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After J Howse comments @ meeting today

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submission_thesis/CH7_Evaluation/copy.tex
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@ -4,7 +4,8 @@
% %
This chapter begins by defining a metric for the complexity of an FMEA analysis task. This chapter defines %begins by defining
a metric for the complexity of an FMEA analysis task.
% %
This concept is called `comparison~complexity' and is a means to assess This concept is called `comparison~complexity' and is a means to assess
the performance of FMMD against current FMEA methodologies. the performance of FMMD against current FMEA methodologies.
@ -123,8 +124,8 @@ of components $G$. %system or {\fg}.
% (except its self of course, that component is already considered to be in a failed state!). % (except its self of course, that component is already considered to be in a failed state!).
% %
%Obviously, f %Obviously, f
For a small number of components and failure modes, we have a smaller number %For a small number of components and failure modes, we have a smaller number
of checks to make than for a complicated larger system. %of checks to make than for a complicated larger system.
% %
% %
\subsection{Formal definitions of entities used in FMEA} \subsection{Formal definitions of entities used in FMEA}
@ -148,7 +149,7 @@ The function $fm$ returns the failure modes of a component,
its signature is %has a component as its domain and the components failure modes % , $fms$, its signature is %has a component as its domain and the components failure modes % , $fms$,
%as its range. % (see equation~\ref{eqn:fm}). %as its range. % (see equation~\ref{eqn:fm}).
$ fm: \mathcal{C} \rightarrow \mathcal{F},$ where $\mathcal{F}$ is the set of all failures. $ fm: \mathcal{C} \rightarrow \mathcal{F},$ where $\mathcal{F}$ is the set of all failures.
We can represent the number of potential failure modes of a component $c$, to be $ | fm(c) | .$ We can represent the number of potential failure modes of a component, $c$, to be $ | fm(c) | .$
%\paragraph{Indexing components with the group $G$.} %\paragraph{Indexing components with the group $G$.}
%If we index all %If we index all
@ -160,8 +161,8 @@ failure mode against all the other components in a system in equation~\ref{eqn:C
Comparison Complexity can be represented by a function $CC$, with its domain as $G$, and Comparison Complexity can be represented by a function $CC$, with its domain as $G$, and
its range as the number of checks---or reasoning stages---to perform to satisfy an XFMEA inspection. its range as the number of checks---or reasoning stages---to perform to satisfy an XFMEA inspection.
Where $\mathcal{G}$ represents the set of all {\fgs} %, and $ \mathbb{Z}^{+} $, Let $\mathcal{G}$ represent the set of all {\fgs} %, and $ \mathbb{Z}^{+} $,
$CC$ is defined by, then $CC$ is defined by,
\begin{equation} \begin{equation}
%$$ %$$
CC:\mathcal{G} \rightarrow \mathbb{Z}^{ }. % could be zero, one component like an op-amp used as a NIBUFF CC:\mathcal{G} \rightarrow \mathbb{Z}^{ }. % could be zero, one component like an op-amp used as a NIBUFF
@ -171,7 +172,7 @@ $CC$ is defined by,
%and, where n is the number of components in the system/{\fg}, %and, where n is the number of components in the system/{\fg},
%and $|fm(c_i)|$ is the number of failure modes %and $|fm(c_i)|$ is the number of failure modes
%in component ${c_i}$. %in component ${c_i}$.
Comparison complexity, $CC$ for a group of components $G$, is given by Comparison complexity, $CC$, for a group of $n$ components $G$, is given by
\begin{equation} \begin{equation}
\label{eqn:CC} \label{eqn:CC}
@ -197,7 +198,7 @@ the equation becomes
CC(G) = K.(|G|-1). CC(G) = K.(|G|-1).
\end{equation} \end{equation}
\subsection{A general formula for counting Comparison Complexity in an FMMD hierarchy}
An FMMD hierarchy consists of many {\fgs} which are subsets of $G$. An FMMD hierarchy consists of many {\fgs} which are subsets of $G$.
%We define the set of all {\fgs} as $\mathcal{FG}$. %We define the set of all {\fgs} as $\mathcal{FG}$.
%Using $FG$ to represent individual {\fgs} %Using $FG$ to represent individual {\fgs}
@ -228,7 +229,7 @@ with the potential divider and the operational amplifier has an $\alpha$ level o
% CC(G) = {|G|}.{|fm(c_n)|}.{(|fg|-1)} . % CC(G) = {|G|}.{|fm(c_n)|}.{(|fg|-1)} .
% %$$ % %$$
% \end{equation} % \end{equation}
\subsection{A general formula for counting Comparison Complexity in an FMMD hierarchy}
An FMMD hierarchy will have reducing numbers of {\fgs} as we progress up the hierarchy. An FMMD hierarchy will have reducing numbers of {\fgs} as we progress up the hierarchy.
In order to calculate its comparison~complexity we need to apply equation~\ref{eqn:CC} to In order to calculate its comparison~complexity we need to apply equation~\ref{eqn:CC} to
@ -239,23 +240,23 @@ We can define an FMMD hierarchy as a set of {\fgs}, $\hh$.
% that returns % that returns
% the sum of all complexity comparison % the sum of all complexity comparison
% applied to {\fgs} at a particular hierarchy level in \hh, % applied to {\fgs} at a particular hierarchy level in \hh,
We define a helper function, %g, We define a helper function, $g$,
that applies $CC$ to all {\fgs} at a particular level, $\xi$ in an FMMD hierarchy {\hh} that applies $CC$ to all {\fgs} at a particular level, $\xi$, in an FMMD hierarchy, {\hh},
and returns the sum of the comparison complexities, and returns the sum of the comparison complexities,
\begin{equation} \begin{equation}
g: \hh \times \mathbb{N} \rightarrow \mathbb{N} . g: \hh \times \mathbb{N} \rightarrow \mathbb{N} .
\end{equation} \end{equation}
%
%$$ %$$
%g(H, i) \rightarrow \forall {\FG}^{\xi} \;where\; ({\xi} = {i}) \wedge ({\FG}^{\xi} \in H) . %g(H, i) \rightarrow \forall {\FG}^{\xi} \;where\; ({\xi} = {i}) \wedge ({\FG}^{\xi} \in H) .
%$$ %$$
%
%IN ENGLISH: A helper function $g$ %IN ENGLISH: A helper function $g$
% %
Where $L$ represents the number of levels in the FMMD hierarchy {\hh} and Let $L$ represent the number of levels in the FMMD hierarchy {\hh} and
$g(\hh,\xi)$ represents the comparison complexity of {\fgs} on the level $\xi$; $g(\hh,\xi)$ represent the comparison complexity of {\fgs} on the level $\xi$.
%and $\hh$ represents an FMMD hierarchy, %and $\hh$ represents an FMMD hierarchy,
we overload the comparison complexity function $CC$, to obtain the comparison complexity of an entire hierarchy thus: We overload the comparison complexity function $CC$, to obtain the comparison complexity of an entire hierarchy thus:
%$$ %$$
\begin{equation} \begin{equation}
\label{eqn:gf} \label{eqn:gf}
@ -266,7 +267,7 @@ we overload the comparison complexity function $CC$, to obtain the comparison co
\subsection{Complexity Comparison Examples} \subsection{Complexity Comparison Examples}
%\pagebreak[4] %\pagebreak[4]
We initially work though the chapter~\ref{sec:chap4} amplifier example, which has two We initially work though the amplifier example from chapter~\ref{sec:chap4}, which has two
stages, the potential divider and then the amplifier. We add the complexities from stages, the potential divider and then the amplifier. We add the complexities from
both these stages to determine how many reasoning paths there were to perform FMMD analysis on the both these stages to determine how many reasoning paths there were to perform FMMD analysis on the
non-inverting amplifier. non-inverting amplifier.
@ -277,8 +278,8 @@ We calculate this using equation~\ref{eqn:CC} thus,
$$CC(potdiv) = \sum_{n=1}^{2} \big( |2| \times (|1|) \big) = 4. $$ $$CC(potdiv) = \sum_{n=1}^{2} \big( |2| \times (|1|) \big) = 4. $$
% %
We next combine the potential divider with an op-amp which has four failure modes We next combine the potential divider with an op-amp which has four failure modes
to form a {\fg} with two components, one with four failure modes and the other (the potential divider) with two. to form a {\fg} with two components, one with four failure modes and the other (the potential divider) with two,
$$CC(invamp) = 2 \times 1 + 4 \times 1 = 6 $$ $$CC(invamp) = 2 \times 1 + 4 \times 1 = 6 . $$
% %
We now add the two calculated complexities to determine the We now add the two calculated complexities to determine the
amount of reasoning paths to analyse the amplifier using FMMD. amount of reasoning paths to analyse the amplifier using FMMD.
@ -291,17 +292,17 @@ Even with this very trivial example, we begin to see benefits of taking a modula
\paragraph{Complexity Comparison for a hypothetical 81 component system.} \paragraph{Complexity Comparison for a hypothetical 81 component system.}
%Even considering a $example$ %Even considering a $example$
A system, $example$, with just 81 components (with these components A system, $example$, with just 81 components, with these components
having 3 failure modes each) would, using equation~\ref{eqn:rd2} have an $CC$ of having 3 failure modes each would, using equation~\ref{eqn:rd2} have a $CC$ of
$$CC(example) = \sum_{n=1}^{81} |3|.(|80|) = 19440 .$$ $$CC(example) = \sum_{n=1}^{81} |3|.(|80|) = 19440 .$$
%
Ensuring all component failure modes are checked against all other components in a system %Ensuring all component failure modes are checked against all other components in a system
-- applying FMEA exhaustively %-- applying FMEA exhaustively
%rigorously %rigorously
-- could be termed %-- could be termed
exhaustive FMEA ({\XFMEA}). %exhaustive FMEA ({\XFMEA}).
The computational order for {\XFMEA} would be polynomial ($O((N)(N-1)K) \approx O(N^2.K)$) (where $K$ is the variable number of failure modes) The computational order for {\XFMEA} would be polynomial ($O((N)(N-1)f) \approx O(N^2.f)$) (where $f$ is the variable number of failure modes)
as discussed in section~\ref{eqn:fmea_single}. as discussed in section~\ref{eqn:fmea_single}.
% %
This order may be acceptable in a computational environment. However, the choosing of {\fgs} and the analysis This order may be acceptable in a computational environment. However, the choosing of {\fgs} and the analysis
@ -345,6 +346,13 @@ use the general formula for comparing the number of checks to make for
If we were to create an example by fixing the number of components in a {\fg} If we were to create an example by fixing the number of components in a {\fg}
and the number of failure modes per component, we can derive formulae and the number of failure modes per component, we can derive formulae
to compare the number of checks to make from an FMMD hierarchy to {\XFMEA}. to compare the number of checks to make from an FMMD hierarchy to {\XFMEA}.
%
%% HEALTH WARNING
%
While real-world analysis models have variable
numbers of failure modes per component type and
different numbers of components in their {\fgs}
a fixed model provides indicative estimates of complexity performance.
%applied to %applied to
%all components in a system. %all components in a system.
@ -369,7 +377,7 @@ If, for the sake of example, we fix the number of components in a {\fg} to three
the number of failure modes per component to three, an FMMD hierarchy the number of failure modes per component to three, an FMMD hierarchy
would look like figure~\ref{fig:three_tree}. would look like figure~\ref{fig:three_tree}.
\subsection{{\XFMEA} FMMD Comparison Example} \subsection{Comparing {\XFMEA} and FMMD: an Example}
Using the diagram in figure~\ref{fig:three_tree}, we have three levels of analysis. Using the diagram in figure~\ref{fig:three_tree}, we have three levels of analysis.
% %
@ -423,22 +431,23 @@ For instance, having four levels
of FMMD analysis, with these fixed numbers, of FMMD analysis, with these fixed numbers,
%(in addition to the top zeroth level) %(in addition to the top zeroth level)
will require 81 base level components. will require 81 base level components.
%
%$$ %$$
Applying equation~\ref{eqn:fmea_state_exp22}, we have
\begin{equation} \begin{equation}
\label{eqn:fmea_state_exp22} \label{eqn:fmea_state_exp22_example}
3^4.(3^4-1).3 = 81.(81-1).3 = 19440 % \\ 3^4.(3^4-1).3 = 81.(81-1).3 = 19440 .% \\
%(N^2 - N).f %(N^2 - N).f
\end{equation} \end{equation}
%$$ %$$
Equation \ref{eqn:fmea_state_exp22} shows that applying XFMEA where components all have three failure modes Equation \ref{eqn:fmea_state_exp22} shows that applying XFMEA where components all have three failure modes
and there are 81 components, would involve 19,440 reasoning paths. and there are 81 components, would involve 19,440 reasoning paths.
Applying equation~\ref{eqn:fmea_state_exp21}, we have
$$ $$
%\begin{equation} %\begin{equation}
% \label{eqn:anscen} % \label{eqn:anscen}
\sum_{n=0}^{3} {3}^{n}.3.3.(2) = 720 \sum_{n=0}^{3} {3}^{n}.3.3.(2) = 720 .
%\end{equation} %\end{equation}
$$ $$
@ -775,9 +784,10 @@ Because of this, the failure mode set $F=fm(R)$ is `unitary~state'.
% %
%Thus because both fault modes cannot be active at the same time, the intersection of $ R_{SHORTED} $ and $ R_{OPEN} $ cannot exist. %Thus because both fault modes cannot be active at the same time, the intersection of $ R_{SHORTED} $ and $ R_{OPEN} $ cannot exist.
% %
The intersection of these failure modes is therefore the empty set, $ R_{SHORTED} \cap R_{OPEN} = \emptyset $, %The intersection of these failure modes is therefore the empty set, $ R_{SHORTED} \cap R_{OPEN} = \emptyset $,
therefore %therefore
$ fm(R) \in \mathcal{U} $. These concepts are expanded in section~\ref{sec:usprob}. %$ fm(R) \in \mathcal{U} $.
These concepts are expanded in section~\ref{sec:usprob}.
@ -905,9 +915,9 @@ less than or equal to 2.
$$ \mathcal{P}_{\le 2} S = \{ \{a,b\},\{b,c\},\{c,a\},\{a\},\{b\},\{c\} \} . $$ $$ \mathcal{P}_{\le 2} S = \{ \{a,b\},\{b,c\},\{c,a\},\{a\},\{b\},\{c\} \} . $$
Note that $\mathcal{P}_{1} S $ (non-empty subsets where cardinality $\leq 1$) for this example is: Note that $\mathcal{P}_{\le 1} S $ (non-empty subsets where cardinality $\leq 1$) for this example is:
$$ \mathcal{P}_{1} S = \{ \{a\},\{b\},\{c\} \} $$. $$ \mathcal{P}_{\le 1} S = \{ \{a\},\{b\},\{c\} \} .$$
\paragraph{Calculating the number of elements in a Cardinality Constrained power-set} \paragraph{Calculating the number of elements in a Cardinality Constrained power-set}
@ -971,7 +981,7 @@ is as we would use them for single failure analysis.
% We use equation~\ref{eqn:ccps} to determine the number of valid combinations. % We use equation~\ref{eqn:ccps} to determine the number of valid combinations.
% %
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:
$$ | P_2 (fm(FG)) | = \frac{5!}{1!(5-1)!} + \frac{5!}{2!(5-2)!} = 15.$$ $$ | P_2 (fm(FG)) | = \frac{5!}{1!(5-1)!} + \frac{5!}{2!(5-2)!} = 15.$$
@ -999,7 +1009,7 @@ $$ \mathcal{P}_{2}(fm(FG)) = \{
\} \}
$$ $$
and whose cardinality is 11. % by inspection whose cardinality is indeed, 11. % by inspection
%$$ %$$
%| %|
%\{ %\{
@ -1023,7 +1033,7 @@ unitary state failure modes.
%\item %\item
Let $C$ be a set of components (indexed by $j \in J$) Let $C$ be a set of components (indexed by $j \in J$)
that are members of the functional group $FG$ that are members of the functional group $FG$
i.e. $ \forall j \in J | C_j \in FG $. i.e. $ \forall j \in J , C_j \in FG $.
%\item %\item
Let $|fm({C}_{j})|$ Let $|fm({C}_{j})|$
@ -1036,7 +1046,7 @@ from all the components in the functional group.
Let $SU$ be the set of failure modes from the {\fg} where all $FG$ is such that Let $SU$ be the set of failure modes from the {\fg} where all $FG$ is such that
components $C_j$ are in components $C_j$ are in
`unitary state' i.e. $(SU = fm(FG)) \wedge (\forall j \in J | fm(C_j) \in \mathcal{U}) $, then `unitary state' i.e. $(SU = fm(FG)) \wedge (\forall j \in J , fm(C_j) \in \mathcal{U}) $, then
%\end{itemize} %\end{itemize}
%} %}
@ -1086,45 +1096,46 @@ reproduced below to verify this.
\end{itemize} \end{itemize}
} }
\begin{equation} \begin{equation}
|{\mathcal{P}_{cc}SU}| = {\sum^{k}_{1..cc} \frac{|{SU}|!}{k!(|{SU}| - k)!}} |{\mathcal{P}_{\le cc}SU}| = {\sum^{k}_{1..cc} \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}
} }
{ {
\begin{equation} \begin{equation}
|{\mathcal{P}_{cc}SU}| = {\sum^{cc}_{k=1} \frac{|{SU}|!}{k!(|{SU}| - k)!}} |{\mathcal{P}_{\le cc}SU}| = {\sum^{cc}_{k=1} \frac{|{SU}|!}{k!(|{SU}| - k)!}}
- {{\sum^{j}_{j \in J} \frac{|FM({C_j})|!}{2!(|FM({C_j})| - 2)!}} } - {{\sum^{j}_{j \in J} \frac{|FM({C_j})|!}{2!(|FM({C_j})| - 2)!}} } .
%\label{eqn:correctedccps2} %\label{eqn:correctedccps2}
\end{equation} \end{equation}
} }
%
%
$|FM(C_j)|$ will always be 2 here, as all the components are resistors and have two failure modes. $|FM(C_j)|$ will always be 2 here, as all the components are resistors and have two failure modes.
%
% %
% Factorial of zero is one ! You can only arrange an empty set one way ! % Factorial of zero is one ! You can only arrange an empty set one way !
%
Populating this equation with $|SU| = 6$ and $|FM(C_j)|$ = 2. Populating this equation with $|SU| = 6$ and $|FM(C_j)|$ = 2.
%is always 2 for this circuit, as all the components are resistors and have two failure modes. %is always 2 for this circuit, as all the components are resistors and have two failure modes.
%
\begin{equation} \begin{equation}
|{\mathcal{P}_{2}SU}| = {\sum^{k}_{1..2} \frac{6!}{k!(6 - k)!}} |{\mathcal{P}_{\le 2}SU}| = {\sum^{k}_{1..2} \frac{6!}{k!(6 - k)!}}
- {{\sum_{1..3} \frac{2!}{2!(2 - 2)!}} } - {{\sum_{1..3} \frac{2!}{2!(2 - 2)!}} }
%\label{eqn:correctedccps2} %\label{eqn:correctedccps2}
\end{equation} \end{equation}
%
$|{\mathcal{P}_{2}SU}|$ is the number of valid combinations of faults to check $|{\mathcal{P}_{2}SU}|$ is the number of valid combinations of faults to check
under the conditions of unitary state failure modes for the components (a resistor cannot fail by being shorted and open at the same time). under the conditions of unitary state failure modes for the components (a resistor cannot fail by being shorted and open at the same time).
%
Expanding the sumations Expanding the summations:
%
%
$$ NoOfTestCasesToCheck = \frac{6!}{1!(6-1)!} + \frac{6!}{2!(6-2)!} - \Big( \frac{2!}{2!(2 - 2)!} + \frac{2!}{2!(2 - 2)!} + \frac{2!}{2!(2 - 2)!} \Big) $$ $$ NoOfTestCasesToCheck = \frac{6!}{1!(6-1)!} + \frac{6!}{2!(6-2)!} -
\Big( \frac{2!}{2!(2 - 2)!} + \frac{2!}{2!(2 - 2)!} + \frac{2!}{2!(2 - 2)!} \Big) , $$
$$ NoOfTestCasesToCheck = 6 + 15 - ( 1 + 1 + 1 ) = 18 $$ %
$$ NoOfTestCasesToCheck = 6 + 15 - ( 1 + 1 + 1 ) = 18 .$$
%
As the test cases are all different and are of the correct cardinalities (6 single faults and (15-3) double) As the test cases are all different and are of the correct cardinalities (6 single faults and (15-3) double)
we can be confident that we have looked at all `double combinations' of the possible faults we can be confident that we have looked at all `double combinations' of the possible faults
in the Pt100 circuit. in the Pt100 circuit.
@ -1207,7 +1218,7 @@ $ fm(C) = \Omega(C) \backslash \{OK\} $
(or expressed as (or expressed as
$ \Omega(C) = fm(C) \cup \{OK\} $). $ \Omega(C) = fm(C) \cup \{OK\} $).
The $OK$ statistical case is the (usually) largest in probability, and is therefore The $OK$ statistical case is usually the largest in probability, and is therefore
of interest when analysing systems from a statistical perspective. of interest when analysing systems from a statistical perspective.
% %
For these examples, the OK state is not represented area proportionately, but is included For these examples, the OK state is not represented area proportionately, but is included
@ -1322,9 +1333,9 @@ $B_1, B_2$ and $B_3$ will now reduce.
We can use the prime character ($\; \prime \;$), to represent the altered value for a failure mode, i.e. We can use the prime character ($\; \prime \;$), to represent the altered value for a failure mode, i.e.
$B_1^\prime$ represents the altered value for $B_1$. $B_1^\prime$ represents the altered value for $B_1$.
Thus Thus
$$ P(B_1^\prime) = B_1 - P(B_1 \cap B_2) - P(B_1 \cap B_3)\; , $$ $$ P(B_1^\prime) = P(B_1) - P(B_1 \cap B_2) - P(B_1 \cap B_3)\; , $$
$$ P(B_2^\prime) = B_2 - P(B_1 \cap B_2) \; and $$ $$ P(B_2^\prime) = P(B_2) - P(B_1 \cap B_2) \; and $$
$$ P(B_3^\prime) = B_3 - P(B_1 \cap B_3) \; . $$ $$ P(B_3^\prime) = P(B_3) - P(B_1 \cap B_3) \; . $$
We now have two new component failure mode $B_4$ and $B_5$, shown in figure \ref{fig:combco3}. We now have two new component failure mode $B_4$ and $B_5$, shown in figure \ref{fig:combco3}.
We can express their probabilities as $P(B_4) = P(B_1 \cap B_3)$ and $P(B_5) = P(B_1 \cap B_2)$. We can express their probabilities as $P(B_4) = P(B_1 \cap B_3)$ and $P(B_5) = P(B_1 \cap B_2)$.
@ -1335,7 +1346,7 @@ We can express their probabilities as $P(B_4) = P(B_1 \cap B_3)$ and $P(B_5) = P
\section{Critiques} \section{Critiques}
\subsection{Problems in choosing membership of functional groups} \subsection{Problems in choosing membership of {\fgs}}
The choice of components for {\fgs} is one to be made by the analyst. The choice of components for {\fgs} is one to be made by the analyst.
The guiding principle it to choose components that are functionally adjacent The guiding principle it to choose components that are functionally adjacent