GATE | GATE-CS-2016 (Set 1) | Question 65

A sender uses the Stop-and-Wait ARQ protocol for reliable transmission of frames. Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps = 1000 bits/second). Size of an acknowledgement is 100 bytes and the transmission rate at the receiver is 8 Kbps. The one-way propagation delay is 100 milliseconds. Assuming no frame is lost, the sender throughput is __________ bytes/second.
(A) 2500
(B) 2000
(C) 1500
(D) 500


Answer: (A)

Explanation:

Total time = Transmission-Time + 2* Propagation-Delay + Ack-Time.
Trans. time = (1000*8)/80*1000 = 0.1 sec
2*Prop-Delay = 2*100ms = 0.2 sec
Ack time = 100*8/8*1000 = 0.1 sec.
Total Time = 0.1 + 0.2 + 0.1 = 0.4 sec.
Throughput = ((L/B)/Total time) * B, 
L = data packet to be sent and
B = BW of sender.

Throughput = L/Total Time
           = 1000/0.4
           = 2500 bytes/sec.

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