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I got answer 12 is it Right or wrong

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Answer should be 11.  
MTU of Router A = 1200B, it can carry 1180 Bytes of Data + 20 Bytes IP Header
But 1180 is not divisible by 8. hence MAX packet will carry 1176B + 20B (IP header)
MTU of Router B = 512B, it can carry 492 Bytes of Data + 20 Bytes IP Header, but same as said above
492 is not divisible by 8, hence Max packet will carry 488B + 20B (IP Header) 

IP Packet is of 4408 B, Means there is 4388 Bytes of Data + 20 Bytes IP. 
At router A there will be 3packets of size 1176 Bytes each and 4th packet will be of size 880, in addition to this each packet will carry 20 Bytes IP header. 
  
At router B, 488B Data + 20B Header. each packet of 1176 B will be divided into three parts and 20B of IP header will be added to each. 
There are three packets of 1180 B,and each is divided into 3 fragements. 3*3 = 9 
Fourth packet which was carrying 880B + 20B IP header will be divided into two parts at router B. 
So total fragemtens will be 11.

PS: MTU header is not considered when we talk about MTU size.

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