Answer:
a) IP datagram has 9 fields (rows) of 4 bytes each. Hence,
length of IP datagram = 9*4 = 36 bytes
This datagram will be stored in the LLC data field of IEEE 802.3 frame. As the length of datagram is 36 bytes, padding will done and 10 padding bytes will be added to fulfill the minimum requirement of 46 bytes.
Hence, length of 802.3 frame = 7 + 1 + 6 + 6 + 2 + 46 + 4 = 72 bytes
b) The length of frame received at the receiver end will be same as sent from sender end i.e. 72 bytes. FCS (Frame Check Sequence) will be used to detect corrupted data.
c) As said above, length of IP datagram will be 36 bytes.
d) Segment length of UDP segment will be 2 bytes as clear from figure 2.14.
e) Total length of TFTP packet = 2 * 4 = 8 bytes
f) At transport layer, UDP segment will be received.
3) [10 marks] Refer to following Figure 2.14 and Figure 16.3. This question is about encapsulation,...
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