VLSM Subnetting Question

billscott92787

Hello,



I have a question in regards to VLSM Subnetting. Here is an example:


Address space: 192.168.1.0 /25


If I am determining the subnet that will be used based on the amount of hosts, and say a specific entity needs 10 hosts total. This tells me that I would need to borrow a total of 4 bits from the host portion of the address (2^4-2 = 16 - 2 = 14 hosts per subnet).


Since the default subnet mask used is /25 or 1111111.11111111.11111111.1000000

Would I include the network bit that has already been borrowed since the default mask is /24? Or would I borrow 4 more bits in addition to the one already borrowed. I'm not sure if I am right, but I would think that I would only borrow 3 more bits for a total of 4 bits. Because if I borrowed 4 in addition to that I would have borrowed a total of 5 bits (2^5-2= 32-2 = 30 hosts per subnet) which means that I would be wasting 20 hosts.

jscimeca715

billscott92787 wrote: »

Hello,



I have a question in regards to VLSM Subnetting. Here is an example:


Address space: 192.168.1.0 /25


If I am determining the subnet that will be used based on the amount of hosts, and say a specific entity needs 10 hosts total. This tells me that I would need to borrow a total of 4 bits from the host portion of the address (2^4-2 = 16 - 2 = 14 hosts per subnet).


Since the default subnet mask used is /25 or 1111111.11111111.11111111.1000000

Would I include the network bit that has already been borrowed since the default mask is /24? Or would I borrow 4 more bits in addition to the one already borrowed. I'm not sure if I am right, but I would think that I would only borrow 3 more bits for a total of 4 bits. Because if I borrowed 4 in addition to that I would have borrowed a total of 5 bits (2^5-2= 32-2 = 30 hosts per subnet) which means that I would be wasting 20 hosts.

This is a good question, one I'm interested in as well. Is this from a book somewhere?

If I was answering it...my gut feeling would be this:

192.168.1.0-192.168.1.128 since that is already being subnetted.

I would take network 192.168.1.128 and use a /28 which would give me a range of .129-.142

I'm probably wrong, but who knows. I'm assuming it means take the four bits after you've mapped out the original network with the remaining hosts you have left.

billscott92787

Yes,


It is from a book. It's an exercise in a work book from Cisco. Routing Protocols and Concepts.


The example is given real world examples of needing hosts, using VLSM, to further subnet the network, based on the original address 192.168.1.0/25 and the amount of hosts needed for a particular LAN, and WAN links, in a contiguous design.

nash0924

I would have to say the same too. 0 to 127 is already allocated to the previous network, 128 will be the next network in line with an increment of 16, since you will use an additional 3 network bits which will leave 4 host bits to satisfy the 14 hosts.

192.168.1.0-192.168.1.128 since that is already being subnetted.

I would take network 192.168.1.128 and use a /28 which would give me a range of .129-.142

Anding to double check?:

Example IP: 192.168.1.140

192.168.00000001.10001100
255.255.11111111.11110000

network ip = 192.168.1.128..

billscott92787

That is definitely the way that I was thinking. Because if we added an additional 4 bits, we would only be able to satisfy 8 hosts (if not reserving one for network and broadcast). I appreciate your input You guys are always helpful here.