Broadcast IP 123.123.2.5 and a subnet mask of 255.255.255.224?

Answer 1

First, convert the relevant portion of the subnet mask to binary: 255.255.255.1110 0000[1], aka /27. Next the last octet of the IP address (slightly more complex, but 5's an easy one): 0000 0101.

That means the network is 123.123.2.0/27, the all-host-bits-1 broadcast address is at 123.123.2.31, and available host addresses are 30 hosts from 123.123.2.1 to 123.123.2.30 .

[1] I skipped the all-ones octets, because they're not really necessary to write out.

If you want it slightly more expanded:

Subnet mask: 1111 1111.1111 1111.1111 1111.1111 1111.1110 0000
IP address: 0111 1011.0111 1011.0000 0010.0000 0101

Network address: 0111 1011.0111 1011.0000 0010.0000 0000

For each bit of the subnet mask that is 1, you take the equivalent bit out of the IP address, and the rest you fill with 0. If you prefer that terminology, you perform a bitwise AND of subnet mask and IP address.

Broadcast address: 0111 1011.0111 1011.0000 0010.0001 1111

For the broadcast address, take the network address and fill the host bits with 1s instead of 0 (this is why they're also called all-0s and all-1s respectively).

Now just convert that binary broadcast address back to decimal: 123.123.2.31
(ProTip: You already know three of those octets because of where you got them from.)
(ProTip: a nibble filled with 1s is 15 or 240 decimal, and a nibble filled with 0s is always 0)

When you have the network address and the broadcast address, you also know the available host range for this subnet: one inward from each end, or 123.123.2.1 to 123.123.2.30 .

Answer 2

No, with a netmask of 255.255.255.224 or /27, Your broadcast address would be:

123.123.2.31

Answer 3

123.45.255.255