Power of 2 - Subnetting again!

kevin31

Hi All,

Sorry I must be really thick I just dont get it can someone explain the above please?

Maths was never a strong point of mine!

thanks

dynamik

That's really not that difficult, even if you're bad at math. All you have to do is put 2 to the power of how many bits you have available to determine how many subnets/hosts you can use.

2^(number of bits)

2^1 = 2

2^5 = 32

2^10 = 1024

To be completely honest, it's much easier to simply memorize the values than it is to do the calculation. Realistically, you only need to know up to 10 or 12 off the top of your head. No one is going to expect you to know how many hosts you can have with 23 bits. Also, one bit less is half of the value while one bit more is double the value. So if you know that 5 bits is 32, you also know that 4 is 16 and 6 is 64. Even if you just memorize 5 is 32 and 10 is 1024, you can quickly double or halve your way to whatever number you're working with.

kevin31

Thanks for your help

How do you get 32 from 2^5? iam very thick clearly cause it just dont make sense!

cisco_trooper

kevin31 wrote:

Thanks for your help

How do you get 32 from 2^5? iam very thick clearly cause it just dont make sense!

This is 2 to the fifth power, ie... 2 * 2 * 2 * 2 * 2.

ConstantlyLearning

kevin31 wrote:

Thanks for your help

How do you get 32 from 2^5? iam very thick clearly cause it just dont make sense!

You just multiple 2 by itself 5 times.

2*2*2*2*2

You use '2' because there are 2 values in binary. 0 & 1

Every time you multiply by 2 you double the previous number, divide by two you have half the previous number.

For the larger numbers I just remember that 2 ^ 8 is 256 and work from there.


Ex. You're given a class B network address with a mask of /25 (255.255.255.12, you have to work out how many subnets available.

Straight away I know the network portion of the classful address is /16. (Class
I take 16 from 25 to get 9 network bits.
I know that 8 bits is 256.
i double that once to get the 9 network bits. Which gives you 512.

512 subnets.

kryolla

just remember it doubles

2^0=1
2^1= 2
2^2 = 4
2^3 = 8
2^4= 16
2^5 = 32
2^6 = 64
2^7 = 128

wbosher

Take a look at kryollas post and look at the numbers 1 2 4 8 16 32 64 128 256...see a pattern? Looks like a binary conversion chart, that's how I think of it and it makes it real easy.

Kaminsky

ConstantlyLearning wrote:

kevin31 wrote:

Thanks for your help

How do you get 32 from 2^5? iam very thick clearly cause it just dont make sense!

You just multiple 2 by itself 5 times.

2*2*2*2*2

You use '2' because there are 2 values in binary. 0 & 1

Every time you multiply by 2 you double the previous number, divide by two you have half the previous number.

For the larger numbers I just remember that 2 ^ 8 is 256 and work from there.


Ex. You're given a class B network address with a mask of /25 (255.255.255.12, you have to work out how many subnets available.

Straight away I know the network portion of the classful address is /16. (Class
I take 16 from 25 to get 9 network bits.
I know that 8 bits is 256.
i double that once to get the 9 network bits. Which gives you 512.

512 subnets.

The number 2 because voltage is either on or off. You really need to jump into the binary. Have a good go over Hex for good measure while your there. These things you musht be able to think in.

worldmac1

One quick question regarding this....Isn't the equation 2^2-2? so with a mask of 240, instead of having 16 subnets or hosts you would actually have 14? I am taking my ICND1 this morning and started having a panic attack. It looks like on subnettingquestions.com this rule is only in effect for finding the amount of hosts and not subnets.

Any help this morning with regards to the way Cisco wants the answer would be a big help!!

Thanks!

mattrgee

When calculating useable hosts you take the amount and minus 2. 1 is for the network number the other is the broadcast address.

So, 172.16.1.1 / 255.255.255.240 gives 12 subnet bits (because its class which is 12^2 = 4096 networks and 4 host bits which is 16 addresses but -2 one for the network address and one for the broadcast which leaves 14 useable addresses.

Good luck man! You'll b

worldmac1

Thanks for the response Mattrgee!!!

That's exactly how I learned it...guess I needed some reinsurance for some pre exam jitters. Thanks again for your help!

High Jitters and High Bandwidth right now...=)

/usr

Is that William Shatner in your profile pic? It looks like him.

Good luck!

/usr

I never "memorized" the powers of 2 for my ICND1 exam. Although I pretty much knew them all, I still couldn't instantaneously rattle off 2^11 if you asked me.

So, I made a chart that helps me a lot.


1 2 3 4 5 6 7 8 9 10
2
2 4 8 16 32 64 128 256 512 1024

Obviously the 2 on the left is the base number. The top row is the power, the bottom is the value of the base number to the power above it.It's hard to get it lined up properly on here.

A lot of folks on the boards (I asked the question) will tell you that you need to have them memorized for the exam. I used the method above on the exam and I did fine. You should use whatever works best for you.

miller811

/usr wrote:

Is that William Shatner in your profile pic? It looks like him.

Good luck!

You must not be a golf fan.....

that would a picture of a drunken (as usual) John Daly.

mattrgee

How did you do?