Re: eval divide and drop decimals.

[Booth Martin <booth@xxxxxxxxxxxx>]

Here is the code I tested and seems to work:

     if        wIncrement <> *zero
     eval      wAmt = wFee *
                  %div(%int(Amt1): %int(wIncrement))
     if        %rem(%int(Amt1): %int(wIncrement)) <> 0
     eval      wAmt = wAmt + wFee
     endif
     endif

Joe, %div & %rem require a whole number, with zero decimal places if I 
understand the manual correctly.  Hence my use of the %int bif, as was 
pointed out originally by several brighter-than-me people.

The idea of using the delta, 1 in this instance, seems to me like it'd 
work but I haven't tested it.



Joe Pluta wrote:
> Jon, this gives the wrong answer for the boundary condition of $500.  In
> that case, you will get eight dollars rather than four.  There is an
> absolute requirement in this case to use the non-zero "delta" figure, as in:
>
>   Fee = 4 * (1 + %div( (650 - delta) : 500))
>
> Personally, I'm not sure the %div will accept that notation, and I don't
> have the time to check it, but the delta is absolutely required.  The
> algorithm I posted earlier using %int was correct, it just used the wrong
> base step (I simply left out the "1 +" portion of your code above).
>
> Thank you
>
> Joe
>
>
> > From: Jon Paris
> >
> >   fee = 4 * (1 + %Div(650: 500))
> >
> > As others have pointed out the algorithm is also incorrect.