推荐回答(5个)
首先确定192.168.10.17-192.168.10.31共有15台主机数
由于可变长子网掩码划分的子网都是2的n次方(n取值为0-7,因为ip地址是由32个8进制组成的,又由3个点分划为4段)
比如c类地址,掩码为255.255.255.0划为2进制为11111111.11111111.11111111.00000000
1表示网段数,0表示主机数
由于你要划分的主机数为15,最接近2的n次方的为4
那么子网掩码则为11111111.11111111.11111111.11111111.11100000
为什么是5个0,是因为2的0次方也算1个
根据2进制换10进制计算为255.255.255.224
所以子网掩码则为255.255.255.224
但是由于ip地址不能全为0,也不能全为1.全为0表示网段,全为1表示广播地址(2进制)
192.168.10.31为全为1,那么我们必须在在扩大一倍主机数
由于0表示主机数,那么最后一段则改为11000000
就有了62个主机数(不能全0,也不能全1)
所以最后的可变长子网掩码为255.255.255.192
ip地址可用的范围为192.168.10.1-192.168.10.62,共有62个地址可以用(不能全0,也不能全1)
明白了不?很简单的.当然我楼上的也可行.
呵呵,这个不难!~
IP地址是32位的二进制地址,分为3个字段(类别字段,网络号字段,主机号字段),一般不考虑类别字段...
IP地址划分子网,意思是将自己的主机号字段中的前若干个比特位分出来作为子网号。
网络掩码是一串连续的1和一串连续的0组成的32位地址。
这是一些首先要清楚的基本的东西...
本题中的202.119.230.0是C类网络,掩码为255.255.255.0,转换为二进制:11111111.11111111.11111111.00000000(其中前三组1为类别号与网络号字段,后面一组0是主机号)
现在要为其划分子网,则需要在其主机号字段内借用一部分0作为子网号字段,到底借几位呢?题目要求划分8个子网,则需要借3位。
【因为2的3次方 = 8 】
再看主机号,这时已经只剩5位了(8-3=5),5位的主机号可以提供30个主机IP。
【2的5次方 - 2 = 30,减2是因为主机号为全0代表所在的子网网络号,不可用;主机号全为1代表所在子网网络的广播地址,也不可用】
所以,题目的解答就出来了!
子网掩码为:255.255.255.224,即为11111111.11111111.11111111.11100000,其中前三组1是类别号和网络号字段,后一组中的前3个1是子网号字段,后面5个0代表主机号字段...
每个子网中可以有30台主机,因为只能存在30个有效主机IP地址。
要是12个子网则需要借用4位来作为子网号,子网掩码为255.255.255.240,即为11111111.11111111.11111111.11110000,后面有一组中的前4个1是子网号字段,后5个0是主机号字段...
每个子网中可以存在14台主机【2的4次方 - 2 = 14,减2的道理同上】
其实1楼的同志回答也算正确,分8个子网时也可以借用4位,而且一般就是这样,但是题目中说“要求每个子网中主机数目最多”,所以借用3位即可!~
呵呵,好好学习,天天向上!~·
这是一个C类网络的子网划分,
因为要划分8个子网,所以掩码为24+3=27位
掩码 225.225.225.11100000/27
225.225.225.224/27
主机数 256除以8等于32,或者是2的5次方等于32
没办法直接划分12个子网,因为子网数是2个N次方,也就是说只能为2,4,8,16,32,64等,12个子网要划分16个子网来满足条件.
掩码 225.225.225.240/28
主机 2的(8-4)次方等于16
这个网络是C类网络,分8个子网的话,需要借主机位4位,子网数实际为16-2=14个,子网掩码用二进制表示为255.255.255.240,每个子网的最多主机数为14个;分12个子网的方法同上
已知ip地址求子网掩码,看视频,简单快速的学习一下
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