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In CSMA/CD method why the propagation delay is $2T_p$?
in Computer Networks edited by
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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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4 Comments

No retransmission takes place from receiver to sender.

Collisions result in a voltage displacement on the cable. This makes it possible for the stations to detect the collisions.
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So confusing
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@DebSujit I think it is not necessary, information reach to the sender because as CSMA/cd firstly sense then transmission takes place but it’s sense by the collision signal. & for 2Tp it is a worst case right? for successful transmission best case can be Tp. suppose in worst case  when a frame is transmit by the station A to Station B to cover all distance and it take approx 1 hour now even one bit is collide from other frame then frame will be lost or discard so suppose at approx 1 hour all bits of frame transmitted except last bit then it takes total transmission delay Tp now at that time suppose collision occur then frame will be discarded and at that place collision signal is generated and sends to all the station then the frame which is discarded it already takes Tp and same for collision signal it take also 1 hour approx that means it also takes Tp so after collision the total transmission delay take 2Tp. i think this will be reason.

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