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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)
What else can I help you with?
How many usable IP addresses are in a 20-bit subnet?
There are 4094 usable IP addresses in a 20-bit subnet.
What is last usable ip address?
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
How many network and host bits are available for class C IP address?
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
What could cause a home computer to get the IP address 169.254.100.88?
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).
As network administrator what is the subnet mask that allows 510 hosts given the IP 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
What is the usable host for 176.34.56.91?
To determine the usable hosts for the IP address 176.34.56.91, we need to know its subnet mask. Assuming a common subnet mask of 255.255.255.0 (or /24), the usable host range would be from 176.34.56.1 to 176.34.56.254, meaning there are 254 usable IP addresses in this subnet. The first address (176.34.56.0) is the network address, and the last address (176.34.56.255) is the broadcast address.
What is the last usable address for subnet 172.16.0.0?
172.16.255.254.
What is the first valid address on a subnet?
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.
What are the first four addresses in the 192.168.1.64?
The first four addresses in the 192.168.1.64 subnet, assuming a typical subnet mask of 255.255.255.0 (or /24), are 192.168.1.1, 192.168.1.2, 192.168.1.3, and 192.168.1.4. These addresses are usable for devices within that subnet. The address 192.168.1.64 itself typically represents the network address for that subnet and is not a usable host address.
Subnet calculation for ip address 172.16.2.130 and subnet mask 255.255.255.128?
network: 172.16.2.128 Broadcast: 172.16.2.255 usable host range: 172.16.2.129 thru .254
Why is the address 172.31.255.255 with a sub net mask of 255.255.255.0 invalid as a host ID?
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
How many subnets and hosts per subnts can you get from 172.30.0.0?
The IP address 172.30.0.0 is a private IP address in the Class B range, which has a default subnet mask of 255.255.0.0 (or /16). If you use the default mask, you can create 65,536 addresses (2^16), allowing for 65,534 usable hosts per subnet (subtracting 2 for the network and broadcast addresses). If you further subnet this address, the number of subnets and hosts per subnet will depend on the subnet mask you choose. For example, using a /24 subnet mask would give you 256 subnets with 254 usable hosts in each.
What is the range of usable hosts for the subnet containing IP address 135.17.20.1?
NetRange: 135.17.0.0 - 135.17.255.255 CIDR: 135.17.0.0/16
How many usable IP addresses are in a 20-bit subnet?
There are 4094 usable IP addresses in a 20-bit subnet.
How many usable host are available given a Class C IP address with the default subnet mask?
with the default subnet mask, the number of clients in a class A network is: 16,777,214
How Do I Solve Custom Subnet Mask Problem 10?
.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.
What is the valid host range for the subnet in which the IP address 192.168.10.206.255.255.255.240 resides?
The host range is 192.168.10.193 - 206; total usable hosts =14.
