If you are given x number of subnet(or remote locations) to create and you need to determine a new subnet mask do the following steps:
1. Find out what your default subnet mask is, if not given. This will let you know how many remaining host bits you have to “borrow from”. For example, if you were given a network ID of 157.3.0.0, this would be a class B address. We know this because the first octet falls in the range of 129-191. Therefore in this example, we will start with a default subnet mask of 255.255.0.0. This will leave us with two octets, or 16 bits to “borrow” from for subnetting.
2. Solve the equation for n, 2n-2 is greater than or equal to the number of networks you desire to gain. For example, if you needed 17 subnetworks… 2n-2 > or = 17. Therefore n=5. 25-2=30. 4 would not work, because 24-2=14. The “n” is equal to the number of bits you will borrow going from left to right from the host bits.
3. Determine the new subnet mask. Since “n” above equals 5, we’ll write it as 11111 Now add enough 0’s to the 1’s to complete the octet or 8 complete bits. Therefore, we will write 11111 as 11111000.
4. Now convert this number to decimal and add to your default mask you started with. 11111000=248, so our new subnet mask is 255.255.248.0
To determine the total number of hosts per subnet, we use the same formula as above, but in this case the “n” = the number of bits leftover for the host ID’s. From the example above we borrowed 5 bits from a possible 16 bits (starting with a default mask of 255.255.0.0 leaves two octets or 16 bits to “borrow”from) Therefore 16-5 = 11 bits for the “n”. 211 = 2048. Therefore, in the above example, we would have 2046 (2n – 2) or (211-2) hosts for each and every subnet.
1 1 1 1 1 1 1 1
128 * 64 * 32 * 16 * 8 * 4 * 2 * 1
If you simply memorize the multiples above, you can easily convert a binary to decimal and vice versa. You don't need no stinkin' calculator!
Use the following chart below to use for your shortcut formula to creating a subnetted mask
128 1 0 0 0 0 0 0 0
192 1 1 0 0 0 0 0 0
224 1 1 1 0 0 0 0 0
240 1 1 1 1 0 0 0 0
248 1 1 1 1 1 0 0 0
252 1 1 1 1 1 1 0 0
254 1 1 1 1 1 1 1 0
255 1 1 1 1 1 1 1 1
|
REMEMBER:
2n-2
Use to create # of Subnetworks needed Where n= number of bits borrowed from the total host bits available,
going from left to right Use to determine # of HOSTS per network after subnetting,
or before Where n= number of TOTAL host bits leftover after borrowing for for new subnetworks |
You have purchased a Class C block of addresses from Internic. The Network ID is 194.1.1.0. You need to segment the network into 6 networks.
a. How many hosts did you originally have? 28-2 = 254
b. What is your new subnet mask? 255.255.255.224
Steps are as follows 1. Solve for 2n-2 > 6. 23 - 2 = 6 therefore there are 3 new network bits being borrowed from the last Octect.
2. New mask = 255.255.255. 11100000
3. Convert above to Decimal and add to default mask = 255.255.255.224
c. How many hosts will there now
be per network?
Note
in Step b. there
are 5 host bits represented by the blue 0’s
1. Solve for 25 -2 = 30. You gain more networks but actually lose hosts!
d. Write down the address ranges for all possible new networks
XXX = subnetted portion ABCD= host
portion
1. 194.1.1. 33 - 194.1.1. 62
194.1.1.00100001 - 194.1.1.00111110
2. 194.1.1. 65 - 194.1.1. 94
194.1.1.01000001 - 194.1.1.01011110
3. 194.1.1. 97 - 194.1.1. 126
194.1.1.01100001 - 194.1.1.01111110
4. 194.1.1. 129 - 194.1.1. 158
194.1.1.10000001 - 194.1.1.10011110
5. 194.1.1. 161 - 194.1.1. 190
194.1.1.10100001 - 194.1.1.10111110
6. 194.1.1. 193 - 194.1.1. 222
194.1.1.11000001 - 194.1.1.11011110
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