class A ip has 16777216 host but valid hosts are only 16777214.(2^24-2)
So answer is 16777214 hosts
There are 2^24 host in class A but (2^24)-2 hosts are valid since the first and last address are reserved .
First octet rule for each class:Class A: 0xxxxxxxClass B: 10xxxxxxClass C: 110xxxxxClass A range is 0 - 1270.0.0.0 and 127.0.0.0 are not "routable" IP addresses. One defines all networks and the other is the loopback. We have a total of 126 usable networks and 16,777,214 usable hostaddresses per network. There are even less if we don't count the private address of 10.0.0.0 (RFC 1918).Class B range is 128 - 191There are 16,384 total networks in this class; that's including the private addresses of 172.16.0.0 - 172.31.0.0 (RFC 1918). There are a total of 65,534 usable host addresses per network.Class C range is 192 - 223There are 2,097,152 total networks in this class; that's including the private addresses of 192.168.0.0 - 192.168.255.0 (RFC 1918). There are 254 usable hosts addresses per network.
16 384 networks are available in Class B network.
If you need to divide it up into the maximum number of subnets containing at least 500 hosts each, you should use a /23 subnet mask. This will provide you with 128 networks of 510 hosts each. If you used a /24 mask, you would be limited to 254 hosts. Similarly, a /22 mask would be wasteful, allowing you 1022 hosts.
By default Class C subnet mask is 255.255.255.0 = 24 bits for network id and 8 bits for host id. in Binary 1111 1111. 1111 1111. 1111 1111. 0000 0000 Here all 1s are Network bits and all 0s are host bits. For this subnet mask you can have 256 hosts. And you can use 254 host and asign IP address to them. By Saurabh
16 000 000
10000
62 hosts.
Set up a series of scope ranges in DHCP, or if you have a class A or class B address use a range that would give you that many addresses total. Then make sure that all the clients use DHCP for their addressing.
NAT
number of hots bits is 16 number of hosts 65 534
How many possible host addresses are there in a Class A range?Class A range is 0 - 1270.0.0.0 and 127.0.0.0 are not "routable" IP addresses. One defines all networks and the other is the loopback. We have a total of 126 usable networks and 16,777,214 usable hostaddresses per network.
There are 2^24 host in class A but (2^24)-2 hosts are valid since the first and last address are reserved .
The default mask class B is 255.255.0.0 and this makes 16 bits available for hosts
First octet rule for each class:Class A: 0xxxxxxxClass B: 10xxxxxxClass C: 110xxxxxClass A range is 0 - 1270.0.0.0 and 127.0.0.0 are not "routable" IP addresses. One defines all networks and the other is the loopback. We have a total of 126 usable networks and 16,777,214 usable hostaddresses per network. There are even less if we don't count the private address of 10.0.0.0 (RFC 1918).Class B range is 128 - 191There are 16,384 total networks in this class; that's including the private addresses of 172.16.0.0 - 172.31.0.0 (RFC 1918). There are a total of 65,534 usable host addresses per network.Class C range is 192 - 223There are 2,097,152 total networks in this class; that's including the private addresses of 192.168.0.0 - 192.168.255.0 (RFC 1918). There are 254 usable hosts addresses per network.
in Class A addresses the first bit identifies the class. The next 7 bits identify the network and the rest are the IP's that belong to that network. However these networks are then broken down using subnets . Class A networks are not given to private clients or small private organisations. So if you want to know how many bits are there for identifying hosts in a Class A network, the answer is 32-8 = 24 bits. However if you mean how many bits are reserved for hosts in your private network space, that depends on your subnet and router. Most routers use NAT to allow multiple hosts to use a single external IP address. I hope this answers your question :)
16 384 networks are available in Class B network.