From ec945d77c6dca6401fa1526c1d02c27ba3736551 Mon Sep 17 00:00:00 2001 From: Robin Date: Sun, 13 Jun 2010 19:54:14 +0100 Subject: [PATCH] overloaded FM function to cope with functional groups as well as components --- .../component_failure_modes_definition.tex | 73 ++++++++++++++----- fzd/fzd.tex | 2 +- 2 files changed, 55 insertions(+), 20 deletions(-) diff --git a/component_failure_modes_definition/component_failure_modes_definition.tex b/component_failure_modes_definition/component_failure_modes_definition.tex index a4c36f5..a86520f 100644 --- a/component_failure_modes_definition/component_failure_modes_definition.tex +++ b/component_failure_modes_definition/component_failure_modes_definition.tex @@ -231,6 +231,41 @@ would have a level of 1. % $$ \bowtie ( FG ) \mapsto DerivedComponent $$ % +\subsection{relationships between functional~groups and failure modes} + + + + +Let the set of all possible components be $\mathcal{C}$ +and let the set of all possible failure modes be $\mathcal{F}$. + +We can define a function $FM$ + +\begin{equation} +FM : \mathcal{C} \mapsto \mathcal{P}\mathcal{F} +\end{equation} + +defined by, where C is a component and F is a set of failure modes. + +$$ FM ( C ) = F $$ + +\paragraph{Finding all failure modes within the functional group} + +For FMMD failure mode analysis we need to consider the failure modes +from all the components in a functional~group as a flat set. +Consider the components in a functional group to be $C$ indexed by j thus $C_j$. +Thflat set of failure modes we are after can be found by applying function $FM$ to all the components +in the functional~group and taking the union of them thus: + +$$ FunctionalGroupAllFailureModes = \bigcup_{j \in \{1...n\}} FM(C_j) $$ + +We can actually overload the notation for the function FM +and define it for the set components within a functional group $FG$ (i.e. where $FG \subset \mathcal{C} $) thus: + +\begin{equation} +FM : FG \mapsto \mathcal{F} +\end{equation} + \section{Unitary State Component Failure Mode sets} @@ -261,19 +296,19 @@ probability theory\cite{probandstat}. Let the set of all possible components to be $\mathcal{C}$ and let the set of all possible failure modes be $\mathcal{F}$. - -We can define a function $FM$ - -\begin{equation} -FM : \mathcal{C} \mapsto \mathcal{F} -\end{equation} - -defined by - -$$ FM ( C ) = F $$ - -i.e. take a given component $C$ and return its set of failure modes $F$. - +% +%We can define a function $FM$ +% +%\begin{equation} +%FM : \mathcal{C} \mapsto \mathcal{F} +%\end{equation} +% +%defined by +% +%$$ FM ( C ) = F $$ +% +%i.e. take a given component $C$ and return its set of failure modes $F$. +% \begin{definition} We can define a set $\mathcal{U}$ which is a set of sets of failure modes, where the component failure modes in each of its members are unitary~state. @@ -415,9 +450,9 @@ For example: were we to have a simple functional group with two components R and $$FM(R) = \{R_o, R_s\}$$ and $$FM(T) = \{T_o, T_s, T_h\}$$. This means that the functional~group $FG=\{R,T\}$ will have a component failure mode set -of $FG_{cfg} = \{R_o, R_s, T_o, T_s, T_h\}$ +of $FM(FG) = \{R_o, R_s, T_o, T_s, T_h\}$ -For a cardinality constrained powerset of 2, because there are 5 error modes ( $|{FG_{cfg}}|=5$), +For a cardinality constrained powerset of 2, because there are 5 error modes ( $|FM(FG)|=5$), applying equation \ref{eqn:ccps} gives :- $$\frac{5!}{1!(5-1)!} + \frac{5!}{2!(5-2)!} = 15$$ @@ -431,12 +466,12 @@ For component R there is only one internal component fault that cannot exist $R_o \wedge R_s$. As a combination ${2 \choose 2} = 1$. For the component $T$ which has three fault modes ${3 \choose 2} = 3$. Thus for $cc == 2$, under the conditions of unitary state failure modes in the components $R$ and $T$, we must subtract $(3+1)$. -The number of combinations to check is thus 11, $|\mathcal{P}_{2}(FG_{cfg})| = 11$, for this example and this can be verified +The number of combinations to check is thus 11, $|\mathcal{P}_{2}(FM(FG))| = 11$, for this example and this can be verified by listing all the required combinations: -$$ \mathcal{P}_{2}(FG_{cfg}) = \{ +$$ \mathcal{P}_{2}(FM(FG)) = \{ \{R_o T_o\}, \{R_o T_s\}, \{R_o T_h\}, \{R_s T_o\}, \{R_s T_s\}, \{R_s T_h\}, \{R_o \}, \{R_s \}, \{T_o \}, \{T_s \}, \{T_h \} \} $$ @@ -466,11 +501,11 @@ that are members of the functional group $FG$ i.e. $ \forall j \in J | C_j \in FG $ \item Let $|FM({C}_{j})|$ indicate the number of mutually exclusive fault modes of each component -\item Let $FG_{cfg}$ be the collection of all failure modes +\item Let $FM(FG)$ be the collection of all failure modes from all the components in the functional group. \item Let $SU$ be a set of failure modes from the functional group, where all contributing components $C_j$ -are guaranteed to be `unitary state' i.e. $(SU = FG_{cfg}) \wedge (\forall j \in J | FM(C_j) \in \mathcal{U}) $ +are guaranteed to be `unitary state' i.e. $(SU = FM(FG)) \wedge (\forall j \in J | FM(C_j) \in \mathcal{U}) $ \end{itemize} %} diff --git a/fzd/fzd.tex b/fzd/fzd.tex index f5c24f0..4fc275a 100644 --- a/fzd/fzd.tex +++ b/fzd/fzd.tex @@ -29,7 +29,7 @@ of the relationships between the contours. { %% Introduction \section{Algorithm Purpose} -This paper discusses a two stage algorithm designed to greatly +This chapter discusses a two stage algorithm designed to greatly reduce the number of Area compare operations required to determine which zones are `available' in an Euler diagram.