The values of parameters for the Stop and Wait ARQ protocol are as given below gate computer science 2017
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The values of parameters for the Stop-and – Wait ARQ protocol are as given below. -gate computer science 2017

Bit rate of the transmission channel = 1Mbps

Propagation delay from sender to receiver = 0.75 ms
 
Time to process a frame = 0.25ms
 
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 that 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 place) . 

A) A number between 86.5 and 87.5
B) A number between 82.4 and 82.5
C) A number between 92.4 and 95.5
D) A number between 96.4 and 97.5


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Answers

A) A number between 86.5 and 87.5
 
Explanation: 
Given, 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 Therefore, useful Data = Total Data – Overhead = 1980 - 20 = 1960 Tt(Data) = (1960*8) / 10^6 = 15.68 millisecond Tt(ACK) = 20*8 / 10^6 = 0.16 millisecond Two way round trip time = 2 * Tp = 2*0.75 = 1.5 millisecond T(precess) = 0.25(for info) + 0.25(for ack) = 0.5 millisecond Efficiency = Useful Time / (Tt(info) + + Tt(ACK) + 2* Tp + Tprocess ) = 15.68 / (15.84 + 0.16 + 2*0.75+ 0.5 ) = 0.8711111 = 87% Option (A) is correct.

This explanation is contributed by Deepak Raj

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