Looking at double sim failures onl;y n the cardinality constrained count

component_failure_modes_definition/component_failure_modes_definition.tex

@@ -204,12 +204,12 @@ We can represent this using a UML diagram in figure \ref{fig:cfg}.
 The symbol $\bowtie$ is used to indicate the analysis process that takes a
 functional group and converts it into a new component.
 
-This can be expresed as ` $ \bowtie ( FG ) \mapsto DerivedComponent $ '.
+This can be expresed as  $ \bowtie ( FG ) \mapsto DerivedComponent $ .
 
 
 \begin{figure}[h]
  \centering
- \includegraphics[width=400pt,bb=0 0 712 286,keepaspectratio=true]{component_failure_modes_definition/cfg.jpg}
+ \includegraphics[width=400pt,bb=0 0 712 286,keepaspectratio=true]{./component_failure_modes_definition/cfg.jpg}
  % cfg.jpg: 712x286 pixel, 72dpi, 25.12x10.09 cm, bb=0 0 712 286
  \caption{UML Meta model for FMMD hierarchy}
  \label{fig:cfg}
@@ -325,8 +325,8 @@ A set of failure modes where only one failure mode
 can be active at one time is termed a `unitary~state' failure mode set.
 \end{definition}
 
-Let the set of all possible components to be $\mathcal{C}$
+Let the set of all possible components to be $ \mathcal{C}$
-and let the set of all possible failure modes be $\mathcal{F}$.
+and let the set of all possible failure modes be $ \mathcal{F}$.
 
 \begin{definition}
 We can define a set $\mathcal{U}$ which is a set of sets of failure modes, where
@@ -339,9 +339,26 @@ Thus if the failure modes of  a component $F$ are unitary~state, we can say $F \
 An example of a component with an obvious set of  ``unitary~state'' failure modes is the electrical resistor.
 
 Electrical resistors can fail by going OPEN or SHORTED.
+%% CUNT 
+%% CUNT For a given resistor R we can apply the 
+%% CUNT the function $fm$ to find its set of failure modes thus  $ fm(R) =  \{R_{SHORTED}, R_{OPEN}\} $.
+%% CUNT A resistor cannot fail with both conditions open and short active at the same time! The conditions
+%% CUNT OPEN and SHORT are thus mutually exclusive.
+%% CUNT Because of this, the failure mode set $F=fm(R)$ is `unitary~state'.
+%% CUNT 
+%% CUNT 
+%% CUNT Thus because both fault modes cannot be active at the same time, the intersection of $ R_{SHORTED} $ and $ R_{OPEN} $ cannot exist.
+%% CUNT 
+%% CUNT  The intersection of these is therefore the empty set, $$ R_{SHORTED} \cap R_{OPEN} \eq \emptyset $$,
+%% CUNT therefore
+%% CUNT $ fm(R) \in \mathcal{U} $.
+%% CUNT 
+%% CUNT 
+
+
  
 For a given resistor R we can apply the 
-the function $fm$ to find its set of failure modes thus  $ fm(R) =  \{R_{SHORTED},R_{OPEN}\} $.
+the function $fm$ to find its set of failure modes thus  $ fm(R) =  \{R_{SHORTED}, R_{OPEN}\} $.
 A resistor cannot fail with both conditions open and short active at the same time! The conditions
 OPEN and SHORT are thus mutually exclusive.
 Because of this, the failure mode set $F=fm(R)$ is `unitary~state'.
@@ -349,11 +366,13 @@ 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 is therefore the empty set, $ R_{SHORTED} \cap R_{OPEN} \eq \emptyset $,
+%%CUNT The intersection of these is therefore the empty set, $$ R_{SHORTED} \cap R_{OPEN} \eq \emptyset $$,
+The intersection of these is therefore the empty set, $$ R_{SHORTED} \cap R_{OPEN} = \emptyset $$,
 therefore
 $ fm(R) \in \mathcal{U} $.
 
 
+
 We can make this a general case by taking a set $F$ (where $f_1, f_2 \in F$) representing a collection
 of component failure modes.
 We can define a boolean function {\ensuremath{\mathcal{ACTIVE}}} that returns 
@@ -559,10 +578,15 @@ components $C_j$ are in
 \end{itemize}
 %}
 
-
+%% CUNT 
+%% CUNT \begin{equation}
+%% CUNT  |{\mathcal{P}_{cc}SU}| = {\sum^{cc}_{k=1}  \frac{|{SU}|!}{k!(|{SU}| - k)!}}  
+%% CUNT - {\sum{j \in J} {|FM({C_{j})}| \choose 2}}} 
+%% CUNT  \label{eqn:correctedccps}
+%% CUNT \end{equation}
 \begin{equation}
   |{\mathcal{P}_{cc}SU}| = {\sum^{cc}_{k=1}  \frac{|{SU}|!}{k!(|{SU}| - k)!}}  
-- \sum^{cc}_{p=2}{{\sum{j \in J} {|FM({C_{j})}| \choose p}}} 
+ - {\sum_{j \in J} {|FM({C_{j})}| \choose 2}} 
   \label{eqn:correctedccps}
 \end{equation}
 
@@ -571,16 +595,22 @@ Expanding the combination in equation \ref{eqn:correctedccps}
 
 \begin{equation}
  |{\mathcal{P}_{cc}SU}| = {\sum^{cc}_{k=1}  \frac{|{SU}|!}{k!(|{SU}| - k)!}}   
-- \sum^{cc}_{p=2}{{\sum{j \in J}   \frac{|FM({C_j})|!}{p!(|FM({C_j})| - p)!}}  }
+- {{\sum_{j \in J}   \frac{|FM({C_j})|!}{2!(|FM({C_j})| - 2)!}}  }
  \label{eqn:correctedccps2}
 \end{equation}
 
 \paragraph{Use of Equation \ref{eqn:correctedccps2} }
 Equation \ref{eqn:correctedccps2} is useful for an automated tool that
-would verify that an `N' simultaneous failures model had complete failure mode coverage.
+would verify that a single or double simultaneous failures model has complete failure mode coverage.
 By knowing how many test cases should be covered, and checking the cardinality
 associated with the test cases, complete coverage would be verified.
 
+\paragraph{N Venn disallowed combinations}
+The general case of equation \ref{eqn:correctedccps2}, involves not just dis-allowing pairs
+of failure modes within components, but also ensuring that combinations across components
+do not involve any pairs of failure modes within the same component.
+A recursive algorithm and proof is described in appendix \ref{chap:vennccps}.
+ 
 \paragraph{Practicality}
 Functional Group may consist, typically of four or five components, which typically
 have two or three failure modes each. Taking a worst case of mutiplying these

component_failure_modes_definition/unitary_state.bc

@@ -12,9 +12,12 @@ define f(x) {
 # determine how many combinations would be dis-allowed
 # from a cardinality constrained powerset
 # given unitary state failure mode conditions
-define uc(k,c,x) {
+define uc(c,k,x) {
  aa = 0;
- for(i=2; i<=k; i++)  aa += c * f(c)/(f(i)*f(c-i));
+ #for(i=2; i<=c; i++)  aa += k * f(k)/(f(i)*f(k-i));
+ if ( c>2 ) {
+	return aa + uc(c-1,k,x);
+ }
  return aa;
 }
 
@@ -22,19 +25,19 @@ define uc(k,c,x) {
 # how many combinations of cardinality k
 # can we have from c number of components
 # with x number of failure modes
-define ccps(k,c,x) {
+define ccps(c,k,x) {
- return f(c*x)/(f(k)*f(c*x-k))
+ return f(k*x)/(f(c)*f(k*x-c))
 }
 
 
 
-define us(k,c,x) {
+define us(c,k,x) {
 	a=0;
-	for(i=1;i<=k;i++) a += ccps(i,c,x);
+	for(i=1;i<=c;i++) a += ccps(i,c,x);
 	# a now holds all combinations 
 	# we must now subtract those combinations
 	# dis-allowed under unitary state conditions.
-	a -= uc(k,c,x);
+	a -= uc(cc,c,x);
 	return a;
 }
 

pt100/pt100.tex

@@ -634,15 +634,15 @@ reproduced below to verify this.
 }
 \begin{equation}
  |{\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)!}}  }
+- {{\sum^{j}_{j \in J}   \frac{|FM({C_j})|!}{2!(|FM({C_j})| - 2)!}}  }
  \label{eqn:correctedccps2}
 \end{equation}
 
 }
 {
 \begin{equation}
- |{\mathcal{P}_{cc}SU}| = {\sum^{k}_{1..cc}  \frac{|{SU}|!}{k!(|{SU}| - k)!}}
+ |{\mathcal{P}_{cc}SU}| = {\sum^{cc}_{k=1}  \frac{|{SU}|!}{k!(|{SU}| - k)!}}
-- \sum^{p}_{2..cc}{{\sum^{j}_{j \in J}   \frac{|FM({C_j})|!}{p!(|FM({C_j})| - p)!}}  }
+- {{\sum^{j}_{j \in J}   \frac{|FM({C_j})|!}{2!(|FM({C_j})| - 2)!}}  }
  %\label{eqn:correctedccps2}
 \end{equation}
 }
@@ -658,7 +658,7 @@ Populating this equation with $|SU| = 6$ and $|FM(C_j)|$ = 2.
 
 \begin{equation}
  |{\mathcal{P}_{2}SU}| = {\sum^{k}_{1..2}  \frac{6!}{k!(6 - k)!}}   
-- \sum^{p}_{2..2}{{\sum^{j}_{1..3}   \frac{2!}{p!(2 - p)!}}  }
+- {{\sum^{j}_{1..3}   \frac{2!}{p!(2 - p)!}}  }
  %\label{eqn:correctedccps2}
 \end{equation}