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How big is the MAC address space? The IPv4 address space? The IPv6 address space?

Short Answer

Expert verified
The MAC address space is 281 trillion, IPv4 is about 4.3 billion, and IPv6 is about 340 undecillion addresses.

Step by step solution

01

Understanding MAC Address Space

MAC addresses are hardware identifiers for network interfaces, typically expressed as six groups of two hexadecimal digits. Since each digit is a hexadecimal (base 16), there are 16 possibilities (0 to 9 and A to F). Therefore, each group can have 16² (256) combinations. Since there are 6 groups, the total number of MAC addresses is found by raising 256 to the power of 6: \[ 2^{48} = 281,474,976,710,656 \] MAC addresses.
02

Understanding IPv4 Address Space

IPv4 addresses are 32-bit numbers typically written as four octets in decimal form, separated by dots. Each octet can range from 0 to 255, giving us 256 possibilities per octet. Therefore, the total number of possible IPv4 addresses is:\[ 2^{32} = 4,294,967,296 \] IPv4 addresses.
03

Understanding IPv6 Address Space

IPv6 addresses are 128-bit numbers, usually written as eight groups of four hexadecimal digits. Each hexadecimal digit has 16 possibilities, and each group (one hex digit pair) consists of 16 bits. Thus, the total IPv6 address space is:\[ 2^{128} = 340,282,366,920,938,463,463,374,607,431,768,211,456 \] IPv6 addresses.

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Key Concepts

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

MAC Address
A MAC (Media Access Control) address is a unique identifier assigned to network interfaces. Think of it as a fingerprint for devices in a network. MAC addresses are crucial in ensuring devices can communicate over a local network, providing a way for networks to know which device is which.
In terms of structure, a MAC address is usually represented as six groups of two hexadecimal digits, for example, AA:BB:CC:DD:EE:FF. With hexadecimal being base 16, each digit has 16 different possibilities, ranging from 0 to 9 and A to F. Given that there are six groups, the number of possible combinations can be calculated by raising 256 to the 6th power, resulting in a total of over 281 trillion potential MAC addresses.
Due to the extensive combinations possible, MAC addresses provide a robust way to identify devices even as more get added to a network, ensuring each remains distinct.
IPv4
IPv4, or Internet Protocol version 4, is one of the core protocols that drive the internet, responsible for directing packets of data from one device to another. It uses a 32-bit addressing scheme, which is written in "dotted decimal" format. This means it's expressed as four numbers separated by dots, like 192.168.1.1.
Each of these four segments is called an octet and can hold a number between 0 and 255, providing 256 possible values (0-255 inclusive) for each octet. Therefore, the total number of unique IPv4 addresses is the result of 256 to the power of 4, or precisely 4.3 billion addresses.
Initially, this seemed ample, but with the explosion of internet-connected devices, IPv4 addresses are running out. Various strategies like NAT (Network Address Translation) have been employed to conserve address space, but a transition to a different protocol has been inevitable.
IPv6
IPv6, or Internet Protocol version 6, was developed primarily to tackle the limitation of IPv4 address exhaustion. It vastly expands the address space, moving from a 32-bit to a 128-bit addressing scheme. Each IPv6 address is represented as eight groups of four hexadecimal digits, separated by colons, making formats like 2001:0db8:85a3:0000:0000:8a2e:0370:7334 quite common.
The 128-bit space allows for an astronomical number of unique addresses: 340 undecillion, to be exact. In number terms, that's 340 followed by 36 zeros. This immense range ensures not only that there are enough addresses for every conceivable need but also that networks can be designed with more flexibility and efficiency.
With IPv6, improvements extend beyond just more address space; it incorporates features such as simplified packet headers and better support for mobile devices. This makes it well suited for the future of networking as the number of connected devices continues to grow.

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Most popular questions from this chapter

In this problem, you will put together much of what you have learned about Internet protocols. Suppose you walk into a room, connect to Ethernet, and want to download a Web page. What are all the protocol steps that take place, starting from powering on your \(\mathrm{PC}\) to getting the Web page? Assume there is nothing in our DNS or browser caches when you power on your PC. (Hint: the steps include the use of Ethernet, DHCP, ARP, DNS, TCP, and HTTP protocols.) Explicitly indicate in your steps how you obtain the IP and MAC addresses of a gateway router.

Suppose nodes \(\mathrm{A}\) and \(\mathrm{B}\) are on the same \(10 \mathrm{Mbps}\) broadcast channel, and the propagation delay between the two nodes is 325 bit times. Suppose CSMA/CD and Ethernet packets are used for this broadcast channel. Suppose node A begins transmitting a frame and, before it finishes, node B begins transmitting a frame. Can A finish transmitting before it detects that B has transmitted? Why or why not? If the answer is yes, then A incorrectly believes that its frame was successfully transmitted without a collision. Hint: Suppose at time \(t=0\) bits, A begins transmitting a frame. In the worst case, A transmits a minimum-sized frame of \(512+64\) bit times. So A would finish transmitting the frame at \(t=512+64\) bit times. Thus, the answer is no, if B's signal reaches A before bit time \(t=512+64\) bits. In the worst case, when does B's signal reach A?

Suppose four active nodes - nodes A, B, C and D-are competing for access to a channel using slotted ALOHA. Assume each node has an infinite number of packets to send. Each node attempts to transmit in each slot with probability \(p\). The first slot is numbered slot 1 , the second slot is numbered slot 2 , and so on. a. What is the probability that node A succeeds for the first time in slot \(5 ?\) b. What is the probability that some node (either A, B, C or D) succeeds in slot 4? c. What is the probability that the first success occurs in slot 3 ? d. What is the efficiency of this four-node system?

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 }}

Suppose the information portion of a packet ( \(D\) in Figure \(5.3\) ) contains 10 bytes consisting of the 8-bit unsigned binary ASCII representation of string "Networking." Compute the Internet checksum for this data.

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