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#
Authors:
James F. Kurose, Keith W. Ross
Chapter:
The Link Layer: Links,access Networks, And Lans
Exercise:
Problems
Question:19 | ISBN:9780132856201 | Edition: 6

Question

Suppose nodes A and B are on the same 10 Mbps broadcast channel, and the propagation delay between the two nodes is 245 bit times. Suppose A and B send Ethernet frames at the same time, the frames collide, and then A and B choose different values of K in the CSMA/CD algorithm. Assuming no other nodes are active, can the retransmissions from A and B collide? For our purposes, it suffices to work out the following example. Suppose A and B begin transmission at t = 0 bit times. They both detect collisions at t = 245 bit times. Suppose K A = 0 and K B = 1. At what time does B schedule its retrans- mission? At what time does A begin transmission? (Note: The nodes must wait for an idle channel after returning to Step 2—see protocol.) At what time does A’s signal reach B? Does B refrain from transmitting at its scheduled time?

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Answer

In the given scenario, nodes A and B are on the same 10 Mbps broadcast channel, and they both start transmitting Ethernet frames at the same time. Let's analyze the sequence of events:

  • Both A and B start transmitting at t = 0 bit times.
  • The frames collide and both A and B detect the collision at t = 245 bit times.
  • Node A chooses K_A = 0, and node B chooses K_B = 1.

Since K_A = 0, node A does not need to wait before retransmitting. Node A schedules its retransmission immediately after the collision is detected, at t = 245 bit times.

Node B, on the other hand, chooses K_B = 1, which means it will wait for one slot time before attempting retransmission.

In Ethernet, a slot time is the time required for a signal to propagate across the maximum distance in the network, which is equivalent to 512 bit times for a 10 Mbps Ethernet.

The propagation delay between nodes A and B is given as 245 bit times.

Therefore, B schedules its retransmission at t = 245 + 512 = 757 bit times.

At t = 757 bit times, B begins transmission. However, since the signal from A is still propagating towards B, B will detect the carrier sense and refrain from transmitting at its scheduled time.

This is because B detects that the channel is still busy due to the ongoing transmission from A.

Once A's signal reaches B, it takes an additional propagation delay of 245 bit times.

Therefore, at t = 757 + 245 = 1002 bit times, B senses that the channel is idle after the propagation delay from A's transmission, and it can proceed with its retransmission.

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