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making it clearer

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Robin 2010-06-05 07:41:07 +01:00
parent 2a5ba2d3d5
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component_failure_modes_definition/component_failure_modes_definition.tex
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@ -377,7 +377,7 @@ from the cardinality constrain powerset number.
\subsubsection{Example: Two Component functional group \\ cardinality Constraint of 2} \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 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 $$FM(T) = \{T_o, T_s, T_h\}$$. $$FM(R) = \{R_o, R_s\}$$ 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 % $FM_{cfg} $ This means that a functional~group $FG=\{R,T\}$ will have a component failure modes set % $FM_{cfg} $
@ -395,7 +395,7 @@ We must then subtract the number of `internal' component fault combinations for
For component R there is only one internal component fault that cannot exist 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 $T$ the component with
three fault modes ${2 \choose 3} = 3$. three fault modes ${2 \choose 3} = 3$.
Thus for $cc == 2$ we must subtract $(3+1)$. 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 for this example and this can be verified The number of combinations to check is thus 11 for this example and this can be verified
by listing all the required combinations: by listing all the required combinations:
% %