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IP Resubnetting

Posted on 2008-06-21
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Last Modified: 2008-06-23
I have the following subnet:
Subnet Address: 192.168.1.0
Subnet Mask: 255.255.255.0
 in order to add more hosts to the network I changed the Mask to :Subnet Mask: 255.255.254.0

I would like to know what would be the new range of IP addresses that are going to be availbale for the hosts using the new subnet 255.255.254.0
Some explanation please.

Thanks
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Question by:jskfan
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15 Comments
 
LVL 5

Assisted Solution

by:sadburger
sadburger earned 400 total points
ID: 21839041
With a subnet mask of 255.255.255.254, 192.168.1.0 is no longer the subnet but instead is a valid host on that network.

The subnet you are working with is 192.168.0.0/23
Subnet 192.168.0.0
Valid host range 192.168.0.1 - 192.168.1.254
Broadcast address for this subnet is 192.168.1.255

See: http://en.wikipedia.org/wiki/Subnetwork for more info
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Author Comment

by:jskfan
ID: 21839151
Can you explain to me how the valid host range now is: 192.168.0.1 - 192.168.1.254 ???
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LVL 4

Assisted Solution

by:raymondzwarts
raymondzwarts earned 400 total points
ID: 21839221
You previously used a subnet of 255.255.255.0 on your network. This meanse that 8 of the 32 bits are used for host addressing (set to 0). The other 24 bits are set to 1 (the network). 8 bits in decimal is 256. Deducting the network and broadcast address gives you a host range of:

192.168.1.1 - 192.168.1.254
network = 192.168.1.0
broadcast = 192.168.1.255

With the new subnetmask of 255.255.254.0 you are using 9 of the 32 bits for host addressing (set to 0). The other 23 bits are set to 1 (the network). 9 bits in decimal is 512. Deducting the network and broadcast address gives you a host range of:

192.168.0.1 - 192.168.1.254
network = 192.168.1.0
broadcast = 192.168.1.254

Raymond
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LVL 4

Expert Comment

by:raymondzwarts
ID: 21839223
typo in previous post.

The second broadcast address = 192.168.1.255
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LVL 4

Expert Comment

by:raymondzwarts
ID: 21839227
second typo (where is the "edit" button ?)
network address = 192.168.0.0
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Author Comment

by:jskfan
ID: 21839423
network address  192.168.0.0
subnet mask 255.255.254.0

how did you make the calculation and you found that the first host in range is 192.168.0.1 and the last will be 192.168.1.254 ????

can you break down you explanation line by line?
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Author Comment

by:jskfan
ID: 21839443
Maybe if you give me also examplantion for the following  examples, might help:
network address  192.168.0.0
subnet mask 255.255.252.0
Valid host IP range???

network address  192.168.0.0
subnet mask 255.255.248..0
Valid host IP range???
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LVL 10

Expert Comment

by:rynox
ID: 21839909
network address  192.168.0.0
subnet mask 255.255.252.0

Host bits     00000000 00000000 00000011 11111111
Subnet bits 11111111 11111111 11111100 00000000

This gives you 1024 addresses, (2^8), which would be 192.168.0.0 - 192.168.3.255
First is the Network address (192.168.0.0) and the last is the Broadcast Address (192.168.3.255)
Leaving you with 1022 usable host addresses (192.168.0.1-192.168.3.254)


network address  192.168.0.0
subnet mask 255.255.252.0

Host bits     00000000 00000000 00000111 11111111
Subnet bits 11111111 11111111 11111000 00000000

This gives you 2048 addresses, (2^9), which would be 192.168.0.0 - 192.168.7.255
First is the Network address (192.168.0.0) and the last is the Broadcast Address (192.168.7.255)
Leaving you with 2046 usable host addresses (192.168.0.1-192.168.7.254)
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LVL 10

Expert Comment

by:rynox
ID: 21839916
Make a mistake in the last one, should be 2^10 and 2^11, should read like this:
(yes I wish there was an edit button too)

network address  192.168.0.0
subnet mask 255.255.252.0

Host bits     00000000 00000000 00000011 11111111
Subnet bits 11111111 11111111 11111100 00000000

This gives you 1024 addresses, (2^10), which would be 192.168.0.0 - 192.168.3.255
First is the Network address (192.168.0.0) and the last is the Broadcast Address (192.168.3.255)
Leaving you with 1022 usable host addresses (192.168.0.1-192.168.3.254)


network address  192.168.0.0
subnet mask 255.255.252.0

Host bits     00000000 00000000 00000111 11111111
Subnet bits 11111111 11111111 11111000 00000000

This gives you 2048 addresses, (2^11), which would be 192.168.0.0 - 192.168.7.255
First is the Network address (192.168.0.0) and the last is the Broadcast Address (192.168.7.255)
Leaving you with 2046 usable host addresses (192.168.0.1-192.168.7.254)
0
 

Author Comment

by:jskfan
ID: 21840467
Where I am still confused is in the host bits of the Third octet:
network address  192.168.0.0 /22
subnet mask 255.255.252.0          11111111.11111111.11111100.00000000

This a class C  and we borrowed 2 bits to expand the host range. Correct?
I don't have a problem with the last Octet which can be 1 to 254:  192.168.0.1-254
but the Third Octed  I thought there should be a formula:
256-252=4 , so we have a block of size 4 starting from 0
192.168.0.0 on the third octet 0 and it is a network ID and the next Network ID will be 4 ex:192.168.4.0
and the broadcast will be 192.168.3.255 (I am not sure about this one)
the host range will be 192168.1.0 to 192.168.2.254

Is that correct?


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LVL 10

Expert Comment

by:rynox
ID: 21841528
This a class C  and we borrowed 2 bits to expand the host range. Correct?
     Correct, but forget about 'Class C', what we are doing here is classless subnetting or 'CIDR'

I don't have a problem with the last Octet which can be 1 to 254:  192.168.0.1-254
but the Third Octed  I thought there should be a formula:
256-252=4 , so we have a block of size 4 starting from 0
192.168.0.0 on the third octet 0 and it is a network ID and the next Network ID will be 4 ex:192.168.4.0
and the broadcast will be 192.168.3.255 (I am not sure about this one)
     This is correct as well, the next range would start at 192.168.4.0 using this subnet mask.

the host range will be 192168.1.0 to 192.168.2.254
Is that correct?
     I'm not sure why you are dropping all the addresses in 192.168.0.x and 192.168.3.x.  The network address is 192.168.0.0 and that is where your IPs start at and they continue all the way to 192.168.3.255, and as with any range, the first IP is the network address and the last is the broadcast address, leaving you wil ALL the IPs inbetween as the usable hosts.
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Author Comment

by:jskfan
ID: 21841646
so the host IP  can't be 192.168.0.0 and can't be 192.168..4.0 but in between except for the broadcast 192.168.3.255

example it can be: 192.168.1.0 ,192.168.1.1 ,192.168.1.2  all the way to 192.168.1.254
               it can be 192.168.2.0 ,192.168.2.1,192.168.2,2 all the way to 192.168.2.254
               it can be 192.168.3.0,192.168.3.1,192.168.3.2 all the way to 192.168.3.254

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LVL 10

Accepted Solution

by:
rynox earned 1200 total points
ID: 21841698
192.168.4.0 is not even in your network, it is the beginning of the next one.

Your WHOLE network is 192.168.0.0-192.168.3.255
You lose the first address as your network address (192.168.0.0)
You lose the last address as the broadcast address (192.168.3.255)

EVERYTING ELSE is a valid IP in the network:
192.168.0.1 - 192.168.0.255
192.168.1.0 - 192.168.1.255
192.168.2.0 - 192.168.2.255
192.168.3.0 - 192.168.0.254

You don't drop the 255's off the end, they are in the middle of the range and are valid IP addresses.
Still not sure why you wnat to ignore all the IPs from 192.168.0.1-192.168.0.255
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Author Comment

by:jskfan
ID: 21842350
Thanks a lot !!!
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LVL 4

Expert Comment

by:raymondzwarts
ID: 21849588
Just a note on compatibility:

Older systems think in the classfull system. This means no 0 or 255 as host IP's as these could be broadcast addresses.
An example of this is Sun Solaris 8. So if you are using old hardware be aware of this especially if you are creating a subnet larger than a class C network (2^8 hosts-2).

Regards,
Raymond Zwarts
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