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 -
- 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
- 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
- Considering frame processing time given is combined both $\text{ACK+Info Frame}$
- 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
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