#CN Stop And Wait Numerical Question - Finding Throughput of Receiver?

Question

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 (1 Kbps = 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 RECIEVER throughout is ________ bytes/ second.

Comments

250?

Answer 1

Sender side 

L = 1000 B = 8000 bytes

B = 80 kbps = 10000 Byte per second

Tp = 100ns

throughput = efficiency * Bandwidth

Tt = 8000 / 80000 =100ms

η = (Tt at sender) / (Tt sender + Tt of ack + 2Tp)

Tack = 100*8 / 8000 = 100ms

η = 100 / ( 100 + 100 + 200) = 0.25

throughput = efficiency * Bandwidth

                  = 0.25 * 80 kpbs = 0.25 * 10000 = 2500 bytes/sec

 

Receiver side:

L = 100B = 800B

B = 8kps = 1000 Bytes per sec

Tp = 100 ms ( same for sender and receiver)

throughput = efficiency * Bandwidth

Tt = 800 / 8000 =100ms

η = (Tt at sender) / (Tt sender + Tt of ack + 2Tp)

Tack = 100*8 / 8000 = 100ms

Tt  at sender = 8000 / 80000 =100ms

η = 100 / ( 100 + 100 + 200) = 0.25

throughput = efficiency * Bandwidth

                  = 0.25 * 8 kpbs = 0.25 * 1000 = 250 bytes/sec

receiver throughput = 250 Bytes/sec

 

 

Comments

@Raja Rawal Awesome brother! Thank you so much.