Question 63

Given parameters:

• IP datagram total size $= 1000$ bytes.
• IP header size $= 20$ bytes.
• Total data payload $= 1000 - 20 = 980$ bytes.
• Link MTU $= 100$ bytes.

Determine fragment payload capacity:

• Each fragment must include an IP header of 20 bytes.
• Maximum data payload possible per fragment $= 100 - 20 = 80$ bytes.
• In IPv4 fragmentation, the data payload of every fragment (except the last) must be a multiple of 8 bytes because the Fragment Offset field is expressed in 8-byte blocks.
• Here, $80$ is an exact multiple of $8$ ($80 = 10 \times 8$). Thus, each fragment can carry at most 80 bytes of payload.

Calculate number of fragments:

• Full fragments required: $\lfloor 980 / 80 \rfloor = 12$ fragments, carrying $12 \times 80 = 960$ bytes of payload.
• Remaining payload: $980 - 960 = 20$ bytes.
• The remaining 20 bytes require 1 final fragment (total size $20 + 20 = 40$ bytes, with $MF = 0$).
• Total number of fragments $= 12 + 1 = 13$.