In a subnet mask, the last octet value of 252 corresponds to the binary representation of 11111100. This means that the first six bits are used for the network part, while the last two bits are for the host part. Therefore, a subnet mask of 255.255.255.252 allows for only two usable IP addresses within that subnet.
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
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).
A 23-bit subnet mask in dotted decimal notation is represented as 255.255.254.0. This is derived from converting the first 23 bits to binary, which results in 11111111.11111111.11111110.00000000. When converted to decimal, the first two octets are 255, the third octet is 254, and the last octet is 0.
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
because these two subnet are reserve
In the IP address 121.234.52.65, the host portion depends on the subnet mask used. If we assume a common subnet mask of 255.255.255.0 (or /24), the host portion would be the last octet, which is 65. This means that the network portion is 121.234.52, while 65 identifies the specific device within that network.
To Know any subnet of this IP you must provide the Subnet mask, otherwise it is impossible to know what is the subnet.
172.16.255.254.
A subnet mask typically appears in the same format as an IP address, consisting of four octets separated by periods (e.g., 255.255.255.0). It identifies the network and host portions of an IP address, with the network part represented by consecutive ones (1s) in binary and the host part by zeros (0s). For example, in the subnet mask 255.255.255.0, the first three octets (255) indicate the network portion, while the last octet (0) indicates the host portion. Subnet masks can also be represented in CIDR notation, such as /24, which signifies that the first 24 bits are used for the network.
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
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.
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.