Question

As the network administrator for a growing ISP, you want to make efficient use of your network addresses. One of the networks addresses IANA assigned to you is a Class C network of 192.168.88.0. You have decided to use the addresses in this Class C network to satisfy the IP address requirements of 18 corporate customers who need between 20 and 24 addresses each. calculate a subnet mask that meets their needs. List the subnet mask and the first four subnetwork addresses the mask will create.

Answer 1

Given IP: 192.168.88

We need to satisfy the IP address requirements of 18 corporate customers who need between 20 and 24 addresses each.
Next power of 2 after 18 = 32
Next power of 2 after 24 = 32

Total network addresses required = 32*32=2^10

Number of bits required =10

Netmask=32-10=22

Therefore:

Address: 192.168.88.0

Netmask: 255.255.252.0 = 22

Hosts/Net: 1022

Subnets

So, our IP address becomes: 192.168.88.0/22

We need to get 20 to 24 addresses for this we will require 5 bits. Out netmask for the subnet = 27.
We will get 32 subnets with 30 hosts per network

Netmask:   255.255.255.224 = 27

The first four subnet address are:

Network:   192.168.88.0/27
Broadcast: 192.168.88.31 
HostMin:   192.168.88.1
HostMax:   192.168.88.30
Hosts/Net: 30

Network:   192.168.88.32/27     
Broadcast: 192.168.88.63
HostMin:   192.168.88.33        
HostMax:   192.168.88.62
Hosts/Net: 30                   

Network:   192.168.88.64/27
Broadcast: 192.168.88.95        
HostMin:   192.168.88.65        
HostMax:   192.168.88.94
Hosts/Net: 30                   

Network:   192.168.88.96/27
Broadcast: 192.168.88.127       
HostMin:   192.168.88.97        
HostMax:   192.168.88.126       
Hosts/Net: 30