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GATE CSE 2017 | Set 1 | Question: 45

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54,097 views
94 94 votes

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).

8 Answers

Best answer
85 85 votes

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:

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:

More Efficiency Concept Reference:

• edited by
55 55 votes
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.
• edited by
17 17 votes

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

3 3 votes

Total time = $T_{t} + T_{p}$ of Sender + $T_{t} + T_{pro} + T_{p}$ of Receiver.

Total time = $16 + 0.75 + 0.16 + 0.25 + 0.75 = 17.91 msec.$

Useful time = $T_{t} (of Sender) = 16 msec$ (because transmission time of ACK is just an overhead)

Transmission efficiency = 16 / 17.91 =89.33%

 

Some people can make the mistake that 1 packet is being sent in 18.16 msec, so efficiency = 1/18.16, but that is wrong.

Efficiency has no unit, so the numerator and denominator must have the same units.

Hence, Efficiency is either $\frac{No- of -packets- sent}{No-of- packets- that- could've -been- sent}$

or

$\frac{time.utilized}{total.time}$

The answer is 86.5 to 89.5 in official answer key.

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