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Let's consider the operation of a learning switch in the context of a network in which 6 nodes labeled A through F are star connected into an Ethernet switch. Suppose that (i) B sends a frame to E, (ii) E replies with a frame to \(\mathrm{B}\), (iii) A sends a frame to B, (iv) B replies with a frame to A. The switch table is initially empty. Show the state of the switch table before and after each of these events. For each of these events, identify the link(s) on which the transmitted frame will be forwarded, and briefly justify your answers.

Short Answer

Expert verified
The switch learns MAC addresses and forwards frames accordingly, optimizing broadcasting over time.

Step by step solution

01

Initialization

Initially, the switch table is empty. This means the switch has no MAC address information. Therefore, when any node sends a frame, the switch doesn't know which port corresponds to the destination MAC address.
02

Event 1: B Sends Frame to E

When node B sends a frame to E, the switch learns the MAC address of node B on the port connected to node B. Since its table is empty, it doesn't have the MAC address of E. Therefore, it broadcasts the frame to all ports except the incoming port. - **Switch Table State After Event 1:** MAC address of B is learned at port connected to B. - **Forwarding:** Frame is forwarded to all ports except the one connected to B.
03

Event 2: E Replies with a Frame to B

Node E sends a frame back to node B. The switch learns the MAC address of node E on the port connected to node E. Now, it has the MAC address of node B from the previous step, so it forwards the frame only to the port corresponding to B. - **Switch Table State After Event 2:** MAC address of B is at port connected to B; MAC address of E is learned at port connected to E. - **Forwarding:** Frame is forwarded only to the port connected to B.
04

Event 3: A Sends Frame to B

Node A sends a frame to B. The switch learns the MAC address of node A on the port connected to A. It already knows B's MAC address, so it forwards the frame only to B. - **Switch Table State After Event 3:** MAC addresses of B, E, and A are learned at their respective ports. - **Forwarding:** Frame is forwarded only to the port connected to B.
05

Event 4: B Replies with a Frame to A

Node B replies to A. The switch already knows both MAC addresses, so it simply forwards the frame to the port connected to A. - **Switch Table State After Event 4:** No new entries are added; the table remains unchanged from after Event 3. - **Forwarding:** Frame is forwarded only to the port connected to A.

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Key Concepts

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

MAC address
In the world of networking, a MAC address is like a unique identifier for your network devices. Think of it as a digital fingerprint. Every device that can connect to a network has a MAC address, ensuring that data sent through a network can be accurately delivered to the right device.

A MAC address is a 48-bit number usually expressed as six groups of two hexadecimal digits. It looks something like "00:1A:2B:3C:4D:5E". This address is assigned by the hardware manufacturer and is typically found on the network interface card (NIC).

MAC addresses are essential because they allow switches and routers to direct data correctly across the network. Without MAC addresses, devices wouldn't be able to communicate effectively. This is why they're a critical component of network communication.
Ethernet switch
An Ethernet switch is a vital component of many network setups, especially in home and business environments. It connects multiple devices to a single network, allowing them to communicate with each other efficiently.

The switch operates by receiving incoming data packets, checking them against its MAC table, and forwarding them to the appropriate port based on their destination MAC address. This ensures that data reaches its intended recipient without cluttering the entire network. This process optimizes bandwidth use and enhances the speed and efficiency of network communications.

Ethernet switches are "intelligent" compared to hubs, which simply broadcast incoming traffic to all ports. This intelligence makes switches a preferred choice in networking for better data management and security.
learning switch
A learning switch is a type of Ethernet switch that can "learn" and remember the MAC addresses of devices on its network. When a new device sends a frame, the switch identifies which port the frame came from and adds the device's MAC address to its MAC table.

This learning process allows the switch to optimize traffic. Initially, if the switch's MAC table is empty, it will broadcast the first incoming frame to all ports (except the sending port) because it doesn't know where to send it. Once it learns the MAC address, it can forward future frames directly to the correct port, reducing unnecessary traffic.

Learning switches continuously update their tables as devices join and leave the network. They help maintain network efficiency by ensuring that data packets are only sent to desired devices, preventing network congestion.
frame forwarding
Frame forwarding is a key function of Ethernet switches and involves directing data frames from their source to their intended destination within the network. This ensures smooth and effective communication between devices.

When a switch receives a frame, it checks the destination MAC address. Based on its MAC table, the switch determines the appropriate port to forward the frame to. If the switch doesn't find the destination MAC address in its table, it typically resorts to broadcasting the frame to all ports except the one it was received on. This is known as "flooding."

Efficient frame forwarding prevents network collisions and data packet loss, improving overall network performance. By ensuring frames reach the correct device, switches manage network resources effectively and keep communications seamless.

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Most popular questions from this chapter

Recall that with the CSMA/CD protocol, the adapter waits \(K \cdot 512\) bit times after a collision, where \(K\) is drawn randomly. For \(K=100\), how long does the adapter wait until returning to Step 2 for a 10 Mbps broadcast channel? For a \(100 \mathrm{Mbps}\) broadcast channel?

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?

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, you will derive the efficiency of a CSMA/CD-like multiple access protocol. In this protocol, time is slotted and all adapters are synchronized to the slots. Unlike slotted ALOHA, however, the length of a slot (in seconds) is much less than a frame time (the time to transmit a frame). Let \(S\) be the length of a slot. Suppose all frames are of constant length \(L=k R S\), where \(R\) is the transmission rate of the channel and \(k\) is a large integer. Suppose there are \(N\) nodes, each with an infinite number of frames to send. We also assume that \(d_{\text {prop }}

Why would the token-ring protocol be inefficient if a LAN had a very large perimeter?

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