In an IPv4 datagram, the $M$ bit is $0$, the value of $HLEN$ is $10$, the value of total length is $400$ and the fragment offset value is $300$. The position of the datagram, the sequence numbers of the first and the last bytes of the payload, respectively are:
$M=0$ meaning no more fragments after this. Hence, its the last fragment.
IHL = internet header length = $10 \times 4 = 40B$ coz $4$ is the scaling factor for this field.
Total Length = $400B$
Payload size = Total length - Header length = $400 - 40 = 360B$
fragment offset = $300 \times 8 = 2400B$ = represents how many Bytes are before this. $8$ is the scaling factor here.
$\therefore$ the first byte # = $2400$
Last byte # = first byte # + total bytes in payload - 1 = $2400 + 360 - 1 = 2759$
Option C is correct.
@amar u are saying that 2400 represents how many Bytes are before this.. then how come it becomes the first byte..?
I think range of sequence number must cover 360 byte. For the last fragmented packet offset range would be [ 300 - 344 ]. So offset field contains number = 300. Only option C satisfies both the conditions of last fragment (MF = 0) and total 360 Byte (400-10*4 byte).i.e. possible sequence number can be [ 2400 - 2759 ] . That's why option C. I don't think initial sequence number assumption (as zero) is valid always. correct if wrong !
What if option C is given as = Last fragment,5000 and 5359. is it correct too ?
there is a similar problem has been given in "Forouzan book"
The reason is Header length is number of 32 bit words. So header length = 10 * 4 = 40
following link is Kvs_Pgt_Question Paper...