VLSM - Calculation

[Yasmeen]

Hi Team, 

          Is It Possible to divide vlsm small block size to big block size ?

I am trying to break my major network 192.168.128.0/20 in to 13 subnets as  follows,

192.168.128.0/20

first block i need - 6 host 

second block - 6 host

Third Block - 16 Host

Fourth Block - 30 Host

Fifth Block - 62 Host

Sixth block - 62 Host

seventh block  - 606 Host

eighth Block -  508 Host

Ninth Block - 303 Host

Tenth Block - 126 Host 

Eleventh  Block - 126 Host

Twelth Block - 120 Host

Thirteenth Block - 126 Host 

The Same Order My ip address need to be arranged, but i am getting as DESCENDING Order, 

I need to divide the Blocks from Small to Big, But i am getting from Big to Small. 

Team please help to Divide as i need.

I Need to Arrange my Network from small to big not as Big network to small one

Thanks,

Regards,

Yasmeen Shaul Hameed.

[Giuseppe Larosa]

Hello Yasmeen,

Given you requirements:

>>

I am trying to break my major network 192.168.128.0/20 in to 13 subnets as follows,

192.168.128.0/20

The rule to use for finding the minimum subnet that can have the requested number of hosts is:

2^H -2

where H is the number of bits of the host portion of the address H = 32 - N where N is the prefix length

The final letters X, Y, Z, W, W2 have been used to accomodate the calculations that follows

first block i need - 6 host ----- >  /29     W2

second block - 6 host     ------>  /29     W2

Third Block - 16 Host      ------> /27   ( notice that /28 provides only 14 hosts)    W

Fourth Block - 30 Host    ------> /27                                                                W

Fifth Block - 62 Host       ------> /26     Z

Sixth block - 62 Host      ------> /26     Z

seventh block - 606 Host  ----- >  /22    X

eighth Block - 508 Host    ----- >  /23     X

Ninth Block - 303 Host    ------> /23       X

Tenth Block - 126 Host     -----> /25       Y

Eleventh Block - 126 Host   ----> /25      Y

Twelth Block - 120 Host  ------> /25      Y

Thirteenth Block - 126 Host  ----> /25    Y

/20 is equivalent to 16 /24 subnets.

However, you take the base address 192.168.128.0/20

You can write in binary the interesting byte byte 3:

128 = 10000000

the /20 is a line between 4th and 5th bit

1000 | 0000.

You start by dividing the address block in four /22 IP subnets

1000|00||00  ------->  192.168.128.0/22

1000|01||00 ------->  192.168.132.0/22

1000|10||00 ------->  192.168.136.0/22

1000|11||00 ------->  192.168.140.0/22

We can start by using the first /22 to accomodate the subnet that requires more hosts.

192.168.128.0/22 ----->  606 hosts.

The we take the next 192.168.132.0/22 and we partition it in two /23

192.168.132.0/23  --->   508 hosts

192.168.134.0/23 ---->  303 hosts

At this point we take the next /22

192.168.136.0/22 and we create 8 /25 subnets

192.168.136.0/25

192.168.136.128/25

192.168.137.0/25

192.168.137.128/25

192.168.138.0/25

192.168.138.128/25

192.168.139.0/25

192.168.139.128/25

The first four are used to accomodate the four subnets that require a /25 prefix length.

192.168.136.0/25

192.168.136.128/25

192.168.137.0/25

192.168.137.128/25

At this point you can use the following /25 subnets with further subnettting to accomodate the remaining subnets

192.168.138.0/25 split:

192.168.138.0/26       Z

192.168.138.64/26     Z

Then we take the next /25

192.168.138.128/25 and we can subnet it into four /27 subnets

Interesting byte is 4th

128 = 10000000     the subnet line is after three digits

100|||00000       192.168.138.128/27     W

101|||00000       192.168.138.160/27     W

110|||00000       192.168.138.192/27     W2

111|||00000       192.168.138.224/27     W2

So you can accomodate all subnets and you leave the last /22 unused 192.168.140.0/22 for future use

Hope to help

Giuseppe

[Yasmeen]

hi Giuseppe Larosa,
Thanks For Quick Reply, My Question is
Is It possible to break the Major Network from smaller to bigger
Like I said in previous Message, I need to Divide top in smaller block size and below should be in big block size

[Yasmeen]

Can U divide this as same order for 192.168.128.0/20

[Yasmeen]

HI,
First block size should be i need as smaller block size after that bigger size, can u pls show me for 192.168.128.0/20

[Giuseppe Larosa]

Hello Yasmeen,

I have given you the methodology you should be able to make different choices.

However, you need to start to subnet to less specific subnets /22 then you choice how to use the four /22 subnets.

Hope to help

Giuseppe

[Richard Burts]

@Giuseppe Larosa has demonstrated the approach that many of us would use, which is to take the address block and break it down into its components and assigning appropriate subnets and masks starting with the largest blocks and working down to smaller blocks. The original poster wants to do it differently and begin with the smallest blocks and working up to the larger blocks. I do not understand why it matters which direction you work, but apparently there is something in the original situation that we do not understand. I do not see any reason why the original poster would not be able to work through the process starting from the smallest blocks and working up to the larger blocks. 

HTH

Rick

[Yasmeen]

HI Giuseppe Larosa,
I have found Solution for our office network, Thanks for your help. My doubt is in vlsm why the block size start from bigger size then smaller size
My plan is to assign initial ip address range for small network after that remaining ip address for larger network
my network 192.168.128.1 to 192.168.128.4 need to assign on core layer devices
after that 192.168.128.10 to 192.168.128.150 for another network, 192.168.128.151 to 192.168.128.255 for another network,
In this case initial set of ip address assigned to department networks. not for core layer devices

[Jon Marshall]

You can't just pick random numbers to start at eg. 192.168.128.10 because subnetting does not work that way ie, you can't use 192.168.128.1 - 4 as one subnet and then use 192.168.128.10 -> 150 for another subnet because it won't work. 

If you really need 140 IPs in a subnet then you need to use the whole 192.168.128.0/24 subnet for those IPs and then you would need another 192.168.x.0/24 subnet for core devices etc. 

When you subnet first work out how many IPs you need and that will tell you what subnet mask to use and also what is left to use for other subnets.

Jon

[Giuseppe Larosa]

Hello Yasmeen,

IP addresses are binary unsigned integers made of four bytes (octets).

The dotted decimal notation has been introduced only to help us as human beings when dealing with IPv4 addresses.

However, when you want to perform operations like subnetting (or the contrary to summarize IP subnets) you need to remember the binary nature of IPv4 addresses.

A.B.C.D ---->   a7.. a0.b7...b0.c7..c0.d7..d0   each octet is made of 8 binary digits called  bits.

for example binary 138 = 128+8+2 = 10001010

subnetting means moving the network boundary to the right and so reducing the host portion of the address.

further subnettting means moving again the network boundary to the right ,futher reducing the host portion of the address.

>> My doubt is in vlsm why the block size start from bigger size then smaller size

It is usually the way to allocate more efficiently IP addresses.

The only valid IP address blocks must be powers of two as the network boundary is always in a given "bit position".

As explained by Jon your proposed "subnetting " is wrong

>>my network 192.168.128.1 to 192.168.128.4 need to assign on core layer devices
after that 192.168.128.10 to 192.168.128.150 for another network, 192.168.128.151 to 192.168.128.255 for another network,

This is not possible.

>> have found Solution for our office network

not the one proposed above because it does not work.

You can use binary math or other tricks to make calculations, but the underlying nature of IPv4 addresses is binary.

Read again my first post in this thread the "|" character has been used to show the subnetting border.

When I use "||" I mean further subnetting and you see it is more on the right then "|".

Some study and some exercises will make you comfortable with IP subnetting and VLSM, all of us had to go via this learning process (and we need to review it from time to time).

Hope to help

Giuseppe

[paul driver · Accepted solution]

Hello

Please attached file for possible subnetting your 192.168.128.0/20 - As the others have mentioned you need to split the original /20 subnet range into smaller subnets and once that subnet has been allocated it CANNOT be used for another network or anything related below it.

In the example posted I have split that original /20 subnet into two /21 subnets both having 2046 hosts each, Then again split those two /21 subnets into Four /22 subnets each supporting 1022 hosts.

Of the first /21 subnet you can see that its two /22 subnets , The first /22 was allocated to 7th block  and the other /22 was further split into two /23 subnets each for 8th -9th blocks , so now that original /21 cannot be used for anything else.

The same logic needs to be applied to the others ip ranges as you go about your ip allocation

first block i need - 6 host <----192.168.138.184/29

second block - 6 host      <----192.168.138.176/29

Third Block - 16 Host       <----192.168.138.160/28

Fourth Block - 30 Host     <----192.168.138.129/27

Fifth Block - 62 Host        <----192.168.138.0/26

Sixth block - 62 Host       <----192.168.138.64/26

seventh block - 606 Host <----192.168.132.0/22

eighth Block - 508 Host    <----192.168.128.0/23

Ninth Block - 303 Host     <----192.168.130.0/23

Tenth Block - 126 Host    <----192.168.136.0/25 126

Eleventh Block - 126 Host <---192.168.136.129/25

Twelth Block - 120 Host    <----192.168.137.0/25

Thirteenth Block - 126 Host <---192.168.137.129/25

192.168.140.0/22 Spare
192.168.139.0/24 Spare