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Beginner

## Subnetting problem - Magic number

Hi,

This is my IP address 174.121.14.165/23

/23  - 255.255.254.0

:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

Magic number is: 256-254=2

:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

Networks:

0

2

4

6

8

10

12

14

16 ...

Where is mistake?

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

Greetings.

3 ACCEPTED SOLUTIONS

Accepted Solutions

Joni,

HTH,

John

HTH, John *** Please rate all useful posts ***
Hall of Fame Cisco Employee

Joni,

The magic number, as you call it, is indeed 2 in the third octet of the netmask. Hence, you need to round down the third octet of the IP address to the nearest integral multiple of 2 when calculating the network address. However, in the IP address 174.121.14.165, the "14" is already an integral multiple of 2 so you do not change it. Right?

Best regards,

Peter

Joni,

In addition to what Peter said, the "magic number" is considered your boundary. For example, if your address the magic number is 2. Multiples of 2 would be the starting network address for the next block:

172.121.2.0

172.121.4.0

172.121.6.0

etc.

The magic number is the starting boundary for that network, and the broadcast address is -1 less than the next starting boundary:

172.121.2.0 - 172.121.3.255 (broadcast) (256 - 1)

The useable addresses are calculated like:

254 (in the second octet) (11111110)

0 = There are 8 bits left in the 4th octet = 00000000

So together it's 11111110.0000000

There are 9 0s in the host portion of your mask, so you get 512 hosts per network. Then you take 512 - 2 leaving you with 510 useable addresses (172.121.2.1 - 172.121.3.254)

HTH,

John

HTH, John *** Please rate all useful posts ***
6 REPLIES 6
Engager

Was your ip not in x.x.14.0?

Joni,

HTH,

John

HTH, John *** Please rate all useful posts ***
Beginner

I know my IP address should be in the range 174.121.14.0 - 172.121.15.255, in my calculation

But I ask myself why, if I calculate correctly all.

Hall of Fame Cisco Employee

Joni,

The magic number, as you call it, is indeed 2 in the third octet of the netmask. Hence, you need to round down the third octet of the IP address to the nearest integral multiple of 2 when calculating the network address. However, in the IP address 174.121.14.165, the "14" is already an integral multiple of 2 so you do not change it. Right?

Best regards,

Peter

Joni,

In addition to what Peter said, the "magic number" is considered your boundary. For example, if your address the magic number is 2. Multiples of 2 would be the starting network address for the next block:

172.121.2.0

172.121.4.0

172.121.6.0

etc.

The magic number is the starting boundary for that network, and the broadcast address is -1 less than the next starting boundary:

172.121.2.0 - 172.121.3.255 (broadcast) (256 - 1)

The useable addresses are calculated like:

254 (in the second octet) (11111110)

0 = There are 8 bits left in the 4th octet = 00000000

So together it's 11111110.0000000

There are 9 0s in the host portion of your mask, so you get 512 hosts per network. Then you take 512 - 2 leaving you with 510 useable addresses (172.121.2.1 - 172.121.3.254)

HTH,

John

HTH, John *** Please rate all useful posts ***
Beginner

Thank you all for  help, now I understand

Thank you!.

greeting from Bosnia

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