Any one give answer with Explanation

Question

Answer 1

the protocol in play here has a minimum header length of 20 Bytes how can you ask someone to go below this value?

Comments

suppose if there is 12 byte minimum header length in network layer, then what will be approach?
I want only approach :)

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kindly make edits to your question; answering like this will violate standards

Answer 2

P to R1 link's MTU is 1024B hence a TCP packet of length 820B will be sent easily .So now the packet is in R1 router . Now R1 will have to fragment the packet because the MTU of R1 to R2 is 256 B . Therefore the possible fragmentation is 8+248 , 8+248 , 8+248 ,8+76 to Router R2 . Now the MTU of R2 to Q is 512B and a packet of size 256B will be sent easily . Hence the no of bytes transferred is  (256 + 256 +256 + 84 )B = 852 B .

Therefore the answer is 852 Bytes .

Comments

TCP segments are encapsulated within Datalink layer frames when it is ready to be dispatched into the physical link .

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TCP segments can directly encapsulated with Datalink layer frame? then what about IP header and mac header? We don't need to add them

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Tcp segment passes the data to network layer . If the MTU of the link that is maximum transmission unit is less than the frame size than IP does the fragmentation of the packets and attaches header to every fragmented packet .

Answer 3

from p-r1 link pkt is 820+12 from r1-r2 link pkts are 248+8,248+8,248+8,88+8 from r2-q link pkts are 256+12, 256+12, 256+12, 96+12 so ans is (256+12+ 256+12+256+12,+96+12)=912 bytes