.

component_failure_modes_definition/component_failure_modes_definition.tex

@@ -495,13 +495,13 @@ Consider the set $S = \{a,b,c\}$.
 
 The powerset of S: 
 
-$$ \mathcal{P} S = \{ \emptyset, \{a,b,c\}, \{a,b\},\{b,c\},\{c,a\},\{a\},\{b\},\{c\} \} $$.
+$$ \mathcal{P} S = \{ \emptyset, \{a,b,c\}, \{a,b\},\{b,c\},\{c,a\},\{a\},\{b\},\{c\} \} .$$
 
 
 $\mathcal{P}_{\le 2} S $ means all non-empty subsets of S where the cardinality of the subsets is 
 less than or equal to 2 or less.
 
-$$ \mathcal{P}_{\le 2} S = \{ \{a,b\},\{b,c\},\{c,a\},\{a\},\{b\},\{c\} \} $$.
+$$ \mathcal{P}_{\le 2} S = \{ \{a,b\},\{b,c\},\{c,a\},\{a\},\{b\},\{c\} \} . $$
 
 Note that $\mathcal{P}_{1} S $ (non-empty subsets where cardinality $\leq 1$) for this example is:
 
@@ -514,9 +514,9 @@ The number of $k$ combinations (each of size $k$) from a set $S$
 with $n$ elements (size $n$) is the binomial coefficient~\cite{probstat} shown in equation \ref{bico}.
 
 \begin{equation}
-C^n_k = {n \choose k} = \frac{n!}{k!(n-k)!}
+C^n_k = {n \choose k} = \frac{n!}{k!(n-k)!} .
 \label{bico}
-\end{equation} .
+\end{equation} 
 
 To find the number of elements in a cardinality constrained subset S with up to $cc$ elements
 in each combination sub-set,
@@ -530,9 +530,9 @@ from $1$ to $cc$ thus
 %
 
 \begin{equation}
- |{\mathcal{P}_{cc}S}| = \sum^{cc}_{k=1} \frac{|{S}|!}{ k! ( |{S}| - k)!}    
+ |{\mathcal{P}_{cc}S}| = \sum^{cc}_{k=1} \frac{|{S}|!}{ k! ( |{S}| - k)!} .
  \label{eqn:ccps}
-\end{equation} .
+\end{equation} 
 
 
 
@@ -556,7 +556,7 @@ $|{n \choose 2}|$ and $|{n \choose 3}|$ for each component in the functional~gro
 \subsubsection{Example: Two Component functional group cardinality Constraint of 2}
 
 For example: suppose we have a simple functional group with two components R and T, of which
-$$fm(R) = \{R_o, R_s\}$$ and $$fm(T) = \{T_o, T_s, T_h\}$$.
+$$fm(R) = \{R_o, R_s\}$$ and $$fm(T) = \{T_o, T_s, T_h\}.$$
 
 This means that the functional~group $FG=\{R,T\}$ will have a component failure mode set 
 of $fm(FG) = \{R_o, R_s, T_o, T_s, T_h\}$
@@ -564,7 +564,7 @@ of $fm(FG) = \{R_o, R_s, T_o, T_s, T_h\}$
 For a cardinality constrained powerset of 2, because there are 5 error modes ( $|fm(FG)|=5$),
 applying equation \ref{eqn:ccps} gives :-
 
-$$ | P_2 (fm(FG)) |  = \frac{5!}{1!(5-1)!} + \frac{5!}{2!(5-2)!}  = 15$$.
+$$ | P_2 (fm(FG)) |  = \frac{5!}{1!(5-1)!} + \frac{5!}{2!(5-2)!}  = 15.$$
 
 This is composed of ${5 \choose 1}$
 five single fault modes, and ${5 \choose 2}$ ten double fault modes.
@@ -585,7 +585,7 @@ $$ \mathcal{P}_{2}(fm(FG)) =  \{
  \}
 $$
 
-And whose cardinality is 11. % by inspection
+and whose cardinality is 11. % by inspection
 %$$ 
 %| 
 %\{
@@ -628,18 +628,18 @@ components $C_j$ are in
 
 \begin{equation}
   |{\mathcal{P}_{cc}SU}| = {\sum^{cc}_{k=1}  \frac{|{SU}|!}{k!(|{SU}| - k)!}}  
- - {\sum_{j \in J} {|FM({C_{j})}| \choose 2}} 
+ - {\sum_{j \in J} {|FM({C_{j})}| \choose 2}} .
   \label{eqn:correctedccps}
-\end{equation} .
+\end{equation} 
 
 Expanding the combination in equation \ref{eqn:correctedccps}
 
 
 \begin{equation}
  |{\mathcal{P}_{cc}SU}| = {\sum^{cc}_{k=1}  \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}
-\end{equation} .
+\end{equation} 
 
 \paragraph{Use of Equation \ref{eqn:correctedccps2} }
 Equation \ref{eqn:correctedccps2} is useful for an automated tool that
@@ -707,7 +707,7 @@ By definition while all components in a system are `working perfectly'
 that system will not exhibit faulty behaviour.
 Thus the statistical sample space $\Omega$ for a component or derived~component $C$ is
 %$$ \Omega = {OK, failure\_mode_{1},failure\_mode_{2},failure\_mode_{3} ... failure\_mode_{N} $$
-$$ \Omega(C) = \{OK, failure\_mode_{1},failure\_mode_{2},failure\_mode_{3}, \ldots ,failure\_mode_{N}\} $$
+$$ \Omega(C) = \{OK, failure\_mode_{1},failure\_mode_{2},failure\_mode_{3}, \ldots ,failure\_mode_{N}\} . $$
 The failure mode set $F$ for a given component or derived~component $C$
 is therefore
 $ fm(C) = \Omega(C) \backslash \{OK\} $