AEC-TOPSIS求解

论文写作或者计算需要帮助可发邮件到 hwstu # sohu.com 把 #替换成@，非免费

流程图

原始数据如下

$$ \begin{array}{c|c|c|c|c|c|c}{M_{11 \times15}} & -PN1 & -PN2 & -PE1 & -PS1 & -PS2 &SN1 &SN2 &SN3 &SE1 &SS1 &RN1 &RN2 &RE1 &RE2 &RS1\\ \hline 2011 &0.654302103 &278.5577711 &0.104246565 &426.623418 &0.484969128 &0.518335321 &0.734225621 &0.050682504 &8.714200492 &8.950944846 &0.850812256 &10.3 &7380.279843 &35115.65191 &176.2\\ \hline 2012 &0.64018787 &230.2072698 &0.100574919 &430.0655108 &0.498760764 &0.524293616 &0.720190933 &0.053800771 &8.311141515 &8.547470721 &0.868804512 &11.1 &8725.395288 &39690.62085 &170.1\\ \hline 2013 &0.634839501 &232.7438152 &0.095228796 &433.091812 &0.510793577 &0.534174192 &0.725746269 &0.053359371 &8.162741259 &8.838568347 &0.889545989 &12.82 &10872.14779 &43857.04467 &144.4\\ \hline 2014 &0.621034465 &198.6498147 &0.094446274 &435.3089733 &0.524130018 &0.545060211 &0.730176133 &0.05345935 &8.324123649 &10.36566358 &0.912120266 &8.67 &12075.052 &47967.53527 &159.4\\ \hline 2015 &0.617366121 &179.1637029 &0.093725645 &438.6801925 &0.540016156 &0.547408526 &0.727055177 &0.052997358 &8.456326593 &10.10966409 &0.932741508 &50.29 &13182.83334 &51652.87565 &157.2\\ \hline 2016 &0.614656439 &134.5724311 &0.093564023 &442.1742592 &0.558923488 &0.547604836 &0.732802092 &0.052527861 &8.333882334 &10.73475734 &0.951164027 &50.54 &14328.14133 &55939.46599 &138.94\\ \hline 2017 &0.605388917 &93.61972547 &0.086820781 &445.1131496 &0.577785857 &0.548797995 &0.737336045 &0.052199574 &8.053402438 &11.01950072 &0.958207523 &62.2 &15557.73277 &61583.50207 &252.86\\ \hline st1 &0.4 &100 &0.05 &400 &0.3 &0.7 &0.8 &0.1 &10 &20 &0.95 &12 &30000 &50000 &150\\ \hline st2 &0.6 &200 &0.1 &600 &0.5 &0.5 &0.7 &0.08 &8 &15 &0.9 &9 &20000 &40000 &100\\ \hline st3 &0.7 &300 &0.2 &800 &0.6 &0.4 &0.6 &0.05 &6 &10 &0.8 &6 &15000 &30000 &50\\ \hline st4 &0.8 &400 &0.3 &1000 &0.7 &0.3 &0.5 &0.02 &4 &5 &0.7 &3 &10000 &20000 &20\\ \hline \end{array} $$

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：$$ n_{ij} = \frac{max(o_{j})-{o_{ij}}}{{max(o_{j})-min(o_{j})}} $$

计算后的归一化矩阵如下

$$ \begin{array}{c|c|c|c|c|c|c}{M_{11 \times15}} & -PN1 & -PN2 & -PE1 & -PS1 & -PS2 &SN1 &SN2 &SN3 &SE1 &SS1 &RN1 &RN2 &RE1 &RE2 &RS1\\ \hline 2011 &0.364 &0.396 &0.783 &0.956 &0.538 &0.546 &0.781 &0.384 &0.786 &0.263 &0.584 &0.123 &0 &0.364 &0.671\\ \hline 2012 &0.4 &0.554 &0.798 &0.95 &0.503 &0.561 &0.734 &0.423 &0.719 &0.236 &0.654 &0.137 &0.059 &0.474 &0.645\\ \hline 2013 &0.413 &0.546 &0.819 &0.945 &0.473 &0.585 &0.752 &0.417 &0.694 &0.256 &0.734 &0.166 &0.154 &0.574 &0.534\\ \hline 2014 &0.447 &0.657 &0.822 &0.941 &0.44 &0.613 &0.767 &0.418 &0.721 &0.358 &0.822 &0.096 &0.208 &0.673 &0.599\\ \hline 2015 &0.457 &0.721 &0.825 &0.936 &0.4 &0.619 &0.757 &0.412 &0.743 &0.341 &0.901 &0.799 &0.257 &0.761 &0.589\\ \hline 2016 &0.463 &0.866 &0.826 &0.93 &0.353 &0.619 &0.776 &0.407 &0.722 &0.382 &0.973 &0.803 &0.307 &0.864 &0.511\\ \hline 2017 &0.487 &1 &0.853 &0.925 &0.306 &0.622 &0.791 &0.402 &0.676 &0.401 &1 &1 &0.362 &1 &1\\ \hline st1 &1 &0.979 &1 &1 &1 &1 &1 &1 &1 &1 &0.968 &0.152 &1 &0.721 &0.558\\ \hline st2 &0.5 &0.653 &0.8 &0.667 &0.5 &0.5 &0.667 &0.75 &0.667 &0.667 &0.775 &0.101 &0.558 &0.481 &0.344\\ \hline st3 &0.25 &0.326 &0.4 &0.333 &0.25 &0.25 &0.333 &0.375 &0.333 &0.333 &0.387 &0.051 &0.337 &0.24 &0.129\\ \hline st4 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0.116 &0 &0\\ \hline \end{array} $$

正极值点构成

$$ \begin{array}{c|c|c|c|c|c|c}{M_{1 \times15}} & -PN1 & -PN2 & -PE1 & -PS1 & -PS2 &SN1 &SN2 &SN3 &SE1 &SS1 &RN1 &RN2 &RE1 &RE2 &RS1\\ \hline \mathbf{Max} &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1\\ \hline \end{array} $$

负极值点构成

$$ \begin{array}{c|c|c|c|c|c|c}{M_{1 \times15}} & -PN1 & -PN2 & -PE1 & -PS1 & -PS2 &SN1 &SN2 &SN3 &SE1 &SS1 &RN1 &RN2 &RE1 &RE2 &RS1\\ \hline \mathbf{Min} &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0\\ \hline \end{array} $$

带权重的距离公式中权重的计算方法——此处处理的为客观权重方法所得权重

采用的是熵权法（EWM）求权重

$$ \begin{array}{c|c|c|c|c|c|c}{M_{2 \times15}} & -PN1 & -PN2 & -PE1 & -PS1 & -PS2 &SN1 &SN2 &SN3 &SE1 &SS1 &RN1 &RN2 &RE1 &RE2 &RS1\\ \hline EWM所得权重 &0.0563 &0.0522 &0.04 &0.0445 &0.0592 &0.0486 &0.0425 &0.0565 &0.0426 &0.0742 &0.0444 &0.1965 &0.1234 &0.0573 &0.0619\\ \hline 权重大小顺序 &8 &9 &15 &11 &5 &10 &14 &7 &13 &3 &12 &1 &2 &6 &4\\ \hline \end{array} $$

带权重的距离公式中距离公式的选择

点击此按钮可以查看选择的距离公式的一般形式

欧几里得距离、欧式距离公式

$$ D_i^+ = \sqrt {\sum_\limits{j=1}^m { \omega_{j}^2 \left({Max(n_j) -n_{ij}} \right)} ^2} $$

$$ D_i^- = \sqrt {\sum_\limits{j=1}^m { \omega_{j}^2 \left({n_{ij}-Min(n_j) } \right)} ^2} $$

计算后的距离矩阵如下

$$Dist=\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &-d^+ &d^-\\ \hline 2011 &0.2344 &0.1088\\ \hline 2012 &0.2269 &0.1119\\ \hline 2013 &0.2164 &0.1152\\ \hline 2014 &0.2203 &0.1212\\ \hline 2015 &0.1318 &0.2001\\ \hline 2016 &0.1268 &0.2046\\ \hline 2017 &0.1118 &0.2462\\ \hline st1 &0.1696 &0.22\\ \hline st2 &0.2021 &0.1368\\ \hline st3 &0.2451 &0.0719\\ \hline st4 &0.2954 &0.0143\\ \hline \end{array} $$

妥协解的公式与基础矩阵

公式

$$ Q_i = \left( 1-k \right) \left(\frac{D_i^+ - Min(D_i^+)}{Max(D_i^+) -Min(D_i^+)} \right) + k\left(\frac{ Max(D_i^-) -D_i^-}{Max(D_i^-) -Min(D_i^-)} \right) $$

$$Q_i 为负向指标，即数值越大，越差$$

$$a_i= \frac{D_i^+ - Min(D_i^+)}{Max(D_i^+) -Min(D_i^+)} \qquad b_i=\frac{ Max(D_i^-) -D_i^-}{Max(D_i^-) -Min(D_i^-)} $$

$$ Q_i = \left( 1-k \right) a_i + k b_i $$

$$base=\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &-a_i &-b_i\\ \hline 2011 &0.6678 &0.5923\\ \hline 2012 &0.627 &0.5792\\ \hline 2013 &0.5697 &0.565\\ \hline 2014 &0.5912 &0.5392\\ \hline 2015 &0.1088 &0.1987\\ \hline 2016 &0.0816 &0.1791\\ \hline 2017 &0 &0\\ \hline st1 &0.3148 &0.1127\\ \hline st2 &0.4922 &0.4718\\ \hline st3 &0.7264 &0.7515\\ \hline st4 &1 &1\\ \hline \end{array} $$

妥协解是基于基础矩阵进行的，有如下特点：

第一、基础矩阵最常用的是曼哈顿距离公式。

第二、Base矩阵与D矩阵的两列是等序等构变换的。所以妥协解$D^-$同vikor的妥协解方向性有变化

基于截距的妥协解、敏感值分析(k值拐点分析)、聚类特征排序分析

$$ Q_i = \left( 1-k \right) a_i + kb_i \quad \quad $$

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。

$$ \left( 1-k \right) a_x + kb_x =\left( 1-k \right) a_y + kb_y $$

$$ a_x-k a_x + kb_x =a_y-k a_y + kb_y $$

$$ a_x- a_y=-k a_y + kb_y +k a_x - kb_x $$

$$ a_x- a_y=(- a_y + b_y + a_x - b_x)k $$

$$ k =\frac{a_x- a_y}{( a_x- a_y + b_y - b_x)} $$

拐点k值分析，敏感值分析

$$Qk_{matrix}=\begin{array}{c|c|c|c|c|c|c}{M_{11 \times5}} &k=0 &k=0.454 &k=0.706 &k=0.778 &k=1\\ \hline 2011 &0.668 &0.633 &0.615 &0.609 &0.592\\ \hline 2012 &0.627 &0.605 &0.593 &0.59 &0.579\\ \hline 2013 &0.57 &0.568 &0.566 &0.566 &0.565\\ \hline 2014 &0.591 &0.568 &0.554 &0.551 &0.539\\ \hline 2015 &0.109 &0.15 &0.172 &0.179 &0.199\\ \hline 2016 &0.082 &0.126 &0.15 &0.157 &0.179\\ \hline 2017 &0 &0 &0 &0 &0\\ \hline st1 &0.315 &0.223 &0.172 &0.157 &0.113\\ \hline st2 &0.492 &0.483 &0.478 &0.476 &0.472\\ \hline st3 &0.726 &0.738 &0.744 &0.746 &0.752\\ \hline st4 &1 &1 &1 &1 &1\\ \hline \end{array} $$

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：

$$Q_{rank}=\begin{array}{c|c|c|c|c|c|c}{M_{11 \times5}} &k=0 &k=0.454 &k=0.706 &k=0.778 &k=1\\ \hline 2011 &9 &9 &9 &9 &9\\ \hline 2012 &8 &8 &8 &8 &8\\ \hline 2013 &6 &6 &7 &7 &7\\ \hline 2014 &7 &6 &6 &6 &6\\ \hline 2015 &3 &3 &3 &4 &4\\ \hline 2016 &2 &2 &2 &2 &3\\ \hline 2017 &1 &1 &1 &1 &1\\ \hline st1 &4 &4 &3 &2 &2\\ \hline st2 &5 &5 &5 &5 &5\\ \hline st3 &10 &10 &10 &10 &10\\ \hline st4 &11 &11 &11 &11 &11\\ \hline \end{array} $$

聚类特征

序号	聚类特征-对应k值区段	Q值排序
1	0<$k$< 0.45442	$2017 \succ 2016 \succ 2015 \succ st1 \succ st2 \succ 2013 \succ 2014 \succ 2012 \succ 2011 \succ st3 \succ st4$
2	0.4544<$k$< 0.70557	$2017 \succ 2016 \succ 2015 \succ st1 \succ st2 \succ 2014 \succ 2013 \succ 2012 \succ 2011 \succ st3 \succ st4$
3	0.7056<$k$< 0.77841	$2017 \succ 2016 \succ st1 \succ 2015 \succ st2 \succ 2014 \succ 2013 \succ 2012 \succ 2011 \succ st3 \succ st4$
4	0.7784<$k$< 1	$2017 \succ st1 \succ 2016 \succ 2015 \succ st2 \succ 2014 \succ 2013 \succ 2012 \succ 2011 \succ st3 \succ st4$

AEC求解过程

排名	要素所占区段
0	2017=1
1	2016=0.7784 　　st1=0.2216
2	2015=0.70559 　　2016=0.2216 　　st1=0.07281
3	st1=0.70559 　　2015=0.29441
4	st2=1
5	2014=0.54558 　　2013=0.45442
6	2013=0.54558 　　2014=0.45442
7	2012=1
8	2011=1
9	st3=1
10	st4=1

排名	最优妥协解
优胜情境	$2017 \succ 2016 \succ 2015 \succ st1 \succ st2 \succ 2014 \succ 2013 \succ 2012 \succ 2011 \succ st3 \succ st4$
劣汰情境	$2017 \succ 2016 \succ 2015 \succ st1 \succ st2 \succ 2014 \succ 2013 \succ 2012 \succ 2011 \succ st3 \succ st4$

劣汰情境的排序为聚类排序中的第2个

取中间值聚类中间值为：(0.45442321131732+0.70557234435066)/2 =0.57999777783399

$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times3}} &a_i &b_i &最优妥协解\\ \hline 2011 &0.6678 &0.5923 &0.624\\ \hline 2012 &0.627 &0.5792 &0.5993\\ \hline 2013 &0.5697 &0.565 &0.567\\ \hline 2014 &0.5912 &0.5392 &0.561\\ \hline 2015 &0.1088 &0.1987 &0.1609\\ \hline 2016 &0.0816 &0.1791 &0.1381\\ \hline 2017 &0 &0 &0\\ \hline st1 &0.3148 &0.1127 &0.1976\\ \hline st2 &0.4922 &0.4718 &0.4804\\ \hline st3 &0.7264 &0.7515 &0.741\\ \hline st4 &1 &1 &1\\ \hline \end{array} $$

优胜情境的排序为聚类排序中的第2个

取中间值聚类中间值为：(0.45442321131732+0.70557234435066)/2 =0.57999777783399