You're right. Committed the simplification.
On Tue, Jul 13, 2010 at 7:05 PM, Masood Mortazavi
wrote:
> In FailureDetector, in ArrivalWindow, in "double p(double t)",
>
> is
>
> return 1 - ( 1 - Math.pow(Math.E, exponent) );
>
> really needed, instead of
>
> return Math.pow(Math.E,
In FailureDetector, in ArrivalWindow, in "double p(double t)",
is
return 1 - ( 1 - Math.pow(Math.E, exponent) );
really needed, instead of
return Math.pow(Math.E, exponent) ;
I believe the integral of the the Exponential Distribution from "t" to
infinity leads to the latter val
