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The packet is sent from source to router with the link having the maximum transmission unit (MTU) as 480 bytes, the header is of size 20 bytes. The intermediate fragment size is X and the last fragment size is Y when the packet size is of 2000 bytes including the header. Calculate the value of X+Y in bytes?

 

what i was getting 672 but they have given 652

1 Answer

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MTU size = 480 byte (including header)

size of the packet =2000 B (including header)

no of fragment = (2000-20)/(480-20) = ceil(4.30)=5

now maximum amount of data the MTU can transmit =(480-20) =460 [20B is for header]

now 460 is not multiple of 8 . fragment size must be multiple of 8.

so 456B is the immediate fragment size.

so packet size is 456 +20 =476B

now last fragment size =1980-4*456

                                      =156B

last packet size =156+20 =176 B

so x+y =652

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