VLSM, magic number and sequence of correct steps

[John Cheetley]

Evening Cisco Techs again

Please let me know if this is in the correct forum..

I understand inverse subnettting but trying to understand some steps for VLSM

Example

170.100.0.0 /16

100 users in 1st allotment

60 users in 2nd allotment

2 in 3rd and 2 in 4th

My workout

170.100.0.0 /16

Mask is 255.255.255.0

Magic number is 128 - 1 = 127

So first subnet range is 170.100.0.1 - 170.100.0.127 /17 with 1 bit borrowed

255.255.255.128 is mask

2nd allotment equates to 64 magic number. I know on binary table 64 is 2nd bit borrowed from left but as we have used the 1st bit already from (128) wasn't sure if that gets included 

170.100.0.128 - 170.100.0.192 /25 (is this correct)

255.255.255.192

I have done more theory but is based on above concept

Thanks again. :) 

[Julio E. Moisa]

Hi

If you are using VLSM for hosts you should use the subnet (2^n)-2 >= # hosts, attached you will find a document that I created a time ago. 

For example

Supernet is: 170.100.0.0 /16

Subnetting for 100 host,

So first identify how many bits you have to subnetting:

Formula:

32 - (CIDR) = # available bits

32 - 16 = 16, so you have 16 bits available to subnetting.

Ok determinate the value of "n" to complete the formula:

(2^n)-2 >= # hosts

(2^n)-2 >= 100

n = 7

(2^7) - 2 >= 100

(128) - 2 >= 100

126 >= 100

ok now "n" will represent the used bits for hosts, now how we are subnetting per hosts these bits are represented as 0 (zeros)

So you will have 7 zeros,  This 7 zeros will be moved from right to left from the current subnet mask:

Current mask: 255.255.0.0

Remember that each octect has 8 bits so if we convert the current subnet mask to binary we have:

255.255.0.0  =  11111111.11111111.00000000.00000000

Ok then select your 7 zeros from the binary (remember right side to left side):

255.255.0.0  =  11111111.11111111.00000000.00000000

The other zeros will be converted to 1 (ones) to find you new mask:

11111111.11111111.11111111.10000000

Now your new subnet mask is:

255.255.255.128  <--- it can be represented as /25 (sum of the all (ones) bits)

And you must follow the same procedure respecting the taken bits and the new subnet masks.

If you want to know the block of the subnets you could use the following formula:

256 - modified octect    (in this case the modified octect is the fourth)

256 - 128 = 128  so your subnet will be moving 128 in 128 on the fourth octect.

170.100.0.0/25    <--- You can use this subnet for the 100 hosts
170.100.0.128/25    <---- You can use this subnet to obtain 

There are many methods to subnetting, 

Hope it is useful

:-)

[Ganesh Hariharan · Accepted solution]

Hi ,

Julio has given depth of explanation on VSLM, so i would suggest to view following links for more practice and clarification...

As , VLSM is a process of dividing an IP network into the subnets of different sizes without wasting IP addresses. When we perform Subnetting, all subnets have the same number of hosts, this is known as FLSM ( Fixed length subnet mask). 

Have a look on below links..

http://www.computernetworkingnotes.com/ccna-study-guide/vlsm-tutorial-with-examples.html

http://www.orbit-computer-solutions.com/variable-length-subnet-mask-vlsm/

http://www.cisco.com/c/en/us/support/docs/ip/routing-information-protocol-rip/13788-3.html

Hope it Helps..

-GI

[guinader]

Hi i know this is an old question but i was searching through and i feel the was a bit of a mix up.   Per slash notation /25  gives you 126 hosts. 

Count the 0s on the binary sequence, of the mask. 2^n -2 = number of hosts. So first range is 

170.100.0.0 /25 valid hosts 0.1- 0.126 with 0.127 broadcast. 

Second you want 60 hosts, so block size of 64 gives your 62 hosts, so from previous block ended at 0.127, your next starting block is 128, or 170.100.0.128 (sub net) for the block of 64 you have /26 mask. That gives you 62 hosts. 

Subnet is .128, and 128+64= 192 (next block) so 191 is your broadcast. And in between is your valid range.  So: 

170.100.0.128 subnet

170.100.0.129 to 170.100.0.190 valid range 62

170.100.0.191 broadcast. That takes care of your first two large blocks. 

Best you need to hosts the slash notation is /30 which gives your 2 hosts, and a block size of 4. So you can calculate from here.  But a hint your 4th subnet broadcast address should be 170.100.0.199. and next block would start at 170.100.0.200.   

Someone correct me if I'm wrong, but reading through the answers no one actually gave the explanation of the block sizes and where they were fitting in the range. This takes hold of 192 hosts spots on your whole network. 

100+60+2+2 =164 but due to blocks size

126+62+2+2=192 hosts. 

P.s. /16 = 255.255.0.0 not 255.255.255.0 that's /24