expanded final equation and gave reason why it is useful
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@ -358,7 +358,7 @@ from $1$ to $cc$ thus
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%
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\begin{equation}
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\begin{equation}
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\#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \frac{\#S!}{k!(\#S-k)!}
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|{\mathcal{P}_{cc}S}| = \sum^{k}_{1..cc} \frac{|{S}|!}{ k! ( |{S}| - k)!}
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\label{eqn:ccps}
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\label{eqn:ccps}
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\end{equation}
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\end{equation}
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@ -377,7 +377,11 @@ from the cardinality constrain powerset number.
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Thus were we to have a simple functional group with two components R and T, of which
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Thus were we to have a simple functional group with two components R and T, of which
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$$FM(R) = \{R_o, R_s\}$$ and $$FM(T) = \{T_o, T_s, T_h\}$$.
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$$FM(R) = \{R_o, R_s\}$$ and $$FM(T) = \{T_o, T_s, T_h\}$$.
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For a cardinality constrained powerset of 2, because there are 5 error modes
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This means that a functional~group $FG=\{R,T\}$ will have a component failure modes set % $FM_{cfg} $
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of $FM_{cfg} = \{R_o, R_s, T_o, T_s, T_h\}$
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For a cardinality constrained powerset of 2, because there are 5 error modes ( $|{FG_{cfg}}|=5$),
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applying equation \ref{eqn:ccps} gives :-
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applying equation \ref{eqn:ccps} gives :-
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$$\frac{5!}{1!(5-1)!} + \frac{5!}{2!(5-2)!} = 15$$
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$$\frac{5!}{1!(5-1)!} + \frac{5!}{2!(5-2)!} = 15$$
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@ -392,41 +396,60 @@ $R_o \wedge R_s$. As a combination ${2 \choose 2} = 1$ . For $T$ the component w
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Thus for $cc == 2$ we must subtract $(3+1)$.
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Thus for $cc == 2$ we must subtract $(3+1)$.
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The number of combinations to check is thus 11 for this example and this can be verified
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The number of combinations to check is thus 11 for this example and this can be verified
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by listing all the required combinations:
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by listing all the required combinations:
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%
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%\vbox{
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%\subsubsection{All Eleven Cardinality Constrained \\ Powerset of 2 Elements Listed}
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%%\tiny
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%\begin{enumerate}
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%\item $\{R_o T_o\}$
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%\item $\{R_o T_s\}$
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%\item $\{R_o T_h\}$
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%\item $\{R_s T_o\}$
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%\item $\{R_s T_s\}$
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%\item $\{R_s T_h\}$
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%\item $\{R_o \}$
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%\item $\{R_s \}$
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%\item $\{T_o \}$
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%\item $\{T_s \}$
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%\item $\{T_h \}$
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%\end{enumerate}
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%%\normalsize
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%}
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%
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\vbox{
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$$ \mathcal{P}_{2}(FG_cfg) = \{
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\subsubsection{All Eleven Cardinality Constrained \\ Powerset of 2 Elements Listed}
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\{R_o T_o\}, \{R_o T_s\}, \{R_o T_h\}, \{R_s T_o\}, \{R_s T_s\}, \{R_s T_h\}, \{R_o \}, \{R_s \}, \{T_o \}, \{T_s \}, \{T_h \}
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%\tiny
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\}
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\begin{enumerate}
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$$
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\item $\{R_o T_o\}$
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\item $\{R_o T_s\}$
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\item $\{R_o T_h\}$
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\item $\{R_s T_o\}$
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\item $\{R_s T_s\}$
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\item $\{R_s T_h\}$
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\item $\{R_o \}$
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\item $\{R_s \}$
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\item $\{T_o \}$
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\item $\{T_s \}$
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\item $\{T_h \}$
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\end{enumerate}
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%\normalsize
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}
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And by inspection
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$$ |\mathcal{P}_{2}(FG_cfg)| = 11 $$
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The cardinality constrained powerset equation \ref{eqn:ccps} corrected for
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The cardinality constrained powerset equation \ref{eqn:ccps} corrected for
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unitary state failure modes can be
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unitary state failure modes can be
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written as a general formula, where C is a set of the components (indexed by j where J
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written as a general formula (see equation \ref{eqn:correctedccps}), where C is a set of the components (indexed by j where J
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is the set of components in the functional~group under analyis) and $\#C$
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is the set of components in the functional~group under analyis) and $|{C}|$
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indicates the number of mutually exclusive fault modes each component has:-
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indicates the number of mutually exclusive fault modes each component has:-
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%$$ \#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \frac{\#S!}{k!(\#S-k)!} $$
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%$$ \#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \frac{\#S!}{k!(\#S-k)!} $$
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\begin{equation}
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\begin{equation}
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\#\mathcal{P}_{cc} S = {\sum^{k}_{1..cc} \frac{\#S!}{k!(\#S-k)!}} - {\sum^{j}_{j \in J} {\#C_{j} \choose cc}}
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|{\mathcal{P}_{cc}S}| = {\sum^{k}_{1..cc} \frac{|{S}|!}{k!(|{S}| - k)!}} - {\sum^{j}_{j \in J} {|{C_{j}}| \choose cc}}
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\label{eqn:correctedccps}
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\label{eqn:correctedccps}
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\end{equation}
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\end{equation}
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Expanding the combination in equation \ref{eqn:correctedccps}
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\begin{equation}
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|{\mathcal{P}_{cc}S}| = {\sum^{k}_{1..cc} \frac{|{S}|!}{k!(|{S}| - k)!}} - {\sum^{j}_{j \in J} \frac{|{C_j}|!}{cc!(|{C_j}| - cc)!}}
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\label{eqn:correctedccps2}
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\end{equation}
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The equation \ref{eqn:correctedccps2} is now useful for an automated tool that
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would verify that a `N' simultaneous failures model had been completly covered
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by knowing how many test case should be covered.
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%$$ \#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \big[ \frac{\#S!}{k!(\#S-k)!} - \sum_{j} (\#C_{j} \choose cc \big] $$
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%$$ \#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \big[ \frac{\#S!}{k!(\#S-k)!} - \sum_{j} (\#C_{j} \choose cc \big] $$
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@ -57,6 +57,7 @@ depth -0.5ex\hfill}\newcommand{\innerhead}[1]{\def\lp@innerhead{#1}}
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\newcommand{\mins}[1]{$#1^{\scriptsize\prime}$} % Minutes symbol
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\newcommand{\mins}[1]{$#1^{\scriptsize\prime}$} % Minutes symbol
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\newcommand{\secs}[1]{$#1^{\scriptsize\prime\prime}$} % Seconds symbol
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\newcommand{\secs}[1]{$#1^{\scriptsize\prime\prime}$} % Seconds symbol
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\newcommand{\key}[1]{\fbox{\sc#1}} % Box for keys
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\newcommand{\key}[1]{\fbox{\sc#1}} % Box for keys
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\newcommand{\modulus}[1]{\ensuremathmode{|#1|}}
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\newcommand{\?}{\_\hspace{0.115em}} % Proper spacing for
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\newcommand{\?}{\_\hspace{0.115em}} % Proper spacing for
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% underscore
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% underscore
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\newcommand{\rev}{PA5}
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\newcommand{\rev}{PA5}
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