Ethernet frame format, MAC address types, minimum and maximum frame sizes

1. ETR token ring efficiency formula?
   Stuck? Peek at 2.3.5 Controlled access: polling, token passing and its efficiency

   • Polling: primary asks each station; η = T_t/(T_t + T_poll); single point of failure.
   • Token passing: only the token holder sends (token holding time).
   • Ring latency = d/v + N·b/R.
   • ETR η = 1/(1 + a/N); DTR η = 1/(1 + a(N+1)/N).

   Read lesson 2.3.5

2. CSMA/CD minimum frame size formula?
   Stuck? Peek at 2.3.3 CSMA and its persistence variants; CSMA/CD; minimum frame size = 2·Tp·B

   • CSMA: sense before sending; collisions still possible within T_p.
   • 1-persistent / non-persistent / p-persistent: what to do when busy.
   • CSMA/CD: listen while sending; on collision jam, stop, back off.
   • L_min = 2·T_p·B; d_max = L_min·v/(2B); i.e. T_t ≥ 2T_p.
   • CSMA/CA (Foundation): wireless avoids collisions instead of detecting them.

   Read lesson 2.3.3

3. What does the CRC in a frame catch for sure?
   Stuck? Peek at 2.2.4 CRC: generator polynomial, modulo-2 division, what CRC catches

   • G of degree r (r + 1 bits); append r zeros, divide modulo 2, send data + r-bit remainder.
   • Receiver divides the whole frame: remainder 0 → accept.
   • Undetected ⇔ error pattern E is a multiple of G.
   • Catches: all single-bit errors (≥ 2 terms), all odd counts (x + 1 | G), all bursts ≤ r; (r+1)-bursts missed w.p. 2⁻⁽ʳ⁻¹⁾, longer 2⁻ʳ.

   Read lesson 2.2.4

Your laptop's wired network jack speaks Ethernet (IEEE 802.3), the dominant wired LAN technology. Every Ethernet frame has the same fields in the same order, and two numbers — 64 and 1518 bytes — appear in almost every GATE question about it. Both numbers follow directly from what you already know: collision detection and buffer sizes.

A program wants to send only 20 bytes of data in one Ethernet frame. Will the frame be about 38 bytes long?

No — 64 bytes. Ethernet requires at least 46 bytes of data so that the frame lasts long enough for collision detection; the sender adds 26 bytes of padding. Header (14) + data with padding (46) + CRC (4) = 64.

The frame, field by field

Ethernet frame (sizes in bytes):

What each field does

Minimum 64 bytes and maximum 1518 bytes

• Minimum frame = 64 bytes (6 + 6 + 2 + 46 + 4), i.e. 512 bits = the CSMA/CD slot time of Lessons 2.3.3–2.3.4 at 10 Mbps. Shorter data is padded to 46 bytes.
• Maximum frame = 1518 bytes (14 + 1500 + 4): the 1500-byte data limit keeps buffers small and stops one station from hogging the medium.
• The largest payload a link can carry is its MTU (maximum transmission unit): 1500 bytes for Ethernet. Phase 3 shows what happens when a packet is bigger than the MTU of the next link.

Unicast, multicast, broadcast addresses

• Unicast: one interface. The least significant bit of the first byte is 0 (the first byte is even).
• Multicast: a group of interfaces. That bit is 1 (the first byte is odd), e.g. 01:00:5E:00:00:01.
• Broadcast: every interface on the LAN: FF:FF:FF:FF:FF:FF (all 48 bits 1).

Putting it together

• An Ethernet frame is a 14-byte header, 46–1500 bytes of data and a 4-byte CRC.
• 64 bytes minimum comes from collision detection; 1500-byte MTU caps the payload.
• First-byte LSB tells unicast from multicast; all-1s is broadcast.

• The 64/1518-byte limits exclude the 8 bytes of preamble and SFD.
• Minimum data is 46 bytes, minimum frame is 64 bytes — do not mix them.
• Multicast is decided by the lowest bit of the first byte, not the last byte.
• MTU 1500 is the payload limit, not the frame limit (1518).

"14 + 46 + 4 = 64; 14 + 1500 + 4 = 1518" — header, data, CRC.

We know what an Ethernet frame looks like. Next (2.4.2): how efficient is classic shared Ethernet under CSMA/CD, and why do modern switched networks no longer collide at all?

1. NAT

   An Ethernet frame carries 38 data bytes. How many padding bytes are added?

   Write your answer before opening the key.

2. NAT

   Size in bytes of an Ethernet frame carrying 1000 data bytes (preamble excluded)?

   Write your answer before opening the key.

3. MCQ

   Which MAC address is multicast?

   1. 00:1B:44:11:3A:B7
   2. 01:00:5E:7F:00:01
   3. 3C:22:FB:0A:91:7E
   4. A8:5E:45:00:00:01
4. MSQ

   Which are true for Ethernet?

   1. Minimum frame 64 bytes
   2. MTU 1500 bytes
   3. CRC is 16 bits
   4. Preamble is counted in the 1518-byte maximum
5. NAT

   Bits on the wire for a minimum-size frame including preamble and SFD?

   Write your answer before opening the key.

• 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.

• Ethernet (IEEE 802.3)
• preamble, SFD
• destination / source address, type/length field
• padding
• FCS (32-bit CRC)
• minimum 64 / maximum 1518 bytes
• MTU (1500 for Ethernet)
• unicast / multicast / broadcast MAC addresses

1. Ethernet min and max frame?
   Reveal answer

   64 and 1518 bytes (preamble excluded).
2. Ethernet MTU?
   Reveal answer

   1500 bytes.
3. How to spot a multicast MAC?
   Reveal answer

   Least significant bit of the first byte is 1 (first byte odd).
4. Padding rule?
   Reveal answer

   Data shorter than 46 bytes is padded to 46.

• Frame bytes = 18 + max(data, 46). — standard Ethernet frame without VLAN tag; preamble/SFD not counted.
• Address type: first byte odd → multicast/broadcast; even → unicast. — MAC addresses written in the usual byte order.