Computer-Networks
August 29, 2024Computer-Networks
August 29, 2024Stop-and-Wait-ARQ
| Question 7 |
In a stop and wait protocol used across a link of bandwidth of 1Mbps, data packets to 1000 bits are transmitted. The round trip time for a bit is 20ms. The link utilization is :
| 0.5 | |
| 0.05 | |
| 0.005 | |
| 5.0 |
Question 7 Explanation:
link utilization = 1/1+2a
a = Tp/Tt
Tp = RTT/2 = 20/2 = 10ns
Tt = 1000bits/106bits/s = 1ms
∴ a = 10/1 = 10
∴ Link utilization = 1/(1+2×10) = 1/21 ≅ 0.05
a = Tp/Tt
Tp = RTT/2 = 20/2 = 10ns
Tt = 1000bits/106bits/s = 1ms
∴ a = 10/1 = 10
∴ Link utilization = 1/(1+2×10) = 1/21 ≅ 0.05
Correct Answer: B
Question 7 Explanation:
link utilization = 1/1+2a
a = Tp/Tt
Tp = RTT/2 = 20/2 = 10ns
Tt = 1000bits/106bits/s = 1ms
∴ a = 10/1 = 10
∴ Link utilization = 1/(1+2×10) = 1/21 ≅ 0.05
a = Tp/Tt
Tp = RTT/2 = 20/2 = 10ns
Tt = 1000bits/106bits/s = 1ms
∴ a = 10/1 = 10
∴ Link utilization = 1/(1+2×10) = 1/21 ≅ 0.05
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