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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 _______.
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19 votes

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 .

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5 votes
5 votes
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
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0 votes

$\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:

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