An IP router with a Maximum Transmission Unit (MTU) of 1500 bytes has received…
2014
An IP router with a Maximum Transmission Unit (MTU) of 1500 bytes has received an IP packet of size 4404 bytes with an IP header of length 20 bytes. The values of the relevant fields in the header of the third IP fragment generated by the router for this packet are
- A.
MF bit: 0, Datagram Length: 1444; Offset: 370
- B.
MF bit: 1, Datagram Length: 1424; Offset: 185
- C.
MF bit: 1, Datagram Length: 1500; Offset: 370
- D.
MF bit: 0, Datagram Length: 1424; Offset: 2960
Attempted by 147 students.
Show answer & explanation
Correct answer: A
Key insight: fragments must carry data in multiples of 8 bytes; the IP header is 20 bytes, and MTU includes the header.
Total data payload = 4404 − 20 = 4384 bytes.
Maximum data per fragment = 1500 − 20 = 1480 bytes (1480 is divisible by 8; 1480/8 = 185).
Two full fragments carry 1480 bytes each: 2 × 1480 = 2960 bytes.
Remaining data for the third fragment = 4384 − 2960 = 1424 bytes; datagram length = 1424 + 20 = 1444 bytes.
Offset for the third fragment (in 8-byte units) = 2960/8 = 370. MF = 0 because it is the last fragment.
Therefore the third fragment's header fields are: MF = 0, Datagram Length = 1444, Offset = 370.
A video solution is available for this question — log in and enroll to watch it.