[libreoffice-users] Re: Missing function: Bankers Rounding

plino wrote:
> I had never heard of bankers rounding before. Interesting concept.
>
> There is no such function in OOo/LO, Excel or Gnumeric...
>
> But you can easily create a function
>
> =IF(A1-INT(A1)=0.5;IF(ISEVEN(INT(A1));A1-0.5;A1+0.5);ROUND(A1))
>
> (Adjust if your separator is a comma instead of a  semi-colon)
>
> What this does is check if the fractional part is 0.5 and adds 0.5 to the
> number if the integer is odd and subtracts if it is even. If  it is not
> 0.5 then it uses the regular Round() function ;)
>
> Hope this helps!

Due to shortcomings in LibreOffice Calc, I had to adjust your formula as
follows:

=IF(ROUND(A1-INT(A1),8)=0.5,IF(MOD(A1-0.5,2),A1+0.5,A1-0.5),ROUND(A1))

and for the more usual case of needing to do a Bankers Round to a penny and
not a dollar, I am actually using this:

=IF(ROUND(100*A1-INT(100*A1),8)=0.5,IF(MOD(100*A1-0.5,2),A1+0.005,A1-0.005),ROUND(A1,2))

The reason I needed to use ROUND... in the test is that LibraOffice Calc
seems to often come up with numbers like .499999999... when evaluating
A1-INT(A1) instead of .5, causing the test to fail when it should work.  

The reason I needed to use MOD... instead of ISEVEN(INT... is that after
closing and opening the file, anywhere Calc needed to evaluate
IF(...IF(ISEVEN(INT... it gave a "#MACRO?" error.  Maybe LibraOffice can't
handle nesting functions that far.  By using MOD, it doesn't nest as far. 
Also, MOD(x,2) returns "0" for even and "1" for odd, so it's really
replacing ISODD..., and so I had to add .5 instead of subtract .5 when
MOD... is "true".

And of course, the reason for multiplying by 100 is to use the same idea for
pennies instead of dollars.

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