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
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
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
The last address in the subnet would be the broadcast address: 192.168.3.255
The first host address is the network address + 1: 192.168.0.1
The last host address is the broadcast address - 1: 192.168.3.254
Since 192.168.3.0 is in the range of host addresses, it is a host address.
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.
IP address: 192.168.3.0
Network mask: 255.255.252.0
Subnet: 192.168.0.0 /22
Network address: 192.168.0.0
Broadcast address: 192.168.3.255
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 ?
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.
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.
So far, so good.if the subnet mask is /22
32
- 22
____
10
2^10 = 1024
1024 - 2 = 1022 host
Is this right way to do ??
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 255.255.252.0why 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 ?
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.
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
IP address: 192.168.3.0
Network mask: 255.255.252.0
Subnet: 192.168.0.0 /22
Network address: 192.168.0.0
Broadcast address: 192.168.3.255
In this subnet, you would have the following host addresses that end in .0:
192.168.1.0
192.168.2.0
192.168.3.0
The range of host addresses is 192.168.0.1 to 192.168.3.254, and the three addresses above fall inside that range.
As far as broadcast address, I agree. There is no binary math that allows for the broadcast address to end in .0
Thanks chopsticks for the diagram and nice trick for beginner
jibtechwill be my sir nice explanation
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