5 bits
254 host as 172.16.32.0/24
Yes - the first bits specify the network, the remaining bits, a host within a network. There is no fixed number of bits for the network; this may vary.Yes - the first bits specify the network, the remaining bits, a host within a network. There is no fixed number of bits for the network; this may vary.Yes - the first bits specify the network, the remaining bits, a host within a network. There is no fixed number of bits for the network; this may vary.Yes - the first bits specify the network, the remaining bits, a host within a network. There is no fixed number of bits for the network; this may vary.
Subnet Mask
255.0.0.0
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
15 or 16.
6
The maximum number of host bits that can be borrowed from a class A address is 22 (technically you could borrow 23 but the resulting network would be useless). A class A address uses 8 bits for its network address and 24 bits for its host addresses. Class A uses a subnet mask of 255.0.0.0 You can only borrow 22 bits (instead of 24) because a valid network requires 4 addresses: A network address, two host addresses and a broadcast address. These networks would result in 30 bits used for the network address and 2 bits used for the host addresses. These networks use a subnet mask of 255.255.255.252
2046 Breakdown: 11111111.11111111.11111000.00000000 /21 - 21 bits in network address represented by ones in binary address above. Leaves 2^11th power host addresses left (the zeros to the right). Equals 2048 host addresses minus the two reserved addresses = 2046
A Network
That leaves us 5 bits for the host (32 - 27 = 5); the size of the network is 25 = 32 IP addresses, that is, addresses 10.150.100.96 - 10.150.100.127. The first and last address are not usable (can't be assigned to hosts), which leaves us with addresses 10.150.100.97 - 10.150.100.126.That leaves us 5 bits for the host (32 - 27 = 5); the size of the network is 25 = 32 IP addresses, that is, addresses 10.150.100.96 - 10.150.100.127. The first and last address are not usable (can't be assigned to hosts), which leaves us with addresses 10.150.100.97 - 10.150.100.126.That leaves us 5 bits for the host (32 - 27 = 5); the size of the network is 25 = 32 IP addresses, that is, addresses 10.150.100.96 - 10.150.100.127. The first and last address are not usable (can't be assigned to hosts), which leaves us with addresses 10.150.100.97 - 10.150.100.126.That leaves us 5 bits for the host (32 - 27 = 5); the size of the network is 25 = 32 IP addresses, that is, addresses 10.150.100.96 - 10.150.100.127. The first and last address are not usable (can't be assigned to hosts), which leaves us with addresses 10.150.100.97 - 10.150.100.126.
The number of bits used to identify the hosts is fixed by the class of the network. Up to 24 bits can make up the host portion of a Class C address.
Write the subnet mask in binary. The zeroes at the end represent the host bits, and therefore, the size of the network. If (for example) you have ten zeroes at the end, you rais 210 = 1024. That's the number of addresses in the network. Of these, the first and the last are reserved for special purposes, and can't be used for host addresses, so the complete calculation (in this case) is 210 - 2.Write the subnet mask in binary. The zeroes at the end represent the host bits, and therefore, the size of the network. If (for example) you have ten zeroes at the end, you rais 210 = 1024. That's the number of addresses in the network. Of these, the first and the last are reserved for special purposes, and can't be used for host addresses, so the complete calculation (in this case) is 210 - 2.Write the subnet mask in binary. The zeroes at the end represent the host bits, and therefore, the size of the network. If (for example) you have ten zeroes at the end, you rais 210 = 1024. That's the number of addresses in the network. Of these, the first and the last are reserved for special purposes, and can't be used for host addresses, so the complete calculation (in this case) is 210 - 2.Write the subnet mask in binary. The zeroes at the end represent the host bits, and therefore, the size of the network. If (for example) you have ten zeroes at the end, you rais 210 = 1024. That's the number of addresses in the network. Of these, the first and the last are reserved for special purposes, and can't be used for host addresses, so the complete calculation (in this case) is 210 - 2.
A Class C IP address has 24 bits for network and 8 bits for host. So to have a subnet mask of 26 bits, you will need to use 2 bits from host part.Number of subnets is given by the formula : 2^(no. of bits used from host part).Hence number of subnets in this case would be = 2^2 = 4.For e.g. if the class C IP address is 200.168.210.0the 4 subnet addresses would be :11001000.10101000.11010010.00000000 = 200.168.210.011001000.10101000.11010010.01000000 = 200.168.210.6411001000.10101000.11010010.10000000 = 200.168.210.12811001000.10101000.11010010.11000000 = 200.168.210.192Note: The digits in bold are the mask bits.
Class C network
255.255.255.0
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.