Help about VLSM

[imran_mcse]

Hi friends. I have used to study different papers for VLSM. But i am confused to understand exact idea behind this. Is there any link where can i find much easy and brief report about VLSM and how to use this pls help me in this regard as u can. I am confused about VLSM.

[Georg Pauwen]

Hello,

in a nutshell, VLSM simply means to use another than the classful mask. To give you an example: let´s say you want to use class B network 172.16.0.0. The default mask is 255.255.0.0, which gives you one subnet with 65534 hosts. Now you need several subnets, let´s say 4. The way to do it is to change the subnet mask to 255.255.192.0, which effectively gives you 4 subnets with 16382 hosts each. That is really all that there is to VLSM.

Check this paper for a good explanantion:

IP network design, part 3: IP addressing and routing

http://searchnetworking.techtarget.com/originalContent/0,289142,sid7_gci803101,00.html

HTH,

Georg

[scottmac]

VLSM also allows you to tailor the number of hosts within a specific address block (each subnet of the block can have variable number of hosts).

The KEY thing to remember is that the host ranges CANNOT overlap.

For example, with a Class C or twenty-four bit mask on the given address, you would not be able to provide twelve subnets with one containing up to 30 hosts.

Normally, you'd subnet at the fourth bit (/28 - 255.255.255.240), giving you

14 subnets (tossing Zero and all ones) with 14 hosts per subnet. But you have a requirement for one network with 30 hosts....

If you work out the subnets up to the eleventh subnet, then slide the mask one bit to the left / MSB side (/27 - 255.255.255.224), you can create at least one subnet with 30 hosts. The dividing line will be the X.X.X.192 subnet: everything below is a /28, everything .192 or above is a /27.

Assuming the address was 192.168.1.0 / 24 - 255.255.255.0

Subnet to 28 bits (/28, 255.255.255.240)

You'd get eleven subnets of fourteen hosts each:

192.168.1.16 (hosts .17 - .30, broadcast .31)

192.168.1.32 (hosts .33 - .47, broadcast .47)

192.168.1.48 (hosts .49 - .62, broadcast .63)

192.168.1.64 (hosts .65 - .78, broadcast .79)

192.168.1.80 (hosts .81 - .94, broadcast .95)

192.168.1.96 (hosts .97 - .110, broadcast .111

192.168.1.112 (hosts .113 - .126, broadcast .127)

192.168.1.128 (hosts .129 - .142, broadcast .143)

192.168.1.144 (hosts .145 - .158, broadcast .159)

192.168.1.160 (hosts .161 - .174, broadcast .175)

192.168.1.176 (hosts .177 - .190, broadcast .191)

At this point, change the mask to /27 - 255.255.255.224

You get the additional subnets, each with 30 hosts:

192.168.1.192 (hosts .193 - .222, broadcast .223)

192.168.1.224 (hosts .225 - .254, broadcast .255)

Note that none of the subnets overlap, and no host addresses overlap. If you create an overlap situation (two overlapping address ranges on the same router),the router will will kick back the address range with an "Address Overlap" error.

The more host bits you start with, the more options you'd have for subnetting / variable-length subnetting.

Hope this helps.

Good Luck

Scott

[keith.campbell]

http://www.kccommunications.com/public/VLSM%20summary%20notes.doc