An Extended Interval Type-2 Fuzzy ORESTE Method for Risk Analysis in FMEA

Abstract

The failure mode and effect analysis (FMEA) is a prospective engineering technique which can prevent risk in advance by identifying, assessing potential failure modes and their effects. However, the traditional FMEA approach possesses the limitations in improperly identifying risk factor weights, inefficiently calculating risk priority number and improperly evaluating failure modes in practice. This paper proposes a hybrid risk analysis framework for FMEA based on interval type-2 fuzzy numbers and ORESTE method. First, the interval type-2 fuzzy numbers are utilized to express the uncertain risk evaluation information provided by experts. Second, the power average (PA) operator for interval type-2 fuzzy numbers is constructed to fuse individual risk assessment information and build the group risk assessment matrix. Then, the extended ORESTE method is introduced to determine the risk priority of each failure mode, in which distance measure of interval type-2 fuzzy numbers is incorporated. Finally, a case of ocean-going fishing vessel in a marine industry is selected to illustrate the application and feasibility of the proposed FMEA approach. A comparative analysis is performed to verify the effectiveness of the extended ORESTE method.

Appendices

Appendix A: Proof of Theorem 1

Proof

Equation (6) can be proved by mathematical induction on k as follows.

• (1) For k = 1, the result of Eq. (6) is obviously.

• (2) Assume that Eq. (6) holds for k = q that is

  $$ PA - IT2TrFNs(a_{1} ,a_{2} , \ldots ,a_{q} ) = \left[ {\begin{array}{*{20}c} {\left( {\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))a_{k1}^{L} } }}{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))} }},\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))a_{k2}^{L} } }}{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))} }},} \right.} \\ {\left. {\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))a_{k3}^{L} } }}{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))} }},\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))a_{k4}^{L} } }}{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))} }};\mathop {\hbox{min} }\limits_{k = 1,2, \ldots ,q} h_{k}^{L} } \right),} \\ {\left( {\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))a_{k1}^{U} } }}{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))} }},\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))a_{k2}^{U} } }}{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))} }},} \right.} \\ {\left. {\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))a_{k3}^{U} } }}{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))} }},\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))a_{k4}^{U} } }}{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))} }};\mathop {\hbox{min} }\limits_{k = 1,2, \ldots ,q} h_{k}^{U} } \right)} \\ \end{array} } \right]. $$

• (3) Then, when $k = q{ + }1$, by the algorithms of IT2TrFN in Definition 2, we have

  $$ \begin{aligned} PA - & IT2TrFNs(a_{1} ,a_{2} , \ldots ,a_{q} ,a_{q + 1} ) \\ & = \left[ {\begin{array}{*{20}c} {\left( {\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))a_{k1}^{L} } }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))} }},\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))a_{k2}^{L} } }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))} }},} \right.} \\ {\left. {\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))a_{k3}^{L} } }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))} }},\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))a_{k4}^{L} } }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))} }};\mathop {\hbox{min} }\limits_{k = 1,2, \ldots ,q} h_{k}^{L} ),} \right)} \\ {\left( {\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))a_{k1}^{U} } }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))} }},\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))a_{k2}^{U} } }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))} }},} \right.} \\ {\left. {\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))a_{k3}^{U} } }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))} }},\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{k} ))a_{k4}^{U} } }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))} }};\mathop {\hbox{min} }\limits_{k = 1,2, \ldots ,q} h_{k}^{U} } \right)} \\ \end{array} } \right] \\ & \; + \left[ {\begin{array}{*{20}c} {\left( {\frac{{(1 + T(a_{q + 1} ))a_{(q + 1)1}^{L} }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{ij}^{k} ))} }},\frac{{(1 + T(a_{q + 1} ))a_{(q + 1)2}^{L} }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{ij}^{k} ))} }},} \right.} \\ {\left. {\frac{{(1 + T(a_{q + 1} ))a_{(q + 1)3}^{L} }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{ij}^{k} ))} }},\frac{{(1 + T(a_{q + 1} ))a_{(q + 1)4}^{L} }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{ij}^{k} ))} }};h_{(q + 1)}^{L} ),} \right)} \\ {\left( {\frac{{(1 + T(a_{q + 1} ))a_{(q + 1)1}^{U} }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))} }},\frac{{(1 + T(a_{q + 1} ))a_{(q + 1)2}^{U} }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))} }},} \right.} \\ {\left. {\frac{{(1 + T(a_{q + 1} ))a_{(q + 1)3}^{U} }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))} }},\frac{{(1 + T(a_{q + 1} ))a_{(q + 1)4}^{U} }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))} }};h_{(q + 1)}^{U} } \right)} \\ \end{array} } \right] \\ & = \left[ {\begin{array}{*{20}c} {\left( {\frac{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))a_{k1}^{L} } }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))} }},\frac{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))a_{k2}^{L} } }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))} }},} \right.} \\ {\left. {\frac{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))a_{k3}^{L} } }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))} }},\frac{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))a_{k4}^{L} } }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))} }};\mathop {\hbox{min} }\limits_{k = 1,2, \ldots ,q + 1} h_{k}^{L} ),} \right)} \\ {\left( {\frac{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))a_{k1}^{U} } }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))} }},\frac{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))a_{k2}^{U} } }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))} }},} \right.} \\ {\left. {\frac{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))a_{k3}^{U} } }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))} }},\frac{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))a_{k4}^{U} } }}{{\sum\nolimits_{k = 1}^{q + 1} {(1 + T(a_{k} ))} }};\mathop {\hbox{min} }\limits_{k = 1,2, \ldots ,q + 1} h_{k}^{U} } \right)} \\ \end{array} } \right] \\ \end{aligned}. $$

  So, when $k = q{ + }1$, Eq. (16) is also right.

  Consequently, according to mathematical induction given by (1) and (2), Eq. (21) holds for all k.

Appendix B: Proof of Theorem 1

Proof

• (1) As $a_{k} = a_{0} (k = 1,2, \ldots ,q)$ , by the algorithms of IT2TrFN in Definition 5, we have

  $$ \begin{aligned} PA - IT2TrFNs(a_{1} ,a_{2} , \ldots ,a_{q} ) = & IT2TrFN - PA(a_{0} ,a_{0} , \ldots ,a_{0} ) \\ & = \left[ {\begin{array}{*{20}c} {\left( {\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{0} ))a_{01}^{L} } }}{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{0} ))} }},\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{0} ))a_{02}^{L} } }}{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{0} ))} }},} \right.} \\ {\left. {\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{0} ))a_{03}^{L} } }}{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{0} ))} }},\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{0} ))a_{04}^{L} } }}{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{0} ))} }};\mathop {\hbox{min} }\limits_{k = 1,2, \ldots ,q} h_{0}^{L} } \right),} \\ {\left( {\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{0} ))a_{01}^{U} } }}{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{0} ))} }},\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{0} ))a_{02}^{U} } }}{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{0} ))} }},} \right.} \\ {\left. {\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{0} ))a_{03}^{U} } }}{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{0} ))} }},\frac{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{0} ))a_{04}^{U} } }}{{\sum\nolimits_{k = 1}^{q} {(1 + T(a_{0} ))} }};\mathop {\hbox{min} }\limits_{k = 1,2, \ldots ,q} h_{0}^{U} } \right)} \\ \end{array} } \right] \\ \end{aligned}. $$

  Since $a_{k} = a_{0}$ for all the k, then we can obtain $T(a_{0} ) = 0$. Consequently,$\begin{aligned} PA - IT2TrFNs(a_{1} ,a_{2} , \ldots ,a_{q} ) = & PA - IT2TrFNs(a_{0} ,a_{0} , \ldots ,a_{0} ) \\ & = \left[ {\begin{array}{*{20}c} {\left( {a_{01}^{L} ,a_{02}^{L} ,} \right.} \\ {\left. {a_{03}^{L} ,a_{04}^{L} ;h_{0}^{L} } \right),} \\ {\left( {a_{01}^{U} ,a_{02}^{U} ,} \right.} \\ {\left. {a_{03}^{U} ,a_{04}^{U} ;h_{0}^{U} } \right)} \\ \end{array} } \right] = a_{0} \\ \end{aligned}$

• (2) Assume that $c_{0} = a_{\hbox{max} } = \hbox{max} (a_{1} ,a_{2} , \ldots ,a_{q} )$ and $d_{0} = a_{\hbox{min} } = \hbox{min} (a_{1} ,a_{2} , \ldots ,a_{q} )$, then we can have

  $$ \begin{aligned} c_{0} = & PA - IT2TrFNs(c_{0} ,c_{0} , \ldots ,c_{0} ) \\ & \le PA - IT2TrFNs(a_{1} ,a_{2} , \ldots ,a_{q} ) \\ & \le PA - IT2TrFNs(d_{0} ,d_{0} , \ldots ,d_{0} ) \\ & = d_{0} \\ \end{aligned}. $$

  Therefore, we can get $a_{\hbox{min} } \le PA - IT2TrFNs(a_{1} ,a_{2} , \ldots ,a_{q} ) \le a_{\hbox{max} }.$

• (3) Since $a_{i} \le b_{j} (i,j = 1,2, \ldots ,q)$ and PA operator is an Augment function, then

  $$ IT2TrFN - PA(a_{1} ,a_{2} , \ldots ,a_{q} ) \le IT2TrFN - PA(b_{1} ,b_{2} , \ldots ,b_{q} ). $$