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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
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!).
%
%Obviously, f
For a small number of components and failure modes, we have a smaller number
of checks to make than for a complicated larger system.
%For a small number of components and failure modes, we have a smaller number
%of checks to make than for a complicated larger system.
%
%
\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 % , $fms$,
%as its range. % (see equation~\ref{eqn:fm}).
$ 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 $G$.}
%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
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}^{+} $,
$CC$ is defined by,
Let $\mathcal{G}$ represent the set of all {\fgs} %, and $ \mathbb{Z}^{+} $,
then $CC$ is defined by,
\begin{equation}
%$$
CC:\mathcal{G} \rightarrow \mathbb{Z}^{ }. % could be zero, one component like an op-amp used as a NIBUFF
@ -171,7 +172,7 @@ $CC$ is defined by,
%and, where n is the number of components in the system/{\fg},
%and $|fm(c_i)|$ is the number of failure modes
%in component ${c_i}$.
Comparison complexity, $CC$ for a group of components $G$, is given by
Comparison complexity, $CC$, for a group of $n$ components $G$, is given by
\begin{equation}
\label{eqn:CC}
@ -197,7 +198,7 @@ the equation becomes
CC(G) = K.(|G|-1).
\end{equation}
\subsection{A general formula for counting Comparison Complexity in an FMMD hierarchy}
An FMMD hierarchy consists of many {\fgs} which are subsets of $G$.
%We define the set of all {\fgs} as $\mathcal{FG}$.
%Using $FG$ to represent individual {\fgs}
@ -228,7 +229,7 @@ with the potential divider and the operational amplifier has an $\alpha$ level o
% CC(G) = {|G|}.{|fm(c_n)|}.{(|fg|-1)} .
% %$$
% \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.
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
% the sum of all complexity comparison
% applied to {\fgs} at a particular hierarchy level in \hh,
We define a helper function, %g,
that applies $CC$ to all {\fgs} at a particular level, $\xi$ in an FMMD hierarchy {\hh}
We define a helper function, $g$,
that applies $CC$ to all {\fgs} at a particular level, $\xi$, in an FMMD hierarchy, {\hh},
and returns the sum of the comparison complexities,
\begin{equation}
g: \hh \times \mathbb{N} \rightarrow \mathbb{N} .
\end{equation}
%
%$$
%g(H, i) \rightarrow \forall {\FG}^{\xi} \;where\; ({\xi} = {i}) \wedge ({\FG}^{\xi} \in H) .
%$$
%
%IN ENGLISH: A helper function $g$
%
Where $L$ represents the number of levels in the FMMD hierarchy {\hh} and
$g(\hh,\xi)$ represents the comparison complexity of {\fgs} on the level $\xi$;
Let $L$ represent the number of levels in the FMMD hierarchy {\hh} and
$g(\hh,\xi)$ represent the comparison complexity of {\fgs} on the level $\xi$.
%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:
%$$
\begin{equation}
\label{eqn:gf}
@ -266,7 +267,7 @@ we overload the comparison complexity function $CC$, to obtain the comparison co
\subsection{Complexity Comparison Examples}
%\pagebreak[4]
We initially work though the chapter~\ref{sec:chap4} amplifier example, which has two
We initially work though the amplifier example from chapter~\ref{sec:chap4}, which has two
stages, the potential divider and then the amplifier. We add the complexities from
both these stages to determine how many reasoning paths there were to perform FMMD analysis on the
non-inverting amplifier.
@ -277,8 +278,8 @@ We calculate this using equation~\ref{eqn:CC} thus,
$$CC(potdiv) = \sum_{n=1}^{2} \big( |2| \times (|1|) \big) = 4. $$
%
We next combine the potential divider with an op-amp which has four failure modes
to form a {\fg} with two components, one with four failure modes and the other (the potential divider) with two.
$$CC(invamp) = 2 \times 1 + 4 \times 1 = 6 $$
to form a {\fg} with two components, one with four failure modes and the other (the potential divider) with two,
$$CC(invamp) = 2 \times 1 + 4 \times 1 = 6 . $$
%
We now add the two calculated complexities to determine the
amount of reasoning paths to analyse the amplifier using FMMD.
@ -291,17 +292,17 @@ Even with this very trivial example, we begin to see benefits of taking a modula
\paragraph{Complexity Comparison for a hypothetical 81 component system.}
%Even considering a $example$
A system, $example$, with just 81 components (with these components
having 3 failure modes each) would, using equation~\ref{eqn:rd2} have an $CC$ of
A system, $example$, with just 81 components, with these components
having 3 failure modes each would, using equation~\ref{eqn:rd2} have a $CC$ of
$$CC(example) = \sum_{n=1}^{81} |3|.(|80|) = 19440 .$$
Ensuring all component failure modes are checked against all other components in a system
-- applying FMEA exhaustively
%
%Ensuring all component failure modes are checked against all other components in a system
%-- applying FMEA exhaustively
%rigorously
-- could be termed
exhaustive FMEA ({\XFMEA}).
The computational order for {\XFMEA} would be polynomial ($O((N)(N-1)K) \approx O(N^2.K)$) (where $K$ is the variable number of failure modes)
%-- could be termed
%exhaustive FMEA ({\XFMEA}).
The computational order for {\XFMEA} would be polynomial ($O((N)(N-1)f) \approx O(N^2.f)$) (where $f$ is the variable number of failure modes)
as discussed in section~\ref{eqn:fmea_single}.
%
This order may be acceptable in a computational environment. However, the choosing of {\fgs} and the analysis
@ -345,6 +346,13 @@ use the general formula for comparing the number of checks to make for
If we were to create an example by fixing the number of components in a {\fg}
and the number of failure modes per component, we can derive formulae
to compare the number of checks to make from an FMMD hierarchy to {\XFMEA}.
%
%% HEALTH WARNING
%
While real-world analysis models have variable
numbers of failure modes per component type and
different numbers of components in their {\fgs}
a fixed model provides indicative estimates of complexity performance.
%applied to
%all components in a system.
@ -369,7 +377,7 @@ If, for the sake of example, we fix the number of components in a {\fg} to three
the number of failure modes per component to three, an FMMD hierarchy
would look like figure~\ref{fig:three_tree}.
\subsection{{\XFMEA} FMMD Comparison Example}
\subsection{Comparing {\XFMEA} and FMMD: an Example}
Using the diagram in figure~\ref{fig:three_tree}, we have three levels of analysis.
%
@ -423,22 +431,23 @@ For instance, having four levels
of FMMD analysis, with these fixed numbers,
%(in addition to the top zeroth level)
will require 81 base level components.
%
%$$
Applying equation~\ref{eqn:fmea_state_exp22}, we have
\begin{equation}
\label{eqn:fmea_state_exp22}
3^4.(3^4-1).3 = 81.(81-1).3 = 19440 % \\
\label{eqn:fmea_state_exp22_example}
3^4.(3^4-1).3 = 81.(81-1).3 = 19440 .% \\
%(N^2 - N).f
\end{equation}
%$$
Equation \ref{eqn:fmea_state_exp22} shows that applying XFMEA where components all have three failure modes
and there are 81 components, would involve 19,440 reasoning paths.
Applying equation~\ref{eqn:fmea_state_exp21}, we have
$$
%\begin{equation}
% \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}
$$
@ -775,9 +784,10 @@ Because of this, the failure mode set $F=fm(R)$ is `unitary~state'.
%
%Thus because both fault modes cannot be active at the same time, the intersection of $ R_{SHORTED} $ and $ R_{OPEN} $ cannot exist.
%
The intersection of these failure modes is therefore the empty set, $ R_{SHORTED} \cap R_{OPEN} = \emptyset $,
therefore
$ fm(R) \in \mathcal{U} $. These concepts are expanded in section~\ref{sec:usprob}.
%The intersection of these failure modes is therefore the empty set, $ R_{SHORTED} \cap R_{OPEN} = \emptyset $,
%therefore
%$ fm(R) \in \mathcal{U} $.
These concepts are expanded in section~\ref{sec:usprob}.
@ -905,9 +915,9 @@ less than or equal to 2.
$$ \mathcal{P}_{\le 2} S = \{ \{a,b\},\{b,c\},\{c,a\},\{a\},\{b\},\{c\} \} . $$
Note that $\mathcal{P}_{1} S $ (non-empty subsets where cardinality $\leq 1$) for this example is:
Note that $\mathcal{P}_{\le 1} S $ (non-empty subsets where cardinality $\leq 1$) for this example is:
$$ \mathcal{P}_{1} S = \{ \{a\},\{b\},\{c\} \} $$.
$$ \mathcal{P}_{\le 1} S = \{ \{a\},\{b\},\{c\} \} .$$
\paragraph{Calculating the number of elements in a Cardinality Constrained power-set}
@ -971,7 +981,7 @@ is as we would use them for single failure analysis.
% We use equation~\ref{eqn:ccps} to determine the number of valid combinations.
%
For a cardinality constrained powerset of 2, because there are 5 error modes ( $|fm(FG)|=5$),
applying equation \ref{eqn:ccps} gives :-
applying equation \ref{eqn:ccps} gives:
$$ | P_2 (fm(FG)) | = \frac{5!}{1!(5-1)!} + \frac{5!}{2!(5-2)!} = 15.$$
@ -999,7 +1009,7 @@ $$ \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
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 $.
i.e. $ \forall j \in J , C_j \in FG $.
%\item
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
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}
%}
@ -1086,45 +1096,46 @@ reproduced below to verify this.
\end{itemize}
}
\begin{equation}
|{\mathcal{P}_{cc}SU}| = {\sum^{k}_{1..cc} \frac{|{SU}|!}{k!(|{SU}| - k)!}}
- {{\sum_{j \in J} \frac{|FM({C_j})|!}{2!(|FM({C_j})| - 2)!}} }
|{\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)!}} } ,
\label{eqn:correctedccps2}
\end{equation}
}
{
\begin{equation}
|{\mathcal{P}_{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)!}} }
|{\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)!}} } .
%\label{eqn:correctedccps2}
\end{equation}
}
%
%
$|FM(C_j)|$ 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 !
%
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.
%
\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)!}} }
%\label{eqn:correctedccps2}
\end{equation}
%
$|{\mathcal{P}_{2}SU}|$ 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).
Expanding the sumations
$$ NoOfTestCasesToCheck = \frac{6!}{1!(6-1)!} + \frac{6!}{2!(6-2)!} - \Big( \frac{2!}{2!(2 - 2)!} + \frac{2!}{2!(2 - 2)!} + \frac{2!}{2!(2 - 2)!} \Big) $$
$$ NoOfTestCasesToCheck = 6 + 15 - ( 1 + 1 + 1 ) = 18 $$
%
Expanding the summations:
%
%
$$ NoOfTestCasesToCheck = \frac{6!}{1!(6-1)!} + \frac{6!}{2!(6-2)!} -
\Big( \frac{2!}{2!(2 - 2)!} + \frac{2!}{2!(2 - 2)!} + \frac{2!}{2!(2 - 2)!} \Big) , $$
%
$$ NoOfTestCasesToCheck = 6 + 15 - ( 1 + 1 + 1 ) = 18 .$$
%
As the test cases are all different and are of the correct cardinalities (6 single faults and (15-3) double)
we can be confident that we have looked at all `double combinations' of the possible faults
in the Pt100 circuit.
@ -1207,7 +1218,7 @@ $ fm(C) = \Omega(C) \backslash \{OK\} $
(or expressed as
$ \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.
%
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.
$B_1^\prime$ represents the altered value for $B_1$.
Thus
$$ P(B_1^\prime) = B_1 - P(B_1 \cap B_2) - P(B_1 \cap B_3)\; , $$
$$ P(B_2^\prime) = B_2 - P(B_1 \cap B_2) \; and $$
$$ P(B_3^\prime) = B_3 - P(B_1 \cap B_3) \; . $$
$$ P(B_1^\prime) = P(B_1) - P(B_1 \cap B_2) - P(B_1 \cap B_3)\; , $$
$$ P(B_2^\prime) = P(B_2) - P(B_1 \cap B_2) \; and $$
$$ P(B_3^\prime) = P(B_3) - P(B_1 \cap B_3) \; . $$
We now have two new component failure mode $B_4$ and $B_5$, shown in figure \ref{fig:combco3}.
We 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}
\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 guiding principle it to choose components that are functionally adjacent