The address of the network formed by USQCollege is 220.127.116.11/20 and the subnet mask is (11111111.11111111.11110000.00000000)2 = (255.255.240.0)10
The first byte of the network address( i.e 160) indicates that it is a class B address( as first byte of class B address ranges from 128 to 191). The default subnet mask of class B is 16 or 255.255.0.0 .Therefore we need extra 20 -16 = 4 1’s to meet the requirement of the subnet as mentioned in the question.
Therefore total number of subnet = 2x (where x is the number of borrowed bits or number of 1’s)=24 = 16
Therefore the number of subnets required is 16.
The maximum number of hosts it can cater for:
The subnet mask for the network is 20.
Therefore maximum number of hosts in this network can be computed as
log2 n = 32(total number of bits in IPv4 address) – subnet mask, where n is the total number of hosts in each subnet.
log2 n = 32-20 = 12
Therefore n = 212 = 4096.
Note that in each subnet there are two addresses kept reserved: one for subnet address and another for broadcast address. Therefore total number of usable addresses in each subnet are 4096 – 2 = 4094.Therefore maximum number of hosts it can refer = 4094.
The subnet mask in decimal notation:
As explained earlier, the subnet mask is 20. Therefore in binary the subnet mask is (11111111.11111111.11110000.00000000)2 (total number of 1’s is 20) = (255.255.240.0)10
Therefore the subnet mask in decimal notation is 255.255.240.0
The subnet address:
The first address is considered as the subnet address. The first address of a subnet can be found by setting (32 – subnet mask) number of last bits of the address (in binary notation) to 0.
The IP address = 18.104.22.168/20 (last 32 -20 = 12 bits are set to 0 which results 10100000.10100000.10100000.00000000).
The first usable host address:
As I mentioned before that 2 addresses are kept reserved for a subnet: one is the subnet address and another is the broadcast address.
22.214.171.124 is used for subnet address. Therefore the first usable address is 126.96.36.199 (or 10100000.10100000.10100000.00000001)
The last usable host address:
Similarly the last address in a network is used as broadcast address. The last address is found by setting 32-20 = 12 bits to 1. Therefore the last address is (10100000.10100000.10101111.11111111)2 = 188.8.131.52. The last usable address is one less than the broadcast address which is 184.108.40.206
The broadcast address:
As explained earlier, the broadcast address the last address in a network. The IP address of this network is 220.127.116.11/20.
The last address is found by setting 32-20 = 12 bits to 1. Therefore the last address is (10100000.10100000.10101111.11111111)2 = 18.104.22.168.
The broadcast address is 22.214.171.124