Worksheet for deciphering an address
1st Octet
2nd Octet
3rd Octet
4th Octet
Result
3rd Octet
00000001
1
4th Octet
01100100
100
Result
IP Address
Subnet Mask
Subnet ID
1st useable
address
Last useable
address
Broadcast
address
How to Use
Start with an address, e.g. 192.168.1.100/25
Fill in the IP address using binary and decimal
IP Address
1st Octet
11000000
192
2nd Octet
10101000
168
192.168.1.100
Next, add the subnet mask the same way. We can se that because it is /25, the subnet mask is using
one bit from the 4th octet (8 + 8 + 8 = 24).
IP Address
Subnet mask
1st Octet
11000000
192
11111111
255
2nd Octet
10101000
168
11111111
2355
3rd Octet
00000001
1
11111111
255
4th Octet
01100100
100
10000000
128
Result
192.168.1.100
255.255.255.128
The subnet ID is next.
IP Address
Subnet mask
Subnet ID
1st Octet
11000000
192
11111111
255
11000000
192
2nd Octet
10101000
168
11111111
2355
10101000
168
3rd Octet
00000001
1
11111111
255
00000001
1
4th Octet
01100100
100
10000000
128
00000000
0
Result
192.168.1.100
255.255.255.128
192.168.1.0
Next we figure out the first and last useable address. Remember, we have already taken away the high
order bit on the 4th octet, so we know the last useable IP address will NOT be 254.
IP Address
Subnet mask
Subnet ID
1st useable
address
Last useable
address
1st Octet
11000000
192
11111111
255
11000000
192
11000000
192
11000000
192
2nd Octet
10101000
168
11111111
2355
10101000
168
10101000
168
10101000
168
3rd Octet
00000001
1
11111111
255
00000001
1
00000001
1
00000001
1
4th Octet
01100100
100
10000000
128
00000000
0
00000001
1
01111110
126
Result
192.168.1.100
255.255.255.128
192.168.1.0
192.168.1.1
192.168.1.127
Remember we cannot use the last address, it’s reserved for broadcast:
IP Address
Subnet mask
Subnet ID
1st useable
address
Last useable
address
Broadcast
address
1st Octet
11000000
192
11111111
255
11000000
192
11000000
192
11000000
192
11000000
192
2nd Octet
10101000
168
11111111
2355
10101000
168
10101000
168
10101000
168
10101000
168
3rd Octet
00000001
1
11111111
255
00000001
1
00000001
1
00000001
1
00000001
1
4th Octet
01100100
100
10000000
128
00000000
0
00000001
1
01111110
126
01111111
127
Result
192.168.1.100
255.255.255.128
192.168.1.0
192.168.1.1
192.168.1.126
192.168.1.127
Note again the high-order bit in the 4th octet is zero, as it is part of the network address, and thereby
reduces the number of useable host addresses.
Using the binary math required to find the subnet number and broadcast address really does help you
understand subnetting to some degree. To get the correct answers faster on the exam, you might want
to avoid all the conversion and binary math.
The following subnet cheat sheet will not solve all of the subnetting questions on the exam. What it will
do, however, is gain valuable time. Since you can't take anything into the exam, the trick is to write the
following chart out on your dry-erase board before you start the exam:
Hosts
Netmask
Number of Subnets
/30
4
255.255.255.252
64
/29
8
255.255.255.248
32
/28
16
255.255.255.240
16
/27
32
255.255.255.224
8
/26
64
255.255.255.192
4
/25
128
255.255.255.128
2
/24
256
255.255.255.0
1
/23
512
255.255.254.0
2
/22
1024
255.255.252.0
4
/21
2048
255.255.248.0
8
/20
4096
255.255.240.0
16
/19
8192
255.255.224.0
32
/18
16384
255.255.192.0
64
/17
32768
255.255.128.0
128
/16
65536
255.255.0.0
256
Doug’s tip, just memorize throug /21. You will find Microsoft uses the 255.255.255.0 where as
Cisco uses /24. Both mean the same thing.
And if you examine the chart very closely, you can actually reproduce it with very little memorization.
Here is the way to do it. First, duplicate column one of the table, which is fairly easy, and then fill in the
second column, which is nothing but multiples of two, starting out at four. (if you aren't good with
multiplication, when you get to the higher numbers, you can actually just write the numbers out to the
side twice and add them together to get the number for the next row).
Next, fill in the netmask for the /24 network and the /16 network, which should also be easy to remember
(if you are about to take the exam and can't remember /24 and /16 netmasks, you might as well hang it
up). The netmask for the /30 network is also fairly easy to remember, but if you forget any of the
netmasks, all that needs to be done is subtract the number of hosts directly to the left of it to get the next
netmask. For example, 255.255.255.252 provides for four hosts (two useable, because zero is the network
and .4 is the broadcast address). If you take 252 and subtract 4 from it, you get the netmask for the next
row, 255.255.255.248. If you take 255.255.255.248 and subtract its 8 hosts, you get 255.255.255.240,
which is the netmask for the next row. This works all the way down to the /24 network.
The only odd netmask to memorize is the /23 netmask, which is 255.255.254.0. That is also not hard to
remember because it is just one off from the .255 directly above it. Note that after the /23 network, all of
the network masks are identical to the /25 - /30 networks, just move them over one octet to the left. So
from /22 to /17, you already have the needed information, just fill it in.
For the fourth column, just put a 1 in the /24 network and put in multiples of two up the chart from the /24
network, as well as down the chart, to the /16 row.
As you can see, once you understand the table, it can be reproduced with very little memorization.