How big is an IP packet frame including headers?

Question

A bit of background.

I'm writing an application that uses UDP. The application will run on a LAN (not internet). I've been assuming that if my MTU is 1500 then thats how big a UDP payload can be, but I'm not sure if the UDP header is meant to fit within that too.

I'm suspecting that if I send a UDP packet with a 1500 byte payload and the machine MTU is 1500 bytes will it end up sending two packets?

Searching the internet for a clear answer here seems harder than it should be, I've seen conflicting information.

Answer 1

------------------------------------------------------------------------------
|Ethernet  | IPv4         |UDP    | Data                   |Ethernet checksum|
------------------------------------------------------------------------------
  14 bytes    20 bytes     8 bytes    x bytes                4 bytes
           \ (w/o options)                               /
            \___________________________________________/
                              |
                             MTU

If your MTU is 1500, you have 1500-20-8 = 1472 bytes for your data.

• If you exceed that, the packets will be fragmented ,i.e. split into more packets.
• There might be more layers involved, e.g. 4 byte a vlan header if you're on top of a vlan ethernet.
• Some routers inbetween you and the destination might add more layers.

Answer 2

Yes your example would not fit in one frame.

The ethernet data payload is 1500 bytes. IPv4 requires a minimum of 20 bytes for its header. Alternatively IPv6 requires a minimum of 40 bytes. UDP requires 8 bytes for its header. That leaves 1472 bytes (ipv4) or 1452 (ipv6) for your data.

More information:

• http://en.wikipedia.org/wiki/Ethernet_II_framing
• http://en.wikipedia.org/wiki/IPv4#Header
• http://en.wikipedia.org/wiki/IPv6_packet#Fixed_header
• http://en.wikipedia.org/wiki/User_Datagram_Protocol