VIKOR-AISM求解过程
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原始矩阵如下：
$$ \begin{array}{c|c|c|c|c|c|c}{M_{8 \times16}} &C1 &C2 &C3 &C4 &C5 &E1 &E2 &E3 &F1 &F2 &F3 &U1 &U2 &D1 &D2 &D3\\
\hline
P1 &0.779 &0.782 &0.701 &0.724 &0.712 &0.756 &0.639 &0.713 &0.838 &0.758 &0.699 &0.71 &0.702 &0.705 &0.754 &0.712\\
\hline
P2 &0.791 &0.751 &0.709 &0.768 &0.732 &0.762 &0.653 &0.702 &0.73 &0.716 &0.595 &0.694 &0.641 &0.701 &0.667 &0.702\\
\hline
P3 &0.736 &0.775 &0.812 &0.728 &0.762 &0.633 &0.475 &0.582 &0.72 &0.701 &0.645 &0.714 &0.706 &0.693 &0.634 &0.674\\
\hline
P4 &0.647 &0.737 &0.722 &0.715 &0.792 &0.78 &0.564 &0.757 &0.768 &0.72 &0.699 &0.75 &0.706 &0.779 &0.712 &0.706\\
\hline
P5 &0.733 &0.806 &0.748 &0.733 &0.696 &0.682 &0.509 &0.689 &0.709 &0.715 &0.554 &0.696 &0.663 &0.671 &0.709 &0.663\\
\hline
P6 &0.734 &0.78 &0.747 &0.714 &0.768 &0.854 &0.758 &0.712 &0.719 &0.706 &0.732 &0.722 &0.729 &0.765 &0.709 &0.765\\
\hline
P7 &0.656 &0.73 &0.74 &0.628 &0.73 &0.692 &0.545 &0.59 &0.595 &0.758 &0.631 &0.706 &0.692 &0.629 &0.628 &0.605\\
\hline
P8 &0.714 &0.733 &0.721 &0.746 &0.65 &0.824 &0.703 &0.632 &0.697 &0.714 &0.761 &0.753 &0.592 &0.842 &0.743 &0.755\\
\hline
\end{array} $$
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
$$ \begin{array}{c|c|c|c|c|c|c}{M_{8 \times16}} &C1 &C2 &C3 &C4 &C5 &E1 &E2 &E3 &F1 &F2 &F3 &U1 &U2 &D1 &D2 &D3\\
\hline
P1 &0.917 &0.684 &0 &0.686 &0.437 &0.557 &0.58 &0.749 &1 &1 &0.7 &0.271 &0.803 &0.357 &1 &0.669\\
\hline
P2 &1 &0.276 &0.072 &1 &0.577 &0.584 &0.629 &0.686 &0.556 &0.263 &0.198 &0 &0.358 &0.338 &0.31 &0.606\\
\hline
P3 &0.618 &0.592 &1 &0.714 &0.789 &0 &0 &0 &0.514 &0 &0.44 &0.339 &0.832 &0.3 &0.048 &0.431\\
\hline
P4 &0 &0.092 &0.189 &0.621 &1 &0.665 &0.314 &1 &0.712 &0.333 &0.7 &0.949 &0.832 &0.704 &0.667 &0.631\\
\hline
P5 &0.597 &1 &0.423 &0.75 &0.324 &0.222 &0.12 &0.611 &0.469 &0.246 &0 &0.034 &0.518 &0.197 &0.643 &0.363\\
\hline
P6 &0.604 &0.658 &0.414 &0.614 &0.831 &1 &1 &0.743 &0.51 &0.088 &0.86 &0.475 &1 &0.638 &0.643 &1\\
\hline
P7 &0.063 &0 &0.351 &0 &0.563 &0.267 &0.247 &0.046 &0 &1 &0.372 &0.203 &0.73 &0 &0 &0\\
\hline
P8 &0.465 &0.039 &0.18 &0.843 &0 &0.864 &0.806 &0.286 &0.42 &0.228 &1 &1 &0 &1 &0.913 &0.938\\
\hline
\end{array} $$
正极值点构成
$$ \begin{array}{c|c|c|c|c|c|c}{M_{1 \times16}} &C1 &C2 &C3 &C4 &C5 &E1 &E2 &E3 &F1 &F2 &F3 &U1 &U2 &D1 &D2 &D3\\
\hline
\mathbf{Zone^+} &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1\\
\hline
\end{array} $$负极值点构成
$$ \begin{array}{c|c|c|c|c|c|c}{M_{1 \times16}} &C1 &C2 &C3 &C4 &C5 &E1 &E2 &E3 &F1 &F2 &F3 &U1 &U2 &D1 &D2 &D3\\
\hline
\mathbf{Zone^-} &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0\\
\hline
\end{array} $$采用的CRITIC方法求权重
$$ \begin{array}{c|c|c|c|c|c|c}{M_{2 \times16}} &C1 &C2 &C3 &C4 &C5 &E1 &E2 &E3 &F1 &F2 &F3 &U1 &U2 &D1 &D2 &D3\\
\hline
CRITIC方法所得权重 &0.0637 &0.0653 &0.058 &0.054 &0.0589 &0.0626 &0.0644 &0.0671 &0.052 &0.0721 &0.0631 &0.0708 &0.0604 &0.0591 &0.0691 &0.0595\\
\hline
权重大小顺序 &7 &5 &14 &15 &13 &9 &6 &4 &16 &1 &8 &2 &10 &12 &3 &11\\
\hline
\end{array} $$
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
$$ \begin{array}{c|c|c|c|c|c|c}{M_{8 \times2}} &期望值 &遗憾值\\
\hline
P1 &0.3462 &0.058\\
\hline
P2 &0.5442 &0.0708\\
\hline
P3 &0.6039 &0.0721\\
\hline
P4 &0.4141 &0.0637\\
\hline
P5 &0.597 &0.0684\\
\hline
P6 &0.3131 &0.0658\\
\hline
P7 &0.7531 &0.0691\\
\hline
P8 &0.4369 &0.0628\\
\hline
\end{array} $$
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
$$Base=\begin{array}{c|c|c|c|c|c|c}{M_{8 \times2}} &a_i &b_i\\
\hline P1 &0.0751 &0\\
\hline P2 &0.5251 &0.9071\\
\hline P3 &0.6607 &1\\
\hline P4 &0.2294 &0.4007\\
\hline P5 &0.6451 &0.741\\
\hline P6 &0 &0.5541\\
\hline P7 &1 &0.7903\\
\hline P8 &0.2812 &0.3406\\
\hline \end{array} $$
拐点k值分析
$$Qk_{matrix}=\begin{array}{c|c|c|c|c|c|c}{M_{8 \times9}} &k=0 &k=0.119 &k=0.42 &k=0.463 &k=0.568 &k=0.599 &k=0.618 &k=0.803 &k=1\\
\hline P1 &0.075 &0.066 &0.044 &0.04 &0.032 &0.03 &0.029 &0.015 &0\\
\hline P2 &0.525 &0.571 &0.685 &0.702 &0.742 &0.754 &0.761 &0.832 &0.907\\
\hline P3 &0.661 &0.701 &0.803 &0.818 &0.854 &0.864 &0.87 &0.933 &1\\
\hline P4 &0.229 &0.25 &0.301 &0.309 &0.327 &0.332 &0.335 &0.367 &0.401\\
\hline P5 &0.645 &0.657 &0.685 &0.69 &0.7 &0.703 &0.704 &0.722 &0.741\\
\hline P6 &0 &0.066 &0.232 &0.257 &0.315 &0.332 &0.342 &0.445 &0.554\\
\hline P7 &1 &0.975 &0.912 &0.903 &0.881 &0.874 &0.87 &0.832 &0.79\\
\hline P8 &0.281 &0.288 &0.306 &0.309 &0.315 &0.317 &0.318 &0.329 &0.341\\
\hline \end{array} $$
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：
$$Q_{rank}=\begin{array}{c|c|c|c|c|c|c}{M_{8 \times9}} &k=0 &k=0.119 &k=0.42 &k=0.463 &k=0.568 &k=0.599 &k=0.618 &k=0.803 &k=1\\
\hline P1 &2 &1 &1 &1 &1 &1 &1 &1 &1\\
\hline P2 &5 &5 &5 &6 &6 &6 &6 &6 &7\\
\hline P3 &7 &7 &7 &7 &7 &7 &7 &8 &8\\
\hline P4 &3 &3 &3 &3 &4 &3 &3 &3 &3\\
\hline P5 &6 &6 &5 &5 &5 &5 &5 &5 &5\\
\hline P6 &1 &1 &2 &2 &2 &3 &4 &4 &4\\
\hline P7 &8 &8 &8 &8 &8 &8 &7 &6 &6\\
\hline P8 &4 &4 &4 &3 &2 &2 &2 &2 &2\\
\hline \end{array} $$
聚类特征
| 序号 |
聚类特征-对应k值区段 |
Q值排序 |
| 1 | 0<$k$< 0.11942 | $P6 \succ P1 \succ P4 \succ P8 \succ P2 \succ P5 \succ P3 \succ P7$ |
|---|
| 2 | 0.1194<$k$< 0.41956 | $P1 \succ P6 \succ P4 \succ P8 \succ P2 \succ P5 \succ P3 \succ P7$ |
|---|
| 3 | 0.4196<$k$< 0.46303 | $P1 \succ P6 \succ P4 \succ P8 \succ P5 \succ P2 \succ P3 \succ P7$ |
|---|
| 4 | 0.463<$k$< 0.56842 | $P1 \succ P6 \succ P8 \succ P4 \succ P5 \succ P2 \succ P3 \succ P7$ |
|---|
| 5 | 0.5684<$k$< 0.59926 | $P1 \succ P8 \succ P6 \succ P4 \succ P5 \succ P2 \succ P3 \succ P7$ |
|---|
| 6 | 0.5993<$k$< 0.61806 | $P1 \succ P8 \succ P4 \succ P6 \succ P5 \succ P2 \succ P3 \succ P7$ |
|---|
| 7 | 0.6181<$k$< 0.80264 | $P1 \succ P8 \succ P4 \succ P6 \succ P5 \succ P2 \succ P7 \succ P3$ |
|---|
| 8 | 0.8026<$k$< 1 | $P1 \succ P8 \succ P4 \succ P6 \succ P5 \succ P7 \succ P2 \succ P3$ |
|---|
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