typically the router that you are using will take 192.168.1.1. It is a usable address, but it is already taken, so the first IP you can use for a device on that network would be 192.168.1.2
the last usable IP is 192.168.1.254 assuming you are using a subnet mask of 255.255.255.0
Keep in mind that if you are giving devices static IP addresses, it is recommended to reserve the IP address in the DHCP server (typically the router in a home network) or assign it outside of the scope of ip addresses. You can do this through the configuration of the router, typically by entering the router's ip into your browser. (make sure you set your wireless settings to wpa otherwise you will have little or no security)
192.168.1.64/29
There are 4094 usable IP addresses in a 20-bit subnet.
The highest usable IP address for non-multicast devices is 223.255.255.254 The highest usable multicast IP address is 239.255.255.254
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
The 169.254/16 network is a reserved range of IP addresses, which a client can assign to itself in case a DHCP lease cannot be obtained (RFC 3927). The idea is that, if the DHCP server fails to respond, the requesting device can self-assign a possibly suitable address (such as 169.254.100.88) in order to obtain any usable IP address. In reality, this scheme doesn't really work, however, as the resulting address is almost always on a not supported subnet, and the self-assignment lacks other important details (such as the gateway address).
N - network bitsH - host bitsIf you are following the first octet rule, this is a class B network with a subnet mask of 255.255.0.0 (N.N.H.H). We can subnet this by "borrowing" some bits from the host portion. There are 16 network bits and 16 host bits. There is a simple formula to calculate the proper subnet mask.2 to the power of what equals at least 510(2^X)? We have a total of 16 host bits to borrow from. 2^1...2^2...2^3...Etc2^9= 512 - 2 = 510 host addressesWe subtract two because the network and broadcast address are not usable addresses. As we can see we need at least 9 host bits to get 510 hosts per subnet.Take 32 and subtract it from the host bits you need. So 32-9=23. Your subnet mask now has 23 network bits instead of 16.In binary the original subnet mask would be 11111111.11111111.00000000.00000000.In binary the new subnet mask is 11111111.11111111.11111110.00000000.If you convert this into dotted decimal form you get 255.255.254.0.TLDR: 172.30.0.0 - 172.30.1.255255.255.254.0
172.16.255.254.
Depends on the subnet. For the subnet of 192.168.1.0, the first usable IP address is 192.168.1.1. This is typical of a default wireless router setup. Valid usable IP addresses under this scenario is 192.168.1.1 to 192.168.1.254.
network: 172.16.2.128 Broadcast: 172.16.2.255 usable host range: 172.16.2.129 thru .254
because that is the last number of the subnet. the last number in a subnet is used as the broadcast domain. the first number is also not usable. an example would be: id 192.168.20.XX subnet 0f 255.255.255.128 192.168.20.0 and 192.168.20.127 may not be used and 192.168.20.128 starts the next subnet making 192.168.20.128 and 192.168.20.255 not usable
NetRange: 135.17.0.0 - 135.17.255.255 CIDR: 135.17.0.0/16
There are 4094 usable IP addresses in a 20-bit subnet.
with the default subnet mask, the number of clients in a class A network is: 16,777,214
0001 0110 001 is the first valid host address of 193.168.22.1.
The host range is 192.168.10.193 - 206; total usable hosts =14.
.Hello everyone, Here's my overview: Hello everyone, I seem to have a problem with this custom subnet masking problem. Number needed usable hosts:60 Network Address:198.100.10.0 I've already figured these: Address Class:C Default Subnet Masks 255.255.255.0 Total Number Of Host Addresses:64 Number Of Usable Address:62. Cannot Figure Out: Total Number Of Subnets, Custom Subnet Masks (I got 255.255.255.240 but checked online and wasn't right). Can someone help me with this problem? Thanks.
It must be an IP address in the same subnet. Other than that, the network designer has the flexibility to assign any IP address in the same subnet. Quite often, the first or last IP addresses in a subnet are used, so if a certain interface on this router has IP address 10.0.5.1, and the network mask is 255.255.255.0, I would strongly suspect that the other router has IP address 10.0.5.2 (10.0.5.0 can't be used for this subnet).For serial (point-to-point) connections, to save address space, quite often a subnet /30 is used (that is, a subnet mask of 255.255.255.252), in which case the other router has the only other usable IP address in the subnet. For example, if this router has IP address 10.0.8.26 and a subnet mask 255.255.255.252, the subnet has addresses in the range 10.0.8.24 - 10.0.8.27, but since the first and last addresses can't be used, the only option for a router or other machine on the other end is 10.0.8.25.It must be an IP address in the same subnet. Other than that, the network designer has the flexibility to assign any IP address in the same subnet. Quite often, the first or last IP addresses in a subnet are used, so if a certain interface on this router has IP address 10.0.5.1, and the network mask is 255.255.255.0, I would strongly suspect that the other router has IP address 10.0.5.2 (10.0.5.0 can't be used for this subnet).For serial (point-to-point) connections, to save address space, quite often a subnet /30 is used (that is, a subnet mask of 255.255.255.252), in which case the other router has the only other usable IP address in the subnet. For example, if this router has IP address 10.0.8.26 and a subnet mask 255.255.255.252, the subnet has addresses in the range 10.0.8.24 - 10.0.8.27, but since the first and last addresses can't be used, the only option for a router or other machine on the other end is 10.0.8.25.It must be an IP address in the same subnet. Other than that, the network designer has the flexibility to assign any IP address in the same subnet. Quite often, the first or last IP addresses in a subnet are used, so if a certain interface on this router has IP address 10.0.5.1, and the network mask is 255.255.255.0, I would strongly suspect that the other router has IP address 10.0.5.2 (10.0.5.0 can't be used for this subnet).For serial (point-to-point) connections, to save address space, quite often a subnet /30 is used (that is, a subnet mask of 255.255.255.252), in which case the other router has the only other usable IP address in the subnet. For example, if this router has IP address 10.0.8.26 and a subnet mask 255.255.255.252, the subnet has addresses in the range 10.0.8.24 - 10.0.8.27, but since the first and last addresses can't be used, the only option for a router or other machine on the other end is 10.0.8.25.It must be an IP address in the same subnet. Other than that, the network designer has the flexibility to assign any IP address in the same subnet. Quite often, the first or last IP addresses in a subnet are used, so if a certain interface on this router has IP address 10.0.5.1, and the network mask is 255.255.255.0, I would strongly suspect that the other router has IP address 10.0.5.2 (10.0.5.0 can't be used for this subnet).For serial (point-to-point) connections, to save address space, quite often a subnet /30 is used (that is, a subnet mask of 255.255.255.252), in which case the other router has the only other usable IP address in the subnet. For example, if this router has IP address 10.0.8.26 and a subnet mask 255.255.255.252, the subnet has addresses in the range 10.0.8.24 - 10.0.8.27, but since the first and last addresses can't be used, the only option for a router or other machine on the other end is 10.0.8.25.
that gives you 16 subnets with 14 usable IPs for hosts that is because one is for subnet and one for broadcas in that subnet for example: 192.168.1.0/28 - subnet number 192.168.1.15 -broadcast number usable IPs for hosts - IPs between them that is 14