SUBNETTING CALCULATION
The scenario of this project can
be referred to Figure 1. Figure 1 shows the logical network flow from a home
network to UTeM facilities. In this figure, a UTeM’s student is trying to access
the Ulearn system located in the UTeM network facilities. The student is using
her modem to connect to the public network before able to connect to the UTeM
facilities. From the UTeM network, the data then was forwarded to IT Center
where the Ulearn System server is located.
Figure
1: Logical Network Flow
Students are
required to suggest private addressing scheme used in UTeM network. Explain how
you calculate the subnetting using FLSM based on the number of the faculties. 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.
Private IP addresses are
- Non-routable addresses - Routers would not deliver packets with private IP addresses
- Free to use without anyone's permission
Why do we need private IP addresses
- There are only 4.3 billion IPv4 addresses
- Private IP address is used to communicate within the network
- They have extended IPv4's life
Class |
Range |
A |
10.0.0.0 – 10.255.255.255 or 10.0.0.0/8 |
B |
172.16.0.0 – 172.31.255.255 or 172.16.0.0/12 |
C |
192.168.0.0 – 192.168.255.255 or 192.168.0.0/16 |
- divides a network into two or more networks
- takes bits from the host portion of the IP address and reserve them to define the subnet address
In short, Subnetting is the process of stealing bits from the HOST part of an IP address in order to divide the larger network into smaller sub-networks called subnets. After subnetting, we end up with NETWORK SUBNET HOST fields. We always reserve an IP address to identify the subnet and another one to identify the broadcast subnet address.
For UTeM network,
To do the subnetting calculation, below shows the five-step subnetting process.
STEP 1: Suggest a private addressing scheme
Referring to Table 1,
192.168.0.0 /16 (Class C)
Subnet Mask - 255.255.0.0
STEP 2: Find the number of required subnets
The number of required subnets is 2. One subnet is for the use of faculties and the other subnet is for the use of IT centre.
By referring to Figure 2, since the number of required bits is 2, the number of bits to borrow is 2.
STEP 3B: Determine Number of Hosts per Subnet
By referring to Figure 2, since the number of required bits is 2, the number of hosts per subnet is 126.
However, the borrowed bits is at third field, so the actual number of hosts per subnet is calculated by 2^14 - 2 = 16382
Prefix size | Network mask | Usable hosts per subnet |
/1 | 128.0.0.0 | 2,147,483,646 |
/2 | 192.0.0.0 | 1,073,741,822 |
/3 | 224.0.0.0 | 536,870,910 |
/4 | 240.0.0.0 | 268,435,454 |
/5 | 248.0.0.0 | 134,217,726 |
/6 | 252.0.0.0 | 67,108,862 |
/7 | 254.0.0.0 | 33,554,430 |
Class A | ||
/8 | 255.0.0.0 | 16,777,214 |
/9 | 255.128.0.0 | 8,388,606 |
/10 | 255.192.0.0 | 4,194,302 |
/11 | 255.224.0.0 | 2,097,150 |
/12 | 255.240.0.0 | 1,048,574 |
/13 | 255.248.0.0 | 524,286 |
/14 | 255.252.0.0 | 262,142 |
/15 | 255.254.0.0 | 131,070 |
Class B | ||
/16 | 255.255.0.0 | 65,534 |
/17 | 255.255.128.0 | 32,766 |
/18 | 255.255.192.0 | 16,382 |
/19 | 255.255.224.0 | 8,190 |
/20 | 255.255.240.0 | 4,094 |
/21 | 255.255.248.0 | 2,046 |
/22 | 255.255.252.0 | 1,022 |
/23 | 255.255.254.0 | 510 |
Class C | ||
/24 | 255.255.255.0 | 254 |
/25 | 255.255.255.128 | 126 |
/26 | 255.255.255.192 | 62 |
/27 | 255.255.255.224 | 30 |
/28 | 255.255.255.240 | 14 |
/29 | 255.255.255.248 | 6 |
/30 | 255.255.255.252 | 2 |
/31 | 255.255.255.254 | 0 |
/32 | 255.255.255.255 | 0 |
STEP 4: Calculate new subnet mask and prefix
By referring to Figure 3, since the number of usable hosts is 16382,
New Subnet Mask = 255.255.192.0
New Prefix = /18
STEP 5: Apply the subnet mask to the IP space
Usage |
Subnet ID |
Usable Range |
Broadcast ID |
Faculties |
192.168.0.0 /18 |
192.168.0.1 /18 – 192.168.127.254 /18 |
192.168.127.255 /18 |
IT Centre |
192.168.128.0 /18 |
192.168.128.1 /18 – 192.168.255.254 /18 |
192.168.255.255 /18 |
The address of Ulearn System Server is located in the 5th available address of IT Centre. By referring to Table 2, the 5th available address is 192.168.128.6. We do not start from 192.168.128.1 because it will set up as router IP Address for that subnet and become default gateway. So, it should start from 192.168.128.2 and so on. The scenario said that the Ulearn System Server is assigned as the 5th available address and make 192.168.128.6 as their assigned IP address.
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