GATE IT 2004 | Question: 82

Question

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

• A. 1 Mbps
• B. 2 Mbps
• C. 5 Mbps
• D. 6 Mbps

Comments

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: $C)$

Ref: https://www.geeksforgeeks.org/computer-network-efficiency-token-ring/

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Not in current syllabus of Gate CS

Answer 1

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

Comments

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.

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

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

=80/124

Effective data rate=80*10/124

=6.45

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How N=10 In your solution

Answer 2

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

Comments

Can you please explain why utilization is Tx/(Tx + 2*Tp) ?

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

Answer 3

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

Comments

What is 10 in tour sollution of Efficieny ?

Answer 4

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 .

Comments

currently token ring is in gate syllabus or not??

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no

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how ?