# Route Summarization **URL:** https://forum.networklessons.com/t/route-summarization/1220 **Category:** Lessons Discussion **Created:** [December 29, 2016, 5:12pm UTC](https://forum.networklessons.com/t/route-summarization/1220 "2016-12-29T17:12:24Z") **Posts on this page:** 1 **Showing post:** 2
### Author: ![gabrielba](https://cdn-forum.networklessons.com/letter_avatar_proxy/v4/letter/g/e79b87/32.png) [@gabrielba](https://forum.networklessons.com/u/gabrielba) #### Post date: [May 17, 2015, 9:13pm UTC](https://forum.networklessons.com/t/route-summarization/1220/2 "2015-05-17T21:13:28Z")
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 192.168.0.0 / 24 192.168.1.0 / 24 192.168.2.0 / 24 192.168.3.0 / 24 192.168.4.0 / 24 The summary address will be 192.168.0.0/21, but when I was writing this example I saw that if I use block size it’s able solve it too. It’s need always think in block size like “powers of 2”. When you have a free time, please, detail below 8 + 8 + 6 = 24 bits Hug and thanks for yours articles. It’s helping a lot 🙂 --- _[View the full topic](https://forum.networklessons.com/t/route-summarization/1220)._