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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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127 votes
127 votes

Answer is C.

Number of fragments  $=\large\Big\lceil\frac{8888}{1480}\Big\rceil = 7$
Offset of last fragment $=\dfrac{(1500 - 20)\times  6} {8}=1110$
(scaling factor of $8$ is used in offset field).

$\text{TCP or UDP header}$ will be added to the DataUnit received from Transport Layer to Network Layer. And fragmentation happens at Network Layer. So no need to add $\text{TCP or UDP}$ header into each fragment.

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44 votes
44 votes

 Answer is : 7 fragments and last fragment offset is 1110

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6 votes
6 votes
UDP data = 8880 bytes
UDP header = 8 bytes
IP Header = 20 bytes

Total Size excluding IP Header = 8888 bytes.

Number of fragments  = ⌈ 8888 / 1480 ⌉ 
                     = 7
Offset of last segment = (1480 * 6) / 8 = 1110 
4 votes
4 votes
1500 is MTU . And header length of ip is 20 . So net data can go is 1500-20 =1480 . and no. of fragment needed  (8880+8)/1472 = 6.something so 7 . and offset 6*1480/8= 1110 .
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