SUBNETTING

 SUBNETTING

Class A Address	N	H	H	H
Class B Address	N	N	H	H
Class C Address	N	N	N	H

 Formula

2^N Where N is equal to number of bits borrowed ( Number to total subnets created).
2^N-2 Number of valid subnets created.
2^H Where H is equal to number of host bits (Number of total hosts per subnet).
2^H-2 Number of valid hosts per subnet.

27	26	25	24	23	22	21	20
128	64	32	16	8	4	2	1

 

To convert a decimal number into binary

	128	64	32	16	8	4	2	1
192=	128	64	0	0	0	0	0	0
192=	1	1	0	0	0	0	0	0

192 = 11000000

Subnetting a Class C Network

You have a class C address of 192.168.10.0/24. You need 4 subnets. What is the valid hosts, broadcast number, subnet mask ?
Step 1- Determine how many H bit you need to borrow to create 4 valid subnet.
2^N -2 ≥ 4
2³-2 ≥ 4 (suppose N =3)
8 -2 ≥ 4
6 ≥ 4 (condition satisfied)
N = 3, so you need to borrow 3 H bits and turn them into N bits.
 Number of valid Subnet = 2³-2 =8 - 2 = 6
Total host bits = 5
Valid Hosts per subnet = 2⁵-2 =32 - 2 = 30

H Bits	H	H	H	H	H	H	H	H
Borrow 3 Bits	N	N	N	H	H	H	H	H

Step 2- Determine the first valid subnet in binary.

001HHHHH	We can not use 000 subnet because it is invalid, so we start with pattern of 001
00100000	All 0s in host portion=subnetwork number
00100001	All 0s in host portion=subnetwork number

 1’st subnetwork = 00100000 =32
1’st valid host = 00100001 =33
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Last valid host = 00111110=62
Broadcast address = 00111111 = 63

Step 3- Determine the second valid subnet in binary.

010HHHHH	010 = second valid subnet
01000000	All 0s in host portion=subnetwork number

 2’nd subnetwork = 01000000 =64
1’st valid host = 01000001 =65
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Last valid host = 01011110=94
Broadcast address = 01011111 = 95

Step 4- Determine the third valid subnet in binary.

011HHHHH	011 = third valid subnet
01100000	All 0s in host portion=subnetwork number

3’rd subnetwork = 01100000 = 96
1’st valid host = 01100001 = 97
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Last valid host = 01111110=126
Broadcast address = 01111111 = 127

Step 5- Determine the third valid subnet in binary.

100HHHHH	100 = forth valid subnet
10000000	All 0s in host portion=subnetwork number

 4’th subnetwork = 10000000 = 128
1’st valid host = 100000001 = 129
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Last valid host = 10011110=158
Broadcast address = 10011111 = 159

Step 6- Determine the fifth valid subnet in binary.

101HHHHH	110 = fifth valid subnet
10100000	All 0s in host portion=subnetwork number

 5’th subnetwork = 10100000 = 160
1’st valid host = 101000001 = 161
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Last valid host = 10111110=190
Broadcast address = 10111111 = 191

Step 7- Determine the sixth valid subnet in binary.

110HHHHH	110 = sixth valid subnet
11000000	All 0s in host portion=subnetwork number

6’th subnetwork = 11000000 = 192
1’st valid host = 110000001 = 193
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Last valid host = 11011110=222
Broadcast address = 11011111 = 223

IP Table

Subnet	Network Address	Valid Hosts Range	Broadcast Address
000 (Invalid)	192.168.10.0	192.16810.1-30	192.168.10.31
001	192.168.10.32	192.16810.33-62	192.168.10.63
010	192.168.10.64	192.16810.65-94	192.168.10.95
011	192.168.10.96	192.16810.95-126	192.168.10.127
100	192.168.10.128	192.16810.129-58	192.168.10.159
101	192.168.10.160	192.16810.161-190	192.168.10.191
110	192.168.10.192	192.16810.193-222	192.168.10.223
111 (Invalid)	192.168.10.224	192.16810.1-30	192.168.10.255