Example
1.
Write the mask or classful
prefix length in binary.
2.
Draw a line to delineate the
significant bits in the assigned
IP address. Cross out the mask
so you can view the significant
bits in the IP address.
3.
IP Address: 192.168.221.37 Subnet Mask /29
4th Octet: 00100101
Assigned Mask: 255.255.255.248 (/29)
4th Octet: 11111000
Split Octet (Binary): 00100 101
Split Mask (Binary): 11111 000
Constructing a Network Addressing Scheme 265
Figure 4-21 Determining the Addressing Scheme, Steps 4??“8
After converting the addresses from binary to decimal, the addresses for the subnets are as
follows:
?– Subnet address: 192.168.221.32
?– First host address: 192.168.221.33
?– Last host address: 192.168.221.38
?– Broadcast address: 192.168.221.39
?– Next subnet address: 192.168.221.40
Notice that the range of the address block, including the subnet address and directed
broadcast address in this example, is from 192.168.221.32 through 192.168.221.39,
which includes eight addresses. The address block is the same size as the number of host
bits (2h = 23 = 8).
Class C Example
In Figure 4-22, we will determine the addressing for a Class C network with a nondefault
mask. Given the address of 192.
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