Total time in Transmission Between two stations

Question

A simple network consist of two nodes X and Y. Each node has a single frame of F = 1500 bits (including all header and preambles) to send to each other. Both nodes attempt to transmit at ‘t = 0’. In case that a node executes the exponential Back-off algorithm. In the algorithm, a node experiencing the n^th collision in a row for given frame, randomly chooses a value of k from {1, . . . 2^m} where m is min (n, 10). The node then wait 8k μs before attempting to retransmit the frame. It is also given that the propagation speed of the signal over the network is 2 × 10⁸ m/s and a data rate of 150 Mbps

Assume that collision occur at time t = 0, and that the node X draws k = 1 and Y draws k = 2. The time taken by X’s packet (from t = 0) to be completely delivered to Y is ________ μs.

I think it should be 36 μs:-

at t = 0:- TT + 2PT here 2PT is the time to detect the collision.

at t = 1. waiting time is 8*1 μs = 8 μs, then TT + PT.

And TT = 10 μs and PT = 4 μs, and thus it is total 36μs. but the answer given is 26μs.

Answer 1

Here answer should be 26µsec.

As given in question at time T=0 : collision occurred.

So collision can occur anywhere on the link. We will consider the worst case, suppose that collision occur at Y. So form there, collision signal needs Tp(propagation time) to reach at X : 4µsec

{ Tp = distance/speed }

[So in this case propagation delay= 800/200,000,000 = 4]

Hence At T=4 : Now X waits for 8µsec.then it send packet

{Here k=1 for X, so X will wait for 8µsec}

Now At T=12 : starts sending packet

And packet needs (Tt + Tp) time to reach at Y.

{ Transmission delay = Frame size/Bandwidth}

[ So here Tt= 10µsec ]

So in this case it takes 14µsec to reach at Y ( i.e 10+4)

So at T = 26 : Complete packet delivered from X to Y.

Here your mistake is taking that formula 2*Tp for collision detection.

This(2*To) formula can be applied if they ask that At time T=0, X sends packet to Y. So in that case we can take worst time for collision detection=2*Tp

But here it is given that "ASSUME THAT COLLISION OCCUR AT TIME T=0" it means collision already occurred at time T=0, we only need to check that what the worst time to transmit collision signal to  X from the point at which collision occurred. So here Tp is the worst time

Comments

At t = 0 both X and Y send the packet, so why don't you considered there TT also? Because collision will occur when both sends there packet at t = 0.

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No, Question is not like that,

Here in this question at time T=0 collision occurres, so don't think about what happened before the collision? When X and Y started transmission? .. No need to think about all that question..

Just start thinking that .. Collision occurres at Time T=0, Now What will happen?

And answer to this question is :1) It send collision signal to X for the point of collision, 2)Then X wait for back-off time and then 3)it start transmitting again...

HERE AS YOU ARE THINKING AT TIME T=0, X AND Y SEND THE PACKET,THATS WRONG , JUST READ THE QUESTION PROPERLY YOU WILL GET THE INFORMATION THAT COLLISION OCCUR AT TIME TIME T=0, DONT WORRY ABOUT WHEN X AND Y TRANSMITTED PACKET...JUST SIMPLY AS GIVEN IN QUESTION "collision occurre at Time T=0"..

Read Question once again and try to understand...