Subnetting example(s) and manual route summarization

1. A network administrator has been provided with the following Class B block of IP addresses:

172.16.0.0/16

It is his task to subnet this Classful network to have (at least) 9 subnets available and maximize the number of hosts available pèr subnet.

A. What is the new netmask?

B. What are the ranges of IP addresses within each subnet?

C.What are the network and broadcast addresses within each subnet?

D.What is the number of IP addresses available for host – assignment in each subnet?

Answers:

A. The netmask will have to be expanded with 4 bits so /16 → /20.

This results in a netmask of 255.255.240.0

B. All addresses start with 172.16

[0.0-15.255] [64.0-79.255] [128.0-143.255] [192.0-207.255]

[16.0-31.255] [80.0-95.255] [144.0-159.255] [208.0-223.255]

[32.0-47.255] [96.0-111.255] [160.0-175.225] [224.0-239.255]

[48.0-63.255] [112.0-127.255] [176.0-191.255] [240.0-255.255]

C.The lowest and highest addresses in each individual subnet, so eg 172.16.0.0 &172.16.15.255, 172.16.16.0 & 172.16.31.255, etc.

D. (2^h-2)=(2^12-2)=4094

E. What would be the effect on the network this design if the router in question is configured with “no subnet-zero”?

The lowest and highest subnetworks 172.16.0.0 – 172.16.15.255 and 172.16.240.0 – 172.16.255.255 would be rendered unusable. There would still be 16 – 2 = 14 subnetworks left, which is sufficient for the needs mentioned above, so there would be no impact on the network design.

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?

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.

A Manual Summarization Example:

A. Manually summarize the following (sub)networks without distributing any route that contains a networks outside of the route summary. More than 1 summarization may be necessary.

192.168.1.0/27

192.168.1.64/28

192.168.1.80/28

192.168.1.96/28

192.168.1.112/29

192.168.1.120/29

B. What would be the effect if 192.168.1.96/28 is not part of the networks to be summarized?

Solution A:

For this we will only analyze the fourth octet. Only the (sub)network – bits will be shown for clarity. All addressess start with 192.168.1.0.

Clearly in trying to find a common denominator, the most common bits can not be found in 1 summarized address since the first network (192.168.1.0/27) does not have a lot in common with the other networks.

So our first summary will be to simply publish 192.168.1.0/27.

For the remaining networks, we can use 192.168.1.64/26, which has the most common host-bits in common. So our routing table entries have changed from the original 6 to just 2.

To check this, we can deduce the range of IP addressess in our summarized network:

192.168.1.64/26 ranges from 192.168.1.64 – 192.168.1.127.

Solution B:

Here the result would be 3 summarizations:

192.168.1.0/27

192.168.1.64/27

And

192.168.112/28.