GATE CS 2015 Set 2 — Question 37
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Computer Networks → Network Layer: Addressing → IPv4 Packet & Fragmentation
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Question
Host A sends a UDP datagram containing 8880 bytes of user data to host B over an Ethernet LAN. Ethernet frames may carry data up to 1500 bytes (i.e. MTU=1500 bytes). Size of UDP header is 8 bytes and size of IP header is 20 bytes. There is no option field in IP header. How many total number of IP fragments will be transmitted and what will be the contents of offset field in the last fragment?
Correct answer
(C) 7 and 1110
Solution
Total data to be transmitted at IP layer = UDP Header + User Data = bytes.
MTU = 1500 bytes. IP Header = 20 bytes.
Maximum payload per IP fragment = bytes.
Since 1480 is divisible by 8 (fragment offset scaling factor), the maximum data in a fragment is 1480 bytes.Number of fragments:
Total payload = 8888 bytes.
Fragment 1: 1480 bytes (Offset 0)
Fragment 2: 1480 bytes (Offset )
Fragment 3: 1480 bytes (Offset )
Fragment 4: 1480 bytes (Offset )
Fragment 5: 1480 bytes (Offset )
Fragment 6: 1480 bytes (Offset )
Total data sent so far = bytes.
Remaining data = bytes.
Fragment 7: 8 bytes (Offset ).Total fragments = 7.
Offset of last fragment = 1110.
MTU = 1500 bytes. IP Header = 20 bytes.
Maximum payload per IP fragment = bytes.
Since 1480 is divisible by 8 (fragment offset scaling factor), the maximum data in a fragment is 1480 bytes.Number of fragments:
Total payload = 8888 bytes.
Fragment 1: 1480 bytes (Offset 0)
Fragment 2: 1480 bytes (Offset )
Fragment 3: 1480 bytes (Offset )
Fragment 4: 1480 bytes (Offset )
Fragment 5: 1480 bytes (Offset )
Fragment 6: 1480 bytes (Offset )
Total data sent so far = bytes.
Remaining data = bytes.
Fragment 7: 8 bytes (Offset ).Total fragments = 7.
Offset of last fragment = 1110.
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