An organisation has been allocated the IPv4 block \(192.168.64.0/22\). It needs subnets for three departments: Admin (50 hosts), HR (30 hosts) and IT (120 hosts). Using the smallest possible subnets for each department, list for each subnet the network address, broadcast address and usable host range. Also state any remaining address space.
A.
IT: 192.168.64.0/25 (network 192.168.64.0, broadcast 192.168.64.128, hosts 192.168.64.1‑192.168.64.127); Admin: 192.168.64.128/26 (network 192.168.64.128, broadcast 192.168.64.192, hosts 192.168.64.129‑192.168.64.191); HR: 192.168.64.192/27 (network 192.168.64.192, broadcast 192.168.64.224, hosts 192.168.64.193‑192.168.64.223); Remaining: 192.168.64.225/27, 192.168.65.0/24, 192.168.66.0/24, 192.168.67.0/24]
B.
IT: 192.168.64.0/24 (network 192.168.64.0, broadcast 192.168.64.255, hosts 192.168.64.1‑192.168.64.254); Admin: 192.168.65.0/26 (network 192.168.65.0, broadcast 192.168.65.63, hosts 192.168.65.1‑192.168.65.62); HR: 192.168.65.64/27 (network 192.168.65.64, broadcast 192.168.65.95, hosts 192.168.65.65‑192.168.65.94); Remaining: 192.168.65.96/27, 192.168.66.0/23]
Correct
C.
IT: 192.168.64.0/25 (network 192.168.64.0, broadcast 192.168.64.127, hosts 192.168.64.1‑192.168.64.126); Admin: 192.168.64.128/26 (network 192.168.64.128, broadcast 192.168.64.191, hosts 192.168.64.129‑192.168.64.190); HR: 192.168.64.192/27 (network 192.168.64.192, broadcast 192.168.64.223, hosts 192.168.64.193‑192.168.64.222); Remaining: 192.168.64.224/27, 192.168.65.0/24, 192.168.66.0/24, 192.168.67.0/24.
D.
Admin: 192.168.64.0/26 (network 192.168.64.0, broadcast 192.168.64.63, hosts 192.168.64.1‑192.168.64.62); HR: 192.168.64.64/27 (network 192.168.64.64, broadcast 192.168.64.95, hosts 192.168.64.65‑192.168.64.94); IT: 192.168.64.96/25 (network 192.168.64.96, broadcast 192.168.64.223, hosts 192.168.64.97‑192.168.64.222); Remaining: 192.168.64.224/27, 192.168.65.0/24, 192.168.66.0/23]