Consider two hosts P and Q connected through a router R. The maximum transfer unit (MTU) value of the link between P and R is 1500 bytes, and between R and Q is 820 bytes. A TCP segment of size 1800 bytes was transferred from P to Q through R, with identification value as 0x1234. Assume that the IP header size is 20 bytes. Further, the packet is allowed to be fragmented. Which of the following statements is/are correct?
A. Two fragments are created at R and the IP header of the second fragment has 'Don't Fragment' (DF) bit set to 1
B. Two fragments are created at R and the IP header of the second fragment has identification 0x1234
C. Three fragments are created at R and the IP header of the second fragment has identification 0x1234
D. Three fragments are created at R and the IP header of the third fragment has 'More Fragment' (MF) bit set to 0
GATE 2021 · Computer Networks · TCP · medium
Answer: C and D are correct: Three fragments are created at R, the second fragment has identification 0x1234, and the third fragment has MF bit set to 0.
Fragmentation at P for MTU=1500 (P to R link): Total IP packet = 1800 + 20 = 1820 bytes. Max data per fragment = 1480 bytes.
Fragment 1 from P: 1480 bytes data, offset=0, MF=1. Total size = 1500 bytes.
Fragment 2 from P: 1800 - 1480 = 320 bytes data, offset=1480/8=185, MF=0. Total size = 340 bytes.
P sends two packets to R: (1500 bytes) and (340 bytes).
Re-fragmentation at R for MTU=820 (R to Q link): Fragment 1 arriving at R: 1480 data bytes, offset=0, MF=1.
R splits this into:
Sub-frag 1a: 800 data, offset=0, MF=1, size=820 bytes.
Sub-frag 1b: 680 data, offset=800/8=100, MF=1, size=700 bytes.
Fragment 2 arriving at R: 320 data bytes, offset=185 (=1480), MF=0.
Since 320 < 800, R sends it as-is:
Sub-frag 2: 320 data, offset=185, MF=0, size=340 bytes.
Total fragments sent by R to Q: 3 fragments.
Verify the options: 3 fragments are created (re-fragmented) at R. All three carry ID = 0x1234 (IP fragments always keep the same identification). The last (3rd) fragment has MF=0 (no more fragments). The 2nd fragment has ID=0x1234 (option C says three fragments and 2nd has ID=0x1234 - CORRECT). Option D says three fragments and 3rd has MF=0 - CORRECT. Option A says two fragments (WRONG). Option B says two fragments (WRONG).