Subnetting question

[The_guroo_2]

Guys can some one explain to me the follwoing

if we have 192.168.1.0/24 address

and we want t furthur subnet that in to /28 (how many addreses we will get)

what is the way to do it and how does the calcultion work number of subnets etc

I cant get my head around

Thanks heaps

[fgasimzade]

Here is a good subnet calculator for you

http://www.subnet-calculator.com/

If you make it /28 or 255.255.255.240, it means you will get 16 addresses. First address will be your network address, last address - broadcast address. After all, you will get 14 hosts per subnet, and 16 subnets overall

[cadet alain]

Hi,

there are 2 ways to do the math:

the long way which is in binary or the short way which is in decimal

192.168.1.0/24 is your network address and that means you have 24 network bits(first 3 octets) and 8 host bits(the last octet).

You want a /28 so you are subnetting so borrowing bits from host part to make them subnet part

How many? 4 because 28-24=4 so how many subnets is this? 2 to the power of 4= 16 subnets

How many hosts in each subnet? (2 to the power of 4 ) minus 2 for broadcast and subnet address= 14

where did this 4 come from : 32-28=4

the binary way:

Now you're gonna have something like this: NNNNNNNN.NNNNNNNN.NNNNNNNN.SSSSHHHH

let's take last octet:

00000000 first subnet  192.168.1.0

00000001 first address in this subnet 192.168.1.1

00001111 broadcast 192.168.1.15

00001110 last address 192.168.1.14

0001000 2nd subnet 192.168.1.16

00010001 first address in this subnet 192.168.1.17

00011111 broadcast 192.168.1.31

00011110 last address in this subnet 192.168.1.30

Or the decimal way

a /28 is 255.255.255.240 so 256-240= 16 is the subnet increment  so

1st subnet 192.168.1.0/28

2nd subnet 192.168.1.16/28

and so on and in first subnet: 192.168.1.1 is first address, 192.168.1.15 is broadcast( next subnet-1) and 192.168.1.14 is last address( broadcast -1)

Regards.

Alain.

[ciscoamit_497]

Hi Frnd,

Your question is quite easy..You have questioned abt the supernetting where we do subnetting of the subnets.

Alain shown you a proper solution in all ways.

And for more detailed knowledge you can refer the attached document.

Regards,

Amit

[Sakun Sharma]

See subnetting basically is used to divided and network into sub-networks.

Subnetting in IPv4 is done via changing subnet mask bits, borrowing 1's from host bit portion.

/ 24 means there are 24 - 1's in subnet mask and rest all are 0's out of 32 bits.

0 - defines hosts bits

1 - defines network bits

24 is subnet mask for Class C IP Address, if you borrow more bits from it, then that network will be subnetted.

As in your example,

Network is - 192.168.1.0

Subnet Mask - 255.255.255.0

When you subnet it into /28 means from last subnet which was in binary 00000000 (eight zeros) you have converted 4 bits to one, so now subnet mask is in binary - 11110000 which is 128 (2 ^ 8) + 64 (2 ^ 7) + 32 (2 ^ 6) + 16 (2 ^ 5) = 240

Complete subnet mask becomes - 255.255.255.240

Number of subnets you got is = 2 raise to power ( ^ ) no of bits you have converted from 0 to 1 = 2 ^ 4 = 16 subnets

Number of hosts per subnet is = 2 raise to power ( ^ ) no of 0 remaining = 2 ^ 4 = 16 hosts per subnet

For verification = 16 x 16 = 256 (total hosts available with /24 subnet mask)

So your subnets will be:

Subnet No - Starting Address - Ending Address

1 - 192.168.1.0 - 192.168.1.15 (192.168.1.0 + 16 = 192.168.1.15)

2 - 192.168.1.16 - 192.168.1.31 (16 to 31 including 16 and 31 = 16 hosts)

3 - 192.168.1.32 - 192.168.1.47

4 - 192.168.1.48 - 192.168.1.63

5 - 192.168.1.64 - 192.168.1.79

6 - 192.168.1.80 - 192.168.1.95

7 - 192.168.1.96 - 192.168.1.111

8 - 192.168.1.112 - 192.168.1.127

9 - 192.168.1.128 - 192.168.1.143

10 - 192.168.1.144 - 192.168.1.159

11 - 192.168.1.160 - 192.168.1.175

12 - 192.168.1.176 - 192.168.1.191

13 - 192.168.1.192 - 192.168.1.207

14 - 192.168.1.208 - 192.168.1.223

15 - 192.168.1.224 - 192.168.1.239

16 - 192.168.1.240 - 192.168.1.255

Usable hosts are always total hosts minus two = 16 - 2 = 14 hosts (1 for Network Address and 1 for Broadcast Address)

[Joseph W. Doherty]

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Posting

Another way to look at this your /24 provides 256 addresses.  The first address we normally set aside as to avoid confusion with the subnet address itself.  The last address is the broadcast address.  (NB:  /31 and /32 special cases)

When you shift the prefix by one value, you impact the address space by a power of two.

Taking a /24 and splitting it into a /25 results in two /25s each with half the addresses (i.e. 128 total with 1 for subnet, 1 for broadcast and 126 for hosts)

Taking a /25 and splitting it into a /26 results in two /26s each with half the addresses (i.e. 64 total with 1 for subnet, 1 for broadcast and 62 for hosts)

If these /26s are from the same /24, we would have four of them (or we could have one /25 and two /26s, if doing classless).

Taking a /26 and splitting into a /27 results in two /27s each with half the addresses (i.e. 32 total with 1 for subnet, 1 for broadcast and 30 for hosts)

If these /27s are from the same /24, we could have eight of them.

Lastly taking a /27 and splitting into a /28 results in two /28s each with half the addresses (i.e. 16 total with 1 for subnet, 1 for broadcast and 14 for hosts)

If these /28s are from the same /24, we would have sixteen of them.