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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

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