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Question

Consider Noisy station that detects transmissions and disrupts them by beginning a competing transmission as soon as it hears the beginning of the transmitted frame, thereby causing a collision. Assume Detector machine, which is on this Ethernet with bandwidth 10 Mbps, detects collision during the transmission of its 12^th bit on the wire (including any preamble). If the speed of the signal in the wire is 2 × 10⁸ meter/second, then the distance (in meters) of the Noisy station from Detector machine is _______

Comments

24?

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Utkarsh Joshi  answer is given as 120

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okay! are you getting the same answer?

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actually i did it this way..dont know whether the approach is correct or not..

Let the distance between the noisy station and detector be d metres. As the collision is detected during the transmission of the 12th bit, so 12 bits have already been transmitted into the network.

So, as T_t>=2*T_p, so 12/(10*10⁶) = (2*d)/(2*10⁸)

SO d=120m.

But dont know whether this approach is correct..Is it correct?

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nice approach

It seems to be right

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why are you considering Tt >= 2Tp?Somoshree Datta 5

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@Utkarsh Joshi

This is Collision Detection Condition 

Tt >= 2Tp

One Important thing if The packet size if less than this then we can't detect the collision

               L >= 2*Tp*B     

 

 

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Utkarsh Joshi  because this is the condition for detecting the collision which needs to be fulfilled..this is the reason why detector m/c is being able to detect the collision

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Somoshree Datta 5 By enforcing this condition the transmitting station can detect the collision on its own. But in this question, we have a detector machine which can sense the collision. So I don't see any reason for satisfying this condition.

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Bro, Ethernet uses CSMA/CD as access control method. So, here Detector machine is CSMA/CD.

So, we have to Apply that condition.

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okay, i agree it can be 120.

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This is the Question from Lan Technology (Ethernet).

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@Somoshree Datta 5 @Utkarsh Joshi

please check this..correct me if i am wrong

In LAN technology 

Why Tt >= 2Tp ?.. because in worst case for collision signal to come back to sender can take 2Tp time and during this time we want sender to be in sending state hence we equate Tt  >= 2Tp  to help Sender understand that collision happened because of him so Stop transmission..But this is for Sender only what about other how can they understand collision ..Hence JAM signal is send when collision signal reached to sender or receiver first.

If we can consider Collision detector as Simple Host in Ethernet which detects collision during its transmission

Hence Tt >= 2TP

 

Answer 1

Using the collision Detection Formula 

L >= 2*Tp*B 

We can derive the distance as 

d <= (L/2)*(V/B )

Here d is the maximum distance up to which we can detect the collision So, 

Here L = 12 , B = 10⁷ , v = 2*10⁸ then 

distance d = 120

Comments

this same approach is also followed my  Somoshree Datta 5

what's the difference ??