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Old Jan 26, 2013, 03:29 PM   #1
ZMacintosh
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Question about determining IP subnet ranges, should be simple

I had a question for those knowledgeable in networking.
Trying to figure out how to determine the steps to get these answers.

So for example a computer has an IP Address of 10.1.5.3 , what is the range of useable IP addresses for this subnet 10.1.5.3/24
So it would be 10.1.5.1 - 10.1.5.254


similar question, you have IP address of 192.168.1.15/24, what other IP addresses are on the same subnet?
So 192.168.1.68 & 192.168.2 would be on the same subnet

Trying to figure out how they use the octets/bits to determine the range.

Read through some CIDR notation info, but im still stuck on it, even though it seems very basic and simple.
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Old Jan 26, 2013, 03:58 PM   #2
ChristianJapan
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Quote:
Originally Posted by ZMacintosh View Post
So for example a computer has an IP Address of 10.1.5.3 , what is the range of useable IP addresses for this subnet 10.1.5.3/24
So it would be 10.1.5.1 - 10.1.5.254
Correct.
The subnet mask in total have 32 bit.
/24 means that the first 24 bits of the IP address define the network address while the last 8 bits define the range/size of that specific subnet.
In case of /24 it would be 8 bits left which makes 1 to 254 valid IP addresses.

Quote:
similar question, you have IP address of 192.168.1.15/24, what other IP addresses are on the same subnet?
So 192.168.1.68 & 192.168.2 would be on the same subnet
It's actually very much the same as you first example with /24

192.168.1.1 to 192.168.1.254 would be valid IP address in 192.168.1/24
192.168.2.xxx would NOT be a valid address in 192.168.1/24

If you want 192.168.1 and 192.168.2 to be in the same subnet you would need a netmask of /22 (for example)

A good presentation with the bit masks is here http://en.wikipedia.org/wiki/Subnetwork
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Last edited by ChristianJapan; Jan 26, 2013 at 04:19 PM.
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Old Jan 26, 2013, 05:03 PM   #3
ZMacintosh
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meant to say 192.168.1.2**

so with the remaining 8 bits
how they calculate to that? binary?
i understand binary easier, always get stuck on this type of stuff.

thanks for the link, reviewing it ab it more now
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Old Jan 26, 2013, 05:30 PM   #4
ChristianJapan
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Quote:
Originally Posted by ZMacintosh View Post
meant to say 192.168.1.2**

so with the remaining 8 bits
how they calculate to that? binary?
i understand binary easier, always get stuck on this type of stuff.

thanks for the link, reviewing it ab it more now
Yes, binary.

Those host number with all bit zero and all bits 1 are reserved. .0 and .255 for example.
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