What notes do you write down when subnetting?

When I'm doing subnet problems on subnettingquestions.com this is what I write down to help me work faster.

128 192 224 240 248 252 254 255

1-126 127-191 192-223

8 16 24 32

128 64 32 16 8 4 2 1

Any suggestions on what else I can add?

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Comments

andy4tech

I write down binary bits 128....64....32....16......8......4......2......1
128..192....224...240..248....252...254...255

SurferdudeHB

is there an efficient way to calculate multiples of 16? Beside writing them down.

hiddenknight821

I'm going to share my secret with you, but you can't share this with anyone without my permission.

The magic number is actually 256.

If you know that you will have 4 subnets, but wonder how many hosts (usable and non-usable) per subnet, all you have to do is:

256 divided by 4 is 64

256 divided by any of the following numbers: (128, 64, 32, 16, 8, 4, 2) will get you what you are looking for. Let me give you another example.

If you are trying to figure out how many subnets are there if you have 32 hosts (usable and non-usable), you can use the same trick. I use this all the time, and it helps.

hiddenknight821

SurferdudeHB wrote: »

is there an efficient way to calculate multiples of 16? Besides writing them down.

I need to know why you want to bother with it. Okay, I lied that 256 is not the "magic" number everyone was talking about, but my trick still works. You need to know what a "magic" number is. In the subnet mask, it's always the first zero after all one in binary. For example,

11111111.11111111.11110000.00000000 which is /20.

Now you see three octets that are either filled with all ones or zeros. Ignore those octets and focus on the one that has both ones AND zeros, which is the third octet in my example.

Like I said, find the first zero, and that would be your magic number. Remember to stay focus on that octet alone. there are 4 1's bits, which is the network portion and there are 4 0's bits that are host portion. There are 16 subnets. Using the trick I shared with you in my previous post, you should have 16 addresses and 14 of them are host. Aha, there's the "magic" number you are looking for. It's 16.

Now, 16 is your magic number. Now let's give ourselves a problem.

172.17.149.115 /20

I'm using the same subnet mask from above. Now, we want to find the host range it belongs to. The problem here is that you are not good at 16 time table, so we would want find the fastest way to find the host range. Well division and rounding is now your best friend, but multiplication should still be in consideration.

One way you can solve it quickly is to divide THE NUMBER in the third octet (remember, I told you to focus on the octet where the increment happens) by the "magic" number. You do not have to bother with the remainder. 149 / 16 = 9.3125. So now, this is where multiplication becomes your best friend again. 16 times 9 has just gotten a whole lot easier, and the answer is 144. So, the host range should be....

172.17.144.1 to 172.17.159.254

You should be smart enough to also perform 16 * 10 calculation to get the end range. Don't forget to subtract by two. 160 will take you to the next subnet address and you cannot use the broadcast address, 172.16.159.255)

SurferdudeHB

hiddenknight821 wrote: »

I need to know why you want to bother with it. Okay, I lied that 256 is not the "magic" number everyone was talking about, but my trick still works. You need to know what a "magic" number is. In the subnet mask, it's always the first zero after all one in binary. For example,

11111111.11111111.11110000.00000000 which is /20.

Now you see three octets that are either filled with all ones or zeros. Ignore those octets and focus on the one that has both ones AND zeros, which is the third octet in my example.

Like I said, find the first zero, and that would be your magic number. Remember to stay focus on that octet alone. there are 4 1's bits, which is the network portion and there are 4 0's bits that are host portion. There are 16 subnets. Using the trick I shared with you in my previous post, you should have 16 addresses and 14 of them are host. Aha, there's the "magic" number you are looking for. It's 16.

Now, 16 is your magic number. Now let's give ourselves a problem.

172.17.149.115 /20

I'm using the same subnet mask from above. Now, we want to find the host range it belongs to. The problem here is that you are not good at 16 time table, so we would want find the fastest way to find the host range. Well division and rounding is now your best friend, but multiplication should still be in consideration.

One way you can solve it quickly is to divide THE NUMBER in the third octet (remember, I told you to focus on the octet where the increment happens) by the "magic" number. You do not have to bother with the remainder. 149 / 16 = 9.3125. So now, this is where multiplication becomes your best friend again. 16 times 9 has just gotten a whole lot easier, and the answer is 144. So, the host range should be....

172.17.144.1 to 172.17.159.254

You should be smart enough to also perform 16 * 10 calculation to get the end range. Don't forget to subtract by two. 160 will take you to the next subnet address and you cannot use the broadcast address, 172.16.159.255)

thx for the reply, I'll give this method a try.

CaySpekko

I use my fingers instead of writing it down, and also use some memorization, but most of the memorization just comes from practice and repetition.
without using my thumbs I start from the pinky of my left going to hand I've memorize the 128 192 224 240 248 252 254 255
I've also memorized these 'magic number' values for subnet increments again starting from my pinky finger on my left hand
128, 64, 32, 16, 8 , 4 ,2 ,1

also using my fingers I can count the cidr mask pretty easy, just start at either 8 16 or 24 then count by ones.

Also I've memorized some of the easier subnet increments:
for 128:
0, 128
64:
and 0, 64, 128, 192
32: 0, 32, 64, 96, 128, 160, 192, 224

As soon as it hits anything with 16 increments or smaller, this is where memorization is just too tedious and it's time to whip out the expo marker, and do what hiddenknight suggested:

ie look at the address, ie 192.168.0.190/28 you know the increment is 16, so you divide 191 by 16, and get 11.8... don't worry about calculating the decimal, just take the whole number and multiply by 16: 16x11 = 176 so you can figure out that the subnet is 192.168.0.176/28 and the last host on that net is 192.168.0.190/28

drew726

I don't really write down anything and do most of it in my head but there is one thing I use to memorize in quickly determining how many valid hosts and subnets there are. In Lammle's book, he said to memorize 2^9 as 512 and I always thought that was a terrible number to memorize because there's no specific pattern. I started using 2^10 as 1024 and thats easy to remember because the number "10" is the first two digits of 1024. I also used 2^6 as 64 and the number "6" is the first digit of 64.

Doubling anything after 1024 is easy if you separate the 1000 from the 24. If you're doing 2^16 then you automatically know that 2^10 is 1024. From there, it is just 1000 * 2^6 which is 64000 and then you add 24 * 2^6, which is 24 * 64 = 1536. Then 64000 + 1536 is just 65536.

The likely hood that you'll have to do 2^16 is pretty rare but I also have a trick for multiplying 24's. Just think of it as 25 instead, which is 24+1, and then obviously for every number you multiple you'll get the non-24 factor added to it. All you have to do is subtract that factor from the total.

For example, at first we did 24 * 64. Instead I would do (24+1) * 64. Now you know that for every "four" 25's it equals 100. So all you gotta do is find out how many 4's are in 64. There are 16 factors of 4 in 64. 16 * 100 = 1600. In the end all you have to do is subtract the original number "64" from 1600 and you get 1600 - 64 = 1536. I don't expect everyone to do this in their head but it once you master doing this, it becomes pretty easy, especially for numbers less than 2^16.

snokerpoker

I always write 128 64 32 16 8 4 2 1 at the top of page so I can look at the subnet mask and determine the increment or to help convert from cider to decimal notation.