VIKOR——AISM联合求解

权重的计算及结果


归一化方法


D-ANP方法的占比


极限超矩阵如下


$$limit W=\begin{array}{c|c|c|c|c|c|c}{M_{16 \times16}} &故事性 &传承性 &寓意性 &情感性 &沟通性 &思考体验 &行为体验 &关联体验 &教育功能 &实用功能 &品质功能 &原创性 &独一性 &美感性 &时尚性 &多样性\\ \hline 故事性 &0.0753 &0.0753 &0.0753 &0.0753 &0.0753 &0.0753 &0.0753 &0.0753 &0.0753 &0.0753 &0.0753 &0.0753 &0.0753 &0.0753 &0.0753 &0.0753\\ \hline 传承性 &0.0644 &0.0644 &0.0644 &0.0644 &0.0644 &0.0644 &0.0644 &0.0644 &0.0644 &0.0644 &0.0644 &0.0644 &0.0644 &0.0644 &0.0644 &0.0644\\ \hline 寓意性 &0.0584 &0.0584 &0.0584 &0.0584 &0.0584 &0.0584 &0.0584 &0.0584 &0.0584 &0.0584 &0.0584 &0.0584 &0.0584 &0.0584 &0.0584 &0.0584\\ \hline 情感性 &0.0616 &0.0616 &0.0616 &0.0616 &0.0616 &0.0616 &0.0616 &0.0616 &0.0616 &0.0616 &0.0616 &0.0616 &0.0616 &0.0616 &0.0616 &0.0616\\ \hline 沟通性 &0.073 &0.073 &0.073 &0.073 &0.073 &0.073 &0.073 &0.073 &0.073 &0.073 &0.073 &0.073 &0.073 &0.073 &0.073 &0.073\\ \hline 思考体验 &0.0556 &0.0556 &0.0556 &0.0556 &0.0556 &0.0556 &0.0556 &0.0556 &0.0556 &0.0556 &0.0556 &0.0556 &0.0556 &0.0556 &0.0556 &0.0556\\ \hline 行为体验 &0.0705 &0.0705 &0.0705 &0.0705 &0.0705 &0.0705 &0.0705 &0.0705 &0.0705 &0.0705 &0.0705 &0.0705 &0.0705 &0.0705 &0.0705 &0.0705\\ \hline 关联体验 &0.0597 &0.0597 &0.0597 &0.0597 &0.0597 &0.0597 &0.0597 &0.0597 &0.0597 &0.0597 &0.0597 &0.0597 &0.0597 &0.0597 &0.0597 &0.0597\\ \hline 教育功能 &0.0678 &0.0678 &0.0678 &0.0678 &0.0678 &0.0678 &0.0678 &0.0678 &0.0678 &0.0678 &0.0678 &0.0678 &0.0678 &0.0678 &0.0678 &0.0678\\ \hline 实用功能 &0.0322 &0.0322 &0.0322 &0.0322 &0.0322 &0.0322 &0.0322 &0.0322 &0.0322 &0.0322 &0.0322 &0.0322 &0.0322 &0.0322 &0.0322 &0.0322\\ \hline 品质功能 &0.0577 &0.0577 &0.0577 &0.0577 &0.0577 &0.0577 &0.0577 &0.0577 &0.0577 &0.0577 &0.0577 &0.0577 &0.0577 &0.0577 &0.0577 &0.0577\\ \hline 原创性 &0.065 &0.065 &0.065 &0.065 &0.065 &0.065 &0.065 &0.065 &0.065 &0.065 &0.065 &0.065 &0.065 &0.065 &0.065 &0.065\\ \hline 独一性 &0.0652 &0.0652 &0.0652 &0.0652 &0.0652 &0.0652 &0.0652 &0.0652 &0.0652 &0.0652 &0.0652 &0.0652 &0.0652 &0.0652 &0.0652 &0.0652\\ \hline 美感性 &0.0609 &0.0609 &0.0609 &0.0609 &0.0609 &0.0609 &0.0609 &0.0609 &0.0609 &0.0609 &0.0609 &0.0609 &0.0609 &0.0609 &0.0609 &0.0609\\ \hline 时尚性 &0.0714 &0.0714 &0.0714 &0.0714 &0.0714 &0.0714 &0.0714 &0.0714 &0.0714 &0.0714 &0.0714 &0.0714 &0.0714 &0.0714 &0.0714 &0.0714\\ \hline 多样性 &0.0613 &0.0613 &0.0613 &0.0613 &0.0613 &0.0613 &0.0613 &0.0613 &0.0613 &0.0613 &0.0613 &0.0613 &0.0613 &0.0613 &0.0613 &0.0613\\ \hline \end{array} $$

指标要素权重如下


$$\omega=\begin{array}{c|c|c|c|c|c|c}{M_{16 \times1}} &D-ANP权重\\ \hline 故事性 &0.07532\\ \hline 传承性 &0.06438\\ \hline 寓意性 &0.05836\\ \hline 情感性 &0.06161\\ \hline 沟通性 &0.07305\\ \hline 思考体验 &0.05565\\ \hline 行为体验 &0.07052\\ \hline 关联体验 &0.05966\\ \hline 教育功能 &0.06778\\ \hline 实用功能 &0.03217\\ \hline 品质功能 &0.05769\\ \hline 原创性 &0.06501\\ \hline 独一性 &0.06516\\ \hline 美感性 &0.06091\\ \hline 时尚性 &0.0714\\ \hline 多样性 &0.06134\\ \hline \end{array} $$$$\omega=\begin{array}{c|c|c|c|c|c|c}{M_{1 \times16}} &故事性 &传承性 &寓意性 &情感性 &沟通性 &思考体验 &行为体验 &关联体验 &教育功能 &实用功能 &品质功能 &原创性 &独一性 &美感性 &时尚性 &多样性\\ \hline D-ANP权重 &0.07532 &0.06438 &0.05836 &0.06161 &0.07305 &0.05565 &0.07052 &0.05966 &0.06778 &0.03217 &0.05769 &0.06501 &0.06516 &0.06091 &0.0714 &0.06134\\ \hline \end{array} $$

原始矩阵


$$ \begin{array}{c|c|c|c|c|c|c}{M_{8 \times16}} &C1 &C2 &C3 &C4 &C5 &E1 &E2 &E3 &F1 &F2 &F3 &U1 &U2 &D1 &D2 &D3\\ \hline P1 &0.779 &0.782 &0.701 &0.724 &0.712 &0.756 &0.639 &0.713 &0.838 &0.758 &0.699 &0.71 &0.702 &0.705 &0.754 &0.712\\ \hline P2 &0.791 &0.751 &0.709 &0.768 &0.732 &0.762 &0.653 &0.702 &0.73 &0.716 &0.595 &0.694 &0.641 &0.701 &0.667 &0.702\\ \hline P3 &0.736 &0.775 &0.812 &0.728 &0.762 &0.633 &0.475 &0.582 &0.72 &0.701 &0.645 &0.714 &0.706 &0.693 &0.634 &0.674\\ \hline P4 &0.647 &0.737 &0.722 &0.715 &0.792 &0.78 &0.564 &0.757 &0.768 &0.72 &0.699 &0.75 &0.706 &0.779 &0.712 &0.706\\ \hline P5 &0.733 &0.806 &0.748 &0.733 &0.696 &0.682 &0.509 &0.689 &0.709 &0.715 &0.554 &0.696 &0.663 &0.671 &0.709 &0.663\\ \hline P6 &0.734 &0.78 &0.747 &0.714 &0.768 &0.854 &0.758 &0.712 &0.719 &0.706 &0.732 &0.722 &0.729 &0.765 &0.709 &0.765\\ \hline P7 &0.656 &0.73 &0.74 &0.628 &0.73 &0.692 &0.545 &0.59 &0.595 &0.758 &0.631 &0.706 &0.692 &0.629 &0.628 &0.605\\ \hline P8 &0.714 &0.733 &0.721 &0.746 &0.65 &0.824 &0.703 &0.632 &0.697 &0.714 &0.761 &0.753 &0.592 &0.842 &0.743 &0.755\\ \hline \end{array} $$

采用的是极差法进行归一化


  正向指标的公式为:$ n_{ij} = \frac{{o_{ij}-min(o_{j})}}{{max(o_{j})-min(o_{j})}} $

  负向指标的公式为:$ n_{ij} = \frac{{o_{ij}-min(o_{j})}}{{max(o_{j})-min(o_{j})}} $

归一化矩阵如下


$$ \begin{array}{c|c|c|c|c|c|c}{M_{8 \times16}} &C1 &C2 &C3 &C4 &C5 &E1 &E2 &E3 &F1 &F2 &F3 &U1 &U2 &D1 &D2 &D3\\ \hline P1 &0.917 &0.684 &0 &0.686 &0.437 &0.557 &0.58 &0.749 &1 &1 &0.7 &0.271 &0.803 &0.357 &1 &0.669\\ \hline P2 &1 &0.276 &0.072 &1 &0.577 &0.584 &0.629 &0.686 &0.556 &0.263 &0.198 &0 &0.358 &0.338 &0.31 &0.606\\ \hline P3 &0.618 &0.592 &1 &0.714 &0.789 &0 &0 &0 &0.514 &0 &0.44 &0.339 &0.832 &0.3 &0.048 &0.431\\ \hline P4 &0 &0.092 &0.189 &0.621 &1 &0.665 &0.314 &1 &0.712 &0.333 &0.7 &0.949 &0.832 &0.704 &0.667 &0.631\\ \hline P5 &0.597 &1 &0.423 &0.75 &0.324 &0.222 &0.12 &0.611 &0.469 &0.246 &0 &0.034 &0.518 &0.197 &0.643 &0.363\\ \hline P6 &0.604 &0.658 &0.414 &0.614 &0.831 &1 &1 &0.743 &0.51 &0.088 &0.86 &0.475 &1 &0.638 &0.643 &1\\ \hline P7 &0.063 &0 &0.351 &0 &0.563 &0.267 &0.247 &0.046 &0 &1 &0.372 &0.203 &0.73 &0 &0 &0\\ \hline P8 &0.465 &0.039 &0.18 &0.843 &0 &0.864 &0.806 &0.286 &0.42 &0.228 &1 &1 &0 &1 &0.913 &0.938\\ \hline \end{array} $$

ViKOR中带权值的公式如下


  期望值计算公式为:$ S_i = \sum_\limits{j=1}^m{ \omega_{j} \left(\frac{Zone_j^+ -n_{ij}}{Zone_j^+ -Zone_j^-} \right)} \quad \quad $

  遗憾值计算公式为:$ R_i = \max_\limits{j=1} { \left( \omega_{j} (\frac{n_{ij}-0}{Zone_j^+ -Zone_j^-} )\right)} \quad \quad $

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
$$ \begin{array}{c|c|c|c|c|c|c}{M_{1 \times16}} &C1 &C2 &C3 &C4 &C5 &E1 &E2 &E3 &F1 &F2 &F3 &U1 &U2 &D1 &D2 &D3\\ \hline \mathbf{Zone^-} &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0\\ \hline \end{array} $$
EWM权重如下
$$ \begin{array}{c|c|c|c|c|c|c}{M_{2 \times16}} &C1 &C2 &C3 &C4 &C5 &E1 &E2 &E3 &F1 &F2 &F3 &U1 &U2 &D1 &D2 &D3\\ \hline EWM所得权重 &0.0617 &0.09255 &0.08727 &0.03327 &0.04357 &0.05415 &0.06774 &0.06789 &0.03998 &0.09462 &0.05303 &0.09413 &0.03966 &0.05994 &0.06789 &0.04262\\ \hline 权重大小顺序 &8 &3 &4 &16 &12 &10 &7 &6 &14 &1 &11 &2 &15 &9 &5 &13\\ \hline \end{array} $$
D-ANP权重如下
$$\omega=\begin{array}{c|c|c|c|c|c|c}{M_{1 \times16}} &故事性 &传承性 &寓意性 &情感性 &沟通性 &思考体验 &行为体验 &关联体验 &教育功能 &实用功能 &品质功能 &原创性 &独一性 &美感性 &时尚性 &多样性\\ \hline 权重 &0.07532 &0.06438 &0.05836 &0.06161 &0.07305 &0.05565 &0.07052 &0.05966 &0.06778 &0.03217 &0.05769 &0.06501 &0.06516 &0.06091 &0.0714 &0.06134\\ \hline \end{array} $$
两者调和权重如下
$$\omega=\begin{array}{c|c|c|c|c|c|c}{M_{1 \times16}} &故事性 &传承性 &寓意性 &情感性 &沟通性 &思考体验 &行为体验 &关联体验 &教育功能 &实用功能 &品质功能 &原创性 &独一性 &美感性 &时尚性 &多样性\\ \hline 权重 &0.06442 &0.08692 &0.08148 &0.03894 &0.04946 &0.05445 &0.0683 &0.06624 &0.04554 &0.08213 &0.05396 &0.08831 &0.04476 &0.06014 &0.06859 &0.04636\\ \hline \end{array} $$

代入权重值等即得(S R)两列矩阵,两列都为负向指标


$$ \begin{array}{c|c|c|c|c|c|c}{M_{8 \times2}} &期望值 &遗憾值\\ \hline P1 &0.367591408 &0.081484873\\ \hline P2 &0.575238293 &0.08830984\\ \hline P3 &0.605084348 &0.082126905\\ \hline P4 &0.443357694 &0.078992808\\ \hline P5 &0.594154295 &0.085401687\\ \hline P6 &0.341124409 &0.074997788\\ \hline P7 &0.752592447 &0.086919565\\ \hline P8 &0.452125823 &0.083572102\\ \hline \end{array} $$

妥协解分析


妥协解公式


公式
$$ Q_i = \left( 1-k \right) \left(\frac{S_i - Min(S_i)}{Max(S_i) -Min(S_i)} \right) + k\left(\frac{R_i - Min(R_i)}{Max(R_i) -Min(R_i)} \right) $$

敏感性分析,排序聚类分析


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

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。

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

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

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

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

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

基础矩阵如下:


$$Base=\begin{array}{c|c|c|c|c|c|c}{M_{8 \times2}} &a_i &b_i\\ \hline P1 &0.0643 &0.4873\\ \hline P2 &0.569 &1\\ \hline P3 &0.6415 &0.5355\\ \hline P4 &0.2485 &0.3001\\ \hline P5 &0.6149 &0.7815\\ \hline P6 &0 &0\\ \hline P7 &1 &0.8956\\ \hline P8 &0.2698 &0.6441\\ \hline \end{array} $$

拐点k值分析


$$Qk_{matrix}=\begin{array}{c|c|c|c|c|c|c}{M_{8 \times8}} &k=0 &k=0.097 &k=0.135 &k=0.174 &k=0.496 &k=0.774 &k=0.805 &k=1\\ \hline P1 &0.064 &0.106 &0.121 &0.138 &0.274 &0.392 &0.405 &0.487\\ \hline P2 &0.569 &0.611 &0.627 &0.644 &0.783 &0.903 &0.916 &1\\ \hline P3 &0.642 &0.631 &0.627 &0.623 &0.589 &0.559 &0.556 &0.536\\ \hline P4 &0.248 &0.253 &0.255 &0.257 &0.274 &0.288 &0.29 &0.3\\ \hline P5 &0.615 &0.631 &0.637 &0.644 &0.698 &0.744 &0.749 &0.782\\ \hline P6 &0 &0 &0 &0 &0 &0 &0 &0\\ \hline P7 &1 &0.99 &0.986 &0.982 &0.948 &0.919 &0.916 &0.896\\ \hline P8 &0.27 &0.306 &0.32 &0.335 &0.455 &0.559 &0.571 &0.644\\ \hline \end{array} $$

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:

$$Q_{rank}=\begin{array}{c|c|c|c|c|c|c}{M_{8 \times8}} &k=0 &k=0.097 &k=0.135 &k=0.174 &k=0.496 &k=0.774 &k=0.805 &k=1\\ \hline P1 &2 &2 &2 &2 &2 &3 &3 &3\\ \hline P2 &5 &5 &5 &6 &7 &7 &7 &8\\ \hline P3 &7 &6 &5 &5 &5 &4 &4 &4\\ \hline P4 &3 &3 &3 &3 &2 &2 &2 &2\\ \hline P5 &6 &6 &7 &6 &6 &6 &6 &6\\ \hline P6 &1 &1 &1 &1 &1 &1 &1 &1\\ \hline P7 &8 &8 &8 &8 &8 &8 &7 &7\\ \hline P8 &4 &4 &4 &4 &4 &4 &5 &5\\ \hline \end{array} $$


聚类特征


序号 聚类特征-对应k值区段 Q值排序
10<$k$< 0.09746$P6 \succ P1 \succ P4 \succ P8 \succ P2 \succ P5 \succ P3 \succ P7$
20.0975<$k$< 0.13508$P6 \succ P1 \succ P4 \succ P8 \succ P2 \succ P3 \succ P5 \succ P7$
30.1351<$k$< 0.17385$P6 \succ P1 \succ P4 \succ P8 \succ P3 \succ P2 \succ P5 \succ P7$
40.1739<$k$< 0.49587$P6 \succ P1 \succ P4 \succ P8 \succ P3 \succ P5 \succ P2 \succ P7$
50.4959<$k$< 0.77397$P6 \succ P4 \succ P1 \succ P8 \succ P3 \succ P5 \succ P2 \succ P7$
60.774<$k$< 0.80496$P6 \succ P4 \succ P1 \succ P3 \succ P8 \succ P5 \succ P2 \succ P7$
70.805<$k$< 1$P6 \succ P4 \succ P1 \succ P3 \succ P8 \succ P5 \succ P7 \succ P2$

区段截取方式Q排序——k值区段截取分析


不多解释了,先点击下面的按钮进去看看就知道了

区段截取方式Q排序——夹逼方式


以0->min max<- 1方式夹逼显示。