RE: eval divide and drop decimals.

This is a perfectly fine algorithm.  In the syntax we've been using to date,
the actual answer for this class of problems is:

   int((x - delta) / y) * z

where delta is any sufficiently small number.  "Sufficiently small" simply
means smaller than the smallest unit of measurement, in this case a penny.
Thus this will ALWAYS work for any amount with a precision of .01 (or
greater):

   Fee = %int( (650 - .001) / 500) * 4.00

In the real world of applied mathematics, using your working set to
determine your boundary conditions is acceptable and in fact often required.
The correct algorithm is the one which gets you the correct results all the
time.

Joe

> From: Scott Klement
> 
> 
> >
> > fee = 4.00 * %int(1 + (650 / (500.00 + .01) )
> 
> With that code, $1000.01 will only be charged for 2 * 4.00 instead of the
> proper 3 * 4.00.   Please stop trying to fix a faulty algorithm by
> inserting different numbers into it.