Worked examples — OSI model — 7 layers, responsibilities, PDU names

Before anything else, one word we lean on constantly:

Look at the picture above: the same yellow payload sits at the centre of every box. Each layer only adds a chalk-blue header (and Layer 2 adds a pink trailer on the right). Nothing is ever removed on the way down — that is the single fact that makes all the "size" examples below work.

The scenario matrix

Every OSI exam question is really one of these cells. Our 8 examples below hit all of them.

Example 1 — Cell A: name → number

1. Identify the job words. "Encryption", "translation", "compression" are the three signature jobs of one specific layer. Why this step? OSI questions are pattern-matching: each verb belongs to exactly one home layer. Learn the verb→layer map and lookups become instant.
2. Map to the layer. Those three verbs are the Presentation layer's responsibilities. Why this step? The Application layer (7) is only the door for user programs; the translating/scrambling happens one floor below.
3. Convert name to number. Counting from the top (7 = Application, 6 = Presentation), Presentation is Layer 6. Why this step? The question asked for a number, so we finish the translation.

Answer: Layer 6 (Presentation).

Example 2 — Cell B: number → PDU name

1. Recall the PDU chain (top→bottom): Data → Data → Data → Segment → Packet → Frame → Bits. Why this step? Layers 7,6,5 all still call it plain "Data"; the name only starts changing at Layer 4. You must know where the first rename happens.
2. Pin Layer 4's output. After Transport adds its header (a TCP header, per TCP vs UDP), the unit is a Segment (or Datagram for UDP). Why this step? The Transport header is what earns the new name — no header, no rename.
3. Name the layer above (Layer 5, Session). At Layer 5 it was still just Data. Why this step? Confirms the boundary: "Data" ends and "Segment" begins exactly at the 5→4 handoff.

Answer: at Layer 4 it is a Segment (UDP: Datagram); at Layer 5 it was still Data.

Example 3 — Cell C: device → layer

1. Hub → Layer 1. A hub only repeats electrical signals to all ports; it reads no addresses at all. Why this step? If a device never inspects any header, it lives at the Physical layer where there are no addresses — just bits.
2. Switch → Layer 2. A switch reads the MAC address in the frame header to forward to the correct port. See MAC Address and ARP. Why this step? Reading MAC = reading the Layer-2 (Data Link) header, so a switch's ceiling is Layer 2.
3. Router → Layer 3. A router reads the IP address to choose a path across networks. See IP Addressing and Subnetting. Why this step? IP lives in the Network (Layer 3) header; understanding it makes the router a Layer-3 device.

Answer: Hub (L1) < Switch (L2) < Router (L3). Full comparison in Routers vs Switches vs Hubs.

Example 4 — Cells D & E: encapsulation down, de-encapsulation up

1. Sender goes DOWN (encapsulation = wrapping). Data (L7-5) → +Transport header → Segment (L4) → +IP header → Packet (L3) → +MAC header/trailer → Frame (L2) → serialized to Bits (L1). Why this step? Going down, each layer prepends its own header, so the unit only grows and gets a new name. This is the mechanism the size examples depend on.
2. Signal crosses the wire as raw bits. Why this step? Layer 1 is the only place actual physics (voltage/light/radio) happens; every layer above is bookkeeping.
3. Receiver goes UP (de-encapsulation = unwrapping). Bits (L1) → assemble → Frame (L2) → strip MAC → Packet (L3) → strip IP → Segment (L4) → strip Transport → Data (L5-7). Why this step? Each layer removes exactly the header its peer added. Symmetry is the whole point of a layered model — a layer only ever talks to the same layer on the other machine.

Answer: Down: Data→Segment→Packet→Frame→Bits. Up: Bits→Frame→Packet→Segment→Data. Perfect mirror.

Example 5 — Cells F & G: overhead arithmetic + zero-byte payload

1. Add every header and trailer to the payload. Encapsulation is purely additive (nothing is removed going down), so: $\text{Frame}=D+h_{\text{TCP}}+h_{\text{IP}}+h_{\text{Eth,hdr}}+h_{\text{Eth,trl}}$ Why this step? From Ex 4, each layer only adds. So total size is just a sum — no subtraction anywhere.
2. Plug in for (a). $\text{Frame}=500+20+20+14+4=558\text{ B}$ Why this step? Direct substitution; this is the number that actually travels.
3. Overhead for (b). Overhead bytes $=558-500=58$. As a fraction of the whole frame: $\frac{58}{558}\times 100\approx 10.39\%$ Why this step? "Overhead" means the bytes you're forced to send that aren't your actual message — a real cost, explained in the parent note.
4. Degenerate case (c): $D=0$. The payload vanishes but the headers/trailer do not: $\text{Frame}=0+20+20+14+4=58\text{ B}$ Why this step? This proves headers exist independently of data — a zero-byte frame (like a bare TCP ACK) is still $58$ B of pure overhead ($100\%$).

Answers: (a) $558$ B, (b) $\approx 10.39\%$, (c) $58$ B (all overhead).

Example 6 — Cell H: limiting case (MTU exceeded → fragmentation)

1. Define the ceiling. MTU is the biggest chunk one frame is allowed to carry. Here $\text{MTU}=1500$ B. Why this step? At the limit, the additive model of Ex 5 hits a wall — you cannot just keep growing one frame forever; the medium caps it.
2. Compare demand to ceiling. We need to move $4000$ B but each packet holds at most $1500$ B. Since $4000>1500$, a single packet is impossible. Why this step? This is the "limiting case" — the exact moment the clean one-frame story breaks and the Network layer must act.
3. Fragment. Layer 3 splits the data into ceiling-sized pieces: $\text{fragments}=\lceil \frac{4000}{1500}\rceil =\lceil 2.67\rceil =3\text{ packets}$ Why this step? The ceiling function $\lceil \rceil$ rounds up because a leftover $1000$ B still needs its own whole packet — you can't send two-thirds of a frame.
4. Check the split. $1500+1500+1000=4000$. ✓ The third packet is a partial $1000$ B. Why this step? Confirms no data is lost and the last fragment absorbs the remainder.

Answer: 3 packets (1500 + 1500 + 1000 B). The receiver's Layer 3 reassembles them.

Example 7 — Cell I: real-world word problem

1. (i) HTTP request → Layer 7 (Application). The browser is a user program; HTTP is its "door" into the network. Why this step? Application is where human-facing protocols live — the request is born here.
2. (ii) TLS encryption → Layer 6 (Presentation). Classic OSI puts encryption/translation at Presentation. Why this step? Encryption is a format transformation, the Presentation layer's signature job (Ex 1).
3. (iii) Port 443 → Layer 4 (Transport). Ports steer data to the right process on the machine. See Ports and Sockets. Why this step? IP finds the machine; the port finds the program (443 = the web server), which is Transport's job.
4. (iv) Destination IP → Layer 3 (Network). IP is the logical address for routing across networks. Why this step? To leave your LAN and cross the Internet you need a Layer-3 address a router understands.
5. (v) Destination MAC → Layer 2 (Data Link). MAC delivers to the correct device on the local link. Why this step? The very next hop is a physical neighbour identified by MAC, not IP — hence the lowest addressing layer.

Answer: (i) L7, (ii) L6, (iii) L4, (iv) L3, (v) L2 — a clean top-to-bottom descent.

Example 8 — Cell J: the exam trap

1. Spot why it feels true. "Transport" sounds like "moving things over distance", so learners assign it the long-haul routing job. Why this step? Naming the trap out loud is how you disarm it — the exam is testing this exact confusion.
2. Recall Transport's real job. Transport is end-to-end: ports, reliability (TCP), flow/error control. It does not know the path. Why this step? Transport treats the network as an abstract pipe — "give this to app X on machine Y, reliably" — and delegates how to lower layers.
3. Assign routing correctly. Choosing a path across networks is Layer 3 (Network) — the router's layer. Why this step? Routing = IP addresses + forwarding decisions, which are Layer-3 concepts (Ex 3).

Answer: FALSE. Routing (path selection) is Layer 3 (Network). Transport (Layer 4) does end-to-end delivery to the right port/process, reliability, and flow control — never path choice.

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