Propagation delay (Tp)
= 225-bit times
= 225 bit / 10 Mbps
= 22.5 x 10-6 sec
= 22.5 μsec
At t = 0,
- Nodes A and B start transmitting their frame.
- Since both, the stations start simultaneously, so collision occurs in the midway.
- Time after which collision occurs = Half of the propagation delay.
- So, time after which collision occurs = 22.5 μsec / 2 = 11.25 μsec.
At t = 11.25 μsec,
- After the collision occurs at t = 11.25 μsec, collided signals start traveling back.
- Collided signals reach the respective nodes after time = Half of propagation delay
- Collided signals reach the respective nodes after time = 22.5 μsec / 2 = 11.25 μsec.
- Thus, at t = 22.5 μsec, collided signals reach the respective nodes.
At t = 22.5 μsec,
- As soon as nodes discover the collision, they immediately release the jamming signal.
- Time taken to finish transmitting the jam signal = 50 bit time = 50 bits/ 10 Mbps = 5 μsec.
Thus,
Time at which the jamming signal is completely transmitted
= 22.5 μsec + 5 μsec
= 27.5 μsec or 275-bit times
