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Suppose nodes A and B are on the same \(10 \mathrm{Mbps}\) 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 \(\mathrm{A}\) and \(\mathrm{B}\) choose different values of \(K\) in the CSMA/CD algorithm. Assuming no other nodes are active, can the retransmissions from \(\mathrm{A}\) and \(\mathrm{B}\) 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 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?

Short Answer

Expert verified
B's retransmission is scheduled at 757 bit times, A transmits at 245. A's signal reaches B at 490, preventing B's transmission.

Step by step solution

01

Calculate B's Backoff Time

Node B chooses a backoff time according to the CSMA/CD protocol. The backoff time is calculated as \( K_B \times 512 \) bit times. Given that \( K_B = 1 \), the backoff time for B is \( 1 \times 512 = 512 \) bit times.
02

Schedule B's Retransmission

After detecting the collision at \( t = 245 \) bit times, node B waits for its backoff time before attempting retransmission. Thus, B schedules its retransmission at \( 245 + 512 = 757 \) bit times.
03

A's Retransmission Time

Node A has chosen \( K_A = 0 \), which means it retransmits as soon as the channel is idle after the collision detection. Because \( K_A = 0 \), the backoff time is \( 0 \times 512 = 0 \) bit times. So, node A can retransmit immediately once the channel is idle at \( t = 245 \) bit times.
04

A's Signal Reaches B

Node A retransmits at \( t = 245 \) bit times, and since the propagation delay is 245 bit times, A's signal will reach B at \( 245 + 245 = 490 \) bit times.
05

Check If B Refrains from Transmitting

At \( t = 490 \) bit times, when A's retransmitted signal reaches B, B is scheduled to retransmit at \( t = 757 \) bit times. Since A's signal indicates an ongoing transmission on the channel when it reaches B, B will refrain from transmitting because the channel is not idle.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Collision Detection
Collision detection is an essential mechanism in Ethernet networks, particularly when using the CSMA/CD (Carrier Sense Multiple Access with Collision Detection) protocol. This protocol is the foundation for managing how data packets are transmitted within a network, reducing the likelihood of data collisions.
In the case of nodes A and B, both attempt to send data at exactly the same time, which results in a collision. This usually happens because both nodes sense that the channel is idle and decide to transmit, but because of the slight propagation delay, they can't detect each other's transmission until the collision occurs.
After the collision, each node must detect that their data collided with another node's data, which is achieved through monitoring the signal on the channel. After detecting the collision, each node will wait a random time before attempting to retransmit. This waiting period decreases further collisions, allowing the channel to be cleared for successful future transmissions.
Backoff Algorithm
The backoff algorithm is a crucial component of the CSMA/CD protocol to manage data collisions efficiently. Once a collision has been detected, the algorithm helps decide how long each node should wait before trying to resend its data.
  • When node A detects a collision, it immediately stops transmission and waits for the channel to be idle before resending. Since its chosen backoff time is zero, node A can try retransmitting almost instantly.
  • Node B, on the other hand, has a backoff time determined by its chosen random value, K. In this scenario, K is 1, so it multiplies this by 512 bit times (the slot time in CSMA/CD) resulting in a 512 bit times backoff period.

    • The randomness in the backoff times helps ensure that the nodes do not retransmit at the same time again. It's a simple yet effective mechanism for reducing collisions after the initial detection.
Propagation Delay
Propagation delay is the time it takes for a signal to travel from one node to another within the network. This is crucial in determining how soon a node can detect a collision and also when a transmission from one node will reach another node.
In the scenario with nodes A and B, the propagation delay is given as 245 bit times. This means once A sends a signal, it takes 245 bit times for the signal to reach B, and vice versa.
This delay is a critical factor in determining when each node detects a collision and how they manage retransmissions. For example, node B plans its retransmission according to this delay and other calculated backoff times to avoid another collision.
Ethernet
Ethernet is a widely-used networking technology for local area networks (LANs). It uses the CSMA/CD protocol to manage communications, especially on older Ethernet systems operating at 10 Mbps.
Ethernet organizes data into frames, which are packets of data that include error-checking information. It ensures that data gets to its destination reliably even in the event of collisions. The protocol is designed for efficiency and to handle multiple devices on the same network channel effectively.
With the use of CSMA/CD, Ethernet ensures that devices take turns to access the network. If a collision occurs, appropriate mechanisms like collision detection and the backoff algorithm reduce the chances of repeated collisions. This makes Ethernet a robust option for network communications even as network traffic increases.

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Most popular questions from this chapter

Why is an ARP query sent within a broadcast frame? Why is an ARP response sent within a frame with a specific destination MAC address?

In this problem, we explore the use of small packets for Voice-over-IP applications. One of the drawbacks of a small packet size is that a large fraction of link bandwidth is consumed by overhead bytes. To this end, suppose that the packet consists of \(P\) bytes and 5 bytes of header. a. Consider sending a digitally encoded voice source directly. Suppose the source is encoded at a constant rate of \(128 \mathrm{kbps}\). Assume each packet is entirely filled before the source sends the packet into the network. The time required to fill a packet is the packetization delay. In terms of \(L\), determine the packetization delay in milliseconds. b. Packetization delays greater than 20 msec can cause a noticeable and unpleasant echo. Determine the packetization delay for \(L=1,500\) bytes (roughly corresponding to a maximum-sized Ethernet packet) and for \(L=50\) (corresponding to an ATM packet). c. Calculate the store-and-forward delay at a single switch for a link rate of \(R=622 \mathrm{Mbps}\) for \(L=1,500\) bytes, and for \(L=50\) bytes. d. Comment on the advantages of using a small packet size.

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