How to calculate create table of Subnets addresses, range of Hosts and Broadcast

Create a table with the subnet addresses, valid host address range, and the subnet's Broadcast address.

A certain organization received from the IP address regulatory authority the address 192.168.1.0, to be used in its internal network, but for various reasons, the company needs 6 subnetworks, with 30 hosts (computers) in each subnetwork.

How to solve this problem?

1st Find out which class belongs to the IP range.

First you need to find out which class this IP range belongs to, see more about IP Protocol Classes at http://www.fabiobmed.com.br/classe-do-protocolo-ip/.

IP ranges 192 to 223 belong to Class C, therefore IP 192.168.1.0 is Class C.

The default Class C IPs Subnet Mask is 255.255.255.0, i.e Host.Host.Host.Net. So the default Subnet Mask of IP 192.168.1.0 is 255.255.255.0

2nd Create a Subnet and Host table

In IP 198.168.1.0. Mask 255.255.255.0 I have the last octet to form the subnet list and the host list. I say Octet because the IP or Mask numbers in binaries correspond to:

  • 255.255.255.0 (decimal) = 11111111.11111111.11111111.00000000 (binary)

255.255.255.0 cannot be used as it is the network address and 255.255.255.255 cannot be used as it is the Broadcast address. What was left was the range from 255.255.255.1 to 255.255.255.254, from these numbers I need to create the subnet and the list of hosts.

  • If I use 1 bit for subnetting (1000000) I will have 2 left7-2 for host list
  • So we will have 21-2 for subnet address and 27-2 for host list
  • That is, a list of 2-2* addresses to the subnet and a list of 128-2* addresses to hosts.

NOTE: O * represents the first and last address that is reserved for the Main Subnet and for Broadcast. These addresses cannot be used.

A list follows.

lista sub-redes e hosts
subnets and hosts list

According to the list above, only option C fits the criterion of 06 subnets with 30 hosts each.

3rd Create the list with the Subnet and Hosts ranges for each Subnet

Subnets

  1. Subnet 000 00000 = 192.168.1.0 The Subnet 000 11111 192.168.1.31
  2. Subnet 001 00000 = 192.168.1.32 The Subnet 000 11111 192.168.1.63
  3. Subnet 010 00000 = 192.168.1.64 The Subnet 000 11111 192.168.1.95
  4. 011 subnet 00000 = 192.168.1.96 The Subnet 000 11111 192.168.1.127
  5. Subnet 100 00000 = 192.168.1.128 The Subnet 000 11111 192.168.1.159
  6. Subnet 101 00000 = 192.168.1.160 The Subnet 000 11111 192.168.1.191
  7. Subnet 110 00000 = 192.168.1.192 The Subnet 000 11111 192.168.1.223
  8. Subnet 111 00000 = 192.168.1.224 The Subnet 000 11111 192.168.1.255

Host List

  1. subnet 192.168.1.0host 192.168.1.1 The 30 | broadcast 192.168.1.31
  2. subnet 192.168.1.32host 192.168.1.33 The 62 | broadcast 192.168.1.63
  3. subnet 192.168.1.64host 192.168.1.65 The 94 | broadcast 192.168.1.95
  4. subnet 192.168.1.96host 192.168.1.97 The 126 | broadcast 192.168.1.127
  5. subnet 192.168.1.128 host 192.168.1.129 The 158 | broadcast 192.168.1.159
  6. subnet 192.168.1.160host 192.168.1.161 The 190 | broadcast 192.168.1.191
  7. subnet 192.168.1.192host 192.168.1.193 The 222 | broadcast 192.168.1.223
  8. subnet 192.168.1.224host 192.168.1.225 The 254 | broadcast 192.168.1.255

That's it, we already have all the Subnet and Hosts addresses for our network.

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7 comments

  1. POSTED!!!The default Class C IPs Subnet Mask is 255.255.255.0, ie Host.Host.Host.Network. So the default Subnet Mask of IP 192.168.1.0 is 255.255.255.0

    ANSWER?

    There is a mistake here when you say; which is Host,Host,Host,Network,
    that the CORRECT state that it is Net,Net,Net,Host treating that the subnet mask 255.255.255.0/24.

  2. The first three octets in the C class mask define the network! that is, it would be Rede.Rede.Rede.Hosts
    192.168.1.0 ; your network is 192.168.1.0 and the last byte is your available hosts.

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