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).
What is the rate at which application data is transferred to host HC? Ignore errors, acknowledgements, and other overheads.
- A.
325.5 Kbps
- B.
354.5 Kbps
- C.
409.6 Kbps
- D.
512.0 Kbps
Attempted by 98 students.
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Correct answer: B
Final answer: ≈ 354.375 kbps (≈ 354.5 Kbps)
Compute the IP packet size leaving host A: TCP header 20 bytes + application data 180 bytes = 200 bytes of IP payload. Add 20-byte IP header => 220 bytes total.
Network B MTU is 100 bytes including the 20-byte IP header, so maximum IP payload per fragment = 80 bytes. Fragment payloads must be multiples of 8 bytes; 80 and 40 satisfy this.
Fragment the 200-byte IP payload into 80, 80, and 40 bytes. Each fragment carries a 20-byte IP header, so fragment sizes on the wire are 100, 100, and 60 bytes.
Total bytes sent over the 512 kbps link = 100 + 100 + 60 = 260 bytes = 2080 bits. Application data per message = 180 bytes = 1440 bits.
Throughput (application bits/s) = 1440 bits / (2080 bits / 512000 bps) = 512000 * 1440 / 2080 = 354375 bps ≈ 354.375 kbps (≈ 354.5 Kbps).
Check other link: A–B carries the original 220-byte packet (1760 bits) over 1 Mbps, which is not the bottleneck. Therefore the application data rate is set by the 512 kbps link after fragmentation: ≈ 354.5 Kbps.