Subnetting Information
This is for IPV4
only. IPV6 is a completely different breed of horse. I won't cover it
here. IPV4 is still in use and will be for awhile.
I hope this is understandable.
Forget about the long division tables to find a
subnet number. Put aside all the so-called short cuts that are easy to get
mixed up and often forgotten.
Stay with BINARY. It is the only sure and safe way to go when working with IP
addressing.
RULES FOR SUBNETTING
1> Borrow at least 2 bits for networks and leave at least 2 bits for
hosts.
2> Usable subnets = 2n bits borrowed - 2
3> Usable hosts = 2n bits left -2
4> Decimal value of last bit borrowed = the increment of the subnet
addresses, host addresses and the first usable subnet.
SUBNET MASKS
Default masks Address
range
Max hosts w/0 bits borrowed -2
A -
255.0.0.0
1 -
126
16,777,214
B - 255.255.0.0
128 -
191
65,534
C - 255.255.255.0 192 -
223
254
Class A 1 - 126 (First bit = 0) 0xxxxxx
| Bits Borrowed |
Subnet Mask |
/ |
Max Subnets (-2) |
Max Hosts (-2) |
| 2 |
255.192.0.0 |
10 |
2 |
4,194,302 |
| 3 |
255.224.0.0 |
11 |
6 |
2,097,150 |
| 4 |
255.240.0.0 |
12 |
14 |
1,048,574 |
| 5 |
255.248.0.0 |
13 |
30 |
524,286 |
| 6 |
255.252.0.0 |
14 |
62 |
262,142 |
| 7 |
255.254.0.0 |
15 |
126 |
131,070 |
| 8 |
255.255.0.0 |
16 |
254 |
65,534 |
| 9 |
255.255.128.0 |
17 |
510 |
32,766 |
| 10 |
255.255.192.0 |
18 |
1,022 |
16,382 |
| 11 |
255.255.224.0 |
19 |
2,046 |
8,190 |
| 12 |
255.255.240.0 |
20 |
4,094 |
4,094 |
| 13 |
255.255.248.0 |
21 |
8,190 |
2,046 |
| 14 |
255.255.252.0 |
22 |
16,382 |
1,022 |
| 15 |
255.255.254.0 |
23 |
32,766 |
510 |
| 16 |
255.255.255.0 |
24 |
65,534 |
254 |
| 17 |
255.255.255.128 |
25 |
131,070 |
126 |
| 18 |
255.255.255.192 |
26 |
262,142 |
62 |
| 19 |
255.255.255.224 |
27 |
524,286 |
30 |
| 20 |
255.255.255.240 |
28 |
1,048,574 |
14 |
| 21 |
255.255.255.248 |
29 |
2,097,150 |
6 |
| 22 |
255.255.255.252 |
30 |
4,194,302 |
2 |
Class B 128 - 191 (first 2 bits = 10) 10xxxxxx
| Bits Borrowed |
Subnet Mask |
/ |
Max Subnets (-2) |
Max Hosts (-2) |
| 2 |
255.255.192.0 |
18 |
2 |
16,382 |
| 3 |
255.255.224.0 |
19 |
6 |
8,190 |
| 4 |
255.255.240.0 |
20 |
14 |
4,094 |
| 5 |
255.255.248.0 |
21 |
30 |
2,046 |
| 6 |
255.255.252.0 |
22 |
62 |
1,022 |
| 7 |
255.255.254.0 |
23 |
126 |
510 |
| 8 |
255.255.255.0 |
24 |
254 |
254 |
| 9 |
255.255.255.128 |
25 |
510 |
126 |
| 10 |
255.255.255.192 |
26 |
1,022 |
62 |
| 11 |
255.255.255.224 |
27 |
2,046 |
30 |
| 12 |
255.255.255.240 |
28 |
4,094 |
14 |
| 13 |
255.255.255.248 |
29 |
8,190 |
6 |
| 14 |
255.255.255.252 |
30 |
16,382 |
2 |
Class C 192- 223 (First 3 bits = 110) 110xxxxx
| Bits Borrowed |
Subnet Mask |
/ |
Max Subnets (-2) |
Max Hosts (-2) |
| 2 |
255.255.255.192 |
26 |
2 |
62 |
| 3 |
255.255.255.224 |
27 |
6 |
30 |
| 4 |
255.255.255.240 |
28 |
14 |
14 |
| 5 |
255.255.255.248 |
29 |
30 |
6 |
| 6 |
255.255.255.252 |
30 |
62 |
2 |
Finding the number of the subnet a given address is on for Class A
(see Long Division for Class B)
What subnet is 122.10.10.156/27 on?
The /27 tells us that 27 bits are used for networks, or 19 bits were
borrowed, this means all of the 2nd and 3rd octets and 3
bits of the 4th octet were borrowed for networks. 122 is the assigned
network address so this is left alone. Work only with the last 3 octets -
10.10.156
Lay it out in binary as follows -
2nd octet _ _ _ _ _ _ _ _ 3rd octet _ _ _ _ _ _ _
_ and the 4th octet _ _ _ (19 bits borrowed)
Put the numbers 10.10.156 into the 2nd, 3rd and 4th octets
like this:
0 0 0 0 1 0 1 0 (=10) . 0 0 0 0 1 0 1 0 (=10) (And as much of the 156
that will fit into the bits that were borrowed from the 4th
octet) 1 0 0 (=128)
Discarding the leading zeros in the 2nd octet, we have:
| 1 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
| 1 |
8 |
4 |
2 |
1 |
5 |
2 |
1 |
6 |
3 |
1 |
8 |
4 |
2 |
1 |
| 6 |
1 |
0 |
0 |
0 |
1 |
5 |
2 |
4 |
2 |
6 |
|
|
|
|
| 3 |
9 |
9 |
4 |
2 |
2 |
6 |
8 |
|
|
|
|
|
|
|
| 8 |
2 |
6 |
8 |
4 |
|
|
|
|
|
|
|
|
|
|
| 4 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
then add the binary number up
16384 |
| 4096 |
| 64 |
| 16 |
| 4 |
|
= 20564 |
122.10.10.156/27 is on the 20564th subnet.
What is the subnet mask of 122.10.10.156/27?
Remember we borrowed 19 bits total, 8 from the 2nd octet, 8 from the 3rd
octet and 3 bits from the 4th octet. The decimal
value of those LAST 3 bits borrowed is 224
(128+64+32). _
_ _ | _ _ _ _ _. The subnet mask would be 255.255.255.224. (If we
borrowed only 3 bits from the 3rd octet, our subnet mask would be 255.255.224.0.
It's still the value of the bits borrowed from the last octet we borrowed from.)
Using Binary math with the subnet mask you will be able to find the wire
address of this subnet.
Subnet mask = 255.255.255.224
Address = 122.10.10.156
We know the wire address starts with 122.10.10 so all we have to add is
the 4th octets
11100000 = 224
10011100 = 156
10000000 = 128
122.10.10.128 is the wire address for 122.10.10.156/27.
A class B address is done the same way. Just use the last 2 octets
Now let's do something a little different but almost the same.
What is the 565th subnet of 122.10.10.156/27?
Again, we know that 19 bits were borrowed, giving us
122. _ _ _ _ _ _ _ _ ._ _ _ _ _ _ _ _ . _ _ _
Starting with the LAST bit borrowed in the 4th octet, plug
in 565
122 . _ _ _ _ _ _ _ _ . 01000110 . 101
We now know that the address starts with 122.0
In the 4th octet, add trailing 0's to fill up the octet.
We now have 122 . 0. 01000110.10100000
Convert octets 3 and 4 to decimal and we have the address:
01000110 = 70
10100000 = 160
The address for the 565th subnet of 122.10.10.156/27 is
122.0.70.160
What is the 4050th subnet of 103.15.25.60/20?
How can we determine if an address is a valid Host, Wire or Broadcast
address?
Let's take a class C address of 193.10.10.50/28
193.10.10._ _ _ _ | _ _ _ _
We know that 4 bits were borrowed from the 4th octet, so plug the '50'
into the whole 4th octet and you have this:
<hosts>
193.10.10. 0011 | 0010
There is a '1' in the host portion of the address, so this makes
193.10.10.50 a valid host address.
(You must have at LEAST a single "1" in the host portion to
make it a valid host address)
If we have 193.10.10.48/28, we get 193.10.10.0011|0000 so this makes
it a Wire address.
If we have 193.10.10.15/28, we get 193.10.10.0000|1111 so this makes it
a Broadcast address.
All 1's in the HOST portion indicates a broadcast address.
All 0's in the HOST portion indicates a wire address.
Now for a class A
120.10.10.50/30
30 bits for networks (22 bits borrowed) and 2 bits for hosts. Plug
the '50' in like before.
We have 120. _ _ _ _ _ _ _ _ . _ _ _ _ _ _ _ _ . _ _ _ _ _ _
| _ _
120
0 0 1 1 0 0 1 0 = 50 = valid host address.
What would be the first usable network address for
120.10.10.50/30?
4 is the decimal value of the LAST bit borrowed so our first usable
network (wire) address would be 120.0.0.4, our first host address on this wire
would be 120.0.0.5, the last host address on this wire would be 120.0.0.6 and
the broadcast address for this wire would be 120.0.0.7. The second wire address
would be 120.0.0.8 (increment of 4 from the first usable wire address.
Private Addresses
A - 10.0.0.0 (one network)
B - 172.16.0.0 - 172.31.0.0 (16 networks)
C - 192.168.0.0 - 192.168.255.0 (256 networks)
Can we subnet private addresses? Yes. You can do the same thing with
private addresses that you can do with public addresses - except go out on the
internet with them. |