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The Transmission Time ($T_{t}$) for a CSMA/CD protocol must be alteast two times the Propagation Time ($2*T_{p}$) i.e. Round Trip Time.

$T_{t}>=2*T_{p}$

Why?

As, in CSMA/CD, a receiver doesn’t send acknowledgment to the sender whether it received a frame or not. Even in the case of a collision, the receiver won’t be able to distinguish if the corrupted frame is for ‘him’ or not. So, it’s entirely the responsibility of the sender to determine if the frame is received by the receiver or not and retransmit in the case of a collision. So, it also has to detect if a collision has occurred or not.

So, what is the solution?

Sender transmits as long as the first bit of the frame is received by the receiver. Why? Because in the worst case the frame gets transmitted almost close to the receiver but faces a collision (bad news) and this collision signal (bad news) also has to get back to the sender, which will cumulatively takes 2 times the propagation time. Thus, sender should keep in contact with the frame transmission in this mean time. That’s why it should keep on transmitting until the possibility of bad news comebacks to him.

But in case the first bit is successfully received by the receiver (good news), there’s no way any other station can transmit data as long as the channel is clear. The sender will not receive any collision signal in $2*T_{p}$ and conclude there’s no collision.

Hence, this $2*T_{p}$ is solely for determining if there’s a collision in the channel by the sender.

Refer to this video (Amazing Explanation): https://youtu.be/sx0UPzztC5o?t=2281

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