Applied Mock Test

Question

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?

Answer 1

Intermediate fragment size can be calculated from the given MTU. Data going through the router is the payload(actual data) + headers. Since the fragment size is required to be a multiple of 8 ,the actual size of data would be  $(MTU - size(header)) - ((MTU-size(header))%8)$.

No. of fragments can be calculated by $\frac{size(packet)-size(headers))}{size(fragment)}$, round it to the next integer value. So in this case a total of 5 packets. Each packet except the last one will be transmitted using the complete MTU size, so the last fragment size would be $(size(packet)-size(headers)) - ((no. of fragments-1)*size(fragment))$, which is  156 in this case. So $X+Y=456+156=612$

Comments

All the fragment size except the last one is = (456+20) = 476

and the last fragment is = (156+20)=176

so, x+y= 652 ????

can you plz check it a bit!!!!