When you borrow bits for a subnet you are intruding into the client portion of an IP address. As a result, you will lose clients in your network because the "borrowed" portion becomes the extended network prefix.
This allows you to separate your bigger network into smaller, logical networks (subnets). The number of bits borrowed will indicate the total number of smaller subnets that you can support in your network.
In each case, regardless of class of address, borrowing 4 bits gives a total of 14 subnets (in the classical sense) and 16 subnets (in Cisco).
you need to reallocate 4 bits to creat 16 subnets. how do I get that? easy! 2 to the power of 4 gives us 16 subnets.
To create ten networks from the 192.168.50.0 subnet, you need to determine how many bits to borrow from the host portion. Since 2^n must be at least 10 to accommodate the networks, you need to borrow 4 bits (since 2^4 = 16, which covers the requirement). This means the new subnet mask will be /28 (or 255.255.255.240), allowing for 16 subnets with 14 usable hosts each.
To calculate a subnet mask, first determine the number of subnets needed and the number of hosts per subnet. Use the formula (2^n \geq \text{number of subnets}) for subnetting and (2^h - 2 \geq \text{number of hosts}) for host calculation, where (n) is the number of bits borrowed for subnets and (h) is the number of bits left for hosts. For example, if you start with a Class C address like 192.168.1.0 and want 4 subnets, you would borrow 2 bits (since (2^2 = 4)), resulting in a subnet mask of 255.255.255.192 (or /26), which allows for 62 hosts per subnet.
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.
that gives you 16 subnets with 14 usable IPs for hosts that is because one is for subnet and one for broadcas in that subnet for example: 192.168.1.0/28 - subnet number 192.168.1.15 -broadcast number usable IPs for hosts - IPs between them that is 14
4 bits
192 is equal to 2 bits borrowed 2^2 = 4 the number of subnets and host are 64 because 2 bits borrowed from the 8 bits of a class C network is 6, therfore 2^6 = 64.
Perhaps you mean 4 bits at a time, I guess its a nibble
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
No. The "byte" is much larger: A "byte" consists of 8 "bits". 4 bytes would equal 32 bits (4 x 8)
Each hexidecimal character represents 4 bits, therefore 256 bits takes 256 / 4 = 64 characters.
1 nibble = 4 bits, so 4 nibbles for 16 bits.