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GO Classes DPP | CN | FRAGMENTATION

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8 8 votes

Consider a network using Ethernet and IPv4 without any options field in the header.
Each Ethernet frame can carry up to $1600 \textbf{ bytes}$ in its data portion. A UDP segment is transmitted.
The UDP payload size is $9000 \textbf{ bytes}$.

Which one of the following choices gives the correct number of IPv4 fragments transmitted and the size of the last fragment (including header)?
 

  1. $5 \mathrm{~fragments} ~1600 \mathrm{~bytes}$
     
  2. $6 \mathrm{~fragments} ~100 \mathrm{~bytes}$
     
  3. $6 \mathrm{~fragments} ~1140 \mathrm{~bytes}$
     
  4. $7 \mathrm{~fragments} ~108 \mathrm{~bytes}$
  • 🚩 Edit necessary | 👮 divyanshu_create | 💬 “Last Freagement is 1148 bytes as 1128 + 20 (udp header size) = 1148 bytes.”
  • 🚩 Edit necessary | 👮 Prashant-G | 💬 “Option C should be 6,1148”

3 Answers

0 0 votes
Ans : Option C


The final fragment contains the IP header and the remainder of the UDP payload, but no UDP header.


This is because the UDP header is only included in the first fragment of the fragmented UDP datagram.

First Fragment: IP Header + UDP Header + Part of UDP Payload

Last Fragment: IP Header + Remaining UDP Payload
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