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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
What is the IP of the router that connects to this router?
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.
Which subnet mask would be assigned to the network address of192.168.32.0 to provide 254 useable host addresses per subnetwork?
255.255.255.0 is the subnet mask that provides 256 addresses of which the first (0) and last (255), the broadcast addresses are excluded, leaving 254 usable addresses.
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
What are subnet broadcast addresses?
This could be a couple of things, but you are likely referring to TCP/IP protocol broadcasts. A broadcast address is when a data packet is sent to a special address to which all nodes on the network are supposed to accept and optionally respond to. While a broadcast address is defined for the entire network, this is not implemented in practice. For an IP subnet, the last address in the subnet's address space is the broadcast. For example, if my subnet is defined as 192.168.1.0/24 (very common for consumer network hardware including Wi-Fi), this defines the following: the network itself is identified as 192.168.1.0, the subnet mask is 255.255.255.0 (24 binary 1s in a row), the usable host addresses are 192.168.1.1 through 192.168.1.254, and the broadcast address is 192.168.1.255.
Which subnet would include the address 192.168.1.96 as a usable host address?
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 would the value of the last octet of the subnet mask be if the CIDR notation for an address is 192.168.1.1627?
This isn't a valid CIDR address, so I assume it is: 192.168.1.162/7 That would yield a subnet mask of 255.255.255.254
How do you find default gateway and subnet mask?
On Windows systems:Click Start > Run. Then type: cmdThe 'DOS' box will open, type: ipconfig /all(Note: there is a SPACE after the word ipconfig) The information displayed will be similar to the below if you have a single address;C:\Documents and Settings\xxxx>ipconfig /allWindows IP ConfigurationHost Name . . . . . . . . . . . . : machinePrimary Dns Suffix . . . . . . . :Node Type . . . . . . . . . . . . : UnknownIP Routing Enabled. . . . . . . . : NoWINS Proxy Enabled. . . . . . . . : NoDNS Suffix Search List. . . . . . : hsd1.ma.comcast.net.Ethernet adapter Local Area Connection:Connection-specific DNS Suffix . : hsd1.ma.comcast.net.Description . . . . . . . . . . . : 3Com EtherLink XL 10/100 PCI TX NIC (3C905B-TX)Physical Address . . . . . . : 00-10-5A-13-53-F7Dhcp Enabled. . . . . . . . . :YesAutoconfiguration Enabled. . . :YesIP Address. . . . . . . . . . . . : 65.12.23.123Subnet Mask. . . . . . . . . . . : 255.255.248.0Default Gateway. . . . . . . . : 66.31.48.1DHCP Server. . . . . . . . . . : 68.87.71.8DNS Servers . . . . . . . . . . . : 68.87.71.22668.87.73.24268.87.64.146Lease Obtained: Friday, April 17, 2009 12:34:21 PMLease Expires: Tuesday, April 21, 2009 12:34:21 PMNote: Type the word: exit > then hit the Enterkey to close the DOS session.On Mac or Linux/Unix systems:Launch the Terminal located in /Applications/Utilities/Type the following command:ifconfigYou will see most of the same info as the Windows example above. Your IP, subnet mask and gateway will all be clearly marked.
What are two invalid IP addresses on each subnet?
The first and last IP address on each network. e.g. In a classful class C network, the IP addresses x.x.x.0 and x.x.x.255 are invalid and cannot be assigned to a network interface card. x.x.x.0 is the network address for the subnet. x.x.x.255 is the network broadcast address for the subnet.
What is the maximum number of subnets in class A using this mask of 255.255.192.0?
Ok good question To subnet any network requires borrowing host addresses The 255.255.192.0 regardless of class says host addresses start at CIDR (Classless Inter Domain Routing Protocol) /18. So if we borrow every available host address space then we have 2^14 = 16,384 possible subnet addresses available, NOT. In reality we have 11111111.11111111.11000000.00000000 or a /18 network. Every network / subnet requires two special reserved addresses. The network or zero address, and the last address in the range which will be assigned as the broadcast address. So we can't borrow all of the bits for sub netting. If we only leave one we will only have two addresses for the hosts, this won't work because we need to reserve two. We have to leave two so we will have 2^2 = 4. We can then give each subnet a network address and a broadcast address and still have 2 usable hosts' addresses. If we do this we only have 2^12 subnets = 4096. Each subnet will only have two usable host addresses and two reserved addresses. See the math confirms that 4096 * 4 = 16384 which is the total number of addresses in the address space we started with.
If your ip address is 182.12.0. 0 what is your subnet mask?
It wouldn't matter, because that IP address wouldn't work. You wouldn't have a 0 as the last digit; that's invalid.
Why first and last subnet cannot be used?
because these two subnet are reserve
