# Failure Mode and Effects Analysis Considering Consensus and Preferences Interdependence
## 5 Case Study
### Implement the Proposed Method  
An FMEA team consisting of five cross-functional experts \(E_{k}\) (\(k\) = 1,2,3,4,5) from different departments is formed to identify the most important failure modes; we suppose that the relative weights of the five team experts are assigned as 0.15, 0.2, 0.25, 0.1, and 0.3 according to their different professional knowledge and expertise. The FMEA team identifies nine potential failure modes through brainstorming, namely, gear abrasion of dynamic head, action invalidation of force motor, non-normal friction of pedrail, leak of hydraulic system, abrasion of feed mechanism, unexpected halt of engine, cavitation erosion of hydraulic pump, failures of hydraulic system induced by hydraulic oil pollution, and nozzle choking of aiguilles. **Step1:** Construct linguistic assessment matrix. In a practical application, the five experts use the linguistic variables to evaluate the nine failure modes on each risk factor and the relative importance of risk factors; the evaluation results are shown in Table 3.  
**Step 2:** Construct a consensus-reaching process. **Step 2.1:** Generally, the threshold value and the maximum number of cycles are determined by the team experts according to the actual situation. In this case study, let \(\gamma=1\), \(\zeta=0.86\), and \(round=3\). **Step 2.2:** Convert all linguistic variables in each linguistic evaluation matrix \(L^{1}\), \(L^{2}\), \(L^{3}\), \(L^{4}\), and \(L^{5}\) into the IVPFNs according to Table 2 to construct assessment matrices \(A^{1}\), \(A^{2}\), \(A^{3}\), \(A^{4}\), and \(A^{5}\). **Step 2.3:** The collective matrix \(C\) is aggregated by the IVPFWA operator to compute the consensus degree. \[C=\left( \begin{array}{ll}\left([0.4896,0.5092],[0.4129,0.5144]\right)&\left( [0.5117,0.6119],[0.3887,0.4890]\right)&\left([0.4010,0.5008],[0.5015,0.6021] \right)\\ \left([0.6528,0.7545],[0.2470,0.3492]\right)&\left([0.6171,0.7186],[0.2830,0. 3849]\right)&\left([0.5307,0.6315],[0.3716,0.4733]\right)\\ \left([0.5877,0.6902],[0.3130,0.4162]\right)&\left([0.4254,0.5252],[0.4786,0. 5797]\right)&\left([0.6645,0.7656],[0.2352,0.3366]\right)\\ \left([0.5025,0.6031],[0.3933,0.5005]\right)&\left([0.4355,0.5352],[0.4704,0. 5720]\right)&\left([0.2651,0.3626],[0.6444,0.7453]\right)\\ \left([0.5858,0.6879],[0.3149,0.4175]\right)&\left([0.4825,0.5826],[0.4183,0. 5186]\right)&\left([0.2788,0.3767],[0.6297,0.7305]\right)\\ \left([0.6692,0.7703],[0.2305,0.3318]\right)&\left([0.5701,0.6720],[0.3307,0. 4331]\right)&\left([0.3422,0.4402],[0.5694,0.6711]\right)\\ \left([0.6287,0.7297],[0.2711,0.3722]\right)&\left([0.7246,0.8310],[0.1712,0. 2781]\right)&\left([0.5442,0.6446],[0.3656,0.4573]\right)\\ \left([0.8310,1.0000],[0.0000,0.1712]\right)&\left([0.4028,0.5025],[0.5005,0. 6012]\right)&\left([0.3393,0.4389],[0.5629,0.6632]\right)\\ \left([0.4542,0.5542],[0.4472,0.5477]\right)&\left([0.4780,0.5780],[0.4229,0. 5233]\right)&\left([0.4639,0.5640],[0.4373,0.5378]\right)\\ \end{array} \right)\]  
**Step 2.4:** Calculate the consensus degree. Based on Equations (11) and (12), the \(CDF\) value is determined as follows. \begin{table}
\begin{tabular}{c c c c c c c c c c c c c c c c c c} \hline \hline \multirow{2}{*}{**Failure Modes**} & \multicolumn{4}{c}{**Severity(S)**} & \multicolumn{4}{c}{**Occurrence(O)**} & \multicolumn{4}{c}{**Detection(D)**} \\ \cline{2-13}  & \(E_{1}\) & \(E_{2}\) &  
\[CDF=\left( \begin{array}{cccccccc}0.8381&0.8288&0.9147&0.9143&0.9715\\ 0.8035&0.8659&0.7972&0.8232&0.8994\\ 0.9204&0.8646&0.7959&0.8604&0.8046\\ 0.8562&0.7725&0.7806&0.8943&0.8834\\ 0.8775&0.7633&0.8981&0.9301&0.8455\\ 0.8420&0.8354&0.8397&0.7942&0.8675\\ 0.8508&0.8364&0.8207&0.8364&0.8921\\ 0.9051&0.7694&0.8459&0.8626&0.9007\\ 0.8830&0.8877&0.8433&0.8644&0.8690\\ \end{array} \right).\]  
Then, according to Equation (13), the \(CDM\) value is calculated as follows. \[CDM=(0.8641,0.8249,0.8374,0.8644,0.8815).\]  
Finally, using Equation (14), the \(CDT\) value is obtained as \(CDT=0.8548\). **Step 2.5:** Due to \(CDT<\zeta\), go to next step.