1. The first octet octet rule states that the highest class C address is 223, and we know that the largest decimal number for an octet is 255. Therefore, there are two types of addresses, one is class D address, which is used for multicast, and the other is class E address, which is used for experiment. The first 4 bits of the class D address are 1110. What is the range of the decimal number of the first octet of a class D address?

The following are the IP address ranges for classes A-E5

kind	first octet byte binary range	first octet byte decimal range
A	00000001~01111110	1~126
B	10000000~10111111	128~191
C	11000000~11011111	192~223
D	11100000~11101111	224~239
E	11110000~11111110	240~254

2. Choose a subnet mask for 10.0.0.0 so that at least 16,000 subnets can be divided, and each subnet has at least 700 host addresses. Choose the subnet mask for 172.27.0.0 so that at least 500 subnets can be divided, and each subnet has at least 100 host addresses.

根据10.0.0.0可以分析出为A类地址
默认掩码为255.0.0.0
根据题目要求需要划分16000个子网，因为
2^14-2=16382>=16000
所以需要借6位来充当网络位
又因为每个子网至少拥有700个主机地址
2^10-2=1022>=700
所以至少需要10位来充当主机位
因为还剩8位主机位使用，所以满足条件
子网掩码为网络位全为1，主机位全为0
11111111 11111111 11111100 00000000
255.255.252.0
10.0.0.0/22

根据172.27.0.0可以分析出为B类地址
默认掩码为255.255.0.0
根据题目要求需要划分500个子网，因为
2^9-2=510>=500
所以借9位来充当网络位
又因为每个子网至少拥有100个主机地址
2^7-2=126>=100
所以至少需要7位来充当主机位
子网掩码为网络位全为1，主机位全为0
11111111 11111111 11111111 10000000
255.255.255.128
172.27.0.0/25

3. If a class C address has 6 subnet bits, then multiple subnets can be divided? How many host addresses per subnet? Is there any practical use for such a subnetting plan?

根据C类地址默认掩码为255.255.255.0
现在有6个子网位
所以子网掩码为255.255.255.252
一个C类地址将有2^6-2=62个子网
每个子网有2^2-2=2个主机地址。

In this mode, a class C address can be used by 62 point-to-point links. A point-to-point link requires only two host addresses—one for each segment of the link. Often used in real networks

4. Subnet the address 192.168.147.0, the subnet mask is 28 bits, please write all the subnets. Try giving the available host addresses for each subnet.

根据192.168.147.0这个IP地址可以看出为C类地址
默认子网掩码为24位，现在子网掩码为28位
子网掩码为255.255.255.240
11000000 10101000 10010011 1111 0000 
所以可以划分2^4-2=14个子网
每个子网可以划分2^4-2=14个主机地址

这些子网是:
11000000 10101000 10010011 0001 0000 = 192.168.147.16
11000000 10101000 10010011 0010 0000 = 192.168.147.32
11000000 10101000 10010011 0011 0000 = 192.168.147.48
11000000 10101000 10010011 0100 0000 = 192.168.147.64
11000000 10101000 10010011 0101 0000 = 192.168.147.80
11000000 10101000 10010011 0110 0000 = 192.168.147.96
11000000 10101000 10010011 0111 0000 = 192.168.147.112
11000000 10101000 10010011 1000 0000 = 192.168.147.128
11000000 10101000 10010011 1001 0000 = 192.168.147.144
11000000 10101000 10010011 1010 0000 = 192.168.147.160
11000000 10101000 10010011 1011 0000 = 192.168.147.176
11000000 10101000 10010011 1100 0000 = 192.168.147.192
11000000 10101000 10010011 1101 0000 = 192.168.147.208
11000000 10101000 10010011 1110 0000 = 192.168.147.224

主机地址以第一个子网为例
11000000 10101000 10010011 0001 0001 = 192.168.147.17
11000000 10101000 10010011 0001 0010 = 192.168.147.18
11000000 10101000 10010011 0001 0011 = 192.168.147.19
11000000 10101000 10010011 0001 0100 = 192.168.147.20
11000000 10101000 10010011 0001 0101 = 192.168.147.21
11000000 10101000 10010011 0001 0110 = 192.168.147.22
11000000 10101000 10010011 0001 0111 = 192.168.147.23
11000000 10101000 10010011 0001 1000 = 192.168.147.24
11000000 10101000 10010011 0001 1001 = 192.168.147.25
11000000 10101000 10010011 0001 1010 = 192.168.147.26
11000000 10101000 10010011 0001 1011 = 192.168.147.27
11000000 10101000 10010011 0001 1100 = 192.168.147.28
11000000 10101000 10010011 0001 1101 = 192.168.147.29
11000000 10101000 10010011 0001 1110 = 192.168.147.30

5. Subnet the address 192.168.147.0, the subnet mask is 29 bits, please write all the subnets. Try giving the available host addresses for each subnet.

根据192.168.147.0这个IP地址可以看出为C类地址
默认子网掩码为24位，现在子网掩码为29位
子网掩码为255.255.255.248
11000000 10101000 10010011 11111 000 
所以可以划分2^5-2=30个子网
每个子网可以划分2^3-2=6个主机地址

子网范围为：192.168.147.8~192.168.147.240
主机范围为：192.168.147.9~192.168.147.14到
		  192.178.147.241~192.168.147.246

详细划分规则看第四题

6. Subnetize the address 172.16.0.0 with a 20-bit subnet mask, and try to give the available host addresses for each subnet (addresses are given in the lowest to highest order).

根据172.16.0.0这个IP地址可以看出为B类地址
默认子网掩码为16位，现在子网掩码为20位
子网掩码为255.255.240.0
11111111 11111111 1111 0000 00000000
所以可以划分2^4-2=14个子网
每个子网可以划分2^12-2=4096个主机地址

子网范围为：172.16.16.0~172.16.224.0
主机范围为：172.16.16.1~172.16.16.31到
		  172.16.224.1~172.16.239.254

详细划分规则看第四题

TCP/IP Routing Technology Volume 1 Configuration Exercise Analysis

The following are the IP address ranges for classes A-E5

2. Choose a subnet mask for 10.0.0.0 so that at least 16,000 subnets can be divided, and each subnet has at least 700 host addresses. Choose the subnet mask for 172.27.0.0 so that at least 500 subnets can be divided, and each subnet has at least 100 host addresses.

3. If a class C address has 6 subnet bits, then multiple subnets can be divided? How many host addresses per subnet? Is there any practical use for such a subnetting plan?

4. Subnet the address 192.168.147.0, the subnet mask is 28 bits, please write all the subnets. Try giving the available host addresses for each subnet.

5. Subnet the address 192.168.147.0, the subnet mask is 29 bits, please write all the subnets. Try giving the available host addresses for each subnet.

6. Subnetize the address 172.16.0.0 with a 20-bit subnet mask, and try to give the available host addresses for each subnet (addresses are given in the lowest to highest order).

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