GATE | Gate IT 2005 | Question 75
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
Explanation: Given in TDM LAN,
Packet size (L) = 100 bits, Bandwidth (B) = 10 Mbps = 10x106 bps, Distance (d) = 1 KM = 1x103 meters, Speed (v) = 2 x 108 m/sec, Throughput of each station (b) = (2/3) Mbps = (2/3)x106 bps
Then, find the number of stations (N) = ?
First determine effective bandwidth of the given system:
Transmission time (Tx) = L/B = (100 bits) / (10x106 bps) = 1x10-5 seconds, And, Propagation time (Tp) = d/v = (1x103 meters) / (2 x 108 m/sec) = (1/2)x10-5 seconds
Now, efficiency of the given system is,
(μ) = (Tx) / (Tx + Tp) (μ) = (1x10-5) / (1x10-5 + (1/2)x10-5) (μ) = (1) / (1 + (1/2)) (μ) = (2/3)
Therefore, effective bandwidth of the given system is,
(B') = (efficiency) * (original bandwidth) (B') = (2/3)*10 Mbps
Hence, let there be N stations, throughput of each station should be (2/3) Mbps. So,
N*(2/3) = (Effective bandwidth) N*(2/3) = (2/3)*10 N = 10
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 10.
Quiz of this Question
Please Login to comment...