On Wed, Sep 19, 2018 at 2:11 AM Toke Høiland-Jørgensen <t...@toke.dk> wrote:
>
> Ruben <ru...@vfn-nrw.de> writes:
>
> > Hey Toke,
> >
> > Am 13.09.2018 21:12 schrieb Toke Høiland-Jørgensen <t...@toke.dk>:
> >
> >
> > Ruben <ru...@vfn-nrw.de> writes:
> >
> > > Hey Toke,
> > >
> > > Thanks for your fast response!
> > >
> > > Am 13.09.2018 12:27 schrieb Toke Høiland-Jørgensen <t...@toke.dk>:
> > >
> > >
> > > Ruben <ru...@vfn-nrw.de> writes:
> > >
> > > > Hey guys,
> > > >
> > > > I've already mentioned this in a response to dtaht on GitHub, but
> > here
> > > > again for everyone:
> > > >
> > > > I was wondering if it's possible to extend the tin statistics by
> > > > packets for backlog.
> > >
> > > Why do you need packets when there's already bytes?
> > >
> > > Easy: dtaht requested a packets graph with ecn marks, which is also
> > > packets, so backlog as bytes do not fit, backlog as packets do.
> > >
> > > The idea was to do a multi-graph which is one graph with combined
> > > stats for all tins and sub graphs for all tins.
> > >
> > > On the main graph a backlog in packets is available, but I would need
> > > to leave out the backlog for the tins, which is somewhat confusing.
> >
> > Why not just do both backlogs in bytes?
> >
> > There's no counter for ecn marked packets in bytes, so it's impossible to
> > implement it that way, too.
> >
> > ECN-Marks is in packets, so everything else need to be in packets as
> > well.
>
> Hmm, so the obvious follow-up question would be "why do you need to have
> backlog and number of drops on the same graph?" :)
A reasonable approximation of backlog in packets is achievable by
dividing by 1000.
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