An IP datagram of size 1000 bytes arrives at a router. The router has to…
2016
An IP datagram of size 1000 bytes arrives at a router. The router has to forward this packet on a link whose MTU (maximum transmission unit) is 100 bytes. Assume that the size of the IP header is 20 bytes.
The number of fragments that the IP datagram will be divided into for transmission is ___________ .
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Correct answer: 13
Key insight: each fragment must fit within the MTU including the 20-byte IP header, and fragment data lengths (except possibly the last) must be multiples of 8 bytes.
Total datagram size = 1000 bytes; IP header = 20 bytes, so payload (data) = 1000 - 20 = 980 bytes.
MTU = 100 bytes, so maximum data per fragment = 100 - 20 = 80 bytes (which is a multiple of 8, so it satisfies the fragmentation offset requirement).
Number of fragments = ceil(980 / 80) = ceil(12.25) = 13.
Fragment breakdown: first 12 fragments carry 80 bytes of data + 20-byte header = 100 bytes each; the last fragment carries the remaining 20 bytes of data + 20-byte header = 40 bytes.
Therefore, the IP datagram will be divided into 13 fragments.
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