# What is the minimum MTU of IPv4 68 bytes or 576 bytes?

I read that 68 bytes is the minimum size of IPv4 datagram every device must be able to forward without further fragmentation.

576 bytes is the minimum size of IPv4 datagram every device has to be able to receive (it can be whole or fragmented).

(according to RFC 791, page 24)

• IPv4 does not have an MTU. MTU is a data-link protocol value, e.g. the ethernet MTU (largest network protocol packet) is 1500. The smallest practical IPv4 packet is probably 68, and the largest IPv4 packet is 65,535. Sep 28, 2021 at 3:42
• Did any answer help you? if so, you should accept the answer so that the question does not keep popping up forever, looking for an answer. Alternatively, you could post and accept your own answer. Dec 23, 2021 at 21:51

576 bytes is the minimum IPv4 packet (datagram) size that any node must be able to handle. (RFC 791: Every internet destination must be able to receive a datagram of 576 octets either in one piece or in fragments to be reassembled.)

68 bytes is IPv4's minimum fragment size that each node must be able to handle. (RFC 791: Every internet module must be able to forward a datagram of 68 octets without further fragmentation.)

RFC 791: The maximum sized datagram that can be transmitted through the next network is called the maximum transmission unit (MTU).

So, you could call the minimum fragment size (68 bytes) the minimum possible MTU for IPv4 as well. "Minimum MTU" = "minimum maximum transmission unit" doesn't make too much sense...

For completeness, the absolute minimum IPv4 packet size is 20 bytes - a packet with the minimum header and no payload. (RFC 791: Internet Header Length is the length of the internet header in 32 bit words, and thus points to the beginning of the data. Note that the minimum value for a correct header is 5.)

• less than 576 bytes of MTU could be go into layer2 payload. Because minimum payload in layer2 is 46 bytes? So minimum fragmentation how 68 bytes in layer3?
– user80210
Sep 28, 2021 at 5:40
• Minimum payload in layer 2 depends on the layer-2 protocol. The minimum MTU size IPv4 can work with is said 68 bytes. 46 byte is the payload size in a minimum Ethernet frame - its maximum payload is 1500 bytes. Of course, all intermediate packet and frame sizes are also allowed - it doesn't make sense to use the minimum unless there's no choice.
– Zac67
Sep 28, 2021 at 16:02

What is the minimum MTU of IPv4 68 bytes or 576 bytes?

That is really an invalid question that is mixing terms from two different network layers. Network (layer-3) protocols, e.g. IPv4, do not have MTUs. An MTU is the Maximum Transmission Unit that a data-link (layer-2) protocol on a specific physical (layer-1) medium can transmit (hence the Transmission in MTU) in its payload (the layer-3 packet).

IPv4 will have a minimum and maximum packet size. The minimum packet size will be the header (20 to 60 octets, with 20 being the real modern header size) plus the payload (UDP has a minimum datagram size of eight). The maximum IPv4 packet size is based on the 16-bit Length field, which means 65,536, and that is larger than any real data-link protocol MTU.

In the modern world, there are no data-link protocols with an MTU as low as 576. It could be possible to configure that on an interface, but why? It would also be possible to get that low with tunnels inside of tunnels, inside of tunnels, but that would be a bad setup that should not be tolerated. In reality, IPv6 requires a minimum MTU in its path of 1280, so that is likely the smallest MTU you will encounter in the real world.

Really, the 576 number is now historical trivia (off-topic here), but a node, such as a router, must be able to accept packets at least that large, so setting an MTU smaller than that is really not allowed.

• I have two confusion. First confusion, for the router, the biggest possible IP header is 60 bytes and the smallest payload is 8 bytes (so, a 68 byte packet). In turn a given end host must be able to receive a final packet of up to 576 bytes before reaching data link layer (which may be made up of a bunch of fragments, depending on the MTU's of routers along the path). Am I right? My second confusion, MTU for data link layer less than 576 bytes possible? If it is possible why we said 576 bytes is minimum?
– user80210
Sep 28, 2021 at 4:54
• Beg to differ - RFC 791 defines The maximum sized datagram that can be transmitted through the next network is called the maximum transmission unit (MTU). Here, datagram refers to the L3 PDU, so MTU is a layer-3 term. The according layer-2 term is the maximum frame size whose SDU/payload is equivalent to the largest L3 PDU = MTU.
– Zac67
Sep 28, 2021 at 5:16