整个处理核心看点有两个:
对期望值S与遗憾值分别用四种不同的妥协解公式求解。
聚类出来的排序是不是一致。
对于四种妥协解,运用AECM得到的最优排序有何不同。
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
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。
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:$$ n_{ij} = \frac{max(o_{j})-{o_{ij}}}{{max(o_{j})-min(o_{j})}} $$
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
$$ \begin{array}{c|c|c|c|c|c|c}{M_{13 \times12}} &A1 &A2 &B1 &B2 &B3 &B4 &C1 &C2 &C3 &C4 & -D1 & -D2\\ \hline M1 &0.702 &0.89 &0.732 &0.488 &0.709 &0.646 &0.371 &0.524 &0.511 &0.951 &0.121 &0.537\\ \hline M2 &0.774 &0.61 &0.524 &0.622 &1 &0.683 &0.914 &0.768 &0.911 &0.72 &0.788 &0.463\\ \hline M3 &1 &0.707 &0.841 &0.732 &0.886 &0.756 &0.543 &0.976 &0.711 &0.793 &0 &0.402\\ \hline M4 &0.44 &0.366 &0.537 &0.659 &0.595 &0.89 &0.314 &0.61 &0.4 &0.61 &0.576 &0.341\\ \hline M5 &0.5 &0.549 &0.659 &0.829 &0.335 &0.415 &0.114 &0.646 &0.178 &0.732 &0.758 &0.561\\ \hline M6 &0.821 &0.756 &0.476 &0.695 &0.354 &0.585 &0.286 &0.439 &0.289 &0.561 &0.697 &0.159\\ \hline M7 &0.75 &0.646 &0.805 &0.622 &0.608 &0.549 &0.771 &0.695 &0.844 &0.707 &0.606 &0.537\\ \hline M8 &0.321 &0.451 &0.598 &0.561 &0.62 &0.78 &1 &0.524 &1 &0.646 &0.606 &0.256\\ \hline M9 &0.643 &0.585 &0.695 &0.659 &0.285 &0.5 &0.143 &0.585 &0.244 &0.5 &0.606 &0.305\\ \hline I &0.976 &1 &1 &1 &0.759 &1 &0.686 &1 &0.8 &1 &1 &1\\ \hline II &0.619 &0.634 &0.634 &0.634 &0.506 &0.634 &0.457 &0.634 &0.533 &0.634 &0.667 &0.683\\ \hline III &0.31 &0.317 &0.317 &0.317 &0.253 &0.317 &0.229 &0.317 &0.267 &0.317 &0.333 &0.366\\ \hline IV &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0\\ \hline \end{array} $$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 $
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。
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
$$D=\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &-期望值S &-遗憾值R\\ \hline M1 &0.4166253 &0.0852883\\ \hline M2 &0.2567691 &0.0415845\\ \hline M3 &0.3175748 &0.0969552\\ \hline M4 &0.4762627 &0.0670459\\ \hline M5 &0.4904031 &0.0866009\\ \hline M6 &0.4983684 &0.0698394\\ \hline M7 &0.3184861 &0.0382327\\ \hline M8 &0.3705105 &0.0576513\\ \hline M9 &0.5326368 &0.0838073\\ \hline I &0.0729687 &0.0307294\\ \hline II &0.3986962 &0.053078\\ \hline III &0.6979584 &0.0754266\\ \hline IV &1 &0.0978946\\ \hline \end{array} $$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。
公式 |
$$ Q_i = \left( 1-k \right) \left(\frac{a_i - Min(a_i)}{Max(a_i) -Min(a_i)} \right) + k\left(\frac{b_i - Min(b_i)}{Max(b_i) -Min(b_i)} \right) $$ |
$$ Q_i = \left( 1-k \right) \left(\frac{ \sqrt {a_i^2 - Min(a_i)^2}} {\sqrt {Max(a_i)^2 -Min(a_i)^2}} \right) + k\left(\frac{\sqrt{ b_i^2 - Min(b_i)^2}}{\sqrt{Max(b_i)^2 -Min(b_i)^2}} \right) $$ |
$$ Q_i = \left( 1-k \right) \left(\frac{ {a_i^2 + Min(a_i)^2 - S_i Min(a_i)}} { {Max(a_i)^2 + Min(a_i)^2 - Max(a_i)Min(a_i)}} \right) + k\left(\frac{ { b_i^2 + Min(b_i)^2 - b_iMin(b_i)}}{{Max(b_i)^2 -Min(b_i)^2 -Max(b_i)Min(b_i)}} \right) $$ |
$$ Q_i = \left( 1-k \right) {a_i} + k b_i $$ |
Q1 |
Q2 |
Q3 |
Q4等于初始SR两列 |
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &- k0 &- k1\\
\hline M1 &0.371 &0.812\\
\hline M2 &0.198 &0.162\\
\hline M3 &0.264 &0.986\\
\hline M4 &0.435 &0.541\\
\hline M5 &0.45 &0.832\\
\hline M6 &0.459 &0.582\\
\hline M7 &0.265 &0.112\\
\hline M8 &0.321 &0.401\\
\hline M9 &0.496 &0.79\\
\hline I &0 &0\\
\hline II &0.351 &0.333\\
\hline III &0.674 &0.665\\
\hline IV &1 &1\\
\hline \end{array} $$
|
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &- k0 &- k1\\
\hline M1 &0.411 &0.856\\
\hline M2 &0.247 &0.301\\
\hline M3 &0.31 &0.989\\
\hline M4 &0.472 &0.641\\
\hline M5 &0.486 &0.871\\
\hline M6 &0.494 &0.675\\
\hline M7 &0.311 &0.245\\
\hline M8 &0.364 &0.525\\
\hline M9 &0.529 &0.839\\
\hline I &0 &0\\
\hline II &0.393 &0.466\\
\hline III &0.696 &0.741\\
\hline IV &1 &1\\
\hline \end{array} $$
|
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &- k0 &- k1\\
\hline M1 &0.159 &0.744\\
\hline M2 &0.056 &0.186\\
\hline M3 &0.089 &0.979\\
\hline M4 &0.212 &0.449\\
\hline M5 &0.225 &0.769\\
\hline M6 &0.233 &0.489\\
\hline M7 &0.09 &0.164\\
\hline M8 &0.124 &0.332\\
\hline M9 &0.268 &0.717\\
\hline I &0.006 &0.126\\
\hline II &0.145 &0.283\\
\hline III &0.474 &0.574\\
\hline IV &1 &1\\
\hline \end{array} $$
|
$$\begin{array}{c|c|c|c|c|c|c}{M_{13 \times2}} &- k0 &- k1\\
\hline M1 &0.417 &0.085\\
\hline M2 &0.257 &0.042\\
\hline M3 &0.318 &0.097\\
\hline M4 &0.476 &0.067\\
\hline M5 &0.49 &0.087\\
\hline M6 &0.498 &0.07\\
\hline M7 &0.318 &0.038\\
\hline M8 &0.371 &0.058\\
\hline M9 &0.533 &0.084\\
\hline I &0.073 &0.031\\
\hline II &0.399 &0.053\\
\hline III &0.698 &0.075\\
\hline IV &1 &0.098\\
\hline \end{array} $$
|
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
序号 | 聚类特征-对应k值区段 | Q值排序 |
---|---|---|
1 | 0<$k$< 0.001123 | $I \succ M2 \succ M3 \succ M7 \succ M8 \succ II \succ M1 \succ M4 \succ M5 \succ M6 \succ M9 \succ III \succ IV$ |
2 | 0.001123<$k$< 0.033284 | $I \succ M2 \succ M7 \succ M3 \succ M8 \succ II \succ M1 \succ M4 \succ M5 \succ M6 \succ M9 \succ III \succ IV$ |
3 | 0.033284<$k$< 0.088905 | $I \succ M2 \succ M7 \succ M3 \succ M8 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |
4 | 0.088905<$k$< 0.118128 | $I \succ M2 \succ M7 \succ M8 \succ M3 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |
5 | 0.118128<$k$< 0.191499 | $I \succ M2 \succ M7 \succ M8 \succ II \succ M3 \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |
6 | 0.191499<$k$< 0.277121 | $I \succ M2 \succ M7 \succ M8 \succ II \succ M3 \succ M4 \succ M1 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |
7 | 0.277121<$k$< 0.277667 | $I \succ M2 \succ M7 \succ M8 \succ II \succ M3 \succ M4 \succ M6 \succ M1 \succ M5 \succ M9 \succ III \succ IV$ |
8 | 0.277667<$k$< 0.308689 | $I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M3 \succ M6 \succ M1 \succ M5 \succ M9 \succ III \succ IV$ |
9 | 0.308689<$k$< 0.325724 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M3 \succ M6 \succ M1 \succ M5 \succ M9 \succ III \succ IV$ |
10 | 0.325724<$k$< 0.380847 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M3 \succ M1 \succ M5 \succ M9 \succ III \succ IV$ |
11 | 0.380847<$k$< 0.52275 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M3 \succ M5 \succ M9 \succ III \succ IV$ |
12 | 0.52275<$k$< 0.542358 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M3 \succ M9 \succ M5 \succ III \succ IV$ |
13 | 0.542358<$k$< 0.547375 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M9 \succ M3 \succ M5 \succ III \succ IV$ |
14 | 0.547375<$k$< 0.56143 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M9 \succ M5 \succ M3 \succ III \succ IV$ |
15 | 0.56143<$k$< 0.571559 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M9 \succ M5 \succ III \succ M3 \succ IV$ |
16 | 0.571559<$k$< 0.573696 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M9 \succ M5 \succ III \succ M3 \succ IV$ |
17 | 0.573696<$k$< 0.588344 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M9 \succ III \succ M5 \succ M3 \succ IV$ |
18 | 0.588344<$k$< 0.673937 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ III \succ M9 \succ M5 \succ M3 \succ IV$ |
19 | 0.673937<$k$< 0.850198 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ III \succ M1 \succ M9 \succ M5 \succ M3 \succ IV$ |
20 | 0.850198<$k$< 1 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ III \succ M9 \succ M1 \succ M5 \succ M3 \succ IV$ |
排名 | 要素所占区段 |
---|---|
0 | I=1 |
1 | M2=0.571559 M7=0.428441 |
2 | M7=0.570436 M2=0.428441 M3=0.001123 |
3 | II=0.691311 M8=0.219784 M3=0.087782 M7=0.001123 |
4 | M8=0.780216 II=0.190561 M3=0.029223 |
5 | M4=0.722333 M3=0.159539 II=0.118128 |
6 | M6=0.674276 M1=0.191499 M4=0.086168 M3=0.048057 |
7 | M1=0.378712 III=0.326063 M4=0.191499 M3=0.055123 M6=0.048603 |
8 | M1=0.279987 M6=0.243837 M9=0.195788 M3=0.161511 III=0.085593 M5=0.033284 |
9 | M5=0.515787 M9=0.281462 M1=0.149802 M6=0.033284 III=0.014648 M3=0.0050169999999999 |
10 | M9=0.52275 M5=0.450929 M3=0.014055 III=0.012266 |
11 | III=0.56143 M3=0.43857 |
12 | IV=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 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M3 \succ M5 \succ M9 \succ III \succ IV$ |
劣汰情境 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M3 \succ M5 \succ M9 \succ III \succ IV$ |
序号 | 聚类特征-对应k值区段 | Q值排序 |
---|---|---|
1 | 0<$k$< 0.001259 | $I \succ M2 \succ M3 \succ M7 \succ M8 \succ II \succ M1 \succ M4 \succ M5 \succ M6 \succ M9 \succ III \succ IV$ |
2 | 0.001259<$k$< 0.039502 | $I \succ M2 \succ M7 \succ M3 \succ M8 \succ II \succ M1 \succ M4 \succ M5 \succ M6 \succ M9 \succ III \succ IV$ |
3 | 0.039502<$k$< 0.104692 | $I \succ M2 \succ M7 \succ M3 \succ M8 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |
4 | 0.104692<$k$< 0.136949 | $I \succ M2 \succ M7 \succ M8 \succ M3 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |
5 | 0.136949<$k$< 0.220036 | $I \succ M2 \succ M7 \succ M8 \succ II \succ M3 \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |
6 | 0.220036<$k$< 0.314211 | $I \succ M2 \succ M7 \succ M8 \succ II \succ M3 \succ M4 \succ M1 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |
7 | 0.314211<$k$< 0.31749 | $I \succ M2 \succ M7 \succ M8 \succ II \succ M3 \succ M4 \succ M6 \succ M1 \succ M5 \succ M9 \succ III \succ IV$ |
8 | 0.31749<$k$< 0.327214 | $I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M3 \succ M6 \succ M1 \succ M5 \succ M9 \succ III \succ IV$ |
9 | 0.327214<$k$< 0.369554 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M3 \succ M6 \succ M1 \succ M5 \succ M9 \succ III \succ IV$ |
10 | 0.369554<$k$< 0.431853 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M3 \succ M1 \succ M5 \succ M9 \succ III \succ IV$ |
11 | 0.431853<$k$< 0.530249 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M3 \succ M5 \succ M9 \succ III \succ IV$ |
12 | 0.530249<$k$< 0.570388 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M3 \succ M5 \succ M9 \succ III \succ IV$ |
13 | 0.570388<$k$< 0.592865 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M3 \succ M9 \succ M5 \succ III \succ IV$ |
14 | 0.592865<$k$< 0.598588 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M9 \succ M3 \succ M5 \succ III \succ IV$ |
15 | 0.598588<$k$< 0.60865 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M9 \succ M5 \succ M3 \succ III \succ IV$ |
16 | 0.60865<$k$< 0.617375 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M9 \succ M5 \succ III \succ M3 \succ IV$ |
17 | 0.617375<$k$< 0.630688 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M9 \succ III \succ M5 \succ M3 \succ IV$ |
18 | 0.630688<$k$< 0.712517 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ III \succ M9 \succ M5 \succ M3 \succ IV$ |
19 | 0.712517<$k$< 0.873164 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ III \succ M1 \succ M9 \succ M5 \succ M3 \succ IV$ |
20 | 0.873164<$k$< 1 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ III \succ M9 \succ M1 \succ M5 \succ M3 \succ IV$ |
排名 | 要素所占区段 |
---|---|
0 | I=1 |
1 | M2=0.530249 M7=0.469751 |
2 | M7=0.52899 M2=0.469751 M3=0.001259 |
3 | II=0.672786 M8=0.222522 M3=0.103433 M7=0.001259 |
4 | M8=0.777478 II=0.190265 M3=0.032257 |
5 | M4=0.68251 M3=0.180541 II=0.136949 |
6 | M6=0.630446 M1=0.220036 M4=0.097454 M3=0.052064 |
7 | M1=0.374839 III=0.287483 M4=0.220036 M3=0.062299 M6=0.055343 |
8 | M1=0.278289 M6=0.274709 M9=0.164659 M3=0.161012 III=0.081829 M5=0.039502 |
9 | M5=0.549673 M9=0.264953 M1=0.126836 M6=0.039502 III=0.013313 M3=0.005723 |
10 | M9=0.570388 M5=0.410825 M3=0.010062 III=0.008725 |
11 | III=0.60865 M3=0.39135 |
12 | IV=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 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M3 \succ M5 \succ M9 \succ III \succ IV$ |
劣汰情境 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M3 \succ M5 \succ M9 \succ III \succ IV$ |
序号 | 聚类特征-对应k值区段 | Q值排序 |
---|---|---|
1 | 0<$k$< 0.000674 | $I \succ M2 \succ M3 \succ M7 \succ M8 \succ II \succ M1 \succ M4 \succ M5 \succ M6 \succ M9 \succ III \succ IV$ |
2 | 0.000674<$k$< 0.027162 | $I \succ M2 \succ M7 \succ M3 \succ M8 \succ II \succ M1 \succ M4 \succ M5 \succ M6 \succ M9 \succ III \succ IV$ |
3 | 0.027162<$k$< 0.051176 | $I \succ M2 \succ M7 \succ M3 \succ M8 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |
4 | 0.051176<$k$< 0.074417 | $I \succ M2 \succ M7 \succ M8 \succ M3 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |
5 | 0.074417<$k$< 0.150938 | $I \succ M2 \succ M7 \succ M8 \succ II \succ M3 \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |
6 | 0.150938<$k$< 0.187949 | $I \succ M2 \succ M7 \succ M8 \succ II \succ M3 \succ M4 \succ M1 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |
7 | 0.187949<$k$< 0.224112 | $I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M3 \succ M1 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |
8 | 0.224112<$k$< 0.226977 | $I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M3 \succ M6 \succ M1 \succ M5 \succ M9 \succ III \succ IV$ |
9 | 0.226977<$k$< 0.230069 | $I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M3 \succ M1 \succ M5 \succ M9 \succ III \succ IV$ |
10 | 0.230069<$k$< 0.301959 | $I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M6 \succ M1 \succ M3 \succ M5 \succ M9 \succ III \succ IV$ |
11 | 0.301959<$k$< 0.392993 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M3 \succ M5 \succ M9 \succ III \succ IV$ |
12 | 0.392993<$k$< 0.405967 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M5 \succ M3 \succ M9 \succ III \succ IV$ |
13 | 0.405967<$k$< 0.453351 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M5 \succ M9 \succ M3 \succ III \succ IV$ |
14 | 0.453351<$k$< 0.486703 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M9 \succ M5 \succ M3 \succ III \succ IV$ |
15 | 0.486703<$k$< 0.559971 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M9 \succ M5 \succ III \succ M3 \succ IV$ |
16 | 0.559971<$k$< 0.589014 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M9 \succ III \succ M5 \succ M3 \succ IV$ |
17 | 0.589014<$k$< 0.603095 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ III \succ M9 \succ M5 \succ M3 \succ IV$ |
18 | 0.603095<$k$< 0.648341 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ III \succ M9 \succ M5 \succ M3 \succ IV$ |
19 | 0.648341<$k$< 0.800042 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ III \succ M1 \succ M9 \succ M5 \succ M3 \succ IV$ |
20 | 0.800042<$k$< 1 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ III \succ M9 \succ M1 \succ M5 \succ M3 \succ IV$ |
排名 | 要素所占区段 |
---|---|
0 | I=1 |
1 | M2=0.603095 M7=0.396905 |
2 | M7=0.602421 M2=0.396905 M3=0.000674 |
3 | II=0.698041 M8=0.250783 M3=0.050502 M7=0.000674 |
4 | M8=0.749217 II=0.227542 M3=0.023241 |
5 | M4=0.812051 M3=0.113532 II=0.074417 |
6 | M6=0.773023 M1=0.150938 M3=0.039028 M4=0.037011 |
7 | M1=0.491446 III=0.351659 M4=0.150938 M3=0.003092 M6=0.002865 |
8 | M9=0.335621 M6=0.19695 M3=0.162924 M1=0.157658 M5=0.08752 III=0.059327 |
9 | M5=0.472451 M9=0.258412 M1=0.199958 III=0.029043 M6=0.027162 M3=0.012974 |
10 | M5=0.440029 M9=0.405967 M3=0.080736 III=0.073268 |
11 | M3=0.513297 III=0.486703 |
12 | IV=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 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ III \succ M9 \succ M5 \succ M3 \succ IV$ |
劣汰情境 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M9 \succ M5 \succ III \succ M3 \succ IV$ |
序号 | 聚类特征-对应k值区段 | Q值排序 |
---|---|---|
1 | 0<$k$< 0.015282 | $I \succ M2 \succ M3 \succ M7 \succ M8 \succ II \succ M1 \succ M4 \succ M5 \succ M6 \succ M9 \succ III \succ IV$ |
2 | 0.015282<$k$< 0.322133 | $I \succ M2 \succ M7 \succ M3 \succ M8 \succ II \succ M1 \succ M4 \succ M5 \succ M6 \succ M9 \succ III \succ IV$ |
3 | 0.322133<$k$< 0.573894 | $I \succ M2 \succ M7 \succ M3 \succ M8 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |
4 | 0.573894<$k$< 0.648978 | $I \succ M2 \succ M7 \succ M8 \succ M3 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |
5 | 0.648978<$k$< 0.765762 | $I \succ M2 \succ M7 \succ M8 \succ II \succ M3 \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |
6 | 0.765762<$k$< 0.841048 | $I \succ M2 \succ M7 \succ M8 \succ II \succ M3 \succ M4 \succ M1 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |
7 | 0.841048<$k$< 0.841412 | $I \succ M2 \succ M7 \succ M8 \succ II \succ M3 \succ M4 \succ M6 \succ M1 \succ M5 \succ M9 \succ III \succ IV$ |
8 | 0.841412<$k$< 0.860395 | $I \succ M2 \succ M7 \succ M8 \succ II \succ M4 \succ M3 \succ M6 \succ M1 \succ M5 \succ M9 \succ III \succ IV$ |
9 | 0.860395<$k$< 0.869579 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M3 \succ M6 \succ M1 \succ M5 \succ M9 \succ III \succ IV$ |
10 | 0.869579<$k$< 0.894625 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M3 \succ M1 \succ M5 \succ M9 \succ III \succ IV$ |
11 | 0.894625<$k$< 0.937958 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M3 \succ M5 \succ M9 \succ III \succ IV$ |
12 | 0.937958<$k$< 0.942387 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M3 \succ M9 \succ M5 \succ III \succ IV$ |
13 | 0.942387<$k$< 0.943476 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M9 \succ M3 \succ M5 \succ III \succ IV$ |
14 | 0.943476<$k$< 0.946435 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M9 \succ M5 \succ M3 \succ III \succ IV$ |
15 | 0.946435<$k$< 0.948488 | $I \succ M2 \succ M7 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M9 \succ M5 \succ III \succ M3 \succ IV$ |
16 | 0.948488<$k$< 0.948913 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M9 \succ M5 \succ III \succ M3 \succ IV$ |
17 | 0.948913<$k$< 0.951752 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ M9 \succ III \succ M5 \succ M3 \succ IV$ |
18 | 0.951752<$k$< 0.966134 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ M1 \succ III \succ M9 \succ M5 \succ M3 \succ IV$ |
19 | 0.966134<$k$< 0.987395 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ III \succ M1 \succ M9 \succ M5 \succ M3 \succ IV$ |
20 | 0.987395<$k$< 1 | $I \succ M7 \succ M2 \succ II \succ M8 \succ M4 \succ M6 \succ III \succ M9 \succ M1 \succ M5 \succ M3 \succ IV$ |
排名 | 要素所占区段 |
---|---|
0 | I=1 |
1 | M2=0.948488 M7=0.051512 |
2 | M7=0.933206 M2=0.051512 M3=0.015282 |
3 | M3=0.558612 M8=0.286501 II=0.139605 M7=0.015282 |
4 | M8=0.713499 II=0.211417 M3=0.075084 |
5 | II=0.648978 M3=0.192434 M4=0.158588 |
6 | M1=0.765762 M6=0.130421 M4=0.07565 M3=0.028167 |
7 | M4=0.765762 M1=0.146795 III=0.033866 M6=0.028531 M3=0.025046 |
8 | M6=0.518915 M5=0.322133 M1=0.074838 M3=0.047762 M9=0.02197 III=0.014382 |
9 | M5=0.621262 M6=0.322133 M9=0.040072 M1=0.012605 III=0.002839 M3=0.001089 |
10 | M9=0.937958 M5=0.056605 M3=0.002959 III=0.002478 |
11 | III=0.946435 M3=0.053565 |
12 | IV=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 | $I \succ M2 \succ M7 \succ M3 \succ M8 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |
劣汰情境 | $I \succ M2 \succ M7 \succ M3 \succ M8 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |
上述四种情况的聚类特征是一样的,即聚类的数目一样,每个聚类排序是一样的。
当某个决策系数k出现两个交点时候,聚类的数目会发生变化。
运用AECM求出的最优排序不同。这是由于聚类的区段值不同导致的。
排名 | 要素所占区段 |
---|---|
0 | I=1 |
1 | M2=0.948488 M7=0.051512 |
2 | M7=0.933206 M2=0.051512 M3=0.015282 |
3 | M3=0.558612 M8=0.286501 II=0.139605 M7=0.015282 |
4 | M8=0.713499 II=0.211417 M3=0.075084 |
5 | II=0.648978 M3=0.192434 M4=0.158588 |
6 | M1=0.765762 M6=0.130421 M4=0.07565 M3=0.028167 |
7 | M4=0.765762 M1=0.146795 III=0.033866 M6=0.028531 M3=0.025046 |
8 | M6=0.518915 M5=0.322133 M1=0.074838 M3=0.047762 M9=0.02197 III=0.014382 |
9 | M5=0.621262 M6=0.322133 M9=0.040072 M1=0.012605 III=0.002839 M3=0.001089 |
10 | M9=0.937958 M5=0.056605 M3=0.002959 III=0.002478 |
11 | III=0.946435 M3=0.053565 |
12 | IV=1 |
上表为专家投票后每个评价对象(要素)在每个层级所占的份额。
排名 | 最优妥协解 |
---|---|
优胜情境 | $I \succ M2 \succ M7 \succ M3 \succ M8 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |
劣汰情境 | $I \succ M2 \succ M7 \succ M3 \succ M8 \succ II \succ M1 \succ M4 \succ M6 \succ M5 \succ M9 \succ III \succ IV$ |