quick subnetting question

hi guys

been going through the cbt nuggets which is helping alot

regarding the below question:

Question: How many subnets and hosts per subnet can you get from the network 172.27.0.0 255.255.254.0?

Answer: 128 subnets and 510 hosts

i know how to find the hosts for this question but can someone please explain how you get 128 subnets?

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Comments

chmorin

You have a class B address which is usually 255.255.0.0 for the address. You add 7 bits to the network side, allowing for 128 subnets. The formula I use is (2^x) where x is the number of network bits used. Similar for the host side, which you have 9 bits for. So you have [(2^x)-2] where x is the number of host bits used, subtracted from two since you cant use the network address or broadcast address in address allocation for hosts.

Make sense?

Way to figure it out on paper is by drawing out the bits and converting to decimal:

1111111 - 7 bits
64-32-16-8-4-2-1 = 127

So you have subnets 0 - 127, which is a total of 128 subnets.

sting_224

Representing the subnet mask 255.255.254.0 in binary we have

255 . 255 . 254 . 0
11111111.11111111.11111110.00000000

From the 172.27.0.0 address we can tell its a class B ip address. We know that the default subnet mask for class B is 255.255.0.0, representing 255.255.0.0 in binary gives us

255 . 255 . 0 . 0
11111111.11111111.00000000.00000000

The network portion is represented by 1s and the host portion is represented by 0s

255.255.0.0 is subnetted by borrowing 7 bits from the host portion of the subnet mask giving us

255 . 255 . 254 . 0
11111111.11111111.11111110.00000000

11111111.11111111.{1111111}0.00000000
7bits

Knowing this simply helps us calculate the number of subnets by using 2^n where n is the number of borrowed bits, this gives us 2^7 = 128

Hope this helps, tried my best, I'm not a cisco guy.

beginner_ccna

excellent, thanks both thats helped out alot and now i can answer the questions of the below website

subnettingquestions.com - Free Subnetting Questions and Answers Randomly Generated Online

wbosher

The method Jeremy uses in CBT Nuggets is by far the easiest way that I've found to subnet. Go over the videos again and again and again, until you've nailed it.

If you're doing ICND1, subnetting is probably the most important thing to learn. You can guarantee that there will be a LOT of questions on subnetting, but then Jeremy has already told you that.

beginner_ccna

sorry to bother you guys again

ive been going through subnettingquestions.com - Free Subnetting Questions and Answers Randomly Generated Online and been getting about 90% correct but come across the below question which i dont know how to get the answer, please could someone explain

Question: What is the broadcast address of the network 192.168.12.192 255.255.255.224?

Answer: 192.168.12.223

when i worked it out, i though the answer would have been, 192.162.12.255

gosh1976

The way I would figure that out is by thinking I have 3 network bits being borrowed which means that the size of the network chunks is 32.

last octet in subnet mask 11100000

128+64+32 = 224

so if you add 32 to the last octet of the network address of 192 you would get 224 which would be the last octet of the next subnet so just subtract one for the broadcast giving you 223 (or you could just add 31!). There's the answer 192.168.12.223

that's probably not the best explanation but I tried

sting_224

What is the broadcast address of the network 192.168.12.192 255.255.255.224?

255.255.255.244 in binary = 11111111.11111111.11111111.11100000

From the last octet 11100000 in the subbet mask, we can tell that there are 8 subnets (2^3) and 30 hosts [(2^5) -2], we minus 2 because of the network and broadcast addresses.

256-224 = 32 as the range of subnets

So the eight possible subnets and their range of addresses are :

192.168.12.0 (192.168.12.1-192.168.12.30) with 192.168.12.31 as broadcast
192.168.12.32 (192.168.12.33-192.168.12.62) with 192.168.12.63 as broadcast
192.168.12.64 (192.168.12.65-192.168.12.94) with 192.168.12.95 as broadcast
192.168.12.96 (192.168.12.97-192.168.12.126) with 192.268.12.127 as broadcast
192.168.12.128 (192.168.12.129-192.168.12.158 with 192.268.12.159 as broadcast
192.168.12.160 (192.168.12.161-192.168.12.190) with 192.268.12.191 as broadcast
192.168.12.192 (192.168.12.193-192.168.12.222) with 192.268.12.223 as broadcast
192.168.12.224 (192.168.12.225-192.168.12.254) with 192.268.12.255 as broadcast

hope this helps

miller811

255.255.255.224

take the intesting octet, the non 255 octect and subtract it from 256

256 - 224 = 32 which equals your block size

255.255.255.128

256-128 = 128 block size

255.255.255.192

256-192 = 64 block size

once you figure out the block size = determine where you network fits in

in this example the network was
192.168.12.192 255.255.255.224
so we know the block size is 32

with this mask, you have to realize that they subnetted 192.168.12.0 to start out with, so that is where you start to solve the problem. Apply the block size and determine the available networks, and determine where your fits in.

192.168.12.0
192.168.12.32
192.168.12.64
192.168.12.96
192.168.12.128
192.168.12.160
192.168.12.192
192.168.12.224

find your network address and the one below it, if one exists

yours is 192.168.12.192
the one below it is 192.168.12.224

your first available host is one above the network address
192.168.12.193
the broadcast address is one below the next network 192.168.12.223
last available host would be one below the broadcast = 192.168.12.222