Today we will discuss:

1. Today we will discuss: • IP Addressing • Subnetting This is probably the most important topic for the exam. Try to memorize the tables and numbers where indicated since they will help immensely on the exam.

2. IP Addressing Fundamentals: • If a device wants to communicate using TCP/IP, it needs an IP address. • Any device that can send and receive IP packets is called an IP host. • IP addresses consist of a 32-bit number, usually written in dotted-decimal notation.

3. IP Address Fundamentals (cont.): • Each byte (8 bits) of the 32-bit IP address is converted to its decimal equivalent (hence, “decimal” of “dotted-decimal notation”). • Each of the decimal numbers in an IP address is called an octet. • The term octet is just a vendor-neutralterm instead of byte.

4. IP Address Fundamentals (cont.): • For example: 168.1.4.25 • The first octet is 168. • The second octet is 1. • The third octet is 4. • The fourth octet is 25. • The RANGE of decimal numbers in each octet is between 0 and 255, inclusive. (256 numbers total)

5. Other IP Terminology • Bit – a bit is one digit; either a 1 or a 0. • Byte – a byte is 8 bits. • Network address – also called the network number, uniquely identifies each network. Every machine on the same network shares that network address as part of its IP address. • Host address – also referred to as a node address, is assigned to and uniquely identifies each machine on a network. • Broadcast address - sent to all nodes on the network

6. Binary to Decimal Conversion

7. Binary • The ability to convert from binary to decimal and back again is very handy when working with IP addresses. • The digits used are limited to either a 1 (one) or a 0 (zero). • Only convert up to 8 bits at a time (an octet).

8. Binary (cont.) • The trick to binary is to learn the decimal values of each bit for the first 8 bits. • Start from the left or right and add. • Just add the decimal values where there is a 1 (one) present, and you will have the decimal value of the octet. • The next slide shows the decimal value of each bit value.

9. Binary to Decimal for 204 Binary = 1100 1100 Decimal = 204 • 64 32 16 8 4 2 1 1 1 0 0 1 1 0 0 ----------------------------------------------------------- • + 64 + 0 + 0 + 8 + 4 + 0 + 0 = 204

10. Memorize • Binary 1111 1111 = Decimal 255 • Binary 0000 0000 = Decimal 0 • Know the values 128-64-32-16-8-4-2-1. • The easiest way to remember it is to start at 1 and multiply by 2. • 1 * 2 = 2 * 2 = 4 * 2 = 8 * 2 = 16 * 2 = 32 * 2 = 64 * 2 = 128.

11. Decimal to Binary • Convert 212 to binary. • Do the math. • Answer: 11010100

12. Another Binary to Decimal **Memorization** Chart (see why this is important in a few slides)

13. Hierarchical IP Addressing Scheme • An IP address consists of 32 bits of information. • The address is broken into four 8-bit (1 byte) groups, converting each octet to decimal values, and separating these values by dots (dotted decimal notation). • Example: 172.16.30.56

14. Hierarchical IP Addressing (cont.) • Two other ways to represent dotted- decimal 172.16.30.56 are: • Binary • 10101100.00010000.00011110.00111000 • Hexadecimal • AC.10.1E.38 • See end of presentation for hex conversion notes and chart.

15. The Classes of Networks

16. Classes of Networks • All addresses in the same class A, B, or C network have the same numeric value NETWORK portion of the address. The rest of the address is called the HOST portion of the address. • When written down, network numbers have all decimal 0s in the host part of the number.

17. Classes of Networks (cont.) • Network number example: • Class A: byte. 0. 0. 0 • Class B: byte. byte. 0. 0 • Class C: byte. byte. byte. 0 • However, network numbers are not actually IP addresses because they cannot be assigned to an interface as an IP address.

18. List of All Possible Valid Network Numbers (memorize)

19. Details of Classes A, B, & C • Before starting to design subnets, you should know what it is that you've been given. Here are three ways of finding out what class your allocation is in. Use whichever you find easiest. • Class A addresses begin with 0xxx binary, or 1 to 126 decimal. (127 is loopback) • Class B addresses begin with 10xx binary, or 128 to 191 decimal. • Class C addresses begin with 110x binary, or 192 to 223 decimal. • If the first bit is 0 it is a Class A address • If the first two bits are 10 it is a Class B address • If the first three bits are 110 it is a Class C address

20. Address of all 0s Address of all 1s Network 127.0.0.1 Means “network address” or “this network segment” Means “broadcast” Reserved for loopback tests. Allows local node to send a test packet to itself without generating network traffic. Reserved IP Addresses

21. Private IP Addresses • These addresses can be used on a private network, but are not routable on the Internet. • With these addresses, use Network Address Translation (NAT) to access the Internet via a legal IP address.

22. SUBNETTING .

23. 21 = 2 22 = 4 23 = 8 24 = 16 25 = 32 26 = 64 27 = 128 28 = 256 Memorize Powers of 2

24. Subnet Masks • A subnet mask is a 32-bit binary number usually written in dotted-decimal format. • The 1s in the subnet mask represent the network (or subnet) part of the IP address. • The 0s represent the host part. • Example: Binary: 11111111.11111111.11111100.00000000 Dotted-decimal: 255.255.252.0 (same number)

25. Subnet Masks (cont.) • Slash notation (/) at the end of an IP address means how many bits are turned on (1s). • Ex: 192.168.10.32 /28 Where /28 is subnet mask 255.255.255.240

26. Default Subnet Mask • Not all networks need subnets, meaning they use the default subnet mask.

27. Five questions to answer when calculating: • How many subnets does the chosen subnet mask produce? • How many valid hosts per subnet are available? • What are the valid subnets? • What’s the broadcast address of each subnet? • What are the valid hosts in each subnet?

28. Easier than it looks: • How many subnets? 2x – 2 = number of subnets. x is the number of masked bits, or the 1s. For example, in 11000000, the number of ones gives us 22 – 2 subnets. In this example, there are 2 subnets.

29. Easy (cont.) 2) How many hosts per subnet? 2y – 2 = number of hosts per subnet. y is the number of unmasked bits, or the 0s. For example, in 11000000, the number of zeros gives us 26 – 2 hosts. In this example, there are 62 hosts per subnet.

30. Easy 3) What are the valid subnets? 256 – subnet mask = block size, or base number. For example, for subnet mask 255.255.192.0, 256 – 192 = 64. 64 is the first subnet. The next subnet would be the base number plus itself, or 64 + 64 = 128, (the second subnet). You keep adding the base number to itself until you reach the value of the subnet mask, which is not a valid subnet because all subnet bits would be turned on (1s).

31. Easy 4) What’s the broadcast address for each subnet? The broadcast address is all host bits turned on, which is the number immediately preceding the next subnet (example in a minute). 5) What are the valid hosts? Valid hosts are the numbers between the subnets, minus the network (subnet) numbers and broadcast numbers.

32. Calculating: Class C, Example 1 Network address = 192.168.10.0 Subnet mask = 255.255.255.192 Just answer the five questions: 1) How many subnets? Since 192 is 2 bits on (11000000), the answer is 22 – 2 = 2 2) How many hosts per subnet? We have 6 host bits off (11000000), so the equation would be 26 – 2 = 62 hosts.

33. Example 1 (cont.) 3) What are the valid subnets? 256-192 = 64, which is the first subnet and also the block size. Keep adding the block size to itself until you reach the subnet mask. 64 + 64 = 128. 128 + 64 = 192, which is invalid because it is the subnet mask (all subnet bits turned on). Our two valid subnets are, then, 64 and 128.

34. Example 1 (cont.) 4) What’s the broadcast address for each subnet? The number right before the value of the next subnet is all host bits turned on and equals the broadcast address. 5) What are the valid hosts? These are the numbers between the subnet and broadcast address.

35. Example 1 Final Result

36. Class C Example 2 Network address = 192.168.10.0 Subnet mask = 255.255.255.224 Follow preceding example. Answers:

37. Subnetting in Your Head Host address = 192.168.10.33 Subnet mask = 255.255.255.240 What subnet and broadcast address is the above IP address a member of? First, answer question 3 of the big 5: 3) What are the valid subnets? 256 – 240 = 16. 16 + 16 = 32. 32 + 16 = 48. The host address is between the 32 and 48 subnets. The subnet is 192.168.10.32, the broadcast address is 192.168.10.47, and the valid host range is 192.168.10.33 – 192.168.10.46

38. Calculating Class B Addresses Network address = 172.16.0.0 Subnet mask = 255.255.192.0 Answer the five questions: • Subnets? 22 – 2 = 2. (192 = 11000000) • Hosts? 214 – 2 = 16,382. (6 bits in the third octet, and 8 in the fourth.) • Valid subnets? 256 – 192 = 64. 64 + 64 = 128. (these are the 2 subnets as stated in question 1.)

39. Class B Example (cont.) 4) Broadcast address for each subnet? See table. 5) Valid hosts? See table.

40. Class B, Example 2 Network address = 172.16.0.0 Subnet mask = 255.255.240.0 Follow prior questions. (240 = 11110000) The table shows first three subnets, valid hosts, and broadcast addresses in a Class B 255.255.240.0 mask:

41. Subnetting in Your Head, Class B Q: What subnet and broadcast address is the IP address 172.16.10.33 255.255.255.224 a member of? A: 256 – 224 = 32. 32 + 32 = 64. 33 is between 32 and 64. However, remember that in Class B addresses the third octet is considered part of the subnet, so the answer would be the 10.32 subnet. The broadcast is 10.63, since 10.64 is the next subnet.

42. In Your Head, Class B Ex. 2 Q: What subnet and broadcast address is the IP address 172.16.90.66 255.255.255.192 a member of? A: 256 – 192 = 64. 64 + 64 = 128. The subnet is 172.16.90.64. The broadcast must be 172.16.90.127, since 90.128 is the next subnet.

43. Subnetting Class A Addresses • Same procedure as with Class B and Class C only you must take into account the 8 additional bits from the second octet.

44. Class A Example Network 10.0.0.0 Subnet mask 255.255.0.0 ( /16 = 16 bits on) • Subnets? 28 -2 = 254. • Hosts? 216 – 2 = 65,534. • Valid subnets? 256 – 255 = 1, 2, 3, etc. (all in the second octet). The subnets would be 10.0.0.0, 10.2.0.0, 10.3.0.0, etc., up to 10.254.0.0.

45. Class A Example (cont.) 4) Broadcast address for each subnet? 10.1.255.255, 10.2.255.255, etc., up to 10.254.255.255. 5) Valid hosts? See table.

46. Appendix(will be added shortly) Hex to Binary to Decimal Conversion Slash Notation (/) Tables Boolean/Binary Calculation Method