GATE CS 2014 Set 3 — Question 38
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Computer Networks → Network Layer: Addressing → IPv4 Packet & Fragmentation
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Question
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
Correct answer
(A) MF bit: 0, Datagram Length: 1444; Offset: 370
Solution
Total packet size = 4404 bytes.
IP Header = 20 bytes.
Data payload = bytes.MTU = 1500 bytes.
Max payload per fragment = bytes.
Since the offset must be a multiple of 8, and 1480 is divisible by 8 (), the max fragment size is valid.Fragment 1:
Bytes 0 to 1479 (1480 bytes).
Offset = 0.
MF = 1.Fragment 2:
Bytes 1480 to 2959 (1480 bytes).
Offset = .
MF = 1.Fragment 3:
Remaining bytes = bytes.
Bytes 2960 to 4383.
Offset = .
Total Length = bytes.
MF = 0 (Last fragment).Thus, MF bit: 0, Datagram Length: 1444, Offset: 370.
IP Header = 20 bytes.
Data payload = bytes.MTU = 1500 bytes.
Max payload per fragment = bytes.
Since the offset must be a multiple of 8, and 1480 is divisible by 8 (), the max fragment size is valid.Fragment 1:
Bytes 0 to 1479 (1480 bytes).
Offset = 0.
MF = 1.Fragment 2:
Bytes 1480 to 2959 (1480 bytes).
Offset = .
MF = 1.Fragment 3:
Remaining bytes = bytes.
Bytes 2960 to 4383.
Offset = .
Total Length = bytes.
MF = 0 (Last fragment).Thus, MF bit: 0, Datagram Length: 1444, Offset: 370.
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