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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
asked in Computer Networks by (19.9k points)  
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3 Answers

+14 votes
Best answer
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.
answered by (2.8k points)  
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how only the transmission time is useful time why not prapogation time...?
+9 votes

Efficiency=useful time/total time

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


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



answered by (7.7k points)  
this is correct.
+8 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


answered by (2.6k points)  
@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 .
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?

total cycle time = T+Tp in TDM then why number of hosts are multiplied to calculate total cycle time

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

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)

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