QUESTION 2

 Referring back to the scenario, our group produce that the network would have several departments. The department based on the list below: 

1)     Information Technology Department

2)     Business and Management Department

3)     Engineering Department

4)     IT Centre

based on all 4 departments above, we need 4 subnets to be used in the network. So, to calculate the subnetting we can use the fixed-length subnet mask (FLSM). An FLSM is a sequence of numbers of unchanging length that streamlines packet routing within the subnets. A subnet can be a defined local area network (LAN).

 

Subnet Calculation (FLSM):

 

We chose a class B IP address because Class B IP addresses are used for medium and large-sized networks in enterprises and organizations. They support up to 65,000 hosts on 16,000 individual networks.

Let that the LAN use the IP-address start with 135.24.0.0

1)     Defined subnet to be borrowed

min(n)

2ⁿ ≥ 4, "n" is the borrowed bits, we pass them from the host portion to the network portion.

Subnet and host = SSHH HHHH

Therefore, borrowed bits will be 2.

 

2)     Find the number of subnets needed

Number of subnets: 2ⁿ 

n = number of bits borrowed

2² = 4 subnets

3)     Find the number of the subnet mask

Subnet mask:

Old mask (16) + 2 (borrowed bits) = /18 - 255.255.192.0

 

4)      Find the number of host per network

Block size = 32 – 18(new mask bits) = 14 host bits

Therefore, 2¹⁴ = 16384 host per network

And 16384 – 2 = 16382 usable host per network.

 

5)     Find the total number of total host address

Number of networks * Number of addresses/network =

4 * 16382 = 65528 hosts

 

6)     Addressing table

Subnet	Department Assigned	Network address	First address	Last address	Broadcast address
Subnet 0	Bus. Mngt Department	135.24.0.0	135.24.0.1	135.24.63.254	135.24.63.255
Subnet 1	Engineering Department	135.24.64.0	135.24.64.1	135.24.127.254	135.24.127.255
Subnet 2	IT Department	135.24.128.0	135.24.128.1	135.24.191.254	135.24.191.255
Subnet 3	IT Centre	135.24.192.0	135.24.192.1	135.24.255.254	135.24.255.255

 

Based on the situation given, we assigned that the subnet network address for the IT centre is 135.24.192.0. Since the question stated to assign the 4th available address as the address of the Online Learning System Server, the IP address assigned is 135.24.192.4.