ALOHAnet — Interactive Protocol Lab

What is ALOHA?

A contention-based MAC protocol where stations transmit without central coordination, handling collisions via random back-off retransmission.

Background · 1970

Origin

Designed by Norman Abramson at the University of Hawaii for satellite and packet-radio networks across the islands. Stations transmit frames whenever data is ready — no prior scheduling, no carrier sensing.

ALOHAnet

Pure ALOHA

Fully Asynchronous

Stations transmit at any instant. On collision, each sender waits a random back-off delay and retransmits. Max theoretical throughput: 18.4% at G = 0.5.

S = G · e^−2G

Slotted ALOHA · 1972

Time-Synchronised

Proposed by Roberts (1972). Time divided into equal slots. Stations begin only at slot boundaries — halves collision window, doubles throughput to 36.8%.

S = G · e^−G

Key Insight: In Pure ALOHA, a frame can collide with any other frame that starts within ±1 frame duration — giving a vulnerable period of 2T_p. Slotted ALOHA forces all frames to start at slot boundaries, so two frames either collide entirely or not at all, reducing the vulnerable period to T_p.

Key Parameters Compared

Parameter	Pure ALOHA	Slotted ALOHA
Time Reference	Continuous	Discrete slots
Transmission Rule	Any instant	Slot boundary only
Vulnerable Period	2 × T_p	1 × T_p
Throughput Formula	S = G · e^−2G	S = G · e^−G
Max Throughput S	≈ 18.4%	≈ 36.8%
Optimal Load G*	0.5	1.0
Clock Sync Required	No	Yes
Collision Window	2 frame durations	1 frame duration
P(Success) optimum p	p = 1/(2k)	p = 1/k
P(Empty slot/period)	e^−2G	e^−G

Core Formulae

Pure ALOHA — Throughput

S = G · e^−2G

Max S = 1/(2e) ≈ 0.184 at G = 0.5

Slotted ALOHA — Throughput

S = G · e^−G

Max S = 1/e ≈ 0.368 at G = 1.0

P(Success) — k Stations

P_s = kp(1−p)^k−1

Optimal p = 1/k → P_s,max = (1−1/k)^k−1

P(Collision) — Slotted

P_c = 1 − e^−G(1+G)

= 1 − P(empty) − P(success)

Pure ALOHA — Protocol Steps

Frame Arrival

A station generates a frame and immediately transmits it onto the shared channel — no carrier sense, no idle check, no wait.

ACK / Collision Detection

If the receiver ACKs within a timeout T_ACK, the transmission succeeded. No ACK → collision assumed. Both frames are destroyed.

Random Back-off

Each colliding station waits a uniformly random T_back drawn from [0, 2^k−1] frame durations, where k is the retry count (binary exponential back-off).

Retransmission

After back-off, the station retransmits the frame. This repeats until ACK received or max retry limit M reached.

Slotted ALOHA — Protocol Steps

Global Clock Synchronisation

All stations are synchronised to a global clock that divides time into equal slots of exactly T_p duration (one frame transmission time).

Slot-Boundary Transmission Only

If a frame arrives mid-slot, the station buffers it and waits for the next slot boundary. No transmission starts in the middle of a slot.

Reduced Collision Window

Two frames either start in the same slot (full collision) or in different slots (no collision). The vulnerable period is halved from 2T_p to T_p.

Slot-Based Back-off & Retry

On collision, each station independently picks a random future slot number to retransmit. This reduces the probability of repeated collisions between the same stations.

S vs G Curve

Theoretical throughput vs. offered load. Slotted ALOHA peaks at exactly double the Pure ALOHA maximum.

Reading the chart: x-axis = offered load G (frames generated per frame time), y-axis = throughput S (successfully delivered frames per frame time). Both curves peak then fall — overloading causes a collision cascade. Amber dashed = Pure ALOHA; Blue solid = Slotted.

Throughput S vs. Offered Load G

Pure ALOHA S = G·e^−2G

Slotted ALOHA S = G·e^−G

Throughput Values at Selected G

G (Offered Load)	Pure S = G·e^−2G	Pure S (%)	Slotted S = G·e^−G	Slotted S (%)	Ratio Slotted/Pure

Pure ALOHA Peak

S = 0.184 at G = 0.5

Channel must be loaded at 50% to achieve maximum utilisation. Collision probability rises quadratically beyond G = 0.5 due to the 2T_p vulnerable period.

Slotted ALOHA Peak

S = 0.368 at G = 1.0

Full channel load (G=1) yields maximum throughput. Synchronised slots eliminate partial collisions — two frames either coincide fully or not at all.

Overload Region G > 1

Collision Cascade

Beyond the peak G, throughput decreases sharply. Heavy retransmissions increase G further — a positive feedback loop that can collapse the channel.

Channel Simulator

Discrete-event simulation of the ALOHA channel. Each slot is independently simulated with a Bernoulli arrival model.

Simulation Model: Each station independently transmits in any given slot (or instant) with probability p = G/N, where G is offered load and N is station count. A slot/period is successful if exactly one station transmits.

Channel Parameters

Live simulation engine

Protocol

Stations 4

Offered Load G 0.5

Time Slots 24

Channel View — All Stations Overlaid

Per-Station Activity Timeline

—

Frames Sent

—

Successful

—

Collisions

—

Throughput S

—Simulation not yet run. Set parameters and click Run.

Animated Slot-by-Slot Replay

Slot: —

Speed

Solved Problems

Eight fully-worked numerical problems covering throughput, probability, load comparison, inverse problems, and channel analysis.

Problem 01 — Pure ALOHA Throughput at Peak

Find the maximum throughput of a Pure ALOHA system and the offered load G at which it occurs.

GivenProtocol: Pure ALOHA | Find max S and optimal G

1Pure ALOHA throughput formula: S = G · e^(−2G)

2Differentiate and set to zero: dS/dG = e^(−2G) − 2G·e^(−2G) = e^(−2G)(1 − 2G) = 0

3Solve: 1 − 2G = 0 → G* = 0.5

4Substitute G = 0.5: S = 0.5 × e^(−1) = 0.5 × 0.3679 = 0.1839

5Maximum: S_max = 1/(2e) ≈ 0.184 → 18.4%

✓ Optimal G = 0.5 | Maximum Throughput S = 1/(2e) ≈ 18.4%

Problem 02 — Slotted ALOHA Full Analysis at G = 1

A Slotted ALOHA system has G = 1.0. Find throughput, P(empty slot), and P(collision).

GivenProtocol: Slotted ALOHA | G = 1.0

1Slotted formula: S = G · e^(−G) = 1.0 × e^(−1) = 0.3679

2P(empty): P(0) = e^(−G) = e^(−1) = 0.3679

3P(success) = S: P(1) = G·e^(−G) = 1×e^(−1) = 0.3679

4P(collision): P(c) = 1 − 0.3679 − 0.3679 = 0.2642

5Verify: 0.3679 + 0.3679 + 0.2642 = 1.0 ✓

✓ S = 36.79% | P(empty) = 36.79% | P(success) = 36.79% | P(collision) = 26.42%

Problem 03 — Load Comparison at Multiple G Values

Compare Pure and Slotted ALOHA throughput at G = 0.25, 0.5, 1.0, and 2.0.

GivenG = {0.25, 0.5, 1.0, 2.0} | Both protocols

1G=0.25 Pure: S = 0.25 × e^(−0.5) = 0.25 × 0.6065 = 0.1516 (15.16%)

2G=0.25 Slotted: S = 0.25 × e^(−0.25) = 0.25 × 0.7788 = 0.1947 (19.47%)

3G=0.5 Pure: S = 0.5 × e^(−1) = 0.1839 (18.39%) ← Peak Pure

4G=0.5 Slotted: S = 0.5 × e^(−0.5) = 0.5 × 0.6065 = 0.3033 (30.33%)

5G=1.0 Pure: S = 1.0 × e^(−2) = 0.1353 (13.53%)

6G=1.0 Slotted: S = 1.0 × e^(−1) = 0.3679 (36.79%) ← Peak Slotted

7G=2.0 Pure: S = 2.0 × e^(−4) = 0.0366 (3.66%)

8G=2.0 Slotted: S = 2.0 × e^(−2) = 0.2707 (27.07%)

✓ Slotted ALOHA always outperforms Pure ALOHA at the same offered load. At G = 2.0: Pure drops to 3.66% while Slotted retains 27.07%.

Problem 04 — Probability of Successful Transmission

5 stations in Slotted ALOHA, each transmitting with p = 0.2 per slot. Find P(exactly one transmits) and the optimal p.

Givenk = 5 | p = 0.2 | Slotted ALOHA

1P(success) formula: P_s = C(k,1) · p · (1−p)^(k−1) = k·p·(1−p)^(k−1)

2Compute (1−p)^(k−1): (1 − 0.2)^4 = (0.8)^4 = 0.4096

3Per station: p · (1−p)^4 = 0.2 × 0.4096 = 0.08192

4All k stations: P_s = 5 × 0.08192 = 0.4096

5Optimal p = 1/k = 1/5 = 0.2 — already at optimal!

6Max P_s at p* = 1/k: (1 − 1/5)^4 = (0.8)^4 = 0.4096

✓ P(successful transmission) = 0.4096 = 40.96% | Optimal p = 1/k = 0.2 ✓ (already optimal)

Problem 05 — Find G Given Throughput (Inverse Problem)

Pure ALOHA throughput is 0.10. Find all possible values of G.

GivenProtocol: Pure ALOHA | S = 0.10

1Equation: 0.10 = G · e^(−2G) — transcendental, two solutions exist.

2S_max = 0.184 > 0.10, so two solutions: one below G=0.5 (stable) and one above (unstable).

3Solution 1 (G < 0.5): G ≈ 0.13: 0.13 × e^(−0.26) = 0.13 × 0.7710 ≈ 0.100 ✓

4Solution 2 (G > 0.5): G ≈ 1.26: 1.26 × e^(−2.52) = 1.26 × 0.0806 ≈ 0.102 ≈ 0.10 ✓

5G₁ ≈ 0.13 is the stable operating point. G₂ ≈ 1.26 is unstable — collisions increase G further.

✓ G₁ ≈ 0.13 (stable, lightly loaded) G₂ ≈ 1.26 (unstable, overloaded)

Problem 06 — Channel Throughput in Kbps

A Pure ALOHA channel has bandwidth 200 kbps and frame size 1000 bits. If G = 0.5, find throughput in frames/sec and in kbps.

GivenBW = 200 kbps | Frame = 1000 bits | G = 0.5 | Pure ALOHA

1Frame transmission time: T_p = 1000 / 200,000 = 5 ms

2Max frames/sec: R = 1 / T_p = 1 / 0.005 = 200 frames/sec

3Throughput S at G = 0.5: S = 0.5 × e^(−1) = 0.1839

4Successful frames/sec: S × R = 0.1839 × 200 = 36.79 frames/sec

5Effective throughput: 36.79 × 1000 = 36,788 bps ≈ 36.79 kbps

✓ Throughput = 36.79 frames/sec = 36.79 kbps (18.4% of 200 kbps channel capacity)

Problem 07 — Slotted ALOHA with N Stations

In Slotted ALOHA with 10 stations each transmitting p = 0.1 per slot, find G, S, P(success), P(empty), P(collision).

Givenk = 10 | p = 0.1 | Slotted ALOHA

1Offered load: G = k × p = 10 × 0.1 = 1.0

2Throughput: S = G × e^(−G) = 1.0 × e^(−1) = 0.3679

3P(success) binomial: P_s = k·p·(1−p)^(k−1) = 10 × 0.1 × (0.9)^9

4Compute (0.9)^9: (0.9)^9 = 0.3874

5P(success): 10 × 0.1 × 0.3874 = 0.3874

6P(empty): (1−p)^k = (0.9)^10 = 0.3487

7P(collision): 1 − 0.3874 − 0.3487 = 0.2639

✓ G = 1.0 | S = 36.79% | P(success) = 38.74% | P(empty) = 34.87% | P(collision) = 26.39%

Problem 08 — Minimum Stations to Achieve Target Throughput

Slotted ALOHA: How many stations k each transmitting p = 0.05 per slot are needed to achieve throughput S ≥ 0.30?

Givenp = 0.05 | Target S ≥ 0.30 | Slotted ALOHA

1G = k × p = k × 0.05. Need S = G·e^(−G) ≥ 0.30

2From S-G curve, G·e^(−G) ≥ 0.30 holds for approximately 0.475 ≤ G ≤ 1.79

3Minimum G = 0.475 → k_min = G/p = 0.475/0.05 = 9.5 → k = 10

4Verify k=10: G = 10×0.05 = 0.5, S = 0.5×e^(−0.5) = 0.3033 ≥ 0.30 ✓

5Verify k=9: G = 9×0.05 = 0.45, S = 0.45×e^(−0.45) = 0.2869 < 0.30 ✗

✓ Minimum stations k = 10 (giving G = 0.5, S = 30.33% ≥ 30%)

∑ Smart Numerical Solver

Protocol

Offered Load G frames per frame-time

No. of Stations k active stations

Tx Probability p per slot per station

Bandwidth BW

channel bandwidth

Frame Size L

size of each frame

Target Throughput S (inverse only) for "find G given S" problems

Target S% S≥? (station count only) minimum required throughput %

What do you want to find?

Protocol Comparison

Advantages, disadvantages, and real-world use cases for Pure vs. Slotted ALOHA.

Pure ALOHA

✓Simple implementation — no clock synchronisation required

✓Works well in distributed, uncoordinated environments

✓Immediate transmission — no slot boundary wait

✗Maximum throughput only 18.4% — large wasted capacity

✗Vulnerable period 2T_p — higher collision probability

✗Unstable under high load — collision cascade possible

✗Partial collisions waste channel time for all senders

Slotted ALOHA

✓Double the max throughput — 36.8% efficiency

✓No partial collisions — cleaner failure modes

✓Better stability at moderate-to-high offered loads

✓Foundation for modern TDMA and LTE random-access

✗Requires precise global clock synchronisation

✗Idle time introduced if frame arrives mid-slot

✗Synchronisation overhead and failure point

Probability Distribution at Peak G

Event	Pure ALOHA (G=0.5, peak)	Slotted ALOHA (G=1.0, peak)
P(success)	0.1839	0.3679
P(empty)	0.3679	0.3679
P(collision)	0.4482	0.2642

Real-World Applications: Pure ALOHA → early satellite networks, RFID tag reading (ISO 18000-6B). Slotted ALOHA → GSM Random Access Channel (RACH), LTE PRACH, Bluetooth Low Energy advertising, IEEE 802.11 DCF (evolved form), satellite VSAT networks.

References

Prescribed Textbook

Computer Networks

Andrew S. Tanenbaum & David Wetherall, 5th Ed. Chapter 4.2.1 — The ALOHA Protocol.

→ Publisher Page

Video Resource

ALOHA Protocols

Video series covering Pure ALOHA, Slotted ALOHA, throughput derivation, and solved numericals.

→ Watch on YouTube

Educational Website

GeeksforGeeks — ALOHA

Website containing definitions, types, formulae and applications.

→ Read Article

Pure & Slotted ALOHA

Origin

Fully Asynchronous

Time-Synchronised

Frame Arrival

ACK / Collision Detection

Random Back-off

Retransmission

Global Clock Synchronisation

Slot-Boundary Transmission Only

Reduced Collision Window

Slot-Based Back-off & Retry

S = 0.184 at G = 0.5

S = 0.368 at G = 1.0

Collision Cascade

Find the maximum throughput of a Pure ALOHA system and the offered load G at which it occurs.

A Slotted ALOHA system has G = 1.0. Find throughput, P(empty slot), and P(collision).

Compare Pure and Slotted ALOHA throughput at G = 0.25, 0.5, 1.0, and 2.0.

5 stations in Slotted ALOHA, each transmitting with p = 0.2 per slot. Find P(exactly one transmits) and the optimal p.

Pure ALOHA throughput is 0.10. Find all possible values of G.

A Pure ALOHA channel has bandwidth 200 kbps and frame size 1000 bits. If G = 0.5, find throughput in frames/sec and in kbps.

In Slotted ALOHA with 10 stations each transmitting p = 0.1 per slot, find G, S, P(success), P(empty), P(collision).

Slotted ALOHA: How many stations k each transmitting p = 0.05 per slot are needed to achieve throughput S ≥ 0.30?

Computer Networks

ALOHA Protocols

GeeksforGeeks — ALOHA

📖 Learn — ALOHA Protocols

What is ALOHA?

Why does it matter?

Two variants

Pure ALOHA — How it works

Vulnerable Period

Why is efficiency so low?

Slotted ALOHA — How it works

Reduced Vulnerable Period

Trade-off

Core Formulae Reference

Variable Definitions

Exam Quick Tips

Development Team

How to Use