GATE CSE 2014 | Set 1 | Question: 26

Question

Consider a token ring network with a length of 2 km having 10 stations including a monitoring station. The propagation speed of the signal is $2 \times10^8m/s$ and the token transmission time is ignored. If each station is allowed to hold the token for $2 µsec$, the minimum time for which the monitoring station should wait (in $µsec$) before assuming that the token is lost is _______.

Comments

minimum time should definitely be equal to ring latency , even if token holding time is given in order to calculate the minimum time i assume token is not held by any of the station and the token has to physically travel the entire trip , only in this case i will get the minimum time = 10usec

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This topic is not part of the current GATE CSE syllabus

Answer 1

Time required to complete one cycle=Tp (Ring Latency) +N*THT 

= 2km/(2*10⁸ m/s) +10*2

= 10 + 10 * 2 = 30 μs

    where THT is token holding time .

Comments

@Arjun Sir,How 10^-5 second equals 10 micro seconds.Plz help Sir.I m stuck.:(. Thanks again Sir

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$1 second = 10^3 millisecond = 10^6 microsecond. Hence 10^-5 second = 10^-5 * 10^6 microsecond = 10 microseconds.$

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I also have the same doubt, here we should consider only 9 stations will hold the token or monitor itself would also hold the token...

Answer 2

My answer is 30 us. Please someone verify.

Comments

yes sir , after 11 yrs :)
Tp = dis / time
2*10^8  =  2000 / time  => 10 usec
for 10 stations => 10*2 => 20 usec
Total => 30 usec

Answer 3

Length = 2 km
Propagation Speed v = 2*10^8 m/s
Token Holding Time = 2 micro sec

Waiting time
= length/speed + (#stations - 1)*(token holding time) to
  length/speed + (#stations)*(token holding time)
= 28 to 30

Answer 4

$\Rightarrow Propagation Time =$$\frac{Distance}{Speed}=\frac{2\,Km}{2 * 10^{8}\, m/s} = 10\, \mu s$

Including the monitoring station there are 10 stations.

Now once the token is released by the monitoring station it should wait atleast the time till the token comes back after travelling 2 Km through other 9 stations present in the ring.

So minimum time to wait is,

$10 \, \mu s$ propagation time for $2\, Km$ distance $+$ $9$ stations hold the token for $2 \mu s$ each.

$\Rightarrow 10 \, \mu s + 2\, \mu s * 9 \, stations \Rightarrow 28 microseconds$

Anyways gate answer key has accepted 30 also as the answer. So instead of 9 if you put 10, it was still accepted.

Answer 5

this part is not in syllabus now after 2016

Comments

Thanks!

Answer 6

Answer: [28, 30]

 

$$\text{Following is an idea about the network topology}$$

$$\text{Monitoring station = 1}$$

$$\text{Non-monitoring stations = 9}$$

 

$$\text{Assuming the monitoring station DOES NOT hold the token.}$$

 

$$\text{Token holding time per station = 2 }\mu s$$

$$\text{Token holding time for 9 stations = 18 } \mu s$$

 

$$\text{Now, if a packet is lost, its information needs to get "propagated" to the monitoring station by the appropriate stations.}$$

$$\text{Maximum propagation delay =} \text{ Distance} / \text{ Propagation Speed} = 2000 m / 2 \times 10^{8} m/s = 10 \mu s$$

 

$$\text{Total wait time before assuming packet is lost = } 18 \mu s + 10 \mu s = 28 \mu s$$

 

$$\text{Assuming the monitoring station also holds the token.}$$

$$\text{Total wait time before assuming packet is lost = } 30 \mu s$$