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Class C Subnetting help


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#1 aaronjwilkinson

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Posted 12 February 2014 - 19:43

i've working in IT for a fair while, but the logic behind subnetting is something i've always struggled with (and mental arithmetic in general)

I've decided to pull my finger out with this and try to get the process hard-wired in my brain due to an exam i'm studying to pass.

Anyway, could someone in the know take a look at the workings in the following link and see if you agree it's accuracy?

Cheers

 

http://imageshack.co...2/6992/qfv5.jpg




#2 +BudMan

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Posted 12 February 2014 - 19:54

here you go, looks like your good to go matches up.
 
 
Subnet	Network Address	Starting Host	End Host	Broadcast	Netmask
0	192.168.0.0	192.168.0.1	192.168.0.30	192.168.0.31	255.255.255.224
1	192.168.0.32	192.168.0.33	192.168.0.62	192.168.0.63	255.255.255.224
2	192.168.0.64	192.168.0.65	192.168.0.94	192.168.0.95	255.255.255.224
3	192.168.0.96	192.168.0.97	192.168.0.126	192.168.0.127	255.255.255.224
4	192.168.0.128	192.168.0.129	192.168.0.158	192.168.0.159	255.255.255.224
5	192.168.0.160	192.168.0.161	192.168.0.190	192.168.0.191	255.255.255.224
6	192.168.0.192	192.168.0.193	192.168.0.222	192.168.0.223	255.255.255.224
7	192.168.0.224	192.168.0.225	192.168.0.254	192.168.0.255	255.255.255.224


#3 OP aaronjwilkinson

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Posted 12 February 2014 - 20:02

Yey, the only mistake i look to have made is calling network 0 network 1

 

I keep forgetting that zero is a REAL number in subnetting haha

 

Cheers :)



#4 Descartes

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Posted 12 February 2014 - 20:03

I keep wondering why the IP team chose a decimal notation over hexadecimal or binary to begin with. At least that's fixed in IPv6 :p



#5 +ChuckFinley

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Posted 12 February 2014 - 20:12

Depends if you are using Cisco (Or other vendors, See appropriate documentation). You can set the ip subnet zero I think. 



#6 sc302

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Posted 12 February 2014 - 20:21

Ok, so lets do this logically

 

32 bit mask is 255.255.255.255 which allows for 1 address per network

31 bit mask is 255.255.255.254 which allows for 2 addresses per network

30 bit mask is 255.255.255.252 which allows for 4 addresses per network

29 bit mask is 255.255.255.248 which allows for 8 addresses per network

28 bit mask is 255.255.255.240 which allows for 16 addresses per network

27 bit mask is 255.255.255.224 which allows for 32 addresses per network

26 bit mask is 255.255.255.192 which allows for 64 addresses per network

 

You want to divide 6 into 255 which gives you roughly 42.  You can go anywhere up to 42 address per network and have your 6 required networks.  If you notice everything is doubling on the network address side...1*2=2, 2*2=4, 4*2=8 on the bit mask side 255-1=254, 254-2=252, 252-4=248, 248-8=240 (see the pattern? the -x is doubling as well).   Basically, the doc you have is correct. 



#7 +riahc3

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Posted 13 February 2014 - 08:25

Hello,

Woah! Handwriting! Havent seen that in a while! Very nice handwriting BTW.

Anyways, here is a good thread also that makes sense to me at least: http://www.techexams...-made-easy.html