Answer the following questions for each of the configuration in the table below.
1. How many subnets?
2. How many hosts per subnet?
3. What are the valid subnets?
4. What are the valid hosts per subnet?
5. What is the broadcast address for each subnet?
IP ADDRESS SUBNET MASK
CONFIGURATION A 192.168.10.0 255.255.255.192
CONFIGURATION B 188.8.131.52 255.255.255.224
CONFIGURATION C 184.108.40.206 255.255.255.248
All the given configurations belong to Class C network, so subnetting involves only the last octet. This answer looks at the Class C specific subnetting computations in different ways.
First, we find out how many bits of host part are used for subnetting in this case.
192 = (128 + 64 + 0 + 0 + 0 + 0 + 0 + 0) OR 11000000 in binary.
Number of bits used for denoting subnetwork (Ns) = 2
Number of bits for denoting hosts in subnetwork (Nh) = 8 - 2 = 6.
1. No. of subnets = 2^Ns - 2 = 2^2 - 2 = 2
2. No. of hosts per subnet = 2^Nh - 2 = 2^6 - 2 = 62
3. To find the first subnet, we compute (256 - subnetmask in fourth octet), and to find further subnets we add this computed value to previous subnet id and continue till we get to the above mentioned subnetmask value.
First subnet id = 256 - 192 = 64
Second subnet id = 64 + 64 = 128
So, the valid subnets are 192.168.10.64 and 192.168.10.128 .
4. Since the bits in host id can not be all zeros or all ones. We compute the valid host ids in each of the above subnets as following.
Solution not only gives detailed explanation along with computations, but also illustrates different ways of subnetting computations.