AECM-VIKOR处理两种不同权重的综合评价求解过程


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流程图



  整个处理分三个部分:

  基于VIKOR的得到期望值S,遗憾值R的计算。由于有两组权重,总共有四列。

  层级拓扑图分析,针对四列的任意组合,可以得到六组对应的对抗哈斯图。

  求出六组妥协值的聚类分布,且假定这六组聚类是等权,直接相加然后通过对抗择优抽取方法,求出最终排序,此处原理认为是无导师学习(自注意力)方式获得解。

原始矩阵如下:


$$ \begin{array}{c|c|c|c|c|c|c}{M_{13 \times12}} &A1 &A2 &B1 &B2 &B3 &B4 &C1 &C2 &C3 &C4 & -D1 & -D2\\ \hline M1 &14.75 &18.25 &15 &10 &112 &13.25 &13 &10.75 &23 &19.5 &29 &9.5\\ \hline M2 &16.25 &12.5 &10.75 &12.75 &158 &14 &32 &15.75 &41 &14.75 &7 &11\\ \hline M3 &21 &14.5 &17.25 &15 &140 &15.5 &19 &20 &32 &16.25 &33 &12.25\\ \hline M4 &9.25 &7.5 &11 &13.5 &94 &18.25 &11 &12.5 &18 &12.5 &14 &13.5\\ \hline M5 &10.5 &11.25 &13.5 &17 &53 &8.5 &4 &13.25 &8 &15 &8 &9\\ \hline M6 &17.25 &15.5 &9.75 &14.25 &56 &12 &10 &9 &13 &11.5 &10 &17.25\\ \hline M7 &15.75 &13.25 &16.5 &12.75 &96 &11.25 &27 &14.25 &38 &14.5 &13 &9.5\\ \hline M8 &6.75 &9.25 &12.25 &11.5 &98 &16 &35 &10.75 &45 &13.25 &13 &15.25\\ \hline M9 &13.5 &12 &14.25 &13.5 &45 &10.25 &5 &12 &11 &10.25 &13 &14.25\\ \hline I &20.5 &20.5 &20.5 &20.5 &120 &20.5 &24 &20.5 &36 &20.5 &0 &0\\ \hline II &13 &13 &13 &13 &80 &13 &16 &13 &24 &13 &11 &6.5\\ \hline III &6.5 &6.5 &6.5 &6.5 &40 &6.5 &8 &6.5 &12 &6.5 &22 &13\\ \hline IV &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &33 &20.5\\ \hline \end{array} $$


  原始矩阵 $ O=[ o_{ij}]_{n \times m}$的特点

  第一、每一列(指标、维度、属性)的性质为如下四大类

  $$\begin{array} {|c|c|c|c|} \hline {类别名称}& {特点} \\ \hline \color{red}{正向指标} &\color{blue}{数值越大越好,数值越小越差。} \\ \hline \color{red}{负向指标} &\color{blue}{数值越大越差,数值越小越好。} \\ \hline \color{red}{振荡性指标} &\color{blue}{数值距离某区段越小越好,距离某区段越大越差。}\\ \hline \color{red}{周期性指标} &\color{blue}{诸如周期性函数 如 sin(2x)} \\ \hline \end{array} $$

  对于振荡性指标与周期性指标,需要先进行转化,转化成正向指标或者负向指标进行处理这样才能保证每一列有严格的可比性。

  第二、每一列(指标、维度、属性)都转化成数值的特征,每一列都是同一量纲。其中描述性的比较需要转化成数值型的比较

  无量纲化式特别注意:为非数值型的属性,则需要转化成数值型的数据进行处理。

  如描述性的属性,即非正规(Non-normal)模糊数,又称为一般性模糊数 (Generalized Fuzzy Numbers)可转化成五分制、十分制、百分制或者特定的数值。该数值可以通过特定的模糊运算进行转化。

  其中五分制是最常见的模糊数值的转化。

$$\begin{array} {|c|c|c|c|} \hline {模糊值} &{5分制}& {表述一}& {表述二} \\ \hline 0.2&1 &\color{red}{非常傻逼} &\color{blue}{垃圾} \\ \hline 0.4&2 &\color{red}{真傻逼} &\color{blue}{有点挫} \\ \hline 0.6&3 &\color{red}{恩} &\color{blue}{还行}\\ \hline 0.8&4 &\color{red}{有点牛逼} &\color{blue}{好} \\ \hline 0.99&5 &\color{red}{真TMD牛逼} &\color{blue}{哇塞} \\ \hline \end{array} $$

  第三、原始矩阵的预处理——在进行规范化之前,原始矩阵的每列必须是正向指标,或者负向指标。

  设某物种最适合的生长的酸碱环境为6.3-7.3区间,偏离此区间成线性的危害。酸碱度分别为8.2,7.1,6.9,5.3,6.1对该物种的危害可进行如下换算。

  $$ \begin{array} {|c|c|c|c|} \hline {PH值} & {计算过程} & {PH值对该生物的危害} \\ \hline {8.2} & \color{red}{8.2-7.3} &\color{blue}{0.9} \\ \hline {7.1} & \color{red}{在区间内为0} &\color{blue}{0} \\ \hline {6.9} &\color{red}{在区间内为0} &\color{blue}{0}\\ \hline {5.3} & \color{red}{6.3-5.3} &\color{blue}{1} \\ \hline {6.1} & \color{red}{6.3-6.1} &\color{blue}{0.2} \\ \hline \end{array} $$

  对于上面的震荡类区间数,周期性数值,一般要转化成正向指标或者负向指标两类

  tips 负向指标在excel处理时候 先敲入空格 再敲入 -号;指标标题尽量不要用中文,用代号即可。指标的方向性极为重要

  第四、虚拟样本,标准刻度,客观标准,行。

  以优秀这个虚拟样本为例,如果存在约定俗成的列,去下限值,比如100分的总分,90到100属于优秀,则优秀该列取值为90;如果无客观标准,可自行定义,注意对于正向指标,优秀的数值必定大于良好的数值


归一方法与归一化矩阵说明


;

  第一、无量纲化、规范化、归一化之间的关系

  无量纲化(nondimensionalize 或者dimensionless)是指通过一个合适的变量替代,将一个涉及物理量的方程的部分或全部的单位移除,以求简化实验或者计算的目的,是科学研究中一种重要的处理思想。

  无量纲化方法选择的标准:

  ① 客观性。无量纲化所选择的转化公式要能够客观地反映指标实际值与指标评价值之间的对应关系。要做到客观性原则,需要评价专家对被评价对象的历史信息做出深入彻底的分析和比较,找出事物发展变化的转折点,才能够确定合适的无量纲化方法。

  ② 可行性。即所选择的无量纲化方法要确保转化的可行性各种方法各有特点,各有千秋,应用时应当加以注意。

  ③ 可操作性。即要确保所选方法具有简便易用的特点,并不是所有的非线性无量纲化方法都比线性无量纲化方法更加精确。评价不是绝对的度量,在不影响被评价对象在评价中的影响程度的前提下,可以使用更为简便的线性无量纲化方法代替曲线关系。

  无量纲化、规范化、归一化是包含关系。无量纲化的概念最广

  归一化方法是指,经过归一化计算后,得到的矩阵,矩阵值都在[0,1]之间。

  第二、常见的六大类归一化方法及注意事项

$$\begin{array} {|c|c|c|c|} \hline {名称} & {正向指标公式} & {负向指标公式} & {说明} \\ \hline 极差法 & n_{ij} = \frac{{o_{ij}-min(o_{j})}}{{max(o_{j})-min(o_{j})}} & n_{ij} = \frac{max(o_{j})-{o_{ij}}}{{max(o_{j})-min(o_{j})}} & \color{red}{最常用} \\ \hline 欧式距离法 & n_{ij} = \frac {o_{ij}} { \sqrt{ {\sum \limits_{i=1}^{n}{o_{ij}^2}} } }= \frac {o_{ij}} { ( {\sum \limits_{i=1}^{n}{o_{ij}^2}} )^{\frac 1 2} } & n_{ij} = 1- \frac {o_{ij}} { \sqrt{ {\sum \limits_{i=1}^{n}{o_{ij}^2}} } }= 1- \frac {o_{ij}} { ( {\sum \limits_{i=1}^{n}{o_{ij}^2}} )^{\frac 1 2} } & \color{red}{ 每一列数据之间差距不宜过大} \\ \hline 均值标准化(Z-score) & \frac{x-\mu }{\sigma } & \frac{\mu-x }{\sigma } & \color{red}{此方法并非归一化方法,需特殊处理} \\ \hline 反三角函数 & \frac{2 \times atan(o_{ij}) }{ \pi} & 1-\frac{2 \times atan(o_{ij}) }{ \pi} & \color{red}{出现零值需要额外处理} \\ \hline 对数压缩数据法 & n_{ij} = \frac{lg({o_{ij}) }}{{ lg (max(o_{j}))}} & n_{ij} = 1- \frac{lg({o_{ij}) }}{{ lg (max(o_{j}))}} & \color{red}{出现零值需要额外处理} \\ \hline sigmoid函数(Logistic函数) & n_{ij} = \frac{1} {1+{e}^{-o_{ij}} } & n_{ij} = 1-\frac{1} {1+{e}^{-o_{ij}} } & \color{red}{机器学习、神经网络等基本用这个} \\ \hline \end{array}$$

  选择归一化方式的时候,一定要对原始矩阵进行预处理。最核心的是每一列的属性一定要是正向指标或者负向指标。每一列数据之间的差异不会出现数量级的差异的时候,一般用极差法来进行归一化。


选择的归一化方法如下


极差法

正向指标公式:$$ n_{ij} = \frac{{o_{ij}-min(o_{j})}}{{max(o_{j})-min(o_{j})}} $$

负向指标公式:$$ n_{ij} = \frac{max(o_{j})-{o_{ij}}}{{max(o_{j})-min(o_{j})}} $$


归一化矩阵相关说明


  规一化矩阵的特点

  第一、归一化矩阵$N$中的指标都是正向指标

  第二、采用极差法得到的归一化矩阵$N$中每一列中的最大值为1,最小值为0。

  第三、归一化矩阵是客观法求权重的基础,熵权法、CRITIC等客观求权重法是在归一化矩阵的基础上求解的。

  第四、归一化矩阵是期望值S,遗憾值R计算的基础,S与R是在归一化矩阵的基础上进行计算的

  归一化矩阵如下

$$ \begin{array}{c|c|c|c|c|c|c}{M_{13 \times12}} &A1 &A2 &B1 &B2 &B3 &B4 &C1 &C2 &C3 &C4 & -D1 & -D2\\ \hline M1 &0.702 &0.89 &0.732 &0.488 &0.709 &0.646 &0.371 &0.524 &0.511 &0.951 &0.121 &0.537\\ \hline M2 &0.774 &0.61 &0.524 &0.622 &1 &0.683 &0.914 &0.768 &0.911 &0.72 &0.788 &0.463\\ \hline M3 &1 &0.707 &0.841 &0.732 &0.886 &0.756 &0.543 &0.976 &0.711 &0.793 &0 &0.402\\ \hline M4 &0.44 &0.366 &0.537 &0.659 &0.595 &0.89 &0.314 &0.61 &0.4 &0.61 &0.576 &0.341\\ \hline M5 &0.5 &0.549 &0.659 &0.829 &0.335 &0.415 &0.114 &0.646 &0.178 &0.732 &0.758 &0.561\\ \hline M6 &0.821 &0.756 &0.476 &0.695 &0.354 &0.585 &0.286 &0.439 &0.289 &0.561 &0.697 &0.159\\ \hline M7 &0.75 &0.646 &0.805 &0.622 &0.608 &0.549 &0.771 &0.695 &0.844 &0.707 &0.606 &0.537\\ \hline M8 &0.321 &0.451 &0.598 &0.561 &0.62 &0.78 &1 &0.524 &1 &0.646 &0.606 &0.256\\ \hline M9 &0.643 &0.585 &0.695 &0.659 &0.285 &0.5 &0.143 &0.585 &0.244 &0.5 &0.606 &0.305\\ \hline I &0.976 &1 &1 &1 &0.759 &1 &0.686 &1 &0.8 &1 &1 &1\\ \hline II &0.619 &0.634 &0.634 &0.634 &0.506 &0.634 &0.457 &0.634 &0.533 &0.634 &0.667 &0.683\\ \hline III &0.31 &0.317 &0.317 &0.317 &0.253 &0.317 &0.229 &0.317 &0.267 &0.317 &0.333 &0.366\\ \hline IV &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0\\ \hline \end{array} $$

正极值点构成
$$ \begin{array}{c|c|c|c|c|c|c}{M_{1 \times12}} &A1 &A2 &B1 &B2 &B3 &B4 &C1 &C2 &C3 &C4 & -D1 & -D2\\ \hline \mathbf{Zone^+} &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1\\ \hline \end{array} $$
负极值点构成
$$ \begin{array}{c|c|c|c|c|c|c}{M_{1 \times12}} &A1 &A2 &B1 &B2 &B3 &B4 &C1 &C2 &C3 &C4 & -D1 & -D2\\ \hline \mathbf{Zone^-} &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0\\ \hline \end{array} $$

求权重方法相关说明


  1、求权重的方法分为主观法,客观法,组合赋权法(如博弈论组合赋权法)

  2、常见的客观法与主观法如表格所示:特别注意,客观法与主观法各有优势,各有特点,不要随便互喷。不能认为主观法只是拍脑袋而鄙视它

  $$ \begin{array} {|c|c|c|c|} \hline {中文名称} & {英文简写} & {简要说明} \\ \hline \color{blue}{变异系数法} & \color{blue}{COV} & \color{blue}{客观} \\ \hline \color{blue}{复相关系数} & \color{blue}{MCC} & \color{blue}{客观} \\ \hline \color{blue}{CRITIC方法} & \color{blue}{CRITIC} & \color{blue}{客观} \\ \hline \color{blue}{熵权法} & \color{blue}{EWM} & \color{blue}{客观} \\ \hline \color{blue}{反熵权法} & \color{blue}{Anti-EWM} & \color{blue}{客观} \\ \hline \color{blue}{主成分分析} & \color{blue}{PCA} & \color{blue}{客观} \\ \hline \color{blue}{因子分析权数法} & \color{blue}{FAM} & \color{blue}{客观} \\ \hline \color{blue}{层次分析法} & \color{blue}{AHP} & \color{red}{主观} \\ \hline \color{blue}{网络分析法} & \color{blue}{ANP} & \color{red}{主观} \\ \hline \color{blue}{决策与实验室方法} & \color{blue}{DEMATEL} & \color{red}{主观} \\ \hline \color{blue}{决策与实验室-网络分析联用方法} & \color{blue}{D-ANP} & \color{red}{主观} \\ \hline \end{array} $$

  3、客观法求得的权重是从归一化矩阵得到,而不是从原始矩阵直接代入公式求出。熵权法中可以允许归一化矩阵中出现0但是不能出现负值,而归一化矩阵中不会有负数。

  4、主观法求得的权重跟归一化矩阵无关,也跟原始矩阵无关。尽量不要用AHP,用滥了,很多老师很鄙视AHP。当然文科生,或者本科生做作业练手用一用还是可以

  5、主观法的适用性有 $ AHP \prec ANP \prec DANP \approx DEMATEL $

  W1采用的是熵权法(EWM)求权重

$$ \begin{array}{c|c|c|c|c|c|c}{M_{2 \times12}} &A1 &A2 &B1 &B2 &B3 &B4 &C1 &C2 &C3 &C4 & -D1 & -D2\\ \hline EWM所得权重 &0.073 &0.0667 &0.0614 &0.0585 &0.0847 &0.0642 &0.1345 &0.065 &0.1091 &0.0607 &0.1283 &0.0938\\ \hline 权重大小顺序 &6 &7 &10 &12 &5 &9 &1 &8 &3 &11 &2 &4\\ \hline \end{array} $$

  W2采用的是DANP法求权重

$$ \begin{array}{c|c|c|c|c|c|c}{M_{2 \times12}} &A1 &A2 &B1 &B2 &B3 &B4 &C1 &C2 &C3 &C4 & -D1 & -D2\\ \hline DANP &0.0765 &0.1207 &0.0781 &0.0598 &0.0728 &0.1243 &0.097 &0.0762 &0.0592 &0.0892 &0.0578 &0.0883\\ \hline 权重大小顺序 &7 &2 &6 &10 &9 &1 &3 &8 &11 &4 &12 &5\\ \hline \end{array} $$

VIKOR的最大化群体效益S和最小化反对意见的个别遗憾R求解原理及相关说明


  闵可夫斯基相关介绍

  闵可夫斯基(Hermann Minkowski,1864-1909),德国数学家,在数论、代数、数学物理和相对论等领域有巨大贡献。他把三维物理空间与时间结合成四维时空(即闵可夫斯基时空)的思想为爱因斯坦的相对论奠定了数学基础。爱因斯坦说闵可夫斯基是他的数学老师。

  闵可夫斯基距离通式为:$({\sum_\limits{i=1}^N|P_i-Q_i|^p})^{\frac{1}{p}}$

$$ \begin{array} {|c|c|c|c|} \hline {距离公式名称} & {对应范数} & {通式} & {说明} \\ \hline 曼哈顿距离 & 1 & ({\sum_\limits{i=1}^N|P_i-Q_i|}) & \color{red}{最常用,常见的求总分方式就是采用此公式,极差法就是此距离公式} \\ \hline 欧几里得距离 & 2 & ({\sum_\limits{i=1}^N|P_i-Q_i|^2})^{\frac{1}{2}} & \color{red}{ TOPSIS采用此距离公式} \\ \hline 切比雪夫距离 & 无穷大 & ({\sum_\limits{i=1}^N|P_i-Q_i|^∞})^{\frac{1}{∞}} & \color{red}{取最大值的方法} \\ \hline \end{array}$$

  VIKOR(VlseKriterijumska Optimizacija I Kompromisno Resenje)是Opricovic(1998)提出一种基于理想解的折中排序方法,通过最大化群体效用和最小化个体遗憾来实现有限备选决策方案的最优排序。其中所谓的最大化群体效用又称为期望值,对应的闵可夫斯基范数为1时候的曼哈顿距离公式;个体遗憾值又称为遗憾值,对应的闵可夫斯基范数为无穷大的切比雪夫距离公式

  一言以蔽之:VIKOR核心就是针对归一化矩阵,通过带权值的范数为1与范数为无穷大闵可夫斯基距离求解出距离

  $$期望值 \quad \quad 闵可夫斯基公式范数为1 \quad \quad({\sum_\limits{i=1}^N|P_i-Q_i|^1})^{\frac{1}{1}} \rightsquigarrow S_i = \sum_\limits{j=1}^m{ \omega_{j} \left(\frac{Max(n_j) -n_{ij}}{Max(n_j) -Min(n_j)} \right)} \quad \quad 曼哈顿距离公式(当前点,到正理想极值点的距离)$$ 

  $$遗憾值 \quad \quad 闵可夫斯基公式范数为无穷大 \quad \quad ({\sum_\limits{i=1}^N|P_i-Q_i|^\infty})^{\frac{1}{\infty}} \rightsquigarrow R_i = \max_\limits{j=1} { \left( \omega_{j} (\frac{Max(n_j) -n_{ij}}{Max(n_j) -Min(n_j)} )\right)} \quad \quad 切比雪夫距离公式(当前点,到正理想极值点的距离)$$ 

  特别的:每个维度(每列)可以采用不同的距离公式。

  特别的:整体可以采用当前点到负理想极值的距离

  特别的:所谓的期望值就是高考,考研中的简单加权求总分。

  极端重要的:求出的每一列一定要判断方向性,即数值越大越优,还是数值越大越差。


当前选择的公式如下:

最大化群体效益
最小化反对意见的个别遗憾
$$ S_i = \sum_\limits{j=1}^m{ \omega_{j} \left(\frac{Max(n_{j}) -n_{ij}}{Max(n_j) -Min(n_j)} \right)} \quad \quad $$ $$ R_i = \max_\limits{j=1} { \left( \omega_{j} (\frac{Max(n_{j}) -n_{ij}}{Max(n_j) -Min(n_j)} )\right)} \quad \quad $$

  计算结果如下

$$D=\begin{array}{c|c|c|c|c|c|c}{M_{13 \times4}} &-期望值S1 &-期望值S2 &-遗憾值R1 &-遗憾值R2\\ \hline M1 &0.45133 &0.37535 &0.11291 &0.06103\\ \hline M2 &0.24533 &0.27974 &0.05037 &0.04743\\ \hline M3 &0.36206 &0.29496 &0.12835 &0.0578\\ \hline M4 &0.49641 &0.46881 &0.09234 &0.07662\\ \hline M5 &0.51234 &0.48947 &0.11927 &0.086\\ \hline M6 &0.51574 &0.48642 &0.09618 &0.07438\\ \hline M7 &0.3178 &0.32905 &0.05061 &0.05614\\ \hline M8 &0.35589 &0.3874 &0.06983 &0.0663\\ \hline M9 &0.55646 &0.52455 &0.11542 &0.08323\\ \hline I &0.08629 &0.06172 &0.04232 &0.03052\\ \hline II &0.40425 &0.39362 &0.0731 &0.05271\\ \hline III &0.70033 &0.69511 &0.10388 &0.08497\\ \hline IV &1 &0.9999 &0.13452 &0.1243\\ \hline \end{array} $$

第二部分,六种组合的基于偏序的对抗哈斯图分析。


  组合数目:这是一个简单的求解,即从n列中随机抽取两列进行组合,其总数如下:

  $Num = \frac {n(n-1)}{2}$

  有4列则有6种组合;有6列就有15种组合;有8列就有28种组合。

  妥协解通式:

  $Q = ( 1-k )f(a) + k f(b)$ 当a与b两列具有相同的方向时。由于初始矩阵具有相同的方向,通常用此方法

  $Q = ( 1-k )f(a) - k f(b)$ 当a与b两列具有不同的方向时。

  f(x)函数的特点与要求

  当k=0时候,$Q=f(a)$ Q的排序必须等于未变换前a列的排序

  当k=1时候,$Q=f(b)$ Q的排序必须等于未变换前b列的排序

  Tips:如果存在着刻度,如优、良、中、差,无论何种几何形变,都有$优 \succ 良 \succ 中 \succ 差$

  Q值哈斯图的特征

  Q的哈斯图定义为:在k等于0到1区间获得的无数列构成的基于偏序的层级拓扑图。

  该哈斯图等于,k=0,k=1时候由两列构成的哈斯图。

  该哈斯图等于,未发生形变时候组成的哈斯图,即初始的a,b两列构成的哈斯图。

    选择的妥协解的公式如下:

公式
$$ Q_i = \left( 1-k \right) \left(\frac{a_i - Min(a_i)}{Max(a_i) -Min(a_i)} \right) + k\left(\frac{b_i - Min(b_i)}{Max(b_i) -Min(b_i)} \right) $$
名称
初始数据
k=0与k=1两列妥协解数据
Q1
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &-S1 &-S2\\ \hline M1 &0.45133 &0.37535\\ \hline M2 &0.24533 &0.27974\\ \hline M3 &0.36206 &0.29496\\ \hline M4 &0.49641 &0.46881\\ \hline M5 &0.51234 &0.48947\\ \hline M6 &0.51574 &0.48642\\ \hline M7 &0.3178 &0.32905\\ \hline M8 &0.35589 &0.3874\\ \hline M9 &0.55646 &0.52455\\ \hline I &0.08629 &0.06172\\ \hline II &0.40425 &0.39362\\ \hline III &0.70033 &0.69511\\ \hline IV &1 &0.9999\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &- k=0 &- k=1\\ \hline M1 &0.39952 &0.3343\\ \hline M2 &0.17405 &0.23239\\ \hline M3 &0.30181 &0.24861\\ \hline M4 &0.44885 &0.43391\\ \hline M5 &0.46629 &0.45594\\ \hline M6 &0.47001 &0.45268\\ \hline M7 &0.25337 &0.28495\\ \hline M8 &0.29506 &0.34714\\ \hline M9 &0.51457 &0.49333\\ \hline I &0 &0\\ \hline II &0.34798 &0.35377\\ \hline III &0.67203 &0.67512\\ \hline IV &1 &1\\ \hline \end{array} $$
Q2
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &-S1 &-R1\\ \hline M1 &0.45133 &0.11291\\ \hline M2 &0.24533 &0.05037\\ \hline M3 &0.36206 &0.12835\\ \hline M4 &0.49641 &0.09234\\ \hline M5 &0.51234 &0.11927\\ \hline M6 &0.51574 &0.09618\\ \hline M7 &0.3178 &0.05061\\ \hline M8 &0.35589 &0.06983\\ \hline M9 &0.55646 &0.11542\\ \hline I &0.08629 &0.04232\\ \hline II &0.40425 &0.0731\\ \hline III &0.70033 &0.10388\\ \hline IV &1 &0.13452\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &- k=0 &- k=1\\ \hline M1 &0.39952 &0.76556\\ \hline M2 &0.17405 &0.08731\\ \hline M3 &0.30181 &0.93307\\ \hline M4 &0.44885 &0.54246\\ \hline M5 &0.46629 &0.83455\\ \hline M6 &0.47001 &0.58419\\ \hline M7 &0.25337 &0.08994\\ \hline M8 &0.29506 &0.29838\\ \hline M9 &0.51457 &0.79282\\ \hline I &0 &0\\ \hline II &0.34798 &0.33382\\ \hline III &0.67203 &0.66764\\ \hline IV &1 &1\\ \hline \end{array} $$
Q3
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &-S1 &-R2\\ \hline M1 &0.45133 &0.06103\\ \hline M2 &0.24533 &0.04743\\ \hline M3 &0.36206 &0.0578\\ \hline M4 &0.49641 &0.07662\\ \hline M5 &0.51234 &0.086\\ \hline M6 &0.51574 &0.07438\\ \hline M7 &0.3178 &0.05614\\ \hline M8 &0.35589 &0.0663\\ \hline M9 &0.55646 &0.08323\\ \hline I &0.08629 &0.03052\\ \hline II &0.40425 &0.05271\\ \hline III &0.70033 &0.08497\\ \hline IV &1 &0.1243\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &- k=0 &- k=1\\ \hline M1 &0.39952 &0.32539\\ \hline M2 &0.17405 &0.18033\\ \hline M3 &0.30181 &0.29092\\ \hline M4 &0.44885 &0.49158\\ \hline M5 &0.46629 &0.59162\\ \hline M6 &0.47001 &0.46766\\ \hline M7 &0.25337 &0.27325\\ \hline M8 &0.29506 &0.3816\\ \hline M9 &0.51457 &0.56204\\ \hline I &0 &0\\ \hline II &0.34798 &0.23665\\ \hline III &0.67203 &0.58066\\ \hline IV &1 &1\\ \hline \end{array} $$
Q4
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &-S2 &-R1\\ \hline M1 &0.37535 &0.11291\\ \hline M2 &0.27974 &0.05037\\ \hline M3 &0.29496 &0.12835\\ \hline M4 &0.46881 &0.09234\\ \hline M5 &0.48947 &0.11927\\ \hline M6 &0.48642 &0.09618\\ \hline M7 &0.32905 &0.05061\\ \hline M8 &0.3874 &0.06983\\ \hline M9 &0.52455 &0.11542\\ \hline I &0.06172 &0.04232\\ \hline II &0.39362 &0.0731\\ \hline III &0.69511 &0.10388\\ \hline IV &0.9999 &0.13452\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &- k=0 &- k=1\\ \hline M1 &0.3343 &0.76556\\ \hline M2 &0.23239 &0.08731\\ \hline M3 &0.24861 &0.93307\\ \hline M4 &0.43391 &0.54246\\ \hline M5 &0.45594 &0.83455\\ \hline M6 &0.45268 &0.58419\\ \hline M7 &0.28495 &0.08994\\ \hline M8 &0.34714 &0.29838\\ \hline M9 &0.49333 &0.79282\\ \hline I &0 &0\\ \hline II &0.35377 &0.33382\\ \hline III &0.67512 &0.66764\\ \hline IV &1 &1\\ \hline \end{array} $$
Q5
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &-S2 &-R2\\ \hline M1 &0.37535 &0.06103\\ \hline M2 &0.27974 &0.04743\\ \hline M3 &0.29496 &0.0578\\ \hline M4 &0.46881 &0.07662\\ \hline M5 &0.48947 &0.086\\ \hline M6 &0.48642 &0.07438\\ \hline M7 &0.32905 &0.05614\\ \hline M8 &0.3874 &0.0663\\ \hline M9 &0.52455 &0.08323\\ \hline I &0.06172 &0.03052\\ \hline II &0.39362 &0.05271\\ \hline III &0.69511 &0.08497\\ \hline IV &0.9999 &0.1243\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &- k=0 &- k=1\\ \hline M1 &0.3343 &0.76556\\ \hline M2 &0.23239 &0.08731\\ \hline M3 &0.24861 &0.93307\\ \hline M4 &0.43391 &0.54246\\ \hline M5 &0.45594 &0.83455\\ \hline M6 &0.45268 &0.58419\\ \hline M7 &0.28495 &0.08994\\ \hline M8 &0.34714 &0.29838\\ \hline M9 &0.49333 &0.79282\\ \hline I &0 &0\\ \hline II &0.35377 &0.33382\\ \hline III &0.67512 &0.66764\\ \hline IV &1 &1\\ \hline \end{array} $$
Q6
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &-R1 &-R2\\ \hline M1 &0.11291 &0.06103\\ \hline M2 &0.05037 &0.04743\\ \hline M3 &0.12835 &0.0578\\ \hline M4 &0.09234 &0.07662\\ \hline M5 &0.11927 &0.086\\ \hline M6 &0.09618 &0.07438\\ \hline M7 &0.05061 &0.05614\\ \hline M8 &0.06983 &0.0663\\ \hline M9 &0.11542 &0.08323\\ \hline I &0.04232 &0.03052\\ \hline II &0.0731 &0.05271\\ \hline III &0.10388 &0.08497\\ \hline IV &0.13452 &0.1243\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &- k=0 &- k=1\\ \hline M1 &0.76556 &0.32539\\ \hline M2 &0.08731 &0.18033\\ \hline M3 &0.93307 &0.29092\\ \hline M4 &0.54246 &0.49158\\ \hline M5 &0.83455 &0.59162\\ \hline M6 &0.58419 &0.46766\\ \hline M7 &0.08994 &0.27325\\ \hline M8 &0.29838 &0.3816\\ \hline M9 &0.79282 &0.56204\\ \hline I &0 &0\\ \hline II &0.33382 &0.23665\\ \hline III &0.66764 &0.58066\\ \hline IV &1 &1\\ \hline \end{array} $$

    六种妥协值的区段截取的排序分析

Q1
Q2
Q3
Q4
Q5
Q6
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &- k0 &- k1\\ \hline M1 &0.4 &0.334\\ \hline M2 &0.174 &0.232\\ \hline M3 &0.302 &0.249\\ \hline M4 &0.449 &0.434\\ \hline M5 &0.466 &0.456\\ \hline M6 &0.47 &0.453\\ \hline M7 &0.253 &0.285\\ \hline M8 &0.295 &0.347\\ \hline M9 &0.515 &0.493\\ \hline I &0 &0\\ \hline II &0.348 &0.354\\ \hline III &0.672 &0.675\\ \hline IV &1 &1\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &- k0 &- k1\\ \hline M1 &0.4 &0.766\\ \hline M2 &0.174 &0.087\\ \hline M3 &0.302 &0.933\\ \hline M4 &0.449 &0.542\\ \hline M5 &0.466 &0.835\\ \hline M6 &0.47 &0.584\\ \hline M7 &0.253 &0.09\\ \hline M8 &0.295 &0.298\\ \hline M9 &0.515 &0.793\\ \hline I &0 &0\\ \hline II &0.348 &0.334\\ \hline III &0.672 &0.668\\ \hline IV &1 &1\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &- k0 &- k1\\ \hline M1 &0.4 &0.325\\ \hline M2 &0.174 &0.18\\ \hline M3 &0.302 &0.291\\ \hline M4 &0.449 &0.492\\ \hline M5 &0.466 &0.592\\ \hline M6 &0.47 &0.468\\ \hline M7 &0.253 &0.273\\ \hline M8 &0.295 &0.382\\ \hline M9 &0.515 &0.562\\ \hline I &0 &0\\ \hline II &0.348 &0.237\\ \hline III &0.672 &0.581\\ \hline IV &1 &1\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &- k0 &- k1\\ \hline M1 &0.334 &0.766\\ \hline M2 &0.232 &0.087\\ \hline M3 &0.249 &0.933\\ \hline M4 &0.434 &0.542\\ \hline M5 &0.456 &0.835\\ \hline M6 &0.453 &0.584\\ \hline M7 &0.285 &0.09\\ \hline M8 &0.347 &0.298\\ \hline M9 &0.493 &0.793\\ \hline I &0 &0\\ \hline II &0.354 &0.334\\ \hline III &0.675 &0.668\\ \hline IV &1 &1\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &- k0 &- k1\\ \hline M1 &0.334 &0.766\\ \hline M2 &0.232 &0.087\\ \hline M3 &0.249 &0.933\\ \hline M4 &0.434 &0.542\\ \hline M5 &0.456 &0.835\\ \hline M6 &0.453 &0.584\\ \hline M7 &0.285 &0.09\\ \hline M8 &0.347 &0.298\\ \hline M9 &0.493 &0.793\\ \hline I &0 &0\\ \hline II &0.354 &0.334\\ \hline III &0.675 &0.668\\ \hline IV &1 &1\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &- k0 &- k1\\ \hline M1 &0.766 &0.325\\ \hline M2 &0.087 &0.18\\ \hline M3 &0.933 &0.291\\ \hline M4 &0.542 &0.492\\ \hline M5 &0.835 &0.592\\ \hline M6 &0.584 &0.468\\ \hline M7 &0.09 &0.273\\ \hline M8 &0.298 &0.382\\ \hline M9 &0.793 &0.562\\ \hline I &0 &0\\ \hline II &0.334 &0.237\\ \hline III &0.668 &0.581\\ \hline IV &1 &1\\ \hline \end{array} $$

第三部分,运用AECM求出最优排序(无导师监督,自注意力方式)


  聚类原理:

  基础矩阵:即任意两列,当决策系数k=0,与决策系数k=1所构成的两列妥协值矩阵。

  妥协解可以写成$ Q_i = \left( 1-k \right) a_i + kb_i \quad \quad $

  拐点问题即变成了求两条线段是否在$[0,1]$值域内有相交的问题

  对于两列矩阵的任意两行$x,y$

  $ \begin{cases} \left( 1-k \right) a_x + kb_x \\ \left( 1-k \right) a_y + kb_y \end{cases} $

  $ k =\frac{a_x- a_y}{( a_x- a_y + b_y - b_x)}$ 其中k必须在$[0,1]$的范围

  聚类分析

  拐点之间其排序是一样的。即进行了聚类

  对抗择优抽取

  对抗择优抽取之前,必须先求解出每一个层级,要素的占比。具体看实例处理


妥协值Q1的最优排序


$$Base=\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &a_i &b_i\\ \hline M1 &0.3995 &0.3343\\ \hline M2 &0.1741 &0.2324\\ \hline M3 &0.3018 &0.2486\\ \hline M4 &0.4488 &0.4339\\ \hline M5 &0.4663 &0.4559\\ \hline M6 &0.47 &0.4527\\ \hline M7 &0.2534 &0.2849\\ \hline M8 &0.2951 &0.3471\\ \hline M9 &0.5146 &0.4933\\ \hline I &0 &0\\ \hline II &0.348 &0.3538\\ \hline III &0.672 &0.6751\\ \hline IV &1 &1\\ \hline \end{array} $$

求解线段在决策区间的交点,k代表决策系数

Q1聚类特征


序号 聚类特征-对应k值区段 Q值排序
10<$k$< 0.064131$I \succ M2 \succ M7 \succ M8 \succ M3 \succ II \succ M1 \succ M4 \succ M5 \succ M6 \succ M9 \succ III \succ IV$
20.064131<$k$< 0.53344$I \succ M2 \succ M7 \succ M3 \succ M8 \succ II \succ M1 \succ M4 \succ M5 \succ M6 \succ M9 \succ III \succ IV$
30.53344<$k$< 0.571407$I \succ M2 \succ M7 \succ M3 \succ M8 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
40.571407<$k$< 0.72572$I \succ M2 \succ M3 \succ M7 \succ M8 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
50.72572<$k$< 0.890527$I \succ M2 \succ M3 \succ M7 \succ M8 \succ M1 \succ II \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
60.890527<$k$< 1$I \succ M2 \succ M3 \succ M7 \succ M1 \succ M8 \succ II \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$

运用AECM对Q1进行求解


排名 要素所占区段
0I=1   
1M2=1   
2M7=0.571407   M3=0.428593   
3M3=0.507276   M7=0.428593   M8=0.064131   
4M8=0.826396   M1=0.109473   M3=0.064131   
5II=0.72572   M1=0.164807   M8=0.109473   
6M1=0.72572   II=0.27428   
7M4=1   
8M5=0.53344   M6=0.46656   
9M6=0.53344   M5=0.46656   
10M9=1   
11III=1   
12IV=1   

上表为针对Q1每个评价对象(要素)在每个层级所占的份额。

排名 最优妥协解
优胜情境$I \succ M2 \succ M7 \succ M3 \succ M8 \succ II \succ M1 \succ M4 \succ M5 \succ M6 \succ M9 \succ III \succ IV$
劣汰情境 $I \succ M2 \succ M7 \succ M3 \succ M8 \succ II \succ M1 \succ M4 \succ M5 \succ M6 \succ M9 \succ III \succ IV$

妥协值Q2的最优排序


$$Base=\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &a_i &b_i\\ \hline M1 &0.3995 &0.7656\\ \hline M2 &0.1741 &0.0873\\ \hline M3 &0.3018 &0.9331\\ \hline M4 &0.4488 &0.5425\\ \hline M5 &0.4663 &0.8346\\ \hline M6 &0.47 &0.5842\\ \hline M7 &0.2534 &0.0899\\ \hline M8 &0.2951 &0.2984\\ \hline M9 &0.5146 &0.7928\\ \hline I &0 &0\\ \hline II &0.348 &0.3338\\ \hline III &0.672 &0.6676\\ \hline IV &1 &1\\ \hline \end{array} $$

求解线段在决策区间的交点,k代表决策系数

Q2聚类特征


序号 聚类特征-对应k值区段 Q值排序
10<$k$< 0.01464$I \succ M2 \succ M7 \succ M8 \succ M3 \succ II \succ M1 \succ M4 \succ M5 \succ M6 \succ M9 \succ III \succ IV$
20.01464<$k$< 0.071533$I \succ M2 \succ M7 \succ M8 \succ M3 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
30.071533<$k$< 0.181072$I \succ M2 \succ M7 \succ M8 \succ II \succ M3 \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
40.181072<$k$< 0.273476$I \succ M2 \succ M7 \succ M8 \succ II \succ M3 \succ M4 \succ M1 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
50.273476<$k$< 0.279872$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M3 \succ M1 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
60.279872<$k$< 0.325276$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M3 \succ M6 \succ M1 \succ M5 \succ M9 \succ III \succ IV$
70.325276<$k$< 0.368394$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M3 \succ M1 \succ M5 \succ M9 \succ III \succ IV$
80.368394<$k$< 0.536416$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ M3 \succ M5 \succ M9 \succ III \succ IV$
90.536416<$k$< 0.552107$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ M3 \succ M9 \succ M5 \succ III \succ IV$
100.552107<$k$< 0.557104$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ M3 \succ M9 \succ III \succ M5 \succ IV$
110.557104<$k$< 0.582427$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ M3 \succ III \succ M9 \succ M5 \succ IV$
120.582427<$k$< 0.602703$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ III \succ M3 \succ M9 \succ M5 \succ IV$
130.602703<$k$< 0.62539$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ III \succ M9 \succ M3 \succ M5 \succ IV$
140.62539<$k$< 0.735666$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ III \succ M9 \succ M5 \succ M3 \succ IV$
150.735666<$k$< 1$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ III \succ M1 \succ M9 \succ M5 \succ M3 \succ IV$

运用AECM对Q2进行求解


排名 要素所占区段
0I=1   
1M2=1   
2M7=1   
3M8=1   
4II=0.928467   M3=0.071533   
5M4=0.726524   M3=0.201943   II=0.071533   
6M6=0.674724   M1=0.181072   M4=0.092404   M3=0.0518   
7M1=0.466072   III=0.264334   M4=0.181072   M6=0.045404   M3=0.043118   
8M1=0.352856   M6=0.265232   M3=0.214033   III=0.153239   M5=0.01464   
9M5=0.521776   M9=0.417985   III=0.025323   M3=0.020276   M6=0.01464   
10M9=0.582015   M5=0.390301   M3=0.022687   III=0.004997   
11III=0.552107   M3=0.37461   M5=0.073283   
12IV=1   

上表为针对Q2每个评价对象(要素)在每个层级所占的份额。

排名 最优妥协解
优胜情境$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ M3 \succ M5 \succ M9 \succ III \succ IV$
劣汰情境 $I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ M3 \succ M5 \succ M9 \succ III \succ IV$

妥协值Q3的最优排序


$$Base=\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &a_i &b_i\\ \hline M1 &0.3995 &0.3254\\ \hline M2 &0.1741 &0.1803\\ \hline M3 &0.3018 &0.2909\\ \hline M4 &0.4488 &0.4916\\ \hline M5 &0.4663 &0.5916\\ \hline M6 &0.47 &0.4677\\ \hline M7 &0.2534 &0.2733\\ \hline M8 &0.2951 &0.3816\\ \hline M9 &0.5146 &0.562\\ \hline I &0 &0\\ \hline II &0.348 &0.2366\\ \hline III &0.672 &0.5807\\ \hline IV &1 &1\\ \hline \end{array} $$

求解线段在决策区间的交点,k代表决策系数

Q3聚类特征


序号 聚类特征-对应k值区段 Q值排序
10<$k$< 0.029136$I \succ M2 \succ M7 \succ M8 \succ M3 \succ II \succ M1 \succ M4 \succ M5 \succ M6 \succ M9 \succ III \succ IV$
20.029136<$k$< 0.069296$I \succ M2 \succ M7 \succ M8 \succ M3 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
30.069296<$k$< 0.267445$I \succ M2 \succ M7 \succ M3 \succ M8 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
40.267445<$k$< 0.459647$I \succ M2 \succ M7 \succ M3 \succ II \succ M8 \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
50.459647<$k$< 0.469476$I \succ M2 \succ M7 \succ II \succ M3 \succ M8 \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
60.469476<$k$< 0.620096$I \succ M2 \succ M7 \succ II \succ M3 \succ M8 \succ M1 \succ M6 \succ M4 \succ M5 \succ M9 \succ III \succ IV$
70.620096<$k$< 0.650139$I \succ M2 \succ M7 \succ II \succ M3 \succ M8 \succ M1 \succ M6 \succ M4 \succ M9 \succ M5 \succ III \succ IV$
80.650139<$k$< 0.721035$I \succ M2 \succ M7 \succ II \succ M3 \succ M1 \succ M8 \succ M6 \succ M4 \succ M9 \succ M5 \succ III \succ IV$
90.721035<$k$< 0.949442$I \succ M2 \succ II \succ M7 \succ M3 \succ M1 \succ M8 \succ M6 \succ M4 \succ M9 \succ M5 \succ III \succ IV$
100.949442<$k$< 1$I \succ M2 \succ II \succ M7 \succ M3 \succ M1 \succ M8 \succ M6 \succ M4 \succ M9 \succ III \succ M5 \succ IV$

运用AECM对Q3进行求解


排名 要素所占区段
0I=1   
1M2=1   
2M7=0.721035   II=0.278965   
3M3=0.390351   M7=0.278965   II=0.261388   M8=0.069296   
4M3=0.609649   M8=0.198149   II=0.192202   
5M8=0.382694   M1=0.349861   II=0.267445   
6M1=0.650139   M8=0.349861   
7M6=0.530524   M4=0.469476   
8M4=0.530524   M6=0.44034   M5=0.029136   
9M5=0.59096   M9=0.379904   M6=0.029136   
10M9=0.620096   M5=0.329346   III=0.050558   
11III=0.949442   M5=0.050558   
12IV=1   

上表为针对Q3每个评价对象(要素)在每个层级所占的份额。

排名 最优妥协解
优胜情境$I \succ M2 \succ M7 \succ II \succ M3 \succ M8 \succ M1 \succ M6 \succ M4 \succ M5 \succ M9 \succ III \succ IV$
劣汰情境 $I \succ M2 \succ M7 \succ II \succ M3 \succ M8 \succ M1 \succ M6 \succ M4 \succ M5 \succ M9 \succ III \succ IV$

妥协值Q4的最优排序


$$Base=\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &a_i &b_i\\ \hline M1 &0.3343 &0.7656\\ \hline M2 &0.2324 &0.0873\\ \hline M3 &0.2486 &0.9331\\ \hline M4 &0.4339 &0.5425\\ \hline M5 &0.4559 &0.8346\\ \hline M6 &0.4527 &0.5842\\ \hline M7 &0.2849 &0.0899\\ \hline M8 &0.3471 &0.2984\\ \hline M9 &0.4933 &0.7928\\ \hline I &0 &0\\ \hline II &0.3538 &0.3338\\ \hline III &0.6751 &0.6676\\ \hline IV &1 &1\\ \hline \end{array} $$

求解线段在决策区间的交点,k代表决策系数

Q4聚类特征


序号 聚类特征-对应k值区段 Q值排序
10<$k$< 0.02675$I \succ M2 \succ M3 \succ M7 \succ M1 \succ M8 \succ II \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
20.02675<$k$< 0.041314$I \succ M2 \succ M3 \succ M7 \succ M8 \succ M1 \succ II \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
30.041314<$k$< 0.043166$I \succ M2 \succ M7 \succ M3 \succ M8 \succ M1 \succ II \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
40.043166<$k$< 0.134373$I \succ M2 \succ M7 \succ M3 \succ M8 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
50.134373<$k$< 0.149289$I \succ M2 \succ M7 \succ M8 \succ M3 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
60.149289<$k$< 0.308684$I \succ M2 \succ M7 \succ M8 \succ II \succ M3 \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
70.308684<$k$< 0.321752$I \succ M2 \succ M7 \succ M8 \succ II \succ M3 \succ M4 \succ M1 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
80.321752<$k$< 0.33841$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M3 \succ M1 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
90.33841<$k$< 0.369054$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M1 \succ M3 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
100.369054<$k$< 0.394938$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M1 \succ M6 \succ M3 \succ M5 \succ M9 \succ III \succ IV$
110.394938<$k$< 0.472639$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ M3 \succ M5 \succ M9 \succ III \succ IV$
120.472639<$k$< 0.567698$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ M3 \succ M9 \succ M5 \succ III \succ IV$
130.567698<$k$< 0.592201$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ M3 \succ M9 \succ III \succ M5 \succ IV$
140.592201<$k$< 0.616398$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ M3 \succ III \succ M9 \succ M5 \succ IV$
150.616398<$k$< 0.635692$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ III \succ M3 \succ M9 \succ M5 \succ IV$
160.635692<$k$< 0.677876$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ III \succ M9 \succ M3 \succ M5 \succ IV$
170.677876<$k$< 0.776822$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ III \succ M9 \succ M5 \succ M3 \succ IV$
180.776822<$k$< 1$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ III \succ M1 \succ M9 \succ M5 \succ M3 \succ IV$

运用AECM对Q4进行求解


排名 要素所占区段
0I=1   
1M2=1   
2M7=0.958686   M3=0.041314   
3M8=0.865627   M3=0.093059   M7=0.041314   
4II=0.850711   M8=0.107623   M1=0.02675   M3=0.014916   
5M4=0.678248   M3=0.172463   II=0.106123   M8=0.02675   M1=0.016416   
6M6=0.605062   M1=0.322046   II=0.043166   M3=0.016658   M4=0.013068   
7M1=0.41161   M4=0.308684   III=0.223178   M3=0.030644   M6=0.025884   
8M6=0.369054   M3=0.247344   M1=0.223178   III=0.160424   
9M9=0.48387   M5=0.472639   III=0.024197   M3=0.019294   
10M9=0.51613   M5=0.417183   M3=0.042184   III=0.024503   
11III=0.567698   M3=0.322124   M5=0.110178   
12IV=1   

上表为针对Q4每个评价对象(要素)在每个层级所占的份额。

排名 最优妥协解
优胜情境$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ M3 \succ M9 \succ M5 \succ III \succ IV$
劣汰情境 $I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ M3 \succ M9 \succ M5 \succ III \succ IV$

妥协值Q5的最优排序


$$Base=\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &a_i &b_i\\ \hline M1 &0.3343 &0.7656\\ \hline M2 &0.2324 &0.0873\\ \hline M3 &0.2486 &0.9331\\ \hline M4 &0.4339 &0.5425\\ \hline M5 &0.4559 &0.8346\\ \hline M6 &0.4527 &0.5842\\ \hline M7 &0.2849 &0.0899\\ \hline M8 &0.3471 &0.2984\\ \hline M9 &0.4933 &0.7928\\ \hline I &0 &0\\ \hline II &0.3538 &0.3338\\ \hline III &0.6751 &0.6676\\ \hline IV &1 &1\\ \hline \end{array} $$

求解线段在决策区间的交点,k代表决策系数

Q5聚类特征


序号 聚类特征-对应k值区段 Q值排序
10<$k$< 0.02675$I \succ M2 \succ M3 \succ M7 \succ M1 \succ M8 \succ II \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
20.02675<$k$< 0.041314$I \succ M2 \succ M3 \succ M7 \succ M8 \succ M1 \succ II \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
30.041314<$k$< 0.043166$I \succ M2 \succ M7 \succ M3 \succ M8 \succ M1 \succ II \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
40.043166<$k$< 0.134373$I \succ M2 \succ M7 \succ M3 \succ M8 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
50.134373<$k$< 0.149289$I \succ M2 \succ M7 \succ M8 \succ M3 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
60.149289<$k$< 0.308684$I \succ M2 \succ M7 \succ M8 \succ II \succ M3 \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
70.308684<$k$< 0.321752$I \succ M2 \succ M7 \succ M8 \succ II \succ M3 \succ M4 \succ M1 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
80.321752<$k$< 0.33841$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M3 \succ M1 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
90.33841<$k$< 0.369054$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M1 \succ M3 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
100.369054<$k$< 0.394938$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M1 \succ M6 \succ M3 \succ M5 \succ M9 \succ III \succ IV$
110.394938<$k$< 0.472639$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ M3 \succ M5 \succ M9 \succ III \succ IV$
120.472639<$k$< 0.567698$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ M3 \succ M9 \succ M5 \succ III \succ IV$
130.567698<$k$< 0.592201$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ M3 \succ M9 \succ III \succ M5 \succ IV$
140.592201<$k$< 0.616398$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ M3 \succ III \succ M9 \succ M5 \succ IV$
150.616398<$k$< 0.635692$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ III \succ M3 \succ M9 \succ M5 \succ IV$
160.635692<$k$< 0.677876$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ III \succ M9 \succ M3 \succ M5 \succ IV$
170.677876<$k$< 0.776822$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ III \succ M9 \succ M5 \succ M3 \succ IV$
180.776822<$k$< 1$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ III \succ M1 \succ M9 \succ M5 \succ M3 \succ IV$

运用AECM对Q5进行求解


排名 要素所占区段
0I=1   
1M2=1   
2M7=0.958686   M3=0.041314   
3M8=0.865627   M3=0.093059   M7=0.041314   
4II=0.850711   M8=0.107623   M1=0.02675   M3=0.014916   
5M4=0.678248   M3=0.172463   II=0.106123   M8=0.02675   M1=0.016416   
6M6=0.605062   M1=0.322046   II=0.043166   M3=0.016658   M4=0.013068   
7M1=0.41161   M4=0.308684   III=0.223178   M3=0.030644   M6=0.025884   
8M6=0.369054   M3=0.247344   M1=0.223178   III=0.160424   
9M9=0.48387   M5=0.472639   III=0.024197   M3=0.019294   
10M9=0.51613   M5=0.417183   M3=0.042184   III=0.024503   
11III=0.567698   M3=0.322124   M5=0.110178   
12IV=1   

上表为针对Q5每个评价对象(要素)在每个层级所占的份额。

排名 最优妥协解
优胜情境$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ M3 \succ M9 \succ M5 \succ III \succ IV$
劣汰情境 $I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ M3 \succ M9 \succ M5 \succ III \succ IV$

妥协值Q6的最优排序


$$Base=\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &a_i &b_i\\ \hline M1 &0.7656 &0.3254\\ \hline M2 &0.0873 &0.1803\\ \hline M3 &0.9331 &0.2909\\ \hline M4 &0.5425 &0.4916\\ \hline M5 &0.8346 &0.5916\\ \hline M6 &0.5842 &0.4677\\ \hline M7 &0.0899 &0.2733\\ \hline M8 &0.2984 &0.3816\\ \hline M9 &0.7928 &0.562\\ \hline I &0 &0\\ \hline II &0.3338 &0.2366\\ \hline III &0.6676 &0.5807\\ \hline IV &1 &1\\ \hline \end{array} $$

求解线段在决策区间的交点,k代表决策系数

Q6聚类特征


序号 聚类特征-对应k值区段 Q值排序
10<$k$< 0.196457$I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ III \succ M1 \succ M9 \succ M5 \succ M3 \succ IV$
20.196457<$k$< 0.246784$I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ III \succ M1 \succ M9 \succ M5 \succ M3 \succ IV$
30.246784<$k$< 0.27724$I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ III \succ M1 \succ M9 \succ M3 \succ M5 \succ IV$
40.27724<$k$< 0.340934$I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ III \succ M9 \succ M3 \succ M5 \succ IV$
50.340934<$k$< 0.478107$I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ III \succ M3 \succ M9 \succ M5 \succ IV$
60.478107<$k$< 0.560401$I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M3 \succ III \succ M9 \succ M5 \succ IV$
70.560401<$k$< 0.5731$I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M1 \succ M6 \succ M3 \succ III \succ M9 \succ M5 \succ IV$
80.5731<$k$< 0.635715$I \succ M2 \succ M7 \succ II \succ M8 \succ M1 \succ M4 \succ M6 \succ M3 \succ III \succ M9 \succ M5 \succ IV$
90.635715<$k$< 0.660637$I \succ M2 \succ M7 \succ II \succ M8 \succ M1 \succ M6 \succ M4 \succ M3 \succ III \succ M9 \succ M5 \succ IV$
100.660637<$k$< 0.663749$I \succ M2 \succ M7 \succ II \succ M8 \succ M1 \succ M6 \succ M3 \succ M4 \succ III \succ M9 \succ M5 \succ IV$
110.663749<$k$< 0.829352$I \succ M2 \succ M7 \succ II \succ M8 \succ M1 \succ M3 \succ M6 \succ M4 \succ III \succ M9 \succ M5 \succ IV$
120.829352<$k$< 0.869497$I \succ M2 \succ M7 \succ II \succ M8 \succ M3 \succ M1 \succ M6 \succ M4 \succ III \succ M9 \succ M5 \succ IV$
130.869497<$k$< 0.870488$I \succ M2 \succ II \succ M7 \succ M8 \succ M3 \succ M1 \succ M6 \succ M4 \succ III \succ M9 \succ M5 \succ IV$
140.870488<$k$< 0.87499$I \succ M2 \succ II \succ M7 \succ M8 \succ M3 \succ M1 \succ M6 \succ M4 \succ M9 \succ III \succ M5 \succ IV$
150.87499<$k$< 0.892602$I \succ M2 \succ II \succ M7 \succ M3 \succ M8 \succ M1 \succ M6 \succ M4 \succ M9 \succ III \succ M5 \succ IV$
160.892602<$k$< 1$I \succ M2 \succ II \succ M7 \succ M3 \succ M1 \succ M8 \succ M6 \succ M4 \succ M9 \succ III \succ M5 \succ IV$

运用AECM对Q6进行求解


排名 要素所占区段
0I=1   
1M2=1   
2M7=0.869497   II=0.130503   
3II=0.67304   M8=0.196457   M7=0.130503   
4M8=0.678533   II=0.196457   M3=0.12501   
5M4=0.5731   M1=0.36365   M3=0.045638   M8=0.017612   
6M6=0.588435   M3=0.165603   M8=0.107398   M1=0.075949   M4=0.062615   
7M6=0.411565   M1=0.283161   III=0.27724   M4=0.024922   M3=0.003112   
8M4=0.339363   M1=0.27724   III=0.200867   M3=0.18253   
9M9=0.470446   III=0.392381   M3=0.137173   
10M9=0.529554   M5=0.246784   III=0.129512   M3=0.09415   
11M5=0.753216   M3=0.246784   
12IV=1   

上表为针对Q6每个评价对象(要素)在每个层级所占的份额。

排名 最优妥协解
优胜情境$I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M3 \succ III \succ M9 \succ M5 \succ IV$
劣汰情境 $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M3 \succ III \succ M9 \succ M5 \succ IV$

六种妥协解占比相加


排名 要素所占区段
0I=6   
1M2=6   
2M7=5.079311   M3=0.511221   II=0.409468   
3M3=1.083745   M7=0.920689   M8=3.061138   II=0.934428   
4M8=1.918324   M1=0.162973   M3=0.900155   II=3.018548   
5II=1.276944   M1=0.91115   M8=0.563279   M4=2.65612   M3=0.592507   
6M1=2.276972   II=0.360612   M6=2.473283   M4=0.181155   M3=0.250719   M8=0.457259   
7M4=2.292838   M1=1.572453   III=0.98793   M6=1.039261   M3=0.107518   
8M5=0.577216   M6=1.91024   M1=1.076452   M3=0.891251   III=0.674954   M4=0.869887   
9M6=0.577216   M5=2.524574   M9=2.236075   III=0.466098   M3=0.196037   
10M9=3.763925   M5=1.800797   M3=0.201205   III=0.234073   
11III=3.636945   M3=1.265642   M5=1.097413   
12IV=6   

上表为六种妥协解每个评价对象(要素)在每个层级所占的份额。

该表有两个特点,每一层所有要素对应的值相加等于6

该表有两个特点,任意一个要素在所有层级的值相加等于6

优胜与劣汰情境的最优解


排名 最优妥协解
优胜情境$I \succ M2 \succ M7 \succ M8 \succ II \succ M3 \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$
劣汰情境 $I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M1 \succ M6 \succ M3 \succ M5 \succ M9 \succ III \succ IV$

扯蛋模型