CSMA/CD

Question

A simple network consist of two nodes X and Z which are 800 m apart. 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 Z draws k = 2. The time taken by X’s packet (from t = 0) to be completely delivered to Z is ________ μs.

Comments

At 0us --collision happens
At 4us--Both stations get to know about collision
At 4us --X draws k=1 and Z draws k=2
At 4us +4 us=8us Last collided bit will leave the station
At 8us+8*1us(wait)=16us X starts transmitting
But My doubt is that Z will also start transmitting at 8us+8*2(wait)=24us and which will again cause collision at t=24us So should X transmit only during 16us to 24us??

joshi please help!!

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i think its 26 micro second

Tx=10 micro second; Tp=4 micro second

collision happened at t=0;

after 8k micro second  retransmission happen as k=1 by station X and k=2 by Y station

8k us=8us---------------1

so X waits again 4 u second to transmit as k=1;---------------2

then tx+tp=10 us + 4 us=14 us ------------3

add 1 +2 + 3 equation = 8us + 4us + 14 us=26 us

...if i m wrong plz correct me.

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2*tp should be added once only to detect collision i think