VIKOR-AISM求解过程
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原始矩阵如下：
$$ \begin{array}{c|c|c|c|c|c|c}{M_{11 \times15}} & -PN1 & -PN2 & -PE1 & -PS1 & -PS2 &SN1 &SN2 &SN3 &SE1 &SS1 &RN1 &RN2 &RE1 &RE2 &RS1\\
\hline
2011 &0.654302103 &278.5577711 &0.104246565 &426.623418 &0.484969128 &0.518335321 &0.734225621 &0.050682504 &8.714200492 &8.950944846 &0.850812256 &10.3 &7380.279843 &35115.65191 &176.2\\
\hline
2012 &0.64018787 &230.2072698 &0.100574919 &430.0655108 &0.498760764 &0.524293616 &0.720190933 &0.053800771 &8.311141515 &8.547470721 &0.868804512 &11.1 &8725.395288 &39690.62085 &170.1\\
\hline
2013 &0.634839501 &232.7438152 &0.095228796 &433.091812 &0.510793577 &0.534174192 &0.725746269 &0.053359371 &8.162741259 &8.838568347 &0.889545989 &12.82 &10872.14779 &43857.04467 &144.4\\
\hline
2014 &0.621034465 &198.6498147 &0.094446274 &435.3089733 &0.524130018 &0.545060211 &0.730176133 &0.05345935 &8.324123649 &10.36566358 &0.912120266 &8.67 &12075.052 &47967.53527 &159.4\\
\hline
2015 &0.617366121 &179.1637029 &0.093725645 &438.6801925 &0.540016156 &0.547408526 &0.727055177 &0.052997358 &8.456326593 &10.10966409 &0.932741508 &50.29 &13182.83334 &51652.87565 &157.2\\
\hline
2016 &0.614656439 &134.5724311 &0.093564023 &442.1742592 &0.558923488 &0.547604836 &0.732802092 &0.052527861 &8.333882334 &10.73475734 &0.951164027 &50.54 &14328.14133 &55939.46599 &138.94\\
\hline
2017 &0.605388917 &93.61972547 &0.086820781 &445.1131496 &0.577785857 &0.548797995 &0.737336045 &0.052199574 &8.053402438 &11.01950072 &0.958207523 &62.2 &15557.73277 &61583.50207 &252.86\\
\hline
st1 &0.4 &100 &0.05 &400 &0.3 &0.7 &0.8 &0.1 &10 &20 &0.95 &12 &30000 &50000 &150\\
\hline
st2 &0.6 &200 &0.1 &600 &0.5 &0.5 &0.7 &0.08 &8 &15 &0.9 &9 &20000 &40000 &100\\
\hline
st3 &0.7 &300 &0.2 &800 &0.6 &0.4 &0.6 &0.05 &6 &10 &0.8 &6 &15000 &30000 &50\\
\hline
st4 &0.8 &400 &0.3 &1000 &0.7 &0.3 &0.5 &0.02 &4 &5 &0.7 &3 &10000 &20000 &20\\
\hline
\end{array} $$
采用的归一方法如下
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：$$ n_{ij} = \frac{max(o_{j})-{o_{ij}}}{{max(o_{j})-min(o_{j})}} $$
归一化矩阵如下
$$ \begin{array}{c|c|c|c|c|c|c}{M_{11 \times15}} & -PN1 & -PN2 & -PE1 & -PS1 & -PS2 &SN1 &SN2 &SN3 &SE1 &SS1 &RN1 &RN2 &RE1 &RE2 &RS1\\
\hline
2011 &0.364 &0.396 &0.783 &0.956 &0.538 &0.546 &0.781 &0.384 &0.786 &0.263 &0.584 &0.123 &0 &0.364 &0.671\\
\hline
2012 &0.4 &0.554 &0.798 &0.95 &0.503 &0.561 &0.734 &0.423 &0.719 &0.236 &0.654 &0.137 &0.059 &0.474 &0.645\\
\hline
2013 &0.413 &0.546 &0.819 &0.945 &0.473 &0.585 &0.752 &0.417 &0.694 &0.256 &0.734 &0.166 &0.154 &0.574 &0.534\\
\hline
2014 &0.447 &0.657 &0.822 &0.941 &0.44 &0.613 &0.767 &0.418 &0.721 &0.358 &0.822 &0.096 &0.208 &0.673 &0.599\\
\hline
2015 &0.457 &0.721 &0.825 &0.936 &0.4 &0.619 &0.757 &0.412 &0.743 &0.341 &0.901 &0.799 &0.257 &0.761 &0.589\\
\hline
2016 &0.463 &0.866 &0.826 &0.93 &0.353 &0.619 &0.776 &0.407 &0.722 &0.382 &0.973 &0.803 &0.307 &0.864 &0.511\\
\hline
2017 &0.487 &1 &0.853 &0.925 &0.306 &0.622 &0.791 &0.402 &0.676 &0.401 &1 &1 &0.362 &1 &1\\
\hline
st1 &1 &0.979 &1 &1 &1 &1 &1 &1 &1 &1 &0.968 &0.152 &1 &0.721 &0.558\\
\hline
st2 &0.5 &0.653 &0.8 &0.667 &0.5 &0.5 &0.667 &0.75 &0.667 &0.667 &0.775 &0.101 &0.558 &0.481 &0.344\\
\hline
st3 &0.25 &0.326 &0.4 &0.333 &0.25 &0.25 &0.333 &0.375 &0.333 &0.333 &0.387 &0.051 &0.337 &0.24 &0.129\\
\hline
st4 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0.116 &0 &0\\
\hline
\end{array} $$
正极值点构成
$$ \begin{array}{c|c|c|c|c|c|c}{M_{1 \times15}} & -PN1 & -PN2 & -PE1 & -PS1 & -PS2 &SN1 &SN2 &SN3 &SE1 &SS1 &RN1 &RN2 &RE1 &RE2 &RS1\\
\hline
\mathbf{Zone^+} &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1\\
\hline
\end{array} $$负极值点构成
$$ \begin{array}{c|c|c|c|c|c|c}{M_{1 \times15}} & -PN1 & -PN2 & -PE1 & -PS1 & -PS2 &SN1 &SN2 &SN3 &SE1 &SS1 &RN1 &RN2 &RE1 &RE2 &RS1\\
\hline
\mathbf{Zone^-} &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0\\
\hline
\end{array} $$采用的是熵权法（EWM）求权重
$$ \begin{array}{c|c|c|c|c|c|c}{M_{2 \times15}} & -PN1 & -PN2 & -PE1 & -PS1 & -PS2 &SN1 &SN2 &SN3 &SE1 &SS1 &RN1 &RN2 &RE1 &RE2 &RS1\\
\hline
EWM所得权重 &0.0563 &0.0522 &0.04 &0.0445 &0.0592 &0.0486 &0.0425 &0.0565 &0.0426 &0.0742 &0.0444 &0.1965 &0.1234 &0.0573 &0.0619\\
\hline
权重大小顺序 &8 &9 &15 &11 &5 &10 &14 &7 &13 &3 &12 &1 &2 &6 &4\\
\hline
\end{array} $$
VIKOR的最大化群体效益和最小化反对意见的个别遗憾
最大化群体效益 |
最小化反对意见的个别遗憾 |
| $$ S_i = \sum_\limits{j=1}^m{ \omega_{j} \left(\frac{Zone_j^+ -n_{ij}}{Zone_j^+ -Zone_j^-} \right)} \quad \quad $$ |
$$ R_i = \max_\limits{j=1} { \left( \omega_{j} (\frac{Zone_j^+ -n_{ij}}{Zone_j^+ -Zone_j^-} )\right)} \quad \quad $$ |
代入权重值等即得(S R)两列矩阵，两列都为负向指标
$$ \begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &期望值 &遗憾值\\
\hline
2011 &0.6067 &0.1724\\
\hline
2012 &0.5845 &0.1697\\
\hline
2013 &0.5633 &0.164\\
\hline
2014 &0.5405 &0.1778\\
\hline
2015 &0.3876 &0.0918\\
\hline
2016 &0.3686 &0.0856\\
\hline
2017 &0.2775 &0.0789\\
\hline
st1 &0.2126 &0.1668\\
\hline
st2 &0.5022 &0.1767\\
\hline
st3 &0.7471 &0.1867\\
\hline
st4 &0.9858 &0.1965\\
\hline
\end{array} $$
妥协解的公式
公式 |
| $$ Q_i = \left( 1-k \right) \left(\frac{S_i - Min(S_i)}{Max(S_i) -Min(S_i)} \right) + k\left(\frac{R_i - Min(R_i)}{Max(R_i) -Min(R_i)} \right) $$ |
截距方式分析k的值——也是常规方法
一般的论文对于下面的公式
$$ Q_i = \left( 1-k \right) \left(\frac{S_i - Min(S_i)}{Max(S_i) -Min(S_i)} \right) + k\left(\frac{R_i - Min(R_i)}{Max(R_i) -Min(R_i)} \right) $$
其中的$k$随便说一下取0.5就拉倒了。这个好比小学生的四舍五入一样天经地义。事实上这个值很有得商榷的。它是一个敏感性有强有弱的范围。
$$ 对于每一行 令a_i =\frac{S_i - Min(S_i)}{Max(S_i) -Min(S_i)} \quad \quad b_i =\frac{R_i - Min(R_i)}{Max(R_i) -Min(R_i)} $$
$$ Q_i = \left( 1-k \right) a_i + kb_i \quad \quad $$
对于 $x,y$样本
$$
\begin{cases}
\left( 1-k \right) a_x + kb_x \\
\left( 1-k \right) a_y + kb_y
\end{cases}
$$
以上问题就变成了求两条线段是否在$[0,1]$值域内有相交的问题，此题属于初中的知识范畴，不再详细描述。
$$
\left( 1-k \right) a_x + kb_x =\left( 1-k \right) a_y + kb_y
$$
$$
a_x-k a_x + kb_x =a_y-k a_y + kb_y
$$
$$
a_x- a_y=-k a_y + kb_y +k a_x - kb_x
$$
$$
a_x- a_y=(- a_y + b_y + a_x - b_x)k
$$
$$
k =\frac{a_x- a_y}{( a_x- a_y + b_y - b_x)}
$$
基础矩阵如下
$$Base=\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &a_i &b_i\\
\hline 2011 &0.5097 &0.7954\\
\hline 2012 &0.4809 &0.7728\\
\hline 2013 &0.4536 &0.7242\\
\hline 2014 &0.4241 &0.8415\\
\hline 2015 &0.2264 &0.1103\\
\hline 2016 &0.2018 &0.0571\\
\hline 2017 &0.084 &0\\
\hline st1 &0 &0.7474\\
\hline st2 &0.3745 &0.8322\\
\hline st3 &0.6913 &0.9169\\
\hline st4 &1 &1\\
\hline \end{array} $$
拐点k值分析
$$Qk_{matrix}=\begin{array}{c|c|c|c|c|c|c}{M_{11 \times12}} &k=0 &k=0.101 &k=0.201 &k=0.226 &k=0.262 &k=0.423 &k=0.453 &k=0.642 &k=0.65 &k=0.786 &k=0.951 &k=1\\
\hline 2011 &0.51 &0.539 &0.567 &0.574 &0.585 &0.631 &0.639 &0.693 &0.696 &0.734 &0.782 &0.795\\
\hline 2012 &0.481 &0.51 &0.54 &0.547 &0.557 &0.604 &0.613 &0.668 &0.671 &0.71 &0.759 &0.773\\
\hline 2013 &0.454 &0.481 &0.508 &0.515 &0.525 &0.568 &0.576 &0.627 &0.63 &0.666 &0.711 &0.724\\
\hline 2014 &0.424 &0.466 &0.508 &0.519 &0.534 &0.601 &0.613 &0.692 &0.696 &0.752 &0.821 &0.841\\
\hline 2015 &0.226 &0.215 &0.203 &0.2 &0.196 &0.177 &0.174 &0.152 &0.151 &0.135 &0.116 &0.11\\
\hline 2016 &0.202 &0.187 &0.173 &0.169 &0.164 &0.141 &0.136 &0.109 &0.108 &0.088 &0.064 &0.057\\
\hline 2017 &0.084 &0.076 &0.067 &0.065 &0.062 &0.048 &0.046 &0.03 &0.029 &0.018 &0.004 &0\\
\hline st1 &0 &0.076 &0.15 &0.169 &0.196 &0.316 &0.339 &0.48 &0.486 &0.588 &0.711 &0.747\\
\hline st2 &0.375 &0.421 &0.466 &0.478 &0.494 &0.568 &0.582 &0.668 &0.672 &0.734 &0.81 &0.832\\
\hline st3 &0.691 &0.714 &0.737 &0.742 &0.75 &0.787 &0.793 &0.836 &0.838 &0.869 &0.906 &0.917\\
\hline st4 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1\\
\hline \end{array} $$
排序分析
上述是负向指标，数值越小越好，每一列数值最小的排第一。因此排序情况如下：
$$Q_{rank}=\begin{array}{c|c|c|c|c|c|c}{M_{11 \times12}} &k=0 &k=0.101 &k=0.201 &k=0.226 &k=0.262 &k=0.423 &k=0.453 &k=0.642 &k=0.65 &k=0.786 &k=0.951 &k=1\\
\hline 2011 &9 &9 &9 &9 &9 &9 &9 &9 &8 &7 &7 &7\\
\hline 2012 &8 &8 &8 &8 &8 &8 &7 &6 &6 &6 &6 &6\\
\hline 2013 &7 &7 &6 &6 &6 &5 &5 &5 &5 &5 &4 &4\\
\hline 2014 &6 &6 &6 &7 &7 &7 &7 &8 &8 &9 &9 &9\\
\hline 2015 &4 &4 &4 &4 &3 &3 &3 &3 &3 &3 &3 &3\\
\hline 2016 &3 &3 &3 &2 &2 &2 &2 &2 &2 &2 &2 &2\\
\hline 2017 &2 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1\\
\hline st1 &1 &1 &2 &2 &3 &4 &4 &4 &4 &4 &4 &5\\
\hline st2 &5 &5 &5 &5 &5 &5 &6 &6 &7 &7 &8 &8\\
\hline st3 &10 &10 &10 &10 &10 &10 &10 &10 &10 &10 &10 &10\\
\hline st4 &11 &11 &11 &11 &11 &11 &11 &11 &11 &11 &11 &11\\
\hline \end{array} $$
聚类特征
| 序号 |
聚类特征-对应k值区段 |
Q值排序 |
| 1 | 0<$k$< 0.10107 | $st1 \succ 2017 \succ 2016 \succ 2015 \succ st2 \succ 2014 \succ 2013 \succ 2012 \succ 2011 \succ st3 \succ st4$ |
|---|
| 2 | 0.1011<$k$< 0.20097 | $2017 \succ st1 \succ 2016 \succ 2015 \succ st2 \succ 2014 \succ 2013 \succ 2012 \succ 2011 \succ st3 \succ st4$ |
|---|
| 3 | 0.201<$k$< 0.22621 | $2017 \succ st1 \succ 2016 \succ 2015 \succ st2 \succ 2013 \succ 2014 \succ 2012 \succ 2011 \succ st3 \succ st4$ |
|---|
| 4 | 0.2262<$k$< 0.26216 | $2017 \succ 2016 \succ st1 \succ 2015 \succ st2 \succ 2013 \succ 2014 \succ 2012 \succ 2011 \succ st3 \succ st4$ |
|---|
| 5 | 0.2622<$k$< 0.42285 | $2017 \succ 2016 \succ 2015 \succ st1 \succ st2 \succ 2013 \succ 2014 \succ 2012 \succ 2011 \succ st3 \succ st4$ |
|---|
| 6 | 0.4229<$k$< 0.45294 | $2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ st2 \succ 2014 \succ 2012 \succ 2011 \succ st3 \succ st4$ |
|---|
| 7 | 0.4529<$k$< 0.64206 | $2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ st2 \succ 2012 \succ 2014 \succ 2011 \succ st3 \succ st4$ |
|---|
| 8 | 0.6421<$k$< 0.6503 | $2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ st2 \succ 2014 \succ 2011 \succ st3 \succ st4$ |
|---|
| 9 | 0.6503<$k$< 0.7864 | $2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ st2 \succ 2011 \succ 2014 \succ st3 \succ st4$ |
|---|
| 10 | 0.7864<$k$< 0.95141 | $2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ 2011 \succ st2 \succ 2014 \succ st3 \succ st4$ |
|---|
| 11 | 0.9514<$k$< 1 | $2017 \succ 2016 \succ 2015 \succ 2013 \succ st1 \succ 2012 \succ 2011 \succ st2 \succ 2014 \succ st3 \succ st4$ |
|---|
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