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What kinds of problems can arise when two hosts on the same Ethernet share the same hardware address? Describe what happens and why that behavior is a problem.

Short Answer

Expert verified
Duplicate MAC addresses cause network conflicts, leading to packet loss, collisions, and potential downtime.

Step by step solution

01

Understand MAC Addresses

Each device on an Ethernet network has a unique Media Access Control (MAC) address assigned to its network interface card (NIC). This address is supposed to be unique to prevent conflicts.
02

Identify the Problem of Same MAC Addresses

When two hosts on the same Ethernet network share the same MAC address, data packets intended for one device may be mistakenly sent to the other device, resulting in a conflict.
03

Describe Network Behavior

When the MAC address conflict occurs, the network switch or router cannot correctly learn and forward network packets. This confusion leads to unpredictable network behavior, such as packet loss, network collisions, and potential downtime.
04

Explain Why It's a Problem

The primary issue with having duplicate MAC addresses is network unreliability. It can disrupt communication, cause data corruption, and compromise network security as data meant for one host could be inadvertently intercepted by another.

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Key Concepts

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

MAC address uniqueness
A MAC address, which stands for Media Access Control address, is a unique identifier assigned to each network interface card (NIC). These addresses, usually represented in a hexadecimal format like '00:1A:2B:3C:4D:5E', are designed to be globally unique. This uniqueness ensures that every device on a network can be distinctly recognized and communicated with.

If two devices share the same MAC address, it leads to a MAC address conflict. Because these addresses are intended to be unique, most devices assume this principle in their operations. A duplicate address disrupts this assumption, leading to network issues.

To maintain network reliability, device manufacturers follow strict processes to ensure each MAC address is unique. However, on rare occasions, bugs or manual misconfigurations can lead to duplicate addresses, causing severe network disruptions.
network packet forwarding
Network packet forwarding is the process of moving data packets from a source to a destination through intermediaries such as switches and routers. Each device within a network uses MAC addresses to identify where data packets should be sent.

When a packet reaches a switch, the switch uses MAC address tables to identify on which port the device with the destination MAC address is located. The switch then forwards the packet to the correct port. In case of a MAC address conflict, the switch gets confused since it can't differentiate between two devices having the same address. This results in:
  • Packets being sent to the wrong device
  • Network collisions
  • Packet loss
This confusion interrupts network operations and leads to significant communication issues. That's why maintaining unique MAC addresses is crucial for proper packet forwarding and overall network efficiency.
network security
MAC address conflicts in an Ethernet network don't just cause operational issues; they can also present serious security concerns. When two devices share the same MAC address, network data may be intercepted by the unintended device, leading to potential data breaches.

For example, sensitive information meant for one device could be accessed by another, violating data privacy. Additionally, an attacker could purposely set their device's MAC address to match another device's MAC address, enabling them to intercept or disrupt the communication to that specific device.

Therefore, maintaining unique MAC addresses contributes not only to network stability but also to the integrity and security of network communications. Network administrators often employ monitoring tools to detect and resolve MAC address conflicts, ensuring both reliable and secure network environments.

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Most popular questions from this chapter

With 1 parity bit we can detect all 1-bit errors. Show that at least one generalization fails, as follows: (a) Show that if messages \(m\) are 8 bits long, then there is no error detection code \(e=e(m)\) of size 2 bits that can detect all 2-bit errors. Hint: Consider the set \(M\) of all 8-bit messages with a single 1 bit; note that any message from \(M\) can be transmuted into any other with a 2 -bit error, and show that some pair of messages \(m_{1}\) and \(m_{2}\) in \(M\) must have the same error code \(e\). (b) Find an \(N\) (not necessarily minimal) such that no 32 -bit error detection code applied to N-bit blocks can detect all errors altering up to 8 bits.

Suppose a \(100-\mathrm{Mbps}\) delayed-release token ring has 10 stations, a ring latency of \(30 \mu \mathrm{s}\), and an agreed-upon TTRT of \(350 \mu \mathrm{s}\). (a) How many synchronous frame bytes could each station send, assuming all are allocated the same amount? (b) Assume stations \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) are in increasing order on the ring. Due to uniform synchronous traffic, the TRT without asynchronous data is \(300 \mu \mathrm{s}\). B sends a \(200-\mu \mathrm{s}(2.5-\mathrm{Kb})\) asynchronous frame. What TRT will \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) then see on their next measurement? Who may transmit such a frame next?

Give some details of how you might augment the sliding window protocol with flow control by having ACKs carry additional information that reduces the SWS as the receiver runs out of buffer space. Illustrate your protocol with a timeline for a transmission; assume the initial sWS and RWS are 4, the link speed is instantaneous, and the receiver can free buffers at the rate of one per second (i.e., the receiver is the bottleneck). Show what happens at \(T=0, T=1, \ldots, T=4 \mathrm{sec}-\) onds.

For a 100-Mbps token ring network with a token rotation time of \(200 \mu\) s that allows each station to transmit one 1-KB packet each time it possesses the token, calculate the maximum effective throughput rate that any one host can achieve. Do this assuming (a) immediate release and (b) delayed release.

Suppose Ethernet physical addresses are chosen at random (using true random bits). (a) What is the probability that on a 1024-host network, two addresses will be the same? (b) What is the probability that the above event will occur on some one or more of \(2^{20}\) networks? (c) What is the probability that of the \(2^{30}\) hosts in all the networks of (b), some pair has the same address? Hint: The calculation for (a) and (c) is a variant of that used in solving the socalled Birthday Problem: Given \(N\) people, what is the probability that two of their birthdays (addresses) will be the same? The second person has probability \(1-\frac{1}{365}\) of having a different birthday from the first, the third has probability \(1-\frac{2}{365}\) of having a different birthday from the first two, and so on. The probability all birthdays are different is thus $$ \left(1-\frac{1}{365}\right) \times\left(1-\frac{2}{365}\right) \times \cdots \times\left(1-\frac{N-1}{365}\right) $$ which for smallish \(N\) is about $$ 1-\frac{1+2+\cdots+(N-1)}{365} $$

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