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On wikipedia there is this paragraph:

Since a pulse's runtime will never exceed slot time (the maximum theoretical time for a frame to travel a network), the NIC waits a minimum of slot time before retransmitting after a collision happened, in order to allow any pulse that was initiated at the time that the waiting NIC was requested to send, to reach all other nodes. By allowing the pulse to reach the waiting NIC, a local collision occurs (ie. while still sending) rather than a late collision occurring (after sending may or may not has ended). By having the collision occur at the NIC (local) and not on the wire (late) CSMA/CD implementation recover the situation by retransmitting later.

What is the waiting NIC referring to? Is it the NIC mentioned in the previous sentence or another NIC which was involved in the collision?

The NIC is "requested to send" by the upper OSI layer (in other words, by the node that it is attached to) ?

Why is the exponential back-off algorithm using multiples of the slot time?

I understand that the slot time is computed such that a node is still transmitting when the signal announcing the collision (jamming signal) arrives at it. But what does this have to do with how much time nodes should wait after a collision occurs?

Another wikipedia article states:

The maximum jam-time is calculated as follows: The maximum allowed diameter of an Ethernet installation is limited to 232 bits. This makes a round-trip-time of 464 bits. As the slot time in Ethernet is 512 bits, the difference between slot time and round-trip-time is 48 bits (6 bytes), which is the maximum "jam-time".

Why should there be a maximum jamming time?

  • Did any answer help you? If so, you should accept the answer so that the question doesn't keep popping up forever, looking for an answer. Alternatively, you could provide and accept your own answer. – Ron Maupin Aug 15 '17 at 17:44
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What is the waiting NIC referring to? Is it the NIC mentioned in the previous sentence or another NIC which was involved in the collision?

That refers to any NICs waiting to send a frame, especially those involved in a collision.

The NIC is "requested to send" by the upper OSI layer (in other words, by the node that it is attached to) ?

The NIC is at OSI layer-1. OSI layer-2 for ethernet is the MAC layer. Layer-3 would be IP. Layer-4 is the transport protocol, e.g. TCP or UDP. Above layer-4 are the application layers (off-topic here). An application will start sending data to the transport layer, that then sends through the network layer to the MAC layer that gives it to the NIC.

Why is the exponential back-off algorithm using multiples of the slot time?

That makes sure that the back-off time is never less than the slot time, and it is quick and easy to calculate.

I understand that the slot time is computed such that a node is still transmitting when the signal announcing the collision (jamming signal) arrives at it. But what does this have to do with how much time nodes should wait after a collision occurs?

That makes sure that all stations have heard the jamming signal, and no station will transmit while the jamming signal is still traversing the link. A host immediately next to the host doing the jamming will hear the signal before other hosts on the link, and it will stop hearing the jamming signal while it is still traversing the link.

Why should there be a maximum jamming time?

Jamming when it is no longer needed will slow collision recovery, and therefore, unnecessarily reduce throughput on the link.

  • Thank you for your answer. I do have some follow-up questions. "That makes sure that all stations have heard the jamming signal, and no station will transmit while the jamming signal is still traversing the link. A host immediately next to the host doing the jamming will hear the signal before other hosts on the link, and it will stop hearing the jamming signal while it is still traversing the link.". Shouldn't half the slot time be enough for all the computers to receive the jamming signal? – user42768 Jan 7 '17 at 21:38
  • If the link is 100 meters, and the jamming station is at one end with the next host only five meters away, then, no. The signal needs to be able to travel the entire 100 meters before it is safe. – Ron Maupin Jan 7 '17 at 21:46
  • But isn't the slot time at least double the time needed for a signal to get from one end to the other? – user42768 Jan 7 '17 at 21:48
  • No. It is the time it takes for a signal to get from one end to the other. You should really read the IEEE 802.3 documents. They are the standard explain everything. – Ron Maupin Jan 7 '17 at 21:51
  • "Slot time is a concept in computer networking. It is at least twice the time it takes for an electronic pulse (OSI Layer 1 - Physical) to travel the length of the maximum theoretical distance between two nodes.". This is taken from the first cited link. And another related question: is the jamming signal sent by the computers that cause the collision (concurrent transmitters) or by any computer that "senses" (different voltage value) the collission? – user42768 Jan 7 '17 at 21:54
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I will answer only the following question, as it was, for me, the most important:

Why is the exponential back-off algorithm using multiples of the slot time?

I thought of this situation: nodes A and B are on opposite ends of the network. They send a message at the same time, and the collision occurs very close to B. Now, B notices the collision and starts its own exponential back-off algorithm sequence. Let's assume that this results in node B backing off. Now, the question is, how long does B need to wait?

To answer the question, consider the same event, but from A's perspective. When the jamming signal reaches A, B has been waiting for half a slot time already. Let's assume that the exponential back-off algorithm sequence, run at node A, results in A not waiting at all. Therefore, A re-sends its original message as soon as the jamming signal reaches it. So, when the re-sent message reaches node B, B has been waiting for an entire slot time. If B decided to wait less than the slot time, a new collision would have occurred, even if the exponential back-off algorithm yielded different results for the two nodes, A and B.

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