A and B are on 10Mbps Ethernet segment and Suppose A and B send frames at the same time
Assume the transmission starts from $T = 0$
the frames collide
When A and B start to transmit simultaneouly at the same time, they will collide somewhere halfway. Now Propagation time of a bit is given = $225$ bit times, so, It depends on how fast A and B transmit and place of collision of their frames depends on their speed of transmission.But no such information is given, we assume both transmit at equal speeds. So, collison happens somewhere in the mid ,i.e, $\frac{225}{2}$ = $112.50$ bit times
How A and B will know that their frames have collided in the halfway ?
After $225$ bit times of PD, both will receive collided frames with double the energy as compared to the frames sent before. So, after $225$ bit times, each of them now know that their packets have collided. But A does not know that B's frame is collided and vice versa. Here, comes the application of JAM signal.
How big is the JAM signal ?
As Bits of JAM signal are not given, we don't know that is $1$ bit sufficient for the jam signal to alert other station that collision has occured or complete $48$ bits signal .
But , we will use $48$ bits JAM signal, as maximum difference between the
$Slot\ Time\ - RTT = 48\ bits$
Hence, Both A and B send and alert each other that collison has occured after $225$ + $48$ = $273$ bit times.
then A and B choose different values of K say for A (k=0) and B (k=1) in the CSMA/CD algorithm
Now, It is given that A wins the back-off race and will send without waiting, but B will wait for $1 * 51.2$ us (or $512$ bit times).
Now, A will take $225$ bit times PD to reach the frame, while B is waiting. So, frame reaches B at $273 + 225 = 498$ bit times.
Can the retransmissions from A and B collide ?
A's frame reaches B at $498$ bit times, while B will start to send its frame at $273 + 512 = 785$ bit times. Hence, no collision now.
