Subnetting - 3rd octet
J_86
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in CCNA & CCENT
Ok, so I seem to be really struggling when subnetting in the 3rd octet for some reason.
I have been watching the CBT nuggets videos, if any of you are familiar with that, but all of those example have been in the 4th octet.
Take for example this question I got from subnettingquestions.com:
How many subnets and hosts per subnet can you get from the network 10.0.0.0 255.255.240.0?
Answer: 4096 subnets and 4094 hosts
So the first step they teach is to break the subnet into binary, but that doesn't seem to work when you are working in the 3rd octet? How would you come up with that answer?
:
I have been watching the CBT nuggets videos, if any of you are familiar with that, but all of those example have been in the 4th octet.
Take for example this question I got from subnettingquestions.com:
How many subnets and hosts per subnet can you get from the network 10.0.0.0 255.255.240.0?
Answer: 4096 subnets and 4094 hosts
So the first step they teach is to break the subnet into binary, but that doesn't seem to work when you are working in the 3rd octet? How would you come up with that answer?
:
Comments
I know that it takes 4 bits to get 240: 128+64+32+16= 240 and there are already 8 bits being borrowed from the second octet. So you have 2^12 = 4096 subnets. That leaves 4 host bits in the third octet and 8 in the last octet so 2^12-2= 4094.
Not sure that helps but that was my thought process when I saw the question.
I hope it further clarifies your concept.
Saeed
Actually, it should be no problem to use binary to solve this problem.
Find the # of subnets/hosts for the following:
10.0.0.0
255.255.240.0
First convert everything down to binary
Looking at the ip address, you know that the 2nd Octet is dedicated to the subnet and along with the bites that are turned on in the 3rd Octet which gets us this:
Everything else that is 0 in the 3rd and 4th Octet goes for the hosts. Remember to subtract 2 for the Network and Broadcast Address:
"Its easier to deceive the masses then to convince the masses that they have been deceived."
-unknown
Ah, now I get it! Thanks guys, I think I was just over thinking it.
Well you should know straight off the bat that an address starting with 10 is a class A address.
The default subnet mask for a class A address is:
255.0.0.0...in binary that is 11111111.00000000.00000000.00000000
255.255.240.0 in binary is...11111111.11111111.11110000.00000000
how many extra 1's are there ? 12
what is 2^12 ? 4096
How many 0's are there ? 12
what is 2^12 - 2....4096-2 = 4094
You need to know what the subnet formulas are where n is the number of bits borrowed from the host portion of the address. In your example you are borrowing 12 bits.
2^n for subnets
2^n - 2 for hosts
Also you really need to know your powers of 2
A different point of view :
As you know, by default 10.0.0.0 belongs to Class A. And by default, subnet mask of Class A is 255.0.0.0
According to this question : the resulted subnet mask is 255.255.240.0
That means, 8 bits from the second octet and the 11110000(number of 1s) 4 bits from the third octet taken.
Besides, the number of subnets is calculated as 8 plus 4 = 12 indicating 2"12 subnets = 4096 subnets
How many zeroes remained? 4 bits on the third octet and 8 bits on the last octet indicating 2"12 -2 = 4094 hosts per subnet
Where does it lead us to in case of block size? as you have seen, third octet shows us the 240. now, 256-240 = 16.
this means block size = 16 is equal to incremental value = 16
So,
if the first subnet is 10.0.0.0
the second will be 10.0.16.0
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until the third octet becomes the 256(i mean which passes to second octet)
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