From 900b58d7ff5e3614486de238bb876a285d494994 Mon Sep 17 00:00:00 2001 From: Robin Clark Date: Sat, 17 Aug 2013 13:00:58 +0100 Subject: [PATCH] went for a ride on the downs.... --- submission_thesis/CH7_Evaluation/copy.tex | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/submission_thesis/CH7_Evaluation/copy.tex b/submission_thesis/CH7_Evaluation/copy.tex index 3549a83..be3acf1 100644 --- a/submission_thesis/CH7_Evaluation/copy.tex +++ b/submission_thesis/CH7_Evaluation/copy.tex @@ -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 \} \} $$