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GATE CSE 2016 | Set 1 | Question: 55

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88 88 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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Best answer
102 102 votes

Answer is 2500 bytes per second.

Throughput is number of bytes we are able to send per second.

Calculate the transmission time of sender $T_{t(\text{Send})},$ calculate one way propagation delay $T_p,$ Calculate the transmission time of receiver $T_{t\text{(Recv)}}$
We get $T_{t(\text{Send})}$ here as $\dfrac{1}{10}$ seconds,

$T_p$ as $\dfrac{1}{10}$ seconds( given in question as $100\ \text{ms}$ ),
$T_{t\text{(Recv)}}$ as $\dfrac{1}{10}$ seconds.

So , total time taken to send a frame from sender to destination,

$=T_{t(\text{Send})}+2\times T_p+T_{t\text{(Recv)}}=\dfrac{4}{10}$ seconds

So, we can send $\text{1000 bytes}$ (frame size) in $\dfrac{4}{10}$ seconds.
in $\text{1 second}$, we can send $\text{2500 bytes}$. So throughput is $\text{2500 bytes}$ per second.

• edited by
84 84 votes
Answer is $2500$.

Sender transmission time $=\dfrac{1000\times 8}{(80\times 1000)}=0.1\text{ sec}=100\text{ ms}$

Receiver transmission time $=\dfrac{100\times 8}{(8\times 1000)}=0.1\text{ sec}=100\text{ ms}$

RTT $=2\times 100=200\text{ ms}$

So, Total time $=400\text{ ms}$

In $400\text{ ms},$ we send only $1000\text{ bytes}$ so,

Throughput $=\dfrac{1000}{(400\times 10^{-3})}=2500\text{ bytes / sec}$
• edited by
18 18 votes
stop and wait arq

sender :

L= 1000 bytes

bandwidth = 80 Kbps

tf= 1000*8/80 * 1000=0.1sec

receiver:

ack. L=100 bytes

bandwidth = 8 Kbps

tf= 100 * 8/8 *1000= 0.1 sec

propogation delay tp= .1 sec

___________________________________________

senders throghput = efficiency * bandwidth

 total time = .1 + .1 + 2* .1 = .4 sec

senders tf = .1 sec

efficiency =  0.1/ .4=.25 = 25%

________________________________________

troghtput of sender = 25 % * 80 Kbps=20000 bps /8= 2500 bytes / sec
14 14 votes
Throughput= datasize/total time

where total time= Transmission time of sender + Transmission timevof receiver + 2* PT

so, total time = 1/10+ 1/10+200×10^-3

                         = 4/10

Now, Throughput = 1000/0.4 =2500 bytes
11 11 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 4 votes

Sender throughput= data/total time 

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