# GATE2014-1-26

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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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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

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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why Tp=10? Tp should be calculated for complete ring.. which is independent of the no. of stations.. then why are you multiplying (length/speed) by 10?
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Ring latency is 2km/(2*10^8 m/s)

= 10-5 s = 10 μs

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oh..thanks..it was my calculation mistake.
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Answer should be 28, because only 9 stations will be holding the token?
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why only 9?
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GATE key says 28 and 30 as answers :)
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Only 9 stations because
Let's say monitoring station release the token at time x, then
1. token will travel the whole ring, time to propagate=10 microseconds
2. token will travel the other 9 stations, which will hold the token for 2 microsec each (18 microsec in all)
3. after that it will reach the monitoring token again. At this point of time it will be expecting the token to come back. that means it is expecting the token to be back at x+28microsecs.

So it will have to wait 28 microsecs before it can infer token is lost.
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yeah .. through this concept itoo get 28us
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Acc to me also it should be 28
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@Arjun Sir,How 10^-5 second equals 10 micro seconds.Plz help Sir.I m stuck.:(. Thanks again Sir
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$1 second = 10^3 millisecond = 10^6 microsecond. Hence 10^-5 second = 10^-5 * 10^6 microsecond = 10 microseconds.$
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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1
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In the diagram shown below, $L1$ is an Ethernet LAN and $L2$ is a Token-Ring LAN. An $IP$ packet originates from sender $S$ and traverses to $R$, as shown. The links within each $\text{ISP}$ and across the two $\text{ISP}$s, are all point-to-point optical links ... of the $\text{TTL}$ field is $32$. The maximum possible value of the $\text{TTL}$ field when $R$ receives the datagram is _______.