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IP Addressing Allan Johnson IPv4 Addressing Review Table of Contents End Slide Show 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. IP Address Classes Class A: 1.0.0.0 to 127.0.0.0 Network Host Host Host 1st Octet Bits: ___ 0 ___ X ___ X ___ X ___ X ___ X ___ X ___ X (The 128 bit is off.) Class B: 128.0.0.0 to 191.255.0.0 Network Network Host Host 1st Octet Bits: ___ 1 ___ 0 ___ X ___ X ___ X ___ X ___ X ___ X (The 128 bit is on and the 64 bit is off.) Class C: 192.0.0.0 to 223.255.255.0 Network Network Network Host 1st Octet Bits: ___ 1 ___ 1 ___ 0 ___ X ___ X ___ X ___ X ___ X (The 128 and 64 bits are on. The 32 bit is off.) Reserved IP Address Classes Multicasting Class D: 224.0.0.0 to 239.0.0.0 1st Octet Bits: ___ 1 ___ 1 ___ 1 ___ 0 ___ X ___ X ___ X ___ X (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 1st Octet Bits: ___ 1 ___ 1 ___ 1 ___ 1 ___ X ___ X ___ X ___ X (The 128, 64, 32, and 16 bit are all on.) 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. 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) Continue With Subnetting Review Skip Subnetting Review 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. Four Subnetting Steps To correctly subnet a given network address into subnet addresses, ask yourself the following questions: 1. 2. 3. 4. 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 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. 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. 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: 2 2x BL 2 2x BB Remember: we need to subtract two hosts to provide for the subnetwork and broadcast addresses. 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 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? 7 2 - 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 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. 1 1 1 128 64 32 16 8 4 2 1 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. 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. 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. Bits Non-Zero Borrowed Octet 1 128 2 192 3 224 4 240 5 248 6 252 7 254 8 255 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 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. 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