CSE GATE 2015 Question No 137 Computer Networks

Question No. 137 CSE | GATE 2015

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?

(A) 6 and 925
(B) 6 and 7400
(C) 7 and 1110
(D) 7 and 8880

Answer : (C) 7 and 1110

GATE 2015 CSE Computer Networks TCP/UDP and Sockets

Solution of Question No 137 of GATE 2015 CSE Paper

User data in UDP datagram = 8880 bytes

UDP header = 8 bytes

IP header = 20 bytes

We need to break (8880+8) bytes into small fragments.

Ethernet frame can hold (1500 - IP header) = 1500 - 20 = 1480 payload of data

1^st fragment 0 to 184 (1480 bytes)

2^nd fragment 185 to 369 (1480 bytes)

3^rd fragment 370 to 554 (1480 bytes)

4^th fragment 555 to 739 (1480 bytes)

5^th fragment 740 to 924 (1480 bytes)

6^th fragment 925 to 1109 (1480 bytes)

7^th fragment 1110 to 1117 (1480 bytes)

$\therefore$ 7 fragments with 1110 as the offset of last fragment.

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GATE Papers >> CSE >> 2015 >> Question No 137

Answer : (C) 7 and 1110

Questions of TCP/UDP and Sockets

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