Computer Networking Question | AL ICT 🥰

(a) Given IP Address Block:

Binary: 10010110.00110010.00000000.00000000

(i) Convert to Dotted Decimal Notation:

Convert each 8-bit segment from binary to decimal:

• 10010110 = 150

• 00110010 = 50

• 00000000 = 0

• 00000000 = 0

So, the IP address block in Dotted Decimal Notation is:
150.50.0.0

(ii) Identify the Class of the IP Address:

• The first octet (150) falls within Class B (128 – 191).

• Class B networks have a default subnet mask of 255.255.0.0 (/16).

(iii) Number of IP Addresses in the Block:

A Class B network with a /16 subnet mask means:

2(32−16)=216=655362^{(32 – 16)} = 2^{16} = 655362(32−16)=216=65536

So, the total number of IP addresses is 65,536.

(iv) First and Last IP Addresses:

• First IP Address (Network Address): 150.50.0.0

• Last IP Address (Broadcast Address): 150.50.255.255

(b) Subnetting to Create 8 Subnets: (i) Number of Bits Needed for Subnetting:

To create 8 subnets, we use the formula:

2n≥82^n \geq 82n≥8 n=3(since 23=8)n = 3 \quad \text{(since \(2^3 = 8\))}n=3(since 23=8)

So, 3 more bits are borrowed from the host portion.

(ii) Number of Hosts per Subnet:

A Class B default has 16 host bits. Borrowing 3 bits leaves us with:

32−(16+3)=13(Host bits left)32 – (16 + 3) = 13 \quad \text{(Host bits left)}32−(16+3)=13(Host bits left)

Total number of hosts per subnet:

213−2=8192−2=81902^{13} – 2 = 8192 – 2 = 8190213−2=8192−2=8190

(We subtract 2 for the Network Address and Broadcast Address.)

(iii) Subnet Mask:

Default Class B subnet mask: 255.255.0.0 (/16)
Adding 3 subnet bits: /19

Subnet Mask in Dotted Decimal: 255.255.224.0

(iv) Subnet Ranges:

Each subnet increments by 32 in the third octet (224 = 256 - 32).

Subnet Table