MadeEasy Test Series 2018: Computer Networks - Lan Technologies

Question

Here i'm not able to get after how much time X and Y will face first collision

Comments

Ya ryt, they are asking the exact time starting from t=0, so we have to even take into account the collision detection time so 4$\mu s$

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I think they had asked from the time from t=0, from the point when collision occurs.

1. Detecting collision = propagation delay / 2 i.e 2 micro seconds.

(As jamming signal travels half of the link to reach source X)

2. Waiting for one time slot of 2* propagation delay i.e 4*2 micro seconds = 8 micro seconds.

(After jamming signal arrives backoff algorithm applied and k value is evaluated randomly)

3. Time taken for transmitting packet onto the link and waiting for its last bit to reach station Y i.e transmission delay + propagation delay = 10 + 4 micro seconds = 14 micro seconds.

Adding 1, 2, 3 we gets total time as 24 micro seconds.

Please correct me if i am wrong.

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since ‘k’ is starting from 1 we need not worry about channel getting cleared.

Answer 1

so initially frame will require tf and tp

after collision x will wait for 8 micro sec

and now after 8 micro sec again tp will required

so T= tf+tp+8 mps+tp= 10+4+8+4=26 micro second

Comments

@Tesla! Isn't the transmission taking again so hence the retransmission take the transmission delay 10 microsecond in extra resulting in 10+4+8+10+4 i'e 36 Microseconds.

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Why do you thin, in any method we use sender keeps frame till ack is received so no retransmission is required

Answer 2

according to data given in the question

Tt = size/B.W

=150/150x10^6 = 10 micro sec

Tp= distance/speed

=800/2x10^6= 4 micro sec

now time taken by X to transfer data from X to Y will be

1. Tt and Tp to load and send data (loading time Tt and sending time Tp) at time t=0 collision  also got detected

2. It will wait 1 round trip time i.e 2Tp for retransmission of frame.

3. Now X=1 and X is ready to send the data it will only require Tp time to do so because data has already been loaded i.e time Tt is not required

now adding all we get

Tt + Tp + 2Tp +Tp = 10+4+8+4

Ans is 26 micro sec