Ethernet, frames and minimum frame size

Why it exists

Ethernet is the wired local network technology: almost every cable that reaches a computer carries Ethernet frames. The classic version is a shared cable using CSMA/CD (medium access); today the cable runs from the host to a switch and there are no collisions, but the frame format, the addresses and the size limits are the same. GATE asks for frame sizes, the reason for the minimum frame, and the efficiency.

Prerequisites

You need CSMA/CD and the minimum frame length (medium access) and the CRC (error detection).

Precise definitions

MAC address. A $48$-bit address, written as six hexadecimal bytes (for example 00:1A:2B:3C:4D:5E), burnt into the network adapter. The first $24$ bits identify the manufacturer (the OUI), the last $24$ are assigned by the manufacturer. Special addresses: FF:FF:FF:FF:FF:FF is the broadcast address (all stations on the LAN); an address whose first transmitted bit is $1$ (the least significant bit of the first byte) is a multicast address. All other addresses are unicast. A station accepts frames addressed to its own address, to broadcast and to the multicast groups it joined.

The Ethernet frame (quoted from the IEEE 802.3 standard):

• the preamble (seven bytes 10101010) and the start frame delimiter (10101011) synchronise the receiver's clock; they are not counted in the frame size;
• destination and source addresses ($6$ bytes each);
• type ($2$ bytes): the upper-layer protocol (for example IPv4 or ARP). In the original IEEE 802.3 frame the field gives the length of the data instead; values above $1500$ are types;
• data: $46$ to $1500$ bytes. The upper limit is the MTU (maximum transmission unit) of Ethernet, $1500$ bytes. If the data are shorter than $46$ bytes the sender pads them (the network layer must know the real length);
• FCS: a $32$-bit CRC over destination, source, type and data.

The frame therefore has $14 + 46 + 4 = 64$ bytes at least and $14 + 1500 + 4 = 1518$ bytes at most. Between frames a sender waits the interframe gap of $96$ bit times ($12$ bytes at the transmission rate).

How it works

Why 64 bytes

At $10$ Mbps a $512$-bit frame lasts $51.2$ µs, the slot time. A sender must still be transmitting when a collision at the far end of the network could be reported back; $51.2$ µs covers the round-trip delay of the longest permitted network, including repeaters and the jam signal. A shorter frame could finish before the collision is noticed, and the sender would not know that it was lost. Hence the minimum data of $46$ bytes (padding if needed), and the maximum network length for a given minimum frame length:

With the bare formula $d = L v / (2R)$ a $512$-bit frame at $10$ Mbps allows $5120$ m; the standard gives $2500$ m, because repeaters, transceivers and the jam signal use up part of the time (quoted). At $100$ Mbps the same frame allows only a tenth of the distance, which is why Fast Ethernet networks are small; Gigabit Ethernet in half-duplex mode extends short frames to $4096$ bits (quoted).

Efficiency

Each frame costs, in addition to its data, the header, the FCS, the preamble with delimiter and the gap, $38$ bytes in total. The efficiency (fraction of the line rate carrying user data, no collisions) is $\text{data} / (\text{data} + 38)$ for data of at least $46$ bytes:

Large frames are efficient ($97.53$ percent for $1500$ bytes) and small ones are not ($54.76$ percent for the minimum, $1.19$ percent for one byte of data padded to $46$). Many exam questions drop the preamble and the gap and use the frame length only: then $1500/1518$ is $98.8$ percent. Read what the question counts.

Shared against switched Ethernet

On a shared (bus or hub) Ethernet all stations are in one collision domain and use CSMA/CD with the efficiency discussed on the previous page. In a switched Ethernet each host has its own cable to a switch; the link is full-duplex (both directions at the same time), there are no collisions, CSMA/CD is not used and the minimum frame constraint is not needed for collision detection (the frame format is kept). A switch forwards a frame only to the port of its destination, learning which address is behind which port (switches and bridges). Ethernet gives unreliable, connectionless service: a frame with a bad FCS is silently dropped and no acknowledgement is sent; recovery is left to higher layers.

Worked examples

Cheat box

Traps and GATE patterns

Connections

The FCS is the CRC of error detection. The size limit of $1500$ bytes forces IP to fragment large packets (IPv4 fragmentation). The destination MAC address is found from an IP address by ARP (ARP, ICMP and DHCP), and a switch forwards frames by MAC address.

PYQs

Consider an Ethernet segment with a transmission
speed of 108 bits/sec and a maximum segment length
of 500 meters. If the maximum speed of propagation
of the signal in the medium is 2×108 meters/sec, then
the minimum frame size (in bits) required for
collision detection is __________.

A network has a data transmission bandwidth of
20×106 bits per second. It uses CSMA/CD in the
MAC layer. The maximum signal propagation time
from one node to another node is 40 microseconds.
The minimum size of a frame in the network is
______ bytes.

In an Ethernet local area network, which one of the
following statements is TRUE?

Consider a CSMA/CD network that transmits data
at a rate of 100 Mbps (108 bits per second) over a 1
km (kilometer) cable with no repeaters. If the
minimum frame size required for this network is
1250 bytes, what is the signal speed (km/sec) in the
cable?

Determine the maximum length of the cable (in km)
for transmitting data at a rate of 500 Mbps in an
Ethernet LAN with frames of size 10,000 bits.
Assume the signal speed in the cable to be 2,00,000
km/s.

Suppose the round trip propagation delay for a 10
Mbps Ethernet having 48-bit jamming signal is 46.4
μs. The minimum frame size is:

Self-check

Revision card

References