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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. $20$

Tt = 10 micro seconds
Tp = 5 micro seconds
In 15 microseconds 100 bits are sent.
Throughput of the whole system  = total data sent per second

15 microseconds = 100 bits
= $\frac{100}{15}*10^6$ per second
we need n systems such that $n*\frac{2}{3}*10^6 \leq \frac{100}{15}*10^6 = 10$
Hence, (C) is the correct choice!

cool!

$T_t = 10 \text{ micro secs}$

$T_p = 5 \text{ micro secs}$

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: $C$

@bikram sir can u explain the meaning of this question even though the max bandwidth is 10mbps then we are getting effective bandwidth =20/3mbps.
As in one time slot only one station accesses the channel why doesn't it make use of whole of the 20/3 throughput/eff. bw instead of just 2/3 throughput?

Actually Each station have 10mbps bandwidth but they cannot use 10mbps bandwidth for whole time period since they are divided into slots(using in time sharing manner)…

we have divided into 10 slots hence even though bandwidth is 10mbps ,we can only use 1 Mbps for each station  hence throughput is coming as 2/3

example –

Bandwidth is 10mbps and divided into 1 slot only so station1 has whole  bandwidth all the time.

Bandwidth is 10mbps and divided into 2 slot only so station1  can use 10mbps BUT only for half time  and for other half station1 will not use (idle) hence station1 get equivalent 5mbps even though 10mbps available to it.

s1(transferring at 10mbps but only for half period,s2 idle)  | s2 (transfewrring at 10bmps but only half peiod,s1 idle) | s1(continue...) |  s2 (continue..)|

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

by

@Tendua 2/3 is the efficiency, but we need 2/3Mbps throughput as per question.
@Anurag_s

May be Beacuse Total Cycle Time is 15Nmicro_sec and In one Cycle a station can send only 10micro_sec thats why for  a single station its useful time is only 10micro_sec out of total cycle time which is 15Nmicro_sec.

that's why (10micro_sec/15Nmicro_sec)
edited
....

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

by

### 1 comment

this is correct.

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

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. $\frac{2}{3}$

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

@nadeshseen-Bro, but Kurose Ross has a different story to it.

Sir, which one is correct?

@Pranavpurkar ...the kurose Ross is completely right ....but @Ayush Upadhyaya 's understanding is completely wrong ...he understood it in a wrong and different way ..