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Robin Clark 2010-11-22 14:20:47 +00:00
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component_failure_modes_definition/component_failure_modes_definition.tex
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@ -495,13 +495,13 @@ Consider the set $S = \{a,b,c\}$.
The powerset of S: The powerset of S:
$$ \mathcal{P} S = \{ \emptyset, \{a,b,c\}, \{a,b\},\{b,c\},\{c,a\},\{a\},\{b\},\{c\} \} $$. $$ \mathcal{P} S = \{ \emptyset, \{a,b,c\}, \{a,b\},\{b,c\},\{c,a\},\{a\},\{b\},\{c\} \} .$$
$\mathcal{P}_{\le 2} S $ means all non-empty subsets of S where the cardinality of the subsets is $\mathcal{P}_{\le 2} S $ means all non-empty subsets of S where the cardinality of the subsets is
less than or equal to 2 or less. less than or equal to 2 or less.
$$ \mathcal{P}_{\le 2} S = \{ \{a,b\},\{b,c\},\{c,a\},\{a\},\{b\},\{c\} \} $$. $$ \mathcal{P}_{\le 2} S = \{ \{a,b\},\{b,c\},\{c,a\},\{a\},\{b\},\{c\} \} . $$
Note that $\mathcal{P}_{1} S $ (non-empty subsets where cardinality $\leq 1$) for this example is: Note that $\mathcal{P}_{1} S $ (non-empty subsets where cardinality $\leq 1$) for this example is:
@ -514,9 +514,9 @@ The number of $k$ combinations (each of size $k$) from a set $S$
with $n$ elements (size $n$) is the binomial coefficient~\cite{probstat} shown in equation \ref{bico}. with $n$ elements (size $n$) is the binomial coefficient~\cite{probstat} shown in equation \ref{bico}.
\begin{equation} \begin{equation}
C^n_k = {n \choose k} = \frac{n!}{k!(n-k)!} C^n_k = {n \choose k} = \frac{n!}{k!(n-k)!} .
\label{bico} \label{bico}
\end{equation} . \end{equation}
To find the number of elements in a cardinality constrained subset S with up to $cc$ elements To find the number of elements in a cardinality constrained subset S with up to $cc$ elements
in each combination sub-set, in each combination sub-set,
@ -530,9 +530,9 @@ from $1$ to $cc$ thus
% %
\begin{equation} \begin{equation}
|{\mathcal{P}_{cc}S}| = \sum^{cc}_{k=1} \frac{|{S}|!}{ k! ( |{S}| - k)!} |{\mathcal{P}_{cc}S}| = \sum^{cc}_{k=1} \frac{|{S}|!}{ k! ( |{S}| - k)!} .
\label{eqn:ccps} \label{eqn:ccps}
\end{equation} . \end{equation}
@ -556,7 +556,7 @@ $|{n \choose 2}|$ and $|{n \choose 3}|$ for each component in the functional~gro
\subsubsection{Example: Two Component functional group cardinality Constraint of 2} \subsubsection{Example: Two Component functional group cardinality Constraint of 2}
For example: suppose we have a simple functional group with two components R and T, of which For example: suppose we have a simple functional group with two components R and T, of which
$$fm(R) = \{R_o, R_s\}$$ and $$fm(T) = \{T_o, T_s, T_h\}$$. $$fm(R) = \{R_o, R_s\}$$ 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 This means that the functional~group $FG=\{R,T\}$ will have a component failure mode set
of $fm(FG) = \{R_o, R_s, T_o, T_s, T_h\}$ of $fm(FG) = \{R_o, R_s, T_o, T_s, T_h\}$
@ -564,7 +564,7 @@ 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 ( $|fm(FG)|=5$), For a cardinality constrained powerset of 2, because there are 5 error modes ( $|fm(FG)|=5$),
applying equation \ref{eqn:ccps} gives :- applying equation \ref{eqn:ccps} gives :-
$$ | P_2 (fm(FG)) | = \frac{5!}{1!(5-1)!} + \frac{5!}{2!(5-2)!} = 15$$. $$ | P_2 (fm(FG)) | = \frac{5!}{1!(5-1)!} + \frac{5!}{2!(5-2)!} = 15.$$
This is composed of ${5 \choose 1}$ This is composed of ${5 \choose 1}$
five single fault modes, and ${5 \choose 2}$ ten double fault modes. five single fault modes, and ${5 \choose 2}$ ten double fault modes.
@ -585,7 +585,7 @@ $$ \mathcal{P}_{2}(fm(FG)) = \{
\} \}
$$ $$
And whose cardinality is 11. % by inspection and whose cardinality is 11. % by inspection
%$$ %$$
%| %|
%\{ %\{
@ -628,18 +628,18 @@ components $C_j$ are in
\begin{equation} \begin{equation}
|{\mathcal{P}_{cc}SU}| = {\sum^{cc}_{k=1} \frac{|{SU}|!}{k!(|{SU}| - k)!}} |{\mathcal{P}_{cc}SU}| = {\sum^{cc}_{k=1} \frac{|{SU}|!}{k!(|{SU}| - k)!}}
- {\sum_{j \in J} {|FM({C_{j})}| \choose 2}} - {\sum_{j \in J} {|FM({C_{j})}| \choose 2}} .
\label{eqn:correctedccps} \label{eqn:correctedccps}
\end{equation} . \end{equation}
Expanding the combination in equation \ref{eqn:correctedccps} Expanding the combination in equation \ref{eqn:correctedccps}
\begin{equation} \begin{equation}
|{\mathcal{P}_{cc}SU}| = {\sum^{cc}_{k=1} \frac{|{SU}|!}{k!(|{SU}| - k)!}} |{\mathcal{P}_{cc}SU}| = {\sum^{cc}_{k=1} \frac{|{SU}|!}{k!(|{SU}| - k)!}}
- {{\sum_{j \in J} \frac{|FM({C_j})|!}{2!(|FM({C_j})| - 2)!}} } - {{\sum_{j \in J} \frac{|FM({C_j})|!}{2!(|FM({C_j})| - 2)!}} } .
\label{eqn:correctedccps2} \label{eqn:correctedccps2}
\end{equation} . \end{equation}
\paragraph{Use of Equation \ref{eqn:correctedccps2} } \paragraph{Use of Equation \ref{eqn:correctedccps2} }
Equation \ref{eqn:correctedccps2} is useful for an automated tool that Equation \ref{eqn:correctedccps2} is useful for an automated tool that
@ -707,7 +707,7 @@ By definition while all components in a system are `working perfectly'
that system will not exhibit faulty behaviour. that system will not exhibit faulty behaviour.
Thus the statistical sample space $\Omega$ for a component or derived~component $C$ is Thus the statistical sample space $\Omega$ for a component or derived~component $C$ is
%$$ \Omega = {OK, failure\_mode_{1},failure\_mode_{2},failure\_mode_{3} ... failure\_mode_{N} $$ %$$ \Omega = {OK, failure\_mode_{1},failure\_mode_{2},failure\_mode_{3} ... failure\_mode_{N} $$
$$ \Omega(C) = \{OK, failure\_mode_{1},failure\_mode_{2},failure\_mode_{3}, \ldots ,failure\_mode_{N}\} $$ $$ \Omega(C) = \{OK, failure\_mode_{1},failure\_mode_{2},failure\_mode_{3}, \ldots ,failure\_mode_{N}\} . $$
The failure mode set $F$ for a given component or derived~component $C$ The failure mode set $F$ for a given component or derived~component $C$
is therefore is therefore
$ fm(C) = \Omega(C) \backslash \{OK\} $ $ fm(C) = \Omega(C) \backslash \{OK\} $