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

This page builds on the parent Ethernet note. If a term feels new, chase it there or into Propagation Delay vs Transmission Delay.

The scenario matrix

Before solving anything, let us list every kind of problem this topic can ask. Think of it as a checklist: if a worked example below covers a row, that row can never trip you up.

#	Case class	What makes it special	Covered by
A	Standard forward — given $B$ , length, speed → find $L_{min}$	The plain "textbook" direction	Ex 1
B	Bandwidth scaling — double $B$ , what happens?	Tests the proportionality $L_{min} \propto B$	Ex 2
C	Reverse solve — given $L_{min}$ , find max cable length	Rearranging the formula backwards	Ex 3
D	Degenerate: zero-length cable ( $T_{p} = 0$ )	Limiting value — what does the rule say when the wire has no length?	Ex 4
E	Padding trap — data field smaller than 46 bytes	Zero/small payload; the pad kicks in	Ex 5
F	Oversize trap — payload above 1500	The other boundary; frame too big	Ex 5
G	Backoff dice — expected/worst wait after $n$ collisions	Randomness, the $2^{n}$ growth, the cap at 10	Ex 6
H	Efficiency word problem — real link, how much wire is "wasted"	Combines $T_{p}$ and $T_{t x}$ into a ratio	Ex 7
I	Exam twist — collision at a specific station position, not the far end	Not-worst-case geometry: partial round trip	Ex 8

We use several quantities throughout. Let us earn each symbol once before any example uses it:

Definition The symbols we will reuse

$B$ = bandwidth = how many bits the wire pushes out per second, in bits/second (bps). Picture a hose: $B$ is how fast water leaves the nozzle.
$L$ = frame length in bits = the total number of bits in the frame we are sending. If a frame is 64 bytes, then $L = 64 \times 8 = 512$ bits. It is just a count of bits.
$T_{p}$ = one-way propagation delay = time for the front edge of a signal to travel from one end of the cable to the other. Picture a runner: $T_{p}$ is how long to cross the field once.
$RTT$ = round-trip time = the time for a signal to go one way and the echo to come back = $2 T_{p}$ . It is just "there and back," which is why the master formula carries a factor of $2$ .
$T_{t x}$ = transmission delay = time to shove the whole frame onto the wire = $\frac{L}{B}$ = (frame length in bits) ÷ (bits per second). We just defined both $L$ and $B$ , so this ratio is now earned.
$η$ (Greek letter "eta") = efficiency = the fraction of one send-cycle that is genuine data-pushing rather than dead waiting. It is a pure number between 0 and 1 (often shown as a percentage). We build it properly in Ex 7 before using it.

The two delays $T_{p}$ and $T_{t x}$ are different things — a full contrast lives in Propagation Delay vs Transmission Delay.

The master formula from the parent note, which every example leans on:

Row A — Standard forward direction

Worked example Ex 1 — Plain textbook problem

(cell A) A shared Ethernet segment runs at $B = 10$ Mbps. The cable is $2500$ m long. Signal travels at $2 \times 1 0^{8}$ m/s. Find the minimum frame size in bits and bytes.

Forecast: guess before computing — will $L_{min}$ be closer to 64 bytes or to 1000 bytes? (Hint: this is roughly the classic 10 Mbps setup.)

Find one-way delay $T_{p}$ . $T_{p} = \frac{distance}{speed} = \frac{2500}{2 \times 1 0 ^{8}} = 12.5 \times 1 0^{- 6} s = 12.5 μ s$ Why this step? Detection depends on how long a signal takes to cross the wire — that is exactly $T_{p}$ .
Double it for the round trip. $RTT = 2 T_{p} = 25 μ s$ Why this step? Worst case: far station starts just as our signal arrives; its garbage must travel all the way back. There and back = $2 T_{p} = RTT$ (the round-trip time we defined above).
Multiply by bandwidth. $L_{min} = B \cdot RTT = 10 \times 1 0^{6} \times 25 \times 1 0^{- 6} = 250 bits$ Why this step? In $25 μ s$ the wire can carry $B \times RTT$ bits; we must still be sending for at least that many.
Convert to bytes. $250 \div 8 = 31.25$ bytes. Why this step? Frame sizes in the Ethernet standard are quoted in bytes (1 byte = 8 bits), so we translate the bit-count into the unit the standard actually uses.

Verify: Units: $bps \times s = bits$ ✓. Note $31.25 < 64$ — so on a real 10 Mbps LAN the standard forces 64 bytes anyway (the standard is built for the worst legal cable, ~2500 m with repeaters ≈ $51.2 μ s$ RTT). Our shorter link only needs 250 bits, so 64 bytes is safely more than enough.

Row B — What if bandwidth doubles?

Worked example Ex 2 — Scaling the bandwidth

(cell B) Same $2500$ m cable, same speed, but now $B = 20$ Mbps (doubled). What is the new $L_{min}$ , and how does it compare to Ex 1?

Forecast: guess — does $L_{min}$ go up, down, or stay the same when the wire gets faster?

$T_{p}$ is unchanged. Distance and speed did not change, so $T_{p} = 12.5 μ s$ , $RTT = 2 T_{p} = 25 μ s$ . Why this step? Propagation depends on physics of the wire, not on how fast we clock bits — bandwidth does not change signal speed.
Recompute with new $B$ . $L_{min} = 20 \times 1 0^{6} \times 25 \times 1 0^{- 6} = 500 bits$ Why this step? $L_{min} = 2 T_{p} B$ : only $B$ changed, and it doubled.

Verify: $500 = 2 \times 250$ — exactly double Ex 1, as $L_{min} \propto B$ predicts. Intuition: a faster wire finishes a fixed frame sooner, so to still be talking during the round trip you need a longer frame. Faster link ⇒ bigger minimum frame. This is why Gigabit Ethernet needs tricks (carrier extension) to keep 64-byte frames legal.

Row C — Solve the formula backwards

Worked example Ex 3 — Reverse: given frame size, find max cable

(cell C) Standard Ethernet uses $L_{min} = 512$ bits (64 bytes) at $B = 10$ Mbps, signal speed $2 \times 1 0^{8}$ m/s. What is the maximum cable length this frame size can protect?

Forecast: guess whether the answer is nearer 100 m, 2500 m, or 100 km.

Start from the formula and isolate $T_{p}$ . $L_{min} = 2 T_{p} B ⟹ T_{p} = \frac{L _{min}}{2 B}$ Why this step? We know $L_{min}$ and $B$ ; the unknown is the distance, which hides inside $T_{p}$ .
Plug in. $T_{p} = \frac{512}{2 \times 10 \times 1 0 ^{6}} = \frac{512}{2 \times 1 0 ^{7}} = 25.6 \times 1 0^{- 6} s = 25.6 μ s$ Why this step? This is the largest one-way delay the 64-byte frame can still detect.
Convert delay to distance. $length = T_{p} \times speed = 25.6 \times 1 0^{- 6} \times 2 \times 1 0^{8} = 5120 m$ Why this step? distance = speed × time; we invert $T_{p} = dist / speed$ .

Verify: Feed 5120 m back forward: $T_{p} = 5120/2 \times 1 0^{8} = 25.6 μ s$ , $L_{min} = 1 0^{7} \times 51.2 \times 1 0^{- 6} = 512$ bits ✓. (The real Ethernet limit is a bit under this because repeaters add delay — but the pure-cable ceiling is ~5.12 km.)

Row D — The degenerate case: no cable at all

Worked example Ex 4 — Zero-length wire

(cell D, limiting value) Two stations sit at the exact same point (cable length $= 0$ ). $B = 10$ Mbps. What does the formula say $L_{min}$ must be?

Forecast: with no distance, is a minimum frame size even needed?

Compute $T_{p}$ . $T_{p} = \frac{0}{2 \times 1 0 ^{8}} = 0 s$ Why this step? A signal crosses zero distance in zero time.
Apply the formula. $L_{min} = 2 \times 0 \times 10 \times 1 0^{6} = 0 bits$ Why this step? $L_{min} = 2 T_{p} B$ , and $T_{p} = 0$ makes the whole product $0$ .

Verify: $L_{min} = 0$ means: with zero distance, both stations hear each other instantly — a collision is detected the moment it starts, so no minimum length is required for detection. This is the limiting sanity check of the whole idea: the minimum frame size exists only because signals take time to travel. Shrink the wire to nothing and the constraint vanishes. (The 64-byte floor in real Ethernet is a fixed standard value, but the underlying need for it $\to 0$ as distance $\to 0$ .)

Row E + F — Payload boundaries and the pad

Worked example Ex 5 — Data too small and data too big

(cells E & F) (E) An application hands Ethernet only 20 bytes of data. What goes on the wire, and how big is the frame? (F) A different app hands Ethernet 1600 bytes. What happens?

Forecast: for (E) guess the final frame size; for (F) guess whether it fits in one frame.

Part E — undersized payload:

Check against the 46-byte floor. $20 < 46$ , so the data field is padded with $46 - 20 = 26$ bytes of padding. Why this step? The frame from Destination-MAC through FCS must reach 64 bytes (the CSMA/CD minimum), and the only stretchy field is Data. So Data is forced up to 46.
Add the fixed fields. $DA 6 + SA 6 + Len/Type 2 + data + pad 46 + FCS 4 = 64 bytes$ Why this step? This is the exact identity that defines why 46 is the minimum data size.

Verify (E): $6 + 6 + 2 + 46 + 4 = 64$ ✓. The receiver uses the Length field (which still says 20) to know only 20 bytes are real data; the 26 pad bytes are discarded. Frame on wire (excluding preamble/SFD) = 64 bytes.

Part F — oversized payload:

Check against the 1500-byte ceiling. $1600 > 1500$ , so it does not fit in one Ethernet frame. Why this step? Max data = 1500 bytes is a hard design limit (bounds buffer/latency).
A higher layer must split the data. IEEE 802.3 Ethernet itself does not fragment at the data-link layer — the frame simply cannot exceed 1500 bytes of payload. Instead, an upper-layer protocol (typically IP) breaks the 1600 bytes into pieces that each fit, e.g. $1500 + 100$ bytes → two frames. Why this step? Each frame must independently obey $46 \leq data \leq 1500$ ; since the link layer offers no fragmentation, the responsibility falls to IP (or whatever protocol sits above).

Verify (F): First frame data $= 1500$ (legal), second frame data $= 100$ (legal, $\geq 46$ so no padding). Total real bytes carried $= 1500 + 100 = 1600$ ✓. Max frame size (DA→FCS) $= 6 + 6 + 2 + 1500 + 4 = 1518$ bytes.

Row G — Binary exponential backoff dice

The figure below is the derivation for this row: each horizontal bar is the "dice" you roll after a given collision, and its length is literally the number of faces on that dice. Watch the bars double as you read downward — that doubling is the "exponential" in binary exponential backoff, and the point where the bars stop growing is the cap at $n = 10$ . Refer back to it as you work the numbers.

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

Figure s01 — Backoff windows as growing dice. Reading top to bottom, each bar is the range of choices for the random number $K$ after collision $n$ : after collision 0 the range is ${0, 1}$ (2 choices), after collision 1 it is ${0, \dots, 3}$ (4 choices), and each later bar is twice as wide as the one above. A wider bar means more possible waits, so a smaller chance two stations pick the same $K$ and collide again. The widening halts at $n = 10$ , where the range freezes at ${0, \dots, 1023}$ .

Worked example Ex 6 — Wait times after collisions

(cell G) Slot time $= 2 T_{p} = 51.2 μ s$ (standard 10 Mbps). Find: (a) the set of choices and expected wait after the 3rd collision, (b) the maximum possible wait after the 12th collision.

Forecast: does the choice set keep doubling forever? Guess the biggest slot number you can ever roll.

Part (a) — after 3rd collision ( $n = 3$ ):

Build the choice set. In the figure, find the bar labelled $n = 3$ : it has $2^{3} = 8$ faces. So $K \in {0, 1, \dots, 2^{n} - 1} = {0, 1, 2, 3, 4, 5, 6, 7}$ . Why this step? After the $n$ -th collision the window is $2^{n}$ wide — a wider bar (bigger dice) ⇒ smaller chance both stations pick the same $K$ , which is exactly what the widening bars in the figure illustrate.
Average the choices. Uniform over $0..7$ : mean $= \frac{0 + 1 + \dots + 7}{8} = \frac{28}{8} = 3.5$ slots. Why this step? Expected wait = average slot count × slot time.
Convert to time. $3.5 \times 51.2 μ s = 179.2 μ s$ .

Verify (a): sum $0..7 = 28$ , $28/8 = 3.5$ ✓; $3.5 \times 51.2 = 179.2 μ s$ ✓.

Part (b) — after 12th collision ( $n = 12$ ):

Apply the cap. In the figure the bars stop widening at $n = 10$ ; beyond it the dice is frozen. So the exponent is $min (n, 10) = min (12, 10) = 10$ , giving $K \in {0, \dots, 2^{10} - 1} = {0, \dots, 1023}$ . Why this step? The window stops growing after 10 collisions — otherwise waits would balloon absurdly, which is why the figure's bars flatten out.
Take the maximum. Largest $K = 1023$ , so max wait $= 1023 \times 51.2 μ s = 52377.6 μ s \approx 52.4$ ms. Why this step? The question asks for the worst possible wait, and a uniform choice over ${0, \dots, 1023}$ has its largest value at $K = 1023$ — the top of the frozen window — so plugging in the biggest slot count gives the longest wait the protocol can ever impose here.

Verify (b): $2^{10} - 1 = 1023$ ✓; $1023 \times 51.2 = 52377.6 μ s$ ✓. Note the station still keeps trying — it only gives up after 16 total collisions, even though the window froze at collision 10.

Row H — Real-world efficiency word problem

Before the example, we must earn the efficiency symbol we will use. We measure one send-cycle of a collision-free transmission as: the time actually pushing bits onto the wire ( $T_{t x}$ ) plus the unavoidable dead time waiting one round-trip ( $2 T_{p}$ ) so the sender can be sure no collision came back. The useful fraction of that cycle is:

Worked example Ex 7 — How much of the wire is "wasted"?

(cell H) A $1$ km, $10$ Mbps segment (speed $2 \times 1 0^{8}$ m/s) sends a full 1518-byte frame. What fraction of the "sending cycle" is genuine data pushing (transmission) versus dead travel time? Use $η = \frac{T _{t x}}{T _{t x} + 2 T _{p}}$ from the box above.

Forecast: guess — is this link mostly efficient (>90%) or mostly wasted?

Transmission time. Frame $= 1518$ bytes $= 1518 \times 8 = 12144$ bits. $T_{t x} = \frac{L}{B} = \frac{12144}{10 \times 1 0 ^{6}} = 1214.4 \times 1 0^{- 6} s = 1214.4 μ s$ Why this step? The efficiency formula needs $T_{t x}$ in its numerator; this is the "useful work" part of the cycle.
Round-trip delay. $T_{p} = \frac{1000}{2 \times 1 0 ^{8}} = 5 \times 1 0^{- 6} s = 5 μ s$ , so $2 T_{p} = 10 μ s$ . Why this step? The formula's denominator adds the $2 T_{p}$ dead time — the mandatory silence the protocol needs for collision safety — so we must compute it.
Combine into $η$ . $η = \frac{T _{t x}}{T _{t x} + 2 T _{p}} = \frac{1214.4}{1214.4 + 10} = \frac{1214.4}{1224.4} = 0.99183 \approx 99.2%$ Why this step? This is exactly the useful-fraction the formula defines: how much of one cycle was real bits versus wasted travel.

Interpretation: $η \approx 99.2%$ — a short cable carrying a big frame is highly efficient; almost none of the cycle is wasted travel time. Sanity contrast: if the frame were tiny (64 bytes = $51.2 μ s$ ) on a long 2500 m cable ( $2 T_{p} = 25 μ s$ ), efficiency would drop to $51.2/76.2 \approx 67%$ — the classic reason shared Ethernet struggles with short frames on long wires.

Verify: $η = 1214.4/1224.4 = 0.99183$ ✓; units cancel ( $μ s / μ s$ = pure number) ✓; and $η < 1$ as required for any real link ✓.

Real-world caveat (simplification flagged): this is the idealized number. Real Ethernet inserts a mandatory Interframe Gap (IFG) of 96 bit-times of silence between frames — at 10 Mbps that is $96/ (10 \times 1 0^{6}) = 9.6 μ s$ . Adding it to the denominator gives $η = \frac{1214.4}{1214.4 + 10 + 9.6} = \frac{1214.4}{1234.0} = 0.98412 \approx 98.4%$ — slightly lower. Exam questions usually want the idealized $99.2%$ unless they explicitly mention the IFG.

Row I — The exam twist: collision not at the far end

The figure shows where on the wire the second station sits — the round-trip is only as long as the distance to that station and back, not the full cable.

Figure s02 — Collision geometry for Ex 8. Station A (lavender, the sender) sits at the left end; station B (coral) is only 600 m along a 2000 m cable; the far end (mint) is 2000 m away. A's signal travels A→B in $3 μ s$ (top arrow), the collision is born at B (star), and the garble echoes B→A in another $3 μ s$ (bottom arrow). The detection round-trip is therefore governed by the 600 m distance to B, not the full 2000 m — that is the whole point of the twist.

Worked example Ex 8 — Collision at a nearer station

(cell I) Stations A and B are $600$ m apart on a $2000$ m, $10$ Mbps cable (speed $2 \times 1 0^{8}$ m/s). A transmits; B (only 600 m away, not at the far end) starts just before A's signal reaches it, causing a collision at B. What is the latest time after A started that A could detect this particular collision?

Forecast: guess whether the answer uses 600 m or the full 2000 m.

Identify the relevant distance. The collision happens at B, so the garbage only has to travel from B back to A: the relevant one-way distance is $600$ m, not the full cable. Why this step? Worst-case detection uses the farthest station, but this problem fixes the colliding station at 600 m. Round-trip is A→B→A over 600 m each way.
One-way delay to B. $t_{A B} = \frac{600}{2 \times 1 0 ^{8}} = 3 \times 1 0^{- 6} s = 3 μ s$ . Why this step? A's signal reaches B in $t_{A B}$ ; the collision is born at that instant.
Echo back to A. The garbled signal returns over the same 600 m in another $3 μ s$ . $t_{d e t ec t} = 2 \times 3 μ s = 6 μ s$ Why this step? Total time = there ( $A \to B$ ) + back ( $B \to A$ ) = $2 t_{A B}$ .

Verify: $2 \times 600/ (2 \times 1 0^{8}) = 6 \times 1 0^{- 6}$ s $= 6 μ s$ ✓. Contrast: if B were at the far end (2000 m), detection could take $2 \times 2000/ (2 \times 1 0^{8}) = 20 μ s$ — that larger number is what sets $L_{min}$ . This is why the standard is built for the worst-case (farthest) station, not any particular collision.

Recall Cover the answers and test yourself

Ex 1 — $L_{min}$ for 2500 m, 10 Mbps? ::: 250 bits = 31.25 bytes (needed), rounded up to the 64-byte standard. Ex 2 — Double the bandwidth: what happens to $L_{min}$ ? ::: It doubles (250 → 500 bits), because $L_{min} \propto B$ . Ex 3 — Max cable length for 512-bit frame at 10 Mbps? ::: 5120 m (5.12 km). Ex 4 — $L_{min}$ when cable length is 0? ::: 0 bits — no minimum needed; detection is instant. Ex 5E — Frame size when app sends 20 bytes? ::: 64 bytes (data padded to 46). Ex 5F — App sends 1600 bytes? ::: Ethernet cannot fragment; a higher layer (IP) splits it, since payload exceeds the 1500-byte max. Ex 6a — Expected wait after 3rd collision (slot 51.2 µs)? ::: 3.5 slots = 179.2 µs. Ex 6b — Max wait after 12th collision? ::: 1023 slots = 52377.6 µs (window capped at n=10). Ex 7 — Efficiency, 1518-byte frame, 1 km link (idealized)? ::: ≈ 99.2%; ≈ 98.4% once the 96-bit IFG is included. Ex 8 — Detection time for collision at a 600 m station? ::: 6 µs (round trip over 600 m).

Connections

Propagation Delay vs Transmission Delay — every example here juggles $T_{p}$ vs $T_{t x}$ ; master that contrast first.
CRC and Error Detection — the FCS field checked in Ex 5's frame layout.
MAC Addresses — the 6-byte DA/SA fields counted in every frame-size sum.
Switched vs Shared Ethernet — Ex 7's efficiency is why switches replaced shared cable.
ALOHA and Slotted ALOHA — the backoff dice of Ex 6 descend from ALOHA's random retry.
CSMA-CA — the wireless cousin where you cannot listen-while-sending, so these detection sums do not apply.