papaer printout 18:35 stillat work

component_failure_modes_definition/component_failure_modes_definition.tex

@@ -294,8 +294,8 @@ we state this formally
 
 That is to say that it is impossible that any pair of failure modes can be active at the same time
 for the failure mode set $F$ to exist in the family of sets $\mathcal{U}$.
-Note where that are more than two failure~modes, 
-by banning any pairs from being active at the same time
+Note where there are more than two failure~modes, 
+by banning any pairs from being active at the same time,
 we have banned larger combinations as well.
 
 
@@ -309,14 +309,15 @@ 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 is useful model this.
+The Powerset concept from Set theory is useful to model this.
 The powerset, when applied to a set S is the set of all subsets of S, including the empty set 
 \footnote{The empty set ( $\emptyset$ ) is a special case for FMMD analysis, it simply means there
 is no fault active in the functional~group under analysis}
 and S itself.
 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. 
- 
+
+\pagebreak[4] 
 \subsection{Cardinality Constrained Powerset }
 \label{ccp}
 
@@ -348,7 +349,7 @@ with $n$ elements (size $n$) is the binomial coefficient
 $$ C^n_k = {n \choose k} = \frac{n!}{k!(n-k)!}$$
 
 To find the number of elements in a cardinality constrained subset S with up to $cc$ elements
-in each comination sub-set,
+in each combination sub-set,
 we need to sum the combinations, 
 %subtracting $cc$ from the final result 
 %(repeated empty set counts)
@@ -372,17 +373,22 @@ the cardinality constrained powerset
 calculation (in equation \ref {eqn:ccps}) would give the correct number of test case combinations to check.
 Because  sets of failure modes in FMMD analysis are constrained to be unitary state,
 the actual number of test cases to check will usually 
-be less than this. This is because combinations of faults with a components failure mode set
-are impossible under the conditions of a unitary state failure mode set.
+be less than this. 
+This is because combinations of faults within a components failure mode set,
+are impossible under the conditions of  unitary state failure mode.
 To correct equation \ref{eqn:ccps} we must subtract the number of component `internal combinations'
 for each component in the functional group under analysis.
+Note we must sequentially subtract using combinations above 1 up to the cardinality constraint. 
+For example, say 
+the cardinality constraint was 3, we would need to subtract both
+$|{n \choose 2}|$ and $|{n \choose 3}|$.
 
 \subsubsection{Example: Two Component functional group \\ cardinality Constraint of 2}
 
 For example: were we to have a simple functional group with two components R and T, of which
 $$FM(R) = \{R_o, R_s\}$$ and $$FM(T) = \{T_o, T_s, T_h\}$$.
 
-This means that a functional~group $FG=\{R,T\}$ will have a component failure modes set 
+This means that the functional~group $FG=\{R,T\}$ will have a component failure mode set 
 of $FG_{cfg} = \{R_o, R_s, T_o, T_s, T_h\}$
 
 For a cardinality constrained powerset of 2, because there are 5 error modes ( $|{FG_{cfg}}|=5$),
@@ -392,18 +398,19 @@ $$\frac{5!}{1!(5-1)!} + \frac{5!}{2!(5-2)!}  = 15$$
 
 This is composed of ${5 \choose 1}$
 five single fault modes, and ${5 \choose 2}$ ten double fault modes.
-However we know that the faults are mutually exclusive for a component.
-We must then subtract the number of `internal' component fault combinations for each component in the functional~group.
+However we know that the faults are mutually exclusive within a component.
+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 
+$R_o \wedge R_s$. As a combination ${2 \choose 2} = 1$. For the component $T$ which has 
   three  fault modes ${3 \choose 2} = 3$.
 Thus for $cc == 2$, under the conditions of unitary state failure modes in the components $R$ and $T$, we must subtract $(3+1)$.
-The number of combinations to check is thus 11, $|\mathcal{P}_{2}(FG_cfg)| = 11$, for this example and this can be verified
+The number of combinations to check is thus 11, $|\mathcal{P}_{2}(FG_{cfg})| = 11$, for this example and this can be verified
 by listing all the required combinations:
 
 
 
-$$ \mathcal{P}_{2}(FG_cfg) =  \{
+$$ \mathcal{P}_{2}(FG_{cfg}) =  \{
  \{R_o T_o\}, \{R_o T_s\}, \{R_o T_h\}, \{R_s T_o\}, \{R_s T_s\}, \{R_s T_h\}, \{R_o \}, \{R_s \}, \{T_o \}, \{T_s \}, \{T_h \}
  \}
 $$
@@ -421,7 +428,7 @@ $$
 \subsubsection{Establishing Formulae for unitary state failure mode \\
 cardinality calculation}
 
-The cardinality constrained powerset in equation \ref{eqn:ccps} can be corrected for 
+The cardinality constrained powerset in equation \ref{eqn:ccps}, can be corrected for 
 unitary state failure modes.
 This is written as a general formula in equation \ref{eqn:correctedccps}. 
 
@@ -431,19 +438,20 @@ where :
 \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 $
-\item Let $|{C}_{j}|$
+\item Let $|FM({C}_{j})|$
 indicate the number of mutually exclusive fault modes of each component
 \item Let $FG_{cfg}$ be the collection of all failure modes
 from all the components in the functional group. 
 \item Let $SU$ be a set of failure modes from the functional group,
 where all contributing components $C_j$ 
-are guaranteed to be `unitary state' i.e. $(SU = FG_{cfg}) \wedge (\forall j \in J | C_j \in \mathcal{U}) $
+are guaranteed to be `unitary state' i.e. $(SU = FG_{cfg}) \wedge (\forall j \in J | FM(C_j) \in \mathcal{U}) $
 \end{itemize}
 %}
 
 
 \begin{equation}
- |{\mathcal{P}_{cc}SU}| = {\sum^{k}_{1..cc}  \frac{|{SU}|!}{k!(|{SU}| - k)!}}  - {\sum^{j}_{j \in J} {|{C_{j}}| \choose cc}} 
+ |{\mathcal{P}_{cc}SU}| = {\sum^{k}_{1..cc}  \frac{|{SU}|!}{k!(|{SU}| - k)!}}  
+- \sum^{p}_{2..cc}{{\sum^{j}_{j \in J} {|FM({C_{j})}| \choose p}}} 
  \label{eqn:correctedccps}
 \end{equation}
 
@@ -451,7 +459,8 @@ Expanding the combination in equation \ref{eqn:correctedccps}
 
 
 \begin{equation}
- |{\mathcal{P}_{cc}SU}| = {\sum^{k}_{1..cc}  \frac{|{SU}|!}{k!(|{SU}| - k)!}}   - {\sum^{j}_{j \in J}   \frac{|{C_j}|!}{cc!(|{C_j}| - cc)!}}  
+ |{\mathcal{P}_{cc}SU}| = {\sum^{k}_{1..cc}  \frac{|{SU}|!}{k!(|{SU}| - k)!}}   
+- \sum^{p}_{2..cc}{{\sum^{j}_{j \in J}   \frac{|FM({C_j})|!}{p!(|FM({C_j})| - p)!}}  }
  \label{eqn:correctedccps2}
 \end{equation}
 
@@ -461,7 +470,7 @@ By knowing how many test cases should be covered, and checking the cardinality
 associated with the test cases, complete coverage would be confirmed.
 
 
-
+\pagebreak[4]
 \section{Component Failure Modes and Statistical Sample Space}
 %\paragraph{NOT WRITTEN YET PLEASE IGNORE}
 A sample space is defined as the set of all possible outcomes.