That's great, but how do you then get that many packets to disk so that you can do something with them?
Presumably you need flash drives, and probably an append-only filesystem?
Presumably you need flash drives, and probably an append-only filesystem?
40G is a little bit more difficult: you need a huge RAID (simple, direct scaling: 32-40 disks) with mechanical disks to achieve the necessary bandwidth, and if you want to use SSDs you will need a lot of them too in order to have enough space to save any meaningful amount of traffic.
I remember seeing papers on on-the-fly compression for network traffic, but IIRC the results were not very impressive and the performance cost was noticeable.
it may be possible to do disk i/o at that high rate e.g. with pci-e or a dedicated appliance for dumping the entire stream. but you would running out of storage pretty fast.
for example, a quick back-of-the-envelope calculation, where you dump packet stream from 4x10gbps cards with minimal 84b size (on ethernet), show that you would exhaust the storage in approx. 4.5 minutes :)
I don't understand your relations here, or the size of the storage you're considering. I did this:
3.5 million packets per second * 84 bytes per packet * 4 interfaces == 101.606 terabytes/day.
Or, at the raw interface rate:
10 Gb/s * 4 interfaces == 432 terabytes/day.
whoopsie, i was considering 1.2tb of storage...
> 3.5 million packets per second
it is actually, 14.88 Mpps.
4 Gigabytes per second times 86400 seconds per day is 345,600 Gigabytes per day.
Roughly: 345 Terabytes per day.
Large, but not stupidly so.
it should be 5GB/s right ? :)
At the lower speeds, things like 8b/10b encoding and Reed-Solomon ECC added enough overhead that dividing by 10 was more accurate than dividing by 8.
on 10gige pipes, at max Ethernet mtu (1500) bytes etc, there is approx. 94% of available bandwidth for user data (accounting for things like inter-frame-gap, crc checksums etc). with jumbo-frames that number goes to 99%.
That would mean that I would need to divide by about roughly nine (8.8 or so).
Sorry, I can't do divide-by-nine quickly in my head. I can do divide by 10 though, and my error is roughly 10%.