Determining attribute weights to improve solution reliability and its application to selecting leading industries

Abstract

In multiple attribute decision analysis, many methods have been proposed to determine attribute weights. However, solution reliability is rarely considered in those methods. This paper develops an objective method in the context of the evidential reasoning approach to determine attribute weights which achieve high solution reliability. Firstly, the minimal satisfaction indicator of each alternative on each attribute is constructed using the performance data of each alternative. Secondly, the concept of superior intensity of an alternative is introduced and constructed using the minimal satisfaction of each alternative. Thirdly, the concept of solution reliability on each attribute is defined as the ordered weighted averaging (OWA) of the superior intensity of each alternative. Fourthly, to calculate the solution reliability on each attribute, the methods for determining the weights of the OWA operator are developed based on the minimax disparity method. Then, each attribute weight is calculated by letting it be proportional to the solution reliability on that attribute. A problem of selecting leading industries is investigated to demonstrate the applicability and validity of the proposed method. Finally, the proposed method is compared with other four methods using the problem, which demonstrates the high solution reliability of the proposed method.

Appendix: Proof of Property 1 and Theorems 1–4

1.1 Proof of Property 1

Property 1

Given $\Delta V(e_i (b_l ))$ ($i = 1, {\ldots }, L, l = 1, {\ldots }, M-1)$ in Eq. (18), it is satisfied that

$$\begin{aligned}&\Delta V(e_i (b_1 ))\ge \cdots \ge \Delta V(e_i (b_{M-1} )), \end{aligned}$$

(20)

$$\begin{aligned}&\Delta V(e_i (b_x ))= \cdots =\Delta V(e_i (b_y )) \end{aligned}$$

$$\begin{aligned}&\hbox {if }V(e_i (b_x ))= \cdots =V(e_i (b_y )) \quad \hbox { for }\quad x<y \quad \hbox { and }x,y\in \{1, \ldots ,M-1\}, \end{aligned}$$

(21)

$$\begin{aligned}&0\le \Delta V(e_i (b_l ))\le M-l, \quad \hbox {and} \end{aligned}$$

(22)

$$\begin{aligned}&0\le Q\left( {e_i } \right) \le M-1. \end{aligned}$$

(23)

Proof

From the condition of $V(e_{i}(b_{1}))$ $\ge {\cdots } \ge V(e_{i}(b_{M}))$ specified in Definition 4, we can deduce that $(V(e_i (b_l ))-V(e_i (b_m )))/2 \ge$ 0 for $m=l$ + 1, ..., $M$. Using Eq. (18), we can further reason that $\Delta V(e_i (b_1 ))=(V(e_i (b_1 ))-V(e_i (b_2 )))/2+\sum _{m=3}^M {(V(e_i (b_1 ))-V(e_i (b_m )))/2} =(V(e_i (b_1 ))-V(e_i (b_2 )))/2+\Delta V(e_i (b_2 )) \ge \Delta V(e_i (b_2 ))$. Similarly, $\Delta V(e_i (b_2 )) \ge$ ... $\ge \Delta V(e_i (b_{M-1} ))$ can be inferred. Therefore, Eq. (20) holds.

Given $x$ and $y$ such that $x < y$ and $x$, $y \in$ {1, ..., $M-1$}, because $V(e_{i}(b_{x})) = V(e_{i}(b_{x+1}))$, we can deduce from Eq. (18) that $\Delta V(e_i (b_x ))=(V(e_i (b_x ))-V(e_i (b_{x+1} )))/2+\sum _{m=x+2}^M {(V(e_i (b_x ))-V(e_i (b_m )))/2} =\sum _{m=x+2}^M {(V(e_i (b_{x+1} ))-V(e_i (b_m )))/2} =\Delta V(e_i (b_{x+1} ))$. Similarly, we can infer from $V(e_{i}(b_{x})) = {\ldots } = V(e_{i}(b_{y}))$ that $\Delta V(e_i (b_x )) = {\ldots } = \Delta V(e_i (b_y ))$, which verifies Eq. (21).

Because $V(e_{i}(b_{l}))$ $\in$ [$-$1, 1] ($i = 1, {\ldots }, L, l = 1, {\ldots }, M-1)$, as presented in Sect. 3.1, $(V(e_i (b_l ))-V(e_i (b_m )))/2$ for $m=l + 1, {\ldots }, M$ is limited to [0, 1]. This can deduce from Eq. (18) that $\Delta V(e_i (b_l ))$ ($l = 1, {\ldots }, M-1)$ is limited to [0, $M-l$] ($l = 1, {\ldots }, M-1)$. As a result, Eq. (22) holds.

Equations (20) and (22) show that $M-1 \ge \Delta V(e_i (b_1 )) \ge$ ... $\ge \Delta V(e_i (b_{M-1} )) \ge$ 0. In this situation, it can be derived that $\sum _{l=1}^{M-1} {\theta _l \cdot \Delta V(e_i (b_l ))} \ge$ 0 and $\sum _{l=1}^{M-1} {\theta _l \cdot \Delta V(e_i (b_l ))} \le \sum _{l=1}^{M-1} {\theta _l \cdot \Delta V(e_i (b_1 ))} =\Delta V(e_i (b_1 )) \le M-1$ on the condition that 0 $\le$ $\theta _{l} \le$ 1 ($l = 1, {\ldots }, M-1)$ and $\sum _{l=1}^{M-1} {\theta _l }$=1, as specified in Definition 4. Therefore, Eq. (23) holds. $\square$

1.2 Proof of Theorem 1

Theorem 1

It is assumed that $\theta _{l}$ ($l = 1, {\ldots }, M-1)$ represents the weight of $\Delta V(e_i (b_l ))$ ($i = 1, {\ldots }, L, l = 1, {\ldots }, M-1)$ for calculating $Q(e_{i})$ in Eq. (19) and $\Delta d$ is defined in Eq. (25). Given the largest weight $\theta _{1}$, $\theta _{l}$ ($l = 2, {\ldots }, M-1)$ is determined by $\theta _1 -(l-1)d_1 +\frac{(l-2)\cdot (l-1)}{2}\Delta d$ using the minimax disparity method based on Assumption 1, i.e., minimizing the maximum disparity between $\theta _{l}$ and $\theta _{l+1}$ ($l = 1, {\ldots }, M-2$), where $d_{1}=\frac{4(M-1)\theta _1 -6}{(M-2)(M-1)}+\varepsilon$, $\Delta d=\frac{6(M-1)\theta _1 -12}{(M-3)(M-2)(M-1)}+\frac{3\varepsilon }{M-3}$, and $\varepsilon$ is a small positive number close to zero.

Proof

The requirement of $(\theta _l -\theta _{l+1} )-(\theta _{l+1} -\theta _{l+2} )=d_l -d_{l+1} =\Delta d >$ 0 ($l = 1, {\ldots }, M-3$) in Assumption 1 can deduce that $\theta _{l}=\theta _1 -(l-1)d_1 +(1+{\ldots }+(l-2))\Delta d=\theta _1 -(l-1)d_1 +\frac{(l-2)\cdot (l-1)}{2}\Delta d$ ($l = 2, {\ldots }, M-1)$. Then, we have

$$\begin{aligned} \sum \limits _{l=1}^{M-1} {\theta _l }&= (M-1)\theta _1 -(1+{\ldots }+(M-1-1))d_1 +\sum \limits _{l=3}^{M-1} {\frac{(l-2)\cdot (l-1)}{2}\Delta d} \\&= (M-1)\theta _1 -\frac{(M-2)\cdot (M-1)}{2}d_1 +\,\frac{1}{2}\sum \limits _{l=3}^{M-1} {(l^{2}-3l+2)\Delta d} \\&= (M-1)\theta _1 -\frac{(M-2)\cdot (M-1)}{2}d_1 \\&\quad +\,\frac{1}{2}\left( \frac{1}{6}(M-1)(M-1+1)(2(M-1)+1)-(1^{2}+2^{2})\right. \\&\quad -\,3\left. \left( \frac{1}{2}(M-1)M-(1+2)\right) +2(M-1-3+1)\right) \Delta d \\&= (M-1)\theta _1 -\frac{(M-2)(M-1)}{2}d_1 +\frac{(M-3)(M-2)(M-1)}{6}\Delta d \\&= 1, \end{aligned}$$

which can further reason that $\Delta d=\frac{6-6(M-1)\theta _1 +3(M-2)(M-1)d_1 }{(M-3)(M-2)(M-1)}$.

The constraint of $\theta _{1} >$ ... $>$ $\theta _{M-1} >$ 0 in Assumption 1 needs that $\theta _{M-1}=\theta _1 -(M-2)d_1 +\frac{(M-3)\cdot (M-2)}{2}\Delta d >$ 0, which is combined with $\Delta d=\frac{6-6(M-1)\theta _1 +3(M-2)(M-1)d_1 }{(M-3)(M-2)(M-1)}$ to deduce that 6-6($M-1)\theta _{1}$+3($M-2$)($M-1)d_{1} >$ 2($M-2$)($M-1)d_{1}$-2($M-1)\theta _{1}$, i.e., $d_{1} > \frac{4(M-1)\theta _1 -6}{(M-2)(M-1)}$. Following the principle of the minimax disparity method, $d_{l}$ ($l = 1, {\ldots }, M-2$) should be minimized. Due to the requirement of $d_l -d_{l+1} =\Delta d >$ 0 ($l = 1, {\ldots }, M-3$) in Assumption 1, the minimum $d_{1}$ results in the minimum $d_{l}$ ($l = 2, {\ldots }, M-2$). Therefore, we obtain the minimal $d_{1}=\frac{4(M-1)\theta _1 -6}{(M-2)(M-1)}+\varepsilon$ with the help of a small positive number $\varepsilon$ close to zero. In this situation, it can be derived from $\Delta d=\frac{6-6(M-1)\theta _1 +3(M-2)(M-1)d_1 }{(M-3)(M-2)(M-1)}$ that $\Delta d=\frac{6(M-1)\theta _1 -12}{(M-3)(M-2)(M-1)}+\frac{3\varepsilon }{M-3}$. $\square$

1.3 Proof of Theorem 2

Theorem 2

On the assumption that $\theta _{l}$ ($l = 1, {\ldots }, M-1)$ represents the weight of $\Delta V(e_i (b_l ))$ ($i = 1, {\ldots }, L, l = 1, {\ldots }, M-1)$ for calculating $Q(e_{i})$ in Eq. (19), the determination of $\theta _{l}$ ($l = 2, {\ldots }, M-1)$ in Theorem 1 based on Assumption 1 requires that $\frac{4-(M-2)(M-1)\varepsilon }{2(M-1)} <$ $\theta _{1} < \frac{3-(M-2)(M-1)\varepsilon }{M-1}$ where $\varepsilon$ is a small positive number close to zero.

Proof

The determination of $\theta _{l}$ ($l = 1, {\ldots }, M-1)$ by Theorem 1 depends on Assumption 1, so the constraints in Eqs. (24)–(26) are satisfied when $\theta _{1}$ is given.

By using Theorem 1, we can know that $d_{1}=\frac{4(M-1)\theta _1 -6}{(M-2)(M-1)}+\varepsilon$ and $\Delta d=\frac{6(M-1)\theta _1 -12}{(M-3)(M-2)(M-1)}+\frac{3\varepsilon }{M-3}$ where $\varepsilon$ is a small positive number close to zero. The requirements of $d_{1} >$ 0 and $\Delta d >$ 0 in Eqs. (24)–(26) deduce that 4($M-1)\theta _{1}$-6+($M-2$)($M-1)\varepsilon$ $>$ 0 and 6($M-1)\theta _{1}$-12+3($M-2$)($M-1)\varepsilon$ $>$ 0, i.e., $\theta _{1} > \frac{6-(M-2)(M-1)\varepsilon }{4(M-1)}$ and $\theta _{1} > \frac{4-(M-2)(M-1)\varepsilon }{2(M-1)}$. As $\frac{6-(M-2)(M-1)\varepsilon }{4(M-1)}<\frac{4-(M-2)(M-1)\varepsilon }{2(M-1)}$ when $\varepsilon$ is sufficiently close to zero, we can obtain that $\theta _{1} > \frac{4-(M-2)(M-1)\varepsilon }{2(M-1)}$. On the other hand, it can be derived from $d_l -d_{l+1} =\Delta d >$ 0 ($l = 2, {\ldots }, M-3$) in Eq. (25) that $d_{l}=d_1 -(l-1)\Delta d >$ 0 ($l = 2, {\ldots }, M-2$), i.e., $d_{1} > (l-1)\Delta d$ ($l = 2, {\ldots }, M-2$). When $d_{M-2}=d_1 -(M-3)\Delta d >$ 0, $d_{1} > (l-1)\Delta d$ ($l = 2, {\ldots }, M-2$) is clearly satisfied. From $d_{M-2} >$ 0, we can obtain that $d_{1}=\frac{4(M-1)\theta _1 -6}{(M-2)(M-1)}+\varepsilon > (M-3)\Delta d=\frac{6(M-1)\theta _1 -12+3(M-2)(M-1)\varepsilon }{(M-2)(M-1)}$, i.e., $\theta _{1} < \frac{3-(M-2)(M-1)\varepsilon }{M-1}$. Therefore, $\theta _{1}$ is limited to ($\frac{4-(M-2)(M-1)\varepsilon }{2(M-1)}$, $\frac{3-(M-2)(M-1)\varepsilon }{M-1})$. $\square$

1.4 Proof of Theorem 3

Theorem 3

Let $\theta _{l}$ ($l = 1, {\ldots }, M-1)$ denote the weight of $\Delta V(e_i (b_l ))$ ($i = 1, {\ldots }, L, l = 1, {\ldots }, M-1)$ for $Q(e_{i})$ in Eq. (19). Given orness degree $\alpha$ such that 0.5 $<$ $\alpha$ $<$ 1, $\theta _{1}$ and $\theta _{l}$ ($l = 2, {\ldots }, M-1)$ can be determined by $\frac{(24\alpha -12)(M-2)+(M-2)(M-1)M\varepsilon }{2(M-1)M}$ and $\theta _1 -(l-1)d_1 +\frac{(l-2)\cdot (l-1)}{2}\Delta d$ ($l = 2, {\ldots }, M-1)$, where $d_{1}=\frac{2(24\alpha -12)(M-2)-6M}{(M-2)(M-1)M}+3\varepsilon$, $\Delta d=\frac{3(24\alpha -12)(M-2)-12M}{(M-3)(M-2)(M-1)M}+\frac{6\varepsilon }{M-3}$, and $\varepsilon$ is a small positive number close to zero.

Proof

According to Eq. (1), the orness degree for $Q(e_{i})$ in Eq. (19) is calculated as

$$\begin{aligned} \alpha&= \frac{1}{(M-1)-1}\sum \limits _{l=1}^{M-1} {(M-1-l)\theta _l } \\&= \frac{1}{M-2}((M-2)\theta _1 +(M-3)(\theta _1 -d_1 )+\sum \limits _{l=3}^{M-1} {(M-1-l)\theta _l } ). \end{aligned}$$

Theorem 1 indicates that $\theta _{l}=\theta _1 -(l-1)d_1 +\frac{(l-2)\cdot (l-1)}{2}\Delta d$ ($l = 2, {\ldots }, M-1)$, with which we can obtain that

$$\begin{aligned} \alpha&= \frac{1}{M-2}((M-2)\theta _1 +(M-3)(\theta _1 -d_1 )+(M-3)(M-1)\theta _1 -\left( \frac{(M-1)M}{2}-3\right) \theta _1 \\&\quad +\left( \frac{1}{6}(M-1)M(2M-1)-5-M\left( \frac{(M-1)M}{2}-3\right) +(M-3)(M-1)\right) d_1\\&\quad +\sum \limits _{l=3}^{M-1} {\frac{1}{2}(M-1-l)(l-2)(l-1)\Delta d)} . \end{aligned}$$

Suppose that $\alpha _{1} = (\frac{1}{6}(M-1)M(2M-1)-5-M(\frac{(M-1)M}{2}-3)+(M-3)(M-1))d_{1}$ and $\alpha _{2}=\sum _{l=3}^{M-1} {\frac{1}{2}(M-1-l)(l-2)(l-1)\Delta d}$, then it is derived that

$$\begin{aligned} \alpha _{1}&= \left( \frac{1}{6}(M-1)M(2M-1)-\frac{1}{2}(M-1)M^{2}+(M-3)(M-1)+3M-5\right) d_1 \\&= \left( \frac{1}{3}M^{3}-\frac{1}{6}M^{2}-\frac{1}{3}M^{2}+\frac{1}{6}M-\frac{1}{2}M^{3}+\frac{3}{2}M^{2}-M-2\right) d_1 \\&= -\frac{1}{6}(M^{3}-6M^{2}+5M+12)d_1 \\&= -\frac{1}{6}(M(M-3)^{2}-4(M-3))d_1 \\&= -\frac{1}{6}(M-3)(M^{2}-3M-4)d_1 \\&= -\frac{1}{6}(M-3)(M-4)(M+1)d_1 \\ \end{aligned}$$

and

$$\begin{aligned} \alpha _{2}&= \frac{\Delta d}{2}\sum \limits _{l=3}^{M-1} {(M-1-l)(l^{2}-3l+2)} \\&= \frac{\Delta d}{2}\sum \limits _{l=3}^{M-1} {((M-1)l^{2}-3l(M-1)+2(M-1)-l^{3}+3l^{2}-2l)} \\&= \frac{\Delta d}{2}\sum \limits _{l=3}^{M-1} {(-l^{3}+(M+2)l^{2}+(1-3M)l+2(M-1))} \\&= \frac{\Delta d}{2}\left( -\left( \frac{(M-1)^{2}M^{2}}{4}-1-2^{3}\right) +(M-2)\left( \frac{(M-1)M(2M-1)}{6}\right. -1-2^{2}\right) \\&\left. +(1-3M)\left( \frac{1}{2}(M-1)M-1-2\right) +2(M-1)(M-1-3+1)\right) \\&= \frac{\Delta d}{2}\Bigg (-\frac{(M-1)^{2}M^{2}}{4}+\frac{(M+2)(M-1)M(2M-1)}{6}+\frac{(1-3M)(M-1)M}{2}\\&\quad +\,2(M-1)(M-3)+(4M-4)\Bigg )\\&= \frac{\Delta d}{2}(M-1)\left( -\frac{(M-1)M^{2}}{4}+\frac{(M+2)M(2M-1)}{6}+\frac{(1-3M)M}{2}+2M-2\right) \\&= \frac{\Delta d}{2}(M-1)\left( \frac{(M-1)}{4}(8-M^{2})+\frac{M}{6}\left( (M+2)(2M-1)+3(1-3M)\right) \right) \\&= \frac{\Delta d}{24}(M-1)(3(M-1)(8-M^{2})+2M(2M^{2}-6M+1))\\&= \frac{\Delta d}{24}(M-1)(M^{3}-9M^{2}+26M-24)\\&= \frac{\Delta d}{24}(M-4)(M-3)(M-2)(M-1). \end{aligned}$$

Thus, we have $\alpha = f(\alpha _{1}, \alpha _{2})=\frac{1}{M-2}((M-2)\theta _1 +(M-3)(\theta _1 -d_1 )+(M-3)(M-1)\theta _1 -(\frac{(M-1)M}{2}-3)\theta _1 + \alpha _{1} + \alpha _{2})=\frac{1}{M-2}(\frac{1}{2}(M-2)(M-1)\theta _1 -\frac{1}{6}(M-3)(M-2)(M-1)d_1 +\frac{1}{24}(M-4)(M-3)(M-2)(M-1)\Delta d)$, which can deduce that

$$\begin{aligned} \theta _{1}=\frac{24\alpha +4(M-3)(M-1)d_1 -(M-4)(M-3)(M-1)\Delta d}{12(M-1)}. \end{aligned}$$

From Theorem 1, we can know that $d_{1}=\frac{4(M-1)\theta _1 -6}{(M-2)(M-1)}+\varepsilon$ and $\Delta d=\frac{6(M-1)\theta _1 -12}{(M-3)(M-2)(M-1)}+\frac{3\varepsilon }{M-3}$. Then, $\theta _{1}$ is calculated as $\frac{(24\alpha -12)(M-2)+(M-2)(M-1)M\varepsilon }{2(M-1)M}$, which is used to determine $d_{1}$ and $\Delta d$ with $d_{1}=\frac{2(24\alpha -12)(M-2)-6M}{(M-2)(M-1)M}+3\varepsilon$ and $\Delta d=\frac{3(24\alpha -12)(M-2)-12M}{(M-3)(M-2)(M-1)M}+\frac{6\varepsilon }{M-3}$. $\square$

1.5 Proof of Theorem 4

Theorem 4

On the assumption that $\theta _{l}$ ($l = 1, {\ldots }, M-1)$ represents the weight of $\Delta V(e_i (b_l ))$ ($i = 1, {\ldots }, L, l = 1, {\ldots }, M-1)$ for calculating $Q(e_{i})$ in Eq. (19), the determination of $\theta _{l}$ ($l = 2, {\ldots }, M-1)$ in Theorem 3 based on Assumption 1 requires that $\frac{8M-12-(M-2)(M-1)M\varepsilon }{12M-24} <$ $\alpha$ $< \frac{6M-8-3(M-2)(M-1)M\varepsilon }{8M-16}$ where $\alpha$ is the orness degree for $Q(e_{i})$ and $\varepsilon$ is a small positive number close to zero.

Proof

Theorems 2 and 3 show that $\frac{4-(M-2)(M-1)\varepsilon }{2(M-1)} <$ $\theta _{1} < \frac{3-(M-2)(M-1)\varepsilon }{M-1}$ and $\theta _{1}=\frac{(24\alpha -12)(M-2)+(M-2)(M-1)M\varepsilon }{2(M-1)M}$.

The inequality $\frac{4-(M-2)(M-1)\varepsilon }{2(M-1)} <$ $\theta _{1}$ can deduce that 24($M-2$)$\alpha$ - 12($M-2$) + ($M-2$)($M-1)M\varepsilon > 4M - (M-2)(M-1)M\varepsilon$, i.e., $\alpha$ $> \frac{8M-12-(M-2)(M-1)M\varepsilon }{12M-24}$. Similarly, from $\theta _{1}=\frac{(24\alpha -12)(M-2)+(M-2)(M-1)M\varepsilon }{2(M-1)M} < \frac{3-(M-2)(M-1)\varepsilon }{M-1}$ it can be derived that 24($M-2)\alpha$ - 12($M-2) + (M-2)(M-1)M \varepsilon < 6M$ - 2($M-2)(M-1)M\varepsilon$, i.e., $\alpha$ $< \frac{6M-8-3(M-2)(M-1)M\varepsilon }{8M-16}$. In addition, we clearly have $\frac{8M-12-(M-2)(M-1)M\varepsilon }{12M-24} >$ 0.5 and $\frac{6M-8-3(M-2)(M-1)M\varepsilon }{8M-16} <$ 1. Therefore, $\alpha$ is limited to ($\frac{8M-12-(M-2)(M-1)M\varepsilon }{12M-24}$, $\frac{6M-8-3(M-2)(M-1)M\varepsilon }{8M-16})$. $\square$