In this lesson we’ll take a look how you can configure summaries. First i’ll show you some examples how to do so in binary and then we’ll take a look at some tricks how you can do it in decimal (which is much faster).

Let’s say we want to create the most optimal summary for the following 4 networks:

- 192.168.0.0 / 24 subnet mask 255.255.255.0
- 192.168.1.0 / 24 subnet mask 255.255.255.0
- 192.168.2.0 / 24 subnet mask 255.255.255.0
- 192.168.3.0 / 24 subnet mask 255.255.255.0

Let’s convert these network addresses to binary:

192.168.0.0 | 11000000 | 10101000 | 00000000 | 00000000 |

192.168.1.0 | 11000000 | 10101000 | 00000001 | 00000000 |

192.168.2.0 | 11000000 | 10101000 | 00000010 | 00000000 |

192.168.3.0 | 11000000 | 10101000 | 00000011 | 00000000 |

Now we have to look how many bits these network addresses have in common. The first and second octets are the same, so that’s 16 bits.

Let’s zoom in on the third octet:

00000000 |

00000001 |

00000010 |

00000011 |

The first 6 bits of the third octet are the same. Now we have enough information to create our summary address.

8 + 8 + 6 = 22 bits

Our summary address will be 192.168.0.0 /22 (subnet mask 255.255.252.0).

Now you have seen how to do this in binary, let’s do it in decimal. There’s a simple trick you can use to calculate this summary.

As you can see we have 4 networks, or when we speak in ‘blocks’ it’s a block of 4. Here’s a formula you can use:

256 – number of networks = subnet mask for summary address.

For example: 256 – 4 networks = 252

The subnet mask will be 255.255.252.0

Another way to look at it is by using the CIDR notation. You know a /24 is a block of 256 addresses. Using a /23 means you have 2 x 256, and a /22 means you have 4 x 256.

Let’s look at another example. Let’s say we want to summarize the following networks:

- 172.16.0.0 / 16 subnet mask 255.255.0.0
- 172.17.0.0 / 16 subnet mask 255.255.0.0
- 172.18.0.0 / 16 subnet mask 255.255.0.0
- 172.19.0.0 / 16 subnet mask 255.255.0.0
- 172.20.0.0 / 16 subnet mask 255.255.0.0
- 172.21.0.0 / 16 subnet mask 255.255.0.0
- 172.22.0.0 / 16 subnet mask 255.255.0.0
- 172.23.0.0 / 16 subnet mask 255.255.0.0

Let’s look at it in binary first. I’ll write down the second octet since the first one is the same for all network addresses:

16 | 00010000 |

17 | 00010001 |

18 | 00010010 |

19 | 00010011 |

20 | 00010100 |

21 | 00010101 |

22 | 00010110 |

23 | 00010111 |

The first 5 bits for all these addresses are the same. The first octet had 8 similar bits so that’s 8 + 5 = 13 bits.

The summary address will be 172.16.0.0 /13 (subnet mask will be 255.248.0.0).

Calculating in binary like this works but it’s slow. Let’s use our trick for this:

256 – number of networks = subnet mask for summary address.

So that’s 256 – 8 = 248. The subnet mask will be 255.248.0.0

We can also find it by just looking at the CIDR notations:

- 172.16.0.0 /16 is one network.
- 172.16.0.0 /15 are two networks.
- 172.16.0.0 /14 are four networks.
- 172.16.0.0 /13 are eight networks.

That’s a lot faster than looking at it in binary.

I hope this example is helpful for you to create summaries. Feel free to share this post! If you have any questions, feel free to leave a comment in our forum.

Rene,

I have a doubt, all examples that you gave are “continuos” networks and an even number of networks.

And when we have networks like below? I just can solve them with binary method. Is there another form?

172.16.10.0/24

172.16.20.0/24

172.16.30.0/24

172.16.40.0/24

172.16.50.0/24

I choosed shortest and highest networks and convert them to binary, so the summary address will be

176.16.0.0/18 a block size 64 networks. I can’t solve it using CIDR notation and block size method.

Another example with an odd number of networks, I can solve it with binary method

19

... Continue reading in our forumHi Gabriel,

Good question, let’s look at these examples. First one:

172.16.10.0/24

172.16.20.0/24

172.16.30.0/24

172.16.40.0/24

172.16.50.0/24

Let’s do it in binary first (in case someone else reads this):

10 = 00001010

20 = 00010100

30 = 00011110

40 = 00101000

50 = 00110010

Only the first 2 bits are the same. Our CIDR notation would be 8 + 8 + 2 = 18 bits and we’ll use network address 172.16.0.0.

This works but it’s slow…you can do it in decimal, just remember the block sizes:

2,4,8,16,32,64,128.

Now you only have to pick a block size that fits all of the netw

... Continue reading in our forumRene,

Thank you so much for your attention!!! Now I can understand better about this subject!

and about:

“The only thing to be aware of is that your summaries include networks that you “don’t have”.”

Yes, There is situation where is not possible to have a summarization so specific like example above, where we had that summarize 5 networks and it was need to use a block size of 8.

I’m really grateful/thankful with your explanations. Thanks, thanks and thanks!!!

Hug

Hey Rene,

I was following you example but got confused on these:

172.1.4.0/25 10101100.00000001.00000100.00000000

172.1.128.0/25 10101100.00000001.10000000.00000000

172.1.5.0/24 10101100.00000001.00000101.00000000

172.1.6.0/24 10101100.00000001.00000110.00000000

172.1.7.0/24 10101100.00000001.00000111.00000000

Particularly the 2nd one 172.1.128.0/25. Following your example using the 3rd octet has a 128 in it, so how would I factor this in as the other octets are using the first 5 bits.

thanks

James

Hi James,

Where did you find this example? It’s not on this page?

Rene