I need the correct solution only. Please provide the correct solution. Hand-writ

Question

I need the correct solution only. Please provide the correct solution. Hand-written also be accepted, but it should be neat and clean, also readable.

An IP datagram 5,600 bytes long (including header with no options) arrives at a router, which determines that the next destination has an MTU of 1,500 bytes. Answer the following questions, showing your calculations and reasoning. a) Assuming that the router decides to fragment the packet into 4 fragments, for each fragment, determine a correct size, and identify the starting byte and ending byte (2.5 marks). b Calculate the fragmentation offset for each fragment (1 mark). c)The total number of bytes from all 4 fragments leaving the router should be greater than the initial datagram size that arrived Explain why this is so (.5 mark).

Explanation / Answer

a) IP datagram=5600 bytes, MTU=1500 bytes, IP Header = 20 bytes. Therefore (5600-20) = 4580 bytes of data.
With an MTU of 1500 bytes, 1500-20=1480 bytes of data may be transmitted in each packet.
Ceiling(4580/1480)= 4 packets are needed.
1st packet: 20 bytes:IP Header, 1480 bytes: Data ==> Correct size:20+1480=1500 bytes, starting bytes : 1 , ending bytes : 1480

2nd packet:   20 bytes:IP Header, 1480 bytes: Data ==> Correct size:20+1480=1500 bytes, starting bytes : 1481, ending bytes : 2960

3rd packet: 20 bytes:IP Header, 1480 bytes: Data ==> Correct size:20+1480=1500 bytes, starting bytes : 2961, ending bytes : 4440

4th packet: 20 bytes:IP Header, 1140 bytes: Data ==> Correct size:20+1140=1160 bytes, starting bytes : 4441 , ending bytes : 5580

B) 1st packet fragmentation offset = 0
2nd packet fragmentation offset = 185
3rd packet fragmentation offset = 370
4th packet fragmentation offset = 555

C) As 20 bytes is always occupied by IP Header in any packet and it is identical in each and every packet, so total size of next datagram is automatically reduced by 20 bytes than previous datagram. Like, total number of bytes in 1st packet = 1500 bytes but starting byte of 2nd packet is just 1481 bytes because of 20 bytes of header, so it makes datagram size of 2nd packet at initially = 1480 bytes. So, the total number of bytes should be greater than the initial datagram size that arrived.

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