Computer-Networks
October 9, 2023Computer-Networks
October 9, 2023Computer-Networks
| Question 30 |
A 2 km long broadcast LAN has 107 bps bandwidth and uses CSMA/CD. The signal travels along the wire at 2×108 m/s. What is the minimum packet size that can be used on this network?
| 50 bytes | |
| 100 bytes | |
| 200 bytes | |
| None of the above |
Question 30 Explanation:
Minimum packet size for a CSMA/CD LAN is the frame which cover whole RTT(round trip time). i.e. Tt = 2Tp
d= 2 km = 2 x 103 m, v = 2 x 108 m/s, B= 107
Tp = d / v = 2 x 103 /(2 x 108 ) seconds = 10-5 seconds
Let L bits be minimum size of frame, then Tt = t L / B = L / 107 seconds
Now, Tt = 2Tp
L/107 = 2 x 10-5 = 200 bits = (200 / 8) bytes = 25 bytes
d= 2 km = 2 x 103 m, v = 2 x 108 m/s, B= 107
Tp = d / v = 2 x 103 /(2 x 108 ) seconds = 10-5 seconds
Let L bits be minimum size of frame, then Tt = t L / B = L / 107 seconds
Now, Tt = 2Tp
L/107 = 2 x 10-5 = 200 bits = (200 / 8) bytes = 25 bytes
Correct Answer: D
Question 30 Explanation:
Minimum packet size for a CSMA/CD LAN is the frame which cover whole RTT(round trip time). i.e. Tt = 2Tp
d= 2 km = 2 x 103 m, v = 2 x 108 m/s, B= 107
Tp = d / v = 2 x 103 /(2 x 108 ) seconds = 10-5 seconds
Let L bits be minimum size of frame, then Tt = t L / B = L / 107 seconds
Now, Tt = 2Tp
L/107 = 2 x 10-5 = 200 bits = (200 / 8) bytes = 25 bytes
d= 2 km = 2 x 103 m, v = 2 x 108 m/s, B= 107
Tp = d / v = 2 x 103 /(2 x 108 ) seconds = 10-5 seconds
Let L bits be minimum size of frame, then Tt = t L / B = L / 107 seconds
Now, Tt = 2Tp
L/107 = 2 x 10-5 = 200 bits = (200 / 8) bytes = 25 bytes
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