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GATE CSE 2015 Set 2 | Question: 52

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$\text{Host A}$ sends a $\text{UDP}$ datagram containing $8880\text{ bytes}$ of user data to $\text{host B}$ over an $\text{Ethernet LAN}.$ Ethernet frames may carry data up to $1500\text{ bytes (i.e. MTU = 1500 bytes)}.$ Size of $\text{UDP}$ header is $8\text{ bytes}$ and size of $\text{IP}$ header is $20\text{ bytes}.$ There is no option field in $\text{IP}$ header. How many total number of $\text{IP}$ fragments will be transmitted and what will be the contents of offset field in the last fragment?

  1. $6$ and $925$
  2. $6$ and $7400$
  3. $7$ and $1110$
  4. $7$ and $8880$
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Corrct Option: C (Remember:UDP header or TCP header is added from TL while sending the packet to the NL, and IP header is added On NL while sending the packet from NL to the DLL ).

It means if Segmentation has to be done, then segmentation will be done then UDP Header or TCP Header will be added to send the packet for NL.

Hence: in this question Fragmentation has to be done, then total MTU is 1500 (including IP header of 20B), so while fragmentation we will subtract IPH.

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