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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

1. 3
2. 5
3. 10
4. 20
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Tt = 10 micro secs

Tp = 5 micro secs

Efficiency of the network = Tt / (Tt + Tp) = 10 / 15 = 2/3

Total throughput available for the entire network = Efficiency * Bandwidth = (2/3) * 10 Mbps = 20/3 Mbps

Let, No. of stations = N (each wants a Throughput of 2/3 Mbps),

N * (2/3 Mbps) = 20/3 Mbps => N = 10

=> 10 stations can be connected in the channel at max.
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how only the transmission time is useful time why not prapogation time...?

Efficiency=useful time/total time

=transmission time/transmission +propagation time=10ms/(10ms+15ms)

=2/3

so Effective bandwith utilized=(2/3)*10Mbps

Let there be N stations,throughput of each station should be 2/3 Mbps according to problem statement

so N *2/3 =Effective bandwith

N*2/3=(2/3)  * 10

N=10

this is correct.

For each station slot time is tx + tp(transmission time+prop. delay)

tx = 100b/10Mbps = 10μs

tp = d/v = 5μs

So slot time is 15μs

If there are N stations then total cycle time is 15Nμ sec

efficiency will be useful time/total time ie (transmission time/total time) = (10μ/15Nμ)

throughtput is eff*bandwidth => (10/15N)*10Mbps = (2/3)Mbps.

Solving this for N gives N as 10