![]() ![]() The reason we subtract 2 above is because the all-ones and all-zeros host numbers are reserved. Maximum Number of hosts = 2**(32 - netmask_length) - 2 This gives you the number of host bits in the address. The easiest way to do this is to subtract the netmask length from 32 (number of bits in an IPv4 address). To find the maximum number of hosts, look at the number of binary bits in the host number above. This time I'll use 1s instead of h, because we need to perform a logical AND on the network address again. just reuse the host mask from the work we did when we calculated the broadcast address of 128.42.5.4/21. This is an equally-valid answer to the constraint, using /23 subnets of 128.42.0.0/21. Since we only want four subnets from the whole 128.42.0.0/21 block, we could use /23 subnets. The lazy way to break 128.42.0.0/21 into four equal subnets: least significant bits), simply subtract 7 from 32 to calculate the minimum subnet prefix for each subnet. Since IPv4 addresses are 32 bits wide, and we are using the host bits (i.e. One needs 7 host bits to contain 100 hosts. How did I know that I need at least a /25 masklength for 100 hosts? Calculate the prefix by backing into the number of host bits required to contain 100 hosts. Finding the required subnet masklength or netmask: Notice that the network address for each subnet borrows host bits from the parent network block. ![]() In this example, we know that you need at least a /25 prefix to contain 100 hosts I chose a /24 because it falls on an octet boundary. ![]() Let's assume we will break 128.42.0.0/21 into 4 subnets that must hold at least 100 hosts each. depending on your constraints, there could be several valid ways to subnet a block of addresses. Many times there isn't one right way to subnet a block. You haven't given enough information to calculate subnets for this network as a general rule you build subnets by reallocating some of the host bits as network bits for each subnet. This means our host bits are the last 11 bits of the IP address, because we find the host mask by inverting the network mask: Host bit mask : 00000000 00000000 00000hhh hhhhhhhh The broadcast address converts all host bits to 1s. Align the bits in both addresses, and perform a logical AND on each pair of the respective bits. The network address is the logical AND of the respective bits in the binary representation of the IP address and network mask. the left-hand-side of the binary number). Then, count the number of contiguous 1 bits, starting at the most significant bit in the first octet (i.e. Calculating the Netmask Length (also called a prefix):Ĭonvert the dotted-decimal representation of the netmask to binary.
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