Please explain how you arrive at the answers so I can get a better understanding of them.
(1) A 4480-octet datagram is to be transmitted and needs to be fragmented because it will pass through an Ethernet with a maximum payload of 1500 octets.
Show the total length, More Flag, and fragment offset values in each of the resulting fragments.
(2) Describe some circumstances where it might be desirable to use source routing rather than let routers make the routing decision.© BrainMass Inc. brainmass.com August 16, 2018, 7:17 pm ad1c9bdddf
1. In a 4480-octet IP datagram, there is a header that consists of 20 octets. Therefore, there are 4460 octets for the data field. An Ethernet frame has 1500 octets, consisting of 20 octet headers and 1480 data octets.
To fit the 4460 data octets carrying 1480 data octets per Ethernet frame, we have:
# of datagrams= 4460 / 1480 = 3, remainder 20 octets.
So we have 3 datagrams ...
This posting contains answers to the given questions.