Subnetting Concept

[Majestics]

Please help me with my sub netting concept...

what 10.2.1.1/9 mean?

[rch1231]

Here is a basic netmask table:

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     

[nalawar]

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)

[Majestics]

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.

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.

[nalawar]

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.