## 5 A Case Study
### Experiment 2  
Assume that (1) the prior probability of each failure is 0.1, and (2) the tests \(T_{4}\), \(T_{8}\), and \(T_{13}\) are passed (denoted as 'Pass', 0), all the other tests fail (denoted as 'Fail', 1). Perform the proposed method on this case. 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: \(T_{4}\) = 0, \(T_{8}\) = 0, \(T_{13}\) = 0; \(T_{1}\), \(T_{2}\), \(T_{3}\), \(T_{5}\), \(T_{6}\), \(T_{7}\), \(T_{9}\), \(T_{10}\), \(T_{11}\), \(T_{12}\), \(T_{14}\), and \(T_{15}\) are all equal to 1. Thus, it can be concluded that the test results of \(T_{4}\), \(T_{8}\) and \(T_{13}\) are qualified and no failure will happen in the correlated fault types \(F_{1}\), \(F_{5}\), \(F_{6}\), \(F_{7}\), \(F_{9}\), and \(T_{10}\). Thus, the simplified matrix can be formulated and shown in Table 10.  
Delete the tests \(T_{4}\), \(T_{7}\), \(T_{8}\), \(T_{10}\), \(T_{12}\), and \(T_{13}\); because of that, (1) \(T_{5}\) is redundant with \(T_{12}\), (2) \(T_{6}\) is redundant with \(T_{7}\), and (3) \(T_{9}\) is redundant with \(T_{10}\). Thus, the simplified matrix of the system can be formulated, as shown in Table 11.  
\begin{table}
\begin{tabular}{l c c c c c c c c c c c c c} \hline \hline
**Fault Type** & \(T_{1}\) & \(T_{2}\) & \(T_{3}\) & \(T_{4}\) & \(T_{5}\) & \(T_{6}\) & \(T_{7}\) & \(T_{8}\) & \(T_{9}\) & \(T_{10}\) & \(T_{11}\) & \(T_{12}\) & \(T_{13}\) & \(T_{14}\) & \(T_{15}\) \\ \hline \(F_{2}\) & 0 & 0 & 1 & 0 & 1 & 0 & 0 Step 3: Formulate the multiple fault dependency matrix. Based on the simplified dependency matrix in Table 11, 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 12.  
Step 4: Fault diagnosis of the system based on the multiple fault dependency matrix. For the multiple fault dependency matrix shown in Table 12, delete the row vector of the fault type with '0', including \(F_{2}\), \(F_{3}\), \(F_{4}\), \(F_{8}\), \(F_{(2,3)}\), \(F_{(2,4)}\), \(F_{(2,8)}\), \(F_{(3,4)}\), and \(F_{(2,3,4)}\). After that, the fault ambiguity group can be formulated, as shown in Table 13.  
Calculating the prior probability of the multiple fault in the fault ambiguity group, the result is as follows:  
\[p\left(f_{(3,8)}\right)=p\left(f_{3}\right)\times p\left(f_{8}\right)=0.1 \times 0.1=0.01,\]  
\[p\left(f_{(2,3,8)}\right)=p\left(f_{2}\right)\times p\left(f_{3}\right)\times p \left(f_{8}\right)=0.1\times 0.1\times 0.1=0.001,\]  
\[p\left(f_{(2,4,8)}\right)=p\left(f_{2}\right)\times p\left(f_{4}\right)\times p \left(f_{8}\right)=0.1\times 0.1\times 0.1=0.001,\]  
\[p\left(f_{(3,4,8)}\right)=p\left(f_{3}\right)\times p\left(f_{4}\right)\times p \left(f_{8}\right)=0.1\times 0.1\times 0.1=0.001,\]  
\[p\left(f_{(2,3,4,8)}\right)=p\left(f_{2}\right)\times p\left(f_{3}\right)\times p \left(f_{4}\right)\times p\left(f_{8}\right)=0.1\times 0.1\times 0.1\times 0.1=0.0001.\]  
According to the calculation results, there are 5 possible reasons for the system failure, which are \(F_{(3,8)}\), \(F_{(2,3,8)}\), \(F_{(2,4,8)}\), \(F_{(3,4,8)}\) or \(F_{(2,3,4,8)}\). In addition, the ranking of the possible failures based on the calculation results of the prior probability is: \(F_{(3,8)}\)\(\succ\)\(F_{(2,3,8)}\)\(=\)\(F_{(2,4,8)}\)\(=\)\(F_{(3,4,8)}\)\(\succ\)\(F_{(2,3,4,8)}\), where \"\(\succ\)\" denotes a higher priority.