Dear McGimpsey,
Sorry, If you use the ROUND(SUM(A2:A4),0), then the same syntax will be
effective.
Thanks,
Selvarathinam.
Peo Sjoblom - 13 Jan 2006 17:28 GMT
So how would that work if you had other decimal values as well, are you
implying that you would round all values to the nearest integer?
what if the values were in A2:A6 with 1 in A1
A2: =ROUND(A1/3,2)
A3: =ROUND(A1/3,2)
A4: =ROUND(A1/3,2)
A5: =ROUND(A1/4,2)
A6: =ROUND(A1/4,2)
using your formula
=ROUND(SUM(A2:A6),0)
it would return 1, that would make the error larger

Signature
Regards,
Peo Sjoblom
> Dear McGimpsey,
>
[quoted text clipped - 3 lines]
> Thanks,
> Selvarathinam.
JE McGimpsey - 13 Jan 2006 17:36 GMT
That's *only* effective if you're dealing in whole dollars. Consider:
A1: $1.03
A2: =ROUND(A1/3, 2) ==> $0.34
A3: =ROUND(A1/3, 2) ==> $0.34
A4: =ROUND(A1/3, 2) ==> $0.34
A5: =ROUND(SUM(A2:A4),0) ==> $1.00
A6: =SUM(A2:A4) ==> $1.02
Simply using ROUND() cannot guarantee that values will balance.
There are many techniques that can be used to "correct" models, but they
need to be evaluated for the particular approach that the model takes.
A better approach might be
A4: =A1-SUM(A2:A3) ==> $0.34
but determining which cell in A2:A4 should get the extra penny is not
always self-evident. And if there were 60 dividends instead of 3, the
remainders could give a very disproportionate answer. For instance:
A1: $1.03
A2: =ROUND(A1/70, 2) ==> $0.01
..
A70: =ROUND(A1/70, 2) ==> $0.01
A71: =A1-SUM(A2:A70) ==> $0.34
> Sorry, If you use the ROUND(SUM(A2:A4),0), then the same syntax will be
> effective.
Selvarathinam - 13 Jan 2006 18:16 GMT
Oops......
Sorry I got it wrong.
Anyway thanks for correcting me.
Regards,
Selvarathinam.
JE McGimpsey - 13 Jan 2006 19:56 GMT
Not a problem - keep posting. I, for one, generally learn better when
I'm wrong in public. Tends to stick a bit more...<g>
> Oops......
>
> Sorry I got it wrong.