Subnetting problem - Magic number

Answered Question
May 13th, 2012

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 ...

Network address: 174.121.12.0  

Broadcast address: 174.121.13.255        (12+Magic number -1 = 12+2-1 =13)

Where is mistake?

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

Please help me,

Greetings.

I have this problem too.
0 votes
Correct Answer by John Blakley about 1 year 11 months ago

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

Correct Answer by Peter Paluch about 1 year 11 months ago

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

Correct Answer by John Blakley about 1 year 11 months ago

Joni,

Your math is fine, but the network address that you have on your interface (your IP) will start at 174.121.14.0 - 172.121.15.255.

HTH,

John

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Correct Answer
John Blakley Sun, 05/13/2012 - 04:53

Joni,

Your math is fine, but the network address that you have on your interface (your IP) will start at 174.121.14.0 - 172.121.15.255.

HTH,

John

trikantejabuka Sun, 05/13/2012 - 05:11

Thanks for the reply,


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

Network Address is: 174.121.12.0

Broadcast address is : 174.121.13.255, which is obviously wrong.


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

Correct Answer
Peter Paluch Sun, 05/13/2012 - 05:38

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

Correct Answer
John Blakley Sun, 05/13/2012 - 05:57

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

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Posted May 13, 2012 at 1:35 AM
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