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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
If internet protocol is 135.210.195.112 what is the subnet mask?
It depends on whether you are using a default subnet mask or you are subnetting the class B network. A default subnet mask would be 255.255.0.0, but if you are subnetting the last 2 octets in the subnet mask could be anything (up to 255 per octet).
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
Why first and last subnet cannot be used?
because these two subnet are reserve
What is the last valid sub net of 172.28.231.4?
To Know any subnet of this IP you must provide the Subnet mask, otherwise it is impossible to know what is the subnet.
What is the last usable address for subnet 172.16.0.0?
172.16.255.254.
What are the steps in subnetting?
A service provider has given you the Class C network range 209.50.1.0. Your company must break the network into 20 separate subnets. Step 1) Determine the number of subnets and convert to binary - In this example, the binary representation of 20 = 00010100. Step 2) Reserve required bits in subnet mask and find incremental value - The binary value of 20 subnets tells us that we need at least 5 network bits to satisfy this requirement (since you cannot get the number 20 with any less than 5 bits -- 10100) - Our original subnet mask is 255.255.255.0 (Class C subnet) - The full binary representation of the subnet mask is as follows: 255.255.255.0 = 11111111.11111111.11111111.00000000 - We must "convert" 5 of the client bits (0) to network bits (1) in order to satisfy the requirements: New Mask = 11111111.11111111.11111111.11111000 - If we convert the mask back to decimal, we now have the subnet mask that will be used on all the new networks -- 255.255.255.248 - Our increment bit is the last possible network bit, converted back to a binary number: New Mask = 11111111.11111111.11111111.1111(1)000 -- bit with the parenthesis is your increment bit. If you convert this bit to a decimal number, it becomes the number "8‟ Step 3) Use increment to find network ranges - Start with your given network address and add your increment to the subnetted octet: 209.50.1.0 209.50.1.8 209.50.1.16 ...etc - You can now fill in your end ranges, which is the last possible IP address before you start the next range 209.50.1.0 -- 209.50.1.7 209.50.1.8 -- 209.50.1.15 209.50.1.16 -- 209.50.1.23 ...etc - You can then assign these ranges to your networks! Remember the first and last address from each range (network / broadcast IP) are unusable
What is the host range for 184.192.125.33?
Not enough information. You also need the subnet mask. For example, if the subnet mask 255.255.255.0 is used (this is quite common), the subnet range is from 0-255 (in the last byte - keep the other bytes as they are), but the first and last of these addresses are reserved for special purposes, and can't be assigned for hosts. Therefore, the host range has addresses 1-254 in the last byte.
A network on the internet has a subnet mask 255.255.240.0 what is the maximum number of hosts it can handle?
Convert the subnet mask to binary: 11111111.11111111.11110000.00000000The zeroes tell you which bytes indicate the host; in this case, 12 zeroes allow for 212 = 4096 different IP addresses within the subnet. Of these, two (the first and the last) are unusable for a host, so you have a maximum of 4094 hosts.
Which network has three most significant bit as 110 with remainder is the first three octets as the network number and last octet as the host number?
Stanbridge College, Windows Server 2008
What is the value for the last number 4.345?
4.345 However, the value for the last DIGIT would be 0.005 or 5 thousandths
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.
