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DNS and IP addressing

DNS and IP addressing. 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?. What is an IP address?. 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.

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DNS and IP addressing

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  1. DNS and IP addressing Instructor: Sam Nanavaty

  2. 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? Instructor: Sam Nanavaty

  3. Instructor: Sam Nanavaty

  4. What is an IP address? Instructor: Sam Nanavaty

  5. 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 Instructor: Sam Nanavaty

  6. # of hosts = 2n-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” Instructor: Sam Nanavaty

  7. ExamplesIdentify Network address, class, and determine if it is valid host address as well. • 222.10.1.1 • 127.12.1.98 • 97.1.255.255 • 197.17.0.255 • 0.12.252.1 • 10.0.1.0 • 220.0.0.254 Instructor: Sam Nanavaty

  8. 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? Instructor: Sam Nanavaty

  9. 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) Instructor: Sam Nanavaty

  10. 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) Instructor: Sam Nanavaty

  11. Subnet mask determination Instructor: Sam Nanavaty

  12. Answers Instructor: Sam Nanavaty

  13. 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 (25 -2) with up to 6 hosts (23-2) 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 1st 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 1st 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) (11111000 – 11111111) 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 Instructor: Sam Nanavaty

  14. 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 Instructor: Sam Nanavaty

  15. 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) Instructor: Sam Nanavaty

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

  17. 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 Instructor: Sam Nanavaty

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

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

  20. 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 Instructor: Sam Nanavaty

  21. ISP 172.16.1.1 /24 192.168.100.17 /28 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 Instructor: Sam Nanavaty

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