Ethernet (IEEE 802.3) — CSMA - CD, frame format: Foundations

Before you can read the parent note, you must own every letter and picture it leans on. We build them in order — each one uses only the ones above it.

1. A "bit" and a "byte" — the atoms of everything

Picture: think of a row of light switches. Each switch is a bit (off = 0, on = 1). Eight switches side by side make one byte.

Why the topic needs it: Ethernet measures frames in both bits (for timing math) and bytes (for the frame table). You will constantly convert: $1 byte = 8 bits$ , so $64 bytes = 512 bits$ . If you can't switch units fluently, the "512 bits = 64 bytes" line looks like magic.

Recall Quick check

How many bits in 46 bytes? ::: $46 \times 8 = 368$ bits.

2. Bandwidth $B$ — how fast bits leave the sender

Picture: imagine bits as beads dropping onto a conveyor belt. $B$ is how many beads per second the machine drops. A faster machine (bigger $B$ ) drops more beads each second.

Figure — Ethernet (IEEE 802.3) — CSMA - CD, frame format

Why the symbol $B$ ? Every timing formula on the parent page needs to know how quickly bits become "wire signal." Without $B$ you cannot turn "512 bits" into "how many seconds."

3. The medium as one shared line — "Multiple Access"

Why the topic needs it: "Multiple Access" (the MA in CSMA/CD) means all stations share this one line equally — there is no boss handing out turns. This is the whole reason a decentralized etiquette (sense, detect, back off) is required. Contrast this with Switched vs Shared Ethernet, where a switch gives each station its own line and collisions vanish.

4. The symbol $L$ — how long is a frame?

Picture: the beads from Section 2 lined up as one message. If the message is $L$ beads long, then $L$ is simply "how many beads I must drop to send the whole thing."

Why it needs a name: The central result $L_{min} = 2 T_{p} B$ is an inequality solved for $L$ . We introduce $L$ before the timing formulas so that every later line already knows what it means: "the thing we are solving for," and $L_{min}$ as "the boundary case where the inequality becomes an equality."

5. Distance, speed, and TIME — where delay comes from

Two different times live on this wire. Confusing them wrecks the whole topic, so we define each with its own picture. This split is the heart of Propagation Delay vs Transmission Delay.

5a. Propagation delay $T_{p}$ — travel time of the signal

Two new symbols enter here, so we name them first:

Picture: you shout across a canyon. The delay before your friend hears the first sound is propagation delay — it depends on how wide the canyon is ( $d$ ) and how fast sound moves ( $v$ ), not on how loudly or how long you shout.

Why $v \approx 2 \times 1 0^{8}$ m/s? Electrical signals in copper travel at roughly two-thirds the speed of light. That number lets us turn "2 km of cable" into "10 microseconds of travel."

5b. Transmission delay $T_{t r an s mi t}$ — time to shove the bits out

Picture: back to the conveyor belt. If a frame is $L$ beads long and the machine drops $B$ beads per second, the time until the last bead leaves is $L / B$ . This is about the length of the message, not the length of the cable.

6. Round-trip, slot time, and the factor of $2$

Why the topic needs it: Every headline formula hides this factor. $L_{min} = 2 T_{p} B$ and $T_{s l o t} = 2 T_{p}$ both come from the same out-and-back picture. If you ever write $T_{p}$ where $2 T_{p}$ belongs, your answer is halved and wrong.

7. The detection rule — and deriving $L_{min}$ step by step

Now we combine the pieces built above. This is the single most important derivation on the whole parent page, so we do it one line at a time.

Now derive $L_{min}$ . Each step says WHAT we do and WHY.

WHAT: replace $T_{s l o t}$ by its definition $2 T_{p}$ (from Section 6). WHY: we want the final answer written in terms of the raw propagation delay $T_{p}$ and bandwidth, not the shorthand $T_{s l o t}$ . $T_{t r an s mi t} \geq 2 T_{p}$
WHAT: replace $T_{t r an s mi t}$ by its definition $L / B$ (from Section 5b). WHY: we want a condition on the frame length $L$ , not on time, so we substitute the formula that contains $L$ . $\frac{L}{B} \geq 2 T_{p}$
WHAT: multiply both sides by $B$ . WHY: to isolate $L$ alone on the left. Since $B > 0$ , the inequality direction is unchanged. $L \geq 2 B T_{p}$
WHAT: read off the smallest allowed $L$ — the boundary where " $\geq$ " becomes " $=$ ". WHY: that boundary is the minimum frame size. $L_{min} = 2 T_{p} B$

So the whole rule is just: time to send $\geq$ time for the crash to come back, rewritten in terms of length.

Recall Quick check

Starting from $T_{t r an s mi t} \geq 2 T_{p}$ , why can we multiply by $B$ without flipping the sign? ::: Because $B$ (bits/second) is always positive, and multiplying an inequality by a positive number keeps its direction.

8. Exponents and $2^{n}$ — the language of backoff

Picture: a doubling staircase. Each collision climbs one step and doubles the number of hiding spots (random wait choices). After collision $n = 2$ there are $2^{2} = 4$ spots ${0, 1, 2, 3}$ ; after $n = 3$ there are $8$ spots.

Why the topic needs it: "Binary exponential backoff" is this staircase. After the $n$ -th collision the station waits $K \times T_{s l o t}$ , where $K$ is picked uniformly from ${0, 1, \dots, 2^{m i n (n, 10)} - 1}$ . More collisions ⇒ bigger window ⇒ smaller chance two stations pick the same wait.

Recall Quick check

After the 3rd collision, from what set is $K$ chosen? ::: ${0, 1, \dots, 2^{3} - 1} = {0, 1, \dots, 7}$ .

64 \times 8 = 512

bits.

State bandwidth

B

in plain words and units

bits pushed onto the wire per second, measured in bits/s (bps).

Difference between

T_{p}

and

T_{t r an s mi t}

T_{p}

= signal travel time (depends on cable length and speed);

T_{t r an s mi t} = L / B

= time to emit all bits (depends on frame length and bandwidth).

What do

d

and

v

stand for

d

= cable length in metres;

v

= signal speed along the copper (~

2 \times 1 0^{8}

m/s).

Why the factor 2 appears in

2 T_{p}

the collision news must travel out to the far end and back — two crossings.

Formula for propagation delay

T_{p} = d / v

, distance over signal speed.

Formula for transmission delay

T_{t r an s mi t} = L / B

Definition of slot time

T_{s l o t} = 2 T_{p}

, the worst-case round-trip time.

Derive

L_{min}

from the detection rule

T_{t r an s mi t} \geq T_{s l o t} = 2 T_{p} \Rightarrow L / B \geq 2 T_{p} \Rightarrow L \geq 2 B T_{p} \Rightarrow L_{min} = 2 T_{p} B

What set does

K

come from after the

n

-th collision

{0, 1, \dots, 2^{m i n (n, 10)} - 1}

Value of

2^{10}

1024

What happens after 16 collisions

the frame is dropped and an error is reported to the upper layer.

Is the preamble counted in the 64-byte minimum

No — only Destination MAC through FCS.

Meaning of "half-duplex"

send or receive, but not both at the same instant.

See the parent topic: Ethernet — CSMA/CD & Frame Format.

Foundations — Ethernet (IEEE 802.3) — CSMA - CD, frame format

1. A "bit" and a "byte" — the atoms of everything

2. Bandwidth $B$ — how fast bits leave the sender

3. The medium as one shared line — "Multiple Access"

4. The symbol $L$ — how long is a frame?

5. Distance, speed, and TIME — where delay comes from

5a. Propagation delay $T_{p}$ — travel time of the signal

5b. Transmission delay $T_{t r an s mi t}$ — time to shove the bits out

6. Round-trip, slot time, and the factor of $2$

7. The detection rule — and deriving $L_{min}$ step by step

8. Exponents and $2^{n}$ — the language of backoff

9. Reading the parent's other symbols

Prerequisite map

Equipment checklist