jskfan
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IP address Classes
I am trying to understand IP address classes , I checked the link below:
http://vlsm-calc.net/ipclasses.php
I am not sure why the top part is saying class C subnet mask is 255.255.255.0 and below Private IP address table it is showing it 255.255.0.0, the same for class B the top part is showing 255.255.0.0 and the Private table is showing 255.240.0.0
Another thing that also confuses me is the Number of the Networks. for instance Class C that has subnet mask 255.255.255.0. How many Networks and hosts we can have.???
To my understanding if we have 192.168.1.X 255.255.255.0 , it means we have on Network since the first 3 octets will not change , but will have (2^8)-2=254 Hosts...
Any explanation will be very much appreciated..
Thanks
http://vlsm-calc.net/ipclasses.php
I am not sure why the top part is saying class C subnet mask is 255.255.255.0 and below Private IP address table it is showing it 255.255.0.0, the same for class B the top part is showing 255.255.0.0 and the Private table is showing 255.240.0.0
Another thing that also confuses me is the Number of the Networks. for instance Class C that has subnet mask 255.255.255.0. How many Networks and hosts we can have.???
To my understanding if we have 192.168.1.X 255.255.255.0 , it means we have on Network since the first 3 octets will not change , but will have (2^8)-2=254 Hosts...
Any explanation will be very much appreciated..
Thanks
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If you had tried either of these examples with a 23 bit mask, you would only be able to have 510 hosts.
OK... Let' say I need 510 hosts when initially having 192.168.1.0/24 Network
How is the Network going to be changed ?
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OK...let's say we start with 192.168.0.0/23
In Binary
11111111.11111111.11111110 .00000000.
how are Hosts going to be positioned on the subnets...?
for instance:
First subnet :
first host will be 192.168.0.1 and the last will be 192.168.0.254
Second Subnet:
First host will be 192.168.1.1 and the last will be 192.168.1.254
It still 254 host per subnet(508 Total), and Only 2 subnets
In Binary
11111111.11111111.11111110
how are Hosts going to be positioned on the subnets...?
for instance:
First subnet :
first host will be 192.168.0.1 and the last will be 192.168.0.254
Second Subnet:
First host will be 192.168.1.1 and the last will be 192.168.1.254
It still 254 host per subnet(508 Total), and Only 2 subnets
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192.168.0.0/23 = 192.168.0.0 255.255.254.0
255.255.254.0= 11111111.11111111.11111110 .00000000
Number of hosts= 00000000.00000000.00000001 .11111111 which gives (2^9)-2=510
I do not see how you came up with:
192.168.0.1 - 192.168.1.254
255.255.254.0= 11111111.11111111.11111110
Number of hosts= 00000000.00000000.00000001
I do not see how you came up with:
192.168.0.1 - 192.168.1.254
ASKER
On the third octet 255.255.254.0, it will give us 2 subnets , so it will be 192.168.0.0 and 192.168.2.0 , 192.168.1.0 should be a Broadcast...
ASKER
192.168.1.0 should be a Broadcast...192.168.1.255 should be a Broadcast...
ASKER
I guess got this part:
256-254=2 , means first Network is 192.168.0.0 and the next will be 192.168.2.0 and The broadcast will be 192.168.1.255
The other part ?? I am trying to figure out 510 hosts ??
can 192.168.0.255 be a host?
can 192.168.1.0 be a host ?
If so then the 2 subnets will have hosts from 192.168.0.1 to 192.168.0.255 (255 hosts) and from 192.168.1.0 to 192.168.1.255 (255 hosts) , that will make 255+255=510
256-254=2 , means first Network is 192.168.0.0 and the next will be 192.168.2.0 and The broadcast will be 192.168.1.255
The other part ?? I am trying to figure out 510 hosts ??
can 192.168.0.255 be a host?
can 192.168.1.0 be a host ?
If so then the 2 subnets will have hosts from 192.168.0.1 to 192.168.0.255 (255 hosts) and from 192.168.1.0 to 192.168.1.255 (255 hosts) , that will make 255+255=510
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I will do some reading and practice and come back...
Thanks
Thanks
It's not clear to me why folks continue to learn about "classes". The whole idea is archaic and unnecessary.
I didn't see any real comments in this regard. Don Johnson said "You are still thinking classfull" which comes closest to making this point.
So, starting at the beginning, learn about subnetting (which I think you have gotten well into by now) and then, if you MUST, "back into" classes for the purposes of academic study. It will be much clearer to you that way I believe.
I didn't see any real comments in this regard. Don Johnson said "You are still thinking classfull" which comes closest to making this point.
So, starting at the beginning, learn about subnetting (which I think you have gotten well into by now) and then, if you MUST, "back into" classes for the purposes of academic study. It will be much clearer to you that way I believe.
It's not clear to me why folks continue to learn about "classes". The whole idea is archaic and unnecessary.Because whether you like it or not, it's still taught and tested. If you're looking at any network training or certification, they still use the concept.
I'm sorry, I should have been clearer:
It's not clear to me why folks continue to teach about "classes".
A suggestion might be that they should also teach germanium transistor technology just to be consistent! :-)
My recommendation, as others have made, is to learn about subnetting from the subnet mask up and, when time allows or demands, THEN learn about the buggy whips called "classes".
It's not clear to me why folks continue to teach about "classes".
A suggestion might be that they should also teach germanium transistor technology just to be consistent! :-)
My recommendation, as others have made, is to learn about subnetting from the subnet mask up and, when time allows or demands, THEN learn about the buggy whips called "classes".
Good luck with that.
ASKER
Thank you Guys.... This tool is much more detailed about subnet and supernet:
http://jodies.de/ipcalc?host=192.168.0.0&mask1=24&mask2=25
http://jodies.de/ipcalc?host=192.168.0.0&mask1=24&mask2=25
That last website has a lot of information; you can try this one also http://www.cronos.est.pr/t cpip/
ASKER
If I understand:
192.168.1.X 255.255.255.0, I have one subnet and 254 hosts from 192.168.1.1 to 192.168.1.254
if I need to have 512 hosts I will have to take one Bit from Network portions , correct ? if so how will I change the 192.168.1.x ??