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Subnetting Concept

Please help me with my sub netting concept...

what 10.2.1.1/9 mean?

Majestics
Practically a Master Poster
619 posts since Jul 2007
Reputation Points: 199
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Here is a basic netmask table:

<pre><code>ID	# Bits	Number Hosts	Usable Hosts	Netmask		Note

6	/6  	67108864	67108862	 252.0.0.0  	

7	/7 	33554432	33554430	254.0.0.0 	

8	/8 	16777216	16777214	255.0.0.0 	Class A Network 

9	/9 	8388608		8388606		255.128.0.0 	

10	/10 	4194304		4194302		255.192.0.0 	

11	/11 	2097152		2097150		255.224.0.0 	

12	/12 	1048576		1048574		255.240.0.0 	

13	/13 	524288		524286		255.248.0.0 	

14	/14 	262144		262142		255.252.0.0 	

15	/15 	131072		131070		255.254.0.0 	

16	/16 	65536		65534		255.255.0.0 	Class B Network 

17	/17 	32768		32766		255.255.128.0 	

18	/18 	16384		16382		255.255.192.0 	

19	/19 	8192		8190		255.255.224.0 	

20	/20 	4096		4094		255.255.240.0 	

21	/21 	2048		2046		255.255.248.0 	

22	/22 	1024		1022		255.255.252.0 	

23	/23 	512		510		255.255.254.0 	

24	/24 	256		254		255.255.255.0 	Class C Network 

25	/25 	128		126		255.255.255.128 	

26	/26 	64		62		255.255.255.192 	

28	/28	16		14		255.255.255.240	

27	/27	32		30		255.255.255.224	

29	/29 	8		6		255.255.255.248 	

30	/30 	4		2		255.255.255.252</code></pre>
rch1231
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959 posts since Sep 2009
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10.2.1.1/9:

First of all, this is an IPv4 address; IPv4 address consists of 32 bits.

The /9 represents the number of bits for the network, meaning that they are reserved for the network, and you can't use them. So by a simple calculation,the number of bits for the hosts in this network is (32 - 9 = 23) and the number of hosts of this network is 2^23 - 2 Hosts ( we subtract 2 because one address is for the network address and the other is the broadcast address for the network)

nalawar
Newbie Poster
14 posts since Nov 2008
Reputation Points: 20
Solved Threads: 1
 

is 192.168.0.1/8 is allowed? I have many tutorial which allow atleast 24 for this....
But an ip calculator calculate the above question.

Majestics
Practically a Master Poster
619 posts since Jul 2007
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Its called Classless Inter-Domain Routing (CIDR). More info here:
http://en.wikipedia.org/wiki/Classless_Inter-Domain_Routing

For more subnetting information:
http://www.punkwalrus.com/howto/subnetting.html

Hope this helps.

thunderstorm98
Veteran Poster
1,117 posts since Nov 2006
Reputation Points: 97
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Hello Majestics,

Theoretically 192.168.0.1/8 is not allowed as a private network, since the network 192.168.x.x/16 is the private network, however if you want to use a private network with /8 you can use the 10.x.x.x/8.

What is meant by private network is a network that will not pass the gateway of the network. i.e. if you have for example a packet that has 192.168.1.10 (suppose that it is a host) as a destination, when it reaches the router or the gateway, it will be redirected to the local network, meaning it will not go through the gateway of the network and to the public network (internet).

But suppose you used 10.x.x.x/8 in your private network, then you have 2^24 - 2 hosts, and that is impossible to have in one network, thus you will have plenty and plenty of unused hosts, which brings us back to the reason of variable length subnet mask (VLSM) and classless Networks.

The reason behind the subnetting is because of the depletion of IPv4 addresses, because of huge network of hosts that was based on classfull networks as a class A network (refer to rch1231 post above).

Hope this helps in understanding subnetting.

nalawar
Newbie Poster
14 posts since Nov 2008
Reputation Points: 20
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This question has already been solved

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