IP Addressing & Subnetting

Lesson 4 · OKSTEM College · AS Computer Science

Work through the concepts, then test yourself with the knowledge check.

Knowledge Check

A network has address 192.168.1.0/24. How many usable host addresses does it have?

256 is the total addresses (2⁸), but 2 are reserved.

Correct — 2⁸ − 2 = 254 usable host addresses.

255 is not a power of 2 minus 2. It's close to 254 but off by one.

128 = 2⁷, which applies to a /25 subnet (7 host bits), not /24.

📖 Quick Recap

/24 means 8 host bits → 2⁸=256 addresses. Subtract 2: network address (all host bits 0) and broadcast (all host bits 1) = 254 usable hosts.

📖 Quick Recap

Usable hosts = 2ʰ − 2 where h = host bits. For /24: 2⁸−2 = 254.

📖 Quick Recap

/24 has 8 host bits → 256 addresses → 254 usable. /25 has 7 host bits → 128 addresses → 126 usable.

The subnet mask 255.255.255.0 in CIDR notation is:

/8 = 255.0.0.0 — only 8 bits for network.

/16 = 255.255.0.0 — 16 network bits.

Correct — 255.255.255.0 has 24 ones followed by 8 zeros.

/32 means all 32 bits are network (single host address).

📖 Quick Recap

/8 means first 8 bits are network. 255.255.255.0 has 24 bits set → /24.

📖 Quick Recap

/16 = 255.255.0.0. 255.255.255.0 has an additional 8 bits set in the third octet → /24.

📖 Quick Recap

/32 = 255.255.255.255, a host route. 255.255.255.0 = /24.