Chapter 1: Network Fundamentals
Every network you will ever touch — from a home Wi-Fi router to a global enterprise WAN — is built on the ideas in this chapter. Cisco expects you to know these cold: what happens at each layer of the models, how bits travel over copper and fiber, how Ethernet frames work, and above all, how to subnet in your head. Subnetting alone can decide whether you pass or fail the exam, so we go deep on it here.
1.1 The OSI Model vs the TCP/IP Model
The OSI model (Open Systems Interconnection) is a 7-layer reference model. Nobody implements it directly, but everyone uses its vocabulary to describe what is happening on a network. The TCP/IP model is the 4-layer model that the real Internet actually runs on. Cisco will ask you to map one to the other, so learn both.
| # | OSI Layer | TCP/IP Layer | What happens there | Example devices / protocols |
|---|---|---|---|---|
| 7 | Application | Application | User-facing services, data formatting | HTTP, DNS, FTP, SMTP |
| 6 | Presentation | Application | Data format, encryption, compression | TLS/SSL, JPEG, ASCII |
| 5 | Session | Application | Sessions between applications | NetBIOS, session setup/teardown |
| 4 | Transport | Transport | End-to-end delivery, ports, reliability | TCP (reliable), UDP (fast) |
| 3 | Network | Internet | Logical addressing, routing between networks | IP, routers, ICMP |
| 2 | Data Link | Network Access | Framing, MAC addressing, LAN delivery | Ethernet, switches |
| 1 | Physical | Network Access | Bits on the wire: cables, hubs, signals | Cables, connectors, hubs |
The OSI model has 7 layers; the TCP/IP model has 4. Layers 5–7 of OSI all collapse into TCP/IP's single Application layer, and OSI Layers 1–2 collapse into TCP/IP's Network Access layer.
Cisco loves asking "at which layer does X operate?" Memorize: switches = Layer 2 (Data Link), routers = Layer 3 (Network), TCP/UDP ports = Layer 4 (Transport), IP addresses = Layer 3, MAC addresses = Layer 2.
Encapsulation and De-encapsulation
When a host sends data, each layer adds its own header (and sometimes a trailer) as the data moves down the stack. This is encapsulation:
- Application — the user data (e.g. a web page request).
- Transport — adds a TCP or UDP header → the unit is now called a segment (TCP) or datagram (UDP). This header carries the source and destination port numbers.
- Network — adds an IP header → now called a packet. This header carries the source and destination IP addresses.
- Data Link — adds an Ethernet header and trailer → now called a frame. The header carries source and destination MAC addresses; the trailer carries the FCS error check.
- Physical — the frame is converted to bits and transmitted as electrical, light, or radio signals.
The receiving host reverses the process — de-encapsulation — stripping each header as the data moves back up the stack.
Remember the unit names in order: data → segment (TCP) → packet → frame → bits. Cisco test questions frequently ask "what is the PDU at Layer 3?" — the answer is packet.
1.2 Cabling Types
Twisted-Pair Copper
Twisted-pair cable contains pairs of copper wires twisted together to cancel electromagnetic interference. UTP (unshielded) is the common type in offices; STP (shielded) adds foil shielding for noisy environments like factories.

| Category | Max speed | Max distance at top speed | Typical use |
|---|---|---|---|
| Cat5e | 1 Gbps | 100 m | Older office LANs |
| Cat6 | 10 Gbps | ~55 m (100 m at 1 Gbps) | Modern office LANs |
| Cat6a | 10 Gbps | 100 m | Data centers, new installs |
If a question mentions 10 Gbps over copper at 100 meters, the answer is Cat6a. Cat6 can only do 10 Gbps for about 55 meters.
Fiber Optics
Fiber carries light instead of electricity, so it is immune to electromagnetic interference and supports far longer distances.
| Single-mode fiber (SMF) | Multi-mode fiber (MMF) | |
|---|---|---|
| Light source | Laser | LED or VCSEL |
| Core size | ~9 microns (narrow) | 50 or 62.5 microns (wide) |
| Distance | Kilometers (10 km+ with standard optics) | Hundreds of meters (e.g. 300–400 m at 10 Gbps on OM3/OM4) |
| Cost | More expensive optics | Cheaper optics |
| Typical use | Long-haul, campus backbones, ISP links | Data centers, building backbones |
Connectors
- RJ45 — 8-pin connector for twisted-pair Ethernet. This is the familiar "network plug."
- LC — small-form-factor fiber connector, the most common in modern equipment.
- SC — larger square fiber connector, common on older gear.
- ST — round bayonet-style fiber connector, mostly legacy.

1.3 Ethernet Concepts
Frames and MAC Addresses
An Ethernet frame has this structure: preamble (synchronization) → destination MAC → source MAC → EtherType/length → payload → FCS (Frame Check Sequence, a 4-byte error-detection trailer).
A MAC address is a 48-bit hardware address written as 12 hexadecimal digits, e.g. AA:BB:CC:DD:EE:FF. The first 24 bits are the OUI (Organizationally Unique Identifier) assigned to the manufacturer; the last 24 bits identify the individual device. The broadcast MAC address is FF:FF:FF:FF:FF:FF — a frame sent there is received by every device on the LAN.
MAC addresses live at Layer 2 and never cross a router. IP addresses live at Layer 3 and identify the device end-to-end across the whole path.
Half Duplex, CSMA/CD, and Full Duplex
- Half duplex — a device can send or receive, but not both at once (like a walkie-talkie). Old hubs worked this way. To manage sharing, Ethernet used CSMA/CD (Carrier Sense Multiple Access with Collision Detection): listen before sending, and if two devices transmit at once (a collision), both back off for a random time and retry.
- Full duplex — a device sends and receives simultaneously (like a phone call). Modern switches give every port its own collision-free path, so collisions no longer occur and CSMA/CD is effectively retired on switched networks.
Speed and Duplex
Ethernet ports negotiate speed (10/100/1000 Mbps and beyond) and duplex automatically via autonegotiation. A duplex mismatch — one side full, the other half — causes late collisions, CRC errors, and painfully slow throughput. When troubleshooting, check both ends:
Switch# show interfaces gigabitethernet0/1 status
Port Name Status Vlan Duplex Speed Type
Gi0/1 connected 10 a-full a-1000 10/100/1000BaseTX
To set speed and duplex manually:
Switch(config)# interface gigabitethernet0/1
Switch(config-if)# speed 1000
Switch(config-if)# duplex full
Cisco loves the duplex mismatch scenario: "users report the link is up but throughput is terrible, and the switch shows many CRC errors." Think duplex mismatch first.
1.4 IPv4 Addressing and Subnetting in Depth
Address Classes and Private Ranges
An IPv4 address is 32 bits, written as four decimal octets (e.g. 192.168.1.10).
| Class | First octet range | Default mask | Use |
|---|---|---|---|
| A | 1–126 | /8 (255.0.0.0) | Very large networks |
| B | 128–191 | /16 (255.255.0.0) | Medium networks |
| C | 192–223 | /24 (255.255.255.0) | Small networks |
| D | 224–239 | — | Multicast |
| E | 240–255 | — | Experimental |
Special addresses: 127.0.0.0/8 is loopback (this device itself); 169.254.0.0/16 is APIPA/link-local (self-assigned when no DHCP server answers).
The private ranges (RFC 1918) are not routable on the public Internet and are reused inside organizations:
10.0.0.0/8(Class A private)172.16.0.0/12(Class B private — 172.16.0.0 through 172.31.255.255)192.168.0.0/16(Class C private)
Know the three private ranges exactly, including that the 172.16 range spans 172.16.x.x through 172.31.x.x (/12). Cisco tests this constantly.
Subnet Mask Math
Subnetting borrows bits from the host portion of an address to create smaller networks. Two formulas run everything:
- Number of subnets = 2^(borrowed bits)
- Usable hosts per subnet = 2^(host bits) − 2 (subtract the network ID and the broadcast address)
The fastest method: the block size (increment) = 256 − (interesting octet of the mask). Count by block sizes to find each subnet's network ID; the broadcast is one address before the next network ID.
Question: For 192.168.10.77 with mask /27, find the subnet mask in dotted decimal, the network ID, the broadcast address, and the valid host range.
Step 1: /27 means 27 network bits. The mask is 11111111.11111111.11111111.11100000 = 255.255.255.224. Step 2: The interesting octet is the 4th (224). Block size = 256 − 224 = 32. Step 3: Count subnets by 32s in the 4th octet: 0, 32, 64, 96, … The address 77 falls in the 64 block. Step 4: Network ID = 192.168.10.64. Broadcast = one before the next block (96 − 1) = 192.168.10.95. Step 5: Valid hosts = everything between: 192.168.10.65 through 192.168.10.94.
Answer: mask 255.255.255.224, network 192.168.10.64, broadcast 192.168.10.95, valid hosts .65–.94.
Question: The address block 172.16.0.0/19 is subnetted from its classful default. How many subnets does it create, and how many usable hosts are in each?
Step 1: 172.16.0.0 is Class B, default mask /16. Borrowed bits = 19 − 16 = 3. Step 2: Number of subnets = 2^3 = 8. Step 3: Host bits = 32 − 19 = 13. Usable hosts = 2^13 − 2 = 8192 − 2 = 8190. Step 4: Mask: /19 = 11111111.11111111.11100000.00000000 = 255.255.224.0.
Answer: 8 subnets, 8,190 usable hosts each, mask 255.255.224.0.
Question: For 10.25.13.200/21, find the network ID, broadcast address, and valid host range.
Step 1: /21 → mask 11111111.11111111.11111000.00000000 = 255.255.248.0. Step 2: Interesting octet is the 3rd (248). Block size = 256 − 248 = 8. Step 3: Count by 8s in the 3rd octet: 0, 8, 16, … The value 13 falls in the 8 block. Step 4: Network ID = 10.25.8.0. Broadcast = 10.25.15.255 (one before 10.25.16.0). Step 5: Valid hosts = 10.25.8.1 through 10.25.15.254.
Answer: network 10.25.8.0, broadcast 10.25.15.255, valid hosts 10.25.8.1–10.25.15.254.
For 172.16.45.123/20, find the subnet mask in dotted decimal, the network ID, the broadcast address, and the valid host range.
Show solution
Step 1: /20 → mask 11111111.11111111.11110000.00000000 = 255.255.240.0. Step 2: Interesting octet is the 3rd (240). Block size = 256 − 240 = 16. Step 3: Count by 16s: 0, 16, 32, 48, … The value 45 falls in the 32 block. Step 4: Network ID = 172.16.32.0. Broadcast = 172.16.47.255. Step 5: Valid hosts = 172.16.32.1 through 172.16.47.254.
Answer: mask 255.255.240.0, network 172.16.32.0, broadcast 172.16.47.255, valid hosts 172.16.32.1–172.16.47.254.
The block 10.0.0.0/11 is subnetted from its classful default. How many subnets are created, and how many usable hosts per subnet? What is the mask in dotted decimal?
Show solution
Step 1: 10.0.0.0 is Class A, default /8. Borrowed bits = 11 − 8 = 3 → subnets = 2^3 = 8. Step 2: Host bits = 32 − 11 = 21 → usable hosts = 2^21 − 2 = 2,097,152 − 2 = 2,097,150. Step 3: /11 = 11111111.11100000.00000000.00000000 = 255.224.0.0.
Answer: 8 subnets, 2,097,150 usable hosts each, mask 255.224.0.0.
Configuring an IPv4 Address
Router(config)# interface gigabitethernet0/0
Router(config-if)# ip address 192.168.10.1 255.255.255.0
Router(config-if)# no shutdown
Router interfaces are shut down by default — no shutdown brings the interface up. Switch access ports are enabled by default. Forgetting no shutdown on a router interface is one of the most common lab mistakes.
1.5 VLSM and Route Summarization Basics
VLSM (Variable Length Subnet Masking) means using different mask lengths inside one address block, so each subnet gets only the addresses it needs. Always allocate largest requirement first, or the big subnets will not fit into the leftover space.
Question: Subnet 192.168.1.0/24 with VLSM for: LAN-A needs 50 hosts, LAN-B needs 25 hosts, LAN-C needs 10 hosts, and two point-to-point WAN links need 2 hosts each.
Step 1: LAN-A (50 hosts). Need host bits where 2^h − 2 ≥ 50 → h = 6 (62 usable) → /26. Block size 64. Assign 192.168.1.0/26 (hosts .1–.62, broadcast .63). Step 2: LAN-B (25 hosts). h = 5 (30 usable) → /27. Block size 32. Next free block: 192.168.1.64/27 (hosts .65–.94, broadcast .95). Step 3: LAN-C (10 hosts). h = 4 (14 usable) → /28. Block size 16. Next free: 192.168.1.96/28 (hosts .97–.110, broadcast .111). Step 4: WAN-1 (2 hosts). h = 2 (2 usable) → /30. Block size 4. Next free: 192.168.1.112/30 (hosts .113–.114, broadcast .115). Step 5: WAN-2 (2 hosts). Next /30: 192.168.1.116/30 (hosts .117–.118, broadcast .119).
Answer: 192.168.1.0/26, 192.168.1.64/27, 192.168.1.96/28, 192.168.1.112/30, 192.168.1.116/30.
Route summarization (aggregation) combines multiple contiguous subnets into one advertised route, shrinking routing tables.
Question: What single summary route covers 192.168.0.0/24, 192.168.1.0/24, 192.168.2.0/24, and 192.168.3.0/24?
Step 1: The first two octets (192.168) are identical. Compare the third octet: 0 = 00000000, 1 = 00000001, 2 = 00000010, 3 = 00000011. Step 2: The common leading bits are 000000 — 6 bits. All four addresses agree on the first 6 bits of the third octet. Step 3: Summary mask = 16 (first two octets) + 6 = /22. Summary network = 192.168.0.0/22.
Answer: 192.168.0.0/22 covers all four subnets (range 192.168.0.0–192.168.3.255).
For summarization questions, find how many leading bits the addresses share — that count plus the identical octets gives the summary mask. If the ranges are not contiguous or do not align on a bit boundary, no clean summary exists.
1.6 IPv6 Fundamentals
IPv6 uses 128-bit addresses written as eight groups of four hexadecimal digits (hextets), e.g. 2001:0db8:0000:0000:0000:ff00:0042:8329. Two shortening rules keep them readable:
- Drop leading zeros in each hextet:
0db8→db8,0042→42. - Replace the longest run of all-zero hextets with
::— but only once per address.
So 2001:0db8:0000:0000:0000:ff00:0042:8329 becomes 2001:db8::ff00:42:8329.
Write the fully shortened form of 2001:0db8:0000:0000:0000:ff00:0042:8329.
Show solution
Step 1: Strip leading zeros per hextet: 2001:db8:0:0:0:ff00:42:8329. Step 2: The longest run of zero hextets is the three consecutive "0" groups — replace with "::" (only once). Answer: 2001:db8::ff00:42:8329.
IPv6 Address Types
| Type | Prefix | Scope and use |
|---|---|---|
| Global unicast | 2000::/3 | Globally routable, like a public IPv4 address |
| Link-local | FE80::/10 | Single link only, never routed; auto-configured on every IPv6 interface |
| Unique local | FC00::/7 | Private addresses for internal use (like RFC 1918) |
| Multicast | FF00::/8 | One-to-many delivery; replaces IPv4 broadcast |
| Loopback | ::1 | This device itself (like 127.0.0.1) |
| Unspecified | :: | "No address yet" placeholder (like 0.0.0.0) |
| Documentation | 2001:db8::/32 | Reserved for examples and documentation — safe to use in labs |
There is no broadcast in IPv6 — multicast replaces it. Every IPv6 interface automatically gets a link-local address (FE80::/10), which routers use to talk to neighbors even before any global address is configured.
EUI-64
A host can build its own 64-bit interface ID from its 48-bit MAC address (EUI-64): split the MAC in half, insert FFFE in the middle, and flip the 7th bit (the U/L bit) of the first byte.
Question: Derive the full IPv6 address for a host with MAC 00:1A:2B:3C:4D:5E on the prefix 2001:db8:abcd:1::/64.
Step 1: Split the MAC: 00:1A:2B | 3C:4D:5E. Insert FFFE in the middle: 00:1A:2B:FF:FE:3C:4D:5E. Step 2: Flip the 7th bit of the first byte. 0x00 = 00000000; flipping the bit with value 2 gives 00000010 = 0x02. Step 3: Interface ID in hextet form: 021A:2BFF:FE3C:4D5E. Step 4: Attach the /64 prefix: 2001:db8:abcd:1:021a:2bff:fe3c:4d5e.
Answer: 2001:db8:abcd:1:021a:2bff:fe3c:4d5e.
A host has MAC address AA:BB:CC:DD:EE:FF. Derive its EUI-64 interface ID.
Show solution
Step 1: Split and insert FFFE: AA:BB:CC:FF:FE:DD:EE:FF. Step 2: Flip the 7th bit of 0xAA. 0xAA = 10101010; flipping the 0x02 bit gives 10101000 = 0xA8. Step 3: Interface ID: A8BB:CCFF:FEDD:EEFF.
Answer: A8BB:CCFF:FEDD:EEFF.
SLAAC vs DHCPv6
- SLAAC (Stateless Address Autoconfiguration): the host listens for Router Advertisements, takes the advertised prefix, and generates its own interface ID (EUI-64 or random). No server needed; the router does not track who got what. Basic SLAAC does not provide DNS server info.
- DHCPv6: stateful DHCPv6 works like DHCP for IPv4 (server assigns full addresses and tracks them); stateless DHCPv6 only supplies extra info such as DNS servers while the host self-assigns its address via SLAAC.
Cisco's favorite IPv6 traps: the U/L bit flip in EUI-64 (0x00 becomes 0x02, not 0x00), that link-local is FE80::/10 and is never routed, and that SLAAC alone gives you no DNS.
Configuring IPv6
Router(config)# ipv6 unicast-routing
Router(config)# interface gigabitethernet0/0
Router(config-if)# ipv6 address 2001:db8:abcd:1::1/64
Router(config-if)# no shutdown
ipv6 unicast-routing is the global command that enables a router to actually route IPv6 — without it, the router will not forward IPv6 packets between interfaces, exactly like ip routing for IPv4 (which is on by default).

1.7 Wireless Principles
RF Bands and Channels
Wi-Fi uses unlicensed radio spectrum in three bands:
| Band | Frequency range | Characteristics |
|---|---|---|
| 2.4 GHz | ~2.4–2.5 GHz | Longest range, slowest, most interference (microwaves, Bluetooth); channels 1–11 in the US |
| 5 GHz | ~5.1–5.8 GHz | Faster, more channels, shorter range, less interference |
| 6 GHz | ~5.9–7.1 GHz | Newest (Wi-Fi 6E/7), widest channels, shortest range, least interference |
In the 2.4 GHz band, channels overlap — only channels 1, 6, and 11 are non-overlapping (in regions using channels 1–11), so neighboring access points should be placed on these three. The 5 and 6 GHz bands offer many more non-overlapping channels and wider channel widths (40, 80, even 160 MHz) for higher throughput.
An office has three access points with overlapping coverage in the 2.4 GHz band (US channels 1–11). Which channels should they use, and why?
Show solution
Use channels 1, 6, and 11 — one per access point. In the 2.4 GHz band each channel is only 5 MHz apart but ~20 MHz wide, so adjacent channels overlap and interfere. Channels 1, 6, and 11 are the only three that do not overlap with each other, giving each AP clean spectrum.
SSID and Wireless Media Access
The SSID (Service Set Identifier) is the network name clients see — up to 32 characters, case-sensitive. Hiding the SSID (not broadcasting it) is not real security; it is trivially discovered.
Because wireless is half-duplex (a radio cannot transmit and receive on the same frequency at once), Wi-Fi uses CSMA/CA (Collision Avoidance) instead of Ethernet's CSMA/CD — the sender first checks the airwaves are clear and may use RTS/CTS handshaking to reserve airtime, avoiding collisions rather than detecting them after the fact.
Wired Ethernet = CSMA/CD (collision detection, half duplex, mostly historical). Wireless = CSMA/CA (collision avoidance). Cisco will absolutely test that contrast.
End-of-Chapter Test
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Answers
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