某公司网络地址是 206.110.64.0/18，被划分成 16 个子网，则每个子网的子网掩码为（），最大可容纳的主机数是（）。
- A.254
- B.512
- C.1022
- D.1024
正确答案及解析
正确答案
解析
本题考查计算机子网的划分，已知该公司网络地址是 206.110.64.0/18（网络位占前18位，主机位占后32-18=14位），要划分成 16 个子网，就需扩展网络位4位（2^4=16），因此网络位由原来的18位扩展到18+4=22位，而主机位由原来的14位缩减到14-4=10位，根据子网掩码网络位全1且主机位全0的定义，划分后每个子网的二进制掩码形式为：
11111111.11111111.111111 00.00000000，十进制表示为：255.255.252.0。
最大可容纳的主机数是（2^10）-2=1022。 这里减去2，即除去每个子网内两个特殊的IP地址，子网网络地址（主机位全1）和子网广播地址（主机位全0）。这两个地址不参与具体主机的分配，因此必须排除。
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