Local Area Networks

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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...o 128.100.15.255 where 128.100.0.0 network address and 128.100.15.255 is broadcast address and rest address can be allocated to host. So total 4096-2=4094 hosts can be there.

• Second college will have a network address of 128.100.16.0 and subnet mask 255.255.240.0(/20)
It will have a range of address from 128.100.16.0 to 128.100.31.255 where 128.100.16.0 network address and 128.100.31.255 is broadcast address and rest address can be allocated to host. So total 4096-2=4094 hosts can be there.

• Third college will have a network address of 128.100.32.0 and subnet mask 255.255.240.0(/20)
It will have a range of address from 128.100.32.0 to 128.100.47.255 where 128.100.32.0 network address and 128.100.47.255 is broadcast address and rest address can be allocated to host. So total 4096-2=4094 hosts can be there.

• 7 relatively small colleges can be given a subnet mask of 255.255.248.0(/21) i.e.
1111111. 1111111.11111000.00000000
Number of zero's in host id = 11
So, number of valid hosts= (2^11-2) = 2046
• First college will have a network address of 128.100.48.0 and subnet mask 255.255.248.0(/21)
It will have a range of address from 128.100.48.0 to 128.100.55.255 where 128.100.48.0 network address and 128.100.55.255 is broadcast address and rest address can be allocated to host. So total 2048-2=2046 hosts can be there.
• Second college will have a network address of 128.100.56.0 and subnet mask 255.255.248.0(/21)
It will have a range of address from 128.100.56.0 to 128.100.63.255 where 128.100.56.0 network address and 128.100.63.255 is broadcast address and rest address can be allocated to host. So total 2048-2=2046 hosts can be there.
• Third college will have a network address of 128.100.64.0 and subnet mask 255.255.248.0(/21)
It will have a range of address from ...