IP Addressing

1. IP Addressing Allan Johnson

2. IPv4 Addressing Review Table of Contents End Slide Show

3. IP Addressing • IP Addressing is a logical addressing scheme at the Network Layer of the OSI Model. • Like all Network Layer addressing schemes (IPX, AppleTalk, DECnet, CLNS, etc.), IP addresses have two parts: • Network—identifies the network or subnet • Host—identifies the device on that network/subnet • An IP Address’ 32 bits are expressed in 4 octets (called dotted-decimal notation). • IP addresses are divided into five class types depending upon the value of bit positions in the first octet.

4. Class A: 1.0.0.0 to 127.0.0.0 Class C: 192.0.0.0 to 223.255.255.0 Class B: 128.0.0.0 to 191.255.0.0 Network Network Network Network Host Network Host Network Host Host Host Host 1 1 0 1 X 0 0 X X X X X X X X X X X X X X X X X 1st Octet Bits: ___ ___ ___ ___ ___ ___ ___ ___ (The 128 bit is off.) 1st Octet Bits: ___ ___ ___ ___ ___ ___ ___ ___ (The 128 and 64 bits are on. The 32 bit is off.) 1st Octet Bits: ___ ___ ___ ___ ___ ___ ___ ___ (The 128 bit is on and the 64 bit is off.) IP Address Classes

5. Multicasting Class D: 224.0.0.0 to 239.0.0.0 1 1 1 0 X X X X 1st Octet Bits: ___ ___ ___ ___ ___ ___ ___ ___ (The 128, 64, and 32 bit are on. The 16 bit is off.) Experimental Class E: 240.0.0.0 to 255.0.0.0 1 1 1 1 X X X X 1st Octet Bits: ___ ___ ___ ___ ___ ___ ___ ___ (The 128, 64, 32, and 16 bit are all on.) Reserved IP Address Classes

6. Class A: 10.0.0.0 (Favored by large enterprises because of its flexibility) Class B: 172.16.0.0 to 172.31.0.0 (In the 3rd Octet, the 128, 64, and 32 bit are off. The 16 bit is on.) Class C: 192.168.0.0 to 192.168.255.0 (256 separate Class C Addresses) Private IP Addresses • Private IP Addresses cannot exist on the public Internet. • Your gateway router uses Name Address Translation (NAT) to give outbound packets a “legitimate” IP source address. • Private Addressing and NAT are discussed later.

7. Continue With Subnetting Review Skip Subnetting Review

8. Why Subnet? • Remember: we are usually dealing with a broadcast topology. • Can you imagine what the network traffic overhead would be like on a network with 254 hosts trying to discover each others MAC addresses? • Subnetting allows us to segment LANs into logical broadcast domains called subnets, thereby improving network performance.

9. Four Subnetting Steps • To correctly subnet a given network address into subnet addresses, ask yourself the following questions: • How many bits do I need to borrow? • What’s the subnet mask? • What’s the “magic number” or multiplier? • What are the first three subnetwork addresses? • Let’s look at each of these questions in detail

10. 1. How many bits to borrow? • First, you need to know how many host bits you have to work with. • Second, you must know either how many subnets you need or how many hosts per subnet you need. • Finally, you need to figure out the number of bits to borrow.

11. 1. How many bits to borrow? • How many host bits do I have to work with? • Depends on the class of your network address. • Class C: 8 host bits • Class B: 16 host bits • Class A: 24 host bits • Remember: you must borrow at least 2 bits for subnets and leave at least 2 bits for host addresses. • 2 bits borrowed allows 22- 2 = 2 subnets • Anyway, that’s how we learned it in our CCNA Curriculum. You will soon discover that subnet zero is actually available for your use.

12. 1. How many bits to borrow? • How many subnets or hosts do I need? • A simple formula: • Host Bits = Bits Borrowed + Bits Left • HB = BB + BL • I need x subnets: • I need x hosts: • Remember: we need to subtract two hosts to provide for the subnetwork and broadcast addresses.

13. 1. How many bits to borrow? • Class C Example: 210.93.45.0 • Design goals specify at least 5 subnets so how many bits do we borrow? • How many bits in the host portion do we have to work with (HB)? Since it’s a Class C, we have 8 bits to work with. • What’s the BB in our HB = BB + BL formula? 8 = BB + BL • 2 to what power will give us at least 5 subnets? 23 - 2 = 6 subnets • How many bits are left for hosts? Since 8 = 3 + BL, then BL = 5 • So how many hosts can we assign to each subnet? 25 - 2 = 30 hosts

14. 1. How many bits to borrow? • Class B Example: 185.75.0.0 • Design goals specify no more than 126 hosts per subnet, so how many bits do we need to leave (BL)? • How many bits in the host portion do we have to work with (HB)? Since it’s a Class B, we have 16 bits to work with. • What’s the BL in our HB = BB + BL formula? 16 = BB + BL • 2 to what power will give us 126 hosts per subnet? 27 - 2 = 126 hosts • How many bits are left for subnets? Since 16 = BB + 7, then BB = 9 • So how many subnets can we have? 29 - 2 = 510 subnets

15. 1 1 1 128 64 32 16 8 4 2 1 2. What’s the subnet mask? • We determine the subnet mask by adding up the decimal value of the bits we borrowed. • In the previous Class C example, we borrowed 3 bits. Below is the host octet showing the bits we borrowed and their decimal values. We add up the decimal value of these bits and get 224. That’s the last non-zero octet of our subnet mask. So our subnet mask is 255.255.255.224 Remember: The subnet mask has all 1s in the network portion.

16. 3. What’s the “magic number?” • To find the “magic number” or the multiplier we will use to determine the subnetwork addresses, we subtract the last non-zero octet from 256. • Note: The “magic number” can also be found by determining the value of the last bit borrowed. • In our Class C example, our subnet mask was 255.255.255.224. 224 is our last non-zero octet. • Our magic number is 256 - 224 = 32 • Note: The last bit borrowed was the 32 bit.

17. Last Non-Zero Octet • Memorize this table. You should be able to: • Quickly calculate the last non-zero octet when given the number of bits borrowed or... • Determine the number of bits borrowed when given the last non-zero octet.

18. 4. What are the subnets? • We now take our “magic number” and use it as a multiplier. • Our Class C address was 210.93.45.0. • We borrowed bits in the fourth octet, so that’s where our multiplier occurs. • 1st subnet: 210.93.45.32 • 2nd subnet: 210.93.45.64 • 3rd subnet: 210.93.45.96 • 4th subnet: 210.93.45.128 • 5th subnet: 210.93.45.160 • 6th subnet: 210.93.45.192

19. Host & Broadcast Addresses • Now you can see why we subtract 2 when determining the number of host addresses. • Let’s look at our 1st subnet: 210.93.45.32 • What is the total range of addresses up to our next subnet, 210.93.45.64? 210.93.45.32 to 210.93.45.63 or 32 addresses • .32 cannot be assigned to a host. Why? Because it is the subnet’s address. • .63 cannot be assigned to a host. Why? Because it is the subnet’s broadcast address. • So our host addresses are .33 - .62 or 30 host addresses--just like we figured out earlier.

20. Practice Your Subnetting!! • If you have not yet mastered subnetting, now is the time to do so. • Semester 5’s curriculum assumes the ability to quickly subnet without pencil & paper! (much like the ability to add and subtract is assumed in Algebra) • You will need to be able to evaluate an addressing scheme quickly just by looking at the address and subnet mask. • Furthermore, Variable Length Subnet Masking (VLSM) becomes much easier if you’ve mastered subnetting. • To practice, simply take any network address/design goal scenario and subnet it!! For example... • 192.168.1.0 with at least 30 subnets • 172.16.0.0 with at least 500 hosts per subnet • 10.0.0.0 with at least 2000 subnets