robin/Robin_PHD
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

describing motivation for cardinality constrained powerset

Browse Source
This commit is contained in:
Robin 2010-06-03 23:15:46 +01:00
parent 7debef2103
commit f469c2bf9f
1 changed files with 46 additions and 15 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

61
component_failure_modes_definition/component_failure_modes_definition.tex
View File

@ -309,10 +309,11 @@ It is an implied requirement of EN298 for instance to consider double simultaneo
To generalise, we may need to consider $N$ simultaneous
failure modes when analysing a functional group. This involves finding
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 and S itself.
The Powerset concept from Set theory when applied to a set S is the set of all subsets of S, including the empty set and S itself
\footnote{The empty set is a special case for FMMD analysis, it simply means there
is no fault active in the functional~group under analysis}.
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'
is proposed and described in the next section. The empty set is a special case for FMMD analysis, it simply means there
is no fault active in the functional~group under analysis.
is proposed and described in the next section.
\subsection{Cardinality Constrained Powerset }
\label{ccp}
@ -320,11 +321,16 @@ is no fault active in the functional~group under analysis.
A Cardinality Constrained powerset is one where sub-sets of a cardinality greater than a threshold
are not included. This theshold is called the cardinality constraint.
To indicate this the cardinality constraint $cc$, is subscripted to the powerset symbol thus $\mathcal{P}_{cc}$.
Consider the set $S = \{a,b,c\}$. $\mathcal{P}_{2} S $ means all subsets of S where the cardinality of the subsets is
less than or equal to 2 or less.
Consider the set $S = \{a,b,c\}$.
The powerset of S:
$$ \mathcal{P} S = \{ 0, \{a,b,c\}, \{a,b\},\{b,c\},\{c,a\},\{a\},\{b\},\{c\} \} $$
$\mathcal{P}_{2} S $ means all subsets of S where the cardinality of the subsets is
less than or equal to 2 or less.
$$ \mathcal{P}_{2} S = \{ \{a,b\},\{b,c\},\{c,a\},\{a\},\{b\},\{c\} \} $$
Note that $\mathcal{P}_{1} S $ for this example is:
@ -352,11 +358,12 @@ from $1$ to $cc$ thus
\begin{equation}
\#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \frac{\#S!}{k!(\#S-k)!}
\label{eqn:ccps}
\end{equation}
\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
calculation give the correct number of test case combinations to check.
@ -367,29 +374,53 @@ be less.
What must actually be done is to subtract the number of component `internal combinations'
from the cardinality constrain powerset number.
Thus were we to have a simple circuit with two components R and T, of which
$FM(R) = {R_o, R_s}$ and $FM(T) = {T_o, T_s, T_h}$.
Thus were we to have a simple functional group with two components R and T, of which
$$FM(R) = \{R_o, R_s\}$$ and $$FM(T) = \{T_o, T_s, T_h\}$$.
For a cardinality constrained powerset of 2, because there are 5 error modes
gives
applying equation \ref{eqn:ccps} gives :-
$$\frac{5!}{1!(5-1)!} + \frac{5!}{2!(5-2)!} = 15$$
This is composed of
5 single fault modes, and ${2 \choose 5}$ ten double fault modes.
This is composed of ${1 \choose 5}$
five single fault modes, and ${2 \choose 5}$ ten double fault modes.
However we know that the faults are mutually exclusive for a component.
We must then subtract the number of `internal' component fault combinations.
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
$R_o \wedge R_s$. As a combination ${2 \choose 2} = 1$ . For $T$ the component with
three fault modes ${2 \choose 3} = 3$.
Thus for $cc == 2$ we must subtract $(3+1)$.
The number of combinations to check is thus 11 for this example and this can be verified
by listing all the required combinations:
Written as a general formula, where C is a set of the components (indexed by j where J
is the set of componets in the functional~group under analyis) and $\#C$
\vbox{
%\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
}
The cardinality constrained powerset equation \ref{eqn:ccps} corrected for
unitary state failure modes can be
written as a general formula, where C is a set of the components (indexed by j where J
is the set of components in the functional~group under analyis) and $\#C$
indicates the number of mutually exclusive fault modes each component has:-
%$$ \#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \frac{\#S!}{k!(\#S-k)!} $$
\begin{equation}
\#\mathcal{P}_{cc} S = {\sum^{k}_{1..cc} \frac{\#S!}{k!(\#S-k)!}} - {\sum^{j}_{j \in J} {\#C_{j} \choose cc}}
\label{eqn:correctedccps}
\end{equation}
@ -416,7 +447,7 @@ $$ F = \Omega(C) \backslash OK $$
The $OK$ statistical case is the largest in probability, and is therefore
of interest when analysing systems from a statistical perspective.
This is of interest to conditional probability calculations
This is of interest for the application of conditional probability calculations
such as Bayes theorem.
\vspace{40pt}