cardinality constrained powerset, modification for unitary state failure mode sets
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fmmdset/Makefile
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fmmdset/Makefile
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#
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# Make the propositional logic diagram a paper
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#
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paper: paper.tex fmmdset_p.tex
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pdflatex paper.tex
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#dvipdf paper
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okular paper.pdf
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# Remove the need for referncing graphics in subdirectories
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#
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fmmdset_p.tex: fmmdset.tex
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cat fmmdset.tex | sed 's/fmmdset\///' > fmmdset_p.tex
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@ -878,7 +878,45 @@ from $1$ to $cc$ thus
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% $$ {\sum}_{k = 1..cc} {\#S \choose k} = \frac{\#S!}{k!(\#S-k)!} $$
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%
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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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\subsection{Actual Number of combinations to check with Unitary State Fault mode sets}
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Where all components analysed only have one fault mode, the cardinality constrained powerset
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calculation give the correct number of test case combinations to check.
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Because set of failure modes is constrained to be unitary state, the acual number will
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be less.
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What must actually be done is to subtract the number of component `internal combinations'
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from the cardinality constrain powerset number.
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Thus were we to have a simple circuit 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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For a cardinality constrained powerset of 2, because there are 5 error modes
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gives $\frac{5!}/{1!(5-1)!} + \frac{5!}{2!(5-2)!} = 15$. OK
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5 single fault modes, and ${2 \choose 5}$ ten double fault modes.
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However we know that the faults are mutually exclusive for a component.
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We must then subtract the number of `internal' component fault combinations.
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For component R there is only one internal component fault that cannot exist
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$R_o \wedge R_s$. As a combination ${2 \choose 2} = 1$ . For $T$ the component with
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three fault modes ${2 \choose 3} = 3$.
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Thus for $cc == 2$ we must subtract $(3+1)$.
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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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is the set of componets under analyis) and $\#C$
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indicates the number of mutually exclusive fault modes the compoent has:-
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%$$ \#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \frac{\#S!}{k!(\#S-k)!} $$
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$$ \#\mathcal{P}_{cc} S = {\sum^{k}_{1..cc} \frac{\#S!}{k!(\#S-k)!}} - {\sum^{j}_{j \in J} {\#C_{j} \choose cc}} $$
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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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\section{Future Ideas}
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\usepackage{fancyhdr}
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\usepackage{tikz}
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\usepackage{amsfonts,amsmath}
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\input{style}
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\input{../style}
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\begin{document}
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\pagestyle{fancy}
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\renewcommand{\footnoterule}{{\small Notes:}}
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%% Robin 01AUG2008
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%%
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%
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