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How long sender needs to transmit to capture the channel in Ethernet when we apply CSMA/CD ? I my book , i have read it sender should be transmitting for at least the 2*PT,where PT is end to end propagation delay.

But i need to understand that whether we need to consider the transmission time that it takes for receiver to transmit the jamming signal on the channel ?

So there are two points

  1. Sender will keep on transmitting till 2*PT. here i assume that as soon as receiver get the first bit of the data it knows about collision and it immediately starts transmitting jamming signal.But when sender gets Jamming signal first but it can detect that collision has occurred

  2. Secondly sender will transmit till (2*PT+Time to transmit Complete jamming signal on the channel),here i assume one way delay is PT,then receiver detects collision as soon as it gets first bit,and then it start transmit the jamming signal and it took receiver some Y time to transmit it on channel and then it reaches sender at PT+Y making effective min. time for sender transmission as PT+PT+Y.

I am not sure if jamming signal is part of data.

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    CSMA/CD is 99.99% obsolete. What is the relevant application of this? "Capturing the channel" is usually not desirable but it's caused by the effect that an active sender is more likely to send additional packets than another one that's been backing off for some time.
    – Zac67 ♦
    Commented Aug 17, 2017 at 17:58
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    Not for practical application. I was learning the concept so got this confusion.
    – rahul sharma
    Commented Aug 17, 2017 at 18:16
  • @rahulsharma Calculations can be found in thread: https://stackoverflow.com/questions/26153333/csma-cd-minimum-frame-size-to-hear-all-collisions
    – user36472
    Commented Aug 18, 2017 at 10:38

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sender should be transmitting for at least the 2*PT,where PT is end to end propagation delay.

Worst case is between senders that are located at opposite ends of a collision domain.

Imagine sender 1 transmitting at one end. Its signal propagates to the far end where sender 2 has just started transmitting as well. Sender 2 detects the collision, aborts its transmission and generates a jam signal.

Now, the jam signal needs to propagate all the way back to sender 1 who needs to still be sending in order to successfully detect the collision. If it had already finished sending the frame it would regard the transmission as successful - the collided frame wouldn't get resent, it'd be just lost.

So, the minimum frame size is required to exceed twice the maximum propagation time multiplied by the link speed.

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