 # Route Summarization

Route Summarization

## Convert to Binary?

First, we convert all of the 172.16.0.0 addresses above to their Binary form.

### 10101100.00010000

The Binary translation of these networks are:

172.16.1.0 ? 10101100.00010000.00000001.00000000
172.16.2.0 ? 10101100.00010000.00000010.00000000
172.16.3.0 ? 10101100.00010000.00000011.00000000

Lets Summarize but ?we need to locate the ?Summary Border.?? To summarize these network addresses, the Summary Border needs to be located. This will be the dividing line between the ?same bits?? and ?different ? bits in the address.? The “same” bits are shown in Orange color??below:

172.16.1.0 ? 10101100.00010000.00000001.00000000
172.16.2.0 ? 10101100.00010000.00000010.00000000
172.16.3.0 ? 10101100.00010000.00000011.00000000

To summarize these addresses we count the number of the ?same? bits to the left of the Summary Border.? In this case, from the farthest left bit to the summary border there are 22 bits.? Therefore, we can summarize these three network addresses as 172.16.0.0 /22.

### 6 thoughts on “Route Summarization”

1. Your way is too correct….but what if you were working with many many subnets and you would like to summarize it all????? this way of converting to binary would not be efficient in this case….how can I summarize a massive number of subnets at once?

• hi rainyer

you have to convert the binary in which octet where we have change for example if we have 192.168.1.1 other is 192.168.31.1

so you have to convert the binary of 3rd octet

00000001
00011111

You will get /16 + /3 = /19 ( /3 first three values are same.

2. Your way is too correct….but what if you were working with many many subnets and you would like to summarize it all????? this way of converting to binary would not be efficient in this case….how can I summarize a massive number of subnets at once?

• hi rainyer

you have to convert the binary in which octet where we have change for example if we have 192.168.1.1 other is 192.168.31.1

so you have to convert the binary of 3rd octet

00000001
00011111

You will get /16 + /3 = /19 ( /3 first three values are same.