Chapter recap
Ethernet frame: preamble 7 + SFD 1 (not counted) | dest 6 | src 6 | type/len 2 | data 46–1500 | CRC 4 → 64–1518 bytes, MTU 1500, pad to 46. First-byte LSB 1 → multicast; all 1s → broadcast. Shared Ethernet η ≈ 1/(1 + 5a); switched full-duplex has no collisions. Hubs: one collision domain; switches: one per interface; routers split broadcast domains.
Merged rule card
From 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.
From 2.4.2 Ethernet efficiency; switched vs shared Ethernet
- 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.
Practice questions
- NAT · 1 mark
Frame size (bytes) for 10 data bytes?
Write your answer before opening the key.
- MCQ · 1 mark
The Ethernet MTU is:
- 1518 bytes
- 1500 bytes
- 64 bytes
- 46 bytes
- NAT · 2 marks
Efficiency (1/(1 + 5a)) for 100 Mbps, 1250-byte frames, 1 km, v = 2 × 10⁸ m/s (3 decimals)?
Write your answer before opening the key.
- MSQ · 2 marks
Which MAC addresses are group (multicast or broadcast) addresses?
- FF:FF:FF:FF:FF:FF
- 33:33:00:00:00:01
- 02:42:AC:11:00:02
- 01:80:C2:00:00:00
- NAT · 1 mark
Three hubs (4 hosts each) are attached to one switch. Collision domains?
Write your answer before opening the key.
- MCQ · 1 mark
Why is the minimum Ethernet frame 64 bytes?
- To fit the CRC
- So a sender is still transmitting when a collision report returns
- Because MAC addresses are 6 bytes
- To align with the MTU
- NAT · 1 mark
With 1/(1 + 6.44a) at a = 0.25, efficiency (3 decimals)?
Write your answer before opening the key.
- MCQ · 1 mark
Which statement about a switch is FALSE?
- Each interface is a separate collision domain.
- It forwards broadcast frames to all interfaces.
- It splits broadcast domains.
- It allows full-duplex links.
Answer key
Try the questions first. Open when you are ready to check.
- 64 — 18 + 46 = 64.
- B — Payload limit 1500.
- A: Maximum frame.
- C: Minimum frame.
- D: Minimum data.
- 0.799..0.801 — T_t = 100 µs, T_p = 5 µs, a = 0.05, η = 1/1.25 = 0.8.
- ABD — First bytes FF, 33, 01 are odd.
- C: 0x02 is even → unicast.
- 3 — Each hub with its hosts is one collision domain.
- B — 512 bits ≥ 2·T_p·B for classic 10 Mbps networks.
- A: CRC is only 4 bytes.
- C: Unrelated.
- D: MTU is the maximum.
- 0.383..0.384 — 1/2.61 = 0.383.
- C — Only routers split broadcast domains.
- A: True.
- B: True.
- D: True.