# [question] subnetting doubt

Member Posts: 10 ■□□□□□□□□□
hello folks,

i have following ip 192.168.3.0 255.255.252.0

Question :
1. I want to know whether this ip is broadcast or network or host address
2. how many total host ?

Thanks
~indra26

• please let me know how do you guys find whether ip is broadcast or network or host address
• Member Posts: 401 ■■■■□□□□□□
Follow the basics of subnetting. The third octet is the interesting one.

256-252 = 4 addresses per "chunk". So you have the following subnets:

192.168.0.0
192.168.4.0
192.168.8.0
192.168.12.0
...and so on.

192.168.3.0 falls between the first two, so it is part of the 192.168.0.0 subnet.

The first address in the subnet would be the network address: 192.168.0.0

Since 192.168.3.0 is in the range of host addresses, it is a host address.
• Member Posts: 401 ■■■■□□□□□□
To find the number of subnets and number of hosts...

192.168.0.0 is a /16 network.

Since there are 32 bits in the full address, there are 16 bits available for subnet and host ranges.

The host bits are identified with the mask. Since a 255.255.252.0 mask is equal to a /22 subnet, 22-16 indicates 6 bits for the subnets.

2^6 = 64, so there are 64 subnets available.

Subtract the subnet bits from the available range. 16-6 = 10 bits.

2^10 = 1024. But you have to subtract two for the network address and the broadcast address.

1024 - 2 = 1022 host addresses.

Subnet: 192.168.0.0 /22
Subnet bits: 6
Host bits: 10
Number of subnets: 64
Number of hosts: 1022
• Thanks for reply first problem resolved.
192 series is in class c the default subnet is /24
i do this way to find total host
total bits is 32
32
- 24
____
8

2^8 = 256
256-2 = 254 host

if the subnet mask is /22
32
- 22
____
10

2^10 = 1024
1024 - 2 = 1022 host
Is this right way to do ??

why did you minus 22-16 suppose if the subnet mask is 27, 27-24 indicates 3 bits for the subnets ?
2^3 = 8 so there are 8 subnets available.
24-3 = 21 bits
2^21 ?
• indra26 wrote: »
why did you minus 22-16 suppose if the subnet mask is 27, 27-24 indicates 3 bits for the subnets ?
2^3 = 8 so there are 8 subnets available.
24-3 = 21 bits
2^21 ?

Where N = Network Portion, H = Host Portion

NNNNNNNN.NNNNNNNN.NNNNNNNN.HHHHHHHH (Classful Network)

NNNNNNNN.NNNNNNNN.NNNNNNNN.NNNHHHHH (Classless Netnetwork, Subnetted the Classful network by borrow 3 bits from the Host Portion)

2^3 = 8 subnets created
2^5 - 2 = 30 hosts in each of the 3 subnets

Hopefully this becomes clearer for you.
• Member Posts: 401 ■■■■□□□□□□
indra26 wrote: »
Thanks for reply first problem resolved.
192 series is in class c the default subnet is /24
i do this way to find total host
total bits is 32
32
- 24
____
8

2^8 = 256
256-2 = 254 host

Slight misconception here. The 192.168.x.x reserved address space is actually a /16 space. It includes all addresses from 192.168.0.0 to 192.168.255.255. The subnets typically created in this range are /24.
192.168.0.0 /24
192.168.1.0 /24
192.168.2.0 /24
... and so on.
if the subnet mask is /22
32
- 22
____
10

2^10 = 1024
1024 - 2 = 1022 host
Is this right way to do ??

So far, so good.
why did you minus 22-16 suppose if the subnet mask is 27, 27-24 indicates 3 bits for the subnets ?
2^3 = 8 so there are 8 subnets available.
24-3 = 21 bits
2^21 ?

This has to do with the "interesting octet" concept. This is the first octet, counting from the left, that does not have a value of 255. Each octet with a value of 255 consists of 8 bits. A mask of /22 is 255.255.252.0
We know the first octet is 8 bits.
We know the second octet is 8 bits.

The third octet is the first octet that does not equal 255, or 8 bits, so it is the "interesting octet". Anything in that octet is a "borrowed bit".

Since the first two octets equal 16 bits, we subtract 16 from the mask /22, to identify the number of borrowed bits. For /22 that is 6 borrowed bits.

Now look at a /27, which is 255.255.255.224.
First octet is 8 bits.
Second octet is 8 bits.
Third octet is 8 bits.
Fourth octet is less than 255, so it is the "interesting octet". We subtract the known bits (8+8+ from the mask (/27), to establish the number of borrowed bits. 27-24 = 3 borrowed bits. Therefore, there are 2^3 subnets available. 2^3 = 8 subnets.

Since there are only 32 bits in an IPv4 address, we subtract the mask from 32, and we get 5 bits. 2^5 = 32 addresses. But, we have to subtract 2 for the network and broadcast addresses, so 32-2 = 30 available host addresses in each of those subnets.
• Member Posts: 401 ■■■■□□□□□□
chopsticks wrote: »
Where N = Network Portion, H = Host Portion

NNNNNNNN.NNNNNNNN.NNNNNNNN.HHHHHHHH (Classful Network)

NNNNNNNN.NNNNNNNN.NNNNNNNN.NNNHHHHH (Classless Netnetwork, Subnetted the Classful network by borrow 3 bits from the Host Portion)

2^3 = 8 subnets created
2^5 - 2 = 30 hosts in each of the 3 subnets

Hopefully this becomes clearer for you.

Good diagram, chopsticks. This may make it a bit clearer.
• Member Posts: 1,539 ■■■■■■□□□□
indra26 wrote: »
i have following ip 192.168.3.0 255.255.252.0

Question :
1. I want to know whether this ip is broadcast or network or host address
Am I missing something here? You will never have a host nor a broadcast having an IP of .0.
• Member Posts: 401 ■■■■□□□□□□
TechGuru80 wrote: »
Am I missing something here? You will never have a host nor a broadcast having an IP of .0.

I am pretty sure you can have a host with a .0, and this particular case is an example. With a network address of 192.168.0.0 /22, you would have 3 different host addresses that end in .0

Subnet: 192.168.0.0 /22
jibtech will be my sir nice explanation 