2,006 views
2 votes
2 votes

Consider a network system consisting of three networks connected with two routers. Network-A has MTU of 1500 bytes, Network-B has MTU of 620 bytes, Network-C has MTU of 1500 bytes (MTU includes header size).

Station-1 needs to send a segment of1380 bytes. The Total size of the packets received at Network-C is _________ bytes if the header size is 20 B.

2 Answers

Best answer
4 votes
4 votes

Segment to be send=$1380$ , which is actually transport layer data .

$\Rightarrow$there will be no fragmentation at Network $A$ as its MTU is $1500$B

$\Rightarrow$on station B, there will be 3 fragments-:

  • $600+20$
  • $600+20$
  • $180+20$

$\Rightarrow$there will be no fragmentation at Network $C$ as its MTU is $1500$B

Hence  Total size of packet recieved at Network$C=620+620+200=1440$


P.S-:It is not necessary that last fragment size should be divisible by $8$.

see the answer here-: 

Every fragment except the last must contain a multiple of 8 bytes of data.

https://stackoverflow.com/questions/7846442/why-the-ip-fragments-must-be-in-multiples-of-8-bytes

selected by
0 votes
0 votes

Segment size = 1380 Bytes ( data = 1360; header = 20 )

Network B MTU = 600 Bytes ( data = 600; header = 20 )

Therefore fragments will be of sizes: ( 600 + 20 ) + ( 600 + 20 ) + ( 160 + 20 )

Total = 1420 Bytes

Related questions

1 votes
1 votes
0 answers
4
Parimal Paritosh asked Jan 21, 2018
224 views
Why can a broadcast IP datagram not pass through any router?