Question

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 to end-to-end propagation delay. Assume a propagation speed of \(2\times10^{8}\) 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

Answer 1

Each slot is equal to transmission time of 100 bits + propagation delay.
Propagation delay = 1 km/2 * 10\(^{8}\)ms = 5\(\mu s\)
\(T_{x}=100/10 \ Mbps =10\mu s\)
Let there are maximum N number of station then Length of cycle is \(=N^{*}(10+5)=15N\mu s\)
In a whole cycle each user transmit for only \(10 \mu s\)
Therefore efficiency is (10/15N)
Throughput of each station (10/15N)*10 Mbps
which is given as 2/3 Mbps
\(N=(10*10*3)/(15*2)=10\)