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As you can see, the network address of 128.42.5.4/21 is 128.42.0.0.

Get the broadcast address by converting all host bits to 1s.

Host bits = log₂(Number-of-hosts) = log₂(100) = 6.643

The reason we subtract 2 above is because the all-ones and all-zeros host numbers are reserved. The all-zeros host number is the network address; the all-ones host number is the broadcast address.

In this case the maximum netmask (minimum hostmask) would be /25.

As you can see, the network address of 128.42.5.4/21 is 128.42.0.0

The broadcast address converts all host bits to 1s...

Host bits = Log₂(Number-of-hosts) = Log₂(100) = 6.643

The reason we subtract 2 above is because the all-ones and all-zeros host numbers are reserved. The all-zeros host number is the network number; the all-ones host number is the broadcast address.

In this case the maximum netmask (minimum hostmask) would be /25

As you can see, the network address of 128.42.5.4/21 is 128.42.0.0

The broadcast address converts all host bits to 1s...

Host bits = Log₂(Number-of-hosts) = Log₂(100) = 6.643

The reason we subtract 2 above is because the all-ones and all-zeros host numbers are reserved. The all-zeros host number is the network number; the all-ones host number is the broadcast address.

In this case the maximum netmask (minimum hostmask) would be /25

As you can see, the network address of 128.42.5.4/21 is 128.42.0.0.

Get the broadcast address by converting all host bits to 1s.

Host bits = log₂(Number-of-hosts) = log₂(100) = 6.643

The reason we subtract 2 above is because the all-ones and all-zeros host numbers are reserved. The all-zeros host number is the network address; the all-ones host number is the broadcast address.

In this case the maximum netmask (minimum hostmask) would be /25.

Finding the minimum netmask which contains two IP addresses:

In this case the minimum netmask would be /25

Finding the maximum netmask (minimum hostmask) which contains two IP addresses:

In this case the maximum netmask (minimum hostmask) would be /25