09-05-2019 07:02 AM
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.
Solved! Go to Solution.
09-10-2019 04:56 AM - edited 09-10-2019 08:39 AM
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
09-05-2019 07:23 AM - edited 09-05-2019 07:55 AM
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
09-05-2019 07:47 AM
09-05-2019 07:50 AM
09-05-2019 07:51 AM
09-05-2019 08:03 AM
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
09-06-2019 06:26 AM
@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
09-09-2019 12:01 AM
09-09-2019 12:20 AM
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
09-09-2019 08:33 AM
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
09-10-2019 04:56 AM - edited 09-10-2019 08:39 AM
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
Discover and save your favorite ideas. Come back to expert answers, step-by-step guides, recent topics, and more.
New here? Get started with these tips. How to use Community New member guide