diff --git a/fmmdset/Makefile b/fmmdset/Makefile new file mode 100644 index 0000000..3a1aa33 --- /dev/null +++ b/fmmdset/Makefile @@ -0,0 +1,16 @@ + +# +# Make the propositional logic diagram a paper +# + + +paper: paper.tex fmmdset_p.tex + pdflatex paper.tex + #dvipdf paper + okular paper.pdf + + +# Remove the need for referncing graphics in subdirectories +# +fmmdset_p.tex: fmmdset.tex + cat fmmdset.tex | sed 's/fmmdset\///' > fmmdset_p.tex diff --git a/fmmdset/fmmdset.tex b/fmmdset/fmmdset.tex index 49eaef2..5455299 100644 --- a/fmmdset/fmmdset.tex +++ b/fmmdset/fmmdset.tex @@ -878,7 +878,45 @@ from $1$ to $cc$ thus % $$ {\sum}_{k = 1..cc} {\#S \choose k} = \frac{\#S!}{k!(\#S-k)!} $$ % -$$ \#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \frac{\#S!}{k!(\#S-k)!} $$ +$$ \#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \frac{\#S!}{k!(\#S-k)!} $$ + + + +\subsection{Actual Number of combinations to check with Unitary State Fault mode sets} + +Where all components analysed only have one fault mode, the cardinality constrained powerset +calculation give the correct number of test case combinations to check. +Because set of failure modes is constrained to be unitary state, the acual number will +be less. + + +What must actually be done is to subtract the number of component `internal combinations' +from the cardinality constrain powerset number. + +Thus were we to have a simple circuit with two components R and T, of which +$FM(R) = {R_o, R_s}$ and $FM(T) = {T_o, T_s, T_h}$. +For a cardinality constrained powerset of 2, because there are 5 error modes +gives $\frac{5!}/{1!(5-1)!} + \frac{5!}{2!(5-2)!} = 15$. OK +5 single fault modes, and ${2 \choose 5}$ ten double fault modes. +However we know that the faults are mutually exclusive for a component. +We must then subtract the number of `internal' component fault combinations. +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 + three fault modes ${2 \choose 3} = 3$. +Thus for $cc == 2$ we must subtract $(3+1)$. + +Written as a general formula, where C is a set of the components (indexed by j where J +is the set of componets under analyis) and $\#C$ +indicates the number of mutually exclusive fault modes the compoent has:- + +%$$ \#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \frac{\#S!}{k!(\#S-k)!} $$ + +$$ \#\mathcal{P}_{cc} S = {\sum^{k}_{1..cc} \frac{\#S!}{k!(\#S-k)!}} - {\sum^{j}_{j \in J} {\#C_{j} \choose cc}} $$ + + + +%$$ \#\mathcal{P}_{cc} S = \sum^{k}_{1..cc} \big[ \frac{\#S!}{k!(\#S-k)!} - \sum_{j} (\#C_{j} \choose cc \big] $$ + \section{Future Ideas} diff --git a/fmmdset/paper.tex b/fmmdset/paper.tex index 4125fa1..9668e7f 100644 --- a/fmmdset/paper.tex +++ b/fmmdset/paper.tex @@ -4,7 +4,7 @@ \usepackage{fancyhdr} \usepackage{tikz} \usepackage{amsfonts,amsmath} -\input{style} +\input{../style} \begin{document} \pagestyle{fancy} diff --git a/fmmdset/style.tex b/fmmdset/style.tex deleted file mode 100644 index a55fcfb..0000000 --- a/fmmdset/style.tex +++ /dev/null @@ -1,138 +0,0 @@ -% -%============= Definition of {asyoulikeit} page style ======================* -% -% Jonathan Burch This is the terse form - 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