GATE CSE 2017 Set 1 | Question: 45

Question

The values of parameters for the Stop-and-Wait ARQ protocol are as given below:

• Bit rate of the transmission channel $= 1$ Mbps.
• Propagation delay from sender to receiver $= 0.75$ ms.
• Time to process a frame $= 0.25$ ms.
• Number of bytes in the information frame $= 1980$.
• Number of bytes in the acknowledge frame $= 20$.
• Number of overhead bytes in the information frame $= 20$.

Assume there are no transmission errors. Then, the transmission efficiency (expressed in percentage) of the Stop-and-Wait ARQ protocol for the above parameters is _____________ (correct to $2$ decimal places).

Comments

i only consider one way propagation, I forgot to consider propagation delay from receiver to sender. I got 93.24. :(

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i got confused to take frame process time on both side or at one end

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i am also getting 89.33 because i considered transmission tym of overhead 20 bytes as useful tyme... dont know wether should we consider that or not..

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@arjun sir.. this question is having lots of diffrent  answers. what should be the correct answer sir? do we need to consider the processing time delay of the acknowledgement frame at sender side?

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The answer to this question is given as 88.... However, the pragy app also considers 89.34 as correct answer. Please check this as this will create a huge difference in rank predictions as you can see most of the answers are 89.34. And answer range provided by Go is 88 to 90, which is very huge range. I strongly believe this should be modified. However it may be the case that there is no perfect answer to this question as of now and we might me waiting for 27th. In that case I am completely fine. But if it's by mistake than this can be corrected.

Thanks

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i think we consider both side propagation delay and ans is===89.33% .the answer lies between 87.25 to 89.50%

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Shouldn't we send overhead bytes with ack also if we are sending overhead bytes with data bytes?

Considering that i am getting 88.8% efficiency.

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@Harish Kumar,

Overhead bytes need not be added in ack.

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frame processing time means time elapsed to process a packet (removing header +copy data+ forming its acknowledgment ) at receiver side. so consider frame processing time for both end.

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updated link: https://nptel.ac.in/content/storage2/courses/Webcoursecontents/IIT%20Kharagpur/Computer%20networks/pdf/M3L3.pdf

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GO to the Root of Concept: Video Explanation with timestamp More Questions on Stop Wait Protocol | GATE PYQs 2016, 2017, 2006, 2023 MIT, Berkeley | With NOTES

Answer 1

Efficieny is usually calculated as, $\dfrac{\text{InfoFrame Transmit Time}}{\text{TotalTime}}$

Efficiency $=\frac{\text{InfoFrame Transmit Time}}{\text{InfoFrame Transmit Time
+InfoFrame Process Time+2$\times$ Prop Delay+AckFrame Transmit Time+AckFrame Process Time}}$

Reference to calculate efficiency formula:

• http://nptel.ac.in/courses/106106091/pdf/Lecture13_StopAndWaitAnalysis.pdf
• http://spinlab.wpi.edu/courses/ece230x/lec14-15.pdf

From the question it is not very clear wether frame processing time is mentioned about $\text{InfoFrame or AckFrame or Combined}.$ It is also explicitly not mentioned wether to consider Frame Processing time for $\text{ACK}$ or not. Thus, following are the different inferences that could be made from the question -

1. As Size of InfoFrame $(1980-2000 \;\text{Bytes})$ is very large as compared to AckFrame $(20\;\text{Bytes})$ one could assume the given processing time is for InfoFrame and processing time for $\text{AckFrame}$ is neglible. The processing time does depend on size of frame for various parameters one of them is checksum calculation.
   Check the below reference for more details -
    http://rp-www.cs.usyd.edu.au/~suparerk/Research/Doc/Stop-and-Wait_Simulation.pdf
2. It is also mentioned in the question that there are no trasmission errors. One can also think as an hint that since frames are successfully transmitted there is no need for $\text{ACK}$ processing at sender Side
3. Considering frame processing time given is combined both $\text{ACK+Info Frame}$
4. Considering frame processing time individually and which is the Ans in Official key ( 86.5 - 87.5 ) 

The below answers could be due to cases $1,2,3 -$
No. of Bytes in the Information frame  $= 1980\;\text{Bytes}$
(Not very clear from question whether it implies total bytes or data bytes )

No of OverHead Bytes $=20\; \text{Bytes}$

Assuming they have explicitly mentioned Overhead bytes -

Total Frame Size $=\text{No of Bytes in the Information frame + No of OverHead Bytes = 2000 B}$

InfoTransmission Time $=\dfrac{\text{InfoFrame Size}}{\text{Bandwidth}}$

$=\dfrac{2000\times 8}{1\times 10^6} = 16\;\text{ms}$

AckTransmissionTime $=\dfrac{20 \times 8}{1 \times 10^6}=0.16\;\text{ms}$

Efficiency $=\dfrac{16}{16+ 2\times 0.75 + 0.25 + 0.16}$

                 $=89.34\%$  ( After round-off )

Assuming bytes in information includes Overhead bytes -

InfoFrameTranmission Time $=15.84$

Efficiency = 89.23 % 

Range could be 87.5 - 89.34

Reference to the similar questions:

• https://gateoverflow.in/43981/isro-2013-41
• https://gateoverflow.in/39696/gate-2016-1-55

More Efficiency Concept Reference:

• http://nptel.ac.in/courses/Webcourse-contents/IIT%20Kharagpur/Computer%20networks/pdf/M3L3.pdf

Comments

what was the answer in the official key?

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As per GATE 2017 official key answer was $86.5$ to $89.5$

If the above link doesn’t work use it’s archive

https://web.archive.org/web/20201229173406if_/https://www.gate.iitg.ac.in/2017answers/CS1.pdf#page=2

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@Arjun @gatecse sir

Official answer is 86.5 to 89.5 (https://gate.iitk.ac.in/doc/papers/2017/cs_2017.pdf) 

see question no. 45 of set 1.

not 86.5 to 87.5(as mention in the above answer).

Answer 2

Transmission efficiency = $\frac{Transmission-time-for-useful-data}{Total-time}$

Useful data = Information frame - Overhead = 1980 - 20 = 1960 B.

Transmission time for useful data = $\frac{1960*8}{10^{6}}$ = 15.68 msec.

Total transmission time for Information frame(with overhead) = $\frac{1980*8}{10^{6}}$ = 15.84 msec.

Total transmission time for Acknowledge frame = $\frac{20*8}{10^{6}}$ = 0.16 msec.

RTT = 2 * Propagation delay = 2 * 0.75 = 1.5 msec.

Time to process frames = 0.25(at receiver for information frame) + 0.25(at sender for acknowledge frame) = 0.5 msec.

$\therefore$ Total time = 15.84 + 0.16 + 1.5 + 0.5 = 18 msec.

$\therefore$ Transmission efficiency = $\frac{15.68}{18}$ = 0.8711 = 87.11%

$\therefore$ 87.11 should be answer.

Comments

HRK they had asked to round-off. SO should be 89.34%

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@Sudeep Yes, I had edited it with 0.5 msec but someone again edited my answer. With 0.5 I'm getting 87.11%.

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Why are we not considering ack frame with info frame?

Answer 3

Time to transmit (Overhead+Data)=(1980*8)/10⁶ =15840  micro.sec

Time to transmit (Data)=(1960*8)/10⁶ =15680  micro.sec

Time to transmit (Ack)=(20*8)/10⁶ =160  micro.sec

One way Processing Delay=250 micro.sec  ,Two way=500   micro sec

One way Propagation Delay=750 micro.sec ,Two way=1500   micro sec

Efficiency=(Time to send Data)/(Total Time)=(15680/(15840+160+500+1500))=15680/18000=0.8711=87.11%

https://www.eecis.udel.edu/~cshen/450419/notes/reliable.pdf

Comments

@arjun sir.. do we take the entire frame into consideration while calculating transmission efficiency for stop and wait or we consider only its useful data part?

Answer 4

solution........