From: Rana Bilal (rab_91@hotmail.com)
Date: Mon Oct 08 2007 - 17:16:35 ART
Folks-
Thanks for all the replies. I'm including some thing interesting here. This
formula is good but I think what really matters is the time and CIR. There
are also some calculations at the bottom.
FRF.12 Frame Relay Fragmentation
FRF.12 allows the mixing of real-time and non-real-time data (e.g. FRF.3.1)
on the same DLCI via the use of fragmentation techniques that can be used by
other FRF implementations and protocols. Before FRF.12 came along, the only
way of performing fragmentation and interleaving was to reduce the MTU size.
The problem with this is that protocols such as IP are reliant upon the
fragmentation bit being set. If it is not then the packet is dropped.
Relying on endpoints to reassemble datagrams adds to the delay. IP MTU size
reduction must be used if the service provider is converting from Frame
Relay to ATM. This because there is no FRF.12 partner in the ATM cloud and
ATM does not support interleaving cells from different packets.
With FRF.12, when a packet is fragmented, a fragmentation header is inserted
that keeps a note of the sequence numbers so that re-assembly can occur at
the other end. If packets arrive out of sequence, they are dropped (unlike
MLP). Fragmenting large frames allows the end stations to interleave
smaller, delay sensitive frames such as encoded voice. This mitigates
against delay and jitter (delay variation). When frames are small, the
chances of Serialisation Delay is reduced i.e. intermediate devices do not
have to wait to receive a large data frame before forwarding it on. The key
here is to set a fragment size that is not too small so as to fragment the
delay sensitive packets. By doing this, delay sensitive packets will not
have a fragmentation header. FRF.12 is ideal for VoIP environments. Below is
a table of link speeds and the delay that the link will add depending on the
fragment size in bytes:
10ms 20ms 30ms 40ms 50ms 100ms 200ms
56kbps 70 140 210 280 350 700 1400
64kbps 80 160 240 320 400 800 n/a
128kbps 160 320 480 640 800 n/a n/a
256kbps 320 640 960 1280 n/a n/a n/a
512kbps 640 1280 n/a n/a n/a n/a n/a
768kbps 1000 n/a n/a n/a n/a n/a n/a
FORMULA:
Frag: 64000* 0.01/8= 80 bytes ----- when the TC is minimum -- 10ms
Frag: 64000* 0.125/8= 1000 bytes -- when the TC is maximum -- 125ms
Frag: CIR * tc/8=
Thanks,
Bilal
>From: Darren Johnson <dazza_johnson@yahoo.co.uk>
>Reply-To: Darren Johnson <dazza_johnson@yahoo.co.uk>
>To: lalit gupta <lalit.tech@gmail.com>, Rana Bilal <rab_91@hotmail.com>
>CC: ccielab@groupstudy.com
>Subject: Re: Frame relay fragmentation.................FRF12
>Date: Sun, 7 Oct 2007 10:40:19 +0100 (BST)
>
>Yeah, this formula works well.
>
>I use 1.25 * speed (in kbps) = x bytes
>
>For example, for a 64kbps line it would be:
>
>1.25 * 64 = 80 bytes....
>
>That or use the command reference where it lists
>recommended fragment sizes depending on speeds...
>
>
>Daz
>
>--- lalit gupta <lalit.tech@gmail.com> wrote:
>
> > HI Bilal,
> >
> >
> >
> > try this Fragment bytes = (0.01*cir)/8
> >
> > rgrds
> > lalit
> >
> > On 10/6/07, Rana Bilal <rab_91@hotmail.com> wrote:
> > >
> > > Folk-
> > >
> > > I'm having some trouble unsderstanding the frame
> > relay fragmentation,
> > > specially the formula to calculate the
> > > fragmentation.....................any
> > > thoughts or ideas will be appreciated....
> > >
> > >
> > > Bilal
> > >
> > >
> >
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