Subnetting Calculation

In this project, students are required to suggest private addressing scheme used in OUM network. Explain how you calculate the subnetting using FLSM based on the number of the faculties. 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 4th available address as the address of the OnlineLearning System Server.

Students are allowed to do the addressing by using any subnetting of any private address class, provided that the explanation and argument for the solution are stated and explained.

  

SUBNETTING CALCULATION

In this section, you are expecting to show the calculation on how to identify the IP address that was assigned for the OnlineLearning System Server based on the information given in the project scenario. Explain the concept and show the calculation to identify the IP address.

 

192.168.0.0/24: Ranges between 192.168.0.0 and 192.168.0.255 with 254 possible hosts

The number following the IP address is the abbreviation for the subnet mask, and the number of ones in the subnet mask when expressed in the binary notation is equal to the number of the abbreviation.

For example, the /16 subnet would have 16 ones in a row, and the rest of the numbers being zeros: 11111111.11111111.00000000.00000000, which equals to 255.255.0.0 when converted from binary into decimal.

 

 

  

Subnet is a division of large IP Networks into multiple Small Logical Networks. Subnet mask is a 32-bit number used to identify the Network portion and the Host portion in the IP Address. There are 2 types of Subnetting, FLSM (Fixed Length Subnet Mask) and VLSM (Variable Length Subnet Mask). In our project we are using FLSM to calculate the subnetting based on the number of the faculties.

FLSM

· Dividing an IP Network with same or equal size.

· The Subnet remains within its class Boundary

· Formula 2^n>=N (Requirements)

· Binary to Decimals

 

Group Size	128	64	32	16	8	4	2	1
Subnet	128	192	224	240	248	252	254	255
CIDR	/25	/26	/27	/28	/29	/30	/31	/32

 

***Each CIDR increment doubles the number of networks***

Private addressing scheme used in OUM network :

Ø 192.168.0.0/24

The number of the faculties:

Ø 5 faculties = subnet

 

STEP 1

If I start with a /24, how many sub-networks could I need to create for 8 IP addresses?

There are 8 /27 networks inside a  /24

 

/27 - /24 = 3

 

To calculate the number of possible subnets, use the formula 2^n, where n equals the number of host bits borrowed. If three host bits are borrowed, then n=3. 2^3=8, so eight subnets are possible if three host bits are borrowed.

STEP 2

Possible Number of Hosts per Subnets

To calculate the number of possible hosts per subnet, use the formula 2^n -2, where h equals the number of host bits. The reason two addresses must be subtracted is because of the network address and the broadcast address.

Determine the number of remaining host bits based on the class of address and the number of host bits borrowed.

Given IP Address Space	C
Number of Host Bits Available	8
Number of Host Bits Borrowed	3
Number of Host Bits Remaining (n)	8-3=5
Number of Possible Hosts	2^5 = 32 32-2 =30
New Subnet Mask	255.255.255.224
New Prefix	/27

 

Where did 224 come from?

Ø 255.255.255.224

Base	2^7	2^6	2^5	2^4	2^3	2^2	2^1	2^0
Place Value	128	64	32	16	8	4	2	1
Binary Value	1	1	1	0	0	0	0	0
Meaning	Network Bit	Network Bit	Network Bit	Host Bit	Host Bit	Host Bit	Host Bit	Host Bit

 

255	255	255	x	x	x
192	168	0	2^7	2^6	2^5
-	-	-	128	64	32

 

Ø Total = 128+64+32 = 224

 

 

 

 

 

STEP 3

To complete the fourth step of the subnetting process, calculating the decimal and prefix values of the new subnet mask.

 

Assigned IP Address Space	192.168.0.0
Class of Assigned IP Address Space	C
Default Subnet Mask (Dotted Decimal Format)	255.255.255.0
Default Subnet Mask in Binary	11111111.11111111.11111111.11111111.00000000
Default Subnet Mask (Prefix Format)	/24
Number of Host Bits Borrowed	3
New Subnet Mask (Dotted Decimal Format)	255.255.255.224
New Subnet Mask in Binary	11111111.11111111.11111111.11100000
New Subnet Mask (Prefix Format)	/27

 

Subnet Number	Network ID	Host ID Range	Broadcast Address
0	192.168.0.0/27	192.168.0.1/27 to 192.168.0.30/27	192.168.0.31/27
1	192.168.0.32/27	192.168.0.32/27 to 192.168.0.62/27	192.168.0.63/27
2	192.168.0.64/27	192.168.0.65/27 to 192.168.0.94/27	192.168.0.95/27
3	192.168.0.96/27	192.168.0.97/27 to 192.168.0.126/27	192.168.0.127/27
4	192.168.0.128/27	192.168.0.129/27 to 192.168.0.158/27	192.168.0.159/27
5	192.168.0.160/27	192.168.0.161/27 to 192.168.0.190/27	192.168.0.191/27

 

Number of Possible Hosts

2^5 = 32 32-2 =30

 

The 4th available address as the address of the OnlineLearning System Server is 192.168.0.165, which is also the IP address of the server. We do not start from 192.168.0.161 since it will be assigned to router IP address as we need to identify the 4^th accessible address. So, it will begin at 192.168.0.162 and so on.