7.1 LAN IP address
7-2
Reference)
For example, if the Subnet mask is 255.255.255.0/24, and is represented in binary
numbers as 11111111 11111111 11111111 00000000, the first 24 bits within the IP
address will be the network address, the last 8 bits will be the host address, and “/24”
will be the network address.
This means that the own IP address is 192.168.1.1, and therefore the IP addresses that
can be allocated will be those within the range of 192.168.1.2 to 192.168.1.254.
x IP address 192.168.1.255;
This is the broadcasting address, and is used for when calling all terminals
within the network.
x IP address 192.168.1.0;
The network address for distinguishing individual networks
Therefore, the range of IP addresses available will be; 192.168.1.2 to 192.168.1.254.
In the case of 210 .89.77 .200 /29, 210 .89.77 .200 will be the network address, /29:
Prefix will be the number of bits for the network. Let's identify the IP addresses that
can be allocated for the host.
The Subnet mask is;
11111111 . 11111111 . 11111111 . 11111000 -> 2 55.255.255.248
255 255 255 248
Since the host address will be represented by the last 3 bits, there are 7 addresses that
can be allocated.
Therefore, the range of IP addresses available is; 210.89.77.200 to 207.
210.89.77.200 is the network address used for identifying the network, and
210.89.77.207 is the broadcasting address, and therefore the addresses available to be
allocated are those within the range of; 210.89.77.201 to 210.89.77.206.
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