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Old 11-22-2010, 12:16 AM   #1
molossus
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why is this subnet wrong?


i would like to know why this subnet is wrong, iam using an online IP calculator
198.32.130.192/25
gives me:
Code:
Address:   198.32.130.192        11000110.00100000.10000010.1 1000000
Netmask:   255.255.255.128 = 25  11111111.11111111.11111111.1 0000000
Wildcard:  0.0.0.127             00000000.00000000.00000000.0 1111111
=>
Network:   198.32.130.128/25     11000110.00100000.10000010.1 0000000 (Class C)
Broadcast: 198.32.130.255        11000110.00100000.10000010.1 1111111
HostMin:   198.32.130.129        11000110.00100000.10000010.1 0000001
HostMax:   198.32.130.254        11000110.00100000.10000010.1 1111110
Hosts/Net: 126
which is wrong, the network shows as 198.32.130.128/25
128???, it is supposed to show as 192/25.

when i run this 198.32.130.192/26 , or /27 or /28 or /29 or /30
it shows ok:
Code:
Address:   198.32.130.192        11000110.00100000.10000010.11 000000
Netmask:   255.255.255.192 = 26  11111111.11111111.11111111.11 000000
Wildcard:  0.0.0.63              00000000.00000000.00000000.00 111111
=>
Network:   198.32.130.192/26     11000110.00100000.10000010.11 000000 (Class C)
Broadcast: 198.32.130.255        11000110.00100000.10000010.11 111111
HostMin:   198.32.130.193        11000110.00100000.10000010.11 000001
HostMax:   198.32.130.254        11000110.00100000.10000010.11 111110
Hosts/Net: 62
online ip calc: http://jodies.de/ipcalc?host=198.32....ask1=26&mask2=
for ubuntu or fedora:
sudo apt-get install ipcalc
su -c 'yum install ipcalc'
 
Old 11-22-2010, 12:55 AM   #2
neonsignal
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The subnet is correct, because you are masking out the second bit of the 192 (C0h) when you specify that only the first 25 bits are the network part of the address.
 
Old 11-22-2010, 01:56 AM   #3
markush
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Hello molossus,

there is everything correct. 198.32.130.128/25 is the networkadress when 198.32.130.192 is an IP-adress within this subnet. When changing the mask to 26 bit the adress 198.32.130.192 is no longer an IP-adress of a machine within the network but the networkadress of the 198.32.130.192/26 subnet.

Markus

Last edited by markush; 11-23-2010 at 06:18 AM. Reason: typo
 
Old 11-23-2010, 05:38 AM   #4
okcomputer44
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Quote:
Originally Posted by markush View Post
Hello molossus,

there is everything correct. 198.32.130.128/25 is the networkadress when 198.32.130.92 is an IP-adress within this subnet. When changing the mask to 26 bit the adress 198.32.130.92 is no longer an IP-adress of a machine within the network but the networkadress of the 198.32.130.92/26 subnet.

Markus
198.32.130.192 you meant, yeah?

0, 64, 128, 192, the networks if the mask /26.

Just to be exact if some one who does not know this and reading it, might misunderstood the whole lot. Actualy the .92 /26 is a host address at the second subnet.

Anyway I know this was a mistyping.

Laz.
 
Old 11-23-2010, 06:17 AM   #5
markush
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Quote:
Originally Posted by okcomputer44 View Post
198.32.130.192 you meant, yeah?...
Well thank you very much, it's a typo, I'll edit my post and change it.

Markus
 
  


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