Chapter 5, Problem 6HA5

Explanation of Solution

Subnet Number	Subnet Address	Range of Host Address	Direct Broadcast Address
0	192.168.111.0	192.168.111.1- 192.168.111.14	192.168.111.15
1	192.168.111.16	192.168.111.17 - 192.168.111.30	192.168.111.31
2	192.168.111.32	192.168.111.33 - 192.168.111.46	192.168.111.47
3	192.168.111.48	192.168.111.49 - 192.168.111.62	192.168.111.63
4	192.168.111.64	192.168.111.65 - 192.168.111.78	192.168.111.79
5	192.168.111.80	192.168.111.81 - 192.168.111.94	192.168.111.95
6	192.168.111.96	192.168.111.97 - 192.168.111.110	192.168.111.111
7	192.168.111.112	192.168.111.113 - 192.168.111.126	192.168.111.127
8	192.168.111.128	192.168.111.129 - 192.168.111.142	192.168.111.143
9	192.168.111.144	192.168.111.145 - 192.168.111.158	192.168.111.159
10	192.168.111.160	192.168.111.161 - 192.168.111.174	192.168.111.175
11	192.168.111.176	192.168.111.177 - 192.168.111.190	192.168.111.191
12	192.168.111.192	192.168.111.193 - 192.168.111.206	192.168.111.207
13	192.168.111.208	192.168.111.209 - 192.168.111.222	192.168.111.223
14	192.168.111.224	192.168.111.225 - 192.168.111.238	192.168.111.239
15	192.168.111.240	192.168.111.241 - 192.168.111.254	192.168.111.255

Explanation:

• The user is assigned with192.168.111.129/28 and the given address belongs to class C network address.
• For class C network, the byte allocation is network.network.netwok.host.
  • The prefix given in class C network address is 28.
• As class C address is given, the user uses 3 bytes for subnet field.
• However, because of a prefix given as 28, the user needs 4 more additional bits to create subnet field...