 # Answered! As the network administrator, you have to assign IP addresses to the devices of USQCollege including the… Question 1:

The address of the network formed by USQCollege is 160.160.160.0/20 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.

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Question 2:

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 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 = 160.160.160.0/20 (last 32 -20 = 12 bits are set to 0 which results 10100000.10100000.10100000.00000000).

As I mentioned before that 2 addresses are kept reserved for a subnet: one is the subnet address and another is the broadcast address.

160.160.160.0 is used for subnet address. Therefore the first usable address is 160.160.160.1 (or 10100000.10100000.10100000.00000001)