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.
Quiz of this Question