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Old 09-18-2011, 09:14 AM   #1
trist007
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A question about CIDR Mask example...

Is this a valid example? and is my thinking correct?

73.19.22.14/17
Code:
11111111.11111111.1/1111111.11111111
   network               host
So 17 bits for network ID
Here's the break down for the value of each bit
Code:
128 64 32 16 8 4 2 1
So 15 bits for host ID which gives us
127 + 256 = 383 ips - 2 which leaves 381 available IP
I reserve 2 IPs for the network ID and broadcast IP.
Code:
Network ID: 73.19.22.14
IP Range: 73.19.22.15 - 73.19.22.255
          73.19.23.0  - 73.19.23.143
Broadcast IP: 73.19.23.144
Now on my IP Range, 73.19.23.0 and 73.19.22.255, can they be an IP or should it have been
Code:
Network ID: 73.19.22.14
IP Range: 73.19.22.15 - 73.19.22.254
          73.19.23.1  - 73.19.23.145
Broadcast IP: 73.19.23.146
Last edited by trist007; 09-18-2011 at 09:25 AM.
 
Old 09-18-2011, 10:11 AM   #2
markush
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Hi,

you'll need not a 17bit mask here but a 23bit mask.
Code:
Networkadress: 73.19.22.0/23 
IP-Range from 73.19.22.1 up to 73.19.23.254
where 73.19.23.255 is the broadcastadress.
this would be a valid subnet.

I think you confused
Quote:
Originally Posted by trist007
127 + 256 = 383 ips - 2 which leaves 381 available IP
with
Code:
128*256 - 2 = 32766 IP-adresses
which is obviously not what you're looking for.

Markus
 
Old 09-18-2011, 10:16 AM   #3
trist007
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Can you explain why I would need a 23 bit mask instead of a 17 bit mask?

IP Range: 73.19.22.1 up to 73.19.23.254
That would only be 73.19.22.1 - 73.19.22.254 and 73.19.22.1 - 73.19.23.254, that's only 255 + 254 IP addresses. How can there be 32766 IP-adresses?

The CIDR Mask is 73.19.22.14/17 not 73.19.22.0/23.

Would 73.19.22.14/17 not be a valid subnet?
Last edited by trist007; 09-18-2011 at 10:27 AM.
 
Old 09-18-2011, 10:39 AM   #4
trist007
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Ah ok I think I understand why I need a 23 bit mask.

So 73.19.22.14/17 is not a valid CIDR because to have a .14 at the end I would only be left with 1 bit for the host
Code:
01010011.00010011.00010110.00001110 
11111111.11111111.1111111 = 23 bits
I see now that the mask I applied above would work with /23
So yes 73.19.22.0/23 would be a valid CIDR mask
Code:
network ID: 73.19.22.0
IP range: 73.19.22.1 - 73.19.22.255 and 73.19.23.1 - 73.19.23.254
Broadcast: 73.19.23.255
Subnet Mask: 255.255.254.0
IPs available: 255 + 254 = 499
So then 73.19.22.255 and 73.19.23.0 are usable IPs?

Am I understanding it better?
Last edited by trist007; 09-18-2011 at 10:47 AM.
 
Old 09-18-2011, 10:42 AM   #5
markush
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Quote:
Originally Posted by trist007 View Post
So 73.19.22.14/17 is not a valid CIDR because to have a .14 at the end I would only be left with 1 bit for the host
it is a valid IP-adress, but not a valid network-adress. But it would be a network with 32766 Adresses and the networkadress would be 73.19.22.0/17.

Markus
 
Old 09-18-2011, 10:45 AM   #6
markush
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Quote:
Originally Posted by trist007
IP range: 73.19.22.1 - 73.19.22.255 and 73.19.23.1 - 73.19.23.254
no, the range would be
Code:
IP range: 73.19.22.1 - 73.19.23.254
what you describe where two subnets
Code:
73.19.22.0/24 and
73.19.23.0/24
Markus
 
Old 09-18-2011, 10:54 AM   #7
trist007
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My previous question where the IP range: 73.19.22.1 - 73.19.23.254
Code:
So then 73.19.22.255 and 73.19.23.0 are usable IPs?
How would the bits look for 73.19.22.0/17?
Code:
01001001.00010011.00010110.00000000
11111111.11111111.1111111 = 23 bits
Wouldn't you need a 23 bit mask though?
Last edited by trist007; 09-18-2011 at 10:56 AM.
 
Old 09-18-2011, 11:00 AM   #8
markush
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with a 17bit mask you would have
Code:
network-adress: 73.19.0.0/17
or
network-adress: 73.19.128.0/17

01001001.00010011.00000000.00000000 
or 
01001001.00010011.10000000.00000000
Markus
 
Old 09-18-2011, 11:05 AM   #9
trist007
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Ah ok I see now, that makes sense. So 73.19.22.0/17 is not possible. What confuses me is when the CIDR notation ends in anything other than 0. But I see now how you can find the possibilities given a certain subnet mask.

My previous question where the IP range: 73.19.22.1 - 73.19.23.254 (in this case this range is considered to be on one subnet correct?)
Code:
So then 73.19.22.255 and 73.19.23.0 are usable IPs?
Last edited by trist007; 09-18-2011 at 11:07 AM.
 
Old 09-18-2011, 11:38 AM   #10
markush
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yes, this would be useable IPs.

Markus
 
Old 09-18-2011, 11:55 AM   #11
trist007
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Thank you Markus, you've been very helpful.
 
  


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