One Private C class IP Block is 192.168.222.0/24
1. 10 PCs in the CSE Department
2. EEE Department at 10 PC
3. BSTE Department at 10 PC
4. BBA Department at 12 PC
5. ENGLISH Department at 5 PC
6. 5 PCs at LAW Department
7. Admission Department at 5 PC
8. Account Department at 5 PC
9. Admin Department at 5 PC
10. IT Department at 5 PC
11. Library on 2 PCs
12. 35 PCs in LAB General
13. In LAB CSE 50 PCs need an internet connection
For this, how do Subnetting your Network
Step-1: First we calculate that we need (10 + 10 + 10 + 12 + 5 + 5 + 5 + 5 + 5 + 5 + 2 + 35 + 50) = 158 IPs and 13 subnetworks.
Step-2: We know that in C class IPv4 address (8: 8: 8: 8 = 32bits) 3 Network 1 Host (N: N: N: H) is used. Since there are three networks in C class, it's default bit is 24 and its default subnet mask is 255.255.255.0. We know that an Octet (128 + 64 + 32 + 16 + 8 + 4 + 2 + 1) = 255 Since three Octets are used in C class, its default Subnet Mask is 255.255.255.0
Step-3: Now, we will start subnetting our 192.168.222.0/24 IP. When doing Subnetting, you have to notice who needs the most IP first and who needs the least IP.
Most (50) IPs are required in LAB CSE.
We know a formula for finding Host IP 2N-2
N = unused host bit
-2 = 1 network address and one broadcast address
If we need an unused host bit for 50 IPs, then we need to increase the bet by 2, since we have 8 bits in an Octet. If it is network-1, we can write 192.168.222.0/26. Our default bit was 24. Since we have increased 2 bits, we have written 26 instead of 24.
Network-1: CSE LAB (50 PCs)
192.168.222.0/26
N = Total Bit - Useable Bit (32 - 26) = 6
2^6 - 2 = 64 - 2 = 62 hosts
Network Address: 192.168.222.0/26
1st Host: 192.168.222.1
Last Host: 192.168.222.62
Broadcast Address: 192.168.222.63
Subnet Mask: 255.255.255.192
We know,
1 bit = 128
2 bits = 192
3 bits = 224
4 bits = 240
5 bits = 248
6 bits = 252
7 bits = 254
8 bits = 255
Since we have used 2 bits, its subnet mask has been 192
Network-2: General LAB (35 PCs)
192.168.222.64/26
2^6 - 2 = 64 - 2 = 62 hosts
Network Address: 192.168.222.64/26
1st Host: 192.168.222.65
Last Host: 192.168.222.126
Broadcast Address: 192.168.222.127
Subnet Mask: 255.255.255.192
Since we have used 2 bits, its subnet mask has been 192
Network-3: BBA Department (12 PCs)
192.168.222.128/28
2^4 - 2 = 16 - 2 = 14 hosts
Network Address: 192.168.222.128/28
1st Host: 192.168.222.129
Last Host: 192.168.222.142
Broadcast Address: 192.168.222.143
Subnet Mask: 255.255.255.240
Since we have used 4 bits, its subnet mask has been 240
Network-4: CSE Department (10 PCs)
192.168.222.144/28
2^4 - 2 = 16 - 2 = 14 hosts
Network Address: 192.168.222.144/28
1st Host: 192.168.222.145
Last Host: 192.168.222.156
Broadcast Address: 192.168.222.159
Subnet Mask: 255.255.255.240
Since we have used 4 bits, its subnet mask has been 240
Network-5: EEE Department (10 PCs)
192.168.222.160/28
2^4 - 2 = 16 - 2 = 14 hosts
Network Address: 192.168.222.160/28
1st Host: 192.168.222.161
Last Host: 192.168.222.174
Broadcast Address: 192.168.222.175
Subnet Mask: 255.255.255.240
Since we have used 4 bits, its subnet mask has been 240
Network-6: BSTE Department (10 PCs)
192.168.222.176/28
2^4-2 = 16 - 2 = 14 hosts
Network Address: 192.168.222.176/28
1st Host: 192.168.222.177
Last Host: 192.168.222.190
Broadcast Address: 192.168.222.191
Subnet Mask: 255.255.255.240
Since we have used 4 bits, its subnet mask has been 240
Network-7: ENGLISH Department (5 PCs)
192.168.222.192/29
2^3 - 2 = 8 - 2 = 6 hosts
Network Address: 192.168.222.192/29
1st Host: 192.168.222.193
Last Host: 192.168.222.198
Broadcast Address: 192.168.222.199
Subnet Mask: 255.255.255.248
Since we have used 5 bits, its subnet mask has been 248
Network-8: LLB Department (5 PCs)
192.168.222.200/29
2^3 - 2 = 8- 2 = 6 hosts
Network Address: 192.168.222.200/29
1st Host: 192.168.222.201
Last Host: 192.168.222.206
Broadcast Address: 192.168.222.207
Subnet Mask: 255.255.255.248
Since we have used 5 bits, its subnet mask has been 248
Network-9: Admission Department (5 PCs)
192.168.222.208/29
2^3 - 2 = 8 - 2 = 6 hosts
Network Address: 192.168.222.208/29
1st Host: 192.168.222.209
Last Host: 192.168.222.214
Broadcast Address: 192.168.222.215
Subnet Mask: 255.255.255.248
Since we have used 5 bits, its subnet mask has been 248
Network-10: Account Department (5 PCs)
192.168.222.216/29
2^3 - 2 = 8 - 2 = 6 hosts
Network Address: 192.168.222.218/29
1st Host: 192.168.222.218
Last Host: 192.168.222.222
Broadcast Address: 192.168.222.223
Subnet Mask: 255.255.255.248
Since we have used 5 bits, its subnet mask has been 248
Network-11: Admin Department (5 PCs)
192.168.222.224/29
2^3 - 2 = 8 - 2 = 6 hosts
Network Address: 192.168.222.224/29
1st Host: 192.168.222.225
Last Host: 192.168.222.222
Broadcast Address: 192.168.222.231
Subnet Mask: 255.255.255.248
Since we have used 5 bits, its subnet mask has been 248
Network-12: IT Department (5 PCs)
192.168.222.232/29
2^3 - 2 = 8 - 2 = 6 hosts
Network Address: 192.168.222.232/29
1st Host: 192.168.222.233
Last Host: 192.168.222.238
Broadcast Address: 192.168.222.239
Subnet Mask: 255.255.255.248
Since we have used 5 bits, its subnet mask has been 248
Network-13: Library (2 PCs)
192.168.222.240/30
2^2 - 2 = 4 - 2 = 2 hosts
Network Address: 192.168.222.240/30
1st Host: 192.168.222.241
Last Host: 192.168.222.242
Broadcast Address: 192.168.222.243
Subnet Mask: 255.255.255.252
Since we have used 6 bits, its subnet mask has been 252
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