How to Identify Subnet & Host bits

1. Subnetting Basics ? ! ?
NO !
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2. Subnetting Basics Made Easy
By:
Michael D. Mann
© Copyright 2011 by : Michael D. Mann 2

3. Subnetting Basics Made Easy
Part 1: Prerequisites
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4. Subnetting Basics
Prerequisite Knowledge
Class 1st Octet Valid Network Total Number of
Range Numbers Networks hosts per
for this network
class
1 -126 1.0.0.0 to 126 224 – 2
A
126.0.0.0 16,777,214
N.H.H.H
128 – 191 128.0.0.0 to 16,384 216 – 2
B
191.255.0.0 65,534
N.N.H.H
192 – 223 192.0.0.0 - 2,097,152 28 – 2
C
223.255.255.0 254
N.N.N.H
N = network octet H = host octet 4

5. Subnetting Basics
Prerequisite Knowledge
Definitions
Network ID : all host bits are set to 0
192.168.32.0000 0000
Ex : 192.168.32.0 /24
Broadcast ID : all host bits are set to 1
192.168.32.1111 1111
Ex : 192.168.32.255 /24
IP Address : any value in between
192.168.32.0000 0001 – 1111 1110
Ex : 192.168.32.1 - .254 /24
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6. Subnetting Basics
Prerequisite Knowledge
Prefix Length Notation
There are 2 ways to express a subnet mask:
Dotted Decimal Notation: Prefix Length Notation:
Class C mask: 255.255.255.0 Class C mask: /24
255.255.255.0 = 1111 1111. 1111 1111. 1111 1111. 0000 0000
Notice that there are 24 blue (network) bits. The PL number (/24)
indicates how many bits are being used to represent the network ID.
Notice how the host bits are always 0 in a subnet mask
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7. Subnetting Basics
Prerequisite Knowledge
The default Class A, B, & C subnet masks
Class A default subnet mask : 255.0.0.0 ( /8 )
Binary: 1111 1111.0000 0000.0000 0000.0000 0000
Class B default subnet mask : 255.255.0.0 ( /16 )
Binary: 1111 1111.1111 1111.0000 0000.0000 0000
Class C default subnet mask : 255.255.255.0 ( /24 )
Binary: 1111 1111.1111 1111.1111 1111.0000 0000
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8. Subnetting Basics
Prerequisite Knowledge
Binary Number System
1 BYTE = 8 BITS : 1010 0001
Each bit position has an associated PLACE VALUE as indicated :
27 26 25 24 23 22 21 20
1 0 1 0 0 0 0 1
128 + 32 + 1 = 16110
Add up the place values wherever you have a “1” bit
This is how you convert from binary to decimal 8

9. Subnetting Basics Made Easy
Part 2: Identifying Subnet and
Host Bits
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10. Subnetting Basics
Identifying Subnet Bits in a Class C IP Address
Example (unsubnetted or classful): 192.168.32.158 /24
Example (subnetted or classless): 192.168.32.158 /28
Step 1: Notice that the prefix length (PL) indicator of the
subnetted example is /28. This number is greater than the
default PL indicator of /24. This tells you that the IP address is
subnetted.
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11. Subnetting Basics
Identifying Subnet Bits in a Class C IP Address
Example: 192.168.32. 158 /28
15810 = 1001 11102
Step 2: Examine the host octet by converting it to a binary
number.
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12. Subnetting Basics
Identifying Subnet Bits in a Class C IP Address
Example: 192.168.32.158 /28 15810 = 1001 11102
Step 3a Subnet Bits Rule: Subnet bits are “assigned” starting
with the high order bits in the host octet.
Step 3b: Subtract the default class C PL indicator from the
current PL indicator to get the number of bits to “assign”.
/28 - /24 = 4 bits must be “borrowed” for use as subnet bits.
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13. Subnetting Basics
Identifying Subnet Bits in a Class C IP Address
Example: 192.168.32.158 /28 15810 = 1001 11102
Step 4: 4 Subnet bits are “borrowed” starting with the high
order bits in the host octet. Color code these bits in blue using
an italicized font and color code the remaining host bits in red:
15810 = 1001 11102
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14. Subnetting Basics
Part 3: Calculating The
Number of Subnets
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15. Subnetting Basics
Calculating The Number of Subnets
Example: 192.168.32.158 /28
15810 = 1001 11102
Procedure: Use the equation 2X where X = the number of
subnet bits. There are four subnet (i.e. blue, italicized ) bits.
Therefore: 24 = 16, so there are 16 subnets available.
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16. Subnetting Basics
Calculating The Number of Subnets
Example: 192.168.32.158 /28 15810 = 1001 11102
The 16 subnets are indicated by all possible combinations of the four subnet
bits. Note that the subnet ID numbers are shown in parentheses next to the
binary.
0000 (ID 0) 1000 (ID 128)
0001 (ID 16) 1001 (ID 144)
0010 (ID 32) 1010 (ID 160)
0011 (ID 48) 1011 (ID 176)
0100 (ID 64) 1100 (ID 192)
0101 (ID 80) 1101 (ID 208)
0110 (ID 96) 1110 (ID 224)
0111 (ID 112) 1111 (ID 240)
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17. Subnetting Basics
Calculating The Number of Subnets
Example: 192.168.32.158 /28 15810 = 1001 11102
The subnet ID numbers shown in parentheses in the previous
slide are the result of binary to decimal conversion using the
method as demonstrated in slide 8. The binary place values are
repeated below for your convenience:
27 = 128 26 = 64 25 = 32 24 = 16 23 = 8 22 = 4 21 = 2 20 = 1
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18. Subnetting Basics
Part 4: Calculating The
Number of Hosts
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19. Subnetting Basics
Calculating The Number of Hosts
Example: 192.168.32.158 /28
15810 = 1001 11102
Procedure: Use the equation 2X - 2 where X = the number of
host bits. There are four host (i.e. red) bits.
Therefore: 24 – 2 = 14, so there are 14 hosts available.
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20. Subnetting Basics
Calculating The Number of Hosts
Example: 192.168.32.158 /28 15810 = 1001 11102
The host ID numbers are given by all possible combinations of
the four host bits:
0000 (ID 0) 1000 (ID 8)
0001 (ID 1) 1001 (ID 9)
0010 (ID 2) 1010 (ID 10)
0011 (ID 3) 1011 (ID 11)
0100 (ID 4) 1100 (ID 12)
0101 (ID 5) 1101 (ID 13)
0110 (ID 6) 1110 (ID 14)
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0111 (ID 7) 1111 (ID 15)

21. Subnetting Basics
Calculating The Number of Hosts
Example: 192.168.32.158 /28 15810 = 1001 11102
Why are the first and last hosts grayed out?
Hint: See slide 5.
0000 (ID 0) 1000 (ID 8)
0001 (ID 1) 1001 (ID 9)
0010 (ID 2) 1010 (ID 10)
0011 (ID 3) 1011 (ID 11)
0100 (ID 4) 1100 (ID 12)
0101 (ID 5) 1101 (ID 13)
0110 (ID 6) 1110 (ID 14)
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0111 (ID 7) 1111 (ID 15)

22. Subnetting Basics
Putting It All Together
Example: 192.168.32.158 /28
15810 = 1001 11102
144 14
The host octet in this class C private IP address
identifies subnet ID #144, which contains host #14.
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23. Subnetting Basics
Be sure to watch the complete Subnetting Basics
presentation found in the Course Documents
section of your Bb course shell.
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