Hello,
I am ashamed to ask this question, because the answer/response might be very simple;
Given this IP address; 192.168.9.0/27  What can be extrapolated from it?
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There is a quick way of calculating
See one of my articles here
The article is focused on simple summarization but I explained subnetting a little to increase understanding.
http://www.experts-exchange.com/Networking/Network_Management/Network_Design_and_Methodology/A_11491-IP-Subnet-Summarization-Simplified.html

The /27 notation you see there does not indicate number of IP addresses, It indicates the number of network bits

255.255.255.255 in bits would be represented as
11111111.11111111.11111111.11111111
This is also a /32 notation.

count the number of 1s from left to right, you will have 32 of them.

The common subnet people use (Class C) has a subnet mask shown below
255.255.255.0
This broken down into bits would be represented as
11111111.11111111.11111111.00000000
Count the number of 1s from the left, you will have 24 of them
This gives you a slash notation of /24

get the picture?
So /27 would look something like this
11111111.11111111.11111111.11100000
Now, our subnet has changed in bit value and we need to convert that to decimal
One Octet is a set of 8 bits
When all turned on, they have this value
11111111
When all turned off, they have this value
00000000

The 8 bits represent the following
128   64  32  16  8  4  2  1

If you add all those numbers together, that gives you 255.
So when all are turned on, you have 255 in the octet of the subnet mask you are working with.

If all turned off, you value there is zero.

Each Subnet mask comprises of 8 octets in IPv4
255.255.255.0
1st Octet (from left) is 11111111 = 255
2nd Octet is 11111111 = 255
3rd Octet is 11111111 = 255
4th Octet is 00000000 = 0

Going back to your /27 notation
11111111.11111111.11111111.11100000000
1st Octet (from left) is 11111111 = 255
2nd Octet is 11111111 = 255
3rd Octet is 11111111 = 255
4th Octet is  11100000 = 128 + 64 + 32 +0 +0 +0 +0 +0 = 224

Your subnet is now 224 meaning 224 out of 255 addresses are part of your network and the rest are hosts.

By the way, 1s represent NETWORK, and 0s represent HOST
subtract 224 from 255, you are left with 31 addresses (mathematically) that can be used as hosts.

I emphasized mathematically because in computer language, 0 is considered a valid digit in a network.
So if we add the 0, that gives us 32 Possible addresses.

Out of the possible addresses, 32 in our case, the network device will automatically assign 1 address to identify the network and use another address to send a message to everyone on the network.

By default, the first addresses is used to identify the network and is designated as NETWORK ID.
In this example, that will be 192.168.1.0

In the same vein, the last addresses is used to broadcast messages to everyone on the network, and is designated as BROADCAST ADDRESS
In this example, that will be 192.168.1.31

Those 2 addresses cannot be assigned to a host on a network, so technically speaking, we only have 32 - 2 possible host addresses which gives you 30.

Your 1st assignable address therefore is 192.168.9.1, and the last assignable is 192.168.9.30

.....And yes
There are tools you can use

http://www.subnet-calculator.com/

All the best
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Network ConsultantCommented:
Yes you can extrapolate the following:

The usable ip addresses on that network are 192.168.9.1 - 30

You can have 30 hosts in that subnet.
Is is a subnetted class C address under RFC 1918 for private addressing.
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Author Commented:
kenboonejr, thank you for the response.  Why is it 1-30 and not 0-26 (making it 27)?
While I know that there is a way of calculating the subnet mask, is there an easy way of looking it up, or calculating it, or you just know based on the fact that it is a Class C.
I'll have other questions base on your initial response, later.
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Author Commented:
Wow, if I could bump the points up to 1000 I would, thanks.  I studies this stuff some time back, this was a good refresher, with added insights.  You do know your Networking.
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Thanks for the points and comments. Greatly appreciated.

All the best
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