See below problem from Kurose & Ross book,

solution a and b from given link

http://www.cs.toronto.edu/~marbach/COURSES/CSC358_S17_2/st3.pdf

The Gateway to Computer Science Excellence

+38 votes

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 acknowledgment 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.

Assuming no frame is lost, the sender throughput is ________ bytes/ second.

+3

Almost same problem given in K&Ross book

See below problem from Kurose & Ross book,

solution a and b from given link

http://www.cs.toronto.edu/~marbach/COURSES/CSC358_S17_2/st3.pdf

See below problem from Kurose & Ross book,

solution a and b from given link

http://www.cs.toronto.edu/~marbach/COURSES/CSC358_S17_2/st3.pdf

+52 votes

Best answer

**Answer**** is 2500 bytes per second.**

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

Calculate the transmission time of sender $T_{t(Send)},$ calculate one way propagation delay $T_p,$ Calculate the transmission time of receiver $T_{t\text{(Recv)}}$

We get $T_{t(Send)}$ here as $\dfrac{1}{10}$ seconds,

$T_p$ as $\dfrac{1}{10}$ seconds( given in question as $100\ ms$ ),

$T_{t\text{(Recv)}}$ as $\dfrac{1}{10}$ seconds.

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

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

+1

I am sorry sir but i feel pathetic of you. Let him answer what he is doing. Language has no relation with technology. Else it would have been colour and not color.

+1

Why do we take 80Kbps as sender transmission time?.....because at receiver R=8Kbps so bottleneck bandwidth of the link should be 8kbps only.Please correct my concepts with this example.

+1

i am in a deep confusion here!! throughput=BW*utilization

when i solve like in this way i am getting 5000 .but why! someone tell me ,what should be the approach to do it in Bw*utilization method!

when i solve like in this way i am getting 5000 .but why! someone tell me ,what should be the approach to do it in Bw*utilization method!

+40

Efficiency = T_{t }/ (T_{t }+2*T_{p}+T_{ack}) = 100 ms / 400 ms = 0.25

Throughput = Efficiency * bandwidth

= 0.25 * 80000/8 bytes per second

= 2500 bytes/ s

+1

transmission time of receiver Tt_Recv = Ack size / trans rate of rcvr = 100 Bytes / 8Kbps =800b/8000bps = 0.1sec.

+1

Sender's utilization is asked and BW is also given. In case the ack frame has to b considered, then we would have taken bottleneck bandwidth.

+1

Hii,

i'm new here and not understood the full answr...?

why we are dividing and multiplying wih 1000??

can any1 tell me the answr with use of term and full explanation....??

thnxxx advnce

i'm new here and not understood the full answr...?

why we are dividing and multiplying wih 1000??

can any1 tell me the answr with use of term and full explanation....??

thnxxx advnce

+1

Why we are considering 'Tt_Recv' ( transmission time of receiver) here????

I dont think that we include transmission time of receiver while calculating transmission time .

Please help!!!!!!!

I dont think that we include transmission time of receiver while calculating transmission time .

Please help!!!!!!!

+1

Abhishek in most of the questions it is written that transmission time of receiver is negligible but here it is not mentioned so you have to go through with the exact formula. In derivation we neglect transmission time of receiver, processing delay at receiver, queuing delay so we get the formula as Transmission time/(Transmission time+2*propagation delay) but here they have mentioned size of acknowledgement packet so we have to add the transmission time of acknowledgement packet in denominator.

+1

Hi Peeyush,

True that in most of the cases transmission time for receiver in most of the case is negligible so we take it as zero,

but now try to understand, in our question it is clearly given 100bytes of Ack it's not negligible at all.

so if A is a sender and B is receiver. A sends his frame of 1000 bytes and as soon as B receives it he will send ACK(100 bytes with bandwith of 8kbps).

so total time in one round travel would be as per question = Tranmission time for sender + tranmission time for receiver(Because he must send the ACK to sender) + propogation delay.

True that in most of the cases transmission time for receiver in most of the case is negligible so we take it as zero,

but now try to understand, in our question it is clearly given 100bytes of Ack it's not negligible at all.

so if A is a sender and B is receiver. A sends his frame of 1000 bytes and as soon as B receives it he will send ACK(100 bytes with bandwith of 8kbps).

so total time in one round travel would be as per question = Tranmission time for sender + tranmission time for receiver(Because he must send the ACK to sender) + propogation delay.

+38 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 $=\dfrac{1000}{(400\times 10^{-3})}=2500\text{ bytes / sec}$

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 $=\dfrac{1000}{(400\times 10^{-3})}=2500\text{ bytes / sec}$

+1

Receiver transmission time =1000×8(80×1000)=0.1 sec=100 ms ?

Your answer is right but Receiver transmission time = (100 * 8 )/ (8 * 1000) = 0.1 sec = 100 ms.

Your answer is right but Receiver transmission time = (100 * 8 )/ (8 * 1000) = 0.1 sec = 100 ms.

+1

i'm new here and not understood the full answr...?

why we are dividing and multiplying wih 1000??

can any1 tell me the answr with use of term and full explanation....??

thnxxx advnce

+1

Why Receiver transmission time has been included??

I dont think that we include transmission time of receiver while calculating transmission time .

Please help!!!!!!!

I dont think that we include transmission time of receiver while calculating transmission time .

Please help!!!!!!!

+7 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

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

+6 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

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

+2 votes

+1

time calculation is like transmission time + propagation time (of sender) which is 0.1 and 0.1 second, and transmission time and propagation time (of receiver) is also 0.1 and 0.1 second.

so total time needed (that is 1 RTT) is 0.1+0.1+0.1+0.1 = 0.4

Therefore, Throughput = Data/ Total time

which is calculated above as 2500 B.

+1 vote

Message Tx Time = 100ms

ACK Tx Time = 100ms

RTT = 200ms

So, efficiency we will 0.25

then throughput would 2500Bytes / sec

ACK Tx Time = 100ms

RTT = 200ms

So, efficiency we will 0.25

then throughput would 2500Bytes / sec

+1 vote

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)

=>$2500Bytes/s$

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)

=>$2500Bytes/s$

- All categories
- General Aptitude 1.9k
- Engineering Mathematics 7.5k
- Digital Logic 2.9k
- Programming and DS 4.9k
- Algorithms 4.4k
- Theory of Computation 6.2k
- Compiler Design 2.1k
- Databases 4.1k
- CO and Architecture 3.4k
- Computer Networks 4.2k
- Non GATE 1.4k
- Others 1.4k
- Admissions 595
- Exam Queries 573
- Tier 1 Placement Questions 23
- Job Queries 72
- Projects 18

50,737 questions

57,284 answers

198,184 comments

104,863 users