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Network+ PBQ Library

Hands-on Performance-Based Questions for N10-009. Subnetting calculators, topology builders, and firewall ACLs — just like the real exam.

D1: Networking Concepts

OSI Model Layer Matching

Drag protocols and network devices (HTTP, TCP, router, switch) to the correct OSI layer. Tests Layer 1-7 knowledge.

OSI model Protocol encapsulation Layer identification
Not attempted Launch Lab
D1: Networking Concepts

Subnetting & IP Addressing Lab

Calculate subnets, network IDs, masks, first/last usable IPs, and broadcast addresses. Assign departments with VLSM.

CIDR/VLSM Subnet calculation IP addressing
Not attempted Launch Lab
D2: Network Implementation

Network Topology Builder

Drag and drop network devices (router, switch, firewall, AP, server, workstation) into the correct zones.

Topology design Device placement Network zones
Not attempted Launch Lab
D4: Network Security

Firewall ACL Configurator

Build firewall rules with direction, protocol, CIDR sources, ports, and implicit deny-all policy.

ACL rules CIDR notation Deny-all policy
Not attempted Launch Lab

Worked Subnetting Example: 192.168.10.130/26

Given this interface address, find the mask, network ID, usable range, broadcast address, and usable host count. The address ends in .130, but that does not make it a network ID; first locate the subnet that contains it.

  1. Convert the prefix. A /26 has 26 network bits and 6 host bits. In the last octet, the two network bits have values 128 and 64, producing mask 255.255.255.192. The six host bits give 64 total addresses per subnet.
  2. Find the containing block. With a 64-address block size, the last-octet boundaries are .0, .64, .128, and .192. Since 130 falls between 128 and 191, the interface belongs to the block beginning at 192.168.10.128. The next block starts at .192, so this block’s final address is .191.
  3. Derive the remaining fields. The first address is the network ID and the last is the broadcast address in this IPv4 host-addressing example. Add one to the network for the first usable address and subtract one from the broadcast for the last usable address. Excluding those two reserved addresses leaves 62 usable host addresses.
Subnet mask255.255.255.192Network ID192.168.10.128
First usable192.168.10.129Last usable192.168.10.190
Broadcast192.168.10.191Usable hosts62

Check the result: .130 lies between .129 and .190, so the original interface is a usable host address. The existing subnetting lab uses the same Network ID, mask, first usable, last usable, and broadcast fields, but starts with a different requirement: divide 192.168.10.0/24 into at least six subnets and assign departments. Derive the prefix from that requirement rather than copying /26, then repeat this process for every row.

Checks That Catch Common Subnetting Mistakes

  • Do not report 64 usable host addresses. The /26 block contains 64 addresses in total, but this host-addressing example reserves the network ID and broadcast address, leaving 62.
  • Do not assume an address ending in .130 is a network ID. Write the block boundaries first; here .130 falls inside .128–.191, so it is a usable host.
  • Check the answer against the original interface. Your own address must sit between the first and last usable values. If it does not, the block size or boundary calculation is wrong.

What the Other Labs Add

  • OSI Model Layer Matching asks you to place HTTP, SSL/TLS, NetBIOS, TCP, a router, a switch, and a hub at their layers, practising protocol and device-role associations.
  • Network Topology Builder places six devices into labelled edge, perimeter, LAN, wireless, server, and user zones, so explain each placement by device function rather than position memory.
  • Firewall ACL Configurator combines direction, protocol, CIDR sources, destinations, ports, and rule order. Its SSH restriction and final deny-all make least-privilege decisions concrete.
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