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 KA = 0 and KB = 1. At what time does B schedule its retransmission? 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?

Answer 1

Node B schedules its retransmission at t = 490 bit times, while Node A waits for an idle channel before beginning transmission again.

Step-by-step explanation:

In this scenario, both nodes A and B start transmission at t = 0 bit times and detect a collision at t = 245 bit times. Node A has KA = 0 and node B has KB = 1. According to the CSMA/CD algorithm, node B will schedule its retransmission at t = 490 bit times, which is twice the propagation delay. Node A will wait for an idle channel after returning to Step 2 before beginning its transmission again. The signal from A will reach B at t = 245 bit times + 245 bit times = 490 bit times. At this time, B will refrain from transmitting since it will detect A's signal on the channel.

Answer 2

In the CSMA/CD algorithm, after a collision, nodes A and B choose different values of K. Node B schedules its retransmission at t=735 bit times, while Node A starts transmission after the signal from Node B reaches it at t=490 bit times. There will be no collision at the retransmission time of Node A.

Step-by-step explanation:

In the given scenario, nodes A and B collided and detect the collision after a propagation delay of 245 bit times. Node A chooses K=0 for retransmission while Node B chooses K=1. With CSMA/CD protocol, after the collision, both nodes have to wait for an idle channel before retransmitting.

Since Node B chose K=1, it waits for an extra slot time compared to Node A. In this case, a slot time is equal to a propagation delay of 245 bit times. Therefore, Node B schedules its retransmission at t=245 + (1 + 1) * 245 = 735 bit times.

On the other hand, Node A starts its retransmission after an idle channel is detected, which is when Node B's signal reaches Node A. The time it takes for Node B's signal to reach Node A is equal to the propagation delay between the two nodes, which is 245 bit times. Therefore, Node A begins transmission at t = 245 + 245 = 490 bit times.

Since Node B's scheduled retransmission time is later than Node A's transmission time, there will be no collision at the retransmission time of Node A. Node B refrains from transmitting at its scheduled time.