We've established that Pure ALOHA's simple 'transmit-at-will' approach creates a vulnerable period of 2T, during which collisions can occur. Now comes the crucial question:

What is the maximum throughput Pure ALOHA can achieve?

The answer—approximately 18.4%—is one of the most famous results in networking theory. It represents a fundamental barrier: no matter how cleverly you tune backoff parameters or optimize frame sizes, Pure ALOHA cannot exceed this efficiency without changing its core operating principle.

This page derives this result rigorously, explaining every step so you understand not just the answer but why this limit exists. You'll see elegant calculus, Poisson statistics, and optimization theory come together to produce a clean, beautiful result: $S_{max} = \frac{1}{2e} \approx 0.184$.

Let's establish our mathematical framework precisely before deriving the throughput equation.

Definitions:

Key Relationship:

Throughput S is a fraction of offered load G. Specifically:

$$S = G \times P(\text{success})$$

Where P(success) is the probability that a given frame transmission does not collide with any other frame.

Traffic Model Assumptions:

1. Poisson Arrivals: Combined traffic (new frames + retransmissions) forms a Poisson process with rate G per frame time
2. Fixed Frame Size: All frames have identical transmission time T
3. Independent Attempts: Each transmission attempt is independent of others
4. Infinite Population: Enough potential transmitters that the Poisson model applies
5. Negligible Propagation Delay: Frames that don't overlap at their start times don't collide

Why Poisson for Total Traffic (not just new arrivals)?

Strictly speaking, retransmissions are not Poisson—they depend on collision history. However, under equilibrium conditions with many stations and reasonable backoff, the aggregate traffic closely approximates Poisson. This assumption is standard and produces results validated by extensive simulation.

Now we derive the fundamental throughput equation for Pure ALOHA.

Step 1: Success Condition

A frame that starts transmission at time t₀ succeeds if and only if:

• No other frame started in the interval (t₀ - T, t₀) — would overlap with our start
• No other frame starts in the interval (t₀, t₀ + T) — would overlap during our transmission

Combining: no other frame can start in the 2T window (t₀ - T, t₀ + T).

Step 2: Probability Calculation

Under Poisson arrivals at rate G per time T:

Expected number of arrivals in time 2T = G × (2T/T) = 2G

Probability of zero arrivals in time 2T (Poisson probability):

$$P(\text{zero arrivals in } 2T) = e^{-2G}$$

Therefore:

$$P(\text{success}) = e^{-2G}$$

Step 3: Throughput Equation

Since throughput is offered load times success probability:

$$\boxed{S = G \cdot e^{-2G}}$$

This is the Pure ALOHA throughput equation—one of the most important equations in random access theory.

Interpretation:

• At low G: $S \approx G$ (almost everything succeeds)
• At high G: $S \approx 0$ (almost everything collides)
• Somewhere in between: there's a maximum

To find where throughput is maximized, we use calculus to find the critical point of S(G).

Step 1: Differentiate S with respect to G

$$S = G \cdot e^{-2G}$$

Using the product rule:

$$\frac{dS}{dG} = e^{-2G} + G \cdot (-2) \cdot e^{-2G}$$

$$\frac{dS}{dG} = e^{-2G}(1 - 2G)$$

Step 2: Set the derivative to zero

$$\frac{dS}{dG} = 0$$

$$e^{-2G}(1 - 2G) = 0$$

Since $e^{-2G} > 0$ for all finite G, we need:

$$1 - 2G = 0$$

$$\boxed{G^* = 0.5}$$

Step 3: Confirm it's a maximum

Second derivative test:

$$\frac{d^2S}{dG^2} = \frac{d}{dG}[e^{-2G}(1-2G)]$$

$$= -2e^{-2G}(1-2G) + e^{-2G}(-2)$$

$$= e^{-2G}[-2(1-2G) - 2]$$

$$= e^{-2G}[-2 + 4G - 2]$$

$$= e^{-2G}(4G - 4)$$

At G* = 0.5:

$$\frac{d^2S}{dG^2}\bigg|_{G=0.5} = e^{-1}(2 - 4) = -2e^{-1} < 0$$

The second derivative is negative, confirming G* = 0.5 is indeed a maximum.

Step 4: Calculate maximum throughput

$$S_{max} = S(0.5) = 0.5 \cdot e^{-2(0.5)} = 0.5 \cdot e^{-1} = \frac{1}{2e}$$

$$\boxed{S_{max} = \frac{1}{2e} \approx 0.1839 \approx 18.4%}$$

Let's develop intuition for why Pure ALOHA is limited to 18.4% efficiency.

The Efficiency Interpretation:

Efficiency of 18.4% means:

• Out of every 100 seconds of channel time, at most ~18.4 seconds carry successful frames
• The remaining ~81.6% is wasted on:
  • Collisions (partially or fully overlapping frames)
  • Idle time (gaps between transmissions)
  • Overhead (retransmissions of collided frames)

Why So Low?

The fundamental problem is the 2T vulnerable period:

1. Each frame 'claims' 2T of vulnerable time for its T of transmission
2. This immediately suggests at most 50% efficiency (T/2T)
3. But even this isn't achievable because random arrivals can't be perfectly scheduled
4. The exponential penalty of Poisson collisions reduces efficiency further
5. Final result: 1/(2e) ≈ 18.4%

The Optimal Load Interpretation:

At G = 0.5 (the optimal point):

• On average, one frame transmission is attempted every 2 frame times
• This seems sparse, but remember: each attempt 'blocks' 2T of vulnerable time
• At G = 0.5, the expected number of interferers per frame is 2G = 1
• This is the 'sweet spot' where collision risk and transmission rate balance

Going above G = 0.5 adds more attempts but increases collisions faster than throughput—a losing trade.

Let's apply the Pure ALOHA throughput equation to concrete scenarios.

Example 1: Light Traffic Network

Scenario: A Pure ALOHA network has 10 stations. Each station generates an average of 5 frames per second. Frame transmission time is 20 ms.

Find: Offered load G, throughput S, and channel efficiency.

Solution:

Total new frame arrivals: λ = 10 × 5 = 50 frames/second

Frame time T = 20 ms = 0.02 seconds

Offered load (assuming equilibrium where retransmissions balance): G = λ × T = 50 × 0.02 = 1.0 per frame time

Actually, for Pure ALOHA at equilibrium: G includes retransmissions. If we assume G ≈ 1.0:

Throughput: S = G × e^{-2G} = 1.0 × e^{-2} = 0.135

Efficiency: 13.5%

Successful frames per second: S / T = 0.135 / 0.02 = 6.75 frames/second

Conclusion: With λ = 50 frames/second desired, only ~7 get through successfully. This system is overloaded!

Example 2: Optimal Load Design

Scenario: Design a Pure ALOHA network to operate at maximum efficiency. Frame size is 1000 bits, channel capacity is 100 Kbps.

Find: Optimal frame rate, maximum successful throughput.

Solution:

Frame transmission time: T = 1000 bits / 100,000 bps = 10 ms

Optimal offered load: G* = 0.5

Optimal attempts per second: G* / T = 0.5 / 0.01 = 50 attempts/second

Maximum throughput: S_max = 1/(2e) ≈ 0.184

Successful frames per second: 0.184 / 0.01 = 18.4 frames/second

Effective data rate: 18.4 × 1000 bits = 18,400 bps = 18.4 Kbps

Conclusion: A 100 Kbps Pure ALOHA channel delivers only ~18.4 Kbps of successful data at best—wasting over 80% of capacity.

The 18.4% efficiency limit has profound implications for network design and capacity planning.

Capacity Planning:

When designing a Pure ALOHA network:

The 5.4× Rule:

To support X bps of actual data, you need approximately 5.4X bps of raw channel capacity:

$$\text{Required Capacity} = \frac{\text{Desired Throughput}}{S_{max}} = \frac{X}{0.184} \approx 5.4X$$

Operating Below Optimum:

In practice, Pure ALOHA networks operate well below G = 0.5 to provide margin:

Operating at G = 0.25 sacrifices only 3% of throughput (15.2% vs 18.4%) but dramatically improves success rate (61% vs 37%). This is often a worthwhile trade for better latency and stability.

For completeness, let's explore some mathematical extensions to the basic throughput analysis.

Average Delay Analysis:

The average number of transmission attempts per successful frame:

$$E[\text{attempts}] = \frac{1}{P(\text{success})} = \frac{1}{e^{-2G}} = e^{2G}$$

At optimal load (G = 0.5): $E[\text{attempts}] = e^1 \approx 2.72$

On average, each frame requires ~2.72 transmission attempts at optimal load.

Delay Distribution:

If backoff is uniformly distributed over [0, K×T] after each collision:

Expected delay per frame = (transmission time) × (expected attempts) + (average backoff time) × (expected collisions)

This leads to:

$$E[\text{delay}] = T \cdot e^{2G} + \frac{KT}{2}(e^{2G} - 1)$$

The Stability Question:

Pure ALOHA can become unstable if offered load G is allowed to grow unbounded. Consider:

1. High G → many collisions → many retransmissions
2. Retransmissions add to offered load
3. Higher offered load → even more collisions
4. System can enter runaway collapse

For stability, the arrival rate λ must satisfy:

$$\lambda < S_{max} = \frac{1}{2e}$$

If new traffic exceeds maximum throughput, the system cannot clear the backlog and will collapse.

Finite Population Models:

With N stations (finite population), the Poisson assumption weakens. More accurate models use:

$$S = N \cdot p \cdot (1-p)^{2(N-1)}$$

where p is per-station transmission probability. This converges to the Poisson result as N → ∞.

We've rigorously derived Pure ALOHA's fundamental efficiency limit. Let's consolidate the key results:

The Key Equations:

$$S = G \cdot e^{-2G}$$

$$G^* = 0.5$$

$$S_{max} = \frac{1}{2e} \approx 18.4%$$

What's Next:

With Pure ALOHA's limits understood, we'll explore how a simple enhancement—time synchronization—can double the maximum throughput. Slotted ALOHA constrains transmissions to begin only at slot boundaries, reducing the vulnerable period from 2T to T and improving efficiency to 36.8%.