How do I find sub range without calculating increments?

Ok, this isn't the first time around but I still can't figure out an efficient way to calculate a subnet range without multiplying the increment till I reach the range.

I.E.

Which subnet does host 192.168.24.204 255.255.255.224 belong to?

Range is 32 bit class C. How do I figure the network where .204 is without multiplying 32 times?

Thank you!

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Comments

dynamik

Just divide by the number of hosts, cut off the remainder/decimal, and then multiply that with the number of hosts.

i.e. /28 will give you increments of 16.

If your IP address is 189, 189/16 = 11.8125. 11*16 = 176.

Essendon

Consider the mask, 255.255.255.224. The "interesting" octet is the last octet. Interesting octet is the rightmost octet that isnt 255.

So subtract 224 from 256 i.e. 256 - 224 = 32. Now multiply 32 with a number so that it is equal to or less than 204 (go as close as you can). So 32 * 6 = 192. 192 is the closest you can get to 204. So the subnet is 192.168.24.192 and the broadcast address is 192.168.24.223 (192+32-1).

If your wondering where I got this method from, it's from Odom's CCNA book.

miller811

it takes math and practice....

1. find the subnet size by examining the mask
2. one ip address above your network address is the first available host address
3. one ip address below the next subnet is the broadcast (use the subnet size to find the boundary)
4. one ip address below the broadcast is the last host address

mella060

These things take time and practice to master. You really need to be fluent in your multiples of 4 , 16 , 32 , 64 etc.

The quickest way to find the answer is to go up by the largest multiple (ie 64) until you get close to the range you are after.

So in your example you are after 192.168.24.204...so going up in multiples of 64...0 , 64 , 128 , 192...and BINGO...there is your answer

The subnet is 192.168.24.192 /27...the next subnet would be 192.168.24.224 /27

Again once you have been doing this stuff for a while you get better at it.

billscott92787

The easiest way is doing the increments to be honest in my opinion, everyone is different though.

32
64
96
128
160
192
224

So we know that the subnet is 192, which has a range of 193 - 222 (223 broadcast)

Firemarshalbill.com

OK let's hope I can do this right on the posting, usually need regular paper

At the top of a plain piece of paper write this

128 64 32 16 8 4 2 1

Below it using decimal to binary math (subtracting the above values from the mask )add up to the mask value in your case it is the forth octet and the value is 224 using 1's to reperesent where the network part is and 0's the hosts

so we have 224 - 128 = 96
96 - 64 = 32
32 - 32 = 0

This would mean the following bit's for our mask (224 from your questions)

128 64 32 16 8 4 2 1

1 1 1 0 0 0 0 0

Now use the last value that has a 1 as the "subnet multiplier"

Quickly looking at it and starting at 0 we get the following networks

0, 32, 64, 96, 128, 160, 192, 224,

The broadcasts would be the network number - 1 and the final one 255

so we get

31,, 63 95 127 159 191 223 and 255

the hosts would be everything in between

If your mask was 240 then the subnet multiplier would be 16 and so on

And as you can see from the replies everyone is different. GOOD LUCK

KPWright

SurferdudeHB wrote: »

still can't figure out an efficient way to calculate a subnet range without multiplying the increment till I reach the range.

SurferDudeHB,

It looks like the consensus is that you already know the most efficient way of getting to the correct subnet

. You either start at the base network address and count up in multiples of the address range, or you divide by the range, chop off the partial bits and multiply by the range again. Looks like there's quite a few ways to write out your work, but the operation seems to be the same.

Handbrake

billscott92787 wrote: »

The easiest way is doing the increments to be honest in my opinion, everyone is different though.

32
64
96
128
160
192
224

So we know that the subnet is 192, which has a range of 193 - 222 (223 broadcast)

Interesting...I have been practicing but always do it the basic way of converting to binary then back to decimal..

For this method
Do you just memorize these numbers?

how does it increment?

i can see 32+64 = 96
96+64 = 160

it just seem odd how u add them up

dynamik

You just go by the host range you're given. He's going in 32-bit increments in the exam, so he has five host bits to work with. If he was using four bits, he would go in 16-bit increments, and so on.

Geek1969

Which subnet does host 192.168.24.204 255.255.255.224 belong to?

I don't know which of the 100 or so ways to subnet I picked this up from but, here is how I did it.

255.255.255.224 is a /27 (3 bits) ....mask=128+64+32=224
(3rd bit is 32)............
Always use the last bit needed to add up to the interesting octet 224 in this case as the increment. Increment is 32

In this case 32 is the increment. Now, find the numbers that you can multiply by 32 that are closest to the host address(204) without going over. Your only choices are 1-8.
would 5 work? 32*5=160....too far away
would 6 work? 32*6=192.....that's pretty close.
would 7 work? 32*7=224....too far,
So, the host is in the 6th subnet. (192) 32*6 = 192(network),193(1st host) We know from 32*7 that 224 is the next subnet, so one less(223) is the broadcast.

As i said, there are 100's of "tricks" etc to learn how to subnet....find one that makes sense for you, and that you can do in your head and stick with it.
One thing that has helped me get faster is learning to count from the left going up in value. ---- 128-192-224-240-248-252-254-255.....as well as going down 128-64-32-16-8-4-2-1.
These numbers never change.....

Hope that made sense.

Satcom

Essendon wrote: »

Consider the mask, 255.255.255.224. The "interesting" octet is the last octet. Interesting octet is the rightmost octet that isnt 255.

So subtract 224 from 256 i.e. 256 - 224 = 32. Now multiply 32 with a number so that it is equal to or less than 204 (go as close as you can). So 32 * 6 = 192. 192 is the closest you can get to 204. So the subnet is 192.168.24.192 and the broadcast address is 192.168.24.223 (192+32-1).

If your wondering where I got this method from, it's from Odom's CCNA book.

I am starting to really understand subnetting with the methods in Odoms book at the End of CCNA.. i just did problem set 5 appendix D off the cd rom 5 25 questions today. I think Im going to do another problem set .

The subnetting video that comes with ICND1 and ICND2 is very good.
The additional problem sets in appendix D I think will bring me to success on the test. I wish they would have included problem sets RP 1-5 on the disk also. You would be able to skip ch12 in book 1 icnd1