1. Instructor: Sam Nanavaty DNS and IP addressing

2. Instructor: Sam Nanavaty How does a router know where to route the information when you simply type in a URL (e.g., www.yahoo.com) in the Internet browser?

3. Instructor: Sam Nanavaty

4. What is an IP address?

5. Instructor: Sam Nanavaty IP addressing 32 bits in length 4 octets Five classes A,B,C,D and E Only A,B and C used for naming devices. D used for multicasting groups E is reserved for experimental purposes

6. Instructor: Sam Nanavaty ClassFormatDefault Network Prefix High Order Bits Usable Network Address range AN.H.H.H Hosts:16.7 m Mask:255.0.0.0 8 bits01.0.0.0 – 126.0.0.0 127.0.0.0 - Loopback BN.N.H.H Hosts:65,534 Mask:255.255.0.0 16 bits10128.0.0.0 – 191.255.0.0 CN.N.N.H Hosts:254 255.255.255.0 24 bits110192.0.0.0 – 223.255.255.0 # of hosts = 2 n -2 (where n = # of bits in host Id) Why? Host ID = all 1’s is not permitted as this refers to broadcast address Host Id = all 0’s is not permitted as this refers to “this network”

7. Instructor: Sam Nanavaty Examples Identify Network address, class, and determine if it is valid host address as well. 1.222.10.1.1 2.127.12.1.98 3.97.1.255.255 4.197.17.0.255 5.0.12.252.1 6.10.0.1.0 7.220.0.0.254

8. Instructor: Sam Nanavaty Subnetting Steal the bits from the host portion of the IP address and add it to the network portion (gives more networks and fewer hosts per network) Can you think of the benefits of subnetting?

9. Instructor: Sam Nanavaty Subnet mask determination Convert IP addr and default mask in to binary Identify your base network addr Determine how many bits are needed to achieve desired number of subnets and extend the 1’s in the subnet mask by this amount Now you know subnet portion and the host portion of the IP address. Make sure that the said mask still provides enough hosts per subnet (including some room for growth)

10. Instructor: Sam Nanavaty Masking rule Mask always is assigned from left to right in the bit order Every mask contains contiguous 1’s –i.e., 255.255.255.192 is correct, however, 255.240.255.192 is incorrect Hosts on the same network, must use the same subnet mask Subnets that are all 0’s or all 1’s are NOT allowed by default (in private environment you may use these however)

11. Instructor: Sam Nanavaty Subnet mask determination IP addressReqd HostsReqd Subnets 192.168.1.0602 172.16.0.0As many as possible 16

12. Instructor: Sam Nanavaty IP addressReqd HostsReqd Subnets Mask 192.168.1.0602 255.255.255.192 192 = 1100 0000 Yields 4 -2 = 2 subnets 172.16.0.0As many as possible 16 255.255.248.0 Any mask > 21 bits will work, however, 21 bits yields most hosts 248 = 11111000 Yields 32-2 = 30 subnets Answers

13. Instructor: Sam Nanavaty Calculating the range of addresses given a subnet mask IP Network address : 192.168.1.0 Subnet Mask: 255.255.255.248 (248 = 11111 000) This yields up to 30 subnets (2 5 -2) with up to 6 hosts (2 3 -2) per subnet 192.168.1.0 – 192.168.1.7 Discard as Subnet Id = 0! (00000000 – 00000111) 192.168.1.8 – 192.168.1.15 (00001000 – 00001111) (discard 1 st and last IP addr as host Id cannot be all 0’s or 1’s) 192.168.1.16 – 192.168.1.23 (00010000 - 00010111) (discard 1 st and last IP addr). 192.168.1.240 – 192.168.1.247 (11110000 – 11110111) (discard 1st and last IP addr) 192.168.1.248 – 192.168.1.255 (discard subnet Id = all 1’s) Remove first and last subnets as well as first and last IP addresses for each subnet The final valid range of addresses are as follows: 192.168.1.9 – 192.168.1.14 192.168.1.17 – 192.168.1.22 192.168.1.241 – 192.168.1.246 (11111000 – 11111111)

14. Instructor: Sam Nanavaty Practice example IP address : 192.168.1.0 Subnet Mask: 255.255.255.224 This yields up to ________ subnets with up to _____ hosts per subnet Now calculate the range of addresses for this subnet mask Next determine the valid IP addresses in this range

15. Instructor: Sam Nanavaty IP Network address : 192.168.1.0 Subnet Mask: 255.255.255.224 (224 = 111 00000) This yields up to 6 subnets with up to 30 hosts per subnet Range : 192.168.1.0 – 192.168.1.31 ( 00000000 – 00011111 ) (discard as subnet ID =0) 192.168.1.32 – 192.168.1.63 ( 00100000 - 00111111 ) (discard addr w/ host =0 and 255) 192.168.1.64 – 192.168.1.95 ( 01000000 - 01011111 ) (discard addr w/ host =0 and 255) 192.168.1.96 – 192.168.1.127 (01100000 - 01111111 ) (discard addr w/ host =0 and 255) 192.168.1.128 – 192.168.1.159 (10000000 - 10011111) (discard addr w/ host =0 and 255) 192.168.1.160 – 192.168.1.191 (10100000 - 10111111 ) (discard addr w/ host =0 and 255) 192.168.1.192 – 192.168.1.223 (11000000 - 11011111) (discard addr w/ host =0 and 255) 192.168.1.224 – 192.168.1.255 (11100000 - 11111111 ) (discard as subnet ID = 255)

16. Instructor: Sam Nanavaty Address Mask A = (D & M) Where: A = 32 bit IP address M = 32 bit address mask D = Destination address

17. Instructor: Sam Nanavaty CIDR Notation 128.10.0.0/16 The number after the / indicates the number of 1’s in the subnet mask. Used to partition addresses: 128.211.0.0/16 ISP 128.211.0.16/28 Network A (128.211.0.17-128.211.0.30) 128.211.0.32/28 Network B IP reserves host address zero (denotes network) IP reserves host address all 1’s for broadcast) IP reserves network prefix 127/8 for loopback

18. Instructor: Sam Nanavaty 172.16.210.0 /22 Which subnet does this address belong to?

19. Instructor: Sam Nanavaty 201.100.5.68 /28 Find the subnet this address belongs to.

20. Instructor: Sam Nanavaty If you are using a subnet mask of 255.255.255.224, can the following IP address be assigned to a host using the given subnet mask. 217.168.166.192

21. Instructor: Sam Nanavaty ISP 192.168.100.17 /28 172.16.1.1 /24 Identify the valid IP address assignment for the workstation 192.168.100.20 255.255.255.240 DG 172.16.1.1 192.168.100.30 255.255.255.240 DG 192.168.100.17 192.168.100.19 255.255.255.248 DG 172.16.1.1

22. Instructor: Sam Nanavaty