3. IP address assignment (18 points) Assuming a university’s starting IP address

Question

3. IP address assignment (18 points)

Assuming a university’s starting IP address is 132.170.0.0 (i.e. the first two bytes of the host addresses must be 132.170). The university has three campuses. The downtown campus needs to accommodate 1000 hosts, the main campus needs to accommodate 4000 hosts, and the west campus needs to accommodate 2,000 hosts. Determine the subnet addresses and masks for each campus starting from downtown campus, then main campus and then west campus so that the IP addresses of downtown campus are smaller than the those of main campus and IP addresses of main campus are smaller than those of the west campus. Provide Starting and ending IP addresses for each subnet and CIDR (Classless Inter-Domain Routing) notation for each subnet.

             Security

(a) We can use Radix-64 code as encryption method (there is no key). Assuming the attacker does not know the Radix-64 algorithm, how 4. Radix-64 (20 points)

(b) Converting this following message to Radix-64 codes (Note: pay attention to the uppercases and lowercases). (15 points) Note: you have to show the intermediate steps. Just showing the final codes receives 0 credit.

effective would this algorithm be against cryptanalysis? (5 points)

Explanation / Answer

As per chegg policy, I solving first one. please post Securty a and b as separate questions

University’s starting address is 132.170.0.0. And its mentioned that first two bytes of each host should be 132.170 So, The mask for this network is /16 . So network address in CIDR for university is 132.170.0.0/16.

This network contains 232-16 = 65536 addresses.

We need to start subnetting from downtown campus. This campus needs 1000 hosts. But as number of hosts in a subnet can only be in power of 2, So we will be assigning 1024 hosts in this network. But our further subnet needs 4000 and 2000 hosts respectively. To accommodate that we need to subnet the network with each subnet having 4000 hosts. This is because if we subnet it first with 1024 hosts, then we will not be able to create a subnet of 4000+ hosts as each other subnet have size of 1024 hosts only.

So we will create a subnet of 4096 hosts ( nearest number to 4000 in power of 2)

Now 232-x = 4096 => 232-x = 212

So 32 – x = 12.

x = 20. So subnet mask needed for downtown campus is 20.

So below are the details of the subnet for downtown campus

Subnet address: - 132.170.0.0/20

Number of addresses = 4096

Broadcast address: - 132.170.15.255

Hosts range: - 132.170.0.1 - 132.170.15.254

Now moving ahead to main campus, this needs 4000 hosts. So we will be assigning 4096 hosts to this network.

So below are the details of the subnet for main campus

Subnet address: - 132.170.16.0/20

Number of addresses = 4096

Broadcast address: - 132.170.31.255

Hosts range: - 132.170.16.1 - 132.170.31.254

Now moving to next subnet which is west campus which needs 2000 hosts. Now again as per reason given In first part, it is clear than we need to assign subnet with equal i.e. 4096 hosts only.

So below are the details of the subnet for west campus

Subnet address: - 132.170.32.0/20

Number of addresses = 4096

Broadcast address: - 132.170.47.255

Hosts range: - 132.170.32.1 - 132.170.47.254

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