# Route Summarization

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

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:

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:

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. Copyright protected by Digiprove © 2015 Rene Molenaar

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## Forum Replies

1. gabrielba says:

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 forum

2. ReneMolenaar says:

Hi 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 forum

3. gabrielba says:

Rene,

“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

4. joburke10 says:

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

5. ReneMolenaar says:

Hi James,

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

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