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GO Classes DPP | CN | IPV4

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In an IPv4 datagram, the $M$ bit is $\mathbf{1}$, the value of $H L E N$ is $\mathbf{5}$, the value of total length is $\mathbf{1 5 0 0}$ and the fragment offset value is $100$. The position of the datagram and the sequence numbers of the first and the last bytes of the payload, respectively, are:

  1. First fragment, $800$ and $2279$
     
  2. Last fragment, $800$ and $2279$
     
  3. Middle fragment, $100$ and $1579$
     
  4. Middle fragment, $800$ and $2279$

1 Answer

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Position: $M=1$ (more fragments) and Offset > 0 , so it is a Middle fragment.

Header Length: $H L E N \times 4=5 \times 4=20$ bytes.

Payload Length: Total Length - Header Length = $1500-20=1480$ bytes.

First Byte: Fragment Offset $\times 8=100 \times 8=800$.

Last Byte: First Byte + Payload Length $-1=800+1480-1=2279$.
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