (a_{{2j}} )))^{2} } } \right). \\ \end{aligned}$$

Therefore, $Z_{1} \cdot Z_{2} \le \left| {Z_{1} } \right|\left| {Z_{2} } \right|$ can be determined. According to the above obtained results, $0 \le \text{Cos} (\alpha (Z_{1} ,\;Z_{2} )) \le 1$ can be proved. In addition, it is easy to prove that (2) and (3) in Theorem 1 are true.

1.2 Proof of Theorem 2

Proof

(1) As described in Theorem 1, $0 \le \text{Cos} (\alpha (Z_{1} ,\;Z_{2} )) \le 1$. Therefore, $0 \le P_{{Z_{2} }} (Z_{1} ) = \left| {Z_{1} } \right|\text{Cos} (\alpha (Z_{1} ,\;Z_{2} )) \le \left| {Z_{1} } \right|$ can be easily proved.

(2) If $Z_{1} = Z_{2}$, then $\text{Cos} (\alpha (Z_{1} ,\;Z_{2} )) = 1$ can be obtained from Theorem 1, and $\left| {Z_{1} } \right| = \left| {Z_{2} } \right|$ can also be determined. Therefore, if $Z_{1} = Z_{2}$, then $P_{{Z_{2} }} (Z_{1} ) = \left| {Z_{1} } \right| = \left| {Z_{2} } \right| = P_{{Z_{1} }} (Z_{2} )$.

(3) If $Z_{1} \le Z_{2}$, then $a_{{1j}} \le a_{{2j}} ,$ $\mu (a_{{1j}} ) \le \mu (a_{{2j}} ),$ $b_{{1l}} \le b_{{2l}} ,$ $\mu (b_{{1l}} ) \le \mu (b_{{2l}} )$ and $F_{{1l}} (a_{{1j}} ) \ge F_{{2l}} (a_{{2j}} )$ can be obtained from Definition 6. Then, the following inequality can be deduced:

$$\begin{aligned} & Z_{1} \cdot Z_{3} = \sum\limits_{{j = 1}}^{n} {(a_{{1j}} \mu (a_{{1j}} ))(a_{{3j}} \mu (a_{{3j}} ))} + \sum\limits_{{l = 1}}^{m} {(b_{{1l}} \mu (b_{{1l}} ))(b_{{3l}} \mu (b_{{3l}} ))} + \sum\limits_{{j = 1}}^{n} {(a_{{1j}} (1 - F_{{1l}} (a_{{1j}} )))(a_{{3j}} (1 - F_{{3l}} (a_{{3j}} )))} \\ & \quad \le Z_{2} \cdot Z_{3} = \sum\limits_{{j = 1}}^{n} {(a_{{2j}} \mu (a_{{2j}} ))(a_{{3j}} \mu (a_{{3j}} ))} + \sum\limits_{{l = 1}}^{m} {(b_{{2l}} \mu (b_{{2l}} ))(b_{{3l}} \mu (b_{{3l}} ))} + \sum\limits_{{j = 1}}^{n} {(a_{{2j}} (1 - F_{{2l}} (a_{{2j}} )))(a_{{3j}} (1 - F_{{3l}} (a_{{3j}} )))} . \\ \end{aligned}$$

Therefore, $P_{{Z_{3} }} (Z_{1} ) = \frac{{Z_{1} \cdot Z_{3} }}{{\left| {Z_{3} } \right|}} \le P_{{Z_{3} }} (Z_{2} ) = \frac{{Z_{2} \cdot Z_{3} }}{{\left| {Z_{3} } \right|}}$ can be proved.

1.3 Proof of Theorem 3

Proof

(1) According to Theorem 2, the following inequalities can be deduced:

$0 \le P_{{Z_{2} }} (Z_{1} ) \le P_{{Z_{2} }} (Z_{1} ) + \left| {\left| {Z_{2} } \right| - P_{{Z_{2} }} (Z_{1} )} \right| \Rightarrow 0 \le \frac{{P_{{Z_{2} }} (Z_{1} )}}{{P_{{Z_{2} }} (Z_{1} ) + \left| {\left| {Z_{2} } \right| - P_{{Z_{2} }} (Z_{1} )} \right|}} \le 1 \Rightarrow 0 \le {\text{NP}}_{{Z_{2} }} (Z_{1} ) \le 1$.

(2) If $Z_{1} = Z_{2}$, then $P_{{Z_{2} }} (Z_{1} ) = \left| {Z_{1} } \right| = \left| {Z_{2} } \right|$ can be obtained from Theorem 2. Therefore, the following equalities can be deduced:

${\text{NP}}_{{Z_{2} }} (Z_{1} ) = \frac{{P_{{Z_{2} }} (Z_{1} )}}{{P_{{Z_{2} }} (Z_{1} ) + \left| {\left| {Z_{2} } \right| - P_{{Z_{2} }} (Z_{1} )} \right|}} = \frac{{P_{{Z_{2} }} (Z_{1} )}}{{P_{{Z_{2} }} (Z_{1} ) + \left| {\left| {Z_{2} } \right| - \left| {Z_{2} } \right|} \right|}} = \frac{{P_{{Z_{2} }} (Z_{1} )}}{{P_{{Z_{2} }} (Z_{1} )}} = 1$.

(3) If $Z_{1} \le Z_{2} \le Z_{3}$, then $a_{{1j}} \le a_{{2j}} \le a_{{3j}} ,$ $\mu (a_{{1j}} ) \le \mu (a_{{2j}} ) \le \mu (a_{{3j}} ),$ $b_{{1l}} \le b_{{2l}} \le b_{{3l}} ,$ $\mu (b_{{1l}} ) \le \mu (b_{{2l}} ) \le \mu (b_{{3l}} )$ and $F_{{1l}} (a_{{1j}} ) \ge F_{{2l}} (a_{{2j}} ) \ge F_{{3l}} (a_{{3j}} )$ can be obtained from Definition 6. Therefore, $\left| {Z_{1} } \right| \le \left| {Z_{2} } \right| \le \left| {Z_{3} } \right|$ can be determined. Then, based on Theorem 2, $0 \le P_{{Z_{3} }} (Z_{1} ) \le \left| {Z_{1} } \right| \le \left| {Z_{3} } \right|$ and $0 \le P_{{Z_{3} }} (Z_{2} ) \le \left| {Z_{2} } \right| \le \left| {Z_{3} } \right|$ can be obtained. Hence, there are

$${\text{NP}}_{{Z_{3} }} (Z_{1} ) = \frac{{P_{{Z_{3} }} (Z_{1} )}}{{P_{{Z_{3} }} (Z_{1} ) + \left| {Z_{3} } \right| - P_{{Z_{3} }} (Z_{1} )}} = \frac{{P_{{Z_{3} }} (Z_{1} )}}{{\left| {Z_{3} } \right|}}\quad {\text{and}}\quad {\text{NP}}_{{Z_{3} }} (Z_{2} ) = \frac{{P_{{Z_{3} }} (Z_{2} )}}{{P_{{Z_{3} }} (Z_{2} ) + \left| {Z_{3} } \right| - P_{{Z_{3} }} (Z_{2} )}} = \frac{{P_{{Z_{3} }} (Z_{2} )}}{{\left| {Z_{3} } \right|}}.$$

Moreover, when $Z_{1} \le Z_{2}$, $P_{{Z_{3} }} (Z_{1} ) \le P_{{Z_{3} }} (Z_{2} )$ can be obtained from Theorem 2. Therefore, ${\text{NP}}_{{Z_{3} }} (Z_{1} ) = \frac{{P_{{Z_{3} }} (Z_{1} )}}{{\left| {Z_{3} } \right|}} \le {\text{NP}}_{{Z_{3} }} (Z_{2} ) = \frac{{P_{{Z_{3} }} (Z_{2} )}}{{\left| {Z_{3} } \right|}}$ can be proved.

1.4 Proof of Theorem 7

Proof

If ${\text{NP}}_{{Z^{\prime}_{1} }} (Z_{1} ) = {\text{NP}}_{{Z^{\prime}_{2} }} (Z_{2} ) = \cdots = {\text{NP}}_{{Z^{\prime}_{s} }} (Z_{s} )$, then ${\text{NP}}_{{Z^{\prime}_{1} }}^{{(1)}} (Z_{1} ) = {\text{NP}}_{{Z^{\prime}_{2} }}^{{(2)}} (Z_{2} ) = \cdots = {\text{NP}}_{{Z^{\prime}_{s} }}^{{(s)}} (Z_{s} ) = {\text{NP}}_{{Z^{\prime}_{i} }} (Z_{i} )$. Therefore, there are

$$\begin{aligned} & {\text{GZCIPA}}((Z_{1} ,\;Z^{\prime}_{1} ),\;(Z_{2} ,\;Z^{\prime}_{2} ), \ldots ,(Z_{s} ,\;Z^{\prime}_{s} )) \\ & \quad = \left( {\sum\limits_{{i = 1}}^{s} {(\pi (C_{{(i)}} ) - \pi (C_{{(i + 1)}} ))\left( {{\text{NP}}_{{Z^{\prime}_{i} }}^{{(i)}} (Z_{i} )} \right)^{\lambda } } } \right)^{{\frac{1}{\lambda }}} = \left( {\sum\limits_{{i = 1}}^{s} {(\pi (C_{{(i)}} ) - \pi (C_{{(i + 1)}} ))\left( {{\text{NP}}_{{Z^{\prime}_{i} }} (Z_{i} )} \right)^{\lambda } } } \right)^{{\frac{1}{\lambda }}} \\ & \quad = \left( {((\pi (C_{{(1)}} ) - \pi (C_{{(2)}} )) + (\pi (C_{{(2)}} ) - \pi (C_{{(3)}} )) + \cdots + (\pi (C_{{(n)}} ) - \pi (C_{{(n + 1)}} )))\left( {{\text{NP}}_{{Z^{\prime}_{i} }} (Z_{i} )} \right)^{\lambda } } \right)^{{\frac{1}{\lambda }}} \\ & \quad = \left( {(\pi (C_{{(1)}} ) - \pi (C_{{(n + 1)}} ))\left( {{\text{NP}}_{{Z^{\prime}_{i} }} (Z_{i} )} \right)^{\lambda } } \right)^{{\frac{1}{\lambda }}} = \left( {({\text{NP}}_{{Z^{\prime}_{i} }} (Z_{i} ))^{\lambda } } \right)^{{\frac{1}{\lambda }}} = {\text{NP}}_{{Z^{\prime}_{i} }} (Z_{i} ). \\ \end{aligned}$$

1.5 Proof of Theorem 8

Proof

If $Z_{i} \le Z^{\prime}_{i} \le Z^{\prime\prime}_{i}$, then ${\text{NP}}_{{Z^{\prime\prime}_{i} }} (Z_{i} ) \le {\text{NP}}_{{Z^{\prime\prime}_{i} }} (Z^{\prime}_{i} )$ can be obtained from Theorem 3. Then, the following inequalities can be deduced:

$$\begin{aligned} & {\text{NP}}_{{Z^{\prime\prime}_{i} }} (Z_{i} ) \le {\text{NP}}_{{Z^{\prime\prime}_{i} }} (Z^{\prime}_{i} ) \Rightarrow ({\text{NP}}_{{Z^{\prime\prime}_{i} }} (Z_{i} ))^{\lambda } \le ({\text{NP}}_{{Z^{\prime\prime}_{i} }} (Z^{\prime}_{i} ))^{\lambda } \\ & \quad \Rightarrow (\pi (C_{i} ) - \pi (C_{{i + 1}} ))({\text{NP}}_{{Z^{\prime\prime}_{i} }} (Z_{i} ))^{\lambda } \le (\pi (C_{i} ) - \pi (C_{{i + 1}} ))({\text{NP}}_{{Z^{\prime\prime}_{i} }} (Z^{\prime}_{i} ))^{\lambda } \\ & \quad \Rightarrow \left( {\sum\limits_{{i = 1}}^{s} {(\pi (C_{i} ) - \pi (C_{{i + 1}} ))({\text{NP}}_{{Z^{\prime\prime}_{i} }} (Z_{i} ))^{\lambda } } } \right)^{{\frac{1}{\lambda }}} \le \left( {\sum\limits_{{i = 1}}^{s} {(\pi (C_{i} ) - \pi (C_{{i + 1}} ))({\text{NP}}_{{Z^{\prime\prime}_{i} }} (Z^{\prime}_{i} ))^{\lambda } } } \right)^{{\frac{1}{\lambda }}} . \\ \end{aligned}$$

Since the fuzzy measure $\pi$ is independent of the location of variables, there are

$$\begin{aligned} & (\pi (C_{i} ) - \pi (C_{{i + 1}} ))({\text{NP}}_{{Z^{\prime\prime}_{i} }} (Z_{i} ))^{\lambda } = (\pi (C_{{(i)}} ) - \pi (C_{{(i + 1)}} ))({\text{NP}}_{{Z^{\prime\prime}_{i} }}^{{(i)}} (Z_{i} ))^{\lambda } , \\ & (\pi (C_{i} ) - \pi (C_{{i + 1}} ))({\text{NP}}_{{Z^{\prime\prime}_{i} }} (Z^{\prime}_{i} ))^{\lambda } = (\pi (C_{{(i)}} ) - \pi (C_{{(i + 1)}} ))({\text{NP}}_{{Z^{\prime\prime}_{i} }}^{{(i)}} (Z^{\prime}_{i} ))^{\lambda } . \\ \end{aligned}$$

Therefore, the following inequalities can be deduced:

$$\begin{aligned} & \left( {\sum\limits_{{i = 1}}^{s} {(\pi (C_{i} ) - \pi (C_{{i + 1}} ))({\text{NP}}_{{Z^{\prime\prime}_{i} }} (Z_{i} ))^{\lambda } } } \right)^{{\frac{1}{\lambda }}} \le \left( {\sum\limits_{{i = 1}}^{s} {(\pi (C_{i} ) - \pi (C_{{i + 1}} ))({\text{NP}}_{{Z^{\prime\prime}_{i} }} (Z^{\prime}_{i} ))^{\lambda } } } \right)^{{\frac{1}{\lambda }}} \\ & \quad \Rightarrow \left( {\sum\limits_{{i = 1}}^{s} {(\pi (C_{{(i)}} ) - \pi (C_{{(i + 1)}} ))({\text{NP}}_{{Z^{\prime\prime}_{i} }}^{{(i)}} (Z_{i} ))^{\lambda } } } \right)^{{\frac{1}{\lambda }}} \le \left( {\sum\limits_{{i = 1}}^{s} {(\pi (C_{{(i)}} ) - \pi (C_{{(i + 1)}} ))({\text{NP}}_{{Z^{\prime\prime}_{i} }}^{{(i)}} (Z^{\prime}_{i} ))^{\lambda } } } \right)^{{\frac{1}{\lambda }}} \\ & \quad \Rightarrow {\text{GZCIPA}}((Z_{1} ,\;Z^{\prime\prime}_{1} ),\;(Z_{2} ,\;Z^{\prime\prime}_{2} ), \ldots ,(Z_{s} ,\;Z^{\prime\prime}_{s} )) \le {\text{GZCIPA}}((Z^{\prime}_{1} ,\;Z^{\prime\prime}_{1} ),\;(Z^{\prime}_{2} ,\;Z^{\prime\prime}_{2} ), \ldots ,(Z^{\prime}_{s} ,\;Z^{\prime\prime}_{s} )). \\ \end{aligned}$$

1.6 Proof of Theorem 9

Proof

Since $\mathop {\min }\limits_{i} ({\text{NP}}_{{Z^{\prime}_{i} }} (Z_{i} )) \le {\text{NP}}_{{Z^{\prime}_{i} }}^{{(i)}} (Z_{i} ) \le \mathop {\max }\limits_{i} ({\text{NP}}_{{Z^{\prime}_{i} }} (Z_{i} ))$, the following inequalities can be deduced:

$$\begin{aligned} & \mathop {\min }\limits_{i} ({\text{NP}}_{{Z^{\prime}_{i} }} (Z_{i} )) \le {\text{NP}}_{{Z^{\prime}_{i} }}^{{(i)}} (Z_{i} ) \Rightarrow (\mathop {\min }\limits_{i} ({\text{NP}}_{{Z^{\prime}_{i} }} (Z_{i} )))^{\lambda } \le ({\text{NP}}_{{Z^{\prime}_{i} }}^{{(i)}} (Z_{i} ))^{\lambda } \\ & \quad \Rightarrow (\pi (C_{{(i)}} ) - \pi (C_{{(i + 1)}} ))(\mathop {\min }\limits_{i} ({\text{NP}}_{{Z^{\prime}_{i} }} (Z_{i} )))^{\lambda } \le (\pi (C_{{(i)}} ) - \pi (C_{{(i + 1)}} ))({\text{NP}}_{{Z^{\prime}_{i} }}^{{(i)}} (Z_{i} ))^{\lambda } \\ & \quad \Rightarrow \left( {\sum\limits_{{i = 1}}^{s} {(\pi (C_{{(i)}} ) - \pi (C_{{(i + 1)}} ))(\mathop {\min }\limits_{i} ({\text{NP}}_{{Z^{\prime}_{i} }} (Z_{i} )))^{\lambda } } } \right)^{{\frac{1}{\lambda }}} \le \left( {\sum\limits_{{i = 1}}^{s} {(\pi (C_{{(i)}} ) - \pi (C_{{(i + 1)}} ))({\text{NP}}_{{Z^{\prime}_{i} }}^{{(i)}} (Z_{i} ))^{\lambda } } } \right)^{{\frac{1}{\lambda }}} \\ & \quad \Rightarrow \mathop {\min }\limits_{i} ({\text{NP}}_{{Z^{\prime}_{i} }} (Z_{i} )) \le {\text{GZCIPA}}((Z_{1} ,\;Z^{\prime}_{1} ),\;(Z_{2} ,\;Z^{\prime}_{2} ), \ldots ,(Z_{s} ,\;Z^{\prime}_{s} )). \\ \end{aligned}$$

Similarly, ${\text{GZCIPA}}((Z_{1} ,\;Z^{\prime}_{1} ),\;(Z_{2} ,\;Z^{\prime}_{2} ), \ldots ,(Z_{s} ,\;Z^{\prime}_{s} )) \le \mathop {\max }\limits_{i} ({\text{NP}}_{{Z^{\prime}_{i} }} (Z_{i} ))$ can be deduced. Therefore, Theorem 9 is true.