Ip range problem

[daniel45793]

If i have the range of 172‪.16.3.0 /25
And i have 172‪.16.3 254

Why 172‪.16.3 254 not in the range of 172‪.16.3.0 /25

And to what ip should i change the 172‪.16.3 254  so it would be in the range
of  =   172‪.16.3.0 /25

Thanks for the halp  (:

[MysticalTh0r]

Hello, 

The IP 172‪.16.3 254 is not in the range 172‪.16.3.0/25 because this range goes from 172‪.16.3.1 to 172‪.16.3.127. If you want to use a /25 network you need to use 172‪.16.3.128/25.

Basically, 172‪.16.3.0/25 is the same as saying 172‪.16.3.0 255.255.255.128 or 11111111.11111111.11111111.10000000

so if you translate the network into binary and overlap network and mask you get:

Network : 10101100.00010000.00000011.00000000

Mask        11111111.11111111.11111111.10000000

------------------------------------------------

And the bytes with "1" on the mask must match and the "0" says "I don't care". So in a /25 the 25 most-relevant bits must match and so the 172‪.16.3.0/25 goes from 172‪.16.3.1 until 172‪.16.3.127. So you need to change your network to 172‪.16.3.128/25 or the mask to /24 making it 172‪.16.3.0/24

HTH.

[daniel45793]

Hi 

Thanks for the replay.  (: 

Can you explain to me how the /24 

fixing this 

[MysticalTh0r]

Just because you make your range of IP wider. When you pass from /25 to /24 you get all the IP from the 172‪.16.3.0, meaning, from 172‪.16.3.1 to 172‪.16.3.254. If you put the /24 into binary like before, you will see that all the last 8 bits are 0, meaning only the 3 first octects must match (172.16.3) and the last can be any number, and this is why an /24 has this range, including your IP.

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[luis_cordova]

Hi @daniel45793 

If you want to keep the mask /25 in your network, you can occupy ip 172.16.3.126

The range for the net/mask you indicate is:

Network: 172.16.3.0

First host: 172.16.3.1

Last host: 172.16.3.126

Broadcast: 172.16.3.127

Regards