IP Range Notation

Question

I have the following IP ranges:

64.233.160.0    64.233.191.255
66.102.0.0  66.102.15.255
66.249.64.0 66.249.95.255
72.14.192.0 72.14.255.255
74.125.0.0  74.125.255.255
209.85.128.0    209.85.255.255
216.239.32.0    216.239.63.255

I need to be able to represent these IP ranges in the format like:

64.233.160/24

I believe this is called CIDR notation. IS this possible? If so, could someone demonstrate how to generate the notations?

Give a try to this site: http://ipcalc.nmonitoring.com/?zmena=4&ip1=64&ip2=233&ip3=160&ip4=0&netmask=18&mask_bits=19&max_subnets=19&max_addresses=19 — maudam, Dec 28 '14 at 16:16
Network address: 64.233.160.0 / 19 Usable IP addresses: 64.233.160.1 - 64.233.191.254 Broadcast: 64.233.191.255 — maudam, Dec 28 '14 at 16:17

score 11 · Accepted Answer · answered Dec 28 '14 at 16:18

11

This is indeed called CIDR notation. You can use the ipcalc Linux tool to convert the ranges to CIDR notation:

mtak@frisbee:~$ ipcalc -r 66.102.0.0  66.102.15.255
deaggregate 66.102.0.0 - 66.102.15.255
66.102.0.0/20

answered Dec 28 '14 at 16:18

mtak

16,894

UfoXp · Answer 2 · 2015-08-02T05:17:38.640

4

Number after slash is just a number of bits set to 1 from left side in a subnet mask for the given IP range. Subnet mask 255.0.0.0 is /8, 255.255.0.0 is /16 and so on. /20 in binary is 1111 1111.1111 1111.1111 0000.0000 0000 and in decimal 255.255.240.0

edited Aug 02 '15 at 05:17

answered Dec 28 '14 at 16:17

UfoXp

211
1
4

How is 66.249.64.0 66.249.95.255 a /16? – mtak Dec 28 '14 at 16:19
Well, ok. After you've posted your answer I've realized that you can fit those more "tightly". – UfoXp Dec 28 '14 at 16:21
1

+1 for an attempt to show how this calculation is done. – kestasx Dec 28 '14 at 18:33
1

For reference, '/' is a slash. '' is a backslash. (Pet peeve.) – cHao Dec 29 '14 at 04:39

score 3 · Answer 3 · answered Dec 29 '14 at 04:15

I'll do it for the last line. 216.239.32.0-216.239.63.255

You need to determine the subnet mask that identifies this network. The interesting octet is the third one since the first two stay the same and the fourth sweeps from 0 to 255.

In the third octet, the network address starts at 32 and the next network starts when it has incremented to 64. It "uses" 32 numbers in that octet for the network address. Therefore it "uses" 1/8 of the 256 possible numbers in that octet.

You could have 8 same sized networks splitting the address space of the interesting octet:

216.239.0.0-216.239.31.255
216.239.32.0-216.239.63.255
216.239.64.0-216.239.95.255
216.239.96.0-216.239.127.255
(... I'll let you continue the sequence)

Now, knowing that you could have 8 unique networks the same size as your network should tell you that you will need 8 unique numbers to represent them. If you are versed in binary math then it should be immediately obvious that it takes 3 bits to represent 8 numbers.

The answer then is the 16 bits from the first two uninteresting octets plus the 3 bits we just determined, so the subnet mask is /19. The CIDR format of the network address is 216.239.32.0/19.

On a side note, within this network you can have 32*256 host addresses--minus one for the broadcast address and minus another for the network address.

IP Range Notation

3 Answers3