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:


$$ \begin{array}{c|c|c|c|c|c|c}{M_{11 \times15}} & -PN1 & -PN2 & -PE1 & -PS1 & -PS2 &SN1 &SN2 &SN3 &SE1 &SS1 &RN1 &RN2 &RE1 &RE2 &RS1\\ \hline 2011 &0.654302103 &278.5577711 &0.104246565 &426.623418 &0.484969128 &0.518335321 &0.734225621 &0.050682504 &8.714200492 &8.950944846 &0.850812256 &10.3 &7380.279843 &35115.65191 &176.2\\ \hline 2012 &0.64018787 &230.2072698 &0.100574919 &430.0655108 &0.498760764 &0.524293616 &0.720190933 &0.053800771 &8.311141515 &8.547470721 &0.868804512 &11.1 &8725.395288 &39690.62085 &170.1\\ \hline 2013 &0.634839501 &232.7438152 &0.095228796 &433.091812 &0.510793577 &0.534174192 &0.725746269 &0.053359371 &8.162741259 &8.838568347 &0.889545989 &12.82 &10872.14779 &43857.04467 &144.4\\ \hline 2014 &0.621034465 &198.6498147 &0.094446274 &435.3089733 &0.524130018 &0.545060211 &0.730176133 &0.05345935 &8.324123649 &10.36566358 &0.912120266 &8.67 &12075.052 &47967.53527 &159.4\\ \hline 2015 &0.617366121 &179.1637029 &0.093725645 &438.6801925 &0.540016156 &0.547408526 &0.727055177 &0.052997358 &8.456326593 &10.10966409 &0.932741508 &50.29 &13182.83334 &51652.87565 &157.2\\ \hline 2016 &0.614656439 &134.5724311 &0.093564023 &442.1742592 &0.558923488 &0.547604836 &0.732802092 &0.052527861 &8.333882334 &10.73475734 &0.951164027 &50.54 &14328.14133 &55939.46599 &138.94\\ \hline 2017 &0.605388917 &93.61972547 &0.086820781 &445.1131496 &0.577785857 &0.548797995 &0.737336045 &0.052199574 &8.053402438 &11.01950072 &0.958207523 &62.2 &15557.73277 &61583.50207 &252.86\\ \hline st1 &0.4 &100 &0.05 &400 &0.3 &0.7 &0.8 &0.1 &10 &20 &0.95 &12 &30000 &50000 &150\\ \hline st2 &0.6 &200 &0.1 &600 &0.5 &0.5 &0.7 &0.08 &8 &15 &0.9 &9 &20000 &40000 &100\\ \hline st3 &0.7 &300 &0.2 &800 &0.6 &0.4 &0.6 &0.05 &6 &10 &0.8 &6 &15000 &30000 &50\\ \hline st4 &0.8 &400 &0.3 &1000 &0.7 &0.3 &0.5 &0.02 &4 &5 &0.7 &3 &10000 &20000 &20\\ \hline \end{array} $$


  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

$$ \begin{array}{c|c|c|c|c|c|c}{M_{11 \times15}} & -PN1 & -PN2 & -PE1 & -PS1 & -PS2 &SN1 &SN2 &SN3 &SE1 &SS1 &RN1 &RN2 &RE1 &RE2 &RS1\\ \hline 2011 &0.364 &0.396 &0.783 &0.956 &0.538 &0.546 &0.781 &0.384 &0.786 &0.263 &0.584 &0.123 &0 &0.364 &0.671\\ \hline 2012 &0.4 &0.554 &0.798 &0.95 &0.503 &0.561 &0.734 &0.423 &0.719 &0.236 &0.654 &0.137 &0.059 &0.474 &0.645\\ \hline 2013 &0.413 &0.546 &0.819 &0.945 &0.473 &0.585 &0.752 &0.417 &0.694 &0.256 &0.734 &0.166 &0.154 &0.574 &0.534\\ \hline 2014 &0.447 &0.657 &0.822 &0.941 &0.44 &0.613 &0.767 &0.418 &0.721 &0.358 &0.822 &0.096 &0.208 &0.673 &0.599\\ \hline 2015 &0.457 &0.721 &0.825 &0.936 &0.4 &0.619 &0.757 &0.412 &0.743 &0.341 &0.901 &0.799 &0.257 &0.761 &0.589\\ \hline 2016 &0.463 &0.866 &0.826 &0.93 &0.353 &0.619 &0.776 &0.407 &0.722 &0.382 &0.973 &0.803 &0.307 &0.864 &0.511\\ \hline 2017 &0.487 &1 &0.853 &0.925 &0.306 &0.622 &0.791 &0.402 &0.676 &0.401 &1 &1 &0.362 &1 &1\\ \hline st1 &1 &0.979 &1 &1 &1 &1 &1 &1 &1 &1 &0.968 &0.152 &1 &0.721 &0.558\\ \hline st2 &0.5 &0.653 &0.8 &0.667 &0.5 &0.5 &0.667 &0.75 &0.667 &0.667 &0.775 &0.101 &0.558 &0.481 &0.344\\ \hline st3 &0.25 &0.326 &0.4 &0.333 &0.25 &0.25 &0.333 &0.375 &0.333 &0.333 &0.387 &0.051 &0.337 &0.24 &0.129\\ \hline st4 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0.116 &0 &0\\ \hline \end{array} $$

正极值点构成
$$ \begin{array}{c|c|c|c|c|c|c}{M_{1 \times15}} & -PN1 & -PN2 & -PE1 & -PS1 & -PS2 &SN1 &SN2 &SN3 &SE1 &SS1 &RN1 &RN2 &RE1 &RE2 &RS1\\ \hline \mathbf{Zone^+} &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1\\ \hline \end{array} $$
负极值点构成
$$ \begin{array}{c|c|c|c|c|c|c}{M_{1 \times15}} & -PN1 & -PN2 & -PE1 & -PS1 & -PS2 &SN1 &SN2 &SN3 &SE1 &SS1 &RN1 &RN2 &RE1 &RE2 &RS1\\ \hline \mathbf{Zone^-} &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0\\ \hline \end{array} $$

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

$$ \begin{array}{c|c|c|c|c|c|c}{M_{2 \times15}} & -PN1 & -PN2 & -PE1 & -PS1 & -PS2 &SN1 &SN2 &SN3 &SE1 &SS1 &RN1 &RN2 &RE1 &RE2 &RS1\\ \hline EWM所得权重 &0.0563 &0.0522 &0.04 &0.0445 &0.0592 &0.0486 &0.0425 &0.0565 &0.0426 &0.0742 &0.0444 &0.1965 &0.1234 &0.0573 &0.0619\\ \hline 权重大小顺序 &8 &9 &15 &11 &5 &10 &14 &7 &13 &3 &12 &1 &2 &6 &4\\ \hline \end{array} $$

  W2采用的是CRITIC方法求权重

$$ \begin{array}{c|c|c|c|c|c|c}{M_{2 \times15}} & -PN1 & -PN2 & -PE1 & -PS1 & -PS2 &SN1 &SN2 &SN3 &SE1 &SS1 &RN1 &RN2 &RE1 &RE2 &RS1\\ \hline CRITIC方法所得权重 &0.0532 &0.0668 &0.0663 &0.0779 &0.0587 &0.06 &0.0658 &0.0601 &0.0637 &0.0626 &0.0728 &0.0884 &0.0677 &0.0701 &0.0659\\ \hline 权重大小顺序 &15 &6 &7 &2 &14 &13 &9 &12 &10 &11 &3 &1 &5 &4 &8\\ \hline \end{array} $$

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

$$D=\begin{array}{c|c|c|c|c|c|c}{M_{11 \times4}} &-期望值S1 &-期望值S2 &-遗憾值R1 &-遗憾值R2\\ \hline 2011 &0.60672 &0.49983 &0.1724 &0.07754\\ \hline 2012 &0.58445 &0.47846 &0.16975 &0.07635\\ \hline 2013 &0.56331 &0.46242 &0.16403 &0.07378\\ \hline 2014 &0.5405 &0.42943 &0.17782 &0.07998\\ \hline 2015 &0.38761 &0.35068 &0.09184 &0.05036\\ \hline 2016 &0.36862 &0.33045 &0.08559 &0.04693\\ \hline 2017 &0.27755 &0.25762 &0.07887 &0.04325\\ \hline st1 &0.21257 &0.1274 &0.16676 &0.075\\ \hline st2 &0.50217 &0.43325 &0.17672 &0.07949\\ \hline st3 &0.74709 &0.71603 &0.18669 &0.08397\\ \hline st4 &0.98582 &0.99222 &0.19646 &0.08836\\ \hline \end{array} $$

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
$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &-S1 &-S2\\ \hline 2011 &0.60672 &0.49983\\ \hline 2012 &0.58445 &0.47846\\ \hline 2013 &0.56331 &0.46242\\ \hline 2014 &0.5405 &0.42943\\ \hline 2015 &0.38761 &0.35068\\ \hline 2016 &0.36862 &0.33045\\ \hline 2017 &0.27755 &0.25762\\ \hline st1 &0.21257 &0.1274\\ \hline st2 &0.50217 &0.43325\\ \hline st3 &0.74709 &0.71603\\ \hline st4 &0.98582 &0.99222\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &- k=0 &- k=1\\ \hline 2011 &0.50973 &0.43064\\ \hline 2012 &0.48094 &0.40594\\ \hline 2013 &0.45359 &0.38738\\ \hline 2014 &0.4241 &0.34924\\ \hline 2015 &0.22637 &0.25819\\ \hline 2016 &0.2018 &0.23479\\ \hline 2017 &0.08403 &0.15058\\ \hline st1 &0 &0\\ \hline st2 &0.37452 &0.35366\\ \hline st3 &0.69126 &0.68064\\ \hline st4 &1 &1\\ \hline \end{array} $$
Q2
$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &-S1 &-R1\\ \hline 2011 &0.60672 &0.1724\\ \hline 2012 &0.58445 &0.16975\\ \hline 2013 &0.56331 &0.16403\\ \hline 2014 &0.5405 &0.17782\\ \hline 2015 &0.38761 &0.09184\\ \hline 2016 &0.36862 &0.08559\\ \hline 2017 &0.27755 &0.07887\\ \hline st1 &0.21257 &0.16676\\ \hline st2 &0.50217 &0.17672\\ \hline st3 &0.74709 &0.18669\\ \hline st4 &0.98582 &0.19646\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &- k=0 &- k=1\\ \hline 2011 &0.50973 &0.79544\\ \hline 2012 &0.48094 &0.77284\\ \hline 2013 &0.45359 &0.72425\\ \hline 2014 &0.4241 &0.84149\\ \hline 2015 &0.22637 &0.1103\\ \hline 2016 &0.2018 &0.05711\\ \hline 2017 &0.08403 &0\\ \hline st1 &0 &0.74741\\ \hline st2 &0.37452 &0.83217\\ \hline st3 &0.69126 &0.91692\\ \hline st4 &1 &1\\ \hline \end{array} $$
Q3
$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &-S1 &-R2\\ \hline 2011 &0.60672 &0.07754\\ \hline 2012 &0.58445 &0.07635\\ \hline 2013 &0.56331 &0.07378\\ \hline 2014 &0.5405 &0.07998\\ \hline 2015 &0.38761 &0.05036\\ \hline 2016 &0.36862 &0.04693\\ \hline 2017 &0.27755 &0.04325\\ \hline st1 &0.21257 &0.075\\ \hline st2 &0.50217 &0.07949\\ \hline st3 &0.74709 &0.08397\\ \hline st4 &0.98582 &0.08836\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &- k=0 &- k=1\\ \hline 2011 &0.50973 &0.76021\\ \hline 2012 &0.48094 &0.73372\\ \hline 2013 &0.45359 &0.67676\\ \hline 2014 &0.4241 &0.81419\\ \hline 2015 &0.22637 &0.15762\\ \hline 2016 &0.2018 &0.08161\\ \hline 2017 &0.08403 &0\\ \hline st1 &0 &0.70391\\ \hline st2 &0.37452 &0.80326\\ \hline st3 &0.69126 &0.90261\\ \hline st4 &1 &1\\ \hline \end{array} $$
Q4
$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &-S2 &-R1\\ \hline 2011 &0.49983 &0.1724\\ \hline 2012 &0.47846 &0.16975\\ \hline 2013 &0.46242 &0.16403\\ \hline 2014 &0.42943 &0.17782\\ \hline 2015 &0.35068 &0.09184\\ \hline 2016 &0.33045 &0.08559\\ \hline 2017 &0.25762 &0.07887\\ \hline st1 &0.1274 &0.16676\\ \hline st2 &0.43325 &0.17672\\ \hline st3 &0.71603 &0.18669\\ \hline st4 &0.99222 &0.19646\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &- k=0 &- k=1\\ \hline 2011 &0.43064 &0.79544\\ \hline 2012 &0.40594 &0.77284\\ \hline 2013 &0.38738 &0.72425\\ \hline 2014 &0.34924 &0.84149\\ \hline 2015 &0.25819 &0.1103\\ \hline 2016 &0.23479 &0.05711\\ \hline 2017 &0.15058 &0\\ \hline st1 &0 &0.74741\\ \hline st2 &0.35366 &0.83217\\ \hline st3 &0.68064 &0.91692\\ \hline st4 &1 &1\\ \hline \end{array} $$
Q5
$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &-S2 &-R2\\ \hline 2011 &0.49983 &0.07754\\ \hline 2012 &0.47846 &0.07635\\ \hline 2013 &0.46242 &0.07378\\ \hline 2014 &0.42943 &0.07998\\ \hline 2015 &0.35068 &0.05036\\ \hline 2016 &0.33045 &0.04693\\ \hline 2017 &0.25762 &0.04325\\ \hline st1 &0.1274 &0.075\\ \hline st2 &0.43325 &0.07949\\ \hline st3 &0.71603 &0.08397\\ \hline st4 &0.99222 &0.08836\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &- k=0 &- k=1\\ \hline 2011 &0.43064 &0.79544\\ \hline 2012 &0.40594 &0.77284\\ \hline 2013 &0.38738 &0.72425\\ \hline 2014 &0.34924 &0.84149\\ \hline 2015 &0.25819 &0.1103\\ \hline 2016 &0.23479 &0.05711\\ \hline 2017 &0.15058 &0\\ \hline st1 &0 &0.74741\\ \hline st2 &0.35366 &0.83217\\ \hline st3 &0.68064 &0.91692\\ \hline st4 &1 &1\\ \hline \end{array} $$
Q6
$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &-R1 &-R2\\ \hline 2011 &0.1724 &0.07754\\ \hline 2012 &0.16975 &0.07635\\ \hline 2013 &0.16403 &0.07378\\ \hline 2014 &0.17782 &0.07998\\ \hline 2015 &0.09184 &0.05036\\ \hline 2016 &0.08559 &0.04693\\ \hline 2017 &0.07887 &0.04325\\ \hline st1 &0.16676 &0.075\\ \hline st2 &0.17672 &0.07949\\ \hline st3 &0.18669 &0.08397\\ \hline st4 &0.19646 &0.08836\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &- k=0 &- k=1\\ \hline 2011 &0.79544 &0.76021\\ \hline 2012 &0.77284 &0.73372\\ \hline 2013 &0.72425 &0.67676\\ \hline 2014 &0.84149 &0.81419\\ \hline 2015 &0.1103 &0.15762\\ \hline 2016 &0.05711 &0.08161\\ \hline 2017 &0 &0\\ \hline st1 &0.74741 &0.70391\\ \hline st2 &0.83217 &0.80326\\ \hline st3 &0.91692 &0.90261\\ \hline st4 &1 &1\\ \hline \end{array} $$

    六种妥协值的区段截取的排序分析

Q1
Q2
Q3
Q4
Q5
Q6
$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &- k0 &- k1\\ \hline 2011 &0.51 &0.431\\ \hline 2012 &0.481 &0.406\\ \hline 2013 &0.454 &0.387\\ \hline 2014 &0.424 &0.349\\ \hline 2015 &0.226 &0.258\\ \hline 2016 &0.202 &0.235\\ \hline 2017 &0.084 &0.151\\ \hline st1 &0 &0\\ \hline st2 &0.375 &0.354\\ \hline st3 &0.691 &0.681\\ \hline st4 &1 &1\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &- k0 &- k1\\ \hline 2011 &0.51 &0.795\\ \hline 2012 &0.481 &0.773\\ \hline 2013 &0.454 &0.724\\ \hline 2014 &0.424 &0.841\\ \hline 2015 &0.226 &0.11\\ \hline 2016 &0.202 &0.057\\ \hline 2017 &0.084 &0\\ \hline st1 &0 &0.747\\ \hline st2 &0.375 &0.832\\ \hline st3 &0.691 &0.917\\ \hline st4 &1 &1\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &- k0 &- k1\\ \hline 2011 &0.51 &0.76\\ \hline 2012 &0.481 &0.734\\ \hline 2013 &0.454 &0.677\\ \hline 2014 &0.424 &0.814\\ \hline 2015 &0.226 &0.158\\ \hline 2016 &0.202 &0.082\\ \hline 2017 &0.084 &0\\ \hline st1 &0 &0.704\\ \hline st2 &0.375 &0.803\\ \hline st3 &0.691 &0.903\\ \hline st4 &1 &1\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &- k0 &- k1\\ \hline 2011 &0.431 &0.795\\ \hline 2012 &0.406 &0.773\\ \hline 2013 &0.387 &0.724\\ \hline 2014 &0.349 &0.841\\ \hline 2015 &0.258 &0.11\\ \hline 2016 &0.235 &0.057\\ \hline 2017 &0.151 &0\\ \hline st1 &0 &0.747\\ \hline st2 &0.354 &0.832\\ \hline st3 &0.681 &0.917\\ \hline st4 &1 &1\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &- k0 &- k1\\ \hline 2011 &0.431 &0.795\\ \hline 2012 &0.406 &0.773\\ \hline 2013 &0.387 &0.724\\ \hline 2014 &0.349 &0.841\\ \hline 2015 &0.258 &0.11\\ \hline 2016 &0.235 &0.057\\ \hline 2017 &0.151 &0\\ \hline st1 &0 &0.747\\ \hline st2 &0.354 &0.832\\ \hline st3 &0.681 &0.917\\ \hline st4 &1 &1\\ \hline \end{array} $$
$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &- k0 &- k1\\ \hline 2011 &0.795 &0.76\\ \hline 2012 &0.773 &0.734\\ \hline 2013 &0.724 &0.677\\ \hline 2014 &0.841 &0.814\\ \hline 2015 &0.11 &0.158\\ \hline 2016 &0.057 &0.082\\ \hline 2017 &0 &0\\ \hline st1 &0.747 &0.704\\ \hline st2 &0.832 &0.803\\ \hline st3 &0.917 &0.903\\ \hline st4 &1 &1\\ \hline \end{array} $$

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


$$Base=\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &a_i &b_i\\ \hline 2011 &0.5097 &0.4306\\ \hline 2012 &0.4809 &0.4059\\ \hline 2013 &0.4536 &0.3874\\ \hline 2014 &0.4241 &0.3492\\ \hline 2015 &0.2264 &0.2582\\ \hline 2016 &0.2018 &0.2348\\ \hline 2017 &0.084 &0.1506\\ \hline st1 &0 &0\\ \hline st2 &0.3745 &0.3537\\ \hline st3 &0.6913 &0.6806\\ \hline st4 &1 &1\\ \hline \end{array} $$

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<$k$< 0.918173$st1 \succ 2017 \succ 2016 \succ 2015 \succ st2 \succ 2014 \succ 2013 \succ 2012 \succ 2011 \succ st3 \succ st4$
20.918173<$k$< 1$st1 \succ 2017 \succ 2016 \succ 2015 \succ 2014 \succ st2 \succ 2013 \succ 2012 \succ 2011 \succ st3 \succ st4$

运用AECM对Q1进行求解


排名 要素所占区段
0st1=1   
12017=1   
22016=1   
32015=1   
4st2=0.918173   2014=0.081827   
52014=0.918173   st2=0.081827   
62013=1   
72012=1   
82011=1   
9st3=1   
10st4=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$st1 \succ 2017 \succ 2016 \succ 2015 \succ st2 \succ 2014 \succ 2013 \succ 2012 \succ 2011 \succ st3 \succ st4$
劣汰情境 $st1 \succ 2017 \succ 2016 \succ 2015 \succ st2 \succ 2014 \succ 2013 \succ 2012 \succ 2011 \succ st3 \succ st4$

妥协值Q2的最优排序


$$Base=\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &a_i &b_i\\ \hline 2011 &0.5097 &0.7954\\ \hline 2012 &0.4809 &0.7728\\ \hline 2013 &0.4536 &0.7242\\ \hline 2014 &0.4241 &0.8415\\ \hline 2015 &0.2264 &0.1103\\ \hline 2016 &0.2018 &0.0571\\ \hline 2017 &0.084 &0\\ \hline st1 &0 &0.7474\\ \hline st2 &0.3745 &0.8322\\ \hline st3 &0.6913 &0.9169\\ \hline st4 &1 &1\\ \hline \end{array} $$

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<$k$< 0.101067$st1 \succ 2017 \succ 2016 \succ 2015 \succ st2 \succ 2014 \succ 2013 \succ 2012 \succ 2011 \succ st3 \succ st4$
20.101067<$k$< 0.200967$2017 \succ st1 \succ 2016 \succ 2015 \succ st2 \succ 2014 \succ 2013 \succ 2012 \succ 2011 \succ st3 \succ st4$
30.200967<$k$< 0.226209$2017 \succ st1 \succ 2016 \succ 2015 \succ st2 \succ 2013 \succ 2014 \succ 2012 \succ 2011 \succ st3 \succ st4$
40.226209<$k$< 0.262163$2017 \succ 2016 \succ st1 \succ 2015 \succ st2 \succ 2013 \succ 2014 \succ 2012 \succ 2011 \succ st3 \succ st4$
50.262163<$k$< 0.422854$2017 \succ 2016 \succ 2015 \succ st1 \succ st2 \succ 2013 \succ 2014 \succ 2012 \succ 2011 \succ st3 \succ st4$
60.422854<$k$< 0.45294$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ st2 \succ 2014 \succ 2012 \succ 2011 \succ st3 \succ st4$
70.45294<$k$< 0.642061$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ st2 \succ 2012 \succ 2014 \succ 2011 \succ st3 \succ st4$
80.642061<$k$< 0.650297$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ st2 \succ 2014 \succ 2011 \succ st3 \succ st4$
90.650297<$k$< 0.786398$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ st2 \succ 2011 \succ 2014 \succ st3 \succ st4$
100.786398<$k$< 0.951409$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ 2011 \succ st2 \succ 2014 \succ st3 \succ st4$
110.951409<$k$< 1$2017 \succ 2016 \succ 2015 \succ 2013 \succ st1 \succ 2012 \succ 2011 \succ st2 \succ 2014 \succ st3 \succ st4$

运用AECM对Q2进行求解


排名 要素所占区段
02017=0.898933   st1=0.101067   
12016=0.773791   st1=0.125142   2017=0.101067   
22015=0.737837   2016=0.226209   st1=0.035954   
3st1=0.689246   2015=0.262163   2013=0.048591   
42013=0.528555   st2=0.422854   st1=0.048591   
52012=0.357939   2013=0.221887   st2=0.219207   2014=0.200967   
62014=0.251973   2011=0.213602   2013=0.200967   2012=0.189121   st2=0.144337   
72012=0.45294   st2=0.213602   2014=0.197357   2011=0.136101   
82011=0.650297   2014=0.349703   
9st3=1   
10st4=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$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ st2 \succ 2012 \succ 2014 \succ 2011 \succ st3 \succ st4$
劣汰情境 $2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ st2 \succ 2012 \succ 2014 \succ 2011 \succ st3 \succ st4$

妥协值Q3的最优排序


$$Base=\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &a_i &b_i\\ \hline 2011 &0.5097 &0.7602\\ \hline 2012 &0.4809 &0.7337\\ \hline 2013 &0.4536 &0.6768\\ \hline 2014 &0.4241 &0.8142\\ \hline 2015 &0.2264 &0.1576\\ \hline 2016 &0.2018 &0.0816\\ \hline 2017 &0.084 &0\\ \hline st1 &0 &0.7039\\ \hline st2 &0.3745 &0.8033\\ \hline st3 &0.6913 &0.9026\\ \hline st4 &1 &1\\ \hline \end{array} $$

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<$k$< 0.106647$st1 \succ 2017 \succ 2016 \succ 2015 \succ st2 \succ 2014 \succ 2013 \succ 2012 \succ 2011 \succ st3 \succ st4$
20.106647<$k$< 0.176657$2017 \succ st1 \succ 2016 \succ 2015 \succ st2 \succ 2014 \succ 2013 \succ 2012 \succ 2011 \succ st3 \succ st4$
30.176657<$k$< 0.244875$2017 \succ st1 \succ 2016 \succ 2015 \succ st2 \succ 2013 \succ 2014 \succ 2012 \succ 2011 \succ st3 \succ st4$
40.244875<$k$< 0.292978$2017 \succ 2016 \succ st1 \succ 2015 \succ st2 \succ 2013 \succ 2014 \succ 2012 \succ 2011 \succ st3 \succ st4$
50.292978<$k$< 0.384623$2017 \succ 2016 \succ 2015 \succ st1 \succ st2 \succ 2013 \succ 2014 \succ 2012 \succ 2011 \succ st3 \succ st4$
60.384623<$k$< 0.41394$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ st2 \succ 2014 \succ 2012 \succ 2011 \succ st3 \succ st4$
70.41394<$k$< 0.604778$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ st2 \succ 2012 \succ 2014 \succ 2011 \succ st3 \succ st4$
80.604778<$k$< 0.613356$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ st2 \succ 2014 \succ 2011 \succ st3 \succ st4$
90.613356<$k$< 0.758495$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ st2 \succ 2011 \succ 2014 \succ st3 \succ st4$
100.758495<$k$< 0.943513$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ 2011 \succ st2 \succ 2014 \succ st3 \succ st4$
110.943513<$k$< 1$2017 \succ 2016 \succ 2015 \succ 2013 \succ st1 \succ 2012 \succ 2011 \succ st2 \succ 2014 \succ st3 \succ st4$

运用AECM对Q3进行求解


排名 要素所占区段
02017=0.893353   st1=0.106647   
12016=0.755125   st1=0.138228   2017=0.106647   
22015=0.707022   2016=0.244875   st1=0.048103   
3st1=0.650535   2015=0.292978   2013=0.056487   
42013=0.55889   st2=0.384623   st1=0.056487   
52012=0.395222   st2=0.220155   2013=0.207966   2014=0.176657   
62011=0.241505   2014=0.237283   2012=0.190838   2013=0.176657   st2=0.153717   
72012=0.41394   st2=0.241505   2014=0.199416   2011=0.145139   
82011=0.613356   2014=0.386644   
9st3=1   
10st4=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$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ st2 \succ 2012 \succ 2014 \succ 2011 \succ st3 \succ st4$
劣汰情境 $2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ st2 \succ 2012 \succ 2014 \succ 2011 \succ st3 \succ st4$

妥协值Q4的最优排序


$$Base=\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &a_i &b_i\\ \hline 2011 &0.4306 &0.7954\\ \hline 2012 &0.4059 &0.7728\\ \hline 2013 &0.3874 &0.7242\\ \hline 2014 &0.3492 &0.8415\\ \hline 2015 &0.2582 &0.1103\\ \hline 2016 &0.2348 &0.0571\\ \hline 2017 &0.1506 &0\\ \hline st1 &0 &0.7474\\ \hline st2 &0.3537 &0.8322\\ \hline st3 &0.6806 &0.9169\\ \hline st4 &1 &1\\ \hline \end{array} $$

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<$k$< 0.167683$st1 \succ 2017 \succ 2016 \succ 2015 \succ 2014 \succ st2 \succ 2013 \succ 2012 \succ 2011 \succ st3 \succ st4$
20.167683<$k$< 0.238097$2017 \succ st1 \succ 2016 \succ 2015 \succ 2014 \succ st2 \succ 2013 \succ 2012 \succ 2011 \succ st3 \succ st4$
30.238097<$k$< 0.245477$2017 \succ st1 \succ 2016 \succ 2015 \succ 2014 \succ 2013 \succ st2 \succ 2012 \succ 2011 \succ st3 \succ st4$
40.245477<$k$< 0.253801$2017 \succ st1 \succ 2016 \succ 2015 \succ 2013 \succ 2014 \succ st2 \succ 2012 \succ 2011 \succ st3 \succ st4$
50.253801<$k$< 0.288382$2017 \succ 2016 \succ st1 \succ 2015 \succ 2013 \succ 2014 \succ st2 \succ 2012 \succ 2011 \succ st3 \succ st4$
60.288382<$k$< 0.321551$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2014 \succ st2 \succ 2012 \succ 2011 \succ st3 \succ st4$
70.321551<$k$< 0.452313$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ st2 \succ 2014 \succ 2012 \succ 2011 \succ st3 \succ st4$
80.452313<$k$< 0.468413$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ st2 \succ 2012 \succ 2014 \succ 2011 \succ st3 \succ st4$
90.468413<$k$< 0.638692$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ st2 \succ 2014 \succ 2011 \succ st3 \succ st4$
100.638692<$k$< 0.677017$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ st2 \succ 2011 \succ 2014 \succ st3 \succ st4$
110.677017<$k$< 0.943574$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ 2011 \succ st2 \succ 2014 \succ st3 \succ st4$
120.943574<$k$< 1$2017 \succ 2016 \succ 2015 \succ 2013 \succ st1 \succ 2012 \succ 2011 \succ st2 \succ 2014 \succ st3 \succ st4$

运用AECM对Q4进行求解


排名 要素所占区段
02017=0.832317   st1=0.167683   
12016=0.746199   2017=0.167683   st1=0.086118   
22015=0.711618   2016=0.253801   st1=0.034581   
3st1=0.655192   2015=0.288382   2013=0.056426   
42013=0.698097   2014=0.245477   st1=0.056426   
52012=0.531587   st2=0.384959   2014=0.076074   2013=0.00738   
62011=0.322983   st2=0.292058   2013=0.238097   2014=0.130762   2012=0.0161   
72012=0.452313   st2=0.322983   2014=0.186379   2011=0.038325   
82011=0.638692   2014=0.361308   
9st3=1   
10st4=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$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ st2 \succ 2014 \succ 2011 \succ st3 \succ st4$
劣汰情境 $2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ st2 \succ 2014 \succ 2011 \succ st3 \succ st4$

妥协值Q5的最优排序


$$Base=\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &a_i &b_i\\ \hline 2011 &0.4306 &0.7954\\ \hline 2012 &0.4059 &0.7728\\ \hline 2013 &0.3874 &0.7242\\ \hline 2014 &0.3492 &0.8415\\ \hline 2015 &0.2582 &0.1103\\ \hline 2016 &0.2348 &0.0571\\ \hline 2017 &0.1506 &0\\ \hline st1 &0 &0.7474\\ \hline st2 &0.3537 &0.8322\\ \hline st3 &0.6806 &0.9169\\ \hline st4 &1 &1\\ \hline \end{array} $$

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<$k$< 0.167683$st1 \succ 2017 \succ 2016 \succ 2015 \succ 2014 \succ st2 \succ 2013 \succ 2012 \succ 2011 \succ st3 \succ st4$
20.167683<$k$< 0.238097$2017 \succ st1 \succ 2016 \succ 2015 \succ 2014 \succ st2 \succ 2013 \succ 2012 \succ 2011 \succ st3 \succ st4$
30.238097<$k$< 0.245477$2017 \succ st1 \succ 2016 \succ 2015 \succ 2014 \succ 2013 \succ st2 \succ 2012 \succ 2011 \succ st3 \succ st4$
40.245477<$k$< 0.253801$2017 \succ st1 \succ 2016 \succ 2015 \succ 2013 \succ 2014 \succ st2 \succ 2012 \succ 2011 \succ st3 \succ st4$
50.253801<$k$< 0.288382$2017 \succ 2016 \succ st1 \succ 2015 \succ 2013 \succ 2014 \succ st2 \succ 2012 \succ 2011 \succ st3 \succ st4$
60.288382<$k$< 0.321551$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2014 \succ st2 \succ 2012 \succ 2011 \succ st3 \succ st4$
70.321551<$k$< 0.452313$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ st2 \succ 2014 \succ 2012 \succ 2011 \succ st3 \succ st4$
80.452313<$k$< 0.468413$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ st2 \succ 2012 \succ 2014 \succ 2011 \succ st3 \succ st4$
90.468413<$k$< 0.638692$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ st2 \succ 2014 \succ 2011 \succ st3 \succ st4$
100.638692<$k$< 0.677017$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ st2 \succ 2011 \succ 2014 \succ st3 \succ st4$
110.677017<$k$< 0.943574$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ 2011 \succ st2 \succ 2014 \succ st3 \succ st4$
120.943574<$k$< 1$2017 \succ 2016 \succ 2015 \succ 2013 \succ st1 \succ 2012 \succ 2011 \succ st2 \succ 2014 \succ st3 \succ st4$

运用AECM对Q5进行求解


排名 要素所占区段
02017=0.832317   st1=0.167683   
12016=0.746199   2017=0.167683   st1=0.086118   
22015=0.711618   2016=0.253801   st1=0.034581   
3st1=0.655192   2015=0.288382   2013=0.056426   
42013=0.698097   2014=0.245477   st1=0.056426   
52012=0.531587   st2=0.384959   2014=0.076074   2013=0.00738   
62011=0.322983   st2=0.292058   2013=0.238097   2014=0.130762   2012=0.0161   
72012=0.452313   st2=0.322983   2014=0.186379   2011=0.038325   
82011=0.638692   2014=0.361308   
9st3=1   
10st4=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$2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ st2 \succ 2014 \succ 2011 \succ st3 \succ st4$
劣汰情境 $2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ st2 \succ 2014 \succ 2011 \succ st3 \succ st4$

妥协值Q6的最优排序


$$Base=\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &a_i &b_i\\ \hline 2011 &0.7954 &0.7602\\ \hline 2012 &0.7728 &0.7337\\ \hline 2013 &0.7242 &0.6768\\ \hline 2014 &0.8415 &0.8142\\ \hline 2015 &0.1103 &0.1576\\ \hline 2016 &0.0571 &0.0816\\ \hline 2017 &0 &0\\ \hline st1 &0.7474 &0.7039\\ \hline st2 &0.8322 &0.8033\\ \hline st3 &0.9169 &0.9026\\ \hline st4 &1 &1\\ \hline \end{array} $$

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<$k$< 1$2017 \succ 2016 \succ 2015 \succ 2013 \succ st1 \succ 2012 \succ 2011 \succ st2 \succ 2014 \succ st3 \succ st4$

运用AECM对Q6进行求解


排名 要素所占区段
02017=1   
12016=1   
22015=1   
32013=1   
4st1=1   
52012=1   
62011=1   
7st2=1   
82014=1   
9st3=1   
10st4=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$2017 \succ 2016 \succ 2015 \succ 2013 \succ st1 \succ 2012 \succ 2011 \succ st2 \succ 2014 \succ st3 \succ st4$
劣汰情境 $2017 \succ 2016 \succ 2015 \succ 2013 \succ st1 \succ 2012 \succ 2011 \succ st2 \succ 2014 \succ st3 \succ st4$

六种妥协解占比相加


排名 要素所占区段
0st1=1.54308   0=0   1=0   2=0   3=0   4=0   
10=0   1=0   st1=0.435606   2=0   3=0   4=0   5=0   6=0   7=0   8=0   9=0   
20=0   1=0   2=0   st1=0.153219   3=0   4=0   5=0   6=0   7=0   8=0   9=0   
30=0   st1=2.650165   1=0   2=0   3=0   4=0   5=0   6=0   7=0   8=0   9=0   
4st2=1.72565   0=0   1=0   st1=1.21793   2=0   3=0   4=0   5=0   6=0   
50=0   st2=1.291107   1=0   2=0   3=0   4=0   5=0   6=0   7=0   8=0   9=0   10=0   11=0   12=0   13=0   
60=0   1=0   2=0   3=0   4=0   st2=0.88217   5=0   6=0   7=0   8=0   9=0   10=0   11=0   12=0   13=0   14=0   15=0   16=0   17=0   
70=0   1=0   st2=2.101073   2=0   3=0   4=0   5=0   6=0   7=0   8=0   9=0   10=0   11=0   12=0   
80=0   1=0   2=0   3=0   4=0   5=0   6=0   7=0   8=0   9=0   
9st3=6   
10st4=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$st1 \succ 6 \succ 7 \succ 9 \succ st2 \succ 8 \succ 13 \succ 12 \succ 14 \succ st3 \succ st4$
劣汰情境 $10 \succ 9 \succ 15 \succ 16 \succ st1 \succ 12 \succ 13 \succ st2 \succ 1 \succ st3 \succ st4$

扯蛋模型