05-01-2018 02:32 PM - edited 03-08-2019 02:51 PM
I'll rate the post of everyone who helps
So, the Network 10.0.0.0/8 was Subnetted into 2^19 Subnets. One of these Subnets, which is the 10.20.30.64/27 was Subnetted into 4 Subnets
I'm trying to find out the Network that the Subnet 195.3.40.0/21 came from. However, it's a Class C IP (/24), so how can it be with a /21 Mask?
- 10.0.0.0/8
- 10.0.0.0/27
- 10.0.0.32/27
- 10.0.0.64/27
…
- 10.0.1.0/27
…
- 10.1.0.0/27
…
- 10.20.0.0/27
…
- 10.20.30.0/27
- 10.20.30.32/27
- 10.20.30.64/27
- 10.20.30.64/29
- 10.20.30.72/29
- 10.20.30.80/29
- 10.20.30.88/29
-10.20.30.96/27
Solved! Go to Solution.
05-01-2018 02:45 PM - edited 05-01-2018 03:06 PM
You can think of it as 0.0.0.0/0 is really the only IPv4 network out there, and it is sub-netted into various smaller divisions. Network classes are these days just a set of guidelines on how to divide it, unless you are working with very legacy software. Just as you can subnet a Class A network, /8 mask, into smaller parts, such as /27, a Class C network can be grouped with other class C networks to make larger ones, like a /21.
195.3.40.0/21 comes from the 0.0.0.0/0 network just like 10.0.0.0/8 does, unless you have some smaller network definition that you are working with in that space.
You might also look up supernet as a term used in contrast to subnet.
05-01-2018 04:57 PM - edited 05-01-2018 05:23 PM
In the specific case of 172.16.100.0 is pretty complicated. What you can say about it is that 172.16.100.0 is only the address of a network if the netmask is /22 or above. At /21, 172.16.100.0 would be a host address in the 172.16.96.0/21 network rather than a network address.
I would answer as if the network address provided was a /24. Simply because only the last byte of the address is 0, and is a reasonable, and easy, assumption for mask given the information.
172.16.100.0/26,
172.16.100.64/26
172.16.100.128/26
172.16.100.192/26
It could also be answered based on a /22 boundary, so you end up with 172.16.100.0/24, 172.16.101.0/24 etc... But to work that out you need to be pretty quick with binary numbers or use a calculator.
I would hope test questions would either have the network addresses be valid for the normal classful netmask for the range, such as 172.16.0.0 being assumed as /16 or be explicit about the netmask. In fact I would hope that they would always be explict about the netmask, as nearly all software provides and requires explicit netmasks, there will be few real world situations where you have to guess at the netmask.
The one that can really trip people up is the 192.168.0.0 network. This is highly ambiguous between the private IP range of 192.168.0.0/16, and the first /24 network in that range of 192.168.0.0/24. It's very likely that ambiguity is why many switches and routers use 192.168.1.0/24 as their default network. The ambiguity is much less for the 172.16.0.0 network, as the two zero bytes tends to indicate /16 to match which the normal classful netmask in that range.
05-02-2018 07:11 AM
05-01-2018 02:45 PM - edited 05-01-2018 03:06 PM
You can think of it as 0.0.0.0/0 is really the only IPv4 network out there, and it is sub-netted into various smaller divisions. Network classes are these days just a set of guidelines on how to divide it, unless you are working with very legacy software. Just as you can subnet a Class A network, /8 mask, into smaller parts, such as /27, a Class C network can be grouped with other class C networks to make larger ones, like a /21.
195.3.40.0/21 comes from the 0.0.0.0/0 network just like 10.0.0.0/8 does, unless you have some smaller network definition that you are working with in that space.
You might also look up supernet as a term used in contrast to subnet.
05-01-2018 04:07 PM
05-01-2018 04:57 PM - edited 05-01-2018 05:23 PM
In the specific case of 172.16.100.0 is pretty complicated. What you can say about it is that 172.16.100.0 is only the address of a network if the netmask is /22 or above. At /21, 172.16.100.0 would be a host address in the 172.16.96.0/21 network rather than a network address.
I would answer as if the network address provided was a /24. Simply because only the last byte of the address is 0, and is a reasonable, and easy, assumption for mask given the information.
172.16.100.0/26,
172.16.100.64/26
172.16.100.128/26
172.16.100.192/26
It could also be answered based on a /22 boundary, so you end up with 172.16.100.0/24, 172.16.101.0/24 etc... But to work that out you need to be pretty quick with binary numbers or use a calculator.
I would hope test questions would either have the network addresses be valid for the normal classful netmask for the range, such as 172.16.0.0 being assumed as /16 or be explicit about the netmask. In fact I would hope that they would always be explict about the netmask, as nearly all software provides and requires explicit netmasks, there will be few real world situations where you have to guess at the netmask.
The one that can really trip people up is the 192.168.0.0 network. This is highly ambiguous between the private IP range of 192.168.0.0/16, and the first /24 network in that range of 192.168.0.0/24. It's very likely that ambiguity is why many switches and routers use 192.168.1.0/24 as their default network. The ambiguity is much less for the 172.16.0.0 network, as the two zero bytes tends to indicate /16 to match which the normal classful netmask in that range.
05-02-2018 12:41 AM
05-02-2018 05:24 AM - edited 05-02-2018 05:25 AM
Hi
It can be originated from a summarization of networks, this network is supported under the interfaces and allowed into the routing table.
Hope it is useful
:-)
05-02-2018 07:11 AM
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