Exercises — OSI model — 7 layers, responsibilities, PDU names

Top → Bottom: Application, Presentation, Session, Transport, Network, Data Link, Physical. Mnemonic: All People Seem To Need Data Processing. (See 4.3.1 OSI model — 7 layers, responsibilities, PDU names for the full table.)

• Transport → Segment (TCP) / Datagram (UDP)
• Network → Packet
• Data Link → Frame
• Physical → Bit
• Application (also Presentation & Session) → Data

Chain bottom→top: Bits → Frames → Packets → Segments → Data.

• MAC address → Layer 2 (Data Link)
• IP address → Layer 3 (Network)
• Port number → Layer 4 (Transport)

Reveal-yourself drill: MAC address lives at layer ::: 2 (Data Link) IP address lives at layer ::: 3 (Network) Port number lives at layer ::: 4 (Transport)

• (a) HTTP → Layer 7 (Application)
• (b) Router → Layer 3 (Network) — it uses IP. See Routers vs Switches vs Hubs.
• (c) Switch → Layer 2 (Data Link) — it uses MAC.
• (d) TLS encryption → Layer 6 (Presentation) in classic OSI.
• (e) Cable/voltage → Layer 1 (Physical).

Port 443 is read at Layer 4 (Transport). Layer 3 delivers the packet to the correct machine (IP), but many programs run on one machine, so Layer 4 uses the port to hand data to the correct process — here, the HTTPS web service. See Ports and Sockets.

Layer 5 (Session) — it manages dialog control and checkpoints, so a broken transfer can resume instead of restarting.

Encapsulation is purely additive — every layer prepends (and L2 also appends a trailer) its own metadata; nothing is removed until the receiver de-encapsulates. So: $\text{Frame}=D+h_T+h_N+h_L=1000+20+20+18=1058\text{ B}$ Overhead bytes $=58$. As a fraction of the frame: $\frac{58}{1058}\times 100\approx 5.48\%$ Look at figure s01: each ring wraps the one inside it, so the outermost band (the frame) is always the biggest. This is why "overhead" exists in networking. See Encapsulation and De-encapsulation.

The frame budget is 1518 B. Subtract every wrapper: $D_{\max }=1518-18-40=1460\text{ B}$ So $D_{\max }===1460==$ bytes — the classic Ethernet MSS. Anything larger must be segmented at Layer 4 into multiple frames.

The name changes because each lower layer adds a new header, producing a new logical unit (Data → Segment → Packet → Frame → Bits). Layers 5–7 do not add a transport-style header that creates a new addressed unit — they hand the same "Data" blob downward — so it keeps the name Data until Transport wraps it into a Segment.

• Switch → up to Layer 2; reads the destination MAC to forward the frame on the local link.
• Router → up to Layer 3; reads the destination IP to choose the next network. It rewrites the L2 frame (new MAC header) for the next hop but keeps the L3 packet's IP addresses.
• PC-B (endpoint) → all the way to Layer 7; it de-encapsulates every wrapper and hands the raw Data to the application.

Figure s02: the switch peeks only into the green (L2) band; the router peeks into the blue (L3) band; only the end host opens the whole onion. See MAC Address and ARP and IP Addressing and Subnetting.

1. (ii) add TCP header → Segment
2. (i) add IP header → Packet
3. (iv) add MAC header/trailer → Frame
4. (iii) encode to bits → Bits

Order = ii → i → iv → iii. This is exactly the encapsulation chain $\text{Data}\stackrel{+\text{TCP}}{\to }\text{Segment}\stackrel{+\text{IP}}{\to }\text{Packet}\stackrel{+\text{MAC}}{\to }\text{Frame}\to \text{Bits}.$

• MAC addresses (L2) → rewritten at every hop. The old frame is stripped and a new frame is built with this hop's source MAC and the next hop's destination MAC.
• IP addresses (L3) → preserved end-to-end (original source IP, final destination IP) — this is why IP is called the "logical/end-to-end" address while MAC is "local/link" only.

OSI's Application + Presentation + Session collapse into the single TCP/IP Application layer. Rough mapping:

• OSI 7/6/5 → TCP/IP Application
• OSI 4 → TCP/IP Transport
• OSI 3 → TCP/IP Internet
• OSI 2/1 → TCP/IP Network Access (or split into Data Link + Physical in the 5-layer version).

OSI is the map; TCP/IP is the road actually built. See TCP-IP Model — 4 vs 5 layers.

No routing happens between them. There is no router, so no device does Layer-3 forwarding. Both hosts still build an L3 packet (IP header) internally, but because both are on the same network, delivery is purely a Layer 2 (MAC) affair over the cable. Layer 3's job — choosing a path across networks — is trivial/absent when there is only one network. This is the degenerate case where the Network layer exists on the endpoints but its routing function is never exercised.

$\text{Frame}=0+20+20+18=58\text{ B}$ Overhead as a fraction of the frame $=58/58===100\%==$. Every single byte on the wire is header — the frame is all wrapper, no gift. This shows why chatty protocols with tiny payloads are so inefficient: the fixed header cost dominates.

With fixed headers $H=h_T+h_N+h_L=58$: ${\lim }_{D\to \infty }\frac{H}{D+H}===0==.$ Why a limit here? We are asking "what happens in the best case of huge transfers?" The header cost is a fixed number of bytes, but the payload grows without bound, so the ratio of fixed overhead to total shrinks toward zero. Big transfers are efficient; the header tax becomes negligible. (Contrast with Q5.3, the other extreme, where the ratio hit 100%.)