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3 votes
3 votes

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5 votes
5 votes

There are 5 fragments.

After first network  
1480+20 
120+20 
After second network 
1480 packet will be divided into 
456+20 
456+20 
456+20 
112+20 
And second fragment is  
120+20 
So total will be 
(120+20)+(456+20)+(456+20)(456+20)+(112+20)=1700 bytes...

so 1700 Bytes transmitted.

The almost same question also asked in gate 2016 set 1 http://quiz.geeksforgeeks.org/gate-gate-cs-2016-set-1-question-63/

also same type of problem you can found from Kuroe & ross book , chapter 4 , No-19p

https://www.chegg.com/homework-help/consider-sending-2400-byte-datagram-link-mtu-700-bytes-suppo-chapter-4-problem-19p-solution-9780133128093-exc  

1 votes
1 votes

mtu is 480.

payload is 460 B.

A sends payload 1600 B to B

so total data A sends to B = (460+20,460+20,460+20 ,220+20) = 1680 B.

Answer:

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