﻿ 基于前景理论的不确定TOPSIS多属性决策方法
 计算机系统应用  2019, Vol. 28 Issue (3): 36-42 PDF

Prospect Theory-Based TOPSIS Method for Multiple Attribute Decision Making with Uncertainty
LIANG Wei, WANG Ying-Ming
Decision Sciences Institute, Fuzhou University, Fuzhou 350116, China
Foundation item: National Natural Science Foundation of China (61773123)
Abstract: For the multi-attribute decision making problem with attribute weights being unknown and attribute values being hesitant fuzzy sets, this study proposes a multi-attribute decision making method based on prospect theory and rough set, which fully considers the influence of decision makers’ psychological risk factors on decision results. Firstly, the positive and negative ideal points are used as reference points to calculate the prospect value function under each attribute and to define a new comprehensive prospect value, and a discernibility matrix is obtained according to the given threshold. Then, according to the discernibility matrix, attribute reduction is performed to determine the attribute weight. Finally, the weighted comprehensive prospect value of each alternative is calculated, and the TOPSIS method is used to rank all the alternatives. An example is given to illustrate the feasibility and effectiveness of the proposed method.
Key words: hesitant fuzzy sets     prospect theory     rough sets     multiple attribute decision making

1 基础理论 1.1 粗糙集的基本知识

RA上的一个等价关系, 即 $\left[ x \right]_B^R = \left\{ y \in S\left| \left( f\left( {x,{\rm{b}}} \right)\right.\right.\right.,$ $\left. \left. f\left( {y,b} \right) \right) \in R,\forall b \in B \right\}$ , 令 ${R_B} = \left\{ {\left[ x \right]_B^R:x \in S} \right\}$ , ${R_B}$ 为对象集S中的所有等价类集合.

$B \subseteq C$ , 则集合X关于B的下近似集和上近似集为:

 $\underline R \left( X \right) = \left\{ {x \in X:\left[ x \right]_B^R \subseteq X} \right\}$
 $\bar R\left( X \right) = \left\{ {x \in X:\left[ x \right]_B^R \cap X \ne \emptyset } \right\}$

 ${r_B}\left( X \right) = \frac{{\left| {\underline R \left( X \right)} \right|}}{{\left| X \right|}}$

$\left( {S,C,A,f} \right)$ 为一个信息系统, ${c_j} \in C$ , 若RC = ${R_{C - \left\{ {{c_j}} \right\}}}$ , 则称属性 ${c_j}$ 在属性集 $C$ 中为冗余属性, 否则属性 ${c_j}$ 在属性集 $C$ 中即为必要属性. 而将冗余属性去除的过程, 称为属性约简. 属性集C中的冗余属性可能不止一个, 所有必要属性所构成的集合称为属性集C的核, 记为 $core\left( C \right)$ .

1.2 犹豫模糊集的基本知识

1.3 前景理论的基本知识

 $v(\Delta x) = \left\{ \begin{array}{l} \Delta {x^\alpha },\Delta x \ge 0\\ - \theta {\left( { - \Delta x} \right)^\beta },\Delta x < 0 \end{array} \right.$

 $v\left( {{h_1}} \right) = \left\{ \begin{array}{l} {\left( {d\left( {{h_1},{h_2}} \right)} \right)^\alpha },{h_1} \ge {h_2}\\ - \theta {\left( {d\left( {{h_1},{h_2}} \right)} \right)^\beta },{h_1} < {h_2} \end{array} \right.$
2 前景理论下犹豫模糊TOPSIS决策方法 2.1 问题描述

2.2 决策方法

Step 1. 构造样本数据的犹豫模糊决策矩阵 $X$ , 首先, 对犹豫模糊数内的所有元素以递增的顺序排列, 将元素个数相对较少的犹豫模糊数按Xu等[20]提出的拓展规则进行拓展, 使所有犹豫模糊集都具有相同的元素个数. 然后为了消除不同量纲对决策结果的影响, 对成本型属性按Zhu[21]提出的方法转化为效益型属性, 即:

 ${x_{ij}} = x_{ij}^c = H\left\{ {1 - \gamma _{ij}^1,1 - \gamma _{ij}^2, \cdots ,1 - \gamma _{ij}^l} \right\}$ (1)

Step 2. 确定属性 ${c_j}$ 的正、负理想点 $x_j^ +$ $x_j^ -$ .

 $x_j^ + = \left\{ {\mathop {\max }\limits_{i = 1}^n \left\langle {x_{ij}^\lambda } \right\rangle } \right\} = \left\{ {{{\left( {x_{ij}^1} \right)}^ + },{{\left( {x_{ij}^2} \right)}^ + }, \cdots ,{{\left( {x_{ij}^l} \right)}^ + }} \right\}$ (2)
 $x_j^ - = \left\{ {\mathop {\min }\limits_{i = 1}^n \left\langle {x_{ij}^\lambda } \right\rangle } \right\} = \left\{ {{{\left( {x_{ij}^1} \right)}^ - },{{\left( {x_{ij}^2} \right)}^ - }, \cdots ,{{\left( {x_{ij}^l} \right)}^ - }} \right\}$ (3)

Step 3. 计算方案 $s_i$ 在属性 ${c_j}$ 下的属性值 ${x_{ij}}$ 分别到 $x_j^ +$ $x_j^ -$ 的距离, 即:

 $d_{ij}^ + = d(x{}_{ij},x_j^ + )$ (4)
 $d_{ij}^ - = d(x{}_{ij},x_j^ - )$ (5)

Step 4. 计算方案 $s_i$ 在各属性下的前景价值函数. 根据前景理论中价值函数的概念可得, 当参考点为正理想解时, 则所有方案相对于正理想解而言, 都是损失的; 而当参考点为负理想解时, 则所有方案相对于负理想解而言, 都是收益的, 则:

 ${v^ - }\left( {d\left( {{x_{ij,}}x_j^ + } \right)} \right) = - \theta {\left( {d\left( {{x_{ij,}}x_j^ + } \right)} \right)^\beta }$ (6)
 ${v^ + }\left( {d\left( {{x_{ij,}}x_j^ - } \right)} \right) = {\left( {d\left( {{x_{ij,}}x_j^ - } \right)} \right)^\alpha }$ (7)

Step 5. 计算方案 $s_i$ 在属性 ${c_j}$ 的综合前景值.

 ${v_{ij}} = \dfrac{{\left| {{v^ + }\left( {d\left( {{x_{ij}},x_j^ - } \right)} \right)} \right|}}{{\left| {\displaystyle\sum\limits_{j = 1}^m {{v^ - }\left( {d\left( {{x_{ij}},x_j^ + } \right)} \right)} } \right|}}$ (8)

Step 6. 利用粗糙集理论进行属性约简, 同时确定属性 ${c_j}$ 的权重 ${w_j}$ .

 ${k_{ij}} = \left\{ \begin{array}{l} 1,{v_{ij \le }} \ge \omega \\ 0,{v_{ij}} < \omega \end{array} \right.$ (9)

S关于B的下近似集定义为:

 $\underline {S} = \left\{ {{s_i}:\left[ {{s_i}} \right]_B^R \subseteq \left[ {{s_i}} \right]_C^R,i \in N} \right\},$

 ${r_B}\left( S \right) = \frac{{\left| {\underline {S} } \right|}}{{\left| S \right|}}$

 ${w_j} = \dfrac{{1 - {r_{core\left( C \right) - \left\{ {{c_j}} \right\}}}\left( S \right)}}{{\displaystyle\sum\limits_{{c_j} \in core\left( C \right)} {\left[ {1 - {r_{core\left( C \right) - \left\{ {{c_j}} \right\}}}\left( S \right)} \right]} }}$ (10)

(1) ${w_l} = 0,{c_l} \in C - core\left( C \right)$

(2) ${w_j} \in \left[ {0,1} \right]$ , 且 $\displaystyle\sum\limits_{{c_j} \in core\left( C \right)} {{w_j} = 1}$

Step 7. 由式(10)所得出的属性权重, 计算各方案的加权综合前景值:

 ${T_i} = \sum\limits_{j = 1}^m {{w_j}{v_{ij}}}$ (11)

3 算例分析 3.1 问题描述

3.2 计算过程

$x_1^ + = \left\{ {0.5,0.8,0.9} \right\}$ , $x_1^ - = \left\{ {0.2,0.3,0.6} \right\}$

$x_2^ + = \left\{ {0.5,0.7,0.8} \right\}$ , $x_2^ - = \left\{ {0.1,0.1,0.4} \right\}$

$x_3^ + = \left\{ {0.4,0.7,0.8} \right\}$ , $x_3^ - = \left\{ {0.1,0.1,0.3} \right\}$

$x_4^ + = \left\{ {0.3,0.4,0.7,0.8} \right\}$ , $x_4^ - = \left\{ {0.1,0.1,0.2,0.3} \right\}$

$x_5^{\rm{ + }} = \left\{ {0.3,0.7,0.8} \right\}$ , $x_5^ - = \left\{ {0.1,0.1,0.3} \right\}$

 ${D^ + } = \left[ {\begin{array}{*{20}{r}} {0.1915}&{0.1000}&{0.1291}&{0.1323}&{0.4655} \\ {0.2646}&{0.1732}&{0.3000}&{0.0707}&{0.2944} \\ {0.0577}&{0.0577}&{0.3873}&{0.3122}&{0.2380} \\ {0.3559}&{0.4761}&{0.2708}&{0.3969}&{0.2646} \\ {0.1414}&{0.3697}&{0.0000}&{0.2179}&{0.0000} \\ {0.2517}&{0.0000}&{0.0816}&{0.0707}&{0.2000} \\ {0.2000}&{0.3697}&{0.3464}&{0.1323}&{0.3416} \\ {0.0577}&{0.4082}&{0.4830}&{0.0000}&{0.1915} \end{array}} \right]$
 ${D^ - } = \left[ {\begin{array}{*{20}{r}} {0.2000}&{0.3786}&{0.3873}&{0.3000}&{0.0000} \\ {0.1414}&{0.4203}&{0.2082}&{0.3708}&{0.1732} \\ {0.4082}&{0.4509}&{0.1000}&{0.1225}&{0.2828} \\ {0.0577}&{0.0000}&{0.2160}&{0.0000}&{0.2582} \\ {0.2646}&{0.1291}&{0.4830}&{0.2000}&{0.4655} \\ {0.1633}&{0.4761}&{0.4082}&{0.3354}&{0.2887} \\ {0.1915}&{0.1291}&{0.1414}&{0.3000}&{0.1414} \\ {0.3559}&{0.0816}&{0.0000}&{0.3969}&{0.2944} \end{array}} \right]$

 ${v^ - } = \left[ {\begin{array}{*{20}{r}} { - 0.5254}&\!\!\!{ - 0.2966}&\!\!\!{ - 0.3714}&\!\!\!{ - 0.3794}&\!\!\!{ - 1.1480} \\ { - 0.6983}&\!\!\!{ - 0.4810}&\!\!\!{ - 0.7799}&\!\!\!{ - 0.2186}&\!\!\!{ - 0.7671} \\ { - 0.1829}&\!\!\!{ - 0.1829}&\!\!\!{ - 0.9765}&\!\!\!{ - 0.8079}&\!\!\!{ - 0.6363} \\ { - 0.9065}&\!\!\!{ - 1.1710}&\!\!\!{ - 0.7127}&\!\!\!{ - 0.9977}&\!\!\!{ - 0.6983} \\ { - 0.4024}&\!\!\!{ - 0.9373}&\!\!\!{0.0000}&\!\!\!{ - 0.5887}&\!\!\!{0.0000} \\ { - 0.6682}&\!\!\!{0.0000}&\!\!\!{ - 0.2481}&\!\!\!{ - 0.2186}&\!\!\!{ - 0.5459} \\ { - 0.5459}&\!\!\!{ - 0.9373}&\!\!\!{ - 0.8852}&\!\!\!{ - 0.3794}&\!\!\!{ - 0.8743} \\ { - 0.1829}&\!\!\!{ - 1.0228}&\!\!\!{ - 1.1860}&\!\!\!{0.0000}&\!\!\!{ - 0.5254} \end{array}} \right]$
 ${v^ + } = \left[ {\begin{array}{*{20}{r}} {0.2426}&{0.4254}&{0.4340}&{0.3466}&{0.0000} \\ {0.1788}&{0.4664}&{0.2513}&{0.4177}&{0.2138} \\ {0.4546}&{0.4961}&{0.1318}&{0.1576}&{0.3291} \\ {0.0813}&{0.0000}&{0.2596}&{0.0000}&{0.3038} \\ {0.3103}&{0.1650}&{0.5271}&{0.2426}&{0.5102} \\ {0.2030}&{0.5204}&{0.4546}&{0.3824}&{0.3351} \\ {0.2335}&{0.1650}&{0.1788}&{0.3466}&{0.1788} \\ {0.4029}&{0.1103}&{0.0000}&{0.4434}&{0.3409} \end{array}} \right]$

 $V = \left[ {\begin{array}{*{20}{r}} {0.0892}&{0.1564}&{0.1595}&{0.1274}&{0.0000} \\ {0.0607}&{0.1584}&{0.0853}&{0.1418}&{0.0726} \\ {0.1631}&{0.1781}&{0.0473}&{0.0565}&{0.1181} \\ {0.0181}&{0.0000}&{0.0579}&{0.0000}&{0.0677} \\ {0.1609}&{0.0856}&{0.2733}&{0.1258}&{0.2646} \\ {0.1208}&{0.3096}&{0.2704}&{0.2275}&{0.1994} \\ {0.0645}&{0.0456}&{0.0494}&{0.0957}&{0.0494} \\ {0.1381}&{0.0378}&{0.0000}&{0.1520}&{0.1169} \end{array}} \right]$

 $K = {\left[ {\begin{array}{*{20}{r}} 1&0&1&0&1&1&0&1 \\ 1&1&1&0&0&1&0&0 \\ 1&0&0&0&1&1&0&0 \\ 1&1&0&0&1&1&1&1 \\ 0&0&1&0&1&1&0&1 \end{array}} \right]^{\rm T}}$

 ${R_C} = \left\{ {\left\{ {{s_1}} \right\},\left\{ {{s_2}} \right\},\left\{ {{s_3}} \right\},\left\{ {{s_4}} \right\},\left\{ {{s_5}} \right\},\left\{ {{s_6}} \right\},\left\{ {{s_7}} \right\},\left\{ {{s_8}} \right\}} \right\}$
 ${R_{C - \left\{ {{c_1}} \right\}}} = \left\{ {\left\{ {{s_1}} \right\},\left\{ {{s_2}} \right\},\left\{ {{s_3}} \right\},\left\{ {{s_4}} \right\},\left\{ {{s_5}} \right\},\left\{ {{s_6}} \right\},\left\{ {{s_7}} \right\},\left\{ {{s_8}} \right\}} \right\}$
 ${R_{C - \left\{ {{c_2}} \right\}}} = \left\{ {\left\{ {{s_1}} \right\},\left\{ {{s_2},{s_7}} \right\},\left\{ {{s_3}} \right\},\left\{ {{s_4}} \right\},\left\{ {{s_5},{s_6}} \right\},\left\{ {{s_8}} \right\}} \right\}$
 ${R_{C - \left\{ {{c_3}} \right\}}} = \left\{ {\left\{ {{s_1}} \right\},\left\{ {{s_2}} \right\},\left\{ {{s_3}} \right\},\left\{ {{s_4}} \right\},\left\{ {{s_5},{s_8}} \right\},\left\{ {{s_6}} \right\},\left\{ {{s_7}} \right\}} \right\}$
 ${R_{C - \left\{ {{c_4}} \right\}}} = \left\{ {\left\{ {{s_1}} \right\},\left\{ {{s_2}} \right\},\left\{ {{s_3}} \right\},\left\{ {{s_4},{s_7}} \right\},\left\{ {{s_5}} \right\},\left\{ {{s_6}} \right\},\left\{ {{s_8}} \right\}} \right\}$
 ${R_{C - \left\{ {{c_5}} \right\}}} = \left\{ {\left\{ {{s_1},{s_6}} \right\},\left\{ {{s_2}} \right\},\left\{ {{s_3}} \right\},\left\{ {{s_4}} \right\},\left\{ {{s_5}} \right\},\left\{ {{s_7}} \right\},\left\{ {{s_8}} \right\}} \right\}$
 ${R_{C - \left\{ {{c_1},{c_2}} \right\}}} = \left\{ {\left\{ {{s_1}} \right\},\left\{ {{s_2},{s_7}} \right\},\left\{ {{s_3}} \right\},\left\{ {{s_4}} \right\},\left\{ {{s_5},{s_6}} \right\},\left\{ {{s_8}} \right\}} \right\}$
 ${R_{C - \left\{ {{c_1},{c_3}} \right\}}} = \left\{ {\left\{ {{s_1},{s_2}} \right\},\left\{ {{s_3}} \right\},\left\{ {{s_4}} \right\},\left\{ {{s_5},{s_8}} \right\},\left\{ {{s_6}} \right\},\left\{ {{s_7}} \right\}} \right\}$
 ${R_{C - \left\{ {{c_1},{c_4}} \right\}}} = \left\{ {\left\{ {{s_1}} \right\},\left\{ {{s_2}} \right\},\left\{ {{s_3}} \right\},\left\{ {{s_4},{s_7}} \right\},\left\{ {{s_5}} \right\},\left\{ {{s_6}} \right\},\left\{ {{s_8}} \right\}} \right\}$
 ${R_{C - \left\{ {{c_1},{c_5}} \right\}}} = \left\{ {\left\{ {{s_1},{s_6}} \right\},\left\{ {{s_2}} \right\},\left\{ {{s_3}} \right\},\left\{ {{s_4}} \right\},\left\{ {{s_5}} \right\},\left\{ {{s_7},{s_8}} \right\}} \right\}$

${r_{core\left( C \right) - \left\{ {{c_2}} \right\}}}\left( S \right) = \dfrac{4}{8}$ , ${r_{core\left( C \right) - \left\{ {{c_3}} \right\}}}\left( S \right) = \dfrac{4}{8}$

${r_{core\left( C \right) - \left\{ {{c_4}} \right\}}}\left( S \right) = \frac{6}{8}$ , ${r_{core\left( C \right) - \left\{ {{c_5}} \right\}}}\left( S \right) = \frac{4}{8}$

${w_2} = \dfrac{{1 - \dfrac{4}{8}}}{{\left( {1 - \dfrac{4}{8}} \right) + \left( {1 - \dfrac{4}{8}} \right) + \left( {1 - \dfrac{6}{8}} \right) + \left( {1 - \dfrac{4}{8}} \right)}} = \dfrac{2}{9}$ , 同理可得, ${w_3} = \dfrac{2}{9}$ , ${w_4} = \dfrac{3}{9}$ , ${w_5} = \dfrac{2}{9}$ .

${T_1} = 0.0779$ , ${T_2} = 0.0824$ ,

${T_3} = 0.0556$ , ${T_4} = 0.0279$ ,

${T_5} = 0.1615$ , ${T_6} = 0.1802$ ,

${T_7} = 0.0538$ , ${T_8} = 0.0766$ .

3.3 比较分析

(1) 文献[14]利用犹豫模糊熵确定属性权重, 计算本文算例, 求得属性权重为 $W = \left( {0.1970,0.1955,0.1910,}\right.$ $\left.{ 0.2095,0.2071} \right)$ . 然后通过计算方案的收益损失比值 ${C_i}$ 对各备选方案进行排序, 结果如下所示:

${C_1} = 1.0548$ , ${C_2} = 1.0967$ , ${C_3} = 0.8541$ ,

${C_4} = 0.2436$ , ${C_5} = 1.1248$ , ${C_6} = 2.1033$ ,

${C_7} = 0.7020$ , ${C_8} = 0.8700$ .

(2) 文献[22]首先利用最大偏差法确定属性的权重 $W = \left( {0.1774,0.2038,\;0.1963,0.2414,0.1469} \right)$ , 再分别计算方案 ${s_i}$ 到正、负理想解的距离, 同时通过距离可以得到各备选方案的相对贴近度 $CI({s_i})$ :

$CI({s_1}) = - 0.9197$ , $CI({s_3}) = - 1.1813$ ,

$CI({s_3}) = - 1.1232$ , $CI({s_4}) = - 3.1385$ ,

$CI({s_5}) = - 0.7105$ , $CI({s_6}) = 0$ ,

$CI({s_7}) = - 2.0168$ , $CI({s_8}) = - 1.5093$ .

4 结论

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