原始矩阵（直接影响矩阵）为

$$Ori=\begin{array}{|c|c|c|c|c|c|c|}\hline {M_{16 \times16}} &C1 &C2 &C3 &C4 &C5 &E1 &E2 &E3 &F1 &F2 &F3 &U1 &U2 &D1 &D2 &D3\\ \hline C1 &0 &7.12 &7.52 &8.92 &7.36 &8.24 &8.56 &8.4 &8.92 &1.28 &1.4 &8.08 &8.92 &2.36 &7.12 &1.04\\ \hline C2 &8.32 &0 &5.12 &8.88 &8.92 &2.16 &2.96 &7.16 &8.92 &1.56 &5.16 &4.12 &5 &5.12 &4.12 &2.68\\ \hline C3 &7.44 &2.32 &0 &8.8 &3.08 &7.12 &1.04 &8.92 &8.56 &0.24 &0.26 &5.52 &6.64 &4.48 &6.52 &3.08\\ \hline C4 &7.04 &5.12 &2.04 &0 &8.76 &7.16 &2.48 &7.28 &8.16 &1.2 &2.12 &5.08 &8.6 &3.12 &6.08 &2.2\\ \hline C5 &8.92 &8.56 &7.96 &7.08 &0 &4.84 &4.76 &6.56 &8.72 &1.76 &2.12 &8.48 &8.8 &4.16 &6 &3.04\\ \hline E1 &2.12 &1.5 &0.92 &7.12 &2.48 &0 &6.2 &5.04 &8.76 &3.28 &4.52 &2.24 &8.72 &3.5 &8.6 &5.64\\ \hline E2 &2 &7.04 &1.08 &8.72 &3.24 &8.2 &0 &7.08 &8.4 &8.6 &8 &2.12 &8.4 &6.2 &6.84 &8.24\\ \hline E3 &7.02 &3.2 &2.5 &8.96 &4.24 &4.88 &5.24 &0 &6.12 &2.24 &6.2 &2 &8.6 &3.52 &4.22 &6.2\\ \hline F1 &8.56 &8.92 &6.52 &8.6 &8.6 &7.2 &8.2 &8.4 &0 &2.2 &2 &1.12 &2 &5.08 &2.04 &6\\ \hline F2 &1.04 &1.52 &0.096 &1.08 &1.12 &3.04 &3.52 &2.04 &1.56 &0 &7.8 &1.52 &1.5 &6 &2.52 &6.5\\ \hline F3 &4.5 &6.04 &1 &1.16 &2.52 &8 &8.04 &8.96 &2.5 &8.25 &0 &1.24 &1.4 &7.04 &7.72 &7.5\\ \hline U1 &7.08 &8.72 &7.08 &6.52 &1.5 &6.28 &3.2 &8 &4.56 &2.04 &4 &0 &8.4 &7.2 &5.56 &3\\ \hline U2 &7.04 &4.96 &6.08 &7.52 &5.02 &7.2 &7 &8.12 &6.5 &2.12 &2 &8.88 &0 &1.24 &0.64 &8.2\\ \hline D1 &2 &1.04 &3.56 &6.96 &6.52 &8.6 &6.96 &6.4 &2.28 &1.02 &8.2 &1.02 &6.5 &0 &7.64 &8.92\\ \hline D2 &6.08 &7.02 &1.12 &8.32 &7 &8.24 &4.04 &8.92 &8.68 &4.12 &3.32 &8.12 &5.08 &8.92 &0 &3.04\\ \hline D3 &1.2 &1 &0.92 &8.76 &2.12 &8.96 &8.24 &7.28 &0.8 &6.52 &7.16 &8.52 &3.2 &7.2 &8.92 &0\\ \hline \end{array} $$

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_{16 \times16}} &C1 &C2 &C3 &C4 &C5 &E1 &E2 &E3 &F1 &F2 &F3 &U1 &U2 &D1 &D2 &D3\\ \hline C1 &0 &0.075 &0.079 &0.094 &0.077 &0.087 &0.09 &0.088 &0.094 &0.013 &0.015 &0.085 &0.094 &0.025 &0.075 &0.011\\ \hline C2 &0.087 &0 &0.054 &0.093 &0.094 &0.023 &0.031 &0.075 &0.094 &0.016 &0.054 &0.043 &0.052 &0.054 &0.043 &0.028\\ \hline C3 &0.078 &0.024 &0 &0.092 &0.032 &0.075 &0.011 &0.094 &0.09 &0.003 &0.003 &0.058 &0.07 &0.047 &0.068 &0.032\\ \hline C4 &0.074 &0.054 &0.021 &0 &0.092 &0.075 &0.026 &0.076 &0.086 &0.013 &0.022 &0.053 &0.09 &0.033 &0.064 &0.023\\ \hline C5 &0.094 &0.09 &0.084 &0.074 &0 &0.051 &0.05 &0.069 &0.092 &0.018 &0.022 &0.089 &0.092 &0.044 &0.063 &0.032\\ \hline E1 &0.022 &0.016 &0.01 &0.075 &0.026 &0 &0.065 &0.053 &0.092 &0.034 &0.047 &0.024 &0.092 &0.037 &0.09 &0.059\\ \hline E2 &0.021 &0.074 &0.011 &0.092 &0.034 &0.086 &0 &0.074 &0.088 &0.09 &0.084 &0.022 &0.088 &0.065 &0.072 &0.087\\ \hline E3 &0.074 &0.034 &0.026 &0.094 &0.045 &0.051 &0.055 &0 &0.064 &0.024 &0.065 &0.021 &0.09 &0.037 &0.044 &0.065\\ \hline F1 &0.09 &0.094 &0.068 &0.09 &0.09 &0.076 &0.086 &0.088 &0 &0.023 &0.021 &0.012 &0.021 &0.053 &0.021 &0.063\\ \hline F2 &0.011 &0.016 &0.001 &0.011 &0.012 &0.032 &0.037 &0.021 &0.016 &0 &0.082 &0.016 &0.016 &0.063 &0.026 &0.068\\ \hline F3 &0.047 &0.063 &0.011 &0.012 &0.026 &0.084 &0.084 &0.094 &0.026 &0.087 &0 &0.013 &0.015 &0.074 &0.081 &0.079\\ \hline U1 &0.074 &0.092 &0.074 &0.068 &0.016 &0.066 &0.034 &0.084 &0.048 &0.021 &0.042 &0 &0.088 &0.076 &0.058 &0.031\\ \hline U2 &0.074 &0.052 &0.064 &0.079 &0.053 &0.076 &0.073 &0.085 &0.068 &0.022 &0.021 &0.093 &0 &0.013 &0.007 &0.086\\ \hline D1 &0.021 &0.011 &0.037 &0.073 &0.068 &0.09 &0.073 &0.067 &0.024 &0.011 &0.086 &0.011 &0.068 &0 &0.08 &0.094\\ \hline D2 &0.064 &0.074 &0.012 &0.087 &0.073 &0.087 &0.042 &0.094 &0.091 &0.043 &0.035 &0.085 &0.053 &0.094 &0 &0.032\\ \hline D3 &0.013 &0.011 &0.01 &0.092 &0.022 &0.094 &0.087 &0.076 &0.008 &0.068 &0.075 &0.089 &0.034 &0.076 &0.094 &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_{16 \times16}} &C1 &C2 &C3 &C4 &C5 &E1 &E2 &E3 &F1 &F2 &F3 &U1 &U2 &D1 &D2 &D3\\ \hline C1 &0.371 &0.409 &0.323 &0.572 &0.413 &0.517 &0.439 &0.557 &0.519 &0.213 &0.288 &0.385 &0.509 &0.34 &0.433 &0.34\\ \hline C2 &0.398 &0.289 &0.265 &0.496 &0.38 &0.396 &0.337 &0.475 &0.45 &0.185 &0.282 &0.302 &0.406 &0.319 &0.352 &0.305\\ \hline C3 &0.364 &0.288 &0.195 &0.467 &0.301 &0.415 &0.295 &0.46 &0.421 &0.157 &0.217 &0.296 &0.397 &0.291 &0.35 &0.287\\ \hline C4 &0.372 &0.328 &0.227 &0.395 &0.364 &0.426 &0.321 &0.458 &0.43 &0.175 &0.244 &0.303 &0.427 &0.288 &0.356 &0.29\\ \hline C5 &0.445 &0.41 &0.321 &0.537 &0.33 &0.47 &0.392 &0.524 &0.499 &0.209 &0.285 &0.38 &0.491 &0.346 &0.41 &0.345\\ \hline E1 &0.29 &0.264 &0.187 &0.424 &0.276 &0.327 &0.33 &0.401 &0.397 &0.188 &0.253 &0.249 &0.389 &0.273 &0.353 &0.304\\ \hline E2 &0.357 &0.377 &0.233 &0.53 &0.349 &0.491 &0.341 &0.51 &0.473 &0.281 &0.346 &0.304 &0.465 &0.363 &0.411 &0.397\\ \hline E3 &0.356 &0.296 &0.217 &0.465 &0.31 &0.397 &0.339 &0.374 &0.395 &0.186 &0.28 &0.264 &0.412 &0.285 &0.332 &0.323\\ \hline F1 &0.41 &0.384 &0.283 &0.517 &0.389 &0.461 &0.401 &0.504 &0.384 &0.203 &0.27 &0.286 &0.399 &0.333 &0.353 &0.352\\ \hline F2 &0.157 &0.152 &0.098 &0.21 &0.15 &0.218 &0.192 &0.22 &0.187 &0.094 &0.202 &0.14 &0.186 &0.199 &0.185 &0.21\\ \hline F3 &0.313 &0.308 &0.187 &0.378 &0.28 &0.413 &0.357 &0.444 &0.344 &0.245 &0.224 &0.241 &0.33 &0.318 &0.36 &0.332\\ \hline U1 &0.386 &0.37 &0.281 &0.48 &0.31 &0.439 &0.342 &0.487 &0.413 &0.193 &0.278 &0.263 &0.442 &0.343 &0.37 &0.315\\ \hline U2 &0.383 &0.337 &0.272 &0.49 &0.338 &0.447 &0.378 &0.487 &0.43 &0.197 &0.259 &0.35 &0.362 &0.287 &0.325 &0.362\\ \hline D1 &0.31 &0.278 &0.225 &0.455 &0.333 &0.442 &0.362 &0.446 &0.364 &0.184 &0.309 &0.259 &0.4 &0.26 &0.376 &0.359\\ \hline D2 &0.407 &0.387 &0.246 &0.537 &0.392 &0.494 &0.383 &0.534 &0.487 &0.232 &0.299 &0.365 &0.448 &0.388 &0.347 &0.345\\ \hline D3 &0.301 &0.279 &0.198 &0.47 &0.291 &0.446 &0.373 &0.454 &0.349 &0.238 &0.306 &0.326 &0.371 &0.336 &0.39 &0.274\\ \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

$$\begin{array}{|c|c|c|c|c|c|c|}\hline {M_{16 \times4}} &Di &Ci &Mi &Ri\\ \hline C1 &6.627 &5.62 &12.247 &1.006\\ \hline C2 &5.637 &5.157 &10.793 &0.48\\ \hline C3 &5.203 &3.759 &8.962 &1.444\\ \hline C4 &5.404 &7.423 &12.827 &-2.019\\ \hline C5 &6.394 &5.206 &11.601 &1.188\\ \hline E1 &4.905 &6.798 &11.703 &-1.893\\ \hline E2 &6.227 &5.579 &11.806 &0.648\\ \hline E3 &5.231 &7.335 &12.566 &-2.104\\ \hline F1 &5.93 &6.542 &12.472 &-0.613\\ \hline F2 &2.798 &3.18 &5.979 &-0.382\\ \hline F3 &5.075 &4.341 &9.416 &0.734\\ \hline U1 &5.714 &4.713 &10.427 &1.001\\ \hline U2 &5.704 &6.434 &12.138 &-0.73\\ \hline D1 &5.362 &4.97 &10.332 &0.392\\ \hline D2 &6.291 &5.704 &11.995 &0.587\\ \hline D3 &5.401 &5.14 &10.541 &0.261\\ \hline \end{array} $$

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

$$D=\begin{array}{c|c|c|c|c|c|c}{M_{16 \times2}} &M &|R|\\ \hline C1 &12.2471 &1.0062\\ \hline C2 &10.7932 &0.48\\ \hline C3 &8.962 &1.4443\\ \hline C4 &12.8269 &2.0191\\ \hline C5 &11.6007 &1.1881\\ \hline E1 &11.7032 &1.8929\\ \hline E2 &11.8064 &0.6482\\ \hline E3 &12.566 &2.1038\\ \hline F1 &12.4722 &0.6128\\ \hline F2 &5.9788 &0.3819\\ \hline F3 &9.4155 &0.7336\\ \hline U1 &10.4274 &1.0005\\ \hline U2 &12.1376 &0.7303\\ \hline D1 &10.3322 &0.3921\\ \hline D2 &11.9951 &0.5865\\ \hline D3 &10.5409 &0.2611\\ \hline \end{array} $$

关系矩阵$A$

$$A=\begin{array} {c|c|c|c|c|c|c|c}{M_{16 \times16}} &C1 &C2 &C3 &C4 &C5 &E1 &E2 &E3 &F1 &F2 &F3 &U1 &U2 &D1 &D2 &D3\\ \hline C1 &1 & & &1 & & & &1 & & & & & & & & \\ \hline C2 &1 &1 & &1 &1 &1 &1 &1 &1 & & & &1 & &1 & \\ \hline C3 & & &1 &1 & &1 & &1 & & & & & & & & \\ \hline C4 & & & &1 & & & & & & & & & & & & \\ \hline C5 & & & &1 &1 &1 & &1 & & & & & & & & \\ \hline E1 & & & &1 & &1 & &1 & & & & & & & & \\ \hline E2 &1 & & &1 & & &1 &1 & & & & &1 & & & \\ \hline E3 & & & & & & & &1 & & & & & & & & \\ \hline F1 & & & &1 & & & &1 &1 & & & & & & & \\ \hline F2 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 & \\ \hline F3 &1 & & &1 &1 &1 & &1 & & &1 &1 & & & & \\ \hline U1 &1 & & &1 &1 &1 & &1 & & & &1 & & & & \\ \hline U2 &1 & & &1 & & & &1 & & & & &1 & & & \\ \hline D1 &1 &1 & &1 &1 &1 &1 &1 &1 & & &1 &1 &1 &1 & \\ \hline D2 &1 & & &1 & & & &1 &1 & & & &1 & &1 & \\ \hline D3 &1 &1 & &1 &1 &1 &1 &1 &1 & & & &1 & &1 &1\\ \hline \end{array} $$

AISM部分求解

关系矩阵到相乘矩阵

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

$$B=\begin{array} {c|c|c|c|c|c|c|c}{M_{16 \times16}} &C1 &C2 &C3 &C4 &C5 &E1 &E2 &E3 &F1 &F2 &F3 &U1 &U2 &D1 &D2 &D3\\ \hline C1 &1 & & &1 & & & &1 & & & & & & & & \\ \hline C2 &1 &1 & &1 &1 &1 &1 &1 &1 & & & &1 & &1 & \\ \hline C3 & & &1 &1 & &1 & &1 & & & & & & & & \\ \hline C4 & & & &1 & & & & & & & & & & & & \\ \hline C5 & & & &1 &1 &1 & &1 & & & & & & & & \\ \hline E1 & & & &1 & &1 & &1 & & & & & & & & \\ \hline E2 &1 & & &1 & & &1 &1 & & & & &1 & & & \\ \hline E3 & & & & & & & &1 & & & & & & & & \\ \hline F1 & & & &1 & & & &1 &1 & & & & & & & \\ \hline F2 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 & \\ \hline F3 &1 & & &1 &1 &1 & &1 & & &1 &1 & & & & \\ \hline U1 &1 & & &1 &1 &1 & &1 & & & &1 & & & & \\ \hline U2 &1 & & &1 & & & &1 & & & & &1 & & & \\ \hline D1 &1 &1 & &1 &1 &1 &1 &1 &1 & & &1 &1 &1 &1 & \\ \hline D2 &1 & & &1 & & & &1 &1 & & & &1 & &1 & \\ \hline D3 &1 &1 & &1 &1 &1 &1 &1 &1 & & & &1 & &1 &1\\ \hline \end{array} $$

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

$$R=\begin{array} {c|c|c|c|c|c|c|c}{M_{16 \times16}} &C1 &C2 &C3 &C4 &C5 &E1 &E2 &E3 &F1 &F2 &F3 &U1 &U2 &D1 &D2 &D3\\ \hline C1 &1 & & &1 & & & &1 & & & & & & & & \\ \hline C2 &1 &1 & &1 &1 &1 &1 &1 &1 & & & &1 & &1 & \\ \hline C3 & & &1 &1 & &1 & &1 & & & & & & & & \\ \hline C4 & & & &1 & & & & & & & & & & & & \\ \hline C5 & & & &1 &1 &1 & &1 & & & & & & & & \\ \hline E1 & & & &1 & &1 & &1 & & & & & & & & \\ \hline E2 &1 & & &1 & & &1 &1 & & & & &1 & & & \\ \hline E3 & & & & & & & &1 & & & & & & & & \\ \hline F1 & & & &1 & & & &1 &1 & & & & & & & \\ \hline F2 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 & \\ \hline F3 &1 & & &1 &1 &1 & &1 & & &1 &1 & & & & \\ \hline U1 &1 & & &1 &1 &1 & &1 & & & &1 & & & & \\ \hline U2 &1 & & &1 & & & &1 & & & & &1 & & & \\ \hline D1 &1 &1 & &1 &1 &1 &1 &1 &1 & & &1 &1 &1 &1 & \\ \hline D2 &1 & & &1 & & & &1 &1 & & & &1 & &1 & \\ \hline D3 &1 &1 & &1 &1 &1 &1 &1 &1 & & & &1 & &1 &1\\ \hline \end{array} $$

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
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline C1&C1,C4,E3&C1 \\\hline C2&C1,C2,C4,C5,E1,E2,E3,F1,U2,D2&C2 \\\hline C3&C3,C4,E1,E3&C3 \\\hline C4&\color{red}{\fbox{C4}}&\color{red}{\fbox{C4}} \\\hline C5&C4,C5,E1,E3&C5 \\\hline E1&C4,E1,E3&E1 \\\hline E2&C1,C4,E2,E3,U2&E2 \\\hline E3&\color{red}{\fbox{E3}}&\color{red}{\fbox{E3}} \\\hline F1&C4,E3,F1&F1 \\\hline F2&C1,C2,C3,C4,C5,E1,E2,E3,F1,F2,F3,U1,U2,D1,D2&F2 \\\hline F3&C1,C4,C5,E1,E3,F3,U1&F3 \\\hline U1&C1,C4,C5,E1,E3,U1&U1 \\\hline U2&C1,C4,E3,U2&U2 \\\hline D1&C1,C2,C4,C5,E1,E2,E3,F1,U1,U2,D1,D2&D1 \\\hline D2&C1,C4,E3,F1,U2,D2&D2 \\\hline D3&C1,C2,C4,C5,E1,E2,E3,F1,U2,D2,D3&D3 \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline C1&C1,C2,E2,F2,F3,U1,U2,D1,D2,D3&C1 \\\hline C2&C2,F2,D1,D3&C2 \\\hline C3&C3,F2&C3 \\\hline C4&C1,C2,C3,C4,C5,E1,E2,F1,F2,F3,U1,U2,D1,D2,D3&C4 \\\hline C5&C2,C5,F2,F3,U1,D1,D3&C5 \\\hline E1&C2,C3,C5,E1,F2,F3,U1,D1,D3&E1 \\\hline E2&C2,E2,F2,D1,D3&E2 \\\hline E3&C1,C2,C3,C5,E1,E2,E3,F1,F2,F3,U1,U2,D1,D2,D3&E3 \\\hline F1&C2,F1,F2,D1,D2,D3&F1 \\\hline F2&\color{blue}{\fbox{F2}}&\color{blue}{\fbox{F2}} \\\hline F3&F2,F3&F3 \\\hline U1&F2,F3,U1,D1&U1 \\\hline U2&C2,E2,F2,U2,D1,D2,D3&U2 \\\hline D1&F2,D1&D1 \\\hline D2&C2,F2,D1,D2,D3&D2 \\\hline D3&\color{blue}{\fbox{D3}}&\color{blue}{\fbox{D3}} \\\hline \end{array} $$
抽取出C4、E3放置上层，删除后剩余的情况如下	抽取出F2,D3放置下层，删除后剩余的情况如下
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline C1&\color{red}{\fbox{C1}}&\color{red}{\fbox{C1}} \\\hline C2&C1,C2,C5,E1,E2,F1,U2,D2&C2 \\\hline C3&C3,E1&C3 \\\hline C5&C5,E1&C5 \\\hline E1&\color{red}{\fbox{E1}}&\color{red}{\fbox{E1}} \\\hline E2&C1,E2,U2&E2 \\\hline F1&\color{red}{\fbox{F1}}&\color{red}{\fbox{F1}} \\\hline F2&C1,C2,C3,C5,E1,E2,F1,F2,F3,U1,U2,D1,D2&F2 \\\hline F3&C1,C5,E1,F3,U1&F3 \\\hline U1&C1,C5,E1,U1&U1 \\\hline U2&C1,U2&U2 \\\hline D1&C1,C2,C5,E1,E2,F1,U1,U2,D1,D2&D1 \\\hline D2&C1,F1,U2,D2&D2 \\\hline D3&C1,C2,C5,E1,E2,F1,U2,D2,D3&D3 \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline C1&C1,C2,E2,F3,U1,U2,D1,D2&C1 \\\hline C2&C2,D1&C2 \\\hline C3&\color{blue}{\fbox{C3}}&\color{blue}{\fbox{C3}} \\\hline C4&C1,C2,C3,C4,C5,E1,E2,F1,F3,U1,U2,D1,D2&C4 \\\hline C5&C2,C5,F3,U1,D1&C5 \\\hline E1&C2,C3,C5,E1,F3,U1,D1&E1 \\\hline E2&C2,E2,D1&E2 \\\hline E3&C1,C2,C3,C5,E1,E2,E3,F1,F3,U1,U2,D1,D2&E3 \\\hline F1&C2,F1,D1,D2&F1 \\\hline F3&\color{blue}{\fbox{F3}}&\color{blue}{\fbox{F3}} \\\hline U1&F3,U1,D1&U1 \\\hline U2&C2,E2,U2,D1,D2&U2 \\\hline D1&\color{blue}{\fbox{D1}}&\color{blue}{\fbox{D1}} \\\hline D2&C2,D1,D2&D2 \\\hline \end{array} $$
抽取出C1、E1、F1放置上层，删除后剩余的情况如下	抽取出C3,F3,D1放置下层，删除后剩余的情况如下
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline C2&C2,C5,E2,U2,D2&C2 \\\hline C3&\color{red}{\fbox{C3}}&\color{red}{\fbox{C3}} \\\hline C5&\color{red}{\fbox{C5}}&\color{red}{\fbox{C5}} \\\hline E2&E2,U2&E2 \\\hline F2&C2,C3,C5,E2,F2,F3,U1,U2,D1,D2&F2 \\\hline F3&C5,F3,U1&F3 \\\hline U1&C5,U1&U1 \\\hline U2&\color{red}{\fbox{U2}}&\color{red}{\fbox{U2}} \\\hline D1&C2,C5,E2,U1,U2,D1,D2&D1 \\\hline D2&U2,D2&D2 \\\hline D3&C2,C5,E2,U2,D2,D3&D3 \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline C1&C1,C2,E2,U1,U2,D2&C1 \\\hline C2&\color{blue}{\fbox{C2}}&\color{blue}{\fbox{C2}} \\\hline C4&C1,C2,C4,C5,E1,E2,F1,U1,U2,D2&C4 \\\hline C5&C2,C5,U1&C5 \\\hline E1&C2,C5,E1,U1&E1 \\\hline E2&C2,E2&E2 \\\hline E3&C1,C2,C5,E1,E2,E3,F1,U1,U2,D2&E3 \\\hline F1&C2,F1,D2&F1 \\\hline U1&\color{blue}{\fbox{U1}}&\color{blue}{\fbox{U1}} \\\hline U2&C2,E2,U2,D2&U2 \\\hline D2&C2,D2&D2 \\\hline \end{array} $$
抽取出C3、C5、U2放置上层，删除后剩余的情况如下	抽取出C2,U1放置下层，删除后剩余的情况如下
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline C2&C2,E2,D2&C2 \\\hline E2&\color{red}{\fbox{E2}}&\color{red}{\fbox{E2}} \\\hline F2&C2,E2,F2,F3,U1,D1,D2&F2 \\\hline F3&F3,U1&F3 \\\hline U1&\color{red}{\fbox{U1}}&\color{red}{\fbox{U1}} \\\hline D1&C2,E2,U1,D1,D2&D1 \\\hline D2&\color{red}{\fbox{D2}}&\color{red}{\fbox{D2}} \\\hline D3&C2,E2,D2,D3&D3 \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline C1&C1,E2,U2,D2&C1 \\\hline C4&C1,C4,C5,E1,E2,F1,U2,D2&C4 \\\hline C5&\color{blue}{\fbox{C5}}&\color{blue}{\fbox{C5}} \\\hline E1&C5,E1&E1 \\\hline E2&\color{blue}{\fbox{E2}}&\color{blue}{\fbox{E2}} \\\hline E3&C1,C5,E1,E2,E3,F1,U2,D2&E3 \\\hline F1&F1,D2&F1 \\\hline U2&E2,U2,D2&U2 \\\hline D2&\color{blue}{\fbox{D2}}&\color{blue}{\fbox{D2}} \\\hline \end{array} $$
抽取出E2、U1、D2放置上层，删除后剩余的情况如下	抽取出C5,E2,D2放置下层，删除后剩余的情况如下
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline C2&\color{red}{\fbox{C2}}&\color{red}{\fbox{C2}} \\\hline F2&C2,F2,F3,D1&F2 \\\hline F3&\color{red}{\fbox{F3}}&\color{red}{\fbox{F3}} \\\hline D1&C2,D1&D1 \\\hline D3&C2,D3&D3 \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline C1&C1,U2&C1 \\\hline C4&C1,C4,E1,F1,U2&C4 \\\hline E1&\color{blue}{\fbox{E1}}&\color{blue}{\fbox{E1}} \\\hline E3&C1,E1,E3,F1,U2&E3 \\\hline F1&\color{blue}{\fbox{F1}}&\color{blue}{\fbox{F1}} \\\hline U2&\color{blue}{\fbox{U2}}&\color{blue}{\fbox{U2}} \\\hline \end{array} $$
抽取出C2、F3放置上层，删除后剩余的情况如下	抽取出E1,F1,U2放置下层，删除后剩余的情况如下
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline F2&F2,D1&F2 \\\hline D1&\color{red}{\fbox{D1}}&\color{red}{\fbox{D1}} \\\hline D3&\color{red}{\fbox{D3}}&\color{red}{\fbox{D3}} \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline C1&\color{blue}{\fbox{C1}}&\color{blue}{\fbox{C1}} \\\hline C4&C1,C4&C4 \\\hline E3&C1,E3&E3 \\\hline \end{array} $$
抽取出D1、D3放置上层，删除后剩余的情况如下	抽取出C1放置下层，删除后剩余的情况如下
$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} & R_{e} & T_{e} \\\hline F2&\color{red}{\fbox{F2}}&\color{red}{\fbox{F2}} \\\hline \end{array} $$	$$\begin{array} {c\|c\|c\|c\|c\|c\|c\|c}{} &Q_{e} & T_{e} \\\hline C4&\color{blue}{\fbox{C4}}&\color{blue}{\fbox{C4}} \\\hline E3&\color{blue}{\fbox{E3}}&\color{blue}{\fbox{E3}} \\\hline \end{array} $$
抽取出F2放置上层，删除后剩余的情况如下	抽取出C4,E3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	C4,E3	C4,E3
第1层	C1,E1,F1	C1
第2层	C3,C5,U2	E1,F1,U2
第3层	E2,U1,D2	C5,E2,D2
第4层	C2,F3	C2,U1
第5层	D1,D3	C3,F3,D1
第6层	F2	F2,D3

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

$$S=\begin{array} {c|c|c|c|c|c|c|c}{M_{16 \times16}} &C1 &C2 &C3 &C4 &C5 &E1 &E2 &E3 &F1 &F2 &F3 &U1 &U2 &D1 &D2 &D3\\ \hline C1 & & & &1 & & & &1 & & & & & & & & \\ \hline C2 & & & & &1 & &1 & & & & & & & &1 & \\ \hline C3 & & & & & &1 & & & & & & & & & & \\ \hline C4 & & & & & & & & & & & & & & & & \\ \hline C5 & & & & & &1 & & & & & & & & & & \\ \hline E1 & & & &1 & & & &1 & & & & & & & & \\ \hline E2 & & & & & & & & & & & & &1 & & & \\ \hline E3 & & & & & & & & & & & & & & & & \\ \hline F1 & & & &1 & & & &1 & & & & & & & & \\ \hline F2 & & &1 & & & & & & & &1 & & &1 & & \\ \hline F3 & & & & & & & & & & & &1 & & & & \\ \hline U1 &1 & & & &1 & & & & & & & & & & & \\ \hline U2 &1 & & & & & & & & & & & & & & & \\ \hline D1 & &1 & & & & & & & & & &1 & & & & \\ \hline D2 & & & & & & & & &1 & & & &1 & & & \\ \hline D3 & &1 & & & & & & & & & & & & & & \\ \hline \end{array} $$

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

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

$$MR=\begin{array}{c|c|c|c|c|c|c}{M_{16 \times2}} &中心度 &原因度\\ \hline C1 &12.2471 &1.0062\\ \hline C2 &10.7932 &0.48\\ \hline C3 &8.962 &1.4443\\ \hline C4 &12.8269 &2.0191\\ \hline C5 &11.6007 &1.1881\\ \hline E1 &11.7032 &1.8929\\ \hline E2 &11.8064 &0.6482\\ \hline E3 &12.566 &2.1038\\ \hline F1 &12.4722 &0.6128\\ \hline F2 &5.9788 &0.3819\\ \hline F3 &9.4155 &0.7336\\ \hline U1 &10.4274 &1.0005\\ \hline U2 &12.1376 &0.7303\\ \hline D1 &10.3322 &0.3921\\ \hline D2 &11.9951 &0.5865\\ \hline D3 &10.5409 &0.2611\\ \hline \end{array} $$

向量的计算采用欧式距离公式

$$d=\begin{array}{c|c|c|c|c|c|c}{M_{16 \times1}} &向量\\ \hline C1 &12.2883\\ \hline C2 &10.8039\\ \hline C3 &9.0776\\ \hline C4 &12.9848\\ \hline C5 &11.6614\\ \hline E1 &11.8553\\ \hline E2 &11.8242\\ \hline E3 &12.7409\\ \hline F1 &12.4873\\ \hline F2 &5.991\\ \hline F3 &9.4441\\ \hline U1 &10.4753\\ \hline U2 &12.1595\\ \hline D1 &10.3397\\ \hline D2 &12.0094\\ \hline D3 &10.5442\\ \hline \end{array} $$

归一化求权重

$$\omega =\begin{array}{c|c|c|c|c|c|c}{M_{16 \times1}} &权重\\ \hline C1 &0.0695\\ \hline C2 &0.0611\\ \hline C3 &0.0514\\ \hline C4 &0.0735\\ \hline C5 &0.066\\ \hline E1 &0.0671\\ \hline E2 &0.0669\\ \hline E3 &0.0721\\ \hline F1 &0.0707\\ \hline F2 &0.0339\\ \hline F3 &0.0535\\ \hline U1 &0.0593\\ \hline U2 &0.0688\\ \hline D1 &0.0585\\ \hline D2 &0.068\\ \hline D3 &0.0597\\ \hline \end{array} $$

归一化求子系统的权重

$$\omega =\begin{array}{c|c|c|c|c|c|c}{M_{5 \times1}} &权重\\ \hline C &0.3216\\ \hline E &0.2061\\ \hline F &0.158\\ \hline U &0.1281\\ \hline D &0.1862\\ \hline \end{array} $$

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：