若要将一个B类的网络172.18.0.0划分为14个子网络。请计算出子网掩码并求出每个子网中最多允
有3个网友回答
教你笨一点的方法。首先你要判断出，借用的网络位是几位。14个子网，2的4次方等于16，就是说要借用4个网络位。这时候求子网掩码，就是原来的B类16位加上4位等于20位。你在草稿上写下，20个1和12个0，然后把二进制转换成点分十进制，就是你要的子网掩码。最大主机数，直接用公式，用掉20个网络，还剩12个主机位，公式就是2的12次方-2..为什么要减2不需要解释了吧。
您需要划分的IP网段为：172.18.0.0
您需要划分的IP子网的子网掩码为：255.255.240.0
您需要划分的子网个数是：14个。
每段子网的网络号是TO前IP的最后一位减1
每段子网的广播地址是TO后IP最后一位加1
1）172.18.0.1 To 172.18.15.254
2）172.18.16.1 To 172.18.31.254
3）172.18.32.1 To 172.18.47.254
4）172.18.48.1 To 172.18.63.254
5）172.18.64.1 To 172.18.79.254
6）172.18.80.1 To 172.18.95.254
7）172.18.96.1 To 172.18.111.254
8）172.18.112.1 To 172.18.127.254
9）172.18.128.1 To 172.18.143.254
10）172.18.144.1 To 172.18.159.254
11）172.18.160.1 To 172.18.175.254
12）172.18.176.1 To 172.18.191.254
13）172.18.192.1 To 172.18.207.254
14）172.18.208.1 To 172.18.223.254
15）172.18.224.1 To 172.18.239.254
16）172.18.240.1 To 172.18.255.254
子网掩码20位，即255.255.240.0，每个子网最多可以容纳3824台主机。
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