Upon completion of this tutorial, you should be able to:
Ethernet is the dominant wired LAN technology, originally developed by Xerox PARC in the 1970s and later standardized by IEEE as 802.3. It has evolved from a shared coaxial cable bus (10BASE5) to switched, full-duplex, point-to-point links with speeds ranging from 10 Mbps to 400 Gbps. This tutorial covers the Ethernet frame format, the CSMA/CD protocol (now largely obsolete but historically important), full-duplex operation, and the various physical layer standards. Understanding Ethernet is crucial for network design, performance analysis, and troubleshooting.
Ethernet was invented by Robert Metcalfe and David Boggs at Xerox PARC in 1973. The first commercial version was DIX Ethernet (Digital, Intel, Xerox) in 1980, which evolved into IEEE 802.3. Key milestones:
Modern Ethernet uses switched, full-duplex links, eliminating collisions and making CSMA/CD obsolete (except in legacy shared-media networks).
Two main frame formats: DIX (Ethernet II) and IEEE 802.3. They differ in the interpretation of the field after the source MAC address.
| Preamble | SFD | Dest MAC | Src MAC | EtherType | Payload | FCS |
|---|---|---|---|---|---|---|
| 7 bytes | 1 byte | 6 bytes | 6 bytes | 2 bytes | 46–1500 bytes | 4 bytes |
Same fields except the EtherType field is replaced by a Length field (indicating the number of payload bytes). For higher-layer protocol identification, an LLC (Logical Link Control) header (802.2) is placed after the Length field, containing DSAP, SSAP, and Control fields.
In practice, most networks use Ethernet II because it is simpler and widely supported.
The minimum frame size (from Dest MAC to FCS) is 64 bytes (including FCS). This ensures CSMA/CD can detect collisions within the maximum network diameter. If payload is less than 46 bytes, padding is added.
Legacy Ethernet used CSMA/CD (Carrier Sense Multiple Access with Collision Detection) over shared coaxial or twisted-pair bus. Nodes sensed carrier, transmitted if idle, and monitored for collisions. If a collision was detected, they sent a jam signal and backed off using binary exponential backoff.
With the advent of switches, each link became point-to-point, and full-duplex operation became possible (transmit and receive simultaneously), eliminating collisions. Today, CSMA/CD is rarely used.
Ethernet specifies various physical layer variants:
| Standard | Speed | Media | Max Distance |
|---|---|---|---|
| 10BASE-T | 10 Mbps | Cat3/Cat5 UTP | 100 m |
| 100BASE-TX | 100 Mbps | Cat5 UTP | 100 m |
| 1000BASE-T | 1 Gbps | Cat5e/Cat6 UTP | 100 m |
| 10GBASE-T | 10 Gbps | Cat6a/Cat7 UTP | 100 m |
| 1000BASE-SX | 1 Gbps | Multimode fiber | 220–550 m |
| 1000BASE-LX | 1 Gbps | Singlemode fiber | 5 km |
| 10GBASE-SR | 10 Gbps | Multimode fiber | 300 m |
| 10GBASE-LR | 10 Gbps | Singlemode fiber | 10 km |
Ethernet efficiency is the ratio of useful payload to total frame size. For a maximum 1500-byte payload, with 18 bytes of header/trailer (excluding preamble and SFD), efficiency = 1500/(1500+18) = 1500/1518 ≈ 98.8%. For smaller frames, efficiency drops significantly.
Throughput = (payload bits / total frame bits) × link speed. For example, at 100 Mbps with 1500-byte frames, throughput ≈ 98.8 Mbps.
Industrial Ethernet (e.g., PROFINET, EtherNet/IP) adds determinism and robustness for factory automation. Automotive Ethernet (e.g., 100BASE-T1) uses single twisted pair for in-vehicle networking, reducing weight and cost.
Question 1: What is the EtherType field used for in an Ethernet II frame?
(b) To indicate the protocol type of the payload (e.g., IPv4, ARP). EtherType values ≥ 0x0600 identify the upper-layer protocol.
Question 2: What is the minimum Ethernet frame size (excluding preamble and SFD) in bytes?
(b) 64 bytes. The minimum frame size from destination MAC to FCS is 64 bytes (including FCS).
Question 3: Which Ethernet standard operates at 10 Gbps over twisted pair copper cabling?
(d) 10GBASE-T. It uses Cat6a or Cat7 cabling.
Question 4: The FCS field in an Ethernet frame contains:
(c) A CRC-32 checksum for error detection. The FCS is used to detect errors in the frame.
Question 5: In full-duplex Ethernet, collisions:
(b) Cannot occur because transmit and receive are on separate channels. Full-duplex links allow simultaneous transmission and reception, and switches eliminate collisions.
Question 6: What is the maximum payload size for a standard Ethernet frame (without VLAN tagging)?
(a) 1500 bytes. The standard Ethernet payload is up to 1500 bytes.
Question 7: Which of the following is true about IEEE 802.3 frame format compared to Ethernet II?
(b) IEEE 802.3 uses a Length field instead of EtherType. The Length field specifies the number of payload bytes.
Question 8: What is the purpose of the preamble in an Ethernet frame?
(b) To synchronize the receiver's clock. The preamble allows the receiver to lock onto the bit stream before the actual frame starts.
Question 9: What is the IEEE standard for Ethernet?
(b) IEEE 802.3. This standard defines Ethernet.
Question 10: How many bytes are in a MAC address used in Ethernet?
(b) 6. MAC addresses are 6 bytes (48 bits).
Question 11: What is the maximum distance for 1000BASE-T (Gigabit Ethernet over copper)?
(a) 100 m. 1000BASE-T supports up to 100 meters over Cat5e or better.
Question 12: Which of the following is a benefit of switched Ethernet over shared Ethernet?
(b) No collisions and higher throughput. Switches create dedicated point-to-point links, eliminating contention.
Question 13: The EtherType value for IPv4 is:
(a) 0x0800. This EtherType indicates IPv4.
Question 14: What is the effect of adding a VLAN tag (802.1Q) to an Ethernet frame?
(a) It increases the maximum frame size to 1522 bytes. The tag adds 4 bytes.
Question 15: Which Ethernet standard is often used for backbone links in data centers with speeds of 40 Gbps?
(b) 40GBASE-SR4. This is a common 40 Gbps standard using multimode fiber.
Question 16: What is the purpose of the "padding" field in an Ethernet frame?
(a) To ensure the frame meets the minimum size requirement. If the payload is less than 46 bytes, padding is added.
Question 17: In CSMA/CD, what does a node do when it detects a collision?
(a) It sends a jamming signal and then uses binary exponential backoff. This ensures all nodes detect the collision and retransmission is delayed.
Question 18: What is the maximum throughput (data rate) of a 100 Mbps Ethernet link if all frames are 64 bytes (minimum size), ignoring preamble and interframe gap?
(d) 100 × (46/64) ≈ 71.9 Mbps. Since payload is 46 bytes out of 64 total, the efficiency is 46/64 ≈ 71.9%. However, this ignores preamble and interframe gap, which further reduce throughput. The correct answer is (d) among the given.
Question 19: Which of the following Ethernet standards uses fiber optic cabling?
(c) 1000BASE-SX. It uses multimode fiber.
Question 20: In Ethernet, what is the interframe gap (IFG) used for?
(a) To allow the receiver to process frames. The IFG is a minimum idle period between frames.
Question 21: What is the typical interframe gap in Ethernet?
(a) 96 bit times. This is the minimum gap between frames.
Question 22: What does the acronym "CSMA/CD" stand for?
(a) Carrier Sense Multiple Access with Collision Detection. This is the original Ethernet access method.
Question 23: Which of the following is a characteristic of 10GBASE-T?
(b) Uses Cat6a or better cabling. 10GBASE-T requires Cat6a or Cat7 for 100 m.
Question 24: What is the role of the Start Frame Delimiter (SFD) in Ethernet?
(a) To indicate the start of the frame after the preamble. The SFD (10101011) signals that the frame is about to begin.
Question 25: If an Ethernet frame has a payload of 1000 bytes, what is the total frame size without preamble and SFD?
(b) 1018 bytes. Total = 6 (dest) + 6 (src) + 2 (type) + 1000 (payload) + 4 (FCS) = 1018 bytes.
Question 26: What is the primary reason for the adoption of full-duplex Ethernet in modern networks?
(b) It doubles the effective bandwidth and eliminates collisions. Full-duplex allows simultaneous transmit and receive, effectively doubling throughput.
Question 27: What is the EtherType value for ARP?
(b) 0x0806. This indicates ARP.
Question 28: Which of the following Ethernet standards uses single-mode fiber for long-distance transmission?
(b) 1000BASE-LX. It uses single-mode fiber for distances up to 5 km.
Question 29: What happens if an Ethernet frame's payload is exactly 1500 bytes?
(b) It is transmitted without padding. 1500 bytes is the maximum payload, so no padding is needed.
Question 30: In a switched Ethernet network, a switch learns MAC addresses by:
(a) Reading the source MAC of incoming frames and associating it with the incoming port. This is called MAC learning.
Exercise 1: Frame Size Calculation
Calculate the total frame size (in bytes) for an Ethernet II frame with a payload of 200 bytes. Include all fields from destination MAC to FCS. What is the efficiency (payload / total)?
Total = 6 (dest) + 6 (src) + 2 (type) + 200 (payload) + 4 (FCS) = 218 bytes.
Efficiency = 200/218 ≈ 91.7%.
Exercise 2: Minimum Frame Padding
A frame has a payload of 30 bytes. How many bytes of padding are needed to meet the minimum frame size? What is the resulting total frame size?
Minimum frame size = 64 bytes (excluding preamble and SFD). Payload = 30 bytes. Header (dest+src+type) = 14 bytes, FCS = 4 bytes. Total without padding = 30+14+4 = 48 bytes. Need 64 - 48 = 16 bytes of padding. Total frame = 64 bytes.
Exercise 3: EtherType Values
Identify the EtherType for the following protocols: IPv4, IPv6, ARP, and 802.1Q VLAN.
Exercise 4: Throughput Calculation
A 1 Gbps Ethernet link transmits frames with a payload of 1500 bytes. Ignoring preamble, SFD, and interframe gap, what is the maximum throughput in Mbps?
Total frame size = 1500 + 18 = 1518 bytes. Efficiency = 1500/1518 ≈ 0.9881. Throughput = 1 Gbps × 0.9881 ≈ 988.1 Mbps.
Exercise 5: Full-Duplex Advantage
In a 100 Mbps half-duplex shared Ethernet segment, the maximum throughput is roughly limited by CSMA/CD to about 50% utilization under high load. In full-duplex switched Ethernet, what is the maximum throughput per link?
In full-duplex, there is no collision and the link can transmit and receive simultaneously, so the maximum throughput is the full link speed in each direction, i.e., 100 Mbps in each direction (200 Mbps aggregate). For data throughput, considering efficiency, ~98.8 Mbps each way.
Exercise 6: VLAN Tagging Impact
How does adding an 802.1Q tag affect the maximum frame size? What is the new maximum payload size if the total frame size must not exceed 1518 bytes?
802.1Q tag adds 4 bytes (2-byte TPID + 2-byte TCI). The original maximum frame size (excluding preamble/SFD) is 1518 bytes. With tagging, the total frame size becomes 1522 bytes. If we want to keep the total under 1518, we would need to reduce payload. However, standard permits 1522 bytes. If we must keep 1518, payload would be 1500 - 4 = 1496 bytes.
Exercise 7: Binary Exponential Backoff
In CSMA/CD, after 3 collisions, what is the range of the backoff timer (in slot times)?
After k collisions, the backoff range is [0, 2^k - 1]. For k=3, range is [0, 7] slots.
Exercise 8: Ethernet Cable Types
Match the following Ethernet standards with their media: 10BASE-T, 1000BASE-SX, 10GBASE-LR, 100BASE-FX.
Exercise 9: Preamble and SFD
What is the purpose of the preamble and SFD? How many bytes are they total?
The preamble (7 bytes) and SFD (1 byte) are used for synchronization and to signal the start of a frame. Total 8 bytes.
Exercise 10: FCS Calculation
If a frame is corrupted, what does the receiver do with the FCS? How does it know?
The receiver computes the CRC-32 over the received frame (excluding preamble/SFD) and compares to the FCS field. If they match, the frame is considered error-free; otherwise, it discards the frame.
Exercise 11: Gigabit Ethernet Distance
What is the maximum distance for 1000BASE-T and 1000BASE-LX?
1000BASE-T: 100 m over Cat5e/6. 1000BASE-LX: 5 km over single-mode fiber (or 550 m over multimode with some limitations).
Exercise 12: Ethernet Switching vs. Hubs
Explain the difference between a hub and a switch in terms of collision domains and frame forwarding.
A hub is a repeater that forwards all frames to all ports, creating a single collision domain. A switch learns MAC addresses and forwards frames only to the intended port, creating separate collision domains and improving performance.
Exercise 13: Interframe Gap
What is the interframe gap in Ethernet and why is it needed?
The interframe gap (IFG) is a minimum idle time of 96 bit times between frames to allow the receiver to process the previous frame and prepare for the next.
Exercise 14: Jumbo Frames
What is a jumbo frame and what are its benefits and drawbacks?
A jumbo frame has a payload larger than 1500 bytes (typically up to 9000 bytes). Benefits: higher efficiency, lower CPU overhead, better throughput for large data transfers. Drawbacks: requires all devices on the path to support it, can increase latency for small packets, and may cause fragmentation if crossing networks that don't support jumbo frames.
Exercise 15: Ethernet Frame Types
How can a receiver distinguish between an Ethernet II frame and an IEEE 802.3 frame?
The receiver checks the field after the source MAC. If the value is ≥ 0x0600, it is an EtherType (Ethernet II). If it is ≤ 0x05FF, it is a Length field (IEEE 802.3), and the payload is followed by an LLC header.
Homework 1: Describe the evolution of Ethernet from 10BASE5 to modern 400 Gbps Ethernet. Highlight key milestones and the improvements in media, speed, and switching.
Ethernet started with 10BASE5 (thick coax) and 10BASE2 (thin coax). Then 10BASE-T brought twisted pair. Fast Ethernet (100BASE-TX) increased speed to 100 Mbps. Gigabit Ethernet (1000BASE-T/SX/LX) raised to 1 Gbps. 10 Gbps, 40 Gbps, 100 Gbps, and now 400 Gbps standards have emerged. Switching replaced shared hubs, providing dedicated bandwidth and full-duplex. Media evolved from coax to UTP to fiber, with longer distances and higher speeds.
Homework 2: Explain in detail the Ethernet frame format for both Ethernet II and IEEE 802.3, including all fields and their sizes. Describe how the EtherType field is used to multiplex different network-layer protocols.
Ethernet II: Preamble (7B), SFD (1B), Dest MAC (6B), Src MAC (6B), EtherType (2B), Payload (46-1500B), FCS (4B). EtherType identifies the protocol (e.g., 0x0800 for IPv4). IEEE 802.3: same but EtherType replaced by Length (2B) indicating payload length. For protocol identification, an LLC header (802.2) follows the Length field, containing DSAP, SSAP, Control. The EtherType allows the receiver to pass the payload to the correct protocol stack.
Homework 3: Calculate the efficiency of an Ethernet link for frame sizes of 64 bytes, 512 bytes, and 1518 bytes (including header/trailer). What is the impact of small frames on throughput?
Efficiency = payload / total frame size. For 64B: payload = 46, total=64 → 71.875%. 512B: payload = 494 (512-18), total=512 → 96.48%. 1518B: payload=1500, total=1518 → 98.81%. Small frames drastically reduce efficiency because header/trailer overhead is significant.
Homework 4: Describe the CSMA/CD protocol in detail, including carrier sense, collision detection, jamming signal, and binary exponential backoff. Why is CSMA/CD no longer used in modern Ethernet?
CSMA/CD: nodes sense carrier; if idle, transmit; if collision detected (by monitoring signal), send jam signal to ensure all nodes detect; then back off using exponential backoff (range doubles after each collision). Modern Ethernet uses switches and full-duplex, which eliminate collisions, making CSMA/CD unnecessary.
Homework 5: Compare and contrast full-duplex and half-duplex Ethernet. How does full-duplex affect the maximum network diameter and performance?
Half-duplex: shared medium, collisions possible, CSMA/CD required, lower throughput (due to contention and collisions). Full-duplex: dedicated point-to-point link, no collisions, simultaneous transmit/receive, doubles effective bandwidth, allows longer distances because no collision detection requirements.
Homework 6: Explain the role of the FCS (Frame Check Sequence) in Ethernet. How is the CRC-32 computed? What happens if the FCS check fails?
FCS is a 4-byte CRC-32 checksum computed over the frame (excluding preamble/SFD). The sender computes and appends it; the receiver recomputes and compares. If mismatch, the frame is considered corrupted and discarded; no retransmission at the link layer – higher layers (e.g., TCP) handle retransmission.
Homework 7: What are the physical layer specifications for 10GBASE-T, including cabling, connectors, and maximum distance? How does it achieve 10 Gbps over copper?
10GBASE-T uses Cat6a or Cat7 unshielded twisted pair (UTP) with RJ45 connectors, maximum distance 100 m. It uses advanced encoding (PAM-16) and error correction (LDPC) to achieve 10 Gbps over 4 pairs.
Homework 8: Describe the concept of VLAN tagging (802.1Q) and its impact on the Ethernet frame. What fields are added and how are they used?
802.1Q adds a 4-byte tag between the source MAC and EtherType/Length. It contains a 2-byte TPID (0x8100) and 2-byte TCI (including 3-bit priority, 1-bit CFI, 12-bit VLAN ID). This allows multiple VLANs on a single physical link (trunk).
Homework 9: What is the interframe gap (IFG) in Ethernet, and how does it affect throughput? Calculate the effective throughput of a 100 Mbps link considering IFG for 1500-byte frames.
IFG is 96 bit times. For 100 Mbps, 96 bits = 0.96 µs. Frame transmission time for 1500B = 12000 bits / 100e6 = 120 µs. Total time per frame = 120 + 0.96 = 120.96 µs. Throughput = (12000 bits) / (120.96 µs) ≈ 99.2 Mbps. So IFG reduces throughput slightly.
Homework 10: Explain how a switch learns MAC addresses and builds its CAM table. What happens when the CAM table is full?
A switch learns MAC addresses by inspecting the source MAC of incoming frames and associating it with the incoming port. It stores this in a CAM table. When forwarding, it looks up the destination MAC and sends the frame only to that port. If the MAC is unknown, it floods to all ports. If the CAM table is full, the switch may flood all frames (like a hub) until entries expire.
Homework 11: Compare Ethernet with other LAN technologies like Token Ring and FDDI. Why did Ethernet become dominant?
Token Ring and FDDI used token-passing, which is deterministic but complex and expensive. Ethernet is simpler, cheaper, and scalable. With the advent of switches, Ethernet's CSMA/CD limitations were overcome, and its cost-effectiveness and continuous speed increases made it dominant.
Homework 12: Describe the purpose and format of the Ethernet preamble and SFD. Why are they considered physical layer fields rather than link layer?
Preamble (7 bytes of 10101010) and SFD (10101011) are used for bit synchronization and frame detection. They are typically considered part of the physical layer because they are not processed by the link layer logic and are stripped before handing the frame to the MAC.
Homework 13: What is the maximum number of VLANs that can be identified using the 12-bit VLAN ID in 802.1Q?
The VLAN ID is 12 bits, so 2^12 = 4096 possible VLANs (0–4095, with 0 and 4095 reserved).
Homework 14: Explain the concept of "cut-through switching" versus "store-and-forward switching" in Ethernet switches. What are the trade-offs?
Store-and-forward receives the entire frame, checks FCS, then forwards; it provides error checking but adds latency. Cut-through reads the destination MAC, starts forwarding immediately, reducing latency but potentially forwarding corrupted frames. Modern switches often use adaptive switching.
Homework 15: Research and describe the Ethernet standard for Power over Ethernet (PoE). How does it work and what are the applications?
PoE (IEEE 802.3af/at) delivers power over Ethernet cabling along with data. It uses phantom power on the data pairs or spare pairs. It negotiates power requirements. Applications: IP cameras, wireless access points, VoIP phones, IoT devices.
This tutorial covered the fundamentals of Ethernet:
Ethernet remains the cornerstone of local area networking, and understanding its operation is essential for network professionals. In the next tutorial, we will explore Ethernet switching and transparent bridging.