1. Binary Lesson 5 Counting

2. Powers of 2 One bit has 2 possible values (2^1) One bit has 2 possible values (2^1) 0 or 1 0 or 1 Two bits have 4 possible values (2^2) Two bits have 4 possible values (2^2) 00 01 11 10 00 01 11 10 Three bits have 8 values (2^3) Three bits have 8 values (2^3) 000 001 010 011 100 101 110 111 000 001 010 011 100 101 110 111 Four bits have 16 values (2^4) Four bits have 16 values (2^4) 0000 0001... 1110 1111 0000 0001... 1110 1111

3. A Byte 8 bits, from 0000000 to 11111111 8 bits, from 0000000 to 11111111 2^8 combinations 2^8 combinations 2x2x2x2x2x2x2x2 = 256 2x2x2x2x2x2x2x2 = 256 In Hex, 00 to FF In Hex, 00 to FF 16 x 16 = 256 16 x 16 = 256 2^8 = 2^4 x 2^4 2^8 = 2^4 x 2^4

4. A Word – 16 Bits 0000 to FFFF 0000 to FFFF 2^16 values = 65,536 2^16 values = 65,536 2^16 = 2^8 x 2^8 = 256 x 256 2^16 = 2^8 x 2^8 = 256 x 256 2^16 = 2^4 x 2^4 x 2^4 x 2^4 2^16 = 2^4 x 2^4 x 2^4 x 2^4 = 16 x 16 x 16 x 16 = 16 x 16 x 16 x 16

5. The Easy Way 2^10 = 1,024, written as 1 K 2^10 = 1,024, written as 1 K So 16 bits have 2^16 values So 16 bits have 2^16 values 2^16 = 2^6 x 2^10 = 2^6 K = 64 K 2^16 = 2^6 x 2^10 = 2^6 K = 64 K Approximately 64,000 Approximately 64,000

6. 32 Bits Range of values from Range of values from 0000:0000 to FFFF:FFFF 0000:0000 to FFFF:FFFF 2^32 values 2^32 values 2^2 x 2^10 x 2^10 x 2^10 2^2 x 2^10 x 2^10 x 2^10 4 K K K = 4 K M = 4 G 4 K K K = 4 K M = 4 G Approximately 4,000,000,000 Approximately 4,000,000,000

7. 64 Bits 0000:0000:0000:0000 to 1111:1111:1111:1111 0000:0000:0000:0000 to 1111:1111:1111:1111 2^64 2^64 2^4 x 2^10 x 2^10 x 2^10 x 2^10 x 2^10 x 2^10 2^4 x 2^10 x 2^10 x 2^10 x 2^10 x 2^10 x 2^10 16 K K K K K K = 16 G G 16 K K K K K K = 16 G G Approximately Approximately 16,000,000,000,000,000,000 16,000,000,000,000,000,000

8. 128 Bits The entire IPv6 address space, from The entire IPv6 address space, from 0000:0000:0000:0000:0000:0000:0000:0000 0000:0000:0000:0000:0000:0000:0000:0000 to to ffff:ffff:ffff:ffff:ffff:ffff:ffff:ffff ffff:ffff:ffff:ffff:ffff:ffff:ffff:ffff 2^128 = 2^8 x 2^120 2^128 = 2^8 x 2^120 2^8 x 2^30 x 2^30 x 2^30 x 2^30 2^8 x 2^30 x 2^30 x 2^30 x 2^30 256 G G G G 256 G G G G Approximately Approximately 256,000,000,000,000,000,000,000,000,000,000,000,000 256,000,000,000,000,000,000,000,000,000,000,000,000

9. Slash Notation fe80::/64 means 64 bits are fixed, so 64 bits vary fe80::/64 means 64 bits are fixed, so 64 bits vary 2^64 addresses 2^64 addresses fe80::/16 means 16 bits are fixed, so 112 bits vary fe80::/16 means 16 bits are fixed, so 112 bits vary 2^112 addresses 2^112 addresses 2620:1:b::/48 means 48 bits are fixed, so 80 bits vary 2620:1:b::/48 means 48 bits are fixed, so 80 bits vary 2^80 addresses 2^80 addresses

10. Binary iClicker Questions

11. How many different nybbles are there? 1 nybble has 4 bits  2  4  16  256  65536

12. How many different bytes are there? 1 byte has 8 bits  16  256  65,536  16,000,000  4,000,000,000

13. How many different words are there? 1 word has 16 bits An approximate value is OK  256  64,000  16,000,000  4,000,000,000  1,000,000,000,000

14. How many addresses are in this range? ff02::/112 (An approximate answer is OK)  1  256  64,000  16,000,000  4,000,000,000

15. How many addresses are in this range? ff02::/96 (An approximate answer is OK)  64,000  4,000,000,000  256,000,000,000,000  16,000,000,000,000,000,000  64,000,000,000,000,000,000,000,000,000

16. How many addresses are in this range? 2610:1:b::/64 (An approximate answer is OK)  64,000  4,000,000,000  256,000,000,000,000  16,000,000,000,000,000,000  64,000,000,000,000,000,000,000,000,000

17. How many addresses are in this range? 2610:1:b::/48 (An approximate answer is OK)  64,000  4,000,000,000  256,000,000,000,000  16,000,000,000,000,000,000  64,000,000,000,000,000,000,000,000,000

18. How many /64 subnets are in a /48 address allocation? (An approximate answer is OK)  16  256  64,000  4,000,000,000  256,000,000,000,000