Consider three IP networks A, B and C. Host HA in network A sends messages…
2004
Consider three IP networks A, B and C. Host HA in network A sends messages each containing 180 bytes of application data to a host HC in network C. The TCP layer prefixes a 20 byte header to the message. This passes through an intermediate network B. The maximum packet size, including 20 byte IP header, in each network is
A : 1000 bytes
B : 100 bytes
C : 1000 bytes
The network A and B are connected through a 1 Mbps link, while B and C are connected by a 512 Kbps link (bps = bits per second).
Assuming that the packets are correctly delivered, how many bytes, including headers, are delivered to the IP layer at the destination for one application message, in the best case ? Consider only data packets.
- A.
200
- B.
220
- C.
240
- D.
260
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Correct answer: D
Answer: 260 bytes
Step-by-step reasoning:
Compute the TCP segment size: 180 bytes of application data + 20 bytes TCP header = 200 bytes of IP payload.
Find the bottleneck MTU: Network B has maximum packet size 100 bytes (this includes a 20-byte IP header). Therefore each IP fragment can carry up to 100 - 20 = 80 bytes of IP payload.
Fragment the 200-byte IP payload into chunks of at most 80 bytes: 80 + 80 + 40.
Add the 20-byte IP header to each fragment, giving fragment sizes: 20+80 = 100 bytes, 20+80 = 100 bytes, and 20+40 = 60 bytes.
Sum of bytes delivered to the IP layer at the destination (only data packets considered) = 100 + 100 + 60 = 260 bytes.
Note: This assumes no retransmissions and that only the data (forward) packets are counted; ACKs are ignored.