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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 $100$ $\text{bits}$ plus the end-to-end propagation delay. Assume a propagation speed of $2 \times 10^8 m/sec$. The length of the LAN is $1$ $\text{km}$ with a bandwidth of $10$ $\text{Mbps}$. The maximum number of stations that can be allowed in the LAN so that the throughput of each station can be $2/3$ $\text{Mbps}$ is
 
  1. $3$
  2. $5$
  3. $10$
  4. $20$
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Best answer
88 votes
88 votes
$T_t = 10 \mu s$

$T_p = 5 \mu s$

Efficiency of the network $=\dfrac{T_t }{(T_t + T_p)}=\dfrac{10}{15}=\dfrac{2}{3}.$

Total throughput available for the entire network $=\text{Efficiency$\times $ Bandwidth}$
$=\dfrac{2}{3}\times 10 \text{ Mbps}=\dfrac{20}{3}\text{ Mbps}$

Let, no. of stations $=N\text{(each wants a throughput of $\dfrac{2}{3}$ Mbps)},$

   $N\times \dfrac{2}{3}\text{ Mbps}=\dfrac{20}{3} \text{ Mbps}\Rightarrow N=10.$

$\Rightarrow 10$ stations can be connected in the channel at max.

Correct Answer: $C$
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19 votes
19 votes

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

Correct answer is 3

13 votes
13 votes

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 

13 votes
13 votes

Length of 1 time slot=Transmission time for 100 bits + propagation time (one-way).

Now

$T_t(100\,bits\,)=\frac{100bits}{10^7bps}=10\mu s$

$T_p=\frac{10^3}{2*10^8}=5 \mu s$

Cycle time or 1 slot time=$15 \mu s$

Now in a TDM based channel access method, each node gets dedicated bandwidth of $\frac{R}{N} bps$ where 

R=Bandwidth of the broadcast medium

N=Total number of nodes in the network

For each node, the efficiency would be = $\frac{10}{15}=\frac{useful\,time}{total\,time}=\frac{2}{3}$

So, for each node the Throughput would be=Efficiency*Available Bandwidth=$\frac{2}{3}*\frac{10\,Mbps}{N}$

and this throghput is given in question i.e. $\frac{2}{3}$

$\frac{2}{3}*\frac{10\,Mbps}{N}=\frac{2}{3}\,Mbps$

N=10. Answer

Answer:

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