
Cisco CCNA 200-301 — Complete Study Guide (2026)
The whole CCNA 200-301 blueprint — network fundamentals & subnetting, switching & VLANs, routing & OSPF, IP services, security, and automation — with subnetting, ACL, and OSPF-cost math worked step by step.
Un CCNA bootcamp/course cuesta $300–$800. Este libro enseña el mismo examen — las mismas reglas, verificadas a los estándares actuales — por un pago único de $14.99 que conservas de por vida.
Ten el libro completo — PDF + EPUB
Las preguntas de práctica y el simulacro siguen gratis. El libro es la mitad de estudio:
- Enseñado capítulo a capítulo — cada sección explicada en orden, no solo preguntas
- Imprímelo y márcalo — una referencia en papel para subrayar y anotar
- Funciona sin conexión — PDF para imprimir, EPUB para el móvil o e-reader
- Todo en un archivo — capítulos, resúmenes y preguntas de práctica juntos
Garantía de devolución de 14 días — ¿no te convence? Escríbenos y te reembolsamos el 100%, sin preguntas. Política de reembolso
El Capítulo 1 es gratis en esta página — pruébalo antes de comprar. Un solo pago desbloquea el libro completo (PDF + EPUB, preguntas con explicaciones, resúmenes por capítulo).
Descarga instantánea PDF + EPUB · pago único, tuyo para siempre · sin suscripción · garantía de reembolso de 14 días · ¿aún lo dudas? lee un capítulo completo gratis abajo.
- Verificado con la fuente oficial(the public Cisco CCNA 200-301 exam topics)
- 520 preguntas de práctica gratis
- Descarga instantánea, tuyo de por vida
Un capítulo completo, tal como viene en el eBook. Desplázate en la ventana para leerlo aquí; sin descargas ni correo.
This domain is the foundation for everything else on the exam. It covers the two reference models (OSI and TCP/IP), device roles, cabling and interfaces, Ethernet switching behavior, IPv4 addressing and — the single highest-yield skill on the whole exam — subnetting, plus the essentials of IPv6. If you can subnet fluently, you will pick up points in this domain and in IP Connectivity both.
The rule: two reference models, seven layers, one mnemonic
Networking is taught with two layered models. The OSI model has seven layers; the TCP/IP model collapses them into four. You must know the OSI layers cold, both by number and by what each does.
| # | OSI Layer | Job | Example PDU / device |
|---|---|---|---|
| 7 | Application | Interface to user apps | HTTP, DNS, DHCP data |
| 6 | Presentation | Format, encrypt, compress | TLS, JPEG, ASCII |
| 5 | Session | Set up / tear down dialogs | RPC, NetBIOS |
| 4 | Transport | End-to-end delivery, ports | Segment — TCP, UDP |
| 3 | Network | Logical addressing, routing | Packet — IP; router |
| 2 | Data Link | Local delivery, MAC, framing | Frame — Ethernet; switch |
| 1 | Physical | Bits on the wire | Cables, signals; hub |
A classic mnemonic, top to bottom: All People Seem To Need Data Processing. The unit of data — the PDU — has a name that changes per layer: Layer 4 = segment, Layer 3 = packet, Layer 2 = frame, Layer 1 = bits. That naming is itself an exam favorite.
The distinction the exam tests most is Layer 2 vs. Layer 3:
- Layer 3 (Network) does logical addressing (IPv4/IPv6) and routing — choosing the best path between networks. Routers live here.
- Layer 2 (Data Link) does physical addressing (MAC) and framing — delivery within a single segment/link. Switches live here.
So a question that says "responsible for logical addressing and determining the best path between networks" is Layer 3, the Network layer — routers, not switches.
The rule: TCP vs. UDP, and how encapsulation works
At Layer 4, two protocols matter:
- TCP is connection-oriented and reliable: it uses a three-way handshake (SYN, SYN-ACK, ACK), sequence numbers, acknowledgments, retransmission, and flow control with a sliding window. Use it when every byte must arrive — web, email, file transfer.
- UDP is connectionless and best-effort: no handshake, no acknowledgment, no retransmission, no flow control — just minimal overhead and low latency. Use it for real-time voice/video (VoIP), DNS queries, DHCP, and TFTP, where speed beats guaranteed delivery.
So a question that says "connectionless, best-effort, no retransmission, suitable for real-time voice" is describing UDP — not TCP (reliable), ICMP (diagnostics/errors), or ARP (MAC resolution).
| Feature | TCP | UDP |
|---|---|---|
| Connection | Connection-oriented (handshake) | Connectionless |
| Reliability | Acknowledged, retransmits | Best-effort, no retransmit |
| Ordering | Sequenced | No ordering |
| Flow control | Windowing | None |
| Overhead | Higher (20-byte header) | Lower (8-byte header) |
| Uses | Web, email, file transfer | VoIP, video, DNS, DHCP, TFTP |
Worked example — identify the protocol from behavior and port
An application opens a session that begins SYN / SYN-ACK / ACK, guarantees byte order, and uses port 443. That is TCP carrying HTTPS. Contrast: a query that fires a single datagram to port 53 with no handshake and no retransmission is UDP carrying DNS. And a lookup to port 67/68 with broadcasts is DHCP over UDP. The exam pairs a behavior ("no handshake, real-time") or a port number with a protocol — knowing both columns of the well-known-ports table lets you answer instantly.
Encapsulation is the process of wrapping data as it moves down the stack: the Transport layer adds a TCP/UDP header (making a segment), the Network layer adds an IP header (packet), the Data Link layer adds a frame header and trailer (frame), and the Physical layer sends bits. De-encapsulation unwraps it going up the stack at the receiver. Each layer talks to its peer layer on the far device.
The rule: switching behavior — learn, forward, flood
A switch builds a MAC address table (also called a CAM table) by learning the source MAC of every frame it receives and associating it with the port it arrived on. When it must forward a frame:
- Known unicast — destination MAC is in the table → forward out that one port.
- Unknown unicast — destination MAC not in the table → flood out all ports except the one it arrived on.
- Broadcast (destination FF:FF:FF:FF:FF:FF) or multicast → flood.
Two key domain concepts follow from this:
- A collision domain is a set of devices that could collide if they transmit at once. Each switch port is its own collision domain (full-duplex switching eliminates collisions). A hub puts everyone in one collision domain.
- A broadcast domain is the set of devices a broadcast reaches. A switch floods broadcasts, so by default all ports are one broadcast domain. A router does not forward broadcasts — so each router interface bounds a broadcast domain, and (as Chapter 2 shows) each VLAN is a separate broadcast domain.
Worked example — counting collision and broadcast domains
A router connects to two switches (one per interface). Switch A has 8 PCs; Switch B has 6 PCs; there are no VLANs beyond the default.
- Collision domains: each switch port (full-duplex) is its own collision domain, and each router-to-switch link is one too. Switch A: 8 PC ports + 1 uplink = 9; Switch B: 6 + 1 = 7. Total 16 collision domains. (If a hub joined several PCs, all of them would collapse into one shared collision domain.)
- Broadcast domains: the router separates the two sides, and each switch (default single VLAN) is one broadcast domain → 2 broadcast domains (one per router interface). Add a VLAN and you add a broadcast domain.
The rule to carry into the exam: switch ports multiply collision domains; routers (and VLANs) multiply broadcast domains; hubs merge collision domains.
The rule: IPv4 addressing anatomy
An IPv4 address is 32 bits, written as four octets (0–255) in dotted decimal, for example 192.168.1.100. A subnet mask (or its CIDR /prefix) marks which leading bits are the network portion and which trailing bits are the host portion. A 1 bit in the mask = network; a 0 bit = host.
Address classes still appear on the exam as vocabulary:
- Class A: 1–126 in the first octet, default /8.
- Class B: 128–191, default /16.
- Class C: 192–223, default /24.
- 127 is loopback; 224–239 is multicast (Class D); 240+ is experimental (Class E).
Private (RFC 1918) ranges — never routed on the public Internet:
10.0.0.0/8172.16.0.0–172.31.255.255(172.16.0.0/12)192.168.0.0/16
Also know: APIPA 169.254.0.0/16 is what a host self-assigns when DHCP fails.
Worked example — classify these addresses
10.55.1.1→ private (inside10.0.0.0/8); needs NAT to reach the Internet.172.20.5.1→ private (inside172.16.0.0–172.31.255.255). But172.32.1.1→ public (just outside the range).192.168.1.1→ private (192.168.0.0/16).169.254.10.10→ APIPA (DHCP failed — a symptom, not a usable Internet address).8.8.8.8→ public (routable). Being able to sort private vs. public at a glance drives NAT and troubleshooting questions.
Subnetting, taught step by step
Subnetting is the most valuable skill on the exam. The good news: it is pure pattern, and four questions answer everything.
For a given prefix /n:
- How many host bits?
H = 32 − n. - How many usable hosts?
2^H − 2(subtract 2 for the network address and the broadcast address). Exception: a /31 point-to-point link gives 2 usable hosts by special rule, and a /32 is a single host route. - What is the block size (increment) in the "interesting" octet?
256 − (mask value in that octet). Subnets step by that block size. - Which subnet does a host fall in? Round the interesting octet down to the nearest multiple of the block size. The network address has the host bits all 0; the broadcast address has them all 1 (the address just below the next subnet); usable hosts are everything in between.
Worked example — convert a prefix to a dotted-decimal mask in binary
What dotted-decimal mask is /28? A mask is 28 ones followed by 4 zeros. Group into octets: 11111111.11111111.11111111.11110000. The first three octets = 255. The last octet 11110000 = 128+64+32+16 = 240. So /28 = 255.255.255.240. Same method for /22: 11111111.11111111.11111100.00000000 → third octet 11111100 = 128+64+32+16+8+4 = 252 → 255.255.252.0. Knowing the eight bit-values (128, 64, 32, 16, 8, 4, 2, 1) lets you build any mask by hand — no table needed in a pinch.
Memorize this mask/prefix table for the last octet — it makes every question a lookup:
| Prefix | Mask (last octet) | Block size | Hosts/subnet (usable) |
|---|---|---|---|
| /24 | 0 | 256 | 254 |
| /25 | 128 | 128 | 126 |
| /26 | 192 | 64 | 62 |
| /27 | 224 | 32 | 30 |
| /28 | 240 | 16 | 14 |
| /29 | 248 | 8 | 6 |
| /30 | 252 | 4 | 2 |
Worked example — network address of a host (the /20 case)
Given 172.16.45.10/20, find the network (subnet) address.
- /20 =
255.255.240.0. The interesting octet is the third (the mask is 240 there). - Block size = 256 − 240 = 16. So third-octet subnets are 0, 16, 32, 48, 64…
- The host's third octet is 45. Round down to the nearest multiple of 16: 32 (because 32 ≤ 45 < 48).
- Zero the host bits (third-octet remainder and the whole fourth octet).
- Network address =
172.16.32.0. (Broadcast would be172.16.47.255, the address just below the next subnet172.16.48.0.)
Worked example — usable hosts on a /26
How many usable host addresses on a /26?
- H = 32 − 26 = 6 host bits.
- Usable = 2^6 − 2 = 64 − 2 = 62.
Worked example — broadcast address of a /27
Find the broadcast address of the subnet containing 192.168.1.100/27.
- /27 → last-octet mask 224 → block size 256 − 224 = 32. Subnets: 0, 32, 64, 96, 128…
- 100 rounds down to 96 → subnet
192.168.1.96. - Next subnet is
192.168.1.128; the broadcast is one below it. - Broadcast =
192.168.1.127. Usable range is.97–.126.
Worked example — the /28 subnet and range
For 192.168.10.75/28, find the subnet address and valid host range.
- /28 → last-octet mask 240 → block size 16. Subnets: 0, 16, 32, 48, 64, 80…
- 75 rounds down to 64 → subnet
192.168.10.64. - Broadcast is one below the next subnet (
.80) →192.168.10.79. - Usable hosts:
192.168.10.65through192.168.10.78(14 addresses = 2^4 − 2).
Worked example — the /30 WAN link
A point-to-point WAN link uses a /30. How many usable hosts?
- H = 32 − 30 = 2 → 2^2 − 2 = 2 usable hosts. Exactly right for a two-router link, which is why /30 is the classic WAN mask (a /31 is the modern two-host alternative).
Worked example — "how many subnets" and "borrow how many bits"
How many /26 subnets fit in a /24? Going from /24 to /26 borrows 26 − 24 = 2 host bits → 2^2 = 4 subnets.
A /24 must be divided into at least 6 subnets — minimum bits to borrow? You need 2^b ≥ 6. 2^2 = 4 (too few); 2^3 = 8 ≥ 6. Borrow 3 bits (giving eight /27 subnets, two spare).
Worked example — "are these two hosts in the same subnet?"
192.168.1.62/26 and 192.168.1.65/26 — same subnet?
- /26 block size = 64. Subnets: 0, 64, 128, 192.
.62rounds down to 0 → subnet192.168.1.0(range .1–.62, broadcast .63)..65rounds down to 64 → subnet192.168.1.64(range .65–.126).- Different subnets. No — and note
.62is the last usable host of the first subnet while.63is its broadcast, a favorite off-by-one trap.
The rule: IPv6 essentials
IPv6 addresses are 128 bits, written as eight groups of four hex digits, with two shortening rules: drop leading zeros in a group, and replace one run of all-zero groups with :: (only once per address). Example: 2001:0db8:0000:0000:0000:0000:0000:0001 → 2001:db8::1.
Key address types (know the prefix):
- Global unicast —
2000::/3— the routable, Internet-facing address (like a public IPv4). - Link-local —
FE80::/10— auto-generated on every IPv6 interface, valid only on the local link, never routed. Used for neighbor discovery and as the next-hop for routing. - Unique local —
FC00::/7(commonlyFD00::/8) — private, site-internal (like RFC 1918). - Multicast —
FF00::/8— IPv6 has no broadcast; it uses multicast instead (e.g.,FF02::1all-nodes,FF02::2all-routers).
So "automatically configured on every interface, valid only on a single link, never routed" is the link-local address, FE80::/10.
Worked example — classify IPv6 addresses by prefix
2001:db8:acad::10→ starts2000::/3→ global unicast (routable).FE80::1→ link-local (on-link only, auto-generated).FD00:1234::99→ inFC00::/7(theFD00::/8half) → unique local (private).FF02::1→FF00::/8→ multicast (this one = all-nodes).::1→ loopback;::→ unspecified.
Read the leading hextet and the type falls out: 2 or 3 = global, FE80 = link-local, FC/FD = unique local, FF = multicast.
IPv6 hosts can build an address automatically with SLAAC (Stateless Address Autoconfiguration), learning the /64 prefix from a router's RA (Router Advertisement) and generating the interface ID themselves — often via EUI-64, which inserts FFFE into the middle of the MAC and flips the 7th bit. NDP (Neighbor Discovery Protocol) replaces IPv4's ARP.
Qué incluye el eBook
¿Por qué comprar el libro si la práctica es gratis?
Nuestras preguntas de práctica y el simulacro cronometrado siguen gratis: nada del sitio se esconde tras este libro. El libro de $14.99 es la mitad de estudio: el material en sí, explicado en orden, en un archivo tuyo.
- Enseñanza sistemática — cada sección del examen explicada capítulo a capítulo, de principio a fin, no solo preguntas
- Imprímelo y márcalo — un PDF listo para papel que puedes resaltar, anotar y llevar a tu mesa de estudio
- Estudia en cualquier lugar, sin conexión — EPUB en tu teléfono o e-reader; sin wifi, sin pestañas
- Todo en un solo lugar — capítulos, resúmenes por capítulo y preguntas de práctica en un archivo
- Tuyo de por vida — pago único de $14.99, descarga instantánea, sin suscripción
Y sin riesgo: Garantía de reembolso de 14 días — ¿no te convence? Escríbenos para un reembolso total, sin preguntas. Consulta la política de reembolso.
Garantía de reembolso de 14 días · reembolso total, sin preguntas.
Compra única, acceso de por vida a la descarga. El eBook es la guía completa de Cisco CCNA 200-301 en PDF y EPUB. Resumen educativo, no asesoría profesional ni legal — confirma siempre las reglas vigentes con la fuente oficial. Última actualización: August 2026.