原始矩阵（直接影响矩阵）为

$$Ori=\begin{array}{|c|c|c|c|c|c|c|}\hline {M_{12 \times12}} &H1 &H2 &H3 &K1 &K2 &K3 &X1 &X2 &X3 &S1 &S2 &S3\\ \hline H1 &0 &2 &1 &0 &9 &0 &2 &9 &1 &0 &2 &1\\ \hline H2 &3 &0 &0 &0 &9 &8 &0 &7 &0 &3 &2 &1\\ \hline H3 &3 &9 &0 &3 &0 &0 &3 &0 &0 &0 &2 &6\\ \hline K1 &3 &9 &4 &0 &0 &0 &1 &5 &0 &0 &2 &5\\ \hline K2 &6 &9 &0 &0 &0 &0 &0 &0 &5 &0 &2 &1\\ \hline K3 &8 &5 &2 &0 &1 &0 &2 &0 &9 &0 &2 &1\\ \hline X1 &3 &0 &0 &0 &0 &0 &0 &0 &0 &0 &2 &1\\ \hline X2 &0 &6 &2 &0 &0 &0 &1 &0 &0 &2 &2 &1\\ \hline X3 &9 &0 &0 &0 &0 &0 &1 &0 &0 &0 &2 &1\\ \hline S1 &0 &0 &3 &3 &3 &3 &0 &0 &0 &0 &0 &1\\ \hline S2 &0 &3 &0 &0 &0 &0 &0 &0 &0 &0 &0 &1\\ \hline S3 &5 &0 &0 &4 &5 &8 &0 &0 &0 &0 &2 &0\\ \hline \end{array} $$

规范直接关系矩阵求解过程 $$ \require{cancel} \require{AMScd} \begin{CD} O @>>>N \\ \end{CD} $$

$$\mathcal{N}=\begin{array}{|c|c|c|c|c|c|c|}\hline {M_{12 \times12}} &H1 &H2 &H3 &K1 &K2 &K3 &X1 &X2 &X3 &S1 &S2 &S3\\ \hline H1 &0 &0.037 &0.018 &0 &0.166 &0 &0.037 &0.166 &0.018 &0 &0.037 &0.018\\ \hline H2 &0.055 &0 &0 &0 &0.166 &0.148 &0 &0.129 &0 &0.055 &0.037 &0.018\\ \hline H3 &0.055 &0.166 &0 &0.055 &0 &0 &0.055 &0 &0 &0 &0.037 &0.111\\ \hline K1 &0.055 &0.166 &0.074 &0 &0 &0 &0.018 &0.092 &0 &0 &0.037 &0.092\\ \hline K2 &0.111 &0.166 &0 &0 &0 &0 &0 &0 &0.092 &0 &0.037 &0.018\\ \hline K3 &0.148 &0.092 &0.037 &0 &0.018 &0 &0.037 &0 &0.166 &0 &0.037 &0.018\\ \hline X1 &0.055 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0.037 &0.018\\ \hline X2 &0 &0.111 &0.037 &0 &0 &0 &0.018 &0 &0 &0.037 &0.037 &0.018\\ \hline X3 &0.166 &0 &0 &0 &0 &0 &0.018 &0 &0 &0 &0.037 &0.018\\ \hline S1 &0 &0 &0.055 &0.055 &0.055 &0.055 &0 &0 &0 &0 &0 &0.018\\ \hline S2 &0 &0.055 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0.018\\ \hline S3 &0.092 &0 &0 &0.074 &0.092 &0.148 &0 &0 &0 &0 &0.037 &0\\ \hline \end{array} $$

综合影响矩阵求解过程 $$\begin{CD} N @>>>T \\ \end{CD} $$

　　综合影响矩阵如下

$T=\mathcal{N}(I-\mathcal{N})^{-1}$

$$T=\begin{array}{|c|c|c|c|c|c|c|}\hline {M_{12 \times12}} &H1 &H2 &H3 &K1 &K2 &K3 &X1 &X2 &X3 &S1 &S2 &S3\\ \hline H1 &0.045 &0.102 &0.028 &0.005 &0.195 &0.021 &0.045 &0.187 &0.041 &0.013 &0.063 &0.035\\ \hline H2 &0.122 &0.081 &0.019 &0.008 &0.21 &0.169 &0.016 &0.161 &0.05 &0.066 &0.069 &0.039\\ \hline H3 &0.104 &0.207 &0.011 &0.066 &0.065 &0.051 &0.064 &0.05 &0.016 &0.013 &0.065 &0.13\\ \hline K1 &0.103 &0.221 &0.085 &0.014 &0.066 &0.051 &0.032 &0.139 &0.016 &0.017 &0.068 &0.116\\ \hline K2 &0.155 &0.196 &0.007 &0.003 &0.063 &0.034 &0.01 &0.051 &0.107 &0.013 &0.061 &0.032\\ \hline K3 &0.207 &0.133 &0.045 &0.006 &0.08 &0.026 &0.052 &0.052 &0.182 &0.009 &0.067 &0.039\\ \hline X1 &0.061 &0.009 &0.002 &0.002 &0.014 &0.005 &0.003 &0.011 &0.003 &0.001 &0.042 &0.022\\ \hline X2 &0.023 &0.133 &0.042 &0.007 &0.032 &0.027 &0.023 &0.022 &0.008 &0.045 &0.05 &0.03\\ \hline X3 &0.178 &0.021 &0.005 &0.002 &0.036 &0.007 &0.026 &0.032 &0.008 &0.002 &0.049 &0.026\\ \hline S1 &0.034 &0.043 &0.064 &0.062 &0.073 &0.067 &0.009 &0.017 &0.019 &0.003 &0.016 &0.036\\ \hline S2 &0.01 &0.061 &0.001 &0.002 &0.014 &0.012 &0.001 &0.01 &0.004 &0.004 &0.005 &0.021\\ \hline S3 &0.149 &0.066 &0.016 &0.077 &0.133 &0.161 &0.015 &0.04 &0.042 &0.005 &0.063 &0.021\\ \hline \end{array} $$

由综合影响矩阵求影响度、被影响度、中心度、原因度 \begin{CD} T @>>>\{D|C|M|R \} \\ \end{CD}

　　影响度、被影响度、中心度与原因度是四种度量要素在系统里影响程度的度量值。都是根据综合影响矩阵计算得出。

求解原理

影响度 $D$	$$ D_i=\sum \limits_{j=1}^{n}{t_{ij}},(i=1,2,3,\cdots,n) $$
被影响度 $C$	$$ C_i=\sum \limits_{j=1}^{n}{t_{ji}},(i=1,2,3,\cdots,n) $$
中心度 $M$	$$ M_i=D_i+C_i $$
原因度 $ R$	$$ R_i=D_i-C_i $$

结果

影响度、被影响度、中心度、原因度

$$\begin{array}{|c|c|c|c|c|c|c|}\hline {M_{12 \times4}} &Di &Ci &Mi &Ri\\ \hline H1 &0.779 &1.191 &1.97 &-0.411\\ \hline H2 &1.008 &1.274 &2.281 &-0.266\\ \hline H3 &0.844 &0.324 &1.168 &0.519\\ \hline K1 &0.93 &0.253 &1.183 &0.676\\ \hline K2 &0.734 &0.98 &1.714 &-0.246\\ \hline K3 &0.898 &0.63 &1.527 &0.268\\ \hline X1 &0.174 &0.298 &0.471 &-0.124\\ \hline X2 &0.44 &0.773 &1.213 &-0.333\\ \hline X3 &0.392 &0.494 &0.886 &-0.101\\ \hline S1 &0.443 &0.191 &0.634 &0.252\\ \hline S2 &0.144 &0.619 &0.763 &-0.474\\ \hline S3 &0.789 &0.547 &1.336 &0.242\\ \hline \end{array} $$

绘制中心度与原因度建构的散点图图表

从两列的决策矩阵到关系矩阵

$$ \begin{CD} D=\left[ d_{ij} \right]_{n \times 2}@>取偏序>>A=\left[a_{ij} \right]_{n \times n} \\ \end{CD} $$

中心度，原因度绝对值组成的决策矩阵D

$$D=\begin{array}{c|c|c|c|c|c|c}{M_{12 \times2}} &M &|R|\\ \hline H1 &1.9702 &0.4115\\ \hline H2 &2.2814 &0.2663\\ \hline H3 &1.1682 &0.5194\\ \hline K1 &1.183 &0.6762\\ \hline K2 &1.714 &0.2461\\ \hline K3 &1.5274 &0.2676\\ \hline X1 &0.4714 &0.124\\ \hline X2 &1.2134 &0.3327\\ \hline X3 &0.886 &0.1014\\ \hline S1 &0.6341 &0.2516\\ \hline S2 &0.7627 &0.4744\\ \hline S3 &1.336 &0.2415\\ \hline \end{array} $$

关系矩阵$A$

$$A=\begin{array} {c|c|c|c|c|c|c|c}{M_{12 \times12}} &H1 &H2 &H3 &K1 &K2 &K3 &X1 &X2 &X3 &S1 &S2 &S3\\ \hline H1 &1 & & & & & & & & & & & \\ \hline H2 & &1 & & & & & & & & & & \\ \hline H3 & & &1 &1 & & & & & & & & \\ \hline K1 & & & &1 & & & & & & & & \\ \hline K2 &1 &1 & & &1 & & & & & & & \\ \hline K3 &1 & & & & &1 & & & & & & \\ \hline X1 &1 &1 &1 &1 &1 &1 &1 &1 & &1 &1 &1\\ \hline X2 &1 & & & & & & &1 & & & & \\ \hline X3 &1 &1 &1 &1 &1 &1 & &1 &1 & & &1\\ \hline S1 &1 &1 &1 &1 & &1 & &1 & &1 &1 & \\ \hline S2 & & &1 &1 & & & & & & &1 & \\ \hline S3 &1 &1 & & &1 &1 & & & & & &1\\ \hline \end{array} $$

AISM部分求解

关系矩阵到相乘矩阵

\begin{CD} A@>A+I>>B \\ \end{CD}

$$B=\begin{array} {c|c|c|c|c|c|c|c}{M_{12 \times12}} &H1 &H2 &H3 &K1 &K2 &K3 &X1 &X2 &X3 &S1 &S2 &S3\\ \hline H1 &1 & & & & & & & & & & & \\ \hline H2 & &1 & & & & & & & & & & \\ \hline H3 & & &1 &1 & & & & & & & & \\ \hline K1 & & & &1 & & & & & & & & \\ \hline K2 &1 &1 & & &1 & & & & & & & \\ \hline K3 &1 & & & & &1 & & & & & & \\ \hline X1 &1 &1 &1 &1 &1 &1 &1 &1 & &1 &1 &1\\ \hline X2 &1 & & & & & & &1 & & & & \\ \hline X3 &1 &1 &1 &1 &1 &1 & &1 &1 & & &1\\ \hline S1 &1 &1 &1 &1 & &1 & &1 & &1 &1 & \\ \hline S2 & & &1 &1 & & & & & & &1 & \\ \hline S3 &1 &1 & & &1 &1 & & & & & &1\\ \hline \end{array} $$

相乘矩阵B到可达矩阵R

\begin{CD} B@>连乘或者幂乘>>R \\ \end{CD}

可达矩阵R

$$R=\begin{array} {c|c|c|c|c|c|c|c}{M_{12 \times12}} &H1 &H2 &H3 &K1 &K2 &K3 &X1 &X2 &X3 &S1 &S2 &S3\\ \hline H1 &1 & & & & & & & & & & & \\ \hline H2 & &1 & & & & & & & & & & \\ \hline H3 & & &1 &1 & & & & & & & & \\ \hline K1 & & & &1 & & & & & & & & \\ \hline K2 &1 &1 & & &1 & & & & & & & \\ \hline K3 &1 & & & & &1 & & & & & & \\ \hline X1 &1 &1 &1 &1 &1 &1 &1 &1 & &1 &1 &1\\ \hline X2 &1 & & & & & & &1 & & & & \\ \hline X3 &1 &1 &1 &1 &1 &1 & &1 &1 & & &1\\ \hline S1 &1 &1 &1 &1 & &1 & &1 & &1 &1 & \\ \hline S2 & & &1 &1 & & & & & & &1 & \\ \hline S3 &1 &1 & & &1 &1 & & & & & &1\\ \hline \end{array} $$

由可达矩阵通过对抗的抽取方式得到一对要素的层级分布 \begin{CD} R @>结果优先抽取>原因优先抽取> \frac {up型要素层级分布 }{down 型要素层级分布} \\ \end{CD}

可达矩阵R层级抽取的过程如下

结果优先——UP型抽取过程	原因优先——DOWN型抽取过程
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline H1&\color{red}{\fbox{H1}}&\color{red}{\fbox{H1}} \\\hline H2&\color{red}{\fbox{H2}}&\color{red}{\fbox{H2}} \\\hline H3&H3,K1&H3 \\\hline K1&\color{red}{\fbox{K1}}&\color{red}{\fbox{K1}} \\\hline K2&H1,H2,K2&K2 \\\hline K3&H1,K3&K3 \\\hline X1&H1,H2,H3,K1,K2,K3,X1,X2,S1,S2,S3&X1 \\\hline X2&H1,X2&X2 \\\hline X3&H1,H2,H3,K1,K2,K3,X2,X3,S3&X3 \\\hline S1&H1,H2,H3,K1,K3,X2,S1,S2&S1 \\\hline S2&H3,K1,S2&S2 \\\hline S3&H1,H2,K2,K3,S3&S3 \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline H1&H1,K2,K3,X1,X2,X3,S1,S3&H1 \\\hline H2&H2,K2,X1,X3,S1,S3&H2 \\\hline H3&H3,X1,X3,S1,S2&H3 \\\hline K1&H3,K1,X1,X3,S1,S2&K1 \\\hline K2&K2,X1,X3,S3&K2 \\\hline K3&K3,X1,X3,S1,S3&K3 \\\hline X1&\color{blue}{\fbox{X1}}&\color{blue}{\fbox{X1}} \\\hline X2&X1,X2,X3,S1&X2 \\\hline X3&\color{blue}{\fbox{X3}}&\color{blue}{\fbox{X3}} \\\hline S1&X1,S1&S1 \\\hline S2&X1,S1,S2&S2 \\\hline S3&X1,X3,S3&S3 \\\hline \end{array} $$
抽取出H1、H2、K1放置上层，删除后剩余的情况如下	抽取出X1,X3放置下层，删除后剩余的情况如下
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline H3&\color{red}{\fbox{H3}}&\color{red}{\fbox{H3}} \\\hline K2&\color{red}{\fbox{K2}}&\color{red}{\fbox{K2}} \\\hline K3&\color{red}{\fbox{K3}}&\color{red}{\fbox{K3}} \\\hline X1&H3,K2,K3,X1,X2,S1,S2,S3&X1 \\\hline X2&\color{red}{\fbox{X2}}&\color{red}{\fbox{X2}} \\\hline X3&H3,K2,K3,X2,X3,S3&X3 \\\hline S1&H3,K3,X2,S1,S2&S1 \\\hline S2&H3,S2&S2 \\\hline S3&K2,K3,S3&S3 \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline H1&H1,K2,K3,X2,S1,S3&H1 \\\hline H2&H2,K2,S1,S3&H2 \\\hline H3&H3,S1,S2&H3 \\\hline K1&H3,K1,S1,S2&K1 \\\hline K2&K2,S3&K2 \\\hline K3&K3,S1,S3&K3 \\\hline X2&X2,S1&X2 \\\hline S1&\color{blue}{\fbox{S1}}&\color{blue}{\fbox{S1}} \\\hline S2&S1,S2&S2 \\\hline S3&\color{blue}{\fbox{S3}}&\color{blue}{\fbox{S3}} \\\hline \end{array} $$
抽取出H3、K2、K3、X2放置上层，删除后剩余的情况如下	抽取出S1,S3放置下层，删除后剩余的情况如下
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline X1&X1,S1,S2,S3&X1 \\\hline X3&X3,S3&X3 \\\hline S1&S1,S2&S1 \\\hline S2&\color{red}{\fbox{S2}}&\color{red}{\fbox{S2}} \\\hline S3&\color{red}{\fbox{S3}}&\color{red}{\fbox{S3}} \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline H1&H1,K2,K3,X2&H1 \\\hline H2&H2,K2&H2 \\\hline H3&H3,S2&H3 \\\hline K1&H3,K1,S2&K1 \\\hline K2&\color{blue}{\fbox{K2}}&\color{blue}{\fbox{K2}} \\\hline K3&\color{blue}{\fbox{K3}}&\color{blue}{\fbox{K3}} \\\hline X2&\color{blue}{\fbox{X2}}&\color{blue}{\fbox{X2}} \\\hline S2&\color{blue}{\fbox{S2}}&\color{blue}{\fbox{S2}} \\\hline \end{array} $$
抽取出S2、S3放置上层，删除后剩余的情况如下	抽取出K2,K3,X2,S2放置下层，删除后剩余的情况如下
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline X1&X1,S1&X1 \\\hline X3&\color{red}{\fbox{X3}}&\color{red}{\fbox{X3}} \\\hline S1&\color{red}{\fbox{S1}}&\color{red}{\fbox{S1}} \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline H1&\color{blue}{\fbox{H1}}&\color{blue}{\fbox{H1}} \\\hline H2&\color{blue}{\fbox{H2}}&\color{blue}{\fbox{H2}} \\\hline H3&\color{blue}{\fbox{H3}}&\color{blue}{\fbox{H3}} \\\hline K1&H3,K1&K1 \\\hline \end{array} $$
抽取出X3、S1放置上层，删除后剩余的情况如下	抽取出H1,H2,H3放置下层，删除后剩余的情况如下
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline X1&\color{red}{\fbox{X1}}&\color{red}{\fbox{X1}} \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline K1&\color{blue}{\fbox{K1}}&\color{blue}{\fbox{K1}} \\\hline \end{array} $$
抽取出X1放置上层，删除后剩余的情况如下	抽取出K1放置下层，删除后剩余的情况如下

抽取方式的结果如下

层级	结果优先——UP型	原因优先——DOWN型
第0层	H1,H2,K1	K1
第1层	H3,K2,K3,X2	H1,H2,H3
第2层	S2,S3	K2,K3,X2,S2
第3层	X3,S1	S1,S3
第4层	X1	X1,X3

计算一般性骨架矩阵 \begin{CD} R @>缩点运算>>R' @>缩边运算>>S' @>以最简菊花链表示回路>>S \\ \end{CD}

一般性骨架矩阵即为不缩点的情况下的最简结构，即边数最少。$S$如下

$$S=\begin{array} {c|c|c|c|c|c|c|c}{M_{12 \times12}} &H1 &H2 &H3 &K1 &K2 &K3 &X1 &X2 &X3 &S1 &S2 &S3\\ \hline H1 & & & & & & & & & & & & \\ \hline H2 & & & & & & & & & & & & \\ \hline H3 & & & &1 & & & & & & & & \\ \hline K1 & & & & & & & & & & & & \\ \hline K2 &1 &1 & & & & & & & & & & \\ \hline K3 &1 & & & & & & & & & & & \\ \hline X1 & & & & & & & & & &1 & &1\\ \hline X2 &1 & & & & & & & & & & & \\ \hline X3 & & &1 & & & & &1 & & & &1\\ \hline S1 & &1 & & & &1 & &1 & & &1 & \\ \hline S2 & & &1 & & & & & & & & & \\ \hline S3 & & & & &1 &1 & & & & & & \\ \hline \end{array} $$

代入一般性骨架矩阵$S$，绘制对抗层级拓扑图

UP型

DOWN型

权重系列求解

$$ \begin{CD} \{M|R \}@>>> d @>>>\omega \\ \end{CD} $$

　　$ \{M|R \}$ 为要素的中心度与原因度

　　$d$　两者合成距离向量

　　$ \omega $ 权重

求解原理

$$d = \sqrt {M^2 + R^2} $$

$$MR=\begin{array}{c|c|c|c|c|c|c}{M_{12 \times2}} &中心度 &原因度\\ \hline H1 &1.9702 &0.4115\\ \hline H2 &2.2814 &0.2663\\ \hline H3 &1.1682 &0.5194\\ \hline K1 &1.183 &0.6762\\ \hline K2 &1.714 &0.2461\\ \hline K3 &1.5274 &0.2676\\ \hline X1 &0.4714 &0.124\\ \hline X2 &1.2134 &0.3327\\ \hline X3 &0.886 &0.1014\\ \hline S1 &0.6341 &0.2516\\ \hline S2 &0.7627 &0.4744\\ \hline S3 &1.336 &0.2415\\ \hline \end{array} $$

向量的计算采用欧式距离公式

$$d=\begin{array}{c|c|c|c|c|c|c}{M_{12 \times1}} &向量\\ \hline H1 &2.0127\\ \hline H2 &2.2969\\ \hline H3 &1.2785\\ \hline K1 &1.3627\\ \hline K2 &1.7316\\ \hline K3 &1.5507\\ \hline X1 &0.4874\\ \hline X2 &1.2582\\ \hline X3 &0.8918\\ \hline S1 &0.6822\\ \hline S2 &0.8982\\ \hline S3 &1.3577\\ \hline \end{array} $$

归一化求权重

$$\omega =\begin{array}{c|c|c|c|c|c|c}{M_{12 \times1}} &权重\\ \hline H1 &0.1273\\ \hline H2 &0.1453\\ \hline H3 &0.0809\\ \hline K1 &0.0862\\ \hline K2 &0.1095\\ \hline K3 &0.0981\\ \hline X1 &0.0308\\ \hline X2 &0.0796\\ \hline X3 &0.0564\\ \hline S1 &0.0432\\ \hline S2 &0.0568\\ \hline S3 &0.0859\\ \hline \end{array} $$

归一化求子系统的权重

$$\omega =\begin{array}{c|c|c|c|c|c|c}{M_{4 \times1}} &权重\\ \hline H &0.3535\\ \hline K &0.2938\\ \hline X &0.1668\\ \hline S &0.1859\\ \hline \end{array} $$

DEMATEL-AISM-Weight

原始矩阵（直接影响矩阵）为

规范直接关系矩阵求解过程 $$ \require{cancel} \require{AMScd} \begin{CD} O @>>>N \\ \end{CD} $$

综合影响矩阵求解过程 $$\begin{CD} N @>>>T \\ \end{CD} $$

由综合影响矩阵求影响度、被影响度、中心度、原因度 \begin{CD} T @>>>\{D|C|M|R \} \\ \end{CD}

求解原理

结果

绘制中心度与原因度建构的散点图图表

从两列的决策矩阵到关系矩阵

AISM部分求解

关系矩阵到相乘矩阵

相乘矩阵B到可达矩阵R

由可达矩阵通过对抗的抽取方式得到一对要素的层级分布 \begin{CD} R @>结果优先抽取>原因优先抽取> \frac {up型要素层级分布 }{down 型要素层级分布} \\ \end{CD}

抽取方式的结果如下

计算一般性骨架矩阵 \begin{CD} R @>缩点运算>>R' @>缩边运算>>S' @>以最简菊花链表示回路>>S \\ \end{CD}

代入一般性骨架矩阵$S$，绘制对抗层级拓扑图

UP型

DOWN型

权重系列求解

求解原理

DEMATEL-AHDT流程图如下：