Binary Lesson 6.1 Classful Subnetting 2

1
Binary Lesson 6.1Classful Subnetting 2
2
Slash Notation

• Consider a class A address of 1.2.3.4
• The network portion is the first octet
• 1
• This situation can be written
• 1.2.3.4/8
• The number after the / is the number of bits in
  the network portion

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Subnet Mask

• Consider a class A address of 1.2.3.4 /8 in
  binary
• The subnet mask is 255.0.0.0
• The network address is 1.0.0.0
• In Binary
• IP 00000001 00000010 00000011 00000100
• Subnt Msk 11111111 00000000 00000000 00000000
• Net Addr 00000001 00000000 00000000 00000000

4
AND

• To find the network address, take a node's IP
  address and do this
• If the Subnet mask bit is 1, keep the IP address
  bit
• If the Subnet mask bit is 0, that bit changes to
  0
• This is called a bitwise AND operation
• IP 00000001 00000010 00000011 00000100
• Subnt Msk 11111111 00000000 00000000 00000000
• Net Addr 00000001 00000000 00000000 00000000

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Class B

• Class B address 147.144.1.212 /16
• The subnet mask is 255.255.0.0
• The network address is 147.144.0.0
• In Binary
• IP 10010011 10010000 00000001 11010100
• Subnt Msk 11111111 11111111 00000000 00000000
• Net Addr 10010011 10010000 00000000 00000000

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Class C

• Class C address 192.168.1.10 /24
• The subnet mask is 255.255.255.0
• The network address is 192.168.1.0
• In Binary
• IP 11000000 10101000 00000001 00001010
• Subnt Msk 11111111 11111111 11111111 00000000
• Net Addr 11000000 10101000 00000001 00000000

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192.168.1.0 /25

• You can split a class C in half with a /25
• IP 192.168.1.10 /25
• Subnet Mask 255.255.255.128
• Network 192.168.1.0
• IP 11000000 10101000 00000001 00001010
• Subnt Msk 11111111 11111111 11111111 10000000
• Net Addr 11000000 10101000 00000001 00000000

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192.168.1.0 /25

• There are only 128 addresses in a /25 block
• Network 192.168.1.0 /25
• 1st Usable 192.168.1.1
• Last Usable 192.168.1.126
• Broadcast 192.168.1.127
• Net Addr 11000000 10101000 00000001 00000000
• First Use 11000000 10101000 00000001 00000001
• Last Use 11000000 10101000 00000001 01111110
• Broadcast 11000000 10101000 00000001 01111111

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192.168.1.128 /25

• There are only 128 addresses in a /25 block
• Network 192.168.1.128 /25
• 1st Usable 192.168.1.129
• Last Usable 192.168.1.254
• Broadcast 192.168.1.255
• Net Addr 11000000 10101000 00000001 10000000
• First Use 11000000 10101000 00000001 10000001
• Last Use 11000000 10101000 00000001 11111110
• Broadcast 11000000 10101000 00000001 11111111

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192.168.1.0 /26

• There are only 64 addresses in a /26 block
• Network 192.168.1.0 /26
• 1st Usable 192.168.1.1
• Last Usable 192.168.1.62
• Broadcast 192.168.1.63
• Net Addr 11000000 10101000 00000001 00000000
• First Use 11000000 10101000 00000001 00000001
• Last Use 11000000 10101000 00000001 00111110
• Broadcast 11000000 10101000 00000001 00111111

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192.168.1.64 /26

• There are only 64 addresses in a /26 block
• Network 192.168.1.64 /26
• 1st Usable 192.168.1.65
• Last Usable 192.168.1.126
• Broadcast 192.168.1.127
• Net Addr 11000000 10101000 00000001 01000000
• First Use 11000000 10101000 00000001 01000001
• Last Use 11000000 10101000 00000001 01111110
• Broadcast 11000000 10101000 00000001 01111111

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192.168.1.128 /26

• There are only 64 addresses in a /26 block
• Network 192.168.1.128 /26
• 1st Usable 192.168.1.129
• Last Usable 192.168.1.190
• Broadcast 192.168.1.191
• Net Addr 11000000 10101000 00000001 10000000
• First Use 11000000 10101000 00000001 10000001
• Last Use 11000000 10101000 00000001 10111110
• Broadcast 11000000 10101000 00000001 10111111

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192.168.1.192 /26

• There are only 64 addresses in a /26 block
• Network 192.168.1.192 /26
• 1st Usable 192.168.1.193
• Last Usable 192.168.1.254
• Broadcast 192.168.1.255
• Net Addr 11000000 10101000 00000001 11000000
• First Use 11000000 10101000 00000001 11000001
• Last Use 11000000 10101000 00000001 11111110
• Broadcast 11000000 10101000 00000001 11111111

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Classful Subnetting iClicker Questions
15
How many subnets containing 9 hosts each can be
made from this network?192.168.1.0/24

1. 4
2. 8
3. 16
4. 32
5. Something else

16
How many subnets containing 90 hosts each can be
made from this network?192.168.1.0/24

1. 4
2. 8
3. 16
4. 32
5. Something else

17
How many subnets containing 9 hosts each can be
made from this network?147.144.1.128/26

1. 4
2. 8
3. 16
4. 32
5. Something else

18
How many subnets containing 2 hosts each can be
made from this network?192.168.1.0/27

1. 4
2. 8
3. 16
4. 32
5. Something else

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How many subnets containing 900 hosts each can be
made from this network?10.0.0.0/16

1. 8
2. 16
3. 32
4. 64
5. Something else

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What subnet contains this host?192.168.1.129/30

1. 192.168.1.126 /30
2. 192.168.1.128 /30
3. 192.168.1.126 /29
4. 192.168.1.130 /30
5. Something else

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What is the smallest subnet containing these two
hosts?192.168.1.1192.168.1.10

1. 192.168.1.0 /24
2. 192.168.1.0 /26
3. 192.168.1.0 /28
4. 192.168.1.0 /30
5. Something else

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What is the smallest subnet containing these two
hosts?10.1.1.211.1.1.1

1. 10.0.0.0 /8
2. 10.0.0.0 /7
3. 10.0.0.0 /6
4. 8.0.0.0 /6
5. Something else

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What is the smallest subnet containing these two
hosts?147.144.128.129147.144.128.193

1. 147.144.128.0 /24
2. 147.144.128.0 /25
3. 147.144.128.128 /25
4. 147.144.128.128 /26
5. Something else

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What is the smallest subnet containing these two
hosts?192.168.1.62192.168.1.65

1. 192.168.1.0 /24
2. 192.168.1.32 /25
3. 192.168.1.0 /25
4. 192.168.1.32 /26
5. Something else