MTTF expressions for K = 1

[cme@ellisun.sw.stratus.com (Carl Ellison)]

Sorry -- I left out the K=1 expressions from that paper.
They're a little more mundane (for Stratus) but of obvious interest here.
These are the most likely to be interesting to us.

Consider, for example, \lambda = 0.5 failures/year and
\mu = 52 repairs/year
for the kind of machines we're talking about (as opposed to Stratus
machines).  The resulting MTTF would be in years.

 - Carl


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N =  2 ;  K =  1

\begin{equation}
\frac { 3\lambda + 1\mu }{ 2\lambda^2 }
\end{equation}


N =  3 ;  K =  1

\begin{equation}
\frac { 11\lambda^2 + 4\lambda\mu + 1\mu^2 }{ 6\lambda^3 }
\end{equation}


N =  4 ;  K =  1

\begin{equation}
\frac { 50\lambda^3 + 18\lambda^2\mu + 5\lambda\mu^2 + 1\mu^3 }{ 24\lambda^4 }
\end{equation}