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280 ??

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
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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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