[j-nsp] M20 FPC throughput

Richard A Steenbergen ras at e-gerbil.net
Mon Aug 15 04:27:49 EDT 2005

On Mon, Aug 15, 2005 at 10:21:38AM +0300, Saku Ytti wrote:
> On (2005-08-15 06:33 +0000), The Drifter wrote:
> > What is the forwarding capacity of a FPC in a M20?
> > Is it 5Gbps, or 3.2 Gbps or 2.5 Gbps or 2.0 Gbps and any supporting 
> > information/link will be indeed helpful
>  It's exactly 4Gbps (32bit*125MHz). But J-CELL's overhead is "16/80 == 20%".
> So that boils down to 4*0.2=0.8Gbps lost to J-CELLs, so 3.2 left for
> packets, and this is packets, not frames, as L2 headers have been stripped
> by now.

With the note that, like all cell technology, one byte extra on the packet 
pushes you over into the next cell. That means it is probably safer to 
think of it in jcells/sec, instead of bandwidth/sec.

So, if my math is right, you get 6.25M jcells/sec (32000000 bytes/sec / 
512 bytes/sec/jcell) per direction per FPC. A 65 byte packet (the worst 
case packet for bandwidth to jcell ratio) will burn 2 jcells, meaning that 
3.125M packets/sec at 65 bytes will fill an FPC.

A 65 byte packet weighs in at 103 bytes after ethernet overhead, (8 bytes 
preamble+sfd, 14 bytes for the eth header, 65 bytes for the payload, 4 
bytes for the fcs, 12 bytes for the ifg), which means that you can stuff 
1000Mbps / 824 bps/packet = 1.21Mpps of 65 byte packets down a GigE at 
line rate.

So the big answer at the end is that it takes 2.575 GigE's sending 65 byte 
packets at their line rate of 1.21Mpps each to exhaust an FPC. Feel free 
to double check my math on that, it's late, but I'm pretty sure thats 
right. Of course that is a non-realistic situation which you would only 
ever encounter on a DoS attack specifically targetting Junipers, but it 
helps to know what the worst case really is.

Richard A Steenbergen <ras at e-gerbil.net>       http://www.e-gerbil.net/ras
GPG Key ID: 0xF8B12CBC (7535 7F59 8204 ED1F CC1C 53AF 4C41 5ECA F8B1 2CBC)

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