Host A is sending data to host B over a full duplex link.

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Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The send and receive window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 byte long and the transmission time for such a packet is 50 ms Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 ms. What is the maximum achievable through put in this communication?

1. 7.69 × 10⁶ bps
2. 11.11 × 10⁶ bps
3. 12.33 × 10⁶ bps
4. 15.00 × 10⁶ bps

Correct Option: B

Now, Transmission time for 1 packet = 50 μs
Transmission time for 5 packet = 5 × 50 μs = 250 μs
Propagation delay = 200 μs
Total time = Transmission time + Propagation Delay
= 250 + 200 = 450 μs = 450 × 10^{– 6} μs
Finally, Maximum Achievable Throughput

=	Size of window	bps
	Total time

=	5000	= 11.11 × 10^-6 bps
	450 × 10^-6