## 5 Experimental Study and Discussion
### Nonadditive Measure Calculation  
First, BWM is used to calculate the initial attribute weights. Attribute set \(A=\{a_{1}\), \(a_{2}\), \(a_{3}\), \(a_{4}\), \(a_{5}\}\). According to the decision maker, \(a_{5}\) is the best attribute (\(a_{b}\)) and \(a_{2}\) is the worst attribute (\(a_{w}\)). The importance ratios of \(a_{b}\) to the other attributes are listed in Table 4, and the importance ratios of other attributes to \(a_{w}\) are listed in Table 5. \begin{table}
\begin{tabular}{c c c c c} \hline \hline  & \(a_{1}\) & \(a_{2}\) & \(a_{3}\) & \(a_{4}\) \\ \hline \(a_{5}\) & 7 & 8 & 3 & 1 \\ \hline \hline \end{tabular}
\end{table} 
Table 4: Importance ratios of \(a_{b}\) to other attributes. \begin{table}
\begin{tabular}{c c c c c c} \hline \hline \multirow{2}{*}{**Case**} & **Aeroengine** & **Aeroengine** & **Aeroengine** & **Other Anomalies** & **Aeroengine Fault Part and** \\  & **Model (\(a_{1}\))** & **Category (\(a_{2}\))** & **Operation State (\(a_{3}\))** & **(\(a_{4}\))** & **Fault Mode (\(a_{5}\))** \\ \hline \(c_{1}\) & XP-6 & Military & Intermediate & Engine explosion & Compressor Disc-Fracture \\ \(c_{2}\) & XP-6 & Military & Intermediate & Engine explosion & Turbine Shaft-Crack \\ \(c_{3}\) & XP-6 & Military & Maximum & \(\backslash\) & Turbine Shaft-Fracture \\ \(c_{4}\) & XP-8 & Military & Engine idling & Abnormal engine sound & Turbine Blade \\  & & & & & & Groove-Fracture \\ \hline \hline \end{tabular}
\end{table} 
Table 2: Four cases in the case base. \begin{table}
\begin{tabular}{c c c c c} \hline \hline  & \(a_{1}\) & \(a_{2}\) & \(a_{3}\) & \(a_{4}\) \\ \hline \(a_{5}\) & 7 & 8 & 3 & 1 \\ \hline \hline \end{tabular}
\end{table} 
Table 5: Importance ratios of other attributes to \(a_{w}\). The optimal attribute weights can then be obtained by solving the following optimization problem:  
\[s.t.\left\{ \begin{array}{l}\min\delta\\ \left|\frac{\omega_{5}}{\omega_{1}}-7\right|\leq\delta,\ \left|\frac{\omega_{5}}{ \omega_{2}}-8\right|\leq\delta,\ \left|\frac{\omega_{5}}{\omega_{3}}-3\right|\leq \delta,\ \left|\frac{\omega_{5}}{\omega_{4}}-1\right|\leq\delta,\\ \frac{\omega_{1}}{\omega_{2}}-2\right|\leq\delta,\ \left|\frac{\omega_{3}}{ \omega_{2}}-4\right|\leq\delta,\ \left|\frac{\omega_{4}}{\omega_{2}}-7\right|\leq \delta,\\ \sum\limits_{j=1}^{5}\omega_{j}=1,\ \omega_{j}\geq 0\end{array} \right. \tag{14}\]  
By using MATLAB programming (Version No. R2019b, License No. 40865358, MathWorks, Beijing, China), we have \(\omega_{1}=0.0632\), \(\omega_{2}=0.0465\), \(\omega_{3}=0.1631\), \(\omega_{4}=0.3255\), \(\omega_{5}=0.4018\), and the minimum deviation \(\delta=0.6411\). The decision maker's attitude in Tables 4 and 5 reveal that the weight relation of the five attributes is \(\omega_{5}>\omega_{4}>\omega_{3}>\omega_{1}>\omega_{2}\). The calculation result is consistent with the decision maker's attitude. After obtaining the initial attribute weights, the interactions between attributes must be considered. According to Equation (9), we can get \(X_{12}=\min\ (2l_{1}/4,2l_{2}/4)=\min\ (0.0316\), \(0.0233)=0.0233\). Here, \(I_{n}\) refers to \(\omega_{n}\). In this experiment study, the attribute number \(n=5\). Similarly, all values for \(X_{jk}\) are calculated and the results are listed in Table 6.  
Second, the interaction interval is divided into subintervals to specify the intensity of the attribute interaction. In the experiment, \(t=5\). Therefore, the interval \([-X_{jk},X_{jk}]\) is divided into five sub-intervals:  
\[[-X_{jk},-\frac{3}{5}X_{jk}],\ [-\frac{3}{5}X_{jk},-\frac{1}{5}X_{jk}],\ [-\frac{1}{5}X_{jk},\ \frac{1}{5}X_{jk}],\ [ \frac{1}{5}X_{jk},\ \frac{3}{5}X_{jk}],\ [\frac{3}{5}X_{jk},\ X_{jk}]\]  
Here, we use \(K_{1}\) to \(K_{5}\) to denote the five subintervals. The decision maker then determines the interaction between any two attributes as a negative/independent/positive relationship. According to the decision maker, the interaction between \(a_{1}\) and \(a_{2}\) is quite negative, thus, the interaction interval between \(a_{1}\) and \(a_{2}\) belongs to \(K_{1}\).