Cisco CCNA (200-301)Network FundamentalsMedium
A network administrator needs to assign a subnet to a new department that will require 60 usable host IP addresses. Which of the following subnet masks, in CIDR notation, would be the smallest block that satisfies this requirement while minimizing wasted addresses?
- A/26
- B/25
- C/24
- D/27
Show answer & explanationAnswer & explanation
Correct answer: A. /26
To find the number of usable hosts, use the formula 2^n - 2, where 'n' is the number of host bits. For 60 usable hosts, we need 2^n >= 62 (60 hosts + network + broadcast). 2^5 = 32 (too small), 2^6 = 64. So, we need 6 host bits. A /26 subnet mask means 26 network bits, leaving 32 - 26 = 6 host bits, which provides 2^6 - 2 = 62 usable hosts. This is the smallest block that meets the requirement.
Why the other options are wrong
- B. /25 provides 2^7 - 2 = 126 usable hosts, which is larger than necessary and wastes more addresses.
- C. /24 provides 2^8 - 2 = 254 usable hosts, which is far too large and wasteful.
- D. /27 provides 2^5 - 2 = 30 usable hosts, which is insufficient for 60 hosts.
Usable Host Calculation
The number of usable host IP addresses in a subnet is calculated as 2^n - 2, where 'n' is the number of host bits.
- The '-2' accounts for the network address and the broadcast address, which cannot be assigned to hosts.
- To find 'n', determine the smallest power of 2 that is greater than or equal to (required hosts + 2).
- The CIDR notation (e.g., /26) indicates the number of network bits; host bits = 32 - network bits.
Memory trick: Hosts need power of 2 minus 2, CIDR gives network, 32 minus that is host.