## 5 A Case Study
### Experiment 1
#### 5.1.1 Experiment Process  
Step 1: Build the fault-test dependency matrix. The dependency matrix is adopted from the literature [28] and presented in Table 5.  
Step 2: Simplify the dependency matrix according to the test results of the system. The test results of the system are: \(T_{1}\) = 0, \(T_{3}\) = 0, \(T_{11}\) = 0; \(T_{2}\), \(T_{4}\), \(T_{5}\), \(T_{6}\), \(T_{7}\), \(T_{8}\), \(T_{9}\), \(T_{10}\), \(T_{12}\), \(T_{13}\), \(T_{14}\), and \(T_{15}\) are all equal to 1. Thus, it can be concluded that the test results of \(T_{1}\), \(T_{3}\), and \(T_{11}\) are qualified and no failure will happen in the correlated fault types \(F_{1}\), \(F_{2}\), \(F_{3}\), \(F_{7}\), \(F_{8}\), \(F_{9}\), and \(T_{10}\). Thus, the simplified matrix can be formulated and shown in Table 6.  
Figure 3: Fault-diagnosis tree of a certain control system.  
\begin{table}
\begin{tabular}{l c c c c c c c c c c c c c} \hline \hline
**Fault Type** & \(\mathbf{T_{1}}\) & \(\mathbf{T_{2}}\) & \(\mathbf{T_{3}}\) & \(\mathbf{T_{4}}\) & \(\mathbf{T_{5}}\) & \(\mathbf{T_{6}}\) & \(\mathbf{T_{7}}\) & \(\mathbf{T_{8}}\) & \(\mathbf{T_{9}}\) & \(\mathbf{T_{10  
Delete the tests \(T_{6}\), \(T_{7}\), \(T_{8}\), \(T_{9}\), \(T_{10}\), \(T_{12}\), and \(T_{13}\); because of that, (1) \(T_{6}\) and \(T_{7}\) are redundant with \(T_{2}\), and (2) \(T_{8}\), \(T_{9}\), \(T_{10}\), and \(T_{13}\) are redundant with \(T_{4}\). Thus, the simplified matrix of the system can be formulated, as shown in Table 7.  
Step 3: Formulate the multiple fault dependency matrix. Based on the simplified dependency matrix in Table 7, perform the logical OR operation on the eigenvectors (row vectors) of the fault to obtain the eigenvectors of the multiple fault group, as shown in Table 8.  
Step 4: Fault diagnosis of the system based on the multiple fault dependency matrix. For the multiple fault dependency matrix shown in Table 8, delete the row vector of the fault type with '0', including \(F_{4}\), \(F_{5}\), \(F_{6}\), and \(F_{(4,5)}\). After that, the fault ambiguity group can be formulated, as shown in Table 9.  
Calculating the prior probability of the multiple fault in the fault ambiguity group, the result is as follows:  
\[p\left(f_{(4,5)}\right)=p\left(f_{4}\right)\times p\left(f_{5}\right)=0.1 \times 0.1=0.01,\]  
\[p\left(f_{(4,6)}\right)=p\left(f_{4}\right)\times p\left(f_{6}\right)=0.1 \times 0.1=0.01,\]  
\[p\left(f_{(4,5,6)}\right)=p\left(f_{4}\right)\times p\left(f_{5}\right)\times p \left(f_{6}\right)=0.1\times 0.1\times 0.1=0.001.\]  
\begin{table}
\begin{tabular}{l c c c c c c c c c c c} \hline \hline
**Fault Type** & \(\mathbf{T_{2}}\) & \(\mathbf{T_{4}}\) & \(\mathbf{T_{5}}\) & \(\mathbf{T_{6}}\) & \(\mathbf{T_{7}}\) & \(\mathbf{T_{8}}\) & \(\mathbf{T_{9}}\) & \(\mathbf{T_{10}}\) & \(\mathbf{T_{12}}\) & \(\mathbf{T_{13}}\) & \(\mathbf{T_{14}}\) & \(\mathbf{T_{15}}\) \\ \hline \(F_{4}\) & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ \(F_{5}\) & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 0 \\ \(F_{6}\) & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline \hline \end{tabular}
\end{table}
Table 6: Simplified dependency matrix of the case study in Experiment 1.  
\begin{table}
\begin{tabular}{l c c c c c} \hline \hline
**Fault Type** & \(\mathbf{T_{2}}\) & \(\mathbf{T_{4}}\) & \(\mathbf{T_{5}}\) & \(\mathbf{T_{14}}\) & \(\mathbf{T_{15}}\) & **Prior Probability** \\ \hline \(F_{4}\) & 1 & 0 & 0 & 1 & 1 & 0.1 \\ \(F_{5}\) & 1 & 1 & 0 & 1 & 0 & 0.1 \\ \(F_{6}\) & 0 & 1 & 1 & 1 & 1 & 0.1 \\ \(F_{(4,5)}\) & 1 & 1 & 0 & 1 & 1 & 0.01 \\ \(F_{(5,6)}\) & 1 & 1 & 1 & 1 & 1 & 0.001 \\ \(F_According to the calculation results, there are three possible reasons for the system failure, which are \(F_{(4,6)}\), \(F_{(5,6)}\) or \(F_{(4,5,6)}\). In addition, the ranking of the possible failures based on the calculation results of the prior probability is: \(F_{(4,6)}=F_{(5,6)}\succ F_{(4,5,6)}\), where \"\(\succ\)\" denotes a higher priority.