An IP packet of size 2050 bytes has arrived at an IP router. Which of the following statements is correct regarding More fragmentation(MF), offset, datagram length?
(IP header is of 20 bytes and maximum transfer unit(MTU) is 250 bytes.)
I – MF bit: 0, Offset:252 , Datagram Length: 22;
II – MF bit: 1, Offset:112, Datagram Length: 224;
III -MF bit: 1, Offset: 196, Datagram Length: 244;
IV – MF bit: 1, Offset: 28, Datagram Length: 250;
(A) I, II are correct.
(B) II, III, and IV are correct.
(C) I and III are correct.
(D) All statements are correct.


Answer: (C)

Explanation: IP packet size = 2050
IP header = 20 bytes.
MTU = 2050
i.e. (20 byte header + 230 byte data)
For the first fragment:
Data will be fragmented like:
0 – 223(20-byte header size ):
MF bit: 1, Offset:0 , Datagram Length: 244;
224 – 447(20-byte header size ):
MF bit: 1, Offset:28 , Datagram Length: 244;
448 – 671(20-byte header size ):
MF bit: 1, Offset:56 , Datagram Length: 244;
672 – 895(20-byte header size ):
MF bit: 1, Offset:84 , Datagram Length: 244;
896 – 1119(20-byte header size ):
MF bit: 1, Offset:112 , Datagram Length: 244;
1120 – 1341(20-byte header size ):
MF bit: 1, Offset:140 , Datagram Length: 244;
1342 – 1567(20-byte header size ):
MF bit: 1, Offset:168 , Datagram Length: 244;
1568 – 1791( 20 byte header size ):
MF bit: 1, Offset:196 , Datagram Length: 244;
1792 – 2015 (20-byte header size ):
MF bit: 1, Offset:224 , Datagram Length: 224;
2015 – 2031(2-byte padding and 20-byte header size ):
MF bit: 0, Offset:252 , Datagram Length: 222;
So, option (C) is correct.

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