Valid Addresses for Server Farm Segment in Figure 2.2

TABLE 2 . 8 Valid Addresses for Ethernet Segment in Figure 2.2
3rd Octet
128 64 32 16 8 4 2 1
4th Octet
128 64 32 16 8 4 2 1
Decimal IP Address
(Last 16 bits in bold)
Subnet mask 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 255.255.255.0
Subnet 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 172.16.1.0
First IP in range 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 172.16.1.1
Last IP in range 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 172.16.1.254
Broadcast 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 172.16.1.255
TABLE 2 . 9 Valid Addresses for Server Farm Segment in Figure 2.2
3rd Octet
128 64 32 16 8 4 2 1
4th Octet
128 64 32 16 8 4 2 1
Decimal IP Address
(Last 16 bits in bold)
Subnet mask 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 255.255.254.0
Subnet 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 172.16.2.0
First IP in range 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 172.16.2.1
Last IP in range 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 172.16.3.254
Broadcast 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 172.16.3.255
52 Chapter 2  IP Addressing
of 1 when 255 is subtracted from 256, which is how default subnets work. Because the previous two
subnets have a third-octet value of 0, we must start in the third octet with at least a value of 1, and
the subnet gets the entire range of addresses with 1 in the third octet.
In addition, notice that the IP address range for the server farm segment is 172.16.2.1–
172.16.3.254. This is because the interesting octet is now the third octet, with a value of 254. Subtracting
this from 256 leaves us with an increment of 2. Remembering to apply this to only the third
octet, we determine we can have values of 0, 2, 4, 6, and so on, in the third octet, but we already have
subnets with 0 and 1 in that octet, so 2 is the first available value. Because the increment is 2, we get
the entire range up to, but not including, the next subnet boundary, which is 4 in the third octet.
In summary, we have