Subnetting Made Easy

Hi all,

I've received an email from one of your members asking me to post up my technique for subnetting as links to external blogs are not allowed due to forum rules. I know that he benefited from it and he wishes to help out others so here goes:

First of all I need you to get rid of all of the negative thoughts surrounding subnetting. Put down all of the books that you have read about the subject and navigate away from other sites claiming to provide an easy way to subnet. This technique requires no charts, just simply the know-how to work with the powers of 2.

We need to start with the fundamentals of IP addressing. An IP address is made up of 32 bits, split into 4 octets (oct = 8, yes?). Some bits are reserved for identifying the network and the other bits are left to identify the host.

There are 3 main classes of IP address that we are concerned with.

Class A Range 0 - 127 in the first octet (0 and 127 are reserved)
Class B Range 128 - 191 in the first octet
Class C Range 192 - 223 in the first octet

Below shows you how, for each class, the address is split in terms of network (N) and host (H) portions.

NNNNNNNN . HHHHHHHH . HHHHHHHH . HHHHHHHH Class A
NNNNNNNN . NNNNNNNN . HHHHHHHH . HHHHHHHH Class B
NNNNNNNN . NNNNNNNN . NNNNNNNN . HHHHHHHH Class C

At each dot I like to think that there is a boundary, therefore there are boundaries after bits 8, 16, 24, and 32. This is an important concept to remember.

We will now look at typical questions that you may see on subnetting. More often than not they ask what a host range is for a specific address or which subnet a certain address is located on. I shall run through examples of each, for each class of IP address.

What subnet does 192.168.12.78/29 belong to?

You may wonder where to begin. Well to start with let's find the next boundary of this address.

Our mask is a /29. The next boundary is 32. So 32 - 29 = 3. Now 2^3 = 8 which gives us our block size i.e. 2 to the power of 3 equals 8.

We have borrowed from the last octet as the 29th bit is in the last octet. We start from zero and count up in our block size. Therefore it follows that the subnets are:-

192.168.12.0
192.168.12.8
192.168.12.16
192.168.12.24
192.168.12.32
192.168.12.40
192.168.12.48
192.168.12.56
192.168.12.64
192.168.12.72
192.168.12.80
.............etc

Our address is 192.168.12.78 so it must sit on the 192.168.12.72 subnet.

What subnet does 172.16.116.4/19 sit on?

Our mask is /19 and our next boundary is 24. Therefore 24 - 19 = 5. The block size is 2^5 = 32.

We have borrowed into the third octet as bit 19 is in the third octet so we count up our block size in that octet. The subnets are:-

172.16.0.0
172.16.32.0
172.16.64.0
172.16.96.0
172.16.128.0
172.16.160.0
.............etc

Our address is 172.16.116.4 so it must sit on the 172.16.96.0 subnet. Easy eh?

What subnet does 10.34.67.234/12 sit on?

Our mask is 12. Our next boundary is 16. Therefore 16 - 12 = 4. 2^4 = 16 which gives us our block size.

We have borrowed from the second octet as bit 12 sits in the second octet so we count up the block size in that octet. The subnets are:-

10.0.0.0
10.16.0.0
10.32.0.0
10.48.0.0
.............etc

Our address is 10.34.67.234 which must sit on the 10.32.0.0 subnet.

Hopefully the penny is starting to drop and you are slapping the side of your head realising that you were a fool to think that subnetting was hard. We will now change the type of question so that we have to give a particular host range of a subnet.

What is the valid host range of of the 4th subnet of 192.168.10.0/28?

Easy as pie! The block size is 16 since 32 - 28 = 4 and 2^4 = 16. We need to count up in the block size in the last octet as bit 28 is in the last octet.

192.168.10.0
192.168.10.16
192.168.10.32
192.168.10.48
192.168.10.64
.................etc

Therefore the 4th subnet is 192.168.10.48 and the host range must be 192.168.10.49 to 192.168.10.62, remembering that the subnet and broadcast address cannot be used.

What is the valid host range of the 1st subnet of 172.16.0.0/17?

/17 tells us that the block size is 2^(24-17) = 2^7 = 128. We are borrowing in the 3rd octet as bit 17 is in the 3rd octet. Our subnets are:-

172.16.0.0
172.16.128.0

The first subnet is 172.16.0.0 and the valid host range is 172.16.0.1 to 172.16.127.254. You must remember not to include the subnet address (172.16.0.0) and the broadcast address (172.16.127.255).

What is the valid host range of the 7th subnet of address 10.0.0.0/14?

The block size is 4, from 16 - 14 = 2 then 22 = 4. We are borrowing in the second octet so count in the block size from 0 seven times to get the seventh subnet.

The seventh subnet is 10.24.0.0. Our valid host range must be 10.24.0.1 to 10.27.255.254 again remembering not to include our subnet (10.24.0.0) and the broadcast address (10.27.255.255).

What if they give me the subnet mask in dotted decimal?

If you're lucky and they give you a mask in dotted decimal format then you should have an even easier time. All you need again is your block size.

Let's say they have given a mask of 255.255.255.248 and you wish to know the block size. Here's the technique:

1. Starting from the left of the mask find which is the first octet to NOT have 255 in it.
2. Subtract the number in that octet from 256 to get your block size e.g. above it is 256 - 248 = block size of 8.
3. Count up from zero in your block size in the octet identified in step 1 as you have learned above (the example above would be in the last octet).

Another example is a mask of 255.255.192.0 - you would simply count up in 256 - 192 = 64 in the third octet.

One more example is 255.224.0.0 - block size is 256 - 224 = 32 in the second octet.

What now?

Now it's time to go and pick up those books again and go straight to the practice questions, completely by-passing any of their techniques. Use my method and you will be laughing!

Happy subnetting!

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Comments

hackmer

miller811 wrote: »

You always start with the largest to smallest.

Thank you for it. I did not know that before.

miller811

hackmer wrote: »

Thank you for it. I did not know that before.

that is the only way you can insure that addresses inside the space are only used once.

/25 = 192.168.1.0 - 192.168.1.127
/26 = 192.168.1.0 - 192.168.1.63
/27 = 192.168.1.0 - 192.168.1.31
/28 = 192.168.1.0 - 192.168.1.15

so if you used say a /28 first and then needed to use a /26 you could potentially reuse addresses or you would have to know to skip ahead 4 /28's to make the addressing correct. Skipping ahead 4 would waste 48 addresses....

Easiest and always advisable to do the largest to smallest and then you will have easily recongnizable address blocks in use, and available for use

binaryhat

I'm having difficulty with subnetting Class B addresses in Lammle's example. In chapter 3 examples 6B, 7B and 8B on Subnetting Practice Examples: Class B Addresses .

In exampl 6B you have a Class B subnet mask of /25. What I'm not clear on is why you get two subnets for each third octet.

In exampl 7B you have a Class B subnet mask of /26. What I'm not clear on is why you get four subnets for each third octet.

In exampl 8B you have a Class B subnet mask of /27. What I'm not clear on is why you get eight subnets for each third octet.

It seems that you are doubling the number of subnets for each third octet (class b address) as you change the subnet mask (starting at /25)

Am I making any sense?

instant000

binaryhat wrote: »

I'm having difficulty with subnetting Class B addresses in Lammle's example. In chapter 3 examples 6B, 7B and 8B on Subnetting Practice Examples: Class B Addresses .

In exampl 6B you have a Class B subnet mask of /25. What I'm not clear on is why you get two subnets for each third octet.

In exampl 7B you have a Class B subnet mask of /26. What I'm not clear on is why you get four subnets for each third octet.

In exampl 8B you have a Class B subnet mask of /27. What I'm not clear on is why you get eight subnets for each third octet.

It seems that you are doubling the number of subnets for each third octet (class b address) as you change the subnet mask (starting at /25)

Am I making any sense?

I'm not familiar with this Lammle's example that you speak of, but I do know that with each additional subnet bit you utilize, you get 2^(bit count) number of networks

Example: at /25, the interesting octet is the fourth octet, with the mask being 255.255.255.128
256-128 = 128
128 x 2 = 256 , there you go, (2 subnets)
or, mathematically speaking, 2^1 = 2

Go ahead and do /26
255.255.255.192
256-192 = 64
64 x 4 = 256, there you go, (4 subnets)
or, mathematically speaking, 2^2 = 4

/27
255.255.255.224
256-224 = 32
32 x 8 = 256, there you go, 8 subnets
or, mathematically speaking, 2^3 = 8

I hope this helps?

EDIT: I didn't explain, I'm only concerned with the 4th octet, as the first 3 octets are already all 1's, covered by /24 mask, so then I just figure out /25, /26, /27, etc. Hope that clarifies things.

miller811

binaryhat wrote: »

I'm having difficulty with subnetting Class B addresses in Lammle's example. In chapter 3 examples 6B, 7B and 8B on Subnetting Practice Examples: Class B Addresses .

In exampl 6B you have a Class B subnet mask of /25. What I'm not clear on is why you get two subnets for each third octet.

In exampl 7B you have a Class B subnet mask of /26. What I'm not clear on is why you get four subnets for each third octet.

In exampl 8B you have a Class B subnet mask of /27. What I'm not clear on is why you get eight subnets for each third octet.

It seems that you are doubling the number of subnets for each third octet (class b address) as you change the subnet mask (starting at /25)

Am I making any sense?

Always look for the interesting octet, the non 255 octet, that is where the subnetting occurs.

binaryhat

instant000 wrote: »

I'm not familiar with this Lammle's example that you speak of,

The examples are in Chapter 3, between pages 130-133.

instant000

binaryhat wrote: »

The examples are in Chapter 3, between pages 130-133.

Cool. This thing looks fun. Maybe I go ahead and read this one (I do have the book, one of those things you buy, and don't go through with ....)

binaryhat

Okay let me expand on my subnetting problem/question:

In example 6B:

172.16.0.0 = Network address
255.255.255.128 = Subnet mask (/25)
Subnets? 2^9 = 512.
Hosts? 2^7 – 2 = 126.
Valid subnets? 256 – 255 = 1. 0, 1, 2, 3, etc.

Now looking at the attachment, for every two subnets its 0.0 0.128 1.0 1.128....

I understand the valid subnets increase by 1 but I'm stuck on why every two subnets use the same number, 0.0, 0.128, 1.0, 1.128 etc? Is it because you are using an extra bit in the fouth octet? Hence 2^1=2 subnets...I'm confused....

subnets.jpg

instant000

binaryhat wrote: »

Okay let me expand on my subnetting problem/question:

In example 6B:

172.16.0.0 = Network address
255.255.255.128 = Subnet mask (/25)
Subnets? 2^9 = 512.
Hosts? 2^7 – 2 = 126.
Valid subnets? 256 – 255 = 1. 0, 1, 2, 3, etc.

Now looking at the attachment, for every two subnets its 0.0 0.128 1.0 1.128....

I understand the valid subnets increase by 1 but I'm stuck on why every two subnets use the same number, 0.0, 0.128, 1.0, 1.128 etc? Is it because you are using an extra bit in the fouth octet? Hence 2^1=2 subnets...I'm confused....

OK, in this same book (I'm reading it now, nice reading)

He goes through his 5 questions of subnetting, and once I review these, the answer will become more clear:

1. How many subnets?
2. How many hosts per subnet?
3. What are the valid subnets?
4. What is the broadcast address?
5. What are the valid hosts?

One thing I like to additionally include is this: What is the block size? Which is like question 1b for me.

,

1. how many subnets? = 2^9 = 512
1b. What is the block size? 2^7 = 128
2. How many hosts per subnet? 2^7-2 = 126
3. What are the valid subnets? Add 128, in the interesting octet, over and over again

ALWAYS start with 0, and increment according to the block size.
So, here you go:
172.16.0.0
172.16.0.128
172.16.1.0 (why? 128+128 = 256, which equals an entire octet)
172.16.1.128
172.16.2.0
... (output omitted)
172.16.254.0
172.16.254.128
172.16.255.0
172.16.255.128

Do the math here, and add up the number of networks you have, they should total 512 (which matches your requirement, of 2^9 power)

4. What is the broadcast address? It's the last address before the next network begins, so you're looking at this:

172.16.0.127
172.16.0.255
172.16.1.127
172.16.1.255
... (output omitted)
172.16.254.127
172.16.254.255
172.16.255.127
172.16.255.255

5. What are the valid hosts?
172.16.0.1 - 172.16.0.126
172.16.0.129-172.16.0.254
172.16.1.1 - 172.16.1.126
172.16.1.129 - 172.16.1.254
... (output omitted)
172.16.254.1 - 172.16.254.126
172.16.254.129 - 172.16.254.254
172.16.255.1 - 172.16.255.126
172.16.255.129 - 172.16.255.254

I hope that this explanation helps you. It's the only way to give you 512 subnets, so it's kinda intuitive, if you think about it that way?

I see that he gave some explanation about incrementing by 1, and also incrementing by .128, and it didn't make sense to me. I can plainly see that adding .128 (to a class

gives me that. Of course, if it was a class C, then you'd just have 2^1 = 2 subnets, and you'd do two increments of 128, and be done already

. I think noting that the address is a Class B is the important thing here, and that is the difference you are seeing in the subnetting result.

If you look through the text, he specifically mentions a reference to doing the Class C example first, in order to help you understand this example.

instant000

binaryhat wrote: »

I'm having difficulty with subnetting Class B addresses in Lammle's example. In chapter 3 examples 6B, 7B and 8B on Subnetting Practice Examples: Class B Addresses .

In exampl 6B you have a Class B subnet mask of /25. What I'm not clear on is why you get two subnets for each third octet.

In exampl 7B you have a Class B subnet mask of /26. What I'm not clear on is why you get four subnets for each third octet.

In exampl 8B you have a Class B subnet mask of /27. What I'm not clear on is why you get eight subnets for each third octet.

It seems that you are doubling the number of subnets for each third octet (class b address) as you change the subnet mask (starting at /25)

Am I making any sense?

Yes, you are making sense.

Of course you are doubling the number of subnets for each third octet

Class B address
Mask - Number of subnets
/24 = 2^8 (256)
/25 = 2^9 (512)
/26 = 2^10 (1024)
/27 = 2^11 (204

Please compare this to the powers of 2 table, on pages 114-115.

You're SUPPOSED to get twice as many networks, as you increase your subnet bit by one each, as you're going up a power of 2, and each time you multiply something by two, you are doubling it.

Also, saying you have a problem with the #6B, #7B, and #8B only makes sense if you also have a problem with the #5B, #1C, #2C, and #3C

I hope this helps. If not, post again, I'm really enjoying this book

instant000

To really get into nitty gritty, I must expose ... binary.
I didn't really find an example in Lammle's book to explain it this way (it could be in there, if so, forgive me, and just take this lesson, he does show binary conversion in his book, and this is the true foundation of subnetting)

... don't fret, it's all cool

Here you go.

172.16.0.0 /25 mask
You know that you have 9 bits of subnetting.
This means that you have 2^9 possible values in your subnet bits.

The "subnet bits" are the bits that you can play with.
172.16.0.0 expressed in binary:
1010 1100.0001 0000.0000 0000.0000 0000
255.255.255.128 expressed in binary:
1111 1111.1111 1111.1111 1111.1000 0000
Note, you have 9 bits of subnetting, let me highlight those here:
1111 1111.1111 1111.1111 1111.1000 0000
What this really means here (if you did not know) is that you can use all 9 of these bits, to create networks with. Most subnetting problems only go across one octet (subnetting a class C using 4th octet is most common) so people fail to catch on to this concept, using "magic number" shortcuts.
But, now that you truly know what this means, you know that you can use ALL of these bit positions, to calculate values with
So, let us concentrate just on these bit values we can manipulate, since there are 2^9 positions, you know plainly that you have 512 subnets to play with. That's all well, fine, and dandy, but by knowing that you actually get to use all of your subnet bits, you really know what's going on here, and I will demonstrate:
First, look at this:
I'm only going to highlight the subnet bits, to try to illustrate the point:
1010 1100.0001 0000.0000 0000.0000 0000 (this would represent the very first subnet (172.16.0.0) NOTE: This is the all zero's subnet.
1010 1100.0001 0000.0000 0000.1000 0000 (This would represent the very next subnet (172.16.0.128
1010 1100.0001 0000.0000 0001.0000 0000 (This would represent the very next subnet (172.16.1.0)
1010 1100.0001 0000.0000 0001.1000 0000 (This would represent the very next subnet (172.16.1.128
1010 1100.0001 0000.0000 0010.0000 0000 (This would represent the very next subnet (172.16.2.0)
1010 1100.0001 0000.0000 0010.1000 0000 (This would represent the very next subnet (172.16.2.128
1010 1100.0001 0000.0000 0011.0000 0000 (This would represent the very next subnet (172.16.3.0)
1010 1100.0001 0000.0000 0011.1000 0000 (This would represent the very next subnet (172.16.3.128
1010 1100.0001 0000.0000 0100.0000 0000 (This would represent the very next subnet (172.16.4.0)
1010 1100.0001 0000.0000 0100.1000 0000 (This would represent the very next subnet (172.16.4.128
(output omitted)
1010 1100.0001 0000.1111 1101.0000 0000 (This would represent the very next subnet (172.16.253.0)
1010 1100.0001 0000.1111 1101.1000 0000 (This would represent the very next subnet (172.16.253.128
1010 1100.0001 0000.1111 1110.0000 0000 (This would represent the very next subnet (172.16.254.0)
1010 1100.0001 0000.1111 1110.1000 0000 (This would represent the very next subnet (172.16.254.128
1010 1100.0001 0000.1111 1111.0000 0000 (This would represent the very next subnet (172.16.255.0)
1010 1100.0001 0000.1111 1111.1000 0000 (This would represent the very last subnet (172.16.255.128 NOTE: This is the all one's subnet.

I think you get the point by now ...

Note: Now that you see it in binary, you can plainly see what is meant by the "all 0's subnet" and the "all 1's subnet", that "ip subnet-zero" is required for. NOTE: ip subnet-zero is enabled by default in Cisco routers for past few years. Back in the day, if you would utilize an all zero's or all one's network, some of your equipment might not know how to interpret it.

I hope this helps?

Have a nice evening!

binaryhat

A-ha! I see it now with the block size and # of subnets.

When counting up in block size for by subnet you can't go over 256.

172.16.0.0 = Network address
255.255.255.128 = Subnet mask (/25)

Subnets = 2^1=2 because you are borrowing an extra bit.

172.16.0.0
172.16.0.128
172.16.1.0
172.16.1.128
....
.........

instant000

binaryhat wrote: »

A-ha! I see it now with the block size and # of subnets.

When counting up in block size for by subnet you can't go over 256.

172.16.0.0 = Network address
255.255.255.128 = Subnet mask (/25)

Subnets = 2^1=2 because you are borrowing an extra bit.

172.16.0.0
172.16.0.128
172.16.1.0
172.16.1.128
....
.........

Huh?

the visual binary example explained it the best.

to get the number of subnets ....

class b address, default mask is /16
the given mask is a /25
This means you have 9 bits of subnetting

2^9 = 512 ... which means that you will have 512 subnets. I have no idea of what you are talking about with 2?

binaryhat

Yes, you get 2^9=512 subnets but every two subnets are the same, hence 2^1=2(extra bit in the 4th octet):

0.0, 0.128, 1.0, 1.128, 2.0, 2.128, 3.0, 3.128....

You actually get two subnets for each third octet value, hence
the 512 subnets. For example, if the third octet is showing subnet 3, the two subnets would actually be 3.0 and 3.128.

I see that he gave some explanation about incrementing by 1, and also incrementing by .128, and it didn't make sense to me. I can plainly see that adding .128 (to a class gives me that. Of course, if it was a class C, then you'd just have 2^1 = 2 subnets, and you'd do two increments of 128, and be done already . I think noting that the address is a Class B is the important thing here, and that is the difference you are seeing in the subnetting result.

In the next example 7B:

every four subnets are the same, hence 2^2=4(2 extra bits in the 4th octet):

0.0, 0.64, 0.128, 0.192, 1.0, 1.64, 1.128, 1.192 (1024 total subnets)

I'm just looking at the patterns and extra bit(s) in the 4th octet when using a class b subnet...

instant000

binaryhat wrote: »

Yes, you get 2^9=512 subnets but every two subnets are the same, hence 2^1=2(extra bit in the 4th octet):

0.0, 0.128, 1.0, 1.128, 2.0, 2.128, 3.0, 3.128....

In the next example 7B:

every four subnets are the same, hence 2^2=4(2 extra bits in the 4th octet):

0.0, 0.64, 0.128, 0.192, 1.0, 1.64, 1.128, 1.192 (1024 total subnets)

I'm just looking at the patterns and extra bit(s) in the 4th octet when using a class b subnet...

OK.

I'm sorry, you appear to be cool, the basic premise is that the number of subnets doubles with each additional masked bit

I'm still in this book, I'm at Chapter 5 now. Not sure if I'll complete it before June, just trying to absorb the material really good. Fortunately, nothing is particularly challenging (yet) LOL.

reppgoa

sorry for the necro, but I just want to express my gratitude for this post. It helped me get subnetting in about 10 minutes

kamay44

Thank you, i completly undertood everything, but now i have this problem.

Question: How many subnets and hosts per subnet can you get from the network 172.29.0.0/25?

Answer: 512 subnets and 126 hosts

Why??? My aswers was 2 subnets and 126 host.

The mask is /25, so we are working in the fourth octect right?? Why do they move to the third one??

j-man

kamay44 wrote: »

Thank you, i completly undertood everything, but now i have this problem.

Question: How many subnets and hosts per subnet can you get from the network 172.29.0.0/25?

Answer: 512 subnets and 126 hosts

Why??? My aswers was 2 subnets and 126 host.

The mask is /25, so we are working in the fourth octect right?? Why do they move to the third one??

Then you haven't completely understood everything in the prior posts.

First rule about subnetting is to find the network portion. Remember your network classes: A, B and C

This is a class B address which means the first two octets are network and the rest are hosts.

Straight away a class B subnet mask will look like

255.255.0.0
or
11111111.11111111.00000000.00000000

Now look at your /25. This tells you 25 bits are used in the mask. In this case you ignore the first 2 octets because it is a class B network which means your subnet mask will now look like

255.255.255.128
or /25 notation (note the bold)
11111111.11111111.11111111.10000000

Count the number of bold to 2^n

2^9 = 512 subnets

everything to the right of the last bold bit will be your hosts minus 2

126 hosts

If you take anything away from this post, please remember that you should first identify the network class (A, B, or C). Those bits will never be counted in figuring out subnets.

COYS!

Thanks for the thread! It'll take at a few minutes off my CCNA time I reckon.

howiehandles

Dude, I swear that I think I love you..as a guy. lol I thought Jeremy explained it well. I've been pounded my brain on this stuff, math isn't my strong suit, and I picked it up well in about 5 minutes. Granted, I need to practice a bunch before I'm confident to test, but you freaking ROCK!

Mason83

This is an absolutely incredible thread and many thanks to the TS for taking the time to type this up. I do have a question and I've get stuck every time I see one of these:

Question: How many subnets are available with the network 172.16.0.0/26?

Answer: 1024

How do I get the answer?

NewManSoon

j-man wrote: »

Then you haven't completely understood everything in the prior posts.

First rule about subnetting is to find the network portion. Remember your network classes: A, B and C

This is a class B address which means the first two octets are network and the rest are hosts.

Straight away a class B subnet mask will look like

255.255.0.0
or
11111111.11111111.00000000.00000000

Now look at your /25. This tells you 25 bits are used in the mask. In this case you ignore the first 2 octets because it is a class B network which means your subnet mask will now look like

255.255.255.128
or /25 notation (note the bold)
11111111.11111111.11111111.10000000

Count the number of bold to 2^n

2^9 = 512 subnets

everything to the right of the last bold bit will be your hosts minus 2

126 hosts

If you take anything away from this post, please remember that you should first identify the network class (A, B, or C). Those bits will never be counted in figuring out subnets.

This post just cleared up so much confusion. Thank you!

cb3dwa

Mason83 wrote: »

This is an absolutely incredible thread and many thanks to the TS for taking the time to type this up. I do have a question and I've get stuck every time I see one of these:

Question: How many subnets are available with the network 172.16.0.0/26?

Answer: 1024

How do I get the answer?

because 2^10 = 1026 subsets and 62 hosts

11111111.11111111.11111111.11000000

chaddsuk

I did a quick youtube video on the subnetting technique I use for exams which I found invaluable at the time (and still continue to use it). Of course I took the time to learn the "and or" binary method first however check out the video if you feel you need to speed things up. (Hope it helps someone)

Subnetting Made Easy - CCNA exam method - Magic number - YouTube

Mason83

cb3dwa wrote: »

because 2^10 = 1026 subsets and 62 hosts

11111111.11111111.11111111.11000000

And right after I typed that question I figured it out

Thank you for replying.

Mike-Mike

The first post in this thread has saved me from thinking I'm stupid, thank you soooooooooo much

Moki99

Okay, I think I'm getting close. The one thing I don't get is how to do some of these math problems quickly on the exam.

For Example:
Question: How many subnets and hosts per subnet can you get from the network 172.25.0.0 255.255.240.0?

2^4 = subnets
2^12 - 2 = Hosts

How am I suppose to do 2^12 during the test with just a pen and paper? It would take forever. If theres a shortcut that everyone uses, I would appreciate it!

miller811

Moki99 wrote: »

Okay, I think I'm getting close. The one thing I don't get is how to do some of these math problems quickly on the exam.

For Example:
Question: How many subnets and hosts per subnet can you get from the network 172.25.0.0 255.255.240.0?

2^4 = subnets
2^12 - 2 = Hosts

How am I suppose to do 2^12 during the test with just a pen and paper? It would take forever. If theres a shortcut that everyone uses, I would appreciate it!

just memorize the powers of 2

2 ^1 = 2
2 ^2 = 4
2 ^3 = 8
2 ^4 = 16
2 ^5 = 32
2 ^6 = 64
2 ^7= 128
2 ^8 = 256
2 ^9 = 512
2 ^ 10 = 1024
2 ^ 11 = 2048
2 ^ 12 = 4096

easy to memorize the first 8, then just double to the number 4 more times in this case

or 2 ^ 10 = 1024 and double two more times.

cyberguypr

miller811 wrote: »

just memorize the powers of 2

easy to memorize the first 8, then just double to the number 4 more times in this case

or 2 ^ 10 = 1024 and double two more times.

Exactly. I memorized the numbers that kind of repeat:

2 ^6 = 64
2 ^ 10 = 1024

From there on just double or half as required.

simonmoon

Moki99 wrote: »

How am I suppose to do 2^12 during the test with just a pen and paper? It would take forever. If theres a shortcut that everyone uses, I would appreciate it!

I've been told you are given 15 minutes to read the EULA before taking the test. Use that 15 minutes to do what math you need to write out a full subnetting chart.

But as for how I would do that quick in my head, well, it's easy to remember /24 is Class C and 256-2 hosts. .240.0 is /20, so I would double from /24 to /20.

/24 = 256-2
/23 = 512-2
/22 = 1024-2
/21 = 2048-2
/20 = 4096-2