[mrtg] Re: .log file values for max in/out on the 5 min intervals

[Alex van den Bogaerdt]

Alex De Kloet wrote:

> I did the following test yesterday afternoon,
> i have taken the counter values from the first
> line of the log file, waited 5 minutes and took them again,
> substracted them and devided them by 300 sec. That gave me
> the exact avr. in en out values of the 2nd line in the .log
> file, while the max values where not right.

If you don't update at *exactly* a multiple of 5 minutes, an interval
is built from two updates (it could even be three).  Each of these
updates is considered in the max-calculation

Tobi once told he considers this a design bug.

> Can somebody explain how the 5 min max in/out values are made ?

10:01 to 10:06:  rate 100
10:06 to 10:11:  rate 200

after normalizing:
10:05 to 10:10:  avg rate (1/5*100+4/5*200), max rate 200

> I am investigating which figures to use to calculate the total uasage
> of the day by adding 288 (5 min interval * 300) values.

Keep it simple.  One day is 86400/300 intervals.  Create an image
with a width of 288 pixels at midnight.  The average displayed at
the bottom is the daily average.  Total throughput is avg * time.

> best regards
> 
> Alex
> The Netherlands

Best regards,
Alex,
The Netherlands.

:)

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