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defining variables b4 equation 3

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Robin Clark 2010-06-07 16:17:24 +01:00
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component_failure_modes_definition/component_failure_modes_definition.tex
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@ -309,8 +309,9 @@ It is an implied requirement of EN298 for instance to consider double simultaneo
To generalise, we may need to consider $N$ simultaneous To generalise, we may need to consider $N$ simultaneous
failure modes when analysing a functional group. This involves finding failure modes when analysing a functional group. This involves finding
all combinations of failures modes of size $N$ and less. all combinations of failures modes of size $N$ and less.
The Powerset concept from Set theory when applied to a set S is the set of all subsets of S, including the empty set The Powerset concept from Set theory is useful model this.
\footnote{The empty set is a special case for FMMD analysis, it simply means there The powerset, when applied to a set S is the set of all subsets of S, including the empty set
\footnote{The empty set ( $\emptyset$ ) is a special case for FMMD analysis, it simply means there
is no fault active in the functional~group under analysis} is no fault active in the functional~group under analysis}
and S itself. and S itself.
In order to consider combinations for the set S where the number of elements in each sub-set of S is $N$ or less, a concept of the `cardinality constrained powerset' In order to consider combinations for the set S where the number of elements in each sub-set of S is $N$ or less, a concept of the `cardinality constrained powerset'
@ -326,7 +327,7 @@ Consider the set $S = \{a,b,c\}$.
The powerset of S: The powerset of S:
$$ \mathcal{P} S = \{ 0, \{a,b,c\}, \{a,b\},\{b,c\},\{c,a\},\{a\},\{b\},\{c\} \} $$ $$ \mathcal{P} S = \{ \emptyset, \{a,b,c\}, \{a,b\},\{b,c\},\{c,a\},\{a\},\{b\},\{c\} \} $$
$\mathcal{P}_{2} S $ means all subsets of S where the cardinality of the subsets is $\mathcal{P}_{2} S $ means all subsets of S where the cardinality of the subsets is
@ -366,14 +367,15 @@ from $1$ to $cc$ thus
\subsection{Actual Number of combinations to check \\ with Unitary State Fault mode sets} \subsection{Actual Number of combinations to check \\ with Unitary State Fault mode sets}
Where all components analysed only have one fault mode, the cardinality constrained powerset Where all the fault modes in $S$ were to be independent,
calculation give the correct number of test case combinations to check. the cardinality constrained powerset
Because set of failure modes is constrained to be unitary state, the acual number will calculation (in equation \ref {eqn:ccps}) would give the correct number of test case combinations to check.
be less. Because sets of failure modes in FMMD analysis are constrained to be unitary state,
the actual number of test cases to check will usually
be less than this. This is because combinations of faults with a components failure mode set
What must actually be done is to subtract the number of component `internal combinations' are impossible under the conditions of a unitary state failure mode set.
from the cardinality constrain powerset number. To correct equation \ref{eqn:ccps} we must subtract the number of component `internal combinations'
for each component in the functional group under analysis.
\subsubsection{Example: Two Component functional group \\ cardinality Constraint of 2} \subsubsection{Example: Two Component functional group \\ cardinality Constraint of 2}
@ -388,38 +390,17 @@ applying equation \ref{eqn:ccps} gives :-
$$\frac{5!}{1!(5-1)!} + \frac{5!}{2!(5-2)!} = 15$$ $$\frac{5!}{1!(5-1)!} + \frac{5!}{2!(5-2)!} = 15$$
This is composed of ${1 \choose 5}$ This is composed of ${5 \choose 1}$
five single fault modes, and ${2 \choose 5}$ ten double fault modes. five single fault modes, and ${5 \choose 2}$ ten double fault modes.
However we know that the faults are mutually exclusive for a component. However we know that the faults are mutually exclusive for a component.
We must then subtract the number of `internal' component fault combinations for each component in the functional~group. We must then subtract the number of `internal' component fault combinations for each component in the functional~group.
For component R there is only one internal component fault that cannot exist 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 $T$ the component with $R_o \wedge R_s$. As a combination ${2 \choose 2} = 1$ . For $T$ the component with
three fault modes ${2 \choose 3} = 3$. 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 $(3+1)$. 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, $|\mathcal{P}_{2}(FG_cfg)| = 11$, for this example and this can be verified
by listing all the required combinations: by listing all the required combinations:
%
%\vbox{
%\subsubsection{All Eleven Cardinality Constrained \\ Powerset of 2 Elements Listed}
%%\tiny
%\begin{enumerate}
%\item $\{R_o T_o\}$
%\item $\{R_o T_s\}$
%\item $\{R_o T_h\}$
%\item $\{R_s T_o\}$
%\item $\{R_s T_s\}$
%\item $\{R_s T_h\}$
%\item $\{R_o \}$
%\item $\{R_s \}$
%\item $\{T_o \}$
%\item $\{T_s \}$
%\item $\{T_h \}$
%\end{enumerate}
%%\normalsize
%}
%
%$$ |\mathcal{P}_{2}(FG_cfg)| = 11 $$
$$ \mathcal{P}_{2}(FG_cfg) = \{ $$ \mathcal{P}_{2}(FG_cfg) = \{
@ -433,21 +414,33 @@ $$
\{ \{
\{R_o T_o\}, \{R_o T_s\}, \{R_o T_h\}, \{R_s T_o\}, \{R_s T_s\}, \{R_s T_h\}, \{R_o \}, \{R_s \}, \{T_o \}, \{T_s \}, \{T_h \} \{R_o T_o\}, \{R_o T_s\}, \{R_o T_h\}, \{R_s T_o\}, \{R_s T_s\}, \{R_s T_h\}, \{R_o \}, \{R_s \}, \{T_o \}, \{T_s \}, \{T_h \}
\} \}
| = 11 $$ | = 11
$$
\subsubsection{Establishing Formulae for unitary state failure mode \\ \subsubsection{Establishing Formulae for unitary state failure mode \\
cardinality calculation} cardinality calculation}
The cardinality constrained powerset equation \ref{eqn:ccps} corrected for The cardinality constrained powerset in equation \ref{eqn:ccps} can be corrected for
unitary state failure modes can be unitary state failure modes.
written as a general formula (see equation \ref{eqn:correctedccps}), where C is a set of the components (indexed by j where J This is written as a general formula in equation \ref{eqn:correctedccps}.
is the set of components in the functional~group under analyis) and $|{C}|$
indicates the number of mutually exclusive fault modes each component has:- %\indent{
where :
\begin{itemize}
\item Let $C$ be a set of components (indexed by $j \in J$)
that are members of the functional group $FG$
i.e. $ \forall j \in J | C_j \in FG $
\item Let $|{C}_{j}|$
indicate the number of mutually exclusive fault modes each component has
\item Let $SU$ be a set of unitary state failure modes from the functional group
nder analysis $SU = FM(FG)$
\end{itemize}
%}
%$$ \#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \frac{\#S!}{k!(\#S-k)!} $$
\begin{equation} \begin{equation}
|{\mathcal{P}_{cc}S}| = {\sum^{k}_{1..cc} \frac{|{S}|!}{k!(|{S}| - k)!}} - {\sum^{j}_{j \in J} {|{C_{j}}| \choose cc}} |{\mathcal{P}_{cc}SU}| = {\sum^{k}_{1..cc} \frac{|{SU}|!}{k!(|{SU}| - k)!}} - {\sum^{j}_{j \in J} {|{C_{j}}| \choose cc}}
\label{eqn:correctedccps} \label{eqn:correctedccps}
\end{equation} \end{equation}
@ -455,16 +448,15 @@ Expanding the combination in equation \ref{eqn:correctedccps}
\begin{equation} \begin{equation}
|{\mathcal{P}_{cc}S}| = {\sum^{k}_{1..cc} \frac{|{S}|!}{k!(|{S}| - k)!}} - {\sum^{j}_{j \in J} \frac{|{C_j}|!}{cc!(|{C_j}| - cc)!}} |{\mathcal{P}_{cc}SU}| = {\sum^{k}_{1..cc} \frac{|{SU}|!}{k!(|{SU}| - k)!}} - {\sum^{j}_{j \in J} \frac{|{C_j}|!}{cc!(|{C_j}| - cc)!}}
\label{eqn:correctedccps2} \label{eqn:correctedccps2}
\end{equation} \end{equation}
Equation \ref{eqn:correctedccps2} is useful for an automated tool that Equation \ref{eqn:correctedccps2} is useful for an automated tool that
would verify that a `N' simultaneous failures model had been completly covered. would verify that a `N' simultaneous failures model had complete failure mode coverage.
By knowing how many test case should be covered, and checking the cardinality By knowing how many test cases should be covered, and checking the cardinality
associated with the test cases complete coverage would be confirmed. associated with the test cases, complete coverage would be confirmed.
%$$ \#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \big[ \frac{\#S!}{k!(\#S-k)!} - \sum_{j} (\#C_{j} \choose cc \big] $$
\section{Component Failure Modes and Statistical Sample Space} \section{Component Failure Modes and Statistical Sample Space}