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5 votes
5 votes

Consider a 10 Mbps token ring LAN with a ring latency of 400 µs. A host that needs to transmit seizes the token. Then it sends a frame of 1000 bytes, removes the frame after it has circulated all around the ring, and finally releases the token. This process is repeated for every frame. Assuming that only a single host wishes to transmit, the effective data rate is

  1. 1 Mbps
  2. 2 Mbps
  3. 5 Mbps
  4. 6 Mbps
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4 Answers

Best answer
12 votes
12 votes

answer will be C
 

transmission time of frame is 800 µs
given that ring latency is 400 µs
since interface delay is not given so it is negligible so Ring latency is equal to propgation time 
it is employing delayed token reinsertion so 
so utilisation of token ring will be 
N*Tt/(PT+N * THT)
in delayed token THT is Tt + Ring Latency so utilisation is  0.5 
so throughput  is utilization * B.W
so it will be 0.5*10 Mbps 
so answer is 5Mbps

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2 votes
2 votes

Answer: C

Transmission Time: 1000*8 bits/10 Mbps = 800 µs
Latency = Propagation Time = 400 µs

Link utilisation = 800/(800+2*400) = 1/2

Effective Data Rate = 1/2 * 10 Mbps = 5 Mbps

2 votes
2 votes
Bw=10Mbps

RL=400 micro sec

L=1000Byte

Efficiency=N*Tt/Tp+N*RL

RL=Tt+Tp;

Efficiency=10*800/400+10*(1200)

Efficiency=0.6452

Effective data rate =Efficiency*Bw=0.6452*10Mbps=6.4521Mbps

Option should be -D
0 votes
0 votes
Actually the keyword use here is LAN so we use general formula of efficiency

e=1/1+2a , a=PD/TD PD=propagation delay , TD=transmission Delay

as PD=400 us . TD=1000*8 / 10 * 10^6=800 us

e=1/1+2*(400/800)=.5

Effective Data Rate = e*Data Rate = .5*10=5 Mbps .
Answer:

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