Question

Suppose a 2800-byte datagram must be transmitted through an intermediate network along its way to its destination. This intermediate network has a maximum transfer unit (MTU) of 1020 bytes. Assume that all of the IP headers are option-less and contain only the minimum 20 bytes. Provide the values of the total length, more flag, and fragmentation offset fields in each of the IP headers for the resulting fragments.

Hint: It is and easier calculation if you use the maximum MTU for the first two

Answer 1

2800 = 2780 +20

MTU = 1020 = 1020 + 20 (for header)

so three packets to be sent 1000 + 1000 + 780

Packet 1: ID=x, Total_len=1020, MF=1, Frag_offset=0
Packet 2: ID=x, Total_len=120, MF=1, Frag_offset=125
Packet 3: ID=x, Total_len=800, MF=0, Frag_offset=250