Skip to content
Related Articles

Related Articles

Improve Article

GATE | GATE-CS-2003 | Question 84

  • Last Updated : 05 Nov, 2020

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
= Transmission delay + 2 * propagation delay 
= 50 microsec + 2 * 200 microsec
= 450 microsec 

Now, Throughput

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

So, option (B) is correct.

Quiz of this Question

Attention reader! Don’t stop learning now.  Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.

Learn all GATE CS concepts with Free Live Classes on our youtube channel.

My Personal Notes arrow_drop_up
Recommended Articles
Page :