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 $$ \require{cancel} \require{AMScd} \begin{CD} O @>>>N \\ \end{CD} $$

$$\mathcal{N}=\begin{array}{|c|c|c|c|c|c|c|}\hline {M_{12 \times12}} &H1 &H2 &H3 &K1 &K2 &K3 &X1 &X2 &X3 &S1 &S2 &S3\\ \hline H1 &0 &0.061 &0.03 &0 &0.273 &0 &0.061 &0.273 &0.03 &0 &0.061 &0.03\\ \hline H2 &0.091 &0 &0 &0 &0.273 &0.242 &0 &0.212 &0 &0.091 &0.061 &0.03\\ \hline H3 &0.091 &0.273 &0 &0.091 &0 &0 &0.091 &0 &0 &0 &0.061 &0.182\\ \hline K1 &0.091 &0.273 &0.121 &0 &0 &0 &0.03 &0.152 &0 &0 &0.061 &0.152\\ \hline K2 &0.182 &0.273 &0 &0 &0 &0 &0 &0 &0.152 &0 &0.061 &0.03\\ \hline K3 &0.242 &0.152 &0.061 &0 &0.03 &0 &0.061 &0 &0.273 &0 &0.061 &0.03\\ \hline X1 &0.091 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0.061 &0.03\\ \hline X2 &0 &0.182 &0.061 &0 &0 &0 &0.03 &0 &0 &0.061 &0.061 &0.03\\ \hline X3 &0.273 &0 &0 &0 &0 &0 &0.03 &0 &0 &0 &0.061 &0.03\\ \hline S1 &0 &0 &0.091 &0.091 &0.091 &0.091 &0 &0 &0 &0 &0 &0.03\\ \hline S2 &0 &0.091 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0.03\\ \hline S3 &0.152 &0 &0 &0.121 &0.152 &0.242 &0 &0 &0 &0 &0.061 &0\\ \hline \end{array} $$

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

　　综合影响矩阵如下

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

$$T=\begin{array}{|c|c|c|c|c|c|c|}\hline {M_{12 \times12}} &H1 &H2 &H3 &K1 &K2 &K3 &X1 &X2 &X3 &S1 &S2 &S3\\ \hline H1 &0.209 &0.328 &0.076 &0.025 &0.444 &0.11 &0.104 &0.403 &0.134 &0.054 &0.178 &0.107\\ \hline H2 &0.395 &0.344 &0.076 &0.036 &0.518 &0.371 &0.072 &0.398 &0.192 &0.146 &0.214 &0.13\\ \hline H3 &0.338 &0.511 &0.057 &0.135 &0.285 &0.196 &0.142 &0.221 &0.107 &0.06 &0.198 &0.275\\ \hline K1 &0.346 &0.557 &0.18 &0.055 &0.299 &0.205 &0.096 &0.372 &0.112 &0.073 &0.211 &0.261\\ \hline K2 &0.396 &0.459 &0.041 &0.021 &0.258 &0.141 &0.05 &0.208 &0.241 &0.054 &0.177 &0.101\\ \hline K3 &0.503 &0.377 &0.105 &0.029 &0.297 &0.125 &0.127 &0.222 &0.367 &0.048 &0.198 &0.122\\ \hline X1 &0.125 &0.046 &0.009 &0.007 &0.055 &0.023 &0.012 &0.045 &0.018 &0.007 &0.084 &0.046\\ \hline X2 &0.121 &0.304 &0.088 &0.027 &0.14 &0.103 &0.057 &0.102 &0.053 &0.094 &0.126 &0.084\\ \hline X3 &0.349 &0.107 &0.023 &0.012 &0.137 &0.044 &0.062 &0.12 &0.043 &0.017 &0.119 &0.067\\ \hline S1 &0.156 &0.182 &0.128 &0.117 &0.206 &0.162 &0.04 &0.099 &0.08 &0.023 &0.077 &0.102\\ \hline S2 &0.048 &0.131 &0.009 &0.008 &0.058 &0.044 &0.009 &0.042 &0.022 &0.014 &0.025 &0.045\\ \hline S3 &0.41 &0.286 &0.065 &0.142 &0.369 &0.338 &0.066 &0.194 &0.161 &0.038 &0.19 &0.096\\ \hline \end{array} $$

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

$$\begin{array}{|c|c|c|c|c|c|c|}\hline {M_{12 \times4}} &Di &Ci &Mi &Ri\\ \hline H1 &2.171 &3.396 &5.567 &-1.225\\ \hline H2 &2.893 &3.632 &6.525 &-0.74\\ \hline H3 &2.525 &0.858 &3.383 &1.667\\ \hline K1 &2.767 &0.612 &3.379 &2.155\\ \hline K2 &2.146 &3.066 &5.212 &-0.92\\ \hline K3 &2.52 &1.862 &4.381 &0.658\\ \hline X1 &0.478 &0.838 &1.316 &-0.36\\ \hline X2 &1.3 &2.426 &3.725 &-1.126\\ \hline X3 &1.1 &1.53 &2.629 &-0.43\\ \hline S1 &1.371 &0.629 &1.999 &0.742\\ \hline S2 &0.456 &1.797 &2.253 &-1.342\\ \hline S3 &2.355 &1.436 &3.791 &0.919\\ \hline \end{array} $$

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

$$D=\begin{array}{c|c|c|c|c|c|c}{M_{12 \times2}} &M &|R|\\ \hline H1 &5.5668 &1.2245\\ \hline H2 &6.5252 &0.7396\\ \hline H3 &3.3828 &1.6673\\ \hline K1 &3.3793 &2.1549\\ \hline K2 &5.2123 &0.9198\\ \hline K3 &4.3812 &0.6581\\ \hline X1 &1.3162 &0.3599\\ \hline X2 &3.7255 &1.1259\\ \hline X3 &2.6292 &0.4302\\ \hline S1 &1.9993 &0.7418\\ \hline S2 &2.2528 &1.3417\\ \hline S3 &3.7911 &0.9194\\ \hline \end{array} $$

关系矩阵$A$

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

AISM部分求解

关系矩阵到相乘矩阵

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

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

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

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

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
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline H1&\color{red}{\fbox{H1}}&\color{red}{\fbox{H1}} \\\hline H2&\color{red}{\fbox{H2}}&\color{red}{\fbox{H2}} \\\hline H3&\color{red}{\fbox{H3}}&\color{red}{\fbox{H3}} \\\hline K1&\color{red}{\fbox{K1}}&\color{red}{\fbox{K1}} \\\hline K2&H1,K2&K2 \\\hline K3&H1,H2,K2,K3&K3 \\\hline X1&H1,H2,H3,K1,K2,K3,X1,X2,X3,S1,S2,S3&X1 \\\hline X2&H1,X2&X2 \\\hline X3&H1,H2,H3,K1,K2,K3,X2,X3,S3&X3 \\\hline S1&H1,H3,K1,K2,X2,S1,S2,S3&S1 \\\hline S2&H3,K1,S2&S2 \\\hline S3&H1,K2,S3&S3 \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline H1&H1,K2,K3,X1,X2,X3,S1,S3&H1 \\\hline H2&H2,K3,X1,X3&H2 \\\hline H3&H3,X1,X3,S1,S2&H3 \\\hline K1&K1,X1,X3,S1,S2&K1 \\\hline K2&K2,K3,X1,X3,S1,S3&K2 \\\hline K3&K3,X1,X3&K3 \\\hline X1&\color{blue}{\fbox{X1}}&\color{blue}{\fbox{X1}} \\\hline X2&X1,X2,X3,S1&X2 \\\hline X3&X1,X3&X3 \\\hline S1&X1,S1&S1 \\\hline S2&X1,S1,S2&S2 \\\hline S3&X1,X3,S1,S3&S3 \\\hline \end{array} $$
抽取出H1、H2、H3、K1放置上层，删除后剩余的情况如下	抽取出X1放置下层，删除后剩余的情况如下
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline K2&\color{red}{\fbox{K2}}&\color{red}{\fbox{K2}} \\\hline K3&K2,K3&K3 \\\hline X1&K2,K3,X1,X2,X3,S1,S2,S3&X1 \\\hline X2&\color{red}{\fbox{X2}}&\color{red}{\fbox{X2}} \\\hline X3&K2,K3,X2,X3,S3&X3 \\\hline S1&K2,X2,S1,S2,S3&S1 \\\hline S2&\color{red}{\fbox{S2}}&\color{red}{\fbox{S2}} \\\hline S3&K2,S3&S3 \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline H1&H1,K2,K3,X2,X3,S1,S3&H1 \\\hline H2&H2,K3,X3&H2 \\\hline H3&H3,X3,S1,S2&H3 \\\hline K1&K1,X3,S1,S2&K1 \\\hline K2&K2,K3,X3,S1,S3&K2 \\\hline K3&K3,X3&K3 \\\hline X2&X2,X3,S1&X2 \\\hline X3&\color{blue}{\fbox{X3}}&\color{blue}{\fbox{X3}} \\\hline S1&\color{blue}{\fbox{S1}}&\color{blue}{\fbox{S1}} \\\hline S2&S1,S2&S2 \\\hline S3&X3,S1,S3&S3 \\\hline \end{array} $$
抽取出K2、X2、S2放置上层，删除后剩余的情况如下	抽取出X3,S1放置下层，删除后剩余的情况如下
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline K3&\color{red}{\fbox{K3}}&\color{red}{\fbox{K3}} \\\hline X1&K3,X1,X3,S1,S3&X1 \\\hline X3&K3,X3,S3&X3 \\\hline S1&S1,S3&S1 \\\hline S3&\color{red}{\fbox{S3}}&\color{red}{\fbox{S3}} \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline H1&H1,K2,K3,X2,S3&H1 \\\hline H2&H2,K3&H2 \\\hline H3&H3,S2&H3 \\\hline K1&K1,S2&K1 \\\hline K2&K2,K3,S3&K2 \\\hline K3&\color{blue}{\fbox{K3}}&\color{blue}{\fbox{K3}} \\\hline X2&\color{blue}{\fbox{X2}}&\color{blue}{\fbox{X2}} \\\hline S2&\color{blue}{\fbox{S2}}&\color{blue}{\fbox{S2}} \\\hline S3&\color{blue}{\fbox{S3}}&\color{blue}{\fbox{S3}} \\\hline \end{array} $$
抽取出K3、S3放置上层，删除后剩余的情况如下	抽取出K3,X2,S2,S3放置下层，删除后剩余的情况如下
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline X1&X1,X3,S1&X1 \\\hline X3&\color{red}{\fbox{X3}}&\color{red}{\fbox{X3}} \\\hline S1&\color{red}{\fbox{S1}}&\color{red}{\fbox{S1}} \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline H1&H1,K2&H1 \\\hline H2&\color{blue}{\fbox{H2}}&\color{blue}{\fbox{H2}} \\\hline H3&\color{blue}{\fbox{H3}}&\color{blue}{\fbox{H3}} \\\hline K1&\color{blue}{\fbox{K1}}&\color{blue}{\fbox{K1}} \\\hline K2&\color{blue}{\fbox{K2}}&\color{blue}{\fbox{K2}} \\\hline \end{array} $$
抽取出X3、S1放置上层，删除后剩余的情况如下	抽取出H2,H3,K1,K2放置下层，删除后剩余的情况如下
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline X1&\color{red}{\fbox{X1}}&\color{red}{\fbox{X1}} \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline H1&\color{blue}{\fbox{H1}}&\color{blue}{\fbox{H1}} \\\hline \end{array} $$
抽取出X1放置上层，删除后剩余的情况如下	抽取出H1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,H3,K1	H1
第1层	K2,X2,S2	H2,H3,K1,K2
第2层	K3,S3	K3,X2,S2,S3
第3层	X3,S1	X3,S1
第4层	X1	X1

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

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

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

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

$$MR=\begin{array}{c|c|c|c|c|c|c}{M_{12 \times2}} &中心度 &原因度\\ \hline H1 &5.5668 &1.2245\\ \hline H2 &6.5252 &0.7396\\ \hline H3 &3.3828 &1.6673\\ \hline K1 &3.3793 &2.1549\\ \hline K2 &5.2123 &0.9198\\ \hline K3 &4.3812 &0.6581\\ \hline X1 &1.3162 &0.3599\\ \hline X2 &3.7255 &1.1259\\ \hline X3 &2.6292 &0.4302\\ \hline S1 &1.9993 &0.7418\\ \hline S2 &2.2528 &1.3417\\ \hline S3 &3.7911 &0.9194\\ \hline \end{array} $$

向量的计算采用欧式距离公式

$$d=\begin{array}{c|c|c|c|c|c|c}{M_{12 \times1}} &向量\\ \hline H1 &5.6999\\ \hline H2 &6.567\\ \hline H3 &3.7714\\ \hline K1 &4.0079\\ \hline K2 &5.2928\\ \hline K3 &4.4304\\ \hline X1 &1.3645\\ \hline X2 &3.8919\\ \hline X3 &2.6642\\ \hline S1 &2.1325\\ \hline S2 &2.6221\\ \hline S3 &3.901\\ \hline \end{array} $$

归一化求权重

$$\omega =\begin{array}{c|c|c|c|c|c|c}{M_{12 \times1}} &权重\\ \hline H1 &0.123\\ \hline H2 &0.1417\\ \hline H3 &0.0814\\ \hline K1 &0.0865\\ \hline K2 &0.1142\\ \hline K3 &0.0956\\ \hline X1 &0.0294\\ \hline X2 &0.084\\ \hline X3 &0.0575\\ \hline S1 &0.046\\ \hline S2 &0.0566\\ \hline S3 &0.0842\\ \hline \end{array} $$

归一化求子系统的权重

$$\omega =\begin{array}{c|c|c|c|c|c|c}{M_{4 \times1}} &权重\\ \hline H &0.3461\\ \hline K &0.2963\\ \hline X &0.1709\\ \hline S &0.1868\\ \hline \end{array} $$

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：