SUBNETTING CALCULATION

 

Students are required to suggest private addressing scheme used in UTeM network.

The number of students of each faculty is within 500 – 1000 students. Each faculty is given their

network address. Assign IT Center as another subnet with their own network address. From this

network address, assigned as the 5th available address as the address of the Ulearn System Server.

 

UTeM network uses Class C private addressing scheme which is 192.168.10.0/24. There are 10 different faculties and 1 IT Center. The number of students of each faculty is within 500 – 1000 students.

Original Network IP: 192.168.10.0 /24

Reserved IP address space: 192.168.10.0 – 192.168.10.255

 Class used: Class C

Default subnet mask: 255.255.255.0 or /24

Class C is used because the IP address’s first octet starts with 192 and it is within the range of 192.0.1.1 to 223.255.254.254.

 

1 - Number of subnet bits search:

Total faculty in UTeM: 10 (excluding IT Center)

Minimum subnet needed: 11

To have 10 faculties, we must find nearest number to 11.

Thus,

            S = subnet bits

            2^S >= 11         2^3 = 8 (can’t be use since it doesn’t fulfil the requirement)

                                    2^4 = 16 (>=11, fulfilled the requirement)

Since 16 is greater than 11, 16 subnets are suitable to use. While 8 cannot be taken because it is smaller than 11 and does not fulfil the requirement needed.

            hence, S = 4 and 2^4 = 16 subnets

 

2 – Check whether it usable for the Project Scenario

Subnets formed: 16

192.168.10.SSSS HHHH

Based on the Subnetting Table:

Custom subnet Mask: 255.255.240.0

Prefix Length: /20

1111 1111. 1111 1111. 1111 0000. 0000 0000

128	64	32	16	8	4	2	1
S	S	S	S	H	H	H	H
1	1	1	1	0	0	0	0

=240

255.255.240.0 is the new subnet mask.

 

Project Scenario requirement: 500 – 1000 students for each faculty

Total hosts usable:

            H = total hosts available – prefix length

            H = 32 – 20 = 12 

            Hence,              usable hosts = 2 ^ H – 2 = 2^12 -2 = 4 094 hosts  

It is sufficient for the students as hosts of each faculty.

The number of hosts that are on this particular subnet (Magic Number): 2^H /256 = 4 096/256 = 16

 

Conclusion: the usable address spaces are enough for hosts (students from each faculty, <= 1000)

3 – Create a table for each Subnet

Subnet	Network Address	First Host Address	Last Host Address	Broadcast Address
1	192.168.10.0	192.168.10.1	192.168.10.14	192.168.10.15
2	192.168.10.16	192.168.10.17	192.168.10.30	192.168.10.31
3	192.168.10.31	192.168.10.33	192.168.10.46	192.168.10.47
4	192.168.10.48	192.168.10.49	192.168.10.62	192.168.10.63
5	192.168.10.64	192.168.10.65	192.168.10.78	192.168.10.79
6	192.168.10.80	192.168.10.81	192.168.10.94	192.168.10.95
7	192.168.10.96	192.168.10.97	192.168.10.110	192.168.10.111
8	192.168.10.112	192.168.10.113	192.168.10.126	192.168.10.127
9	192.168.10.128	192.168.10.129	192.168.10.142	192.168.10.143
10	192.168.10.144	192.168.10.145	192.168.10.158	192.168.10.159
11	192.168.10.160	192.168.10.161	192.168.10.174	192.168.10.175
12	192.168.10.176	192.168.10.177	192.168.10.190	192.168.10.191
13	192.168.10.192	192.168.10.193	192.168.10.206	192.168.10.207
14	192.168.10.208	192.168.10.209	192.168.10.222	192.168.10.223
15	192.168.10.224	192.168.10.225	192.168.10.238	192.168.10.239
16	192.168.10.240	192.168.10.241	192.168.10.254	192.168.10.255

 

4 – Assign which address from the table are available for Ulearn System Server

Total number of faculty in UTeM: 10

            Thus, 11^th subnet was assigned for the IT Center’s Network Address which is 192.168.10.160

 

Based on the Project Scenario, the Network Address for ULearn System Server:

            5^th available address: 192.168.10.166

It has been concluded that Ulearn System server IP address is 192.168.10.166 because 192.168.10.161 will be set up as the router IP address for that subnet and become the default gateway. So, it will start from 192.168.10.162 and so on.