What is the "in-the-wire" size of a ethernet frame? 1518 or 1542?
- by chrisapotek
According to the table here, it says that MTU = 1500 bytes and that the payload part is 1500 - 42 bytes or 1458 bytes (<- this is actually wrong!). Now on top of that you have to add IPv4 and UDP headers, which are 28 bytes (20 IP + 8 UDP). That leaves my maximum possible application message to as 1430 bytes! But by looking for this number in the Internet I see 1472 instead. Am I doing this calculation wrong here?
All I want to find out is the maximum application message I can send over the wire without the risk of fragmentation. It is definitely not 1500 because that includes the frame headers. Can someone help?
The confusion is the the PAYLOAD can actually be as large as 1500 bytes and that's the MTU. So now what is the size in-the-wire for a payload of 1500? From that table it can be as big as 1542 bytes.
So the maximum app messages I can send is 1472 (1500 - 20 (ip) - 8 (udp)) for a maximum in the wire size of 1542. It amazes me how things can get so complicated when they are actually simple. And I have not clue how someone came up with the number 1518 if the table says 1542.
