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

Parent: Ethernet (IEEE 802.3) — CSMA/CD & Frame Format

Symbols we reuse (all built in the parent, restated so line one stands alone):

Definition The quantities in play

$B$ = bandwidth = how many bits the wire pushes per second (bits/s). Think of it as the width of the pipe.
$L$ = ==frame length in bits==. This is the master unit for every $L / B$ formula, because $B$ is in bits/s. When a problem gives a size in bytes, we convert with $1$ byte $= 8$ bits before using it in $L / B$ . We will always write "bits" or "bytes" explicitly.
$d$ = cable length = physical distance between the two ends of the wire, in metres.
$v$ = signal speed = how fast a bit travels along the cable, in m/s (typically $2 \times 1 0^{8}$ m/s).
$T_{p}$ = one-way propagation delay $= d / v$ = time for the first bit to travel from one end to the other. Depends only on $d$ and $v$ , not on $B$ .
$T_{t x}$ = transmission delay $= L / B$ = time to push all $L$ bits onto the wire (with $L$ in bits).
slot time $= 2 T_{p}$ = the worst-case round trip; the unit of the backoff clock.

Level 1 — Recognition

Exercise 1.1 (L1)

The bytes below are Ethernet fields shuffled. Order them from first-on-wire to last, and give each field's size in bytes: FCS, Source MAC, Preamble, Data+Pad, Destination MAC, SFD, Length/Type.

Recall Solution 1.1

Wire order (mnemonic "Please Send Dad Some Lovely Data First"):

Order	Field	Bytes
1	Preamble	7
2	SFD (Start Frame Delimiter)	1
3	Destination MAC (DA)	6
4	Source MAC (SA)	6
5	Length/Type	2
6	Data + Pad	46–1500
7	FCS (CRC-32)	4

WHAT / WHY: Preamble+SFD lead so the receiver clock can lock on before real data arrives; FCS trails so it covers everything before it. See MAC Addresses for the two 6-byte address fields and CRC and Error Detection for the FCS.

Exercise 1.2 (L1)

State what each letter of CSMA/CD means, in one phrase each.

Recall Solution 1.2

CS — Carrier Sense: listen before you talk.
MA — Multiple Access: everyone shares one medium equally.
CD — Collision Detection: keep listening while sending; abort on garble.

Exercise 1.3 (L1)

Fill the blanks: minimum Ethernet frame (DA→FCS) is ____ bytes; maximum is ____ bytes; the preamble+SFD are ____ counted in these limits.

Recall Solution 1.3

Minimum $= 64$ bytes, maximum $= 1518$ bytes, and preamble+SFD are not counted (they sit before the frame proper).

Level 2 — Application

Exercise 2.1 (L2)

A $10$ Mbps Ethernet has worst-case round-trip time $51.2 μ s$ . Compute $L_{min}$ in bits and bytes.

Recall Solution 2.1

WHY the formula: to detect a collision, the sender must still be transmitting when the far-end echo returns; that echo takes the full round trip. So $T_{t x} \geq 2 T_{p}$ , i.e. $L / B \geq 2 T_{p}$ (with $L$ in bits), giving $L_{min} = B \cdot (round trip)$ . $L_{min} = 10 \times 1 0^{6} \times 51.2 \times 1 0^{- 6} = 512 bits = \frac{512}{8} = 64 bytes .$ This is exactly the 64-byte minimum.

Exercise 2.2 (L2)

Cable length $2 km$ , signal speed $2 \times 1 0^{8} m/s$ , bandwidth $B = 100$ Mbps. Find $L_{min}$ in bytes.

Recall Solution 2.2

Step 1 — one-way delay $T_{p}$ (why first: detection needs the round trip, built from the one-way): $T_{p} = \frac{d}{v} = \frac{2000 m}{2 \times 1 0 ^{8} m/s} = 10 \times 1 0^{- 6} s = 10 μ s .$ Step 2 — round trip (why $\times 2$ : worst case is a collision at the far end echoing back): $2 T_{p} = 20 μ s .$ Step 3 — minimum length (result is in bits since $B$ is bits/s; then convert): $L_{min} = B \cdot 2 T_{p} = 100 \times 1 0^{6} \times 20 \times 1 0^{- 6} = 2000 bits = \frac{2000}{8} = 250 bytes .$

Exercise 2.3 (L2)

An application wants to send exactly $30$ bytes of payload. What must the sender do, and what is the resulting frame size (DA→FCS)?

Recall Solution 2.3

$30$ bytes is below the $46$ -byte data minimum. WHY: the header+trailer overhead is $6 + 6 + 2 + 4 = 18$ bytes (recall DA = Destination Address, SA = Source Address, FCS = Frame Check Sequence), so data must reach $46$ to hit the $64$ -byte floor. The sender pads the data field from $30$ up to $46$ bytes (adding $16$ pad bytes). (All quantities here are in bytes.) $D A 6 + S A 6 + L e n 2 + d a t a + p a d 46 + F C S 4 = 64 bytes.$

Level 3 — Analysis

Exercise 3.1 (L3)

A 10 Mbps LAN gives 64-byte minimum frames. An engineer upgrades to 100 Mbps but keeps the same cable (same length, same $T_{p}$ ). If she keeps the 64-byte minimum frame, does collision detection still work? What are her two escape routes?

Recall Solution 3.1

The problem: $L_{min} = 2 T_{p} B$ . She multiplied $B$ by $10$ but left $L$ and $T_{p}$ fixed. Now the required minimum is $2 T_{p} B^{'} = 10 \times (old L_{min}) = 640$ bytes, but her frames are only $64$ bytes. A $64$ -byte ( $= 512$ -bit) frame at $100$ Mbps finishes in $512/1 0^{8} = 5.12 μ s$ , but the round trip still needs $51.2 μ s$ — she stops transmitting long before the collision echo returns, so CD silently fails. Two escape routes (both preserve $2 T_{p} B$ ):

Shrink the cable so $T_{p}$ drops by $10 \times$ (round trip back to $5.12 μ s$ ) — this is exactly why Fast Ethernet has a smaller maximum segment length.
Raise the minimum frame to $640$ bytes (carrier extension) so the sender stays on the wire long enough. Modern answer: go full-duplex/switched and abolish collisions altogether — see Switched vs Shared Ethernet.

Exercise 3.2 (L3)

Two stations A and B at opposite ends of a shared cable each want to send. A starts at time $t = 0$ . The one-way delay is $T_{p}$ . Give the exact instants (in terms of $T_{p}$ ) at which: (a) B could still start and collide; (b) A learns of the collision.

The space–time diagram below carries the argument. The horizontal axis is space (station A at the left end of the cable, station B at the right); the vertical axis is time, increasing upward. A black line sloping up-and-right is A's signal travelling out toward B; the red line sloping up-and-left is B's colliding signal echoing back toward A. The three tick marks on A's world-line label the instants $t = 0$ , $t = T_{p}$ , and $t = 2 T_{p}$ .

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

Recall Solution 3.2

Trace the red signal path in the figure.

A's first bit reaches B at $t = T_{p}$ .
(a) B collides if it starts any time in the window $0 \leq t < T_{p}$ — i.e. before A's signal has reached it, so B senses idle. The latest disaster is B starting at $t = T_{p}^{-}$ (just before arrival).
(b) B's colliding signal, launched at $\approx T_{p}$ , travels back and reaches A at $t \approx 2 T_{p}$ . So A only learns of the collision at $t = 2 T_{p}$ . Conclusion: A must still be transmitting at $t = 2 T_{p}$ , hence $T_{t x} \geq 2 T_{p}$ . This is the geometric root of the whole minimum-frame rule.

Level 4 — Synthesis

Exercise 4.1 (L4)

Two stations collide twice in a row. Using binary exponential backoff (slot time $= 51.2 μ s$ ), give the range of $K$ after the 2nd collision and the expected wait. Then compute the probability that both stations pick the same $K$ and collide a third time.

Recall Solution 4.1

After the $n$ -th collision $K$ is uniform on ${0, \dots, 2^{m i n (n, 10)} - 1}$ . Here $n = 2 \Rightarrow K \in {0, 1, 2, 3}$ (four values). Expected wait: $\overset{ˉ}{K} = \frac{0 + 1 + 2 + 3}{4} = 1.5, wait = 1.5 \times 51.2 μ s = 76.8 μ s .$ Same- $K$ probability: B independently picks one of the $4$ values; probability it matches A is $P (match) = \frac{1}{4} = 0.25.$ WHY doubling helps: at $n = 1$ the window was ${0, 1}$ and $P = \frac{1}{2}$ ; doubling the window halves the collision chance. That geometric decay is the point of exponential backoff.

Exercise 4.2 (L4)

Sketch (as a table) how the backoff window grows from $n = 1$ up to $n = 12$ , and identify where and why it stops changing and where the station gives up.

Recall Solution 4.2

Window size after collision $n$ is $2^{m i n (n, 10)}$ values, i.e. $K \in {0, \dots, 2^{m i n (n, 10)} - 1}$ .

$n$	window size	$K$ range	$P (2 stations match)$
1	2	0–1	$1/2$
2	4	0–3	$1/4$
3	8	0–7	$1/8$
4	16	0–15	$1/16$
5	32	0–31	$1/32$
6	64	0–63	$1/64$
7	128	0–127	$1/128$
8	256	0–255	$1/256$
9	512	0–511	$1/512$
10	1024	0–1023	$1/1024$
11	1024	0–1023	$1/1024$ (capped)
12	1024	0–1023	$1/1024$ (capped)
16	1024	0–1023	give up: abort, report to upper layer

Cap at $n = 10$ : the window freezes at $0..1023$ ; growing forever would make waits absurdly long for no extra benefit.
Give up at $n = 16$ : after 16 attempts the station reports failure to the upper layer — the medium is presumed hopelessly congested or broken. (The window is still frozen at $0..1023$ from $n = 10$ onward; the give-up is a separate rule about retry count.)

Level 5 — Mastery

Exercise 5.1 (L5)

You are designing a new shared-medium LAN. Requirements: bandwidth $B = 1$ Gbps ( $1 0^{9}$ bits/s), signal speed $v = 2 \times 1 0^{8}$ m/s, and you insist on keeping the classic 64-byte (512-bit) minimum frame for backward compatibility. Derive the maximum cable length $d_{ma x}$ that still permits reliable CSMA/CD collision detection. Then argue what real Gigabit Ethernet did instead.

Recall Solution 5.1

Restate the underlying constraint first (why it still applies here): CSMA/CD only works if the sender is still transmitting when the worst-case collision echo returns — otherwise it finishes, believes success, and never sees the collision. That echo takes the full round trip $2 T_{p}$ . So exactly as in the parent, we demand $T_{t x} \geq 2 T_{p}, T_{t x} = \frac{L}{B}, T_{p} = \frac{d}{v} .$ Now solve for $d$ ( $L$ in bits, $d$ = cable length, $v$ = signal speed): $\frac{L}{B} \geq \frac{2 d}{v} ⟹ d \leq \frac{L v}{2 B} = d_{ma x} .$ Plug in $L = 512$ bits, $v = 2 \times 1 0^{8}$ m/s, $B = 1 0^{9}$ bits/s: $d_{ma x} = \frac{512 \times 2 \times 1 0 ^{8}}{2 \times 1 0 ^{9}} = \frac{1.024 \times 1 0 ^{11}}{2 \times 1 0 ^{9}} = 51.2 m .$ Interpretation: at Gigabit speed a 512-bit frame drains off the wire so fast that the collision domain must be under ≈51 m — uselessly short for a building. What real Gigabit Ethernet did:

Carrier extension — pad the minimum transmission to $512$ bytes $= 4096$ bits (not 512 bits), so $L$ grows by $8 \times$ and $d_{ma x}$ grows by the same $8 \times$ : $d_{ma x}^{'} = 8 \times 51.2 = 409.6 m (one-way theoretical reach) .$ But $d_{ma x}^{'}$ is the one-way cable length that consumes the entire round-trip budget with zero margin, and it also ignores real-world delays inside repeaters/transceivers and safety margin. The IEEE budget spends roughly half of that theoretical reach on those fixed overheads, leaving a practical half-duplex segment of ≈200 m — which is why Gigabit's carrier-extended collision domain is quoted at about that, not 409.6 m.
Abandon CSMA/CD entirely — run full-duplex over switches, where each link is point-to-point and collisions are impossible (Switched vs Shared Ethernet). This is what everyone actually deploys; CSMA/CD is now historical.

Exercise 5.2 (L5)

Compare CSMA/CD (wired) with the wireless approach in CSMA-CA: in one crisp sentence each, say why detection-during-transmission works on a cable but not on radio, and what CA does instead.

Recall Solution 5.2

CSMA/CD (cable): a station can compare the voltage it sends with the voltage it reads; a mismatch = collision — so it detects collisions while transmitting.
CSMA/CA (radio): its own transmit power drowns out any faint colliding signal, so it cannot listen-while-sending — instead it avoids collisions using inter-frame spacing, random backoff before every send, and optional RTS/CTS handshakes.
Ancestor of both: ALOHA and Slotted ALOHA, which had no carrier sensing at all.

Connections

Propagation Delay vs Transmission Delay — the two delays every problem here balances.
Switched vs Shared Ethernet — why modern Ethernet retires CSMA/CD.
CSMA-CA · ALOHA and Slotted ALOHA · CRC and Error Detection · MAC Addresses