04-10-2006 12:16 AM - edited 03-03-2019 02:44 AM
I have brushed upon vlsm, Where say you can have multiple /30's out of a /27, can anyone give me an easy example of this ?
thanks
04-10-2006 12:21 AM
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
04-10-2006 12:25 AM
can you explain how you did this please, thanks very much indeed !!
04-10-2006 12:33 AM
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
04-10-2006 04:34 AM
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 ?
04-10-2006 04:43 AM
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
04-10-2006 06:03 PM
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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