You must understand and remember this table well in order to master the
shortcut.
Note: You must
borrow at least 2 bits and must leave at least 2 bits
The 'Subnet Table'
| Bits
Borrowed (N) |
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
| Bit
Value |
128
|
64
|
32
|
16
|
8
|
4
|
2
|
1
|
| Subnet
Mask |
128
|
192
|
224
|
240
|
248
|
252
|
254
|
255
|
Number
of Subnets
((2^N)-2) |
0
|
2
|
6
|
14
|
30
|
62
|
126
|
254
|
Stop: Before continuing, please read this very important article regarding the use of the ((2^N)-2) method of subnetting.
If using the (2^N) method as defined in RFC 1878, your table would look like this instead.
| Bits
Borrowed (N) |
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
| Bit
Value |
128
|
64
|
32
|
16
|
8
|
4
|
2
|
1
|
| Subnet
Mask |
128
|
192
|
224
|
240
|
248
|
252
|
254
|
255
|
Number
of Subnets
(2^N) |
2
|
4
|
8
|
16
|
32
|
64
|
128
|
256
|
The 'subnet table'
is commonly seen in lecture notes or certification guides but what the
author/lecturer did not tell you is how to derive this table on the fly. Its
actually quite simple, lets look at it line by line
1) Bits borrowed,
this is the easy one, just remember that the table consists of only 8
columns.
2) Bits Value,
remember by heart that the first value starts with 128 and the subsequent
values are divided by two.
3) Subnet Mask,
this line tells you what the subnet mask would be, to get the figures,
add up the corresponding bits value and all of the values prior to it.
128 + 0 (there is
no prior value) =128
128 + 64 = 192
192 + 32 = 224
224 + 16 = 240
240 + 8 = 248
248 + 4 = 252
252 + 2 + 254
254 + 1 + 255
4) Number of Subnets,
tells you how many subnet you'll get if you use the subnet mask. Just look at the corresponding N value at the
top and you can derive the figures.
Once you understand
how to derive the 'subnet table', spend some time practicing. I would
advise you to draw out the table once you are in the exam room (before
starting the actual exam) it will take you less than a minute.
How to tackle the
questions
There are only a few different ways that Microsoft or Cisco can phrase their
questions, lets take a look at some examples,
Question Type 1:
If you are to determine the subnet mask based on a number of hosts and
an IP address
Example:
You are assigned an IP address of 172.30.0.0 and you need 1000 hosts
on your network, what is your subnet mask.
Step one: Determine
the number of bits needed for the hosts.
In this scenario, we need ten bits as 2^10 = 1024 (the question asks
for 1000 hosts only)
Step two: Determine
the number of bits left for the subnet.
32 - (number of bits needed for the host) which is 32-10 = 22 bits
Step three: Determine
the number of bits actually borrowed.
We take the number of bits left for the subnet and minus as many 8s
as possible as each 8 represents 1 octal. Therefore 22 - 8 - 8 = 6 bits
were borrowed
| Bits
Borrowed (N) |
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
| Bits
Value |
128
|
64
|
32
|
16
|
8
|
4
|
2
|
1
|
| Subnet
Mask |
128
|
192
|
224
|
240
|
248
|
252
|
254
|
255
|
Number
of Subnets
(2N-2) |
0
|
2
|
6
|
14
|
30
|
62
|
126
|
254
|
With reference to
the subnet table, 6 bits would have a subnet of 255.255.252.0 . Take note
that a total of two 8s were subtracted off, therefore the first two octal
would be 255.255.x.x and the 3rd octal was 6 bits borrowed which leaves
with 255.255.252.x.
Simple?
Question Type 2:
If you were given an IP address of 172.30.0.0 and you need 15 subnets
| Bits
Borrowed (N) |
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
| Bits
Value |
128
|
64
|
32
|
16
|
8
|
4
|
2
|
1
|
| Subnet
Mask |
128
|
192
|
224
|
240
|
248
|
252
|
254
|
255
|
Number
of Subnets
(2N-2) |
0
|
2
|
6
|
14
|
30
|
62
|
126
|
254
|
With reference to
the subnet table, the subnet mask should be 255.255.248.0. 172.30.0.0
is a Class B address and the subnet should be 255.255.0.0.
Question Type 3:
You are assigned an IP address of 172.30.0.0 and you need 55 subnets,
how many hosts do you have per subnet?
Step One: Determine
the number of bits used for the subnet.
| Bits
Borrowed (N) |
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
| Bits
Value |
128
|
64
|
32
|
16
|
8
|
4
|
2
|
1
|
| Subnet
Mask |
128
|
192
|
224
|
240
|
248
|
252
|
254
|
255
|
Number
of Subnets
(2N-2) |
0
|
2
|
6
|
14
|
30
|
62
|
126
|
254
|
According to the chart,
the closest match to 55 subnet would be 62 and therefore, the number of
bits borrowed for the subnet is 6. Since 172.30.0.0 is a Class B, we would
need to add another 16 bits to the 6 making it 22 bits in total.
Step Two: Determine
the number of bits used for the host.
Number of bits used
for the hosts is 32 - (number of bits used for the subnet) which is 22
= 10 bits.
2^10-2 = 1022, therefore
there are a total of 1022 usable hosts in each subnet.
The key to mastering this shortcut is the same as with any other mathematical
question - pratice.
Good luck on your next exam
By Adam Chee W.S.
(MCP ID 2915249)
http://www.adamchee.cjb.net
