Super User is a question and answer site for computer enthusiasts and power users. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

To create a route summarised address and subnet mask for the following addresses:

  • 192.168.204.0/24
  • 192.168.205.0/24
  • 192.168.206.0/24
  • 192.168.207.0/24
  • 192.168.208.0/24

First I will write out the binary form of the addresses up to and including the changing octet. 11000000.10101000.11001100

11000000.10101000.11001101

11000000.10101000.11001110

11000000.10101000.11001111

11000000.10101000.11010000

From the list, I counted from the left how many bits are the same in each address, as can be seen the first 19 bits for each address are the same so that gives us the subnet mask in slash notation. So the summarised address and subnet mask is 192.168.204.0/19.

Would this be correct?

share|improve this question
up vote 1 down vote accepted

One incorrect step:

11000000.10101000.110 is the common part. You then revert it to binary notation and get 192.168.192.0/19.

Basically, X.X.X.0/19 means that there are 2^(32-19) = 2^13 IP addresses in given subnetwork. As a rule, subnetworks cannot intersect. So for 19 bit mask, you have:

  • 192.168.0.0/19
  • 192.168.32.0/19
  • 192.168.64.0/19
  • 192.168.96.0/19
  • 192.168.128.0/19
  • 192.168.160.0/19
  • 192.168.192.0/19
  • 192.168.224.0/19

Total 8 /19 networks in 192.168.0.0/16 family (19-16=3, 2^3=8).

share|improve this answer
    
thank you, done a few practise ones online there using your format and got them all correct thank you. – user68062 Jun 1 '11 at 20:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .