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+4 votes
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asked in Computer Networks by Boss (7.1k points) 4 27 82 | 66 views
280 ??

2 Answers

+2 votes
Best answer

Given , total length of packet   =  1500 B

           header length               =  10 * 4   =  40 B [ As HLEN field value = 10 ]

          So payload (data) length   =   1500  -  40   =  1460 B

Now fragmentation is done into fragment sizes of 200 B [ As MTU of router given =  200 B ]

We should know that the header of packet is there in each fragment with some changes e.g. in flags and offset fields..

So payload length  in each fragment   =   200  -  40   =   160 B

Hence number of offsets in each fragment  =  160 / 8    =  20

Now as 160 is a multiple of 8 , hence it is allowed as it is else we have to take nearest multiple of 8 which is less than 160..To accomodate this , we do padding in last fragment..Payload in each fragment should be a multiple of 8 because 1 offset = 8 bytes by convention..

So number of fragments needed   =  ceil ( 1460 / 160 )   =  10

Hence offset of last fragment       =  Offset of 1st fragment + (number of fragments - 1) * (no of offsets in each fragment)

                                                  =  100  +  9 * 20

                                                  =   280

Hence offset of last fragment        =   280

answered by Veteran (87k points) 15 57 292
selected by

offset of last fragment       =  Offset of 1st fragment + (number of fragments - 1) * (no of offsets in each fragment)

 Why the offset of 1st fragment(old offset value) is added ?

Offset is just like an index in the array..If the start index of the array is  100(say) , so we need to calculate subsequent offsets accordingly..
offset of first fragment is always 0 right
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