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VLSM

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

6 REPLIES
Purple

Re: VLSM

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

Community Member

Re: VLSM

can you explain how you did this please, thanks very much indeed !!

Purple

Re: VLSM

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

Community Member

Re: VLSM

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 ?

Purple

Re: VLSM

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

Re: VLSM

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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