Question

If you send a 2500 byte datagram into a link with an MTU of 700 bytes stamped with an identification number of 422, how many fragments are generated, what values do the identification, offset and fragment flag take on and what is the size of each fragment. in simple language please.

Answer 1

SOLUTION:-

Assume that IP Header = 20 bytes

Original Datagram = 2500 bytes, Now we will subtract 20 bytes for header that will leave 2500 - 20 = 2480 bytes

MTU = 700 bytes, 700 - 20 = 680 bytes of data can be transmitted in each packet.

Therefore, ceiling(2480 / 680) = 4 packets are required to transmit the data.

Offset is measured in 8 byte blocks. Each fragment will have Identification number 422. Each fragment except the last one will be of size 700 bytes (including IP header). The last datagram will be of size 460 bytes (including IP header). [ ((700-20)*3) + (460-20) = 2480 bytes].

The offsets of the 4 fragments will be 0, 85, 170, 255. Each of the first 3 fragments will have flag=1, the last fragment will have flag=0 .

1st fragment = 0

2nd fragment = 680 / 8 = 85

3rd fragment = 85 + 85 = 170

4th fragment = 170 + 85 = 255

========================================================================