Recall warm-up
- Minimum Ethernet frame size?
Stuck? Peek at 2.4.1 Ethernet frame format, MAC address types, minimum and maximum frame sizes
- Frame: preamble 7 + SFD 1 | dest 6 | src 6 | type/len 2 | data 46–1500 | CRC 4.
- Frame 64–1518 bytes (preamble excluded); pad data to 46.
- MTU = 1500 bytes.
- First-byte LSB 1 → multicast; FF:…:FF → broadcast; else unicast.
- 64 bytes = 512 bits = CSMA/CD slot time at 10 Mbps.
- A at range 2 and B at range 4: P(A wins)?
Stuck? Peek at 2.3.4 Binary exponential backoff; probability of winning
- After the n-th collision: K uniform in {0, …, 2^min(n,10) − 1}; wait K × slot; abort after 16.
- Slot = 512 bit times (51.2 µs at 10 Mbps).
- P(A wins) = #{K_A < K_B} / (a·b); both at range 2: win 1/4, tie 1/2.
- Capture: A at range 2 vs B at range 4 → A wins 5/8.
- Send-and-wait utilisation in terms of a?
Stuck? Peek at 1.3.3 RTT, bandwidth-delay product, utilisation and a = Tp/Tt
- RTT = 2·Tp for tiny signals (add Tt / queuing if stated).
- BDP = B × Tp bits in flight; B × RTT before any reply.
- a = Tp / Tt.
- Send-and-wait: U = Tt/(Tt + 2Tp) = 1/(1 + 2a); throughput = U·B.
- To keep the link busy: about 1 + 2a units in flight.
The Hook
Classic Ethernet had every computer clamped onto one coaxial cable — one big classroom with CSMA/CD as the speaking rule. Today each computer has its own cable to a switch, and collisions have vanished. GATE asks both: the efficiency formula for the shared case, and how switches change collision and broadcast domains.
Predict First
On a shared Ethernet, does efficiency get better or worse when the cable gets longer, with the same frame size and rate?
,Worse. A longer cable means a larger Tp, so a = Tp/Tt grows: collisions take longer to notice and more time is lost in each contention period.
Core Concept
Efficiency of shared (CSMA/CD) Ethernet
Efficiency = fraction of time the channel carries successful frames when many stations are busy. The standard approximation:
As a → 0 (short cable or long frames), η → 1; as a grows, η falls.
Textbook variants
| Source | Formula | η at a = 0.1 |
|---|---|---|
| Kurose (course default) | 1/(1 + 5a) | 0.667 |
| Tanenbaum (2e) | 1/(1 + 5.44a) | 0.648 |
| Forouzan | 1/(1 + 6.44a) | 0.608 |
Shared vs switched Ethernet
- Shared (bus or hub): one medium, CSMA/CD needed, everyone in one collision domain, half-duplex.
- Switched: each host has a dedicated full-duplex link to a switch interface; the switch stores and forwards frames (Lesson 1.2.2) only toward the right interface. No collisions, so CSMA/CD is not needed, and each link gets its full rate.
Collision domains and broadcast domains
| Device | Collision domains | Broadcast domains |
|---|---|---|
| Hub / repeater | does not split (1 for all interfaces) | does not split |
| Switch / bridge | one per interface | does not split (forwards broadcasts) |
| Router | one per interface | one per interface (stops broadcasts) |
Putting it together
- Shared Ethernet efficiency falls with a: η ≈ 1/(1 + 5a).
- Switches give each interface its own collision-free, full-duplex link.
- Switches split collision domains; routers split broadcast domains.
Visual
Worked Examples
Shared Ethernet: 10 Mbps, 1000-byte frames, longest path 2500 m, v = 2 × 10⁸ m/s. Efficiency with 1/(1 + 5a)?
Show the full solution
Tt = 8000 / 10⁷ = 0.8 ms. Tp = 2500 / 2×10⁸ = 12.5 µs. a = 0.015625. η = 1/(1 + 0.078125) = 0.9275.
For a = 0.1, compute the efficiency with each textbook form.
Show the full solution
1/(1 + 0.5) = 0.667; 1/(1 + 0.544) = 0.648; 1/(1 + 0.644) = 0.608.
A router connects to a switch. The switch has two hubs attached (3 hosts on each) and two hosts attached directly. Count collision domains and broadcast domains among these devices (router's other side ignored).
Show the full solution
Collision domains: hub 1 group, hub 2 group, each directly attached host's link (2), and the switch–router link (1) → 1 + 1 + 2 + 1 = 5. Broadcast domains: the switch forwards broadcasts, so everything behind the router's one interface is 1 broadcast domain.
GATE Traps
- Switched full-duplex Ethernet has no collisions — CSMA/CD efficiency formulas do not apply to it.
- Switches do not split broadcast domains; routers do.
- A hub with many hosts is still one collision domain.
Memory Hook
"Hubs share the floor, switches give everyone a private lane, routers build walls against shouting."
Connect the Dots
Phase 2 is complete: our frame can cross one link reliably and fairly. The packet inside still needs a route across many networks. Next (Phase 3): the network layer — what does the address that stays fixed from laptop to server look like?
Self-Test
- NAT
Shared Ethernet with a = 0.2. Efficiency using 1/(1 + 5a)?
Write your answer before opening the key.
- MCQ
Which device splits broadcast domains?
- Hub
- Repeater
- Switch
- Router
- NAT
A switch has 8 interfaces, each connected to one host. How many collision domains?
Write your answer before opening the key.
- MSQ
Which are true of switched full-duplex Ethernet?
- No collisions occur.
- CSMA/CD is not needed.
- All interfaces are one collision domain.
- Broadcast frames still reach every interface.
- NAT
Shared Ethernet: T_t = 51.2 µs, T_p = 5.12 µs. Efficiency with 1/(1 + 5a) (3 decimals)?
Write your answer before opening the key.
Rule Card
- Shared Ethernet efficiency η = 1/(1 + 5a) (course default); variants 1/(1 + 5.44a), 1/(1 + 6.44a).
- Switched full-duplex: no collisions, no CSMA/CD, full rate per link.
- Collision domains: hub = 1; switch/router = 1 per interface.
- Broadcast domains: only routers split them.
Concept Ledger
- Ethernet efficiency 1/(1 + 5a) and variants
- shared vs switched Ethernet
- collision domain
- broadcast domain
Answer Vault
Try the questions first. Open when you are ready to check.
Warm-up answers
- 64 bytes.
- 5/8.
- 1/(1 + 2a).
Self-test answers
- 0.5 — 1/(1 + 1) = 0.5.
- D — Routers do not forward link-layer broadcasts.
- A: Hubs repeat everything.
- B: Repeaters repeat everything.
- C: Switches forward broadcasts to all interfaces.
- 8 — One per switch interface.
- ABD — Dedicated full-duplex links; broadcasts are flooded.
- C: Each interface is its own collision domain.
- 0.666..0.667 — a = 0.1 → 1/1.5 = 0.667.
Flashcards
- Shared Ethernet efficiency (default)?
Reveal answer
1/(1 + 5a). - Which device splits broadcast domains?
Reveal answer
A router. - Collision domains on an 8-interface switch?
Reveal answer
8.
Exam shortcuts
- η = 1/(1 + 5a) for shared Ethernet. — CSMA/CD with many active stations; course default unless the question states another constant.
- Counting domains: each switch/router interface = one collision domain (a hub and everything on it = one); each router interface = one broadcast domain. — hubs, switches and routers only (no VLANs).