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IP Address and Subnetting

IP Address and Subnetting. Review. Session overview. Settle down Review Qs - IP addressing _Basic Lecture/interactive discussion Revisit IP addressing _Basic Complete any unanswered questions Student presentation Task 1 Problem definition, analysis and design(5 minutes/group)

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IP Address and Subnetting

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  1. IP Address and Subnetting Review RD-CSY2001-2009/10

  2. Session overview • Settle down • Review Qs - IP addressing _Basic • Lecture/interactive discussion • Revisit IP addressing _Basic • Complete any unanswered questions • Student presentation Task 1 • Problem definition, analysis and design(5 minutes/group) • Class discussion/ questions (10 minutes) • Group Feedback • Homework : Review Qs - Subnetting RD-CSY2001-2009/10

  3. Review QsIP Addressing-Basics Available on the module web page Time: 10 minutes RD-CSY2001-2009/10

  4. Lecture Overview • IPv4 address • Classful addressing • Private and Public IP addresses • Subnet • Need to subnet • Subnet Class C address • Task RD-CSY2001-2009/10

  5. Review: IPv4 Address Classes • IP v4 addresses are 32 bits long, given as a.b.c.d • IP addresses are divided into five classes, identified by the first group of numbers in the dotted decimal notation as • Class Range • A 0-127 • B 128-191 • C 192-223 • D 224-239 • E 240-255 • Addresses from classes A, B, C are assignable 8 8 16 24 32 Network ID 0 Host ID Class A 10 Class B 110 Class C MulticastAddresses 1110 Class D Reserved for experiments 1111 Class E RD-CSY2001-2009/10

  6. Generally, IP addresses have two parts Network (Net id) Host ID Netid and Hostid in a given IP address are identified by Subnet mask Default subnet masks are Class A : 255.0.0.0 Class B : 255.255.0.0 Class C : 255.255.255.0 Review - Subnet mask 1st octet 2nd octet 3rd octet 4th octet Network Class A Host Host Host Class B Network Host Host Network Network Network Network Host Class C RD-CSY2001-2009/10

  7. Special IP Addresses • Loopback address • 127.0.0.0 • Network address • IP address with all host bits set to 0 • Example: 172.16.0.0 • Broadcast address • IP address with all host bits set to 1 • Example: 172.16.255.255 RD-CSY2001-2009/10

  8. Public and private IP addresses • Public IP addresses • Unique • Used to connect to Internet. • Use of an address class depends on number of hosts / networks, required to be connected • Private IP addresses • Use to conserve public IP addresses • Three special ranges, one each in class A, B and C. RD-CSY2001-2009/10

  9. Private IP addresses • Assigned to hosts that do not connect directly to the Internet • Three blocks are available, one each from • Class A • Class B • Class C addresses • Addresses need to be ‘translated’ for connecting hosts to the Internet . RD-CSY2001-2009/10

  10. Problems with traditional (Classful) IP Addressing • Inefficient Address Usage • In danger of running out of classes A and B • Why? • Class C too small for most domains • Very few class A – very careful about giving them out • Class B poses greatest problem • Class B sparsely populated • But users refuse to give it back • Need simple way to reduce the number of network numbers assigned RD-CSY2001-2009/10

  11. Some solutions to overcome IP address problem • Use Private Addresses • Dynamic allocation of addresses • DHCP • Subnet the given address • Use Classless IP addressing schemes (CIDR) • Use larger address space • IPv6 uses 128 bit address (32 bits for IPv4 addresses) RD-CSY2001-2009/10

  12. Task: Assign IP addresses Device ? Router Assign IP addresses to above network using appropriate subnet mask: Class A Class B Class C RD-CSY2001-2009/10

  13. Organizations have multiple networks which are independently managed Subnetting allows us to break LANs into small sub-networks Sub-networks created by borrowing bits from host-id. from the given IP address What are the maximum number of bits that can be borrowed in a Class C address? Class B address? University Network Engineering School Business School Library Why Subnet? RD-CSY2001-2009/10

  14. Designing addressing scheme for an Inter-network. • When designing an address scheme, assign addresses to hosts, network devices and the router interface RD-CSY2001-2009/10

  15. Document: Addressing Maps RD-CSY2001-2009/10

  16. How to: Subnetting • Steps • Know how many Different Networks are required • Borrow bits from the host portion of the IP address • Find New Subnet Mask. • Calculate the number of sub-networks and the hosts available corresponding to borrowed bits • Find the sub-network boundary • Network Address • Find the broadcast address. • Let’s look at each of these steps in detail RD-CSY2001-2009/10

  17. 1. How many bits to borrow? • How many host bits CAN/DO I have to borrow to create required subnets • Depends on the class of your network address. • How do you find the IP address class? • First octet of IP address • What are the host bits for the default subnet mask? • Class C: • 8 host bits • Class B: • 16 host bits • Class A: • 24 host bits RD-CSY2001-2009/10

  18. Example: How many bits to borrow? • Class C Address: 210.93.45.0 • Requirement: At least 5 subnets • how many bits do we borrow (Bits Borrow (BB))? • How many bits in the host portion (HB) do we have for default mask? Since it’s a Class C, we have 8 bits to work with. • 2 to what power will give us at least 5 subnets? 23 - 2 = 6 subnets • How many bits are left for hosts? Bits left = Bits available – bits borrowed 5 = 8-3 • Assignable host addresses 25 - 2 = 30 hosts One network address, one broadcast address RD-CSY2001-2009/10

  19. 1 1 1 128 64 32 16 8 4 2 1 2. What’s the new subnet mask? • We determine the new 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 (128+64+32). NEW subnet mask is255.255.255.224(as against default subnet mask of 255.255.255.0) RD-CSY2001-2009/10

  20. 3. What’s the “magic number?” • 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. RD-CSY2001-2009/10

  21. Assigning sub-network addresses • 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 RD-CSY2001-2009/10

  22. 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. RD-CSY2001-2009/10

  23. 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. Last Non-Zero Octet RD-CSY2001-2009/10

  24. Revisit Review QsIP Addressing-Basics Complete/correct answers Time: 10 minutes RD-CSY2001-2009/10

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