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quick subnetting question
beginner_ccna
Member Posts: 59 ■■□□□□□□□□
in CCNA & CCENT
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?
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?
Comments
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Optionschmorin Member Posts: 1,446 ■■■■■□□□□□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.Currently PursuingWGU (BS in IT Network Administration) - 52%| CCIE:Voice Written - 0% (0/200 Hours)mikej412 wrote:Cisco Networking isn't just a job, it's a Lifestyle. -
Optionssting_224 Member Posts: 43 ■■□□□□□□□□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. -
Optionsbeginner_ccna Member Posts: 59 ■■□□□□□□□□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
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Optionswbosher Member Posts: 422The 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. -
Optionsbeginner_ccna Member Posts: 59 ■■□□□□□□□□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 -
Optionsgosh1976 Member Posts: 441The 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 -
Optionssting_224 Member Posts: 43 ■■□□□□□□□□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 -
Optionsmiller811 Member Posts: 897255.255.255.224
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 = 192.168.12.222I don't claim to be an expert, but I sure would like to become one someday.
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