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LECTURE # 20

IP ADDRESSING

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Binary

2

 All digital electronics use a binary

method for communication.

 Binary can be expressed using only two

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Converting Binary to

Decimal

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 First, moving from right to left, create a

chart that starts at the decimal number 1 and then double it 7 times.

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Converting Binary to

Decimal

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 Given a binary number, place the number

under the chart (right justified).

128 64 32 16 8 4 2 1

1 0 1 1 0 0 1 1

•Add the numbers together to arrive at a final decimal amount.

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Converting Decimal to

Binary

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 Find the largest number that is equal to

or less than the number you are

converting to binary. If our example

number is 220, the largest number that is equal to or less than 220 is 128. Place a 1 under that space on the chart.

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Converting Decimal to

Binary

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 Next, subtract that number from the

original decimal number. Subtracting 128 from 220 gives us 92.

 Repeat this process until we have a

subtracted result of 0.

128 64 32 16 8 4 2 1

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Counting in Binary

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 0 + 1 = 1

 1 + 1 = 10 (carry the 1)

 10 + 1 = 11

 11 + 1 = 100

 100 + 1 = 101

 101 + 1 = 110

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Binary Counting Chart

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1 1 11 1011

2 10 12 1100

3 11 13 1101

4 100 14 1110

5 101 15 1111

6 110 16 10000

7 111 17 10001

8 1000 18 10010

9 1001 19 10011

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Addressing Principles

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 Logical addresses in TCP/IP are 32 bits in

length.

 To make addressing more humanly

manageable, the 32 bits are broken into four 8 bit octets.

 We separate the octets by using a period

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Dotted-decimal notation

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TCP/IP Host

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 A host is a device that has a network

interface card connected to a network.

 If a device has two network interfaces, it

should be considered two separate hosts.

 Each host that is attached to a TCP/IP

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TCP/IP Addresses

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 IP Addresses divided into two parts

(network address and host address) :

 Network ID or Net ID

 Analogous to a street address.

 Host ID

 Analogous to a house or building number.

 TCP/IP Addresses are hierarchical addresses

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Example 1

Example 1

Change the following IP addresses from binary notation to dotted-decimal notation.

a. 10000001 00001011 00001011 11101111

b. 11111001 10011011 11111011 00001111

Solution

Solution

We replace each group of 8 bits with its equivalent decimal number and add dots for separation:

a. 129.11.11.239

b. 249.155.251.15

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Example 2

Example 2

Change the following IP addresses from dotted-decimal notation to binary notation.

a. 111.56.45.78

b. 75.45.34.78

Solution

Solution

We replace each decimal number with its binary equivalent

a. 01101111 00111000 00101101 01001110

b. 01001011 00101101 00100010 01001110

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In classful addressing, the address

space is divided into five classes: A, B,

C, D, and E.

Note

Note

:

:

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Finding the class in binary notation

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Finding the address class

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Example 3

Example 3

Find the class of each address:

a. 000000001 00001011 00001011 11101111

b. 111111110011 10011011 11111011 00001111

Solution

Solution

a. The first bit is 0; this is a class A address.

b. The first 4 bits are 1s; this is a class E address.

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Finding the Class in decimal Notation

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Example 4

Example 4

Find the class of each address: a. 227.12.14.87

b. 252.5.15.111 c. 134.11.78.56

Solution

Solution

a.The first byte is 227 (b/w 224 and 239); the class is D. b.The first byte is 252 (b/w 240 and 255)the class is E. c.The first byte is 134 (b/w 128 and 191); the class is B.

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Example

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 A 00001010 00000010 00000000 00100101  B 10101100 00010000 00001010 00000001  C 10000001 00110100 00000110 00000001

Solution

 A 10.2.0.37

 B 172.16.10.1

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Netid and Hostid

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Class A Address

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 First bit will always be a 0.

 Remaining bits can be either 0s or 1s.

 Range of first octet is 00000001 to

01111111

 Network addresses cannot be all 0s.  127 is reserved for loopback testing.

 126 valid Class A network addresses

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Host ID Portion

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 Cannot be all 0s

 If host portion is all 0s, represents a

network address.

 Cannot be all 1s

 If host portion is all 1s, represents a

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Blocks in class A

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Class B Address

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 First two bits will always be a 10.

 Remaining bits can be either 0s or 1s.

 Range of first octet is 10000000 to

10111111

 Range of networks 128.0.y.z to

191.255.y.z

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Blocks in class B

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Class C Address

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 First three bits will always be a 110.

 Remaining bits can be either 0s or 1s.

 Range of class C addresses is

11000000.x.y.z to 11011111.x.y.z 192.0.0.z to 223.255.255.z.

 2,097,152 valid Class C network

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Blocks in class C

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Class D Address

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 First octet in binary is defined as

1110xxxx, replacing x’s with whatever we wish.

 Range of Class D addresses is from

224.x.y.z to 239.x.y.z.

 Used for multicasting – method of

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Class E Address

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 First octet is 1111xxxx, replacing x’s

with whatever we wish.

 Address ranges from 240.x.y.z to

255.x.y.z.

 Experimental address range that is not

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A network address is different from a

netid. A network address has both

netid and hostid,

with 0s for the hostid.

Note

Note

:

:

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Network address

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Example 5

Example 5

Given the address 23.56.7.91, find the network address.

Solution

Solution

The class is A. Only the first byte defines the netid. We can find the network address by replacing the hostid bytes (56.7.91) with 0s. Therefore, the network address is 23.0.0.0.

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Example 6

Example 6

Given the address 132.6.17.85, find the network address.

Solution

Solution

The class is B. The first 2 bytes defines the netid. We can find the network address by replacing the hostid bytes (17.85) with 0s.

Therefore, the network address is 132.6.0.0.

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Example 7

Example 7

Given the network address 17.0.0.0, find the class.

Solution

Solution

The class is A because the netid is only 1 byte.

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Addressing Guidelines

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 Network ID Cannot Be 127

 127 is reserved for lookback functions

 Network ID and Host ID Cannot Be 255 (All Bits Set to 1)

 255 is a broadcast address

 Network ID and Host ID Cannot Be 0 (All Bits Set to 0)

 0 means “this network only”

 Example: 145.20.0.0 refers to Class B network 145.20.0.0

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Assigning Network IDs

39

1

1 22 33

124.x.y.z

124.x.y.z 192.121.73.192.121.73.zz 131.107.131.107.y.zy.z

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Assigning Host IDs

40 Router Router 124.0.0.27 124.0.0.27 124.0.0.28 124.0.0.28 124.0.0.29 124.0.0.29 131.107.0.27 131.107.0.27 131.107.0.28 131.107.0.28 131.107.0.29 131.107.0.29 192.121.73. 1 192.121.73. 1 124.0.0.1

124.0.0.1 192.121.73. 2

192.121.73.

2

131.107.0.1 131.107.0.1

124.x.y.z

124.x.y.z 192.121.73.192.121.73.zz 131.107.0.131.107.0.zz

1

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Address Class Summary

41 [[ Number Number of Networks of Networks Number Number of Networks of Networks 126 126 16,384 16,384 2,097,152 2,097,152

Number of Hosts Number of Hosts

per Network per Network Number of Hosts

Number of Hosts

per Network per Network 16,777,214 16,777,214 65,534 65,534 254 254 Class A Class A Class A Class A Class B Class B Class B Class B Class C Class C Class C Class C Range of Range of Network IDs Network IDs (First Octet) (First Octet) Range of Range of Network IDs Network IDs (First Octet) (First Octet)

1 – 126

1 – 126

128 – 191

128 – 191

192 – 223

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Reserved IP Addresses

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 Defined in RFC 1918.

 Class A

 10.0.0.0 to 10.255.255.255

 Class B

 172.16.0.0 to 172.31.255.255

 Class C

 192.168.0.0 to 192.168.255.255

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Reserved IP Addresses

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 169.254.0.0 to 169.254.255.255

 used for operating systems that

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 Europe :

194.0.0.0 – 195.255.255.255

 North America:

198.0.0.0 – 199.255.255.255

 Central & South America:

200.0.0.0 – 201.255.255.255

 Asia & Pacific:

202.0.0.0 – 203.255.255.255

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