First keep in mind that a range is a set of arbitrary contiguous IP address and does not always match a CIDR network.
In your case it does, though.
So what you need is to find the smallest 'supernet` of the two addresses.
To do so you can refer to my answer here.
Doing the math for your range we get:
220.127.116.11 = 01111000.01001100.00000000.00000000
Yes, we can calculate the smallest supernet that contains 2 given networks (or any number of networks).
The easiest way to do this is in binary (yes really):
First convert your two addresses in binary:
10.1.192.0 = 00001010.00000001.11000000.00000000
10.1.240.0 = 00001010.00000001.11110000.00000000
As you can see, the part that is common to this two ...
Generally this network address 192.168.1.0/24 has total 254 usable hosts that means network 18.104.22.168/24 is representing total 254 hosts if any another networks want to communicate to this 254 host . Route entry is configured with as destination as 192.168 .1.0 so traffic will route to all 254 hosts if not we have to write 254 route entries which make ...
When your isp provided with four static public ips .
1 - you can use for egress interfàe ip for your perimeter làyer3 device
You will configure default route pointing towards isp gateway in perimeter làyer3 device
Ip route 0.0.0.0 0 0.0.0.0 pointing towards isp gateway
Remaining three ip address are your inventory . Your càn use for destination nàtting ...