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A router with MTU of 1024 bytes has received an IP packet of size 4240 bytes with an IP header of 20 bytes. The value of MF and offset of the 3rd fragment is ________
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## MF = 1 , OFFSET = 250 …

 

IP packet of size 4240 bytes, actual data/payload size=4240-20=4220 bytes(which is to be transferred) ..

Next MTU is 1024 & 1024<4220. So we'll use fragmentation.

Each fragment will have its own header, MTU for data/payload to be transferred = 1024-20=1004..

Fragment data should be in multiples of 8 and 1004 is not a multiple of 8, So we choose the nearest multiple of 8 i.e 1000…

Hence, we'll send fragments as- 1000(data) + 20(header)

First fragment- 1000+20 , Fragmentation offset(F.O)=0 ,

Remaining data=4220-1000=3220, MF bit=1

Second fragment-1000+20 , Fragmentation offset(F.O)=0+1000/8=125 ,Remaining data=3220-1000=2220, MF bit=1

Third fragment-1000+20 , Fragmentation offset(F.O)=125 +1000/8=250 ,Remaining data=2220-1000=1220, MF bit=1

Fourth fragment-1000+20 , Fragmentation offset(F.O)=250 +1000/8=375,Remaining data=1220-1000=220, MF bit=1

Fifth fragment-220+20 , Fragmentation offset(F.O)=375 +1000/8=500,Remaining data=220-220=0, MF bit=0 ..

 

## https://gateoverflow.in/206387/Fragmentation

 

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