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Subnetting Question

Posted on 2003-11-29
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Last Modified: 2010-04-11
im having trouble understanding subnetting. i hope u can help me
i have a host of
211.10.16.85 and subnet mask of 255.255.255.240

i want to know how many valid subnets and hosts are availabe on this subnetted network and what is the range of valid IP addresses for each subnet.

i have worked out the answer but i dont know if its correct.

also wot is the subnet networks address ans subnet broadcast address?

any help wud be very much appreciated
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Question by:gerrymcd
4 Comments
 
LVL 32

Expert Comment

by:LucF
ID: 9842431
Hi gerrymcd,

Check out this site: http://www.telusplanet.net/public/sparkman/netcalc.htm

Greetings,

LucF
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LVL 18

Accepted Solution

by:
chicagoan earned 125 total points
ID: 9842636
There's a stand-alone calculator here:
http://www.solarwinds.net/Tools/Free_tools/Subnet_Calc/index.htm
 see http://www.ralphb.net/IPSubnet/subnet.html for a tutorial


One of the easiest ways is to use a table for the commonly used (and practival masks)

Net bits
 mask     total-addresses
  255.255.240.0  4096
 
  255.255.248.0  2048
 
 255.255.252.0 1024
 
 255.255.254.0 512
 
 255.255.255.0 256
 
 255.255.255.128 128
 
 255.255.255.192 64
 
 255.255.255.224 32
 
 255.255.255.240 16
 
 255.255.255.248 8
 
 255.255.255.252 4
 
Here you see your mask produces a subnet size of 16 addresses
so the ranges  would be
0      15
16      31
32      47
48      63
64      79
80      95
96      111
112      127
128      143
144      159
160      175
176      191
192      207
208      223
224      239
240      255

and your network falls on 80      95
the first number (80) being the network and the last (95) the broadcast address which gives you 14 usable addresses.



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Expert Comment

by:pgilchrist
ID: 9844525
The first thing to do is convert your IP address and network mask to binary (several network calculators are available for this or you can simply use Windows Calculator to convert between base 10/Dec and base 2/Bin):

211.10.16.85        => 11010011.00001010.00010000.01010101
255.255.255.240   => 11111111.11111111.11111111.11110000

Performing a bitwise AND on the host address and the network mask will give you the first (network) address:

11010011.00001010.00010000.01010000 => 211.10.16.80

Performing a bitwise OR on the host address and the INVERTED subnet mask will give you the last (directed broadcast) address:

11010011.00001010.00010000.01011111 => 211.10.16.95

This will give you all the details for the subnet on which your host resides.  Other subnets (even in the same class C network of 211.10.16.0) will not necessarily have the same subnet mask.
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Expert Comment

by:pixxy
ID: 9856195
211.10.16.0 is a public class C IP address. The address 211.10.16.85 is a host address belonging to a network, created by subnetted 211.10.16.0 with a /28 bit network mask.

To explain the /28:

255.255.255.255 would be 8bits + 8bits + 8bits  + 8bits = /32

So,

255.255.255.240 = 8bits + 8bits + 8bits + 4bits = 28bits for the network and 4 left over for hosts = /28

211.10.16.0 with a mask of 255.255.255.240 (/28) will create networks like this:

211.10.16.0
211.10.16.16
211.10.16.32
.. and so on.

The address 211.10.16.85 belongs to this network:

211.10.16.80 is the network address
211.10.16.81 to 211.10.16.95 are for hosts (14 in total)
211.10.16.96 is the broadcast address for this network

You could further subnet the network 211.10.16.80 with a /30 network mask, creating networks like this:

211.10.16.80
211.10.16.84
211.10.16.88
.. and so on

The address 211.10.16.85 would then belongs to this network:

211.10.16.84 is the network address
211.10.16.85 to 211.10.16.86 are for hosts (2 in total)
211.10.16.87 is the broadcast address for this network


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