Understanding the Requirements
* Network Address: 192.168.1.0/24
* The /24 notation means a subnet mask of 255.255.255.0, providing 256 total addresses (192.168.1.0-192.168.1.255).
* Usable hosts: 256 - 2 (network and broadcast) = 254.
* Goal: Create3 subnets, each with30 hosts.
* Each subnet needs enough addresses to accommodate 30 hosts, plus 2 reserved addresses (network and broadcast) per subnet.
* Total addresses per subnet = 30 (hosts) + 2 (network/broadcast) =32 addresses.
Subnetting Basics (Networking Fundamentals)
* Subnet Mask: Determines how many bits are borrowed from the host portion to create subnets.
* Original Mask: /24 (255.255.255.0) = 24 network bits, 8 host bits.
* Formulae:
* Number of subnets = 2^(number of borrowed bits).
* Number of addresses per subnet = 2^(remaining host bits).
* Usable hosts per subnet = 2^(remaining host bits) - 2.
We need:
* At least3 subnets.
* At least32 addresses per subnet(to fit 30 hosts + 2 reserved).
Step-by-Step Analysis
* Determine Addresses Needed per Subnet:
* 32 addresses is a power of 2 (2^5 = 32).
* This means each subnet requires 5 host bits (since 2^5 = 32 total addresses, and 32 - 2 = 30 usable hosts).
* Calculate Remaining Bits:
* Original network has 8 host bits (/24).
* If 5 bits are left for hosts, we borrow: 8 - 5 =3 bitsfor subnetting.
* New Subnet Mask:
* Original mask: /24 (24 network bits).
* Borrow 3 bits: 24 + 3 =/27.
* /27 = 255.255.255.224 (binary: 11111111.11111111.11111111.11100000).
* Verify Requirements:
* Number of Subnets: 2^3 =8 subnets(meets the requirement of at least 3).
* Addresses per Subnet: 2^5 =32 addresses.
* Usable Hosts per Subnet: 32 - 2 =30 hosts(exactly meets the requirement).
* Subnet Breakdown:
* Increment: 256 - 224 =32(each subnet increments by 32 in the fourth octet).
* Subnets:
* 192.168.1.0-192.168.1.31 (Network: .0, Broadcast: .31, Hosts: .1-.30)
* 192.168.1.32-192.168.1.63 (Network: .32, Broadcast: .63, Hosts: .33-.62)
* 192.168.1.64-192.168.1.95 (Network: .64, Broadcast: .95, Hosts: .65-.94)
* (And 5 more subnets up to 192.168.1.255.)
* Three subnets fit perfectly with 30 hosts each.
Evaluating the Options
* A. 255.255.255.128 (/25):
* Borrow 1 bit: 24 + 1 = /25.
* Subnets: 2^1 = 2 (not enough, need 3).
* Host bits: 7 (2^7 = 128 addresses, 126 hosts).
* Why Not: Only 2 subnets, fails the requirement.
* B. 255.255.255.192 (/26):
* Borrow 2 bits: 24 + 2 = /26.
* Subnets: 2^2 = 4 (meets 3).
* Host bits: 6 (2^6 = 64 addresses, 62 hosts).
* Why Not: 62 hosts exceeds 30, but it's overkill; /27 is more efficient and still valid.
* C. 255.255.255.224 (/27):
* Borrow 3 bits: 24 + 3 = /27.
* Subnets: 2^3 = 8 (meets 3).
* Host bits: 5 (2^5 = 32 addresses, 30 hosts).
* Why Yes: Perfectly fits 3 subnets with exactly 30 hosts each.
* D. 255.255.255.240 (/28):
* Borrow 4 bits: 24 + 4 = /28.
* Subnets: 2^4 = 16 (meets 3).
* Host bits: 4 (2^4 = 16 addresses, 14 hosts).
* Why Not: Only 14 hosts per subnet, fails the 30-host requirement.
Why /27 (255.255.255.224) is Best
* It providesexactly 30 usable hosts per subnet, avoiding waste while meeting the minimum requirement.
* It allows8 subnets, exceeding the need for 3, ensuring flexibility.
* The study guide emphasizes efficient subnet design, and /27 balances host count and subnet availability.
CompTIA Network+ Context
* Networking Fundamentals: Subnetting is a core skill, requiring understanding of CIDR, binary conversion, and address allocation.
* Example from Study Guide: Similar problems calculate subnet masks for specific host counts, reinforcing /27 as a common solution for ~30 hosts.