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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

3 votes
3 votes
ans c)
1 votes
1 votes

Given Fragment offset = 300, M = 0 So Last Fragment

For Fragment offset scaling factor is 8 so Total 300*8= 2400B ahead of Last fragment which are ranging from (0 to 2399)

So Last Fragment seq no starts from 2400

Now Given Total length = 400B In this Header size = 10*4=40B (Scaling factor of IP Header = 4)

Data = Total length - header length = 400-40 =360B

So Sequence number of last byte = 2759 

Option C

 

0 votes
0 votes
The M bit means 'More Fragment' and a value of 0 indicates that this packet is the last
fragment (no more fragment). So, options B and D are invalid.

The HLEN field refers to header length which is mentioned as 10. This means that the length of the header is 40 bytes (10 x 4).

The total length mentioned is 400 which includes this 40 bytes header i.e. 360B payload. The frame offset value of 300 implies that 2400 bytes (300 x 8 bytes) were sent in the previous frame.

So, the sequence number of the first byte of payload will be 2400 and the sequence number of the last byte will be 2400 + 359= 2759.

so option d)

 

TWO THINGS TO REMEMBER

SCALING FACTOR HLEN= X4

FRAGMENT OFFSET= X8
Answer:

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