Assume a datagram of length $5*10^3$ bytes needs to pass through five networks to reach its destination. The MTU’s of each network is 1000, 820, 850, 950 and 900 respectively. Then at destination how many datagrams have to reached and what is the offset value of 3rd fragment after fragmentation?
- $7, 200$
- $6, 400$
- $7, 400$
- $6, 200$
According to me all the options are wrong as the MTU of the first router is 1000, that is we can send datagram of size 980, so we are having 5 packets of $976 B$ and 1 datagram of $100B$. Now when this comes to the router having MTU 820B we need to send 2 datagram for each of $976B$ , and 1 datagram for that $100B$ packet. So total we need to send $11$ $datgrams$ right?
The answer they have provided is this→
But I believe fragmenting the datagram based on the smallest MTU is not even possible in case of IPv4, this feature is included in IPv6 only.