# Very basic subnetting doubt

Mar 14th, 2008

Hello

I am currently preparing for my CCNA exam and am almost done with it. I will be taking it in few days time but this one subnetting problem has been nailing me constantly.

If i have an ip address 10.0.0.0/20, then will the subnets range from 10.0.0.0 to 10.0.240.0 or 10.0.0.0 to 10.255.240.0?

Another example: if the ip address is 10.0.0.0/26, will the subnets range from 10.0.0.0 to 10.0.0.192 or from 10.0.0.0 to 10.255.255.192?

I know how to calculate the host bits, subnet bits, # of hosts and subnets n all related stuff but still i cant figure out the solution to above mentioned problem.

Thank You

Regards

Hi There

If you have the network 10.0.0.0/20, this means that the first 20 bits define the netowrk/subnet, leaving 12 bits for host addressing.

So for this network your subnet mask is 255.255.240.0. So the octet of the mask in which subnetting is taking place is the 3rd octet.

Therefore your subnet spans the range 10.0.0.0 to 10.0.15.255

If you have the entire 10.0 range, then you could have 16 subnets, each with 4094 usable IP addresses. Each subnet number would be 16 higher then the previous one (in the third octet).

e.g.

10.0.0.0 - 10.0.15.255

10.0.16.0 - 10.0.31.255

10.0.32.0 - 10.0.47.255

10.0.48.0 - 10.0.63.255

.... ....

.... ....

10.0.240.0 - 10.0.255.255

In example 2 you have a network 10.0.0.0.26. This means that the first 26 bits define the network/subnet, leaving 6 bits for host addressing.

So in this example your netmask is 255.255.255.192. So the octet in which subneting is taking place is the last octet.

Therefore in this example your subnet range is 10.0.0.0 to 10.0.0.63.

Again if you had the entire 10.0.0.x range you would have 4 subnets, each with 62 available host IP addresses. Each subnet number would be 64 higher then the previous subnet (in the last octet)

e.g.

10.0.0.0 - 10.0.0.63

10.0.0.64 - 10.0.0.127

10.0.0.128 - 10.0.0.191

10.0.0.192 - 10.0.0.255

HTH and does not cause even more confusion so close to your exam.

Best Regards,

Michael

Overall Rating: 4.5 (2 ratings)

## Replies

Jon Marshall Fri, 03/14/2008 - 01:35

Hi

Firstly, good luck with your CCNA exam, hope it goes well.

On to the question. Lets use the 10.0.0.0/26 example you give.

/26 = 255.255.255.192

If the octet = 255 in the subnet mask then it does not change on the IP address so straight away we can see that 10.255.255.192 will simply not work with a /26 subnet mask.

Breaking it down further

10.0.0.0 255.255.255.192

The first 3 octets in the subnet mask are 255.255.255. So following on from the above explanation all subnets will begin with

10.0.0.

then we just need to work out the subnets. Quick way to do it in your head.

256 - 192 = 64. So your networks go up in 64's ie.

10.0.0.0/26

10.0.0.64/26

10.0.0.128/26

etc...

10.0.0.0/20 =

10.0.0.0 255.255.240.0

256 - 240 = 16 so

10.0.0.0/20

10.0.16.0/20

10.0.32.0/20

etc...

Hope this makes sense

Jon

keeleym@o2.ie Fri, 03/14/2008 - 01:36

Hi There

If you have the network 10.0.0.0/20, this means that the first 20 bits define the netowrk/subnet, leaving 12 bits for host addressing.

So for this network your subnet mask is 255.255.240.0. So the octet of the mask in which subnetting is taking place is the 3rd octet.

Therefore your subnet spans the range 10.0.0.0 to 10.0.15.255

If you have the entire 10.0 range, then you could have 16 subnets, each with 4094 usable IP addresses. Each subnet number would be 16 higher then the previous one (in the third octet).

e.g.

10.0.0.0 - 10.0.15.255

10.0.16.0 - 10.0.31.255

10.0.32.0 - 10.0.47.255

10.0.48.0 - 10.0.63.255

.... ....

.... ....

10.0.240.0 - 10.0.255.255

In example 2 you have a network 10.0.0.0.26. This means that the first 26 bits define the network/subnet, leaving 6 bits for host addressing.

So in this example your netmask is 255.255.255.192. So the octet in which subneting is taking place is the last octet.

Therefore in this example your subnet range is 10.0.0.0 to 10.0.0.63.

Again if you had the entire 10.0.0.x range you would have 4 subnets, each with 62 available host IP addresses. Each subnet number would be 64 higher then the previous subnet (in the last octet)

e.g.

10.0.0.0 - 10.0.0.63

10.0.0.64 - 10.0.0.127

10.0.0.128 - 10.0.0.191

10.0.0.192 - 10.0.0.255

HTH and does not cause even more confusion so close to your exam.

Best Regards,

Michael

pankaj_goyal Fri, 03/14/2008 - 03:26

Thank You both Michel and Jon for the posts. This certainly has helped me clear my long lingering doubt.

Thank You for the wishes.

keeleym@o2.ie Fri, 03/14/2008 - 04:58

Hi There

Glad that ny post helped. Thansk for the rating. Much appreciated.

Best Regards & Good luck with the exam,

Michael