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58 votes
58 votes
A sender uses the Stop-and-Wait $\text{ARQ}$ protocol for reliable transmission of frames. Frames are of size $1000$ bytes and the transmission rate at the sender is $80\;\textsf{Kbps} (1\textsf{Kbps} = 1000\;\text{ bits/second}).$  Size of an acknowledgment is $100$ bytes and the transmission rate at the receiver is $8\;\textsf{Kbps.}$ The one-way propagation delay is $100$ milliseconds.

Assuming no frame is lost, the sender throughput is ________ bytes/ second.
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8 votes
8 votes
Throughput = efficiency * bandwidth.

Sender's throughput = Sender's efficiency * Sender's bandwidth.

 

Sender's throughput  $=\frac{T_t}{T_t + T_{t'}+T_p+T_{p'}}*80Kbps$

=> $\frac{100msec}{100msec + 100msec+100msec+100msec}*80Kbps$
 

=> $\frac{1}{4}*80Kbps$
 

=> $20Kbps$
 

=> $\frac{20*1000}{8}Bytes/s$ (The question defines 1Kbps = 1000 bits per sec)
 

=>$2500Bytes/s$
4 votes
4 votes

Sender throughput= data/total time 

2 votes
2 votes
Message Tx Time = 100ms

ACK Tx Time = 100ms

RTT = 200ms

So, efficiency we will 0.25

then throughput would 2500Bytes / sec
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