instructor: sam nanavaty dns and ip addressing. instructor: sam nanavaty how does a router know...

Instructor: Sam Nanavaty

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.

• D used for multicasting groups

• E is reserved for experimental purposes


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”


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


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?


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)


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)


Subnet mask determination

IP address Reqd Hosts Reqd Subnets

192.168.1.0 60 2

172.16.0.0 As many as possible

16


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


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)


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


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)


Address MaskA = (D & M)

Where:

A = 32 bit IP address

M = 32 bit address mask

D = Destination address


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


172.16.210.0 /22 Which subnet does this address belong to?


• 201.100.5.68 /28

Find the subnet this address belongs to.


• 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


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 dns and ip addressing. instructor: sam nanavaty how does a router know...

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