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went for a ride on the downs....

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Robin Clark 2013-08-17 13:00:58 +01:00
parent 56ec258b72
commit 900b58d7ff
1 changed files with 3 additions and 3 deletions
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6
submission_thesis/CH7_Evaluation/copy.tex
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@ -987,7 +987,7 @@ is as we would use them for single failure analysis.
For a cardinality constrained powerset of 2, because there are 5 error modes ( $|fm(FG)|=5$),
applying equation \ref{eqn:ccps} gives:
$$ | P_2 (fm(FG)) | = \frac{5!}{1!(5-1)!} + \frac{5!}{2!(5-2)!} = 15.$$
$$ | P_{\le 2} (fm(FG)) | = \frac{5!}{1!(5-1)!} + \frac{5!}{2!(5-2)!} = 15.$$
This is composed of ${5 \choose 1}$
five single fault modes, and ${5 \choose 2}$ ten double fault modes.
@ -1002,13 +1002,13 @@ $R_o \wedge R_s$. As a combination ${2 \choose 2} = 1$. For the component $T$ wh
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, $|\mathcal{P}_{2}(fm(FG))| = 11$, for this example and this can be verified
The number of combinations to check is thus 11, $|\mathcal{P}_{\le 2}(fm(FG))| = 11$, for this example and this can be verified
by listing all the required combinations:
% Because there are only two components, this is simply the cross product
% of fm(R) and fm(T) but this does not hold for larger {\fgs}...
$$ \mathcal{P}_{2}(fm(FG)) = \{
$$ \mathcal{P}_{\le 2}(fm(FG)) = \{
\{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 \}
\}
$$