1. CSIS 3723

2.  We need to create some logic to the environment  We want to keep like devices together  We want to make money leasing the use of the space  Security

3.  When designing the Internet it was decided that not all devices would need or want to be addressable from the Internet but the will still need to communicate using the network  Because of this private address space was created  These addresses are not accessible from the Internet without the network administrator doing something to give them a Internet address (NAT)  These addresses can be accessible in your intranet (corporate space)

4.  RFC 1918 defines these ◦ 10.0.0.0 - 10.255.255.255 (10.0.0.0/8) ◦ 172.16.0.0 – 172.31.255.255 (172.16.0.0/12) ◦ 192.168.0.0 – 192.168.255.255 (192.168.0.0/16)  These are the IP address spaces that can be used internally in an enterprise

5.  RFC states a “link local” block ◦ 169.254.0.0 – 169.254.255.255 (169.254.0.0/16) ◦ To be used when a device can not get an IP address through DHCP  Also reserves lowest Class B ◦ 128.0.0.0 -128.0.255.255 (128.0.0.0/16) ◦ Not able to be used under old class system but can be assigned to someone  Also defines loop back space (RFC 1700) ◦ 127.0.0.0 – 127.255.255.255 (127.0.0.0/8) ◦ Used for a machine to communicate internally  Also defines multicast address space (RFC 5771) ◦ 224.0.0.0 – 239.255.255.255 (224.0.0.0/4)  So you should never use these IP address spaces!

6. 150.134.10.0/24 150.134.10.10 150.134.10.30192.168.1.12 Internet 192.168.1.0/24

7.  What are the IP addresses for the subnet 192.168.0.0/24?  192.168.0.0 through 192.168.0.255  Anything after the 24 th most significant bit can change and be in the same subnet 110000001010100000000000 11000000101010000000000011111111

8.  We use the CIDR as a binary number  Every most significant bit is a one the rest are zero  So a /24 would be: 11111111 00000000 Subnet Mask 255 00000000

9.  What would subnet mask be for /20? 11111111 1111000000000000 255 24000000000 11111111 1286432168421 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255

10. 19216800 11000000101010000000000010001010 /24 11111111 00000000 IP Address Subnet Mask Logical AND 110000001010100000000000 Logical AND 0 0 1 1 0 1 00010001 Bitwise AND is used. Logical AND is done on each bit between the IP address and the subnet mask. If the result matches the network it is in the same subnet 1921680138 Network Destination Address Network 110000001010100000000000

11. 19216800 10010110100001100000101000011100 /24 11111111 00000000 IP Address Subnet Mask Logical AND 10010110100001100000101000000000 1501341028 110000001010100000000000 Network Destination Address Network No match different subnet

12.  If we look at just one octet we see a pattern MaskBinaryRatio 00000 1:256 1281000 00002:128 1921100 00004:64 2241110 00008:32 2401111 000016:16 2481111 100032:8 2521111 110064:4 2541111 1110128:2 2551111 256:1

13. 192168100/24 If I start with: This is the last octet: 1281000 00002:128 00000 1:256 If we change it to: We would get two networks: 192168100/25 19216810128/25

14. Lets look at what happens when the number change in the last octet 1000000000000001 MaskLast Octet of IP address 00010010 01011010 10010100 As long as this bit does not become a one in the IP address it is in the first subnet First Subnet Second Subnet 10010100

15. What if an ISP owns a block of IP addresses like: ISP's block 11001000 00010111 00010000 00000000 200.23.16.0/20 If I had 8 customers that want to buy subnets how could I change the subnet mask to get 8 subnets? 11001000 00010111 00010000 00000000 200.23.16.0/20 11001000 00010111 00010000 00000000 Each place I move I get a multiple of two

16. 11001000 00010111 00010000 00000000 To get 8 in binary I would need 3 binary numbers 111 421 4 + 2 + 1 = 7 ??? We start count from zero

17. ISP's block 11001000 00010111 00010000 00000000 200.23.16.0/20 Organization 0 11001000 00010111 00010000 00000000 200.23.16.0/23 Organization 1 11001000 00010111 00010010 00000000 200.23.18.0/23 Organization 2 11001000 00010111 00010100 00000000 200.23.20.0/23... ….. …. …. Organization 7 11001000 00010111 00011110 00000000 200.23.30.0/23

18. Organization 0 11001000 00010111 00010000 00000000 200.23.16.0/24 Organization 1 11001000 00010111 00010001 00000000 200.23.17.0/24 Organization 2 11001000 00010111 00010010 00000000 200.23.18.0/24... ….. …. …. Organization 16 11001000 00010111 00011111 00000000 200.23.31.0/24 What if I needed 11 subnets??? ISP's block 11001000 00010111 00010000 00000000 200.23.16.0/20 We need to create 16 subnets to get 11

19. What if I needed subnets that can have 56 hosts??? ISP's block 11001000 00010111 00010000 00000000 200.23.16.0/20 Where can we move from the right to get a number larger the 56 ??? 11111111 1286432168421 32 + 16 + 8 + 4 + 2 + 1 = 63 (plus one for zero) is 64

20. Organization 0 11001000 00010111 00010000 00000000 200.23.16.0/26 Organization 1 11001000 00010111 00010000 01000000 200.23.16.64/26 Organization 2 11001000 00010111 00010000 10000000 200.23.16.128/26... ….. …. …. Organization 64 11001000 00010111 00011111 11000000 200.23.31.192/26 What if I needed subnets that can have 56 hosts??? ISP's block 11001000 00010111 00010000 00000000 200.23.16.0/20 We need to create 64 subnets each having 64 IP addresses

21.  On the subnet 200.23.16.0/26  We only get 61 usable IP address  One is used for the network ◦ 200.23.16.0 not used  One must be used for the router interface on the subnet ◦ 200.23.16.1 is usual used as the router IP address but does not have to (could be 200.23.16.62 or any other host IP address)  One must be used for the broadcast address ◦ Is always the last IP address in the subnet (200.23.16.63)