3.1k views
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. $20$
edited | 3.1k views
+3

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!

$T_t = 10 \text{ micro secs}$

$T_p = 5 \text{ micro secs}$

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.
edited
+1
how only the transmission time is useful time why not prapogation time...?
0
+4

how only the transmission time is useful time why not prapogation time ?

efficiency or throughput  formula = transmission time / ( transmission time + propagation time )

here we say , Total time = transmission time + propagation time

--------

Now Throughput means  the amount of data ( the whole message ) moved successfully from one place to another in a given time period . it is measured as bits per seconds bps ( how many bits transmit per seconds )or Kbps , Mbps  etc.. so it is measurement of whole number of bits ...

Where ,

Transmission time = This is the amount of time required to transmit all of the packet's bits into the link.

Propagation time = This is the time taken by a single bit ( MSB in general ) to reach from sender(A) to receiver(B).

As Transmission time is the time to transmit a whole packet( or total number of bits ) over a link , we consider it as useful time ..

and for total time we consider Transmission time + Propagation time .

0
Sir why are we multiplying by n.. In TDM medium access at each time slot is allotted to one station.So at any time the throughput of station should be equal to that of the link.. please correct me where I am going wrong.
+1

jaig

see this line :

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

If there are N stations and throughput of each station is 2/3 Mbps according to problem statement then maximum throughput = N * 2/3

as it asking for maximum value, we need to multiply with N here..

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ok now i got it. thank u sir
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Why are we not taking throughput = TT/TT+2*PT . as we know total time is TT+ 2*PT where 2*PT is RTT time  please answer this i am very much confused.
+5
This is TDM, here total cycle time is sum of transmission time and propagation time. In case of other sliding protocol mechanisms, we also consider the time for the data as well as ack to propagate and hence 2* propagation delay in 1 cycle
+1
thanks :-)

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

0
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

0
@arjun sir . if i just find the efficency if one sytem is there it will be tt/tt+tp where tt is transmissin time and tp is propogation delay . so it will be 10/15 which is equal to 2/3 which means efficency with one system is 2/3 then it should be one system only. i think so .
+2
some doubt..in 1 cycle ie 15*N msec..why useful time is 10 micro sec..? shouldn't each station transmit..1 packet each n transmission time as 10*N ms?
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total cycle time = T+Tp in TDM then why number of hosts are multiplied to calculate total cycle time

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@Tendua 2/3 is the efficiency, but we need 2/3Mbps throughput as per question.
0
@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)

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. $\frac{2}{3}$

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