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Another subnetting headache
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| Rudeboy 2003-02-03, 3:11 am |
| I have just read a guide to subnetting (that I think came from 3com) recommended in a previous post and the examples in the back have me a little confused.
The example sites your network is 132.45.0.0/16 and you require 8 subnets. The guide sites 3 subnet bits 2^3=8 which gives you 132.45.0.0/19 132.45.32.0/19 etc to 132.45.224.0/19 - 8 subnets.
Now this is where I get confused. Freak says that the formula is ((2^3)-2)= 6 subnets.
I believe that you cant use the first or last subnet as the first has subnet bits of 000 and the last has subnet bits of 111. So you end up with 6 usuable subnets.
Am I right in my thinking?
Sid (my head hurts) | |
| IronMaiden 2003-02-03, 9:12 am |
| The subnet mask of 255.255.224.0 (19 mask bits, 3 subnet bits) will create only 6 subnets -- 2^3-2=6.
You should use a subnet mask of 255.255.240.0 (20 mask bits, 4 subnet buts). This will create 14 usable subnets. 2^4-2=14.
Here is another good subnetting tutorial:
http://www.jrksoftware.com/subnetting.html | |
| jeff_j_black 2003-02-03, 5:29 pm |
| Most likely, from and exam standpoint, you can assume that the All Zeros and All Ones subnets can be excluded, hence the 2^X-2 formula.
More modern interpretations and implementations allow for these subnets to be used.
Cisco Website: Ip Zero Subnets
Again, from an exam standpoint assume 2^X-2. | |
| Rudeboy 2003-02-03, 9:12 pm |
| Thanks for your help.
Sid |
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