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29,930 views
48 votes
48 votes

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:

  1. Last fragment, $2400$ and $2789$
  2. First fragment, $2400$ and $2759$
  3. Last fragment, $2400$ and $2759$
  4. Middle fragment, $300$ and $689$
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7 Answers

Best answer
85 votes
85 votes

$M=0$ 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.

edited by
28 votes
28 votes
payload =total length-header

             =400-40

              =360

M bit is 0 so it is last fragment

offset is 300

so packet's first bit  300*8=2400

last bit =2400+359=2759

so ans is c
26 votes
26 votes

there is a similar problem has been given in "Forouzan book" 

http://erdos.csie.ncnu.edu.tw/~ccyang/TCPIP/TCPIPSlide.html

8 votes
8 votes
Since M=0; It is indication of last fragment.

Data length=Total Length-header length

Data length=400-10*4=360

Fragments offset=300

Number of data byte ahead from this fragments is=300*8=2400

Last byte of last frame is=2400+360=2760

Last byte=2759(Because we starting our counting from 0---5---59)
Answer:

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