GATE | GATE-CS-2003 | Question 84

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 bytes long and the transmission time for such a packet is 50 µs. Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 us. What is the maximum achievable throughput in this communication?
(A) 7.69 × 106 bytes per second
(B) 11.11 × 106 bytes per second
(C) 12.33 × 106 bytes per second
(D) 15.00 × 106 bytes per second


Answer: (B)

Explanation:

Network throughput ≈ Window size / roundtrip time

Roundtrip time = 2 × Packet delivery time + processing delay
= ransmission delay+2*propagation delay 
=50microsec+2*200microsec=450microsec

Now Throughput = ((5*1000*bytes)/450microsec) = 11.1111 * 106 bytes per second


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