GATE | GATE-CS-2015 (Set 2) | Question 65
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 may 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
UDP data = 8880 bytes UDP header = 8 bytes IP Header = 20 bytes Total Size excluding IP Header = 8888 bytes. Number of fragments = ⌈ 8888 / 1480 ⌉ = 7 Refer the Kurose book slides on IP (Offset is always scaled by 8) Offset of last segment = (1480 * 6) / 8 = 1110