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How many contiguous class c addresses are required to extend the host address by four bits?

15 or 16.


How many host bits are necessary to assign addresses to 62 hosts?

6


What is the maximum number of bits that can be borrowed with class b?

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


How many host address are there for an address with 21 bits for the network address?

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


What is a group of host called that have identical bit pattern in high order bits of their addresses?

A Network


What is the range of usable IP addresses for 10.150.100.96 27?

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.


A network with 6 bits remaining for the host portion will have how many usable hosts and 8203?

A network with 6 bits remaining for the host portion can accommodate (2^6 = 64) total addresses. However, two addresses are reserved: one for the network address and one for the broadcast address. Therefore, the number of usable hosts in this network will be (64 - 2 = 62).


Which two statements describe classful IP addresses?

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.


Calculate the number of hosts for a specified network?

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.


How many individual systems can be subnet with netmask 255.255.240.0?

A netmask of 255.255.240.0 corresponds to a /20 subnet, which provides 12 bits for host addresses (32 total bits minus 20 bits for the network). This allows for (2^{12} - 2 = 4096 - 2 = 4094) usable individual systems, accounting for the network and broadcast addresses that cannot be assigned to hosts.


How many subnets can you have in a class C network with a 26 bit mask?

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.


Your network address is 172.25.114.250 what is the host range?

If you mean the range of IP addresses in the subnet, additional information is required. Specifically, the subnet mask.