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pkhatri Mon, 04/10/2006 - 00:21
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Here's one:


Given 10.1.1.32/27, you can split it up into the following:


10.1.1.32/30

10.1.1.36/30

10.1.1.40/30

10.1.1.44/30

10.1.1.48/30

10.1.1.52/30

10.1.1.56/30

10.1.1.60/30


So you get 8 /30s out of a /27.


Paresh

carl_townshend Mon, 04/10/2006 - 00:25
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can you explain how you did this please, thanks very much indeed !!

pkhatri Mon, 04/10/2006 - 00:33
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Sure...


We start off with 10.1.1.32/27


Let's just consider the last octet in binary:

32 = 00100000


The last octet of the mask of /27 is:

224 = 11100000


The last octet of the mask of /30 is:

252 = 11111100


Now, putting them together


00100000 = 32

11100000 = 224

11111100 = 252


Now, the three bits that are ones in the 252 mask that are zeros in the 224 mask are bits 3-5 (counting from left to right, starting at 0). Since we are only subnetting within the /27, these bits are the only ones that change, so we change them one bit at a time, giving us:


11100000 = 224

11100100 = 228

11101000 = 232

11101100 = 236

11110000 = 240

11110100 = 244

11111000 = 248

11111100 = 252


Note that bits 3-5 have taken on every possible combination that 3 bits can take ...


That is how you work it ...


Pls do remember to rate posts...



Paresh




carl_townshend Mon, 04/10/2006 - 04:34
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thanks for that, I am nearly there on this, Is it basically squeezing out more networks from an already subnetted address? Is there an easier way of doing this ?

pkhatri Mon, 04/10/2006 - 04:43
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It's all about flexibility, mate...


Say you have subnetted a /24 into 8 /27s. Now you need some /30s. Instead of subnetting the whole /24 into /30s, you sub-subnet one of those /27s, leaving the others intact. So your /24 now consists of 7 /27s and 8 /30s.


I'm afraid you have to learn to do this in binary before you can get comfortable with it. It looks a lot more complicated than it really is...


Pls do rate the post if it helps...



Paresh

Mohamed Sobair Mon, 04/10/2006 - 18:03
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Hi Carl,


It's not as difficult as u think it is, but every one might has his own calculations for subnetting, In the end, there is one standard for it, here is an example and explanation:-


lets say we have: 10.10.10.0/27 (that means it's a subnet of 32 IP's).


the available IP's range is from (10.10.10.1----10.10.10.30)


the ip 10.10.10.31 is considered broadcast ip.

the ip 10.10.10.32 is considered the next subnetwork.


so, now, we u spilit it up to /30s, it should be as follows:-


10.10.10.0/32 (that means it's a subnet of 4 ip's)


the available IP's range is from(10.10.10.1----10.10.10.2)


the ip 10.10.10.3 is considered broadcast ip


the ip 10.10.10.4 is the next subnetwork.


Now:


10.10.10.4/30 (that means it's a subnet of 4)


available ip's is (from 10.10.10.5-----10.10.10.6)


the ip 10.10.10.7 is considered broadcast ip (not usable)


the ip 10.10.10.8 is the next subnetwork.


And So On till we reach:


10.10.10.28/30 subnet


Available ip addresses (from 10.10.10.29-to-10.10.10.30)


the ip address 10.10.10.31 is considered a broadcast address (not usable)


the ip 10.10.10.31 is a next network ID (next subnetwork).


so simply, now we devided 10.10.10.0/27 to multiple 10.10.10.0/30s subnets as follows:


10.10.10.0/30

10.10.10.4/30

10.10.10.8/30

10.10.10.12/30

10.10.10.16/30

10.10.10.20/30

10.10.10.24/30

10.10.10.28/30


Note: the more lower /(bits) of subnet the more bigger lenght of usable IP's and vice verca


for example: Any ip address consists of 32 bits;


/32 highest bits means (255.255.255.255)(1 ip address available)

/30 high bits means (255.255.255.252)(2 ip addresses available)

/28 low bits meas (255.255.255.240) (14 ip addresses available)

/26 lower bits means (255.255.255.192) (62 ip addresses available).


And so on.....


pls rate the post if it helps,


Mohamed

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