function will return the maximally possible value of p=100%.

Background:

 The maximum fval, f(100%), is approximately 244, i.e. the scaled value of fval
 should never exceed 244E6, which fits easily into u32. The problem is the scaling
 by 10^6, since additionally R(TT) is in microseconds.
 This is resolved by breaking the division into two stages: the first stage
 computes fval=(s*10^6)/R, stores that into u64; the second stage computes
 fval = (fval*10^6)/X_recv and complains if overflow is reached for u32.
 This case is safe since the TFRC reverse-lookup routine then returns p=100%.

Signed-off-by: Gerrit Renker <gerrit@erg.abdn.ac.uk>
Acked-by: Ian McDonald <ian.mcdonald@jandi.co.nz>
Signed-off-by: Arnaldo Carvalho de Melo <acme@mandriva.com>
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