A quick subnet method
On your Brain dump sheet make a table showing the decimal
values of the first 8 binary positions.
An easy way to do this is start at the right side with 1 (one) and
each position to the left is double the previous position, like this:
128 64 32 16 8 4 2 1
Then starting at the leftmost column write 128 under the 128, then
add each previous column, one at a time, and place the result
under the number you added on. Your table looks like this:
128+64=192,192+32=224,224+16=240, etc
128 64 32 16 8 4 2 1
128 192 224 240 248 252 254 255
Now, to get the subnet mask, you take the number of subnets
needed and using the TOP ROW starting from the left find the first
number that will divide INTO your number of subnets. (It doesn't
have to be an even division).
Then count the number of columns back to one (towards the
right). Move to the leftmost column (128) and count THAT
SAME NUMBER of columns to the right again. The number in
the second row of THAT column is your subnet mask.
An example:
Say you need 18 subnets. The first number, starting from the left,
that will go into 18 is 16;
X
128 64 32 16 8 4 2 1
128 192 224 240 248 252 254 255
Count the columns to the right, down to one; (five columns)
X(1) (2) (3) (4) (5)
128 64 32 16 8 4 2 1
128 192 224 240 248 252 254 255
Now go back to 128, and count five columns to the right;
(1) (2) (3) (4) (5)
128 64 32 16 8 4 2 1
128 192 224 240 248 252 254 255
X
Put an X in that column.
248 is your subnet mask!
A class A address subnet becomes 255.248.0.0
A class B address subnet becomes 255.255.248.0
And a class C address subnet is 255.255.255.248
Now, the number of hosts per subnet is easy also.
In this example you used 5 bits for your subnet mask (five
columns), so you have three bits left over in that octet. Add 3 and
8 times the number of octets to the right. Take 2 to THAT power,
subtract 2, and you have the number of hosts per subnet.
Using the example above for a class b address; 3+8(1 octet)=11.
Take 2 to the 11th power gives 2048.
Subtract 2 (for all zeros and all ones); 2048-2 = 2046 The
number of hosts per subnet.
Another example;
This time lets use a class C address of 195.13.24.0. And you
need 8 subnets. We will also figure the host address ranges.
The first number from the left that will go into 8 is 8;
X(1) (2) (3) (4)
128 64 32 16 8 4 2 1
128 192 224 240 248 252 254 255
X
(1) (2) (3) (4)
Count the columns down to one (4 columns), count down four
from 128, giving 240 as the subnet mask.
So our class C subnet mask for 8 subnets is 255.255.255.248
Number of hosts; 4 bits left over+8*zero(no octets)=4, 2 to the
4th=16,16-2=14 hosts per subnet.
Now to get the host IP addresses you start with the decimal
number in row one and keep adding it to itself till you get to the
mask number;
16, 16+16=32, 32+16=48, 48+16=64, 64+16=80, 80+16=96,
96+16=112, 112+16=128, 128+16=144,
144+16=160, 160+16=178, 178+16=192, 192+16=208,
208+16=224, 224+16=240.
We can't use 240 as a host ID.
These numbers give your STARTING host subnet addresses.
So our host addressing ranges are;
195.13.24.16 - 195.13.24.31
195.13.24.32 - 195.13.24.47
195.13.24.48 - 195.13.24.63
195.13.24.64 - 195.13.24.79
195.13.24.80 - 195.13.24.95
195.13.24.96 - 195.13.24.111
195.13.24.112 - 195.13.24.127
195.13.24.128 - 195.13.24.143
195.13.24.144 - 195.13.24.159
195.13.24.160 - 195.13.24.177
195.13.24.178 - 195.13.24.191
195.13.24.192 - 195.13.24.207
195.13.24.208 - 195.13.24.223
195.13.24.224 - 195.13.24.239
Now, we only needed 8 subnets. That's up to 195.13.24.128.
The other six can be counted on for GROWTH!
Now a class A example. InterNIC address 101.0.0.0. Subnets
needed 5. The table on the braindump sheet would look like this.
4 goes into 5, then count three columns to the right.
128 64 32 16 8 4 2 1
128 192 224 240 248 252 254 255
X
Then count three columns right from 128. Subnet mask is
255.224.0.0.
Host address ranges are ;
101.32.1.1 - 101.63.254.254
101.64.1.1 - 101.95.254.254
101.96.1.1 - 101.127.254.254
101.128.1.1 - 101.159.254.254
101.160.1.1 - 101.191.254.254
101.192.1.1 - 101.223.254.254 (one for growth)
Five subnets with 5+8(2)=21, 2 to the 21st=2,097,152, subtract
2;
2,097,152-2=2,097,150 hosts per subnet. With another subnet
for growth.