Consider an IP packet with a length of 4,500 bytes that includes a 20-byte IPv4 header and a 40-byte TCP header

Q. Consider an IP packet with a length of 4,500 bytes that includes a 20-byte IPv4 header and a 40-byte TCP header. The packet is forwarded to an IPv4 router that supports a Maximum Transmission Unit (MTU) of 600 bytes. Assume that the length of the IP header in all the outgoing fragments of this packet is 20 bytes. Assume that the fragmentation offset value stored in the first fragment is 0.

The fragmentation offset value stored in the third fragment is

Ans: 144

Sol:

 MTU = 600 bytes and IP Header = 20 bytes
So, Payload will be 600 – 20 = 580 bytes

580 is not multiple of 8, but we know fragment size should be multiple of 8. So fragment size = 576 bytes

K^th fragmentation offset value = Fragment Size * (K^th fragment – 1) / Scaling Factor
Offset value of 3^rd fragment = 576 * (3 – 1) / 8 = 144