Minimum MTU size on Ethernet and minimum Ethernet frame size

Question

I am a bit confused about the understanding of the relationship between minimum MTU of MTU and minimum Ethernet frame size.

We know that the maximum MTU of Ethernet is 1500 Bytes which makes the maximum frame size equal to 1518 Bytes (1500+18) or 1522 Bytes (1500+22).

We also know that the minimum MTU is 576 Bytes.

So confusing part and my question is, why isn't the minimum frame size 594 Bytes(576+18) or 598 Bytes (576+22)? And instead, the minimum frame size is 64 Bytes.

I feel I'm missing something here and that's why I don't understand the relation between these two.

Answer 1

You are confusing a few things.

The MTU - maximum transmission unit - is the largest IP packet an underlying protocol or a link can carry - the maximum IP PDU size.

For Ethernet, the largest standard frame is 1518 bytes. 18 bytes are frame overhead, so the largest payload (L2 SDU) is 1500 bytes, so that's the MTU (maximum L3 PDU) over Ethernet as well.

Ethernet's minimum frame size is 64 bytes, but that has no relation to the MTU. (IP packets smaller than 64-18=46 bytes simply require padding.)

576 bytes is the smallest unfragmented IPv4 packet size that an underlying protocol or link needs to support. Since Ethernet's maximum payload size is larger than 576 bytes, that is of now consequence here.

Answer 2

"We also know that the minimum MTU is 576 Bytes."

No, we do not know that because it is not true. I am not sure where you got such an idea.

The MTU is the maximum payload of the data-link protocol. In the case of ethernet, the IEEE standard says that is 1500 bytes. The minimum payload size is 46 bytes (42 bytes for tagged frames), but that is not a minimum MTU because the MTU is the maximum payload size, not the minimum. Remember that IP (whether IPv4 or IPv6) is not the only network-layer protocol, and ethernet is not the only data-link protocol in use for IP. Ethernet and IP were developed by different people for different reasons, and the standards are maintained by different groups, even today.

576 bytes is the minimum datagram size that hosts must be able to process for IPv4, and 1280 bytes for IPv6, but that is not the smallest MTU for ethernet. Most hosts can process IP datagrams as small as 28 bytes for IPv4 and UDP (IPv4 header of 20 bytes, UDP header of 8 bytes, and 0 data bytes) and 48 bytes for IPv6 (IPv6 header of 40 bytes, UDP header of 8 bytes and, 0 data bytes).

Answer 3

Ethernet as defined by IEEE 802.3 has no minimum MTU. However, there will be practical limits to the smallest frame one can send. (64bytes, which is a function of CSMA/CD at 100m.) It does specify a maximum of 1500 -- jumbo frames are not 802.3 compliant, and due to backwards compatibility, never will be.

The 576 number comes from a great deal of misunderstanding of IPv4. Specifically, RFC791 states that an IPv4 system must be prepared to handle a datagram of up to 576bytes, no matter how it manages to get there. (i.e. it could've been reassembled from a dozen ATM cells.) That's the largest datagram you can assume an IPv4 host can accept; any host that says it cannot handle packets that large is violating RFC791. One can, of course, send datagrams smaller than that, and even larger if such capability has been negotiated. (MSS, PMTUd, in-band, etc.) IP being layer-3, has nothing to do with the various layer-2's that may be carrying it.