In a TDM medium access control bus LAN, each station is assigned one time slot…
2005
In a TDM medium access control bus LAN, each station is assigned one time slot per cycle for transmission. Assume that the length of each time slot is the time to transmit 100 bits plus the end-to-end propagation delay. Assume a propagation speed of 2 x 108 m/sec. The length of the LAN is 1 km with a bandwidth of 10 Mbps. The maximum number of stations that can be allowed in the LAN so that the throughput of each station can be 2/3 Mbps is
- A.
3
- B.
5
- C.
10
- D.
20
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Correct answer: C
Solution:
Compute slot length: time to send 100 bits = 100 / 10,000,000 = 0.00001 s (10 µs). Propagation delay = 1000 m / (2 × 10^8 m/s) = 0.000005 s (5 µs). Therefore slot length = 0.00001 + 0.000005 = 0.000015 s (15 µs).
Throughput per station in TDM = bits per slot / (N × slot length) = 100 / (N × 0.000015) bits/s.
Set required throughput = 2/3 Mbps = 666,666.67 bps and solve for N: 100 / (N × 0.000015) = 666,666.67.
Rearrange: N = 100 / (666,666.67 × 0.000015) = 100 / 10 = 10.
Check: per-station throughput with 10 stations = 100 / (10 × 0.000015) = 666,666.67 bps = 2/3 Mbps.
Answer: 10 stations.
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