1.7 - IP addressing and subnetting
Network+ N10-009 objective 1.7 covers IP addressing and subnetting, a math-heavy topic worth practising. In CIDR, a /27 subnet provides 30 usable host addresses (32 total minus network and broadcast), and each extra host bit doubles the count. Route summarization combines several small subnets into one larger route advertisement to shrink routing tables. For IPv6, SLAAC lets a host build its own address from a router-advertised prefix plus its interface identifier, with no DHCP needed. Transition mechanisms like NAT64 let IPv6-only clients reach IPv4-only servers. You should also know private ranges, APIPA, subnet masks, VLSM and IPv6 address types. Expect scenario questions that ask for the usable hosts in a given prefix, or which addressing or transition technique fits - so be ready to subnet quickly and recognise SLAAC, summarization and NAT64.
/27 = 30 usable hosts (each host bit doubles it). Combine subnets into one advert = route summarization. IPv6 self-config from router prefix = SLAAC. IPv6-only to IPv4-only = NAT64.
Practice questions
1. Which of these is a private (RFC 1918) IPv4 range?
- 11.0.0.0/8
- 10.0.0.0/8 (correct answer)
- 172.15.0.0/16
- 192.169.0.0/16
RFC 1918 private ranges: 10.0.0.0/8, 172.16.0.0/12, 192.168.0.0/16. The others (11.x, 172.15.x, 192.169.x) are public.
2. What prefix identifies an IPv6 link-local address?
- 2000::/3
- ff00::/8
- fe80::/10 (correct answer)
- ::1/128
Link-local IPv6 addresses start with fe80::/10 and are used on a single link (auto-configured). 2000::/3 is global unicast, ff00::/8 is multicast, ::1 is loopback.
3. How many usable host addresses does a /27 subnet provide?
- 62
- 14
- 30 (correct answer)
- 32
A /27 leaves 5 host bits (32-27), so 2^5 = 32 addresses minus network and broadcast = 30 usable hosts.
4. Which subnet mask corresponds to CIDR prefix /26?
- 255.255.255.192 (correct answer)
- 255.255.255.224
- 255.255.255.240
- 255.255.255.128
/26 sets 26 leading 1s, giving 255.255.255.192 (the last octet 11000000 = 192). /27 is .224, /28 is .240, /25 is .128.
5. Which address is the broadcast address of 192.168.10.0/26?
- 192.168.10.127
- 192.168.10.63 (correct answer)
- 192.168.10.255
- 192.168.10.64
/26 makes 64-address blocks; the first is .0-.63, so the broadcast is 192.168.10.63. .64 starts the next subnet.
6. How many equal /28 subnets fit inside a single /24 network?
- 16 (correct answer)
- 8
- 4
- 32
Going from /24 to /28 borrows 4 bits (28-24), giving 2^4 = 16 subnets of 16 addresses each.