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48 IP Subnetting
The first three octets of the address make up the network number (class “C”). You want to have two separate
networks.
Divide the network 192.168.1.0 into two separate subnets by converting one of the host ID bits of the IP
address to a network number bit. The “borrowed” host ID bit can be either “0” or “1” thus giving two
subnets; 192.168.1.0 with mask 255.255.255.128 and 192.168.1.128 with mask 255.255.255.128.
In the following charts, shaded/bolded last octet bit values indicate host ID bits
“borrowed” to form network ID bits. The number of “borrowed” host ID bits
determines the number of subnets you can have. The remaining number of host ID
bits (after “borrowing”) determines the number of hosts you can have on each
subnet.
Chart K-5 Subnet 1
NETWORK NUMBER LAST OCTET BIT VALUE
IP Address 192.168.1. 0
IP Address (Binary) 11000000.10101000.00000001.
00000000
Subnet Mask 255.255.255. 128
Subnet Mask (Binary) 11111111.11111111.11111111.
10000000
Subnet Address: 192.168.1.0 Lowest Host ID: 192.168.1.1
Broadcast Address: 192.168.1.127 Highest Host ID: 192.168.1.126
Chart K-6 Subnet 2
NETWORK NUMBER LAST OCTET BIT VALUE
IP Address 192.168.1. 128
IP Address (Binary) 11000000.10101000.00000001.
10000000
Subnet Mask 255.255.255. 128
Subnet Mask (Binary) 11111111.11111111.11111111.
10000000
Subnet Address: 192.168.1.128 Lowest Host ID: 192.168.1.129
Broadcast Address: 192.168.1.255 Highest Host ID: 192.168.1.254
The remaining 7 bits determine the number of hosts each subnet can have. Host IDs of all zeros represent the
subnet itself and host IDs of all ones are the broadcast address for that subnet, so the actual number of hosts
available on each subnet in the example above is 2
7
– 2 or 126 hosts for each subnet.