overloaded FM function to cope with functional groups as well as components

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 $FM$  
+%We can define a function $FM$  
-
+%
-\begin{equation}
+%\begin{equation}
-FM : \mathcal{C} \mapsto \mathcal{F}
+%FM : \mathcal{C} \mapsto \mathcal{F}
-\end{equation}
+%\end{equation}
-
+%
-defined by
+%defined by
-
+%
-$$  FM ( C ) = F $$
+%$$  FM ( C ) = F $$
-
+%
-i.e. take a given component $C$ and return its set of failure modes $F$.
+%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}
 %}
 

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.