Software-Engineering
August 24, 2024ER-Model
August 25, 2024Access-Control-Methods
| Question 1 |
A broadcast channel has 10 nodes and total capacity of 10 Mbps. It uses polling for medium access. Once a node finishes transmission, there is a polling delay of 80 μs to poll the next node. Whenever a node is polled, it is allowed to transmit a maximum of 1000 bytes. The maximum throughput of the broadcast channel is
| 1 Mbps | |
| 100/11 Mbps | |
| 10 Mbps | |
| 100 Mbps |
Question 1 Explanation:
Efficiency = Transmission time(Tt)/Transmission time+Polling delay
(Tt) = 1000bytes/10×106bits/sec = 800μs
Polling delay = 80 μs
Efficiency = 800/800+80 = 800/880 = 10/11
Maximum throughput is
= 10/11 × 10 Mbps
= 100/11 Mbps
(Tt) = 1000bytes/10×106bits/sec = 800μs
Polling delay = 80 μs
Efficiency = 800/800+80 = 800/880 = 10/11
Maximum throughput is
= 10/11 × 10 Mbps
= 100/11 Mbps
Correct Answer: B
Question 1 Explanation:
Efficiency = Transmission time(Tt)/Transmission time+Polling delay
(Tt) = 1000bytes/10×106bits/sec = 800μs
Polling delay = 80 μs
Efficiency = 800/800+80 = 800/880 = 10/11
Maximum throughput is
= 10/11 × 10 Mbps
= 100/11 Mbps
(Tt) = 1000bytes/10×106bits/sec = 800μs
Polling delay = 80 μs
Efficiency = 800/800+80 = 800/880 = 10/11
Maximum throughput is
= 10/11 × 10 Mbps
= 100/11 Mbps