Calculating Host/Network Portions – Detailed Guide

✅ IP Address Structure

An IPv4 address is a 32-bit numeric address, divided into 4 octets, each consisting of 8 bits.

🔸 Example (Decimal & Binary):

Decimal: 192.168.10.5
Binary:  11000000.10101000.00001010.00000101

✅ Network and Host Portions

An IPv4 address has two main parts:

Network Portion: Identifies the network.
Host Portion: Identifies individual devices.

✅ IP Classes & Default Masks:

Class	Address Range	Default Mask	Network/Host Portions
A	0.0.0.0 – 127.255.255.255	255.0.0.0 (/8)	N.H.H.H
B	128.0.0.0 – 191.255.255.255	255.255.0.0 (/16)	N.N.H.H
C	192.0.0.0 – 223.255.255.255	255.255.255.0 (/24)	N.N.N.H

✅ Subnet Masks and CIDR Notation

A subnet mask separates network and host parts explicitly.

Example: 192.168.10.5/24 means 24 bits for network and 8 bits for hosts.

✅ Calculating Network Portion

Perform a bitwise AND operation between the IP address and subnet mask:

IP:   11000000.10101000.00001010.00000101 (192.168.10.5)
Mask: 11111111.11111111.11111111.00000000 (255.255.255.0)
------------------------------------------------ (AND)
Net:  11000000.10101000.00001010.00000000 (192.168.10.0)

Network Address: 192.168.10.0

✅ Calculating Host Portion

The number of usable hosts per subnet is calculated by:

Hosts per subnet = (2^{host bits}) – 2

Example: A /24 subnet has 8 host bits:

Hosts = (2⁸) – 2 = 254 usable IPs

✅ Network and Broadcast Addresses

Network Address: First IP (all host bits = 0).
Broadcast Address: Last IP (all host bits = 1).

Example:

Subnet: 192.168.10.0/24
Usable IPs: 192.168.10.1 – 192.168.10.254
Broadcast: 192.168.10.255

✅ Subnetting (Bits Borrowing)

Subnetting involves borrowing host bits for subnet creation:

Example: Divide 192.168.20.0/24 into 4 subnets:

Borrowed Bits	Subnets Created	Hosts per Subnet	Subnet Mask	CIDR
2 bits	4	62	255.255.255.192	/26

✅ Practical Example Scenario

Given IP: 172.16.50.123/20. Find network, broadcast, and host range.

Subnet mask: 255.255.240.0
Block size: 256 – 240 = 16
50 belongs to range: 48–63

Results:

Network: 172.16.48.0
Broadcast: 172.16.63.255
Usable IP range: 172.16.48.1 – 172.16.63.254

✅ Common Mistakes

Miscalculating borrowed bits.
Forgetting network and broadcast addresses.
Confusing subnet mask with wildcard mask.

✅ Tools and Commands

Subnet Calculators: subnet-calculator.com
Windows: ipconfig /all
Linux/macOS: ifconfig or ip addr show

📗 Summary Table

Item	Calculation/Definition
Network Address	IP AND subnet mask
Broadcast Address	Network address + host bits = 1
Usable Host Range	Network +1 to Broadcast -1
Hosts per subnet	(2^{host bits}) - 2

Calculating Host/Network Portions Quiz

1. How many bits are there in an IPv4 address?

A. 16 bits B. 32 bits C. 64 bits D. 128 bits

Correct answer is B. IPv4 addresses are 32 bits long, divided into four 8-bit octets.

2. Which subnet mask is the default for a Class C network?

A. 255.0.0.0 B. 255.255.0.0 C. 255.255.255.255 D. 255.255.255.0

Correct answer is D. The default subnet mask for Class C is 255.255.255.0.

3. What is the purpose of the subnet mask?

A. To separate the network and host portions of an IP address B. To encrypt IP addresses C. To assign IP addresses dynamically D. To convert IP addresses to MAC addresses

Correct answer is A. The subnet mask defines which part of the IP address identifies the network and which part identifies hosts.

4. What is the network address when performing an AND operation on IP 192.168.10.5 and mask 255.255.255.0?

A. 192.168.10.5 B. 192.168.0.0 C. 192.168.10.0 D. 192.168.255.255

Correct answer is C. The network address is the result of bitwise AND between IP and subnet mask, here 192.168.10.0.

5. How do you calculate the number of usable hosts in a subnet?

A. 2^(number of network bits) B. 2^(number of host bits) minus 2 C. 2^(total bits) minus 2 D. Number of host bits multiplied by 2

Correct answer is B. The formula is 2^(host bits) minus 2 to account for network and broadcast addresses.

6. Given the subnet mask /26, how many host bits are left?

A. 6 bits B. 26 bits C. 8 bits D. 24 bits

Correct answer is A. With a /26 mask, 26 bits are for network, leaving 6 bits for hosts.

7. What is the broadcast address for the network 192.168.10.0/24?

A. 192.168.10.0 B. 192.168.10.1 C. 192.168.10.254 D. 192.168.10.255

Correct answer is D. Broadcast address sets all host bits to 1, which for /24 is 192.168.10.255.

8. For the IP 172.16.50.123/20, what is the network address?

A. 172.16.50.0 B. 172.16.48.0 C. 172.16.48.0 D. 172.16.63.255

Correct answer is C. With a /20 mask, network is 172.16.48.0 (block size 16 in the third octet).

9. What is the usable IP range for the subnet 192.168.1.0/24?

A. 192.168.1.0 to 192.168.1.255 B. 192.168.1.1 to 192.168.1.254 C. 192.168.1.1 to 192.168.1.255 D. 192.168.1.0 to 192.168.1.254

Correct answer is B. Usable IPs are from network +1 to broadcast -1, here 192.168.1.1 to 192.168.1.254.

10. How many subnets are created when borrowing 2 bits from the host portion?

A. 4 subnets B. 2 subnets C. 8 subnets D. 16 subnets

Correct answer is A. Number of subnets = 2^borrowed_bits = 2^2 = 4 subnets.

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