Please Help! faster SUBNETTING

jknight117

Hi folks,

I am having a hard time grasping the method involved with finding how many subnets within a given mask at a pace that would be fast enough to take the CCENT Exam. FYI I suck at math!

example:

How many subnets and hosts per subnet can you get from the network 172.31.0.0 255.255.252.0?

The process I take is 252 on the third octet is 6bits borrowed for subnets and 10 bits for Hosts.

11111111.11111111.11111100.00000000
nnnnnnnn.nnnnnnnn.nnnnnnhh.hhhhhhhh

leaving 1024-2 = 1022 Hosts

Then 6 bits would have my increment number of 4.

How would I determine how many subnets Without writing down the following 64 times.

1. 172.31.0.0 to 172.31.3.255
2. 172.31.4.0 to 172.31.7.255
3. 172.31.8.0 to 172.31.11.255
...
64. times!
There is no way anyone does this in the exam and passes.

Your secrets would be appreciated.

wallpaper_01

Hey,

Read this:

http://www.techexams.net/forums/ccna-ccent/38772-subnetting-made-easy.html

You will be subnetting off the top of your head within the next hour guaranteed if you get this down

devils_haircut

I think you might be sweating the CCENT a little too hard

The only thing I have committed to memory is that /24 = 255.255.255.0, or 256 available addresses (since you can have .0-.255).

In a /30 network, or 255.255.255.252 as in your example, I know that a /30 has 4 addresses. 4 goes into 256 (the number of unique values for the fourth octet) exactly 64 times, which means we have 64 subnets of 4 addresses each. Now since you wrote down a Class B network of 172.31.0.0, you would actually have 16,384 subnets because you are looking at BOTH the third and fourth octets, not just the fourth. So you would take your 64 subnets from the fourth octet and multiple that by 256, since your third octet can also be anything from .0-.255, which is 256 more values. So 256*64=16,384.

Each time you move up by a network number (/24 to /25, or /25 to /26) you decrease the number of addresses by half of the previous network number.

Example:
A /24 network has 256 addresses
/25 has 128 addresses
/26 has 64
/27 has 32

...and so on and so on. Hopefully I'm making sense. If not, I'll check back again when I'm home from work.

aftereffector

It's easy! Just write down a table like the one I attached here. I started at /24 and went down from there, but you can easily go up as well. If you memorize the /24 row, you can get the rest of them by just dividing or multiplying by 2. The picture explains it a lot better than I can:

xnx

The way i've done it is memorised 128, 192, 224, 240, 248, 252, 254, 255

.128 = increments of 128's
.192 = increments of 64's
.224 = increments of 32's
.240 = increments of 16's
.248 = increments of 8's
.252 = increments of 4's

Now you can apply this to any of the 4 octets:

e.g /18 = increments of 64 in 3rd octet
e.g /12 = incrememts of 16 in 2nd octet
e.g /25 = increments of 128 in 4th octet

jknight117

Thanks everyone, didn't think so much users would help me out with this. its great!

I don't mean to ignore anyones method but I believe xnx method helps me the most.

After a few questions on subnettingquestions.com its starting to make sense.

Again Thank you everyone!!

Xyro

255.255.252.0 = X.X.6.0 = 2 to the power of 6 = 64

(252 = 6 bits)
(Xs = default SM bits)