"The Maximum Transmission Unit (MTU) is the largest possible frame size of a communications Protocol Data Unit (PDU) on an OSI Model Layer 2 data network." In today's network the standard MTU for Layer 3 IP packet is 1500 bytes. Meanwhile, the standard MTU for Layer 2 Ethernet frame is 1514 bytes ( 6 bytes source MAC + 6 bytes destination MAC + 2 bytes EtherType + 1500 bytes IP packet). For the Dot1Q trunk frame, extra 4 bytes for Dot1Q tag is added. So up to here, we understand that there are two types of MTUs. MTU for layer 2 frames and MTU for layer 3 packets.
Click to Expand |
Ok, now we understand L2 MTU vs L3 MTU. Another interesting question comes up. When a L3 IP packet passing through a router, where the MTU will be checked and where the fragmentation will happen? The inbound interface or the outbound interface. In order to find out the answer of this question, let's do a simple Lab as the following topology.
Click to Expand |
Let's send another IP packet with size 1500 from R3 to R1. This time, we can see the packet has been fragmented into two by R2.
Click to Expand |
by doing this lab, we understand that the IP MTU is checked by the outgoing interface of the router. If fragmentation is required, the packet is fragmented by the outgoing interface of the router. So if the DF bit is set, then the packet will be dropped on the outbound interface as well.
Reference:
Very nice explanation. But after fragmented the packets of 1500 bytes it changes to 1314 and 234 bytes? How it is calculated could be please elaborate?
ReplyDeleteHi Dipan, nice catch! The 1314 and 234 I originally put in were actually the L2 frame size. It was 1300 (20 byte IP header + payload) + 14 (L2 header) & 220 (20 byte IP header + payload) + 14 (L2 header). But I think in the diagram, it's better to put all in IP packet only. So the diagram has been updated.
ReplyDeleteThank you!