# #Gate 2016

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A sender uses the stop and wait ARQ protocol for reliable transmissions of frames. Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps(1Kbps=1000bits/sec). Size of an acknowledgement is 100 bytes and the transmission rate at the receiver is 8 Kbps. The one way propogation delay is 100 milliseconds.

Assuming no frame is lost, the sender throughput is ______bytes/second.
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2500Bps
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Bhai solution?

Throughput of sender = size of frame sent by sender/total time

Total time = t1+t2+t3+t4

Where

t1=transmission time of sender to place the frame of 1000B on the channel

t2=propagation time of the frame to reach the receiver

t3=transmission time of receiver to put the acknowledgement of 100B on the channel

t4=propagation time for the acknowledge to reach the sender

t2=t4=100ms (given)

t1=L/B,  L=1000B and B=80kbps

So t1=100ms

And t3=La/B' , La=100B and B'=8kbps

t3=100ms

Total time= 4*100ms =400ms

Throughput= 1000B/400ms = 2500B/s

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Thank you.

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