Chapters 4 & 5 Addressing Starting with Ch 5

1. Chapters 4 & 5 Addressing Starting with Ch 5 Lecture

2. IP Addresses • Internetworking Protocol (IP) of the Network Layer is responsible for uniquely identifying all devices and connections on the Internet • The unique identifier is called an IP address • IP address consist of 32 bits (for version 4) • Keep in mind that, if a single device had multiple connections to the Internet, you would need an IP address for each connection • Address space is 232 = 4,294,967,296 32-bit addresses • In theoretical terms, 4,294,967,296 connections can be made to the Internet (not really true in real life) Lecture

3. 0111 0101 1001 0101 0001 1101 1110 1010 75 95 1D EA 75951DEA IP Addresses • The IP Address has 3 notations: Binary, Dotted-decimal and Hexadecimal • Binary: 4 Octets: 01110101 10010101 00011101 11101010 • Dotted-Decimal (or dot notation): • For Dotted-Decimal, each number can range from 0 to 255 • Hexadecimal: Lecture

4. EXAMPLES Change the following IP address from binary notation to dotted-decimal notation: 10000001 00001011 00001011 11101111 Solution 129.11.11.239 Change the following IP address from dotted-decimal notation to binary notation: 111.56.45.78 Solution 01101111 00111000 00101101 01001110 Find the error, if any, in the following IP address: 111.56.045.78 Solution There are no leading zeroes in dotted-decimal notation (045). Change the following IP addresses from binary notation to hexadecimal notation: 10000001 00001011 00001011 11101111 810B0BEF16 Solution Lecture

5. IP Addresses: Classful Addressing • When IP addressing was first started, it used a concept called “classful addressing”. A newer concept called “classless addressing” is slowly replacing it though. • Regarding “classful addressing”, the address space is divided into five classes: A, B, C, D and E. Lecture

6. Finding the class in binary notation Finding the class in decimal notation Lecture

7. EXAMPLES Solution Find the class of the address: 00000001 00001011 00001011 11101111 The first bit is 0. This is a class A address. Solution Find the class of the address: 11000001 10000011 00011011 11111111 The first 2 bits are 1; the third bit is 0. This is a class C address. Solution Find the class of the address: 227.12.14.87 The first byte is 227 (between 224 and 239); the class is D. Lecture

8. Netid and hostid • A, B and C class-addresses are divided into network id and host id • For Class A, Netid=1 byte, Hostid = 3 bytes • For Class B, Netid=2 bytes, Hostid = 2 bytes • For Class C, Netid=3 bytes, Hostid = 1 byte Lecture

9. Blocks in class A • Class A has 128 blocks or network ids • First byte is the same (netid), the remaining 3 bytes can change (hostids) • Network id 0 (first), Net id 127 (last) and Net id 10 are reserved – leaving 125 ids to be assigned to organizations/companies • Each block contains 16,777,216 addresses – this block should be used by large organizations. How many Host can be addressed ???? • The first address in the block is called the “network address” – defines the network of the organization Example • Netid 73 is assigned • Last address is reserved • Recall: routers have addressees Lecture

10. Blocks in class B • Class B is divided into 16,384 blocks (65,536 addresses each) • 16 blocks are reserved • First 2 bytes are the same (netid), the remaining 2 bytes can change (hostids) • For example, Network id 128.0 covers addresses 128.0.0.0 to 128.0.255.255 • Network id 191.225 is the last netid for this block Example • Netid 180.8 is assigned • Last address is reserved • Recall: routers have addresses Lecture

11. Blocks in class C • Class C is divided into 2,097,152 blocks (256 addresses each) • 256 blocks are reserved • First 3 bytes are the same (netid), the remaining 1 byte can change (hostids) • For example, Network id 192.0.0 covers addresses 192.0.0.0 to 192.0.0.255 Lecture

12. Class D addresses are used for multicasting; there is only one block in this class. Class E addresses are reservedfor special purposes; most of the block is wasted. Lecture

13. Network Addresses The network address is the first address. The network address defines the network to the rest of the Internet. Given the network address, we can find the class of the address, the block, and the range of the addresses in the block Given the network address 17.0.0.0, find the class, the block, and the range of the addresses. The class is A because the first byte is between 0 and 127. The block has a netid of 17. The addresses range from 17.0.0.0 to 17.255.255.255. Solution Given the network address 132.21.0.0, find the class, the block, and the range of the addresses. The class is B because the first byte is between 128 and 191. The block has a netid of 132.21. The addresses range from 132.21.0.0 to 132.21.255.255. Solution Given the network address 220.34.76.0, find the class, the block, and the range of the addresses. The class is C because the first byte is between 192 and 223. The block has a netid of 220.34.76. The addresses range from 220.34.76.0 to 220.34.76.255. Solution Lecture

14. REMINDER: Go Over Converting Binary to Decimal and Vice Versa Lecture

15. Converting Number Systems - Review Base-10 The decimal number system is based on power of the base 10. For example, for the number 1259, the 9 is in the 10^0 column - 1s column the 5 is in the 10^1 column - 10s column the 2 is in the 10^2 column - 100s column the 1 is in the 10^3 column - 1000s column 1259 is 9 X 1 = 9 + 5 X 10 = 50 + 2 X 100 = 200 + 1 X 1000 = 1000 ----- 1259 Base-2 (Binary) The Binary number system uses the same mechanism and concept however, the base is 2 versus 10  The place values for binary are based on powers of the base 2: … 2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0 128 64 32 16 8 4 2 1 Lecture

16. Converting Number Systems - Review So, the binary number 10110011 can be converted to a decimal number 1 X 1 = 1 (right most bit or position) 1 X 2 = 2 0 X 4 = 0 0 X 8 = 0 1 X 16 = 16 1 X 32 = 32 0 X 64 = 0 1 X 128 = 128 (left most bit or position) ------ 179 in decimal To convert from decimal to binary requires a different method called the division/remainder method. The idea is to repeatedly divide the decimal number and resulting quotients by 2. The answer will be the remainders. Example: convert 155 to binary (Start from the top and work down) 155/2 Q = 77, R = 1 (Start) 77/2 Q = 38, R = 1 38/2 Q = 19, R = 0 19/2 Q = 9, R = 1 9/2 Q = 4, R = 1 4/2 Q = 2, R = 0 2/2 Q = 1, R = 0 1/2 Q = 0, R = 1 (Stop) Answer is 10011011. Be careful to place the digits in the correct order. Lecture

17. Converting Number Systems - Review Check the answer (10011011) : 1 X 1 = 1 1 X 2 = 2 0 X 4 = 0 1 X 8 = 8 1 X 16 = 16 0 X 32 = 0 0 X 64 = 0 1 X 128 = 128 ----- 155 Base-16 (Hex) The hexadecimal number system is based 16, and uses the same mechanisms and conversion routines we have already examined. The place values for hexadecimal are based on powers of the base 16 The digits for 10-15 are the letters A - F (A is 10, …….., F is 15) …….. 16^3 16^2 16^1 16^0 4096 256 16 1 The binary number 10110011 can be converted to hexadecimal by grouping bits into groups of 4 bits: 1011 0011 B 3 Lecture