# [mrtg] Re: Total of bytes

Alex van den Bogaerdt alex at ergens.op.HET.NET
Fri Aug 23 16:09:18 MEST 2002

```Eduardo Gargiulo wrote:

> My questions is how could I know the total of bytes transferred (In and
> Out of my LAN) based upon monthly average (14.2Kb/sec and 10.3Kb/sec)?
> Is it right to do the following ...
>
> 10.3Kb/sec * 3600 sec = 37080 Kb in an hour

Yup, it is.  (I didn't check the numbers; the method is OK).

> 37080 * 24 = 889920 Kb in a day

Indeed.  Or: 10.3Kb/sec * 86400 sec = 889920 Kb per day.

The averages you get are probably based on 400 samples of
2 hours each.  The total time span is thus 400 * 7200 = 2880000
seconds (33 days 8 hours).  If you happen to have work hours on
both the start and the end of this interval, the averages will
be slightly larger than normal.

> This number is too big (I think) so I'm looking for some technical
> explanation about my misunderstanding or error ;)

MRTG numbers usually are not too big.  In fact, due to rounding you
probably get numbers slightly *smaller* than reality.  But don't
forget what I wrote in the previous paragraph.

Are you sure you understand the difference between kilobits (Kb)
and kilobytes (KB)?  Maybe you think bytes while you get presented
bits?

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