原始矩阵（直接影响矩阵）为

$$Ori=\begin{array}{|c|c|c|c|c|c|c|}\hline {M_{12 \times12}} &R1 &R2 &R3 &K1 &K2 &K3 &X1 &X2 &X3 &T1 &T2 &T3\\ \hline R1 &0 &2 &1 &0 &0 &5 &2 &9 &1 &0 &2 &1\\ \hline R2 &3 &0 &0 &0 &0 &5 &0 &7 &7 &7 &7 &7\\ \hline R3 &3 &9 &0 &6 &6 &5 &6 &6 &6 &6 &6 &6\\ \hline K1 &3 &9 &4 &0 &0 &5 &1 &0 &8 &4 &8 &5\\ \hline K2 &6 &9 &0 &0 &0 &5 &0 &0 &5 &0 &2 &1\\ \hline K3 &8 &5 &2 &0 &1 &0 &2 &0 &9 &0 &2 &1\\ \hline X1 &3 &0 &0 &0 &0 &5 &0 &0 &0 &0 &2 &1\\ \hline X2 &0 &6 &2 &0 &0 &5 &1 &0 &0 &2 &2 &1\\ \hline X3 &9 &0 &0 &0 &0 &5 &1 &0 &0 &0 &2 &1\\ \hline T1 &0 &0 &3 &3 &3 &5 &0 &0 &0 &0 &0 &1\\ \hline T2 &0 &3 &0 &0 &0 &5 &0 &0 &0 &0 &0 &1\\ \hline T3 &5 &0 &0 &0 &0 &5 &0 &0 &0 &0 &2 &0\\ \hline \end{array} $$

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

$$\mathcal{N}=\begin{array}{|c|c|c|c|c|c|c|}\hline {M_{12 \times12}} &R1 &R2 &R3 &K1 &K2 &K3 &X1 &X2 &X3 &T1 &T2 &T3\\ \hline R1 &0 &0.023 &0.012 &0 &0 &0.059 &0.023 &0.106 &0.012 &0 &0.023 &0.012\\ \hline R2 &0.035 &0 &0 &0 &0 &0.059 &0 &0.082 &0.082 &0.082 &0.082 &0.082\\ \hline R3 &0.035 &0.106 &0 &0.07 &0.07 &0.059 &0.07 &0.07 &0.07 &0.07 &0.07 &0.07\\ \hline K1 &0.035 &0.106 &0.047 &0 &0 &0.059 &0.012 &0 &0.094 &0.047 &0.094 &0.059\\ \hline K2 &0.07 &0.106 &0 &0 &0 &0.059 &0 &0 &0.059 &0 &0.023 &0.012\\ \hline K3 &0.094 &0.059 &0.023 &0 &0.012 &0 &0.023 &0 &0.106 &0 &0.023 &0.012\\ \hline X1 &0.035 &0 &0 &0 &0 &0.059 &0 &0 &0 &0 &0.023 &0.012\\ \hline X2 &0 &0.07 &0.023 &0 &0 &0.059 &0.012 &0 &0 &0.023 &0.023 &0.012\\ \hline X3 &0.106 &0 &0 &0 &0 &0.059 &0.012 &0 &0 &0 &0.023 &0.012\\ \hline T1 &0 &0 &0.035 &0.035 &0.035 &0.059 &0 &0 &0 &0 &0 &0.012\\ \hline T2 &0 &0.035 &0 &0 &0 &0.059 &0 &0 &0 &0 &0 &0.012\\ \hline T3 &0.059 &0 &0 &0 &0 &0.059 &0 &0 &0 &0 &0.023 &0\\ \hline \end{array} $$

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

　　综合影响矩阵如下

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

$$T=\begin{array}{|c|c|c|c|c|c|c|}\hline {M_{12 \times12}} &R1 &R2 &R3 &K1 &K2 &K3 &X1 &X2 &X3 &T1 &T2 &T3\\ \hline R1 &0.014 &0.04 &0.017 &0.001 &0.002 &0.076 &0.028 &0.112 &0.025 &0.007 &0.035 &0.02\\ \hline R2 &0.061 &0.018 &0.008 &0.004 &0.005 &0.092 &0.006 &0.091 &0.095 &0.087 &0.095 &0.091\\ \hline R3 &0.079 &0.142 &0.012 &0.074 &0.076 &0.115 &0.079 &0.091 &0.107 &0.089 &0.105 &0.096\\ \hline K1 &0.07 &0.126 &0.053 &0.006 &0.007 &0.099 &0.021 &0.021 &0.12 &0.062 &0.118 &0.079\\ \hline K2 &0.093 &0.116 &0.004 &0.001 &0.002 &0.082 &0.006 &0.02 &0.078 &0.01 &0.041 &0.025\\ \hline K3 &0.116 &0.07 &0.026 &0.002 &0.014 &0.028 &0.03 &0.02 &0.119 &0.008 &0.04 &0.024\\ \hline X1 &0.044 &0.007 &0.002 &0 &0.001 &0.065 &0.003 &0.005 &0.008 &0.001 &0.027 &0.014\\ \hline X2 &0.015 &0.081 &0.027 &0.003 &0.004 &0.074 &0.016 &0.01 &0.017 &0.032 &0.036 &0.023\\ \hline X3 &0.116 &0.009 &0.003 &0 &0.001 &0.071 &0.017 &0.013 &0.01 &0.001 &0.03 &0.016\\ \hline T1 &0.016 &0.018 &0.039 &0.038 &0.039 &0.072 &0.006 &0.006 &0.018 &0.006 &0.012 &0.02\\ \hline T2 &0.01 &0.04 &0.002 &0 &0.001 &0.064 &0.002 &0.004 &0.01 &0.004 &0.006 &0.016\\ \hline T3 &0.067 &0.007 &0.003 &0 &0.001 &0.066 &0.003 &0.008 &0.009 &0.001 &0.028 &0.003\\ \hline \end{array} $$

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

$$\begin{array}{|c|c|c|c|c|c|c|}\hline {M_{12 \times4}} &Di &Ci &Mi &Ri\\ \hline R1 &0.377 &0.699 &1.075 &-0.322\\ \hline R2 &0.652 &0.674 &1.326 &-0.022\\ \hline R3 &1.066 &0.197 &1.263 &0.869\\ \hline K1 &0.783 &0.13 &0.913 &0.652\\ \hline K2 &0.477 &0.153 &0.629 &0.324\\ \hline K3 &0.497 &0.904 &1.402 &-0.407\\ \hline X1 &0.178 &0.218 &0.395 &-0.04\\ \hline X2 &0.338 &0.401 &0.74 &-0.063\\ \hline X3 &0.289 &0.617 &0.906 &-0.328\\ \hline T1 &0.29 &0.308 &0.598 &-0.018\\ \hline T2 &0.16 &0.573 &0.733 &-0.413\\ \hline T3 &0.196 &0.428 &0.624 &-0.232\\ \hline \end{array} $$

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

$$D=\begin{array}{c|c|c|c|c|c|c}{M_{12 \times2}} &M &|R|\\ \hline R1 &1.0753 &0.3222\\ \hline R2 &1.3258 &0.0216\\ \hline R3 &1.2625 &0.8689\\ \hline K1 &0.9132 &0.6524\\ \hline K2 &0.6293 &0.3237\\ \hline K3 &1.4015 &0.4068\\ \hline X1 &0.3954 &0.04\\ \hline X2 &0.7398 &0.0632\\ \hline X3 &0.9056 &0.3285\\ \hline T1 &0.5975 &0.0184\\ \hline T2 &0.7331 &0.4127\\ \hline T3 &0.6238 &0.2317\\ \hline \end{array} $$

关系矩阵$A$

$$A=\begin{array} {c|c|c|c|c|c|c|c}{M_{12 \times12}} &R1 &R2 &R3 &K1 &K2 &K3 &X1 &X2 &X3 &T1 &T2 &T3\\ \hline R1 &1 & &1 & & &1 & & & & & & \\ \hline R2 & &1 & & & &1 & & & & & & \\ \hline R3 & & &1 & & & & & & & & & \\ \hline K1 & & &1 &1 & & & & & & & & \\ \hline K2 & & &1 &1 &1 &1 & & &1 & &1 & \\ \hline K3 & & & & & &1 & & & & & & \\ \hline X1 &1 & &1 &1 &1 &1 &1 &1 &1 & &1 &1\\ \hline X2 &1 & &1 &1 & &1 & &1 &1 & & & \\ \hline X3 & & &1 &1 & &1 & & &1 & & & \\ \hline T1 &1 &1 &1 &1 &1 &1 & &1 &1 &1 &1 &1\\ \hline T2 & & &1 &1 & & & & & & &1 & \\ \hline T3 &1 & &1 &1 &1 &1 & & &1 & &1 &1\\ \hline \end{array} $$

AISM部分求解

关系矩阵到相乘矩阵

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

$$B=\begin{array} {c|c|c|c|c|c|c|c}{M_{12 \times12}} &R1 &R2 &R3 &K1 &K2 &K3 &X1 &X2 &X3 &T1 &T2 &T3\\ \hline R1 &1 & &1 & & &1 & & & & & & \\ \hline R2 & &1 & & & &1 & & & & & & \\ \hline R3 & & &1 & & & & & & & & & \\ \hline K1 & & &1 &1 & & & & & & & & \\ \hline K2 & & &1 &1 &1 &1 & & &1 & &1 & \\ \hline K3 & & & & & &1 & & & & & & \\ \hline X1 &1 & &1 &1 &1 &1 &1 &1 &1 & &1 &1\\ \hline X2 &1 & &1 &1 & &1 & &1 &1 & & & \\ \hline X3 & & &1 &1 & &1 & & &1 & & & \\ \hline T1 &1 &1 &1 &1 &1 &1 & &1 &1 &1 &1 &1\\ \hline T2 & & &1 &1 & & & & & & &1 & \\ \hline T3 &1 & &1 &1 &1 &1 & & &1 & &1 &1\\ \hline \end{array} $$

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

$$R=\begin{array} {c|c|c|c|c|c|c|c}{M_{12 \times12}} &R1 &R2 &R3 &K1 &K2 &K3 &X1 &X2 &X3 &T1 &T2 &T3\\ \hline R1 &1 & &1 & & &1 & & & & & & \\ \hline R2 & &1 & & & &1 & & & & & & \\ \hline R3 & & &1 & & & & & & & & & \\ \hline K1 & & &1 &1 & & & & & & & & \\ \hline K2 & & &1 &1 &1 &1 & & &1 & &1 & \\ \hline K3 & & & & & &1 & & & & & & \\ \hline X1 &1 & &1 &1 &1 &1 &1 &1 &1 & &1 &1\\ \hline X2 &1 & &1 &1 & &1 & &1 &1 & & & \\ \hline X3 & & &1 &1 & &1 & & &1 & & & \\ \hline T1 &1 &1 &1 &1 &1 &1 & &1 &1 &1 &1 &1\\ \hline T2 & & &1 &1 & & & & & & &1 & \\ \hline T3 &1 & &1 &1 &1 &1 & & &1 & &1 &1\\ \hline \end{array} $$

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
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline R1&R1,R3,K3&R1 \\\hline R2&R2,K3&R2 \\\hline R3&\color{red}{\fbox{R3}}&\color{red}{\fbox{R3}} \\\hline K1&R3,K1&K1 \\\hline K2&R3,K1,K2,K3,X3,T2&K2 \\\hline K3&\color{red}{\fbox{K3}}&\color{red}{\fbox{K3}} \\\hline X1&R1,R3,K1,K2,K3,X1,X2,X3,T2,T3&X1 \\\hline X2&R1,R3,K1,K3,X2,X3&X2 \\\hline X3&R3,K1,K3,X3&X3 \\\hline T1&R1,R2,R3,K1,K2,K3,X2,X3,T1,T2,T3&T1 \\\hline T2&R3,K1,T2&T2 \\\hline T3&R1,R3,K1,K2,K3,X3,T2,T3&T3 \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline R1&R1,X1,X2,T1,T3&R1 \\\hline R2&R2,T1&R2 \\\hline R3&R1,R3,K1,K2,X1,X2,X3,T1,T2,T3&R3 \\\hline K1&K1,K2,X1,X2,X3,T1,T2,T3&K1 \\\hline K2&K2,X1,T1,T3&K2 \\\hline K3&R1,R2,K2,K3,X1,X2,X3,T1,T3&K3 \\\hline X1&\color{blue}{\fbox{X1}}&\color{blue}{\fbox{X1}} \\\hline X2&X1,X2,T1&X2 \\\hline X3&K2,X1,X2,X3,T1,T3&X3 \\\hline T1&\color{blue}{\fbox{T1}}&\color{blue}{\fbox{T1}} \\\hline T2&K2,X1,T1,T2,T3&T2 \\\hline T3&X1,T1,T3&T3 \\\hline \end{array} $$
抽取出R3、K3放置上层，删除后剩余的情况如下	抽取出X1,T1放置下层，删除后剩余的情况如下
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline R1&\color{red}{\fbox{R1}}&\color{red}{\fbox{R1}} \\\hline R2&\color{red}{\fbox{R2}}&\color{red}{\fbox{R2}} \\\hline K1&\color{red}{\fbox{K1}}&\color{red}{\fbox{K1}} \\\hline K2&K1,K2,X3,T2&K2 \\\hline X1&R1,K1,K2,X1,X2,X3,T2,T3&X1 \\\hline X2&R1,K1,X2,X3&X2 \\\hline X3&K1,X3&X3 \\\hline T1&R1,R2,K1,K2,X2,X3,T1,T2,T3&T1 \\\hline T2&K1,T2&T2 \\\hline T3&R1,K1,K2,X3,T2,T3&T3 \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline R1&R1,X2,T3&R1 \\\hline R2&\color{blue}{\fbox{R2}}&\color{blue}{\fbox{R2}} \\\hline R3&R1,R3,K1,K2,X2,X3,T2,T3&R3 \\\hline K1&K1,K2,X2,X3,T2,T3&K1 \\\hline K2&K2,T3&K2 \\\hline K3&R1,R2,K2,K3,X2,X3,T3&K3 \\\hline X2&\color{blue}{\fbox{X2}}&\color{blue}{\fbox{X2}} \\\hline X3&K2,X2,X3,T3&X3 \\\hline T2&K2,T2,T3&T2 \\\hline T3&\color{blue}{\fbox{T3}}&\color{blue}{\fbox{T3}} \\\hline \end{array} $$
抽取出R1、R2、K1放置上层，删除后剩余的情况如下	抽取出R2,X2,T3放置下层，删除后剩余的情况如下
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline K2&K2,X3,T2&K2 \\\hline X1&K2,X1,X2,X3,T2,T3&X1 \\\hline X2&X2,X3&X2 \\\hline X3&\color{red}{\fbox{X3}}&\color{red}{\fbox{X3}} \\\hline T1&K2,X2,X3,T1,T2,T3&T1 \\\hline T2&\color{red}{\fbox{T2}}&\color{red}{\fbox{T2}} \\\hline T3&K2,X3,T2,T3&T3 \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline R1&\color{blue}{\fbox{R1}}&\color{blue}{\fbox{R1}} \\\hline R3&R1,R3,K1,K2,X3,T2&R3 \\\hline K1&K1,K2,X3,T2&K1 \\\hline K2&\color{blue}{\fbox{K2}}&\color{blue}{\fbox{K2}} \\\hline K3&R1,K2,K3,X3&K3 \\\hline X3&K2,X3&X3 \\\hline T2&K2,T2&T2 \\\hline \end{array} $$
抽取出X3、T2放置上层，删除后剩余的情况如下	抽取出R1,K2放置下层，删除后剩余的情况如下
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline K2&\color{red}{\fbox{K2}}&\color{red}{\fbox{K2}} \\\hline X1&K2,X1,X2,T3&X1 \\\hline X2&\color{red}{\fbox{X2}}&\color{red}{\fbox{X2}} \\\hline T1&K2,X2,T1,T3&T1 \\\hline T3&K2,T3&T3 \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline R3&R3,K1,X3,T2&R3 \\\hline K1&K1,X3,T2&K1 \\\hline K3&K3,X3&K3 \\\hline X3&\color{blue}{\fbox{X3}}&\color{blue}{\fbox{X3}} \\\hline T2&\color{blue}{\fbox{T2}}&\color{blue}{\fbox{T2}} \\\hline \end{array} $$
抽取出K2、X2放置上层，删除后剩余的情况如下	抽取出X3,T2放置下层，删除后剩余的情况如下
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline X1&X1,T3&X1 \\\hline T1&T1,T3&T1 \\\hline T3&\color{red}{\fbox{T3}}&\color{red}{\fbox{T3}} \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline R3&R3,K1&R3 \\\hline K1&\color{blue}{\fbox{K1}}&\color{blue}{\fbox{K1}} \\\hline K3&\color{blue}{\fbox{K3}}&\color{blue}{\fbox{K3}} \\\hline \end{array} $$
抽取出T3放置上层，删除后剩余的情况如下	抽取出K1,K3放置下层，删除后剩余的情况如下
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline X1&\color{red}{\fbox{X1}}&\color{red}{\fbox{X1}} \\\hline T1&\color{red}{\fbox{T1}}&\color{red}{\fbox{T1}} \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline R3&\color{blue}{\fbox{R3}}&\color{blue}{\fbox{R3}} \\\hline \end{array} $$
抽取出X1、T1放置上层，删除后剩余的情况如下	抽取出R3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	R3,K3	R3
第1层	R1,R2,K1	K1,K3
第2层	X3,T2	X3,T2
第3层	K2,X2	R1,K2
第4层	T3	R2,X2,T3
第5层	X1,T1	X1,T1

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

$$S=\begin{array} {c|c|c|c|c|c|c|c}{M_{12 \times12}} &R1 &R2 &R3 &K1 &K2 &K3 &X1 &X2 &X3 &T1 &T2 &T3\\ \hline R1 & & &1 & & &1 & & & & & & \\ \hline R2 & & & & & &1 & & & & & & \\ \hline R3 & & & & & & & & & & & & \\ \hline K1 & & &1 & & & & & & & & & \\ \hline K2 & & & & & & & & &1 & &1 & \\ \hline K3 & & & & & & & & & & & & \\ \hline X1 & & & & & & & &1 & & & &1\\ \hline X2 &1 & & & & & & & &1 & & & \\ \hline X3 & & & &1 & &1 & & & & & & \\ \hline T1 & &1 & & & & & &1 & & & &1\\ \hline T2 & & & &1 & & & & & & & & \\ \hline T3 &1 & & & &1 & & & & & & & \\ \hline \end{array} $$

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

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

$$MR=\begin{array}{c|c|c|c|c|c|c}{M_{12 \times2}} &中心度 &原因度\\ \hline R1 &1.0753 &0.3222\\ \hline R2 &1.3258 &0.0216\\ \hline R3 &1.2625 &0.8689\\ \hline K1 &0.9132 &0.6524\\ \hline K2 &0.6293 &0.3237\\ \hline K3 &1.4015 &0.4068\\ \hline X1 &0.3954 &0.04\\ \hline X2 &0.7398 &0.0632\\ \hline X3 &0.9056 &0.3285\\ \hline T1 &0.5975 &0.0184\\ \hline T2 &0.7331 &0.4127\\ \hline T3 &0.6238 &0.2317\\ \hline \end{array} $$

向量的计算采用欧式距离公式

$$d=\begin{array}{c|c|c|c|c|c|c}{M_{12 \times1}} &向量\\ \hline R1 &1.1226\\ \hline R2 &1.326\\ \hline R3 &1.5326\\ \hline K1 &1.1223\\ \hline K2 &0.7077\\ \hline K3 &1.4594\\ \hline X1 &0.3974\\ \hline X2 &0.7425\\ \hline X3 &0.9633\\ \hline T1 &0.5978\\ \hline T2 &0.8413\\ \hline T3 &0.6654\\ \hline \end{array} $$

归一化求权重

$$\omega =\begin{array}{c|c|c|c|c|c|c}{M_{12 \times1}} &权重\\ \hline R1 &0.0978\\ \hline R2 &0.1155\\ \hline R3 &0.1335\\ \hline K1 &0.0978\\ \hline K2 &0.0617\\ \hline K3 &0.1271\\ \hline X1 &0.0346\\ \hline X2 &0.0647\\ \hline X3 &0.0839\\ \hline T1 &0.0521\\ \hline T2 &0.0733\\ \hline T3 &0.058\\ \hline \end{array} $$

归一化求子系统的权重

$$\omega =\begin{array}{c|c|c|c|c|c|c}{M_{4 \times1}} &权重\\ \hline R &0.3468\\ \hline K &0.2866\\ \hline X &0.1832\\ \hline T &0.1833\\ \hline \end{array} $$

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：