2. A network administrator has been provided with the following block of class A IP addresses: 12.24.0.0/21.
He has been asked to create at least 3 subnets with equal size netmasks, and the number of the hosts per subnet are 31, 13 and 47. He prefers to use small subnets over large subnets in order to assign hosts to new subnets in case of future needs.
A. What is the new netmask?
B. What are the ranges of the first three and the last subnet?
C.What are the network and broadcast addresses within these subnets?
D.What is the number of IP addresses available for host – assignment in these subnets?
E. Manually summarize the first 4 subnets
F. What would be the effect of using “no subnet-zero” in the router’s configuration?
Solution:
A. In this case the administrator prefers to create more subnets with less hosts than vice versa. Hence he will determine the minimum amount of hosts bits needed for subnetting.
The largest subnet requires 47 hosts or more.
The smallest number of hosts bits that fulfills this requirement is 6, since (2^6-2)=62 hosts / subnet. This leaves 26 bits for the netmask, so the netmask will expand from /21 → /26, which corresponds to a new netmask of: 255.255.255.192.
B. The number of bit available for subnetting is 5. This will allow us to create 2^5 = 32 subnets.
A few will be shown here:
All IP addresses start with: 12.24.
[0.0 – 0.63] [1.0 – 1.63] etc etc…
[0.64 – 0.127] [1.64 – 1.127]
[0.128 – 0.191] [1.128 – 1.191]
[0.192 – 0.255] [1.192 – 1.255]
The last subnetwork is: 12.24.[ 7.192– 7.255]
C. Again the lowest and highest addresses in each individual subnet, eg 12.24.0.0 & 12.24.0.63.
D. The number of IP addresses available for host – assignment per subnet is: (2^6-2)=62.
E. Manually summarize the first 4 subnets:
12.24.0.0 → 12 . 24 . 0000 0000 . 0 0 x x x x x x
12.24.0.64 → 12 . 24 . 0000 0000 . 0 1 x x x x x x
12.24.0.128 → 12 . 24 . 0000 0000 . 1 0 x x x x x x
12.24.0.192 → 12 . 24 . 0000 0000 . 1 1 x x x x x x
←———-24———-→
Here the summarized route would be: 12.24.0.0/24
F. Out of a possible 2^5=32 subnets, only 30 would be available, since the lowest and highest subnets would be rendered unusable. In this case the number of subnets remaining would still be sufficient.
