Host A wants to transfer 5000 bytes of data to Host B. On the network path there are two routers. First router R1, has MTU(Maximum Transformation Unit) of 1900 bytes. Second router R2, has MTU of 900 bytes.
Answer
1. In Case of 1st router it has MTU of 1900B and data is
5000B>1900B so fragmentation is needed.
So 20B of header info will be added to each packed being
sent.
So in one packed we can send 1900 - 20 = 1880B of data,
So we will need 5000/1880 = 2 full packets of 1880B data and one
packet having data of 1240B
So Router 1 will send 3 packets.
2. Packet 1: More_Flag = true, offset_value = 0, data_size =
1880B
Packet 2: More_Flag = true, offset_value = 1881,
data_size = 1880B
Packet 3: More_Flag = false, offset_value = 3761,
data_size = 1240B
3. So as router B has MTU of 900B so it can send 900-20B = 880B
of data, so fragmentation is needed for the
packets receiving from router A.
So it will need
So total of 8 packets
4. Following would be packets of router 2.
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