Chapter 1 · CCNA 200-301 Study Guide · Based on the CCNA 200-301 v2.0 exam blueprint (effective February 2027)

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 LayerTCP/IP LayerWhat happens thereExample devices / protocols
7ApplicationApplicationUser-facing services, data formattingHTTP, DNS, FTP, SMTP
6PresentationApplicationData format, encryption, compressionTLS/SSL, JPEG, ASCII
5SessionApplicationSessions between applicationsNetBIOS, session setup/teardown
4TransportTransportEnd-to-end delivery, ports, reliabilityTCP (reliable), UDP (fast)
3NetworkInternetLogical addressing, routing between networksIP, routers, ICMP
2Data LinkNetwork AccessFraming, MAC addressing, LAN deliveryEthernet, switches
1PhysicalNetwork AccessBits on the wire: cables, hubs, signalsCables, connectors, hubs
Key point

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.

Exam tip

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:

  1. Application — the user data (e.g. a web page request).
  2. 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.
  3. Network — adds an IP header → now called a packet. This header carries the source and destination IP addresses.
  4. 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.
  5. 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.

Key point

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.

close-up of RJ45 ethernet connectors, one plugged into a wall jack and one unplugged showing the eight gold pins
close-up of RJ45 ethernet connectors, one plugged into a wall jack and one unplugged showing the eight gold pinsPhoto: Pixabay via https://www.pexels.com/photo/415043/
CategoryMax speedMax distance at top speedTypical use
Cat5e1 Gbps100 mOlder office LANs
Cat610 Gbps~55 m (100 m at 1 Gbps)Modern office LANs
Cat6a10 Gbps100 mData centers, new installs
Exam tip

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 sourceLaserLED or VCSEL
Core size~9 microns (narrow)50 or 62.5 microns (wide)
DistanceKilometers (10 km+ with standard optics)Hundreds of meters (e.g. 300–400 m at 10 Gbps on OM3/OM4)
CostMore expensive opticsCheaper optics
Typical useLong-haul, campus backbones, ISP linksData centers, building backbones

Connectors

side-by-side comparison of fiber optic cable connectors (LC, SC, ST) next to a standard RJ45 copper connector
side-by-side comparison of fiber optic cable connectors (LC, SC, ST) next to a standard RJ45 copper connectorPhoto: Brett Sayles via https://www.pexels.com/photo/1624895/

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.

Key point

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

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
Exam tip

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).

ClassFirst octet rangeDefault maskUse
A1–126/8 (255.0.0.0)Very large networks
B128–191/16 (255.255.0.0)Medium networks
C192–223/24 (255.255.255.0)Small networks
D224–239—Multicast
E240–255—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:

Exam tip

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:

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.

Worked calculation

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.

Worked calculation

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.

Worked calculation

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.

Try it yourself

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.

Try it yourself

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
Key point

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.

Worked calculation

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.

Worked calculation

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).

Exam tip

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:

  1. Drop leading zeros in each hextet: 0db8 → db8, 0042 → 42.
  2. 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.

Try it yourself

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

TypePrefixScope and use
Global unicast2000::/3Globally routable, like a public IPv4 address
Link-localFE80::/10Single link only, never routed; auto-configured on every IPv6 interface
Unique localFC00::/7Private addresses for internal use (like RFC 1918)
MulticastFF00::/8One-to-many delivery; replaces IPv4 broadcast
Loopback::1This device itself (like 127.0.0.1)
Unspecified::"No address yet" placeholder (like 0.0.0.0)
Documentation2001:db8::/32Reserved for examples and documentation — safe to use in labs
Key point

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.

Worked calculation

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.

Try it yourself

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

Exam tip

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
Key point

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).

network engineer configuring a router via console cable, laptop screen showing CLI
network engineer configuring a router via console cable, laptop screen showing CLIPhoto: Field Engineer via https://www.pexels.com/photo/442150/

1.7 Wireless Principles

RF Bands and Channels

Wi-Fi uses unlicensed radio spectrum in three bands:

BandFrequency rangeCharacteristics
2.4 GHz~2.4–2.5 GHzLongest range, slowest, most interference (microwaves, Bluetooth); channels 1–11 in the US
5 GHz~5.1–5.8 GHzFaster, more channels, shorter range, less interference
6 GHz~5.9–7.1 GHzNewest (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.

Try it yourself

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.

Exam tip

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

Q1 At which OSI layer do IP addresses operate? A. Layer 2 — Data Link B. Layer 3 — Network C. Layer 4 — Transport D. Layer 7 — Application

Q2 A host sends data and the Transport layer adds its header. What is the resulting PDU called (for TCP)? A. Packet B. Frame C. Segment D. Datagram

Q3 You need 10 Gbps over twisted-pair copper at a distance of 90 meters. Which cable category is required? A. Cat5e B. Cat6 C. Cat6a D. Cat7

Q4 Which fiber type uses a laser light source and supports multi-kilometer distances? A. Multi-mode fiber B. Single-mode fiber C. Both, depending on the connector D. Neither — fiber is limited to 550 meters

Q5 What is the valid host range for 192.168.20.100/26? A. 192.168.20.65–192.168.20.126 B. 192.168.20.64–192.168.20.127 C. 192.168.20.65–192.168.20.94 D. 192.168.20.97–192.168.20.126

Q6 Which of these is a valid private IPv4 address? A. 172.15.10.5 B. 172.32.0.10 C. 172.20.5.100 D. 192.169.1.1

Q7 What single summary route covers 10.10.8.0/24 through 10.10.11.0/24? A. 10.10.8.0/22 B. 10.10.0.0/22 C. 10.10.8.0/23 D. 10.10.8.0/21

Q8 Which IPv6 address type is automatically configured on every IPv6-enabled interface and is never routed beyond the local link? A. Global unicast B. Unique local C. Link-local (FE80::/10) D. Multicast

Q9 A host with MAC address 0C:11:22:33:44:55 uses EUI-64. What is the first hextet of its interface ID? A. 0C11 B. 0E11 C. 8C11 D. 0C10

Q10 Three access points with overlapping coverage operate in the 2.4 GHz band. Which channel plan avoids interference? A. Channels 1, 2, 3 B. Channels 1, 6, 11 C. Channels 5, 6, 7 D. Any three channels, since 2.4 GHz channels do not overlap

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Answers

Answer key

Q1: B — IP addressing and routing happen at Layer 3 (Network). MAC addresses are Layer 2; ports are Layer 4.
Q2: C — TCP adds its header at the Transport layer, producing a segment . (UDP produces a datagram, the Network layer produces a packet, and the Data Link layer produces a frame.)
Q3: C — Cat6a supports 10 Gbps to 100 meters. Cat6 only reaches about 55 meters at 10 Gbps, and Cat5e tops out at 1 Gbps.
Q4: B — Single-mode fiber uses a laser and spans kilometers. Multi-mode uses LED/VCSEL sources and is limited to hundreds of meters.
Q5: A — /26 → mask 255.255.255.192, block size 64. Address 100 falls in the 64 block: network 192.168.20.64, broadcast 192.168.20.127, valid hosts .65–.126.
Q6: C — 172.20.5.100 falls inside the 172.16.0.0/12 private range (172.16.0.0–172.31.255.255). 172.15.x.x and 172.32.x.x are outside it, and 192.169.x.x is public (only 192.168.x.x is private).
Q7: A — Third octets 8 (00001000), 9, 10, 11 (00001011) share the first 6 bits → /16+6 = /22, network 10.10.8.0/22 (range 10.10.8.0–10.10.11.255).
Q8: C — Link-local addresses (FE80::/10) are auto-configured on every IPv6 interface and never leave the local link. Unique local addresses are private but routable within an organization.
Q9: B — EUI-64 flips the 7th (U/L) bit of the first byte: 0x0C = 00001100 → flipping the 0x02 bit gives 00001110 = 0x0E. The first hextet is 0E11 (from 0E:11:22…).
Q10: B — In the 2.4 GHz band only channels 1, 6, and 11 are non-overlapping (US channels 1–11); all other combinations interfere.

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