Suppose you are the network manager for Central University, a medium-size university with 13,000 students. The university has 10 separate colleges (e.g., business, arts, journalism), 3 of which are relatively large (300 faculty and staff members, 2,000 students, and 3 buildings) and 7 of which are relatively small (200 faculty and staff, 1000 students, and 1 building). In addition, there are another 2,000 staff members who work in various administration departments (e.g library, maintenance, finance) spread over another 10 buildings. There are 4 residence halls that house a total of 2,000 students. Suppose the university has the 128.100.xxx.xxx address range on the Internet. BRIEFLY describe how would you assign the IP addresses the various subnets? How would you control the process by which IP addresses are assigned to individual computers? You will have to make some assumptions to answer both questions, so be sure to state your assumptions.
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The network address of university is 184.108.40.206. This is a class B network.
Number of valid host in this network is =256*256 -2 = 65534
Since university has 10 collge there should be subnetting.
While using subnetting we need to give a network address and a subnet mask.
For example say network address 220.127.116.11. Now, we are using subnet mask 255.255.255.0 i.e 11111111.11111111. 11111111.00000000 for subnettting
Subnet mask has 2 parts -:
i) Net id-which represents network id and remains constant per network.
Here in class B subnet mask first 16 bit represent Net id.
ii) Host id- Represents Node id and can change from 0 to 255 per network. Here in class B next 16 bits represents host id.
Now in a subnet, Number of valid host = ((2^y)-2), where y = number of zero bits in host id.
In the above example number of valid host = (2^8)-2=254. Since 2 addresses are used for network address and broad cast address, therefore 265-2=254 hosts.
Number of valid small network = 2^x, where x represents number of 1's in host id.
In the above example, number of valid small network in (subnet) in network 18.104.22.168 is (2^8) = 256. This 256 network each network will have 256 address.
Now let come to the problem:
• We are using variable length subnet mask (VLSM) since different subnet has different number of nodes.
Since three colleges are very large we will give them a subnet mask of 255.255.240.0
Number of zero's in host id = 12
So number of valid host in each of these subnet = (2^12-2) =4094
• First college will have a network address of 22.214.171.124 and subnet mask 255.255.240.0(/20)
It will have a range of address from 126.96.36.199 ...
The local area networks are examined.