## 3 Levels of Analysis
### Discussion on the AHP Result  
The above method yields the following:  
\[C= \begin{pmatrix}0.155&0.262&0.077&0.398&0.108\end{pmatrix}  \tag{6}\]  
Figure 7: Orthogonal test AHP model. \[S= \begin{pmatrix}0.039&0&\cdots&0\\ 0&0.039&\cdots&0\\ \cdots&\cdots&\cdots&\cdots\\ 0&0&\cdots&0.039\end{pmatrix} _{5\times 5} \tag{7}\] \[A= \begin{pmatrix}5.236&0&0&0&0\\ 6.579&0&0&0&0\\ 6.473&0&0&0&0\\ 7.122&0&0&0&0\\ 0&8.084&0&0&0\\ 0&6.742&0&0&0\\ 0&5.685&0&0&0\\ 0&4.899&0&0&0\\ 0&0&6.533&0&0\\ 0&0&6.846&0&0\\ 0&00&6.104&0&0\\ 0&0&5.906&0&0\\ 0&0&0&3.780&0\\ 0&0&0&5.756&0\\ 0&0&0&7.252&0\\ 0&0&0&8.621&0\\ 0&0&0&0&5.546\\ 0&0&0&0&6.524\\ 0&0&0&0&6.482\\ 0&0&0&6.856\end{pmatrix}  \tag{8}\]  
Therefore, the degree of influence of the various factors on the peak intensity \(\sigma_{1}\) was as follows:  
\[\sigma_{1}=ASC^{T}= \begin{pmatrix}0.039&0&\cdots&0\\ 0&0.039&\cdots&0\\ \cdots&\cdots&\cdots&\cdots\\ 0&0&\cdots&0.039\end{pmatrix} _{5\times 5} \begin{pmatrix}0.0317\\ 0.0389\\ 0.0431\\ 0.0826\\ 0.0689\\ 0.0581\\ 0.0501\\ 0.0197\\ 0.0206\\ 0.0183\\ 0.0177\\ 0.0587\\ 0.0893\\ 0.1126\\ 0.1338\\ 0.0234\\ 0.0275\\ 0.0289\\ 0.0289\\ 0.0273\end{pmatrix} = \begin{pmatrix}0.039&0&\cdots&0\\ 0.0391&0.0389\\ 0.0431&0.0826\\ 0.0689\\ 0.0581\\ 0.0581\\ 0.0501\\ 0.0197\\ 0.0206\\ 0.0183\\ 0.0177\\ 0.0587\\ 0.0893\\ 0.1126\\ 0.1338\\ 0.0234\\ 0.0275\\ 0.0289\\ 0.02873\end{pmatrix} = \begin{pmatrix}0.0501\\ 0.0197\\ 0.0206\\ 0.0177\\ 0.0587\\ 0.0893\\ 0.1126\\ 0.1338\\ 0.0234\\ 0.0275\\ 0.0289\\ 0.0273\end{pmatrix}  \tag{9}\]\[\omega_{1}=ASC^{T}= \begin{pmatrix}3.728&0&0&0&0\\ 5.406&0&0&0&0\\ 4.351&0&0&0&0\\ 4.873&0&0&0&0\\ 0&5.823&0&0&0\\ 0&4.790&0&0&0\\ 0&4.044&0&0&0\\ 0&3.700&0&0&0\\ 0&0&4.737&0&0\\ 0&0&4.954&0&0\\ 0&00&4.473&0&0\\ 0&0&4.194&0&0\\ 0&0&0&0&2.500\\ 0&0&0&0&3.827\\ 0&0&0&0&5.171\\ 0&0&0&0&6.860\\ \end{pmatrix}\begin{pmatrix}0.054&0&\cdots&0\\ 0.054&\cdots&0\\ \cdots&\cdots&\cdots&\cdots\\ 0&0&\cdots&0.054\\ 0&0&\cdots&0.054\\ 0&0&\cdots&0.054\\ 0&0&4.279&0\\ 0&0&5.523&0\\ 0&0&4.194&0\\ 0&0&0&2.500\\ 0&0&0&3.827\\ 0&0&0&5.171\\ 0&0&0&6.860\\ \end{pmatrix} \begin{pmatrix}0.054&0&\cdots&0\\ 0.054&\cdots&0\\ \cdots&\cdots&\cdots&\cdots\\ 0&0&\cdots&0.054\\ 0&0&\cdots&0.054\\ 0&0&4.279&0\\ 0&0&5.523&0\\ 0&0&4.194&0\\ 0&0&0&2.500\\ 0&0&0&3.827\\ 0&0&0&0&5.1  
Based on the above deviation definition method, the calculation process of the significance level of the five-factor four-level orthogonal test reported in this paper was as follows:  
The sum of the squared deviations \(S_{T}\) was:  
\[S_{T}=\sum\limits_{i=1}^{16}x_{i}^{2}-\frac{1}{16}\left(\sum\limits_{i=1}^{16}x _{i}\right)^{2} \tag{13}\]  
Let \(\eta=\frac{1}{16}\left(\sum\limits_{i=1}^{16}x_{i}\right)^{2}\). From the data listed in Table 3, we obtained \(\eta\)\(=645.63\); then, the sum of the squared deviations of each factor \(S_{A},S_{B},\ldots,S_{E}\) was expressed as:  
\[ \begin{array}{l}S_{A}=\frac{1}{4}\sum\limits_{i=1}^{4}T_{Ai}^{2}-\frac{1}{16 }\left(\sum\limits_{i=1}^{16}x_{i}\right)^{2}\\ S_{B}=\frac{1}{4}\sum\limits_{i=1}^{4}T_{Bi}^{2}-\frac{1}{16}\left(\sum\limits _{i=1}^{16}x_{i}\right)^{2}\\ \ldots\ldots\ldots\\ S_{E}=\frac{1}{4}\sum\limits_{i=1}^{4}T_{EI}^{2}-\frac{1}{16}\left(\sum \limits_{i=1}^{16}x_{i}\right)^{2}\end{array}  \tag{14}\]  
In the formula: \(T_{Ai},T_{Bi},\cdot\cdot\cdot,T_{EI}\) were the sum of the test values of each factor at a certain level, e.g., \(T_{Ai}\) was the sum of the four groups of test values at the level of factor 1. The sum of squares of the error effect deviation \(S_{e}\) was:  
\[S_{e}=S_{T}-(S_{A}+S_{B}+\cdots+S_{E}) \tag{15}\]  
The sum of the squared deviations of the error effect reflected the fluctuation value of the test error in this set of data. From Equations (13)-(15), the sum of the squared deviations corresponding to each factor was obtained. For example, the calculation process of \(S_{A}\) was:  
\[T_{A1}=(3.789+5.694+5.35+6.11)=20.943 \tag{16}\]  
\[S_{A}=(20.9432+26.3152+25.892+28.4892)/4-645.63=7.621\]  
Table 7 presents the rest of the calculation results.