Sure, let’s take a look. Here is the list of subnets once again:
The first thing you’ll notice is the 184.108.40.206/25 subnet. The third octet is numerically “far away” compared to the others. You can see this more clearly if we look at the addresses in binary:
Notice that all the non-bold bits are the same for all addresses. The first bit (from left to right) that differs between all of the values is the 17th bit, and that takes place in the 220.127.116.11 address.
So if we want to summarize all of these, we must use a /16 subnet, because only the first 16 bits are the same. So, the summarization in binary would be:
10101100.00000001.00000000.00000000 with a subnet mask of
In dotted decimal, that would be 18.104.22.168/16
However, this summarization would include many more subnets than just these five. Specifically, this summarization includes the range from 22.214.171.124 to 126.96.36.199. Such a large range may include subnets that exist elsewhere. It is for this reason that it is best practice to summarize into two different subnets, one including only the 188.8.131.52/25 network, and another including the rest. Here are the binary addresses once again, but without the 184.108.40.206/25 network:
This time you can see that the 22 leftmost bits are the same, and there are differences only from the 23rd bit. For this reason, we can summarize these like so:
10101100.00000001.00000100.00000000 with a subnet mask of
Or in decimal, 220.127.116.11/22 which is what Rene came up with in his post.
So ultimately, the summary of the original question would come down to two summarizations:
18.104.22.168/22 and 22.214.171.124/25. This would be the most efficient result.
I hope this has been helpful!