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making it clearer

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Robin 2010-06-05 07:36:35 +01:00
parent 185f7958fd
commit 2a5ba2d3d5
1 changed files with 8 additions and 2 deletions
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component_failure_modes_definition/component_failure_modes_definition.tex
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@ -375,6 +375,8 @@ be less.
What must actually be done is to subtract the number of component `internal combinations' What must actually be done is to subtract the number of component `internal combinations'
from the cardinality constrain powerset number. from the cardinality constrain powerset number.
\subsubsection{Example: Two Component functional group \\ cardinality Constraint of 2}
Thus were we to have a simple functional group with two components R and T, of which Thus were we to have a simple functional group with two components R and T, of which
$$FM(R) = \{R_o, R_s\}$$ and $$FM(T) = \{T_o, T_s, T_h\}$$. $$FM(R) = \{R_o, R_s\}$$ and $$FM(T) = \{T_o, T_s, T_h\}$$.
@ -426,6 +428,10 @@ And by inspection
$$ |\mathcal{P}_{2}(FG_cfg)| = 11 $$ $$ |\mathcal{P}_{2}(FG_cfg)| = 11 $$
\subsubsection{Establishing Formulae for unitary state failure mode \\
cardinality calculation}
The cardinality constrained powerset equation \ref{eqn:ccps} corrected for The cardinality constrained powerset equation \ref{eqn:ccps} corrected for
unitary state failure modes can be unitary state failure modes can be
written as a general formula (see equation \ref{eqn:correctedccps}), where C is a set of the components (indexed by j where J written as a general formula (see equation \ref{eqn:correctedccps}), where C is a set of the components (indexed by j where J
@ -446,10 +452,10 @@ Expanding the combination in equation \ref{eqn:correctedccps}
\label{eqn:correctedccps2} \label{eqn:correctedccps2}
\end{equation} \end{equation}
The equation \ref{eqn:correctedccps2} is now useful for an automated tool that Equation \ref{eqn:correctedccps2} is useful for an automated tool that
would verify that a `N' simultaneous failures model had been completly covered. would verify that a `N' simultaneous failures model had been completly covered.
By knowing how many test case should be covered, and checking the cardinality By knowing how many test case should be covered, and checking the cardinality
associated with the test cases complete coverage could be confirmed. associated with the test cases complete coverage would be confirmed.
%$$ \#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \big[ \frac{\#S!}{k!(\#S-k)!} - \sum_{j} (\#C_{j} \choose cc \big] $$ %$$ \#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \big[ \frac{\#S!}{k!(\#S-k)!} - \sum_{j} (\#C_{j} \choose cc \big] $$