put proof for comparison equations in

PLT

opamp_circuits_C_GARRETT/Makefile

@@ -1,6 +1,6 @@
 
 
-PNG_DIA = circuit1_dag.png mvampcircuit.png pd.png invamp.png shared_component.png tree_abstraction_levels.png
+PNG_DIA = circuit1_dag.png mvampcircuit.png pd.png invamp.png shared_component.png tree_abstraction_levels.png three_tree.png
 
 
 

opamp_circuits_C_GARRETT/opamps.tex

@@ -813,7 +813,7 @@ This can be simplified if we can determine the total number of failure modes in
 equation~\ref{eqn:rd} becomes $$ RD(fg) = fT.(|fg|-1).$$
 
 \pagebreak[4]
-\subsection{Reasoning Distance Examples}
+\subsection{Reasoning Distance Examples}(c-1)
 
 The potential divider discussed in section~\ref{potdivfmmd} has a four failure modes and two components and therefore has an $RD$ of 4.
 $$RD(potdiv) = \sum_{n=1}^{2} |2|.(|1|) = 4 $$
@@ -823,5 +823,96 @@ having 3 failure modes each, we would have an $RD$ of
 
 $$RD(fictitious) = \sum_{n=1}^{81} |3|.(|80|) = 19440 .$$
 
+This would be the polynomial ($O(N^2)$) result of applying FMEA rigorously (we could term this
+Rigorous FMEA (RFMEA).
+
+
+\pagebreak[4]
+\subsection{Using the concept of Reasoning Distance to compare RFMEA with FMMD}
+
+\begin{figure}
+ \centering
+ \includegraphics[width=400pt,keepaspectratio=true]{./three_tree.png}
+ % three_tree.png: 851x385 pixel, 72dpi, 30.02x13.58 cm, bb=0 0 851 385
+ \caption{FMMD Hierarchy with $(|fg| = 3) \wedge (|fm(c)| = 3)$}
+ \label{fig:three_tree}
+\end{figure}
+
+Because components have variable numbers of failure modes,
+ and {\fgs} have variable numbers of components it is difficult to
+come up with a general formula for comparing the number of checks to make for
+RFMEA and FMMMD.
+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 represent the number of checks to make.
+ 
+Consider $k$ to be the number of components in a {\fg} (i.e. $k=|fg|$), 
+$f$ is the number of failure modes per component (i.e. $f=|fm(c)|$), and 
+$L$ to be the number of levels in the hierarchy of an FMMD analysis.
+We can represent the number of failure scenarios to check in an FMMD
+with equation~\ref{eqn:anscen}.
+ 
+\begin{equation}
+  \label{eqn:anscen}
+  \sum_{n=0}^{L} {k}^{n}.k.f.(k-1) 
+\end{equation}
+ 
+The thinking behind equation~\ref{eqn:anscen}, is that for each level of analysis -- counting down from the top --
+there are ${k}^{n}$ {\fgs} within each level; we need to apply RFMEA to each {\fg} on the level.
+The number of checks to make for RFMEA is number of components $k$ multiplied by the number of failure modes $f$
+checked against the remaining components in the {\fg} $(k-1)$.
+
+If, for the sake of example we fix the number of components in a {\fg} to three and 
+the number of failure modes per component to three, an FMMD hierarchy
+would look like figure~\ref{fig:three_tree}.
+
+\subsection{Worked Example}
+
+Using the diagram in figure~\ref{fig:three_tree}, we have three levels of analysis.
+Starting at the top, we have a {\fg} with three derived components, each of which has
+three failure modes.
+Thus the number of checks to make in the top level is $3^0.3.2.3=18$.
+On the level below that, we have three {\fgs} each with a 
+an identical number of checks, $3^1.3.2.3=56$.{\fg}
+On the level below that we have nine {\fgs}, $3^2.3.2.3=168$.
+Adding these together gives $242$ checks to make to perform RFMEA \textbf{within}
+{\fgs}.
+
+If we were to take the system represented in  figure~\ref{fig:three_tree}, and 
+apply RFMEA on it as a whole system, we can use equation~\ref{eqn:rd},
+$ RD(fg) = \sum_{n=1}^{|fg|} |fm(c_n)|.(|fg|-1)$, where $|fg|$ is 27, $fm(c_n)$ is 3
+and $(|fg|-1)$ is 26.
+This gives:
+$ RD(fg) = \sum_{n=1}^{27} |3|.(|27|-1) = 2106$
+
+In order to get general equations with which to compare RFMEA with FMMD
+we can re-write equation~\ref{eqn:rd} in terms of the number of levels 
+in an FMMD hierarchy. The number of components in is number of components
+in a {\fg} raised to the power of the level plus one.
+Thus we re-write equation~\ref{eqn:rd} as:
+ 
+
+\begin{equation}
+  \label{eqn:fmea_state_exp21}
+  \sum_{n=1}^{k^{L+1}}.(k^{L+1}-1).f  \; , % \\
+  %(N^2 - N).f 
+\end{equation}
+
+or
+
+\begin{equation}
+  \label{eqn:fmea_state_exp22}
+  k^{L+1}.(k^{L+1}-1).f  \;. % \\
+  %(N^2 - N).f 
+\end{equation}
+
+We can now use  equation~\ref{eqn:anscen} and \ref{eqn:fmea_state_exp22} to compare (for fixed sizes of $|fg|$ and $|fm(c)|$)
+the two approaches, for the work required to perform rigorous checking.
+
+
+\subsection{Exponential squared to Exponential}
+
+can I say that ?
+
 
 \end{document}