1. Review Aug-141RD-CSY3021

2.  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 Aug-142RD-CSY3021

3. Available on the module web page Time: 10 minutes Aug-143RD-CSY3021

4.  IPv4 address ◦ Classful addressing ◦ Private and Public IP addresses  Subnet ◦ Need to subnet ◦ Subnet Class C address ◦ Task Aug-144RD-CSY3021

5. 5 Network ID Host ID 8 16 Class A 32 Class B 10 Class C 110 Multicast Addresses Class D 1110 Reserved for experiments Class E 1111 248  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  ClassRange  A0-127  B128-191  C192-223  D224-239  E240-255  Addresses from classes A, B, C are assignable 0 Aug-14

6. RD-CSY30216  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 Network Host Network Host Network Host 1st octet2nd octet3rd octet4th octet Class A Class B Class C Aug-14

7. RD-CSY30217  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 Aug-14

8. RD-CSY30218  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. Aug-14

9. RD-CSY30219  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. ClassRange A 10.0.0.0 – 10.255.255.255 B 172.16.0.0 – 172.31.255.255 C 192.168.0.0 – 192.168.255.255 Aug-14

10. RD-CSY302110  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 Aug-14

11.  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) Aug-1411RD-CSY3021

12. 12 Assign IP addresses to above network using appropriate subnet mask: Class A Class B Class C Device ? Router Aug-14

13. RD-CSY3021  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 Business School Library Engineering School Aug-1413

14. RD-CSY302114  When designing an address scheme, assign addresses to hosts, network devices and the router interface Aug-14

15. RD-CSY302115Aug-14

16. RD-CSY3021  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 Aug-1416

17. RD-CSY3021  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 Aug-1417

18. RD-CSY3021  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? 2 3 - 2 = 6 subnets ◦ How many bits are left for hosts? Bits left = Bits available – bits borrowed 5 = 8-3 ◦ Assignable host addresses 2 5 - 2 = 30 hosts One network address, one broadcast address Aug-1418

19. RD-CSY3021  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. 128 64 32 16 8 4 2 1 111 We add up the decimal value of these bits and get 224 (128+64+32). NEW subnet mask is 255.255.255.224 (as against default subnet mask of 255.255.255.0) Aug-1419

20. RD-CSY3021  magic number : total_value_of_non-zero_octet – new_subnet_mask  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. Aug-1420

21. RD-CSY3021  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 Aug-1421

22. RD-CSY3021  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. Aug-1422

23. RD-CSY3021  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. Aug-1423

24. Complete/correct answers Time: 10 minutes Aug-1424RD-CSY3021

25. Task 1: Basic home network Objectives covered 1. Problem analysis and definition 2. Design Aug-1425RD-CSY3021