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+4 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
in Computer Networks by Boss (16.3k points)
recategorized by | 2.8k views

sends a frame of 1000 bytes, removes the frame after it has circulated all around the ring, and finally releases the token.

This signifies token passing technique used is DELAYED TOKEN REINSERTION( DTR )

In this case,

THT = Tt + RL = Tt + Tp + N*b

(In most cases, bit delay is 0)

So, THT = Tt + Tp [ RL = Tp ]

where Tt = transmission delay

Tp = propagation delay

Token Holding Time (THT) –The maximum time a token frame can be held by a station is known as THT.
No station can hold the token beyond THT.

Efficieny, e = Useful time/ Cycle time

useful time = N*Tt

cycle time = Tp + (THT*N)

N = no. of stations in ring

$\eta \ = \frac{N*T_t}{T_p +(THT*N)}$

N = 1 ( only 1 station is sending )

$T_t \ =\ 800 \mu s$

THT = (800 + 400)$\mu s$

$= 1200\mu s$

$\eta \ = \frac{1*800}{400 +(1200*1)}$

$\eta \ = \frac{800}{1600}$

$\eta \ = \frac{1}{2}$

BW = $\eta \ * R$

$= \frac{1}{2} \ *\ 10Mbps$

$= 5Mbps$

Answer: $C)$


4 Answers

+11 votes
Best answer

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

by Active (2.5k points)
selected by

Ring latency = propagation delay + interface delay

So I believe they have given the ring latency including interface delay. So we can't say interface delay is not given .It is included there in the ring latency itself.


efficiency: N*Tt/(PT+N * THT)



Effective data rate=80*10/124


How N=10 In your solution
+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

by Boss (33.9k points)
Can you please explain why utilization is Tx/(Tx + 2*Tp) ?

I feel it should be Tx/(Tx + Tp) since it's token ring.
+2 votes

RL=400 micro sec






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

Option should be -D
by Boss (10.2k points)
What is 10 in tour sollution of Efficieny ?
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


Effective Data Rate = e*Data Rate = .5*10=5 Mbps .
by Active (4.1k points)
currently token ring is in gate syllabus or not??

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