12-08-2017 05:41 PM - edited 03-08-2019 01:03 PM
Slowly getting an handle on subnetting...I can sort of work a Class C problem ok, but class A or B, especially when you put a subnet mask that has bits that throw it out of the class of the given IP address confused the crap out of me and I don't know how to work thru the magic number, 1st subnet, range, hosts per subnet, and valid subnets like I can sort of do with Class C. I've looked at a few online resources and am currently using http://dw.nh4mi.com/subnetting/ to try to get up to speed, but I need to have an explanation for working with class A and B, because my answers haven't been correct yet for those classes.
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12-10-2017 05:09 AM
Hi
You are welcome :-)
You are correct about:
192.168.0.192 jump to 192.168.1.0 basically you will be sum 64 on the 4th octet:
192.168.0.192 +
64
---------------
192.168.0.256 <--- 256 is not a valid value in an IPv4 address (maximum is 255), so you jump to the next network segment 192.168.1.0/26
To be honest I dont think a subnetting class A or B will be deployed on an exam, it is a long process but no body knows hehehe.
You dont need to memorize the values just the formulas, remember it is practice and analize how it works.
:-)
12-08-2017 07:25 PM - edited 12-09-2017 05:36 AM
Hi
It is not complex but you must understand the formulas:
** There are many ways to subnetting, this is my way **
Available bits to subnetting:
32 - CIDR = Available bits to subnetting
To get X number of subnets:
(2^n) = X number of subnets (request)
Where "n" is the amount of bits used to resolve the formula
To get X number of hosts:
((2^n)-2) = X number of subnets (request)
Obtain new subnet mask:
n = bits
When you are subnetting to get subnets (network segment) the "n" value is subtracted from the available bits and enabled with ones (1) from left to right, once the ones are enabled the rest is completed with zeros, and the inverse when you are subnetting for host (host segment), "n" value is represented with zeros (0) from right to left, once the zeros are enabled the rest is completed with ones.
Block or size of networks, basically it is the new network segments
256 - alterated octets
Now for example: subnetting a Class B
172.16.0.0/16 ; we need 1000 subnets
Step 1) Identify Available bits to subnetting:
Formula
32 - CIDR
so
32 - 16 = 16 ; we have 16 bits to subnetting
Step 2) Identify the request, in this case: get subnets
Formula
(2^n) = X number of subnets (request)
so
2^n = 1000
Remember this is a exponential so we need to find a value to get 1000 or a little bit more than 1000 (not exceeding drastically)
n = 10 ; remember the "n" value represents bits, they will be subtracted from the available bits ( we will see it later)
2^10 = 1000
1024 = 1000 Bingo! exceeding 24 is not a big deal so we can use this value.
Step 3) Obtain the new subnet mask, it is obtained with the "n"
n = 10 bits
Available bits - n
16 - 10 = 6
My current subnet mask is:
/16 = 255.255.0.0 = 11111111.11111111.00000000.00000000
Remember each octect (we have 4 in a IPv4 structure) is built with 8 bits, ones (1) represents active and network segment and zeros (0) represents inactive bits and host segment.
Ok we have used 10 bits so our new subnet mask is:
11111111.11111111.11111111.11000000
10 ones and 6 zeros, the previous bits are not modified
So the new subnet mask is: 255.255.255.192 = /26
Alright, next step.
Step 4) Block or size of subnets
256 - alterated octets, which octets were modified? easy = 3rd and 4th, why? because we had:
255.255.0.0 and now we have 255.255.255.192
So they were alterated, ok we proceed getting the block or size of the new networks:
256 - 255 = 1 ; so the 3rd octect will be increased 1 in 1
256 - 192 = 64 ; so the 4th octect will be increased 64 in 64
Alright so we have:
172.16.0.0/26
172.16.0.64/26
172.16.0.128/26
172.16.0.192/26
172.16.1.0/26
172.16.1.64/26
172.16.1.128/26
172.16.1.192/26
172.16.2.0/26
172.16.2.64/26
.
.
.
.
The same process to get the 1000 new subnets
For Class A is basically the same but "n" will be other number probably 17 or + and it could alterated the 2nd octet, but the procedure is the same.
Hope it is useful
:-)
12-09-2017 03:55 PM - edited 12-09-2017 04:02 PM
that was helpful, thank you. My way was a confused mix of different ways I think...Do you suppose this is the absolute fastest way to work subnetting questions for an exam? It looks like it is to me...
One thing that messed me up I think, was not writing out the subnets first like you showed in your example, then filling in the range and broadcast, that's easy on the eye and fast. When you do it that way , it's easier to see that even with a magic number of 64 like your example, you still have to move over an octect for the next subnet range...
172.16.0.192/26
172.16.1.0/26 ....that's one of the parts that confused me..I didn't know where to go from 0.192, but now that I see it written out, there's no other option but to start in on the octet to the left with 1.0, since .192 + the magic number 64, would go over the 255 limit..., and somehow I guess I hadn't even gotten to the point of figuring out the new mask (although maybe I did it with another way without realizing it)...I think I'm also seeing that it may be a good idea to memorize the powers of two to at least 10 maybe?
I'm not to the point yet were I can 'see into the numbers' as our instructor put it, to prepare for figuring out test questions on VLSM although we have already talked about it (I'm behind the curve lol, but trying to play catch-up)
Thanks for the info again!
12-10-2017 05:09 AM
Hi
You are welcome :-)
You are correct about:
192.168.0.192 jump to 192.168.1.0 basically you will be sum 64 on the 4th octet:
192.168.0.192 +
64
---------------
192.168.0.256 <--- 256 is not a valid value in an IPv4 address (maximum is 255), so you jump to the next network segment 192.168.1.0/26
To be honest I dont think a subnetting class A or B will be deployed on an exam, it is a long process but no body knows hehehe.
You dont need to memorize the values just the formulas, remember it is practice and analize how it works.
:-)
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