Instructor: Sam Nanavaty
DNS and IP addressing
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?
Instructor: Sam Nanavaty
Instructor: Sam Nanavaty
What is an IP address?
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
Instructor: Sam Nanavaty
Class Format Default Network
Prefix
High Order Bits
Usable Network Address range
A N.H.H.H
Hosts:16.7 m
Mask:255.0.0.0
8 bits 0 1.0.0.0 – 126.0.0.0
127.0.0.0 - Loopback
B N.N.H.H
Hosts:65,534
Mask:255.255.0.0
16 bits 10 128.0.0.0 –
191.255.0.0
C N.N.N.H
Hosts:254
255.255.255.0
24 bits 110 192.0.0.0 –
223.255.255.0
# 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 addressHost Id = all 0’s is not permitted as this refers to “this network”
Instructor: Sam Nanavaty
ExamplesIdentify 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
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?
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)
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)
Instructor: Sam Nanavaty
Subnet mask determination
IP address Reqd Hosts Reqd Subnets
192.168.1.0 60 2
172.16.0.0 As many as possible
16
Instructor: Sam Nanavaty
IP address Reqd Hosts Reqd Subnets
Mask
192.168.1.0 60 2 255.255.255.192
192 = 1100 0000
Yields 4 -2 = 2 subnets
172.16.0.0 As 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
Instructor: Sam Nanavaty
Calculating the range of addresses given a subnet mask
IP Network address : 192.168.1.0Subnet 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)
Remove first and last subnets as well as first and last IP addresses for each subnetThe final valid range of addresses are as follows:
192.168.1.9 – 192.168.1.14192.168.1.17 – 192.168.1.22
192.168.1.241 – 192.168.1.246
(11111000 – 11111111)
Instructor: Sam Nanavaty
Practice example
IP address : 192.168.1.0Subnet 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 maskNext determine the valid IP addresses in this range
Instructor: Sam Nanavaty
IP Network address : 192.168.1.0Subnet 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
Address MaskA = (D & M)
Where:
A = 32 bit IP address
M = 32 bit address mask
D = Destination address
Instructor: Sam Nanavaty
CIDR Notation128.10.0.0/16
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
172.16.210.0 /22 Which subnet does this address belong to?
Instructor: Sam Nanavaty
• 201.100.5.68 /28
Find the subnet this address belongs to.
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
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
Instructor: Sam Nanavaty
Instructor: Sam Nanavaty
Instructor: Sam Nanavaty
Instructor: Sam Nanavaty
Instructor: Sam Nanavaty
Instructor: Sam Nanavaty
Instructor: Sam Nanavaty
Instructor: Sam Nanavaty
Instructor: Sam Nanavaty