said by dsiemon:let me clarify that I'm trying to figure out the number of bytes added to an IP packet.
Well then, let's go question-by-question:
said by dsiemon:More questions:
- Does payload rate include 802.3 overhead or just the encoding on the line? I'm guessing the later based on some experiments I've done.
It includes the full 802.3 overhead;
payload in this case is
IP packets in 802.3 frames.
said by dsiemon:- Is the full 802.3 header transmitted over the VDSL link?
- 7 bytes preamble
- 1 byte start of frame
- 6 byte D MAC
- 6 byte S MAC
- 2 byte Ethertype
- 2 byte 802.1Q tag it's 4 and always will be
- 4 byte FCS
- 12 byte Interframe gap
Total: 40 bytes
+ 8 for PPPoE gives me a total per packet overhead of 48 bytes.
You're comparing apples to oranges. 8 bytes for PPPoE gives you a total packet overhead of 8 bytes. That's it. Nothing more. At all. Period. There are 40 bits added to the
frame, but you also seem to be under the impression that the frame is subject to the IP MTU which it is not. Wireline frames don't care about what protocol is running atop them.
This means your 1492 + 8 byte IP packet becomes (1500+7+1+6+6+2+4+12=) 1538 bytes on the wire.
So if you're looking for
per-packet overhead, it's 8 bytes, just like I said in the beginning. The rate of 1492-byte IP packets on the WAN is given to you by the payload rate of your modem. When you're doing calculations for shaping,
ignore 802.3 overhead (as most guides from commercial appliances will tell you). Shaping is done a layers higher than data link, acknowledging that framing will always be there and doesn't actually add anything to the packet. Nothing useful comes from caring about 802.3 overhead (not even for small packets).
I talked to the person who managed our massive Sandvine boxes here, and he said that you're wasting your time worrying about 802.3 stuff for small-packet performance; he says to "just treat them like any other packet because that's what every other device does."
HTH.