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Lesson 1/73 — OSI model, encapsulation, PDU names, TCP/IP stack

Why it matters

Every network fault maps to a specific layer. If the symptom is "web page times out" the root cause could be a dying physical cable (layer 1), a bad ARP entry (layer 2), a missing route (layer 3), or a blocked TCP port (layer 4). Pinpointing the layer before troubleshooting saves hours of random checks.

Each layer's PDU & device

Common pitfall

Symptom: Browser reports "Connection refused", yet ping 10.1.1.1 succeeds.
Root cause: ping uses ICMP at layer 3 — it reaches the host. "Connection refused" is a TCP RST at layer 4 — the application port is not listening. The OS itself is alive; the service died.
Fix: Test the port: nc -zv 10.1.1.1 443. If closed, start the web server or check process.

1. Question 1. A router forwards a packet from one subnet to another. At which OSI layer does the router operate for this decision?
   Reveal answer 1

   Layer 3 (Network). The router reads the destination IP address to choose the outgoing interface — a network-layer decision. Switches use MAC addresses (layer 2).

2. Question 2. SSL/TLS encryption happens at which layer of the TCP/IP model? (Hint: it is not a transport-layer function.)
   Reveal answer 2

   Application layer — more precisely the Presentation layer (OSI 6), which is bundled into the TCP/IP Application layer. Transport security operates above TCP port 443 but the encryption is a session/presentation function from the OSI perspective.

3. Question 3. A host sends an ARP request. What is the PDU type (frame, packet, segment, bit) and at which layer does ARP operate?
   Reveal answer 3

   ARP is carried inside an Ethernet frame, making it a frame-level PDU. It operates at layer 2 (Data Link) — it resolves IP (L3) to MAC (L2), so it straddles layers but is encapsulated as a frame and never leaves the broadcast domain.

Lesson 2/73 — IPv4 addressing: classes, private/public, broadcast vs multicast

Why it matters

An IPv4 address is not just a number — it carries a network prefix and a host suffix. The boundary between the two, set by the subnet mask, controls:

• How many subnets a business can carve from its address block
• How many hosts fit inside each subnet
• Whether a destination is local (same subnet) or needs a router

Poor planning wastes addresses (too large a mask = unused space) or forces renumbering later (too small = run out of host IPs).

The 32-bit structure

Every IPv4 address is a 32‑bit integer displayed in dotted‑decimal:

192.168.1.10 → 11000000.10101000.00000001.00001010
 <--- network portion (mask) ---> <------- host portion ------->

Legacy classes

Classful addressing is obsolete — CIDR replaced it in 1993 — but the ranges still appear in route‑origin policies and legacy discussions.

Note: 127.0.0.0/8 is loopback — packets to any 127.x.x.x never leave the host.

Private vs. public (RFC 1918)

Three blocks cannot be routed on the public Internet. Border routers drop traffic with a private source IP unless NAT is active:

Other reserved: 169.254.0.0/16 (APIPA — Windows self‑assigns when DHCP fails). 100.64.0.0/10 (Carrier‑Grade NAT, RFC 6598).

Broadcast & the mathematics

Every subnet has a broadcast address: all host bits set to 1. Sending to it reaches every host on that subnet.

Example: 10.10.0.0/16

Network:    00001010.00001010.00000000.00000000  = 10.10.0.0
Mask:       11111111.11111111.00000000.00000000  (16 bits fixed)
Broadcast:  00001010.00001010.11111111.11111111  = 10.10.255.255 (flip host bits)
--------------------
Usable hosts: 2^(32−16) − 2 = 65,536 − 2 = 65,534

Two types:
Limited broadcast (255.255.255.255) — never forwarded by a router.
Directed broadcast (10.10.255.255) — can be forwarded unless no ip directed‑broadcast is set, which is the default on modern IOS.

Multicast

Class D addresses (224.0.0.0 – 239.255.255.255) send to a group, not a single host or all hosts. Routers use PIM to build distribution trees; hosts join groups with IGMP.

No router forwards multicast by default — ip multicast‑routing must be enabled.

Common pitfall

Symptom: Two hosts on the same switch cannot ping.
Check: One has 169.254.1.25 (APIPA — no DHCP), the other is 192.168.1.10. Different subnets, no router between them.
Fix: Static IP on same subnet, or restore DHCP server reachability.

IOS verification commands

Router# show ip interface brief
Interface     IP-Address          OK? Method Status Protocol
Gig0/0        192.168.1.1/24      YES manual up     up

Router# show ip interface gig0/0
  Broadcast address is 255.255.255.255
  Directed broadcast forwarding is disabled

1. Question 1. What is the broadcast address for 172.20.0.0/19? Show the binary.
   Reveal answer 1

   /19 mask = 19 bits fixed → 13 host bits
   172.20.0.0  = 10101100.00010100.00000000.00000000
   broadcast   = 10101100.00010100.00011111.11111111  (flip all host bits)
               = 172.20.31.255

2. Question 2. A host 10.0.5.25/8 pings 10.255.255.255. Is the destination a directed broadcast or a unicast address from this host's perspective?
   Reveal answer 2

   Directed broadcast for the 10.0.0.0/8 network. The host sees 10.255.255.255 — all host bits are 1. The packet will reach all hosts on the 10.0.0.0/8 subnet if the intermediate routers permit directed‑broadcast forwarding (disabled by default).

3. Question 3. Multicast group 239.1.1.1 — is it routable on the public Internet? What protocol pair makes it work inside an enterprise?
   Reveal answer 3

   No. 239.x.y.z is in the Administratively Scoped range (239.0.0.0/8) — it is never routed beyond an organization's boundary. Within the enterprise: IGMP (hosts tell the router they want the group) and PIM (routers build a distribution tree toward the source or the Rendezvous Point).

Lesson 3/73 — Subnetting: binary math, /24 slicing, usable host ranges

Why it matters

A /24 (255.255.255.0) gives 254 hosts. Most LANs need far fewer, so you subnet it to create multiple smaller networks. Every subnet consumes address space — the math must be exact or you run out of addresses or route table memory.

Subnet binary math

A /24 means 24 fixed network bits; the remaining 8 host bits can vary.

Mask /24 = 11111111.11111111.11111111.00000000

Split one /24 into four /26s (borrow 2 host bits):

Mask /26 = 11111111.11111111.11111111.11000000
                ^-- 2 borrowed bits = 2^(32-26) = 64 addresses per subnet
                2 subnets worth? Actually 4 subnets (192,193,194,195 high octet slices):
                192.168.0.0/26  (hosts .0 - .63,   bcast .63)
                192.168.0.64/26  (hosts .64 - .127,  bcast .127)
                192.168.0.128/26 (hosts .128 - .191, bcast .191)
                192.168.0.192/26 (hosts .192 - .255, bcast .255)

Usable hosts = 2^(host bits) − 2 (subtract network & broadcast). So /26 → 64 − 2 = 62 usable hosts per subnet.

The 5-step subnetting algorithm

1. Determine required hosts → use the smallest power of 2 block that fits (add ≥50% headroom for growth).
2. Compute host bits *h*: find smallest *h* where 2^h − 2 ≥ required.
3. Mask = 32 − h. First subnet's network = 0 in host octets.
4. Subnet size = 2^h. Increment each subsequent subnet by that size.
5. Broadcast = first IP + (size − 1). Usable range = network+1 … broadcast−1.

Worked example

Given network 10.0.0.0/24, create 7 subnets for 7 VLANs. Each VLAN hosts 20 devices.

Step 1: need 20 hosts. 2^h − 2 ≥ 20 → h = 5 (30 hosts). 7 subnets needed.
Step 2: 2^3 = 8 ≥ 7 → borrow 3 bits. Mask = 32−5 = /27.
Step 3: block size = 2^5 = 32.
        Subnet 1: 10.0.0.0/27   (hosts .1–.30,  bcast .31)
        Subnet 2: 10.0.0.32/27  (hosts .33–.62, bcast .63)
        Subnet 3: 10.0.0.64/27  (hosts .65–.94, bcast .95)
        Subnet 4: 10.0.0.96/27  (hosts .97–.126,bcast .127)
        Subnet 5: 10.0.1.0/27   (hosts .1–.30)
        Subnet 6: 10.0.1.32/27  (hosts .33–.62)
        Subnet 7: 10.0.1.64/27  (hosts .65–.94)

IOS verification

Switch# show ip interface brief
Switch# show running-config interface vlan 10
Switch# ping 10.0.0.1 source vlan 10

1. Question 1. Given 172.16.0.0/16, you must create subnets that each support at least 500 hosts. What is the new mask in dotted-decimal?
   Reveal answer 1

   500 hosts → h where 2^h − 2 ≥ 500 → h = 9 (510 hosts).
   Mask = 32 − 9 = /23
   /23 dotted = 255.255.254.0

2. Question 2. For subnet 192.168.10.0/27, what are the usable host range and the broadcast address?
   Reveal answer 2

   Mask /27 → block 32. Network = 192.168.10.0. Broadcast = 192.168.10.0 + 31 = 192.168.10.31. Usable range 192.168.10.1 – 192.168.10.30.

3. Question 3. A network admin needs 5 subnets from a /24. They choose a /29 mask. How many subnets does this actually create, and is the design valid for 5 subnets?
   Reveal answer 3

   /29 → borrow 5 bits → 2^5 = 32 subnets, each holding 6 hosts (2^3−2). Valid count (32 ≥ 5) but only 6 hosts per subnet — too small for most VLANs. The admin over-subnetted. A /28 would give 6 usable hosts; a /27 gives 14 hosts — check host needs first.

Lesson 4/73 — VLSM: variable-length subnet masks, real-world address plans

Why it matters

VLSM = allocate subnets of different sizes from the same block. Fixed subnetting (one mask) wastes addresses. VLSM is the backbone of an efficient RFC 1918 plan and of route summarization.

Core principle

Allocate largest subnet first, then carve the remainder. Order by decreasing block size — otherwise you create fragmentation holes your next subnet can't use.

Worked example

Block 10.0.0.0/24 must serve:

• LAN A — 100 hosts
• LAN B — 50 hosts
• LAN C — 20 hosts
• Point-to-point — 2 hosts (×4 links)
• Loopback pool — /32 (×3)

1) 100 hosts → /25 (block 128) — 10.0.0.0/25 → usable .1–.126, bcast .127
2) 50 hosts  → /26 (block 64)  — 10.0.0.128/26 → .129–.190
3) 20 hosts  → /27 (block 32)  — 10.0.0.192/27 → .193–.222
   Remainder from .224 is two /29s (block 8):
4) P2P (2 h) → /30 (block 4)  — 10.0.0.224/30 → .225–.226
   remaining .228–.255 = six /30s available for future P2Ps
   loopbacks: use /32s from the tail: 10.0.0.253 - 255

Summarization with VLSM

Route summarization collapses many subnets into one shorter prefix. Rule: the block must align on a power‑of‑two boundary matching the summary.

Summarize 10.0.0.0/25, 10.0.0.128/26, 10.0.0.192/27
 → 10.0.0.0/25 covers .0–.127 ❌ does NOT cover .128+
 → Use 10.0.0.0/24 = whole block. Summary = 10.0.0.0/24.

Common mistake: summarizing 172.16.1.0/24 + 172.16.2.0/24 as 172.16.0.0/23 — valid only if 172.16.0.0/24 doesn't exist or is also summarized.

IOS

Router# show ip route
R    10.0.0.0/24 is subnetted, 4 subnets
     10.0.0.0/25 is subnetted
     10.0.0.128/26 is subnetted
S    10.0.0.0/24 [1/0] via 192.168.1.1   ← static summary
Router# show ip protocol
Routing Protocol is "ospf 1"
   Area 0  Summary routes:
   10.0.0.0/24  (summary network)

1. Question 1. Block 192.168.10.0/24. Subnet A needs 120 hosts, B needs 60, C needs 28. Draw the VLSM plan with masks + ranges.
   Reveal answer 1

   A: 120 hosts → /25 block 128 → 192.168.10.0/25 (.1–.126, bcast .127)
   B: 60 hosts  → /26 block 64  → 192.168.10.128/26 (.129–.190, bcast .191)
   C: 28 hosts  → /27 block 32  → 192.168.10.192/27 (.193–.222, bcast .223)
   Remainder 192.168.10.224/27 is free reserve / /30s.

2. Question 2. Which single summary covers 10.4.8.0/24 through 10.4.15.0/24 with the smallest prefix length?
   Reveal answer 2

   Cover 8 subnets (/24) → need 3 borrowed bits → /21. Verify alignment: 10.4.8.0 is divisible by 8 in the third octet (8 = 0b1000) → aligned. Summary = 10.4.8.0/21 (covers .8.0–.15.255).

3. Question 3. An OSPF area has three /24s: 172.16.0.0/24, 172.16.1.0/24, 172.16.2.0/24. A fourth link needs 172.16.3.0/24. Why should you summarize the first three — and at which device?
   Reveal answer 3

   Summarizing reduces the OSPF LSDB and SPF domain: one inter-area (type‑3) LSA instead of three. Summarize at the ABR connecting this area to the backbone. The ABR advertises 172.16.0.0/22, hiding .0/.1/.2 specifics. The new .3 link is included in that same summary since 172.16.0.0/22 already spans .0–.3.