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Having ARP table entries time out after 10-15 minutes is an attempt at a reasonable compromise. Describe the problems that can occur if the timeout value is too small or too large.

Short Answer

Expert verified
Too small timeout increases network traffic; too large causes stale entries and potential communication failures.

Step by step solution

01

Understanding ARP Table

The ARP table maps IP addresses to MAC addresses. Devices use it to know where to send packets within a local network.
02

Timeout Value Too Small

A very small timeout value (e.g., less than a minute) means the ARP entries will expire quickly. This could result in frequent ARP requests, increasing network traffic and possibly causing delays.
03

Timeout Value Too Large

A very large timeout value (e.g., several hours) means the ARP entries seldom expire. This could lead to stale entries, where devices might attempt to use outdated MAC addresses, causing communication failures.
04

Finding a Balance

An ARP timeout value of 10-15 minutes aims to balance between minimizing network traffic (by reducing frequent ARP requests) and ensuring the ARP table does not contain stale entries.

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Key Concepts

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

Address Resolution Protocol
The Address Resolution Protocol (ARP) is a critical network protocol used for mapping an Internet Protocol (IP) address to a physical machine (MAC) address that resides within the same local network. When a device wants to communicate with another device, it needs the MAC address associated with an IP address. ARP bridges this gap.

How ARP Works:

- When Device A needs to communicate with Device B, it checks its ARP table for Device B’s MAC address.
- If the ARP entry exists, Device A can immediately use the MAC address to direct the message.
- If the entry does not exist, Device A broadcasts an ARP request on the local network.
- Device B responds with its MAC address, which Device A then stores in its ARP table.

This process allows seamless communication between devices on a local network. However, since network conditions and device statuses change, ARP table entries need to be managed efficiently.
Network Traffic
Network traffic refers to the amount of data moving across a network at any given time. Efficient management of network traffic is crucial for maintaining optimum network performance.

How ARP Affects Network Traffic:

- ARP requests generate traffic on the network, as they are broadcast to all devices within the local network.
- A smaller ARP entry timeout value will lead to frequent ARP requests, increasing network traffic. More ARP requests mean higher bandwidth usage, which can potentially slow down network performance.
- Conversely, a larger timeout value reduces the frequency of ARP requests, thus lessening network traffic. However, stale ARP entries, resulting from a too-large timeout, can cause other issues.

Balancing network traffic involves tweaking ARP timeout values. The 10-15 minute timeout is a compromise to ensure that we manage ARP traffic without overwhelming the network or dealing with outdated information.
Communication Failures
Communication failures in a network can stem from outdated ARP table entries among other issues. Maintaining accurate ARP table entries is vital for successful communication between devices.

How ARP Timeout Influences Communication:

- With a very short timeout value, ARP entries refresh frequently, ensuring up-to-date MAC addresses for IPs. However, this can flood the network with ARP requests, causing delays and congestion.
- If the ARP timeout value is too large, devices might try to communicate using stale MAC addresses that no longer match the IPs, resulting in failed message delivery.

Finding a suitable ARP timeout value is essential for minimizing communication failures. Network administrators often aim for a 10-15 minutes interval to reduce the likelihood of outdated information without overly increasing traffic. This setting helps maintain a healthy balance between stability and efficiency within the network.

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Most popular questions from this chapter

An ISP with a class \(\mathrm{B}\) address is working with a new company to allocate it a portion of address space based on CIDR. The new company needs IP addresses for machines in three divisions of its corporate network: Engineering, Marketing, and Sales. These divisions plan to grow as follows: Engineering has 5 machines as of the start of year 1 and intends to add 1 machine every week; Marketing will never need more than 16 machines; and Sales needs 1 machine for every two clients. As of the start of year 1, the company has no clients, but the sales model indicates that by the start of year 2 , the company will have six clients and each week thereafter gets one new client with probability \(60 \%\), loses one client with probability \(20 \%\), or maintains the same number with probability \(20 \%\). (a) What address range would be required to support the company's growth plans for at least seven years if marketing uses all 16 of its addresses and the sales and engineering plans behave as expected? (b) How long would this address assignment last? At the time when the company runs out of address space, how would the addresses be assigned to the three groups? (c) If CIDR addressing were not available for the seven-year plan, what options would the new company have in terms of getting address space?

What aspect of IP addresses makes it necessary to have one address per network interface, rather than just one per host? In light of your answer, why does IP tolerate point-to-point interfaces that have nonunique addresses or no addresses?

Some signalling errors can cause entire ranges of bits in a packet to be overwritten by all 0 s or all 1s. Suppose all the bits in the packet including the Internet checksum are overwritten. Could a packet with all 0s or all 1s be a legal IPv4 packet? Will the Internet checksum catch that error? Why or why not?

Suppose host \(\mathrm{A}\) is sending to a multicast group; the recipients are leaf nodes of a tree rooted at A with depth \(N\) and with each nonleaf node having \(k\) children; there are thus \(k^{N}\) recipients. (a) How many individual link transmissions are involved if A sends a multicast message to all recipients? (b) How many individual link transmissions are involved if A sends unicast messages to each individual recipient? (c) Suppose A sends to all recipients, but some messages are lost and retransmission is necessary. Unicast retransmissions to what fraction of the recipients is equivalent, in terms of individual link transmissions, to a multicast retransmission to all recipients?

Why do you think IPv4 has fragment reassembly done at the endpoint, rather than at the next router? Why do you think IPv6 abandoned fragmentation entirely? Hint: Think about the differences between IP-layer fragmentation and link-layer fragmentation.

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