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CMI 2019
August 31, 2024
CMI 2019
August 31, 2024

Stop-and-Wait-ARQ

Question 9
What will be the efficiency of a Stop and Wait protocol, if the transmission time for a frame is 20ns and the propagation time is 30ns?
A
20%
B
25%
C
40%
D
66%
Question 9 Explanation: 
→Assuming that the transmission time for the acknowledgement and the processing time at nodes are negligible.
→The time taken by a sender to send a single packet is equal to transmission time (data) + 2 * propagation delay , out of which only transmission time is useful .
→Then, efficiency =transmission time / transmission time + 2 * propagation delay.
→Given data is transmission time is 20ns
→Propagation delay(time) is 30ns.
→Efficiency =20+(20+2*30)=20/80=0.25 which is equivalent to 25%
Correct Answer: B
Question 9 Explanation: 
→Assuming that the transmission time for the acknowledgement and the processing time at nodes are negligible.
→The time taken by a sender to send a single packet is equal to transmission time (data) + 2 * propagation delay , out of which only transmission time is useful .
→Then, efficiency =transmission time / transmission time + 2 * propagation delay.
→Given data is transmission time is 20ns
→Propagation delay(time) is 30ns.
→Efficiency =20+(20+2*30)=20/80=0.25 which is equivalent to 25%
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