>The last value less than or equal to zero will always win as best_tin.
>If all tdiff values are positive then best_time will end up as the lowest 
>value however.
I agree with xnoreq. The algorithm as it is will not always return the least 
overdue tin. It returns either the lowest positive tdiff (if all tdiff values 
are positive), or the lowest negative one (if not all tdiff values are 
positive). This could be accomplished just with "if (tdiff<=best_time)" as 
xnoreq pointed out.
George
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