AEC-VOTE求解过程
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流程图
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CRITIC方法求权重
$$ \begin{array}{c|c|c|c|c|c|c}{M_{2 \times15}} & -PN1 & -PN2 & -PE1 & -PS1 & -PS2 &SN1 &SN2 &SN3 &SE1 &SS1 &RN1 &RN2 &RE1 &RE2 &RS1\\
\hline
CRITIC方法所得权重 &0.0532 &0.0668 &0.0663 &0.0779 &0.0587 &0.06 &0.0658 &0.0601 &0.0637 &0.0626 &0.0728 &0.0884 &0.0677 &0.0701 &0.0659\\
\hline
权重大小顺序 &15 &6 &7 &2 &14 &13 &9 &12 &10 &11 &3 &1 &5 &4 &8\\
\hline
\end{array} $$
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
$$ \begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &期望值 &遗憾值\\
\hline
2011 &0.4998 &0.0775\\
\hline
2012 &0.4785 &0.0763\\
\hline
2013 &0.4624 &0.0738\\
\hline
2014 &0.4294 &0.08\\
\hline
2015 &0.3507 &0.0504\\
\hline
2016 &0.3304 &0.0469\\
\hline
2017 &0.2576 &0.0432\\
\hline
st1 &0.1274 &0.075\\
\hline
st2 &0.4332 &0.0795\\
\hline
st3 &0.716 &0.084\\
\hline
st4 &0.9922 &0.0884\\
\hline
\end{array} $$
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
$$Base=\begin{array}{c|c|c|c|c|c|c}{M_{11 \times2}} &a_i &b_i\\
\hline 2011 &0.4306 &0.7602\\
\hline 2012 &0.4059 &0.7337\\
\hline 2013 &0.3874 &0.6768\\
\hline 2014 &0.3492 &0.8142\\
\hline 2015 &0.2582 &0.1576\\
\hline 2016 &0.2348 &0.0816\\
\hline 2017 &0.1506 &0\\
\hline st1 &0 &0.7039\\
\hline st2 &0.3537 &0.8033\\
\hline st3 &0.6806 &0.9026\\
\hline st4 &1 &1\\
\hline \end{array} $$
拐点k值分析
$$Qk_{matrix}=\begin{array}{c|c|c|c|c|c|c}{M_{11 \times13}} &k=0 &k=0.176 &k=0.21 &k=0.217 &k=0.274 &k=0.288 &k=0.321 &k=0.413 &k=0.429 &k=0.601 &k=0.641 &k=0.934 &k=1\\
\hline 2011 &0.431 &0.489 &0.5 &0.502 &0.521 &0.526 &0.536 &0.567 &0.572 &0.629 &0.642 &0.739 &0.76\\
\hline 2012 &0.406 &0.464 &0.475 &0.477 &0.496 &0.5 &0.511 &0.541 &0.547 &0.603 &0.616 &0.712 &0.734\\
\hline 2013 &0.387 &0.438 &0.448 &0.45 &0.467 &0.471 &0.48 &0.507 &0.512 &0.561 &0.573 &0.658 &0.677\\
\hline 2014 &0.349 &0.431 &0.447 &0.45 &0.477 &0.483 &0.498 &0.541 &0.549 &0.629 &0.647 &0.784 &0.814\\
\hline 2015 &0.258 &0.24 &0.237 &0.236 &0.231 &0.229 &0.226 &0.217 &0.215 &0.198 &0.194 &0.164 &0.158\\
\hline 2016 &0.235 &0.208 &0.203 &0.202 &0.193 &0.191 &0.186 &0.171 &0.169 &0.143 &0.137 &0.092 &0.082\\
\hline 2017 &0.151 &0.124 &0.119 &0.118 &0.109 &0.107 &0.102 &0.088 &0.086 &0.06 &0.054 &0.01 &0\\
\hline st1 &0 &0.124 &0.148 &0.153 &0.193 &0.203 &0.226 &0.291 &0.302 &0.423 &0.451 &0.658 &0.704\\
\hline st2 &0.354 &0.433 &0.448 &0.451 &0.477 &0.483 &0.498 &0.539 &0.547 &0.624 &0.642 &0.774 &0.803\\
\hline st3 &0.681 &0.72 &0.727 &0.729 &0.741 &0.745 &0.752 &0.772 &0.776 &0.814 &0.823 &0.888 &0.903\\
\hline st4 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1\\
\hline \end{array} $$
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:
$$Q_{rank}=\begin{array}{c|c|c|c|c|c|c}{M_{11 \times13}} &k=0 &k=0.176 &k=0.21 &k=0.217 &k=0.274 &k=0.288 &k=0.321 &k=0.413 &k=0.429 &k=0.601 &k=0.641 &k=0.934 &k=1\\
\hline 2011 &9 &9 &9 &9 &9 &9 &9 &9 &9 &8 &7 &7 &7\\
\hline 2012 &8 &8 &8 &8 &8 &8 &8 &7 &6 &6 &6 &6 &6\\
\hline 2013 &7 &7 &6 &5 &5 &5 &5 &5 &5 &5 &5 &4 &4\\
\hline 2014 &5 &5 &5 &5 &6 &6 &7 &7 &8 &8 &9 &9 &9\\
\hline 2015 &4 &4 &4 &4 &4 &4 &3 &3 &3 &3 &3 &3 &3\\
\hline 2016 &3 &3 &3 &3 &2 &2 &2 &2 &2 &2 &2 &2 &2\\
\hline 2017 &2 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1\\
\hline st1 &1 &1 &2 &2 &2 &3 &3 &4 &4 &4 &4 &4 &5\\
\hline st2 &6 &6 &6 &7 &7 &6 &6 &6 &6 &7 &7 &8 &8\\
\hline st3 &10 &10 &10 &10 &10 &10 &10 &10 &10 &10 &10 &10 &10\\
\hline st4 &11 &11 &11 &11 &11 &11 &11 &11 &11 &11 &11 &11 &11\\
\hline \end{array} $$
聚类特征
AEC求解过程
排名 |
要素所占区段 |
0 | 2017=0.82378 st1=0.17622 |
---|
1 | 2016=0.726062 2017=0.17622 st1=0.097718 |
---|
2 | 2015=0.679062 2016=0.273938 st1=0.047 |
---|
3 | st1=0.613554 2015=0.320938 2013=0.065508 |
---|
4 | 2013=0.717245 2014=0.217247 st1=0.065508 |
---|
5 | 2012=0.570874 st2=0.351695 2014=0.070663 2013=0.006768 |
---|
6 | 2011=0.358658 st2=0.289647 2013=0.210479 2014=0.125416 2012=0.0158 |
---|
7 | 2012=0.413326 st2=0.358658 2014=0.18795 2011=0.040066 |
---|
8 | 2011=0.601276 2014=0.398724 |
---|
9 | st3=1 |
---|
10 | st4=1 |
排名 |
最优妥协解 |
优胜情境 | $2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ st2 \succ 2014 \succ 2011 \succ st3 \succ st4$ |
---|
劣汰情境 | $2017 \succ 2016 \succ 2015 \succ st1 \succ 2013 \succ 2012 \succ st2 \succ 2014 \succ 2011 \succ st3 \succ st4$ |
劣汰情境的排序为聚类排序中的第9个
取中间值聚类中间值为:(0.42912590573016+0.60127649706287)/2 =0.51520120139651
$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times3}} &a_i &b_i &最优妥协解\\
\hline 2011 &0.4306 &0.7602 &0.6004\\
\hline 2012 &0.4059 &0.7337 &0.5748\\
\hline 2013 &0.3874 &0.6768 &0.5365\\
\hline 2014 &0.3492 &0.8142 &0.5888\\
\hline 2015 &0.2582 &0.1576 &0.2064\\
\hline 2016 &0.2348 &0.0816 &0.1559\\
\hline 2017 &0.1506 &0 &0.073\\
\hline st1 &0 &0.7039 &0.3627\\
\hline st2 &0.3537 &0.8033 &0.5853\\
\hline st3 &0.6806 &0.9026 &0.795\\
\hline st4 &1 &1 &1\\
\hline \end{array} $$优胜情境的排序为聚类排序中的第9个
取中间值聚类中间值为:(0.42912590573016+0.60127649706287)/2 =0.51520120139651
$$\begin{array}{c|c|c|c|c|c|c}{M_{11 \times3}} &a_i &b_i &最优妥协解\\
\hline 2011 &0.4306 &0.7602 &0.6004\\
\hline 2012 &0.4059 &0.7337 &0.5748\\
\hline 2013 &0.3874 &0.6768 &0.5365\\
\hline 2014 &0.3492 &0.8142 &0.5888\\
\hline 2015 &0.2582 &0.1576 &0.2064\\
\hline 2016 &0.2348 &0.0816 &0.1559\\
\hline 2017 &0.1506 &0 &0.073\\
\hline st1 &0 &0.7039 &0.3627\\
\hline st2 &0.3537 &0.8033 &0.5853\\
\hline st3 &0.6806 &0.9026 &0.795\\
\hline st4 &1 &1 &1\\
\hline \end{array} $$
扯蛋模型