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We've established that Slotted ALOHA's synchronization reduces the vulnerable period from 2T to T. Now we'll rigorously prove the consequence:
Slotted ALOHA achieves maximum throughput of exactly 1/e ≈ 36.8%
This result—precisely double Pure ALOHA's 18.4%—emerges directly from halving the vulnerable period. The mathematics is elegant: replacing the exponent -2G with -G in the throughput equation shifts the optimal point from G=0.5 to G=1.0 and doubles S_max from 1/(2e) to 1/e.
This page completes our mathematical analysis of the ALOHA protocol family, providing the rigorous derivation, extensive numerical examples, and the complete comparative framework for understanding random access efficiency.
By the end of this page, you will be able to derive Slotted ALOHA's throughput equation, prove that optimal load is G=1.0, calculate the exact maximum efficiency of 1/e ≈ 36.8%, and confidently compare the complete ALOHA family with rigorous mathematical backing.
Let's derive the Slotted ALOHA throughput equation from first principles.
Setup:
Step 1: Success Condition
A frame transmitted in a slot succeeds if and only if it is the only frame in that slot. Unlike Pure ALOHA, there's no possibility of partial overlap—frames in different slots never interfere.
Step 2: Calculate Success Probability
Under Poisson arrivals with rate G per slot:
Expected number of frames in one slot = G
Probability of exactly zero other frames in the same slot:
$$P(\text{success}) = P(\text{0 other arrivals}) = e^{-G}$$
Note: This assumes the slot contains our frame plus potentially others. Since we're conditioning on our frame being present, we need zero additional frames.
Step 3: Derive Throughput Equation
Throughput S = (frames per slot) × (probability each succeeds) = G × e^{-G}
$$\boxed{S = G \cdot e^{-G}}$$
Alternatively, throughput equals the probability that a slot contains exactly one frame:
$$S = P(\text{exactly 1 frame per slot}) = G \cdot e^{-G}$$
(This is the Poisson probability of exactly 1 arrival when the mean is G.)
Comparison with Pure ALOHA:
| Protocol | Throughput Equation | Exponent |
|---|---|---|
| Pure ALOHA | S = G·e^{-2G} | -2G |
| Slotted ALOHA | S = G·e^{-G} | -G |
The factor of 2 in Pure ALOHA's exponent comes from its 2T vulnerable period. Eliminating this factor via synchronization is the entire source of Slotted ALOHA's improvement.
Both ALOHA variants have throughput of the form S = G·e^(-αG), where α is the vulnerable period in frame times. Pure ALOHA: α=2. Slotted ALOHA: α=1. This unified framework makes the comparison crystal clear.
Now we find the optimal load and maximum throughput using calculus.
Step 1: Differentiate S with respect to G
$$S = G \cdot e^{-G}$$
Using the product rule:
$$\frac{dS}{dG} = e^{-G} + G \cdot (-1) \cdot e^{-G} = e^{-G}(1 - G)$$
Step 2: Set the derivative to zero
$$\frac{dS}{dG} = 0$$ $$e^{-G}(1 - G) = 0$$
Since e^{-G} > 0 for all finite G:
$$1 - G = 0$$ $$\boxed{G^* = 1.0}$$
The optimal offered load is exactly 1 frame attempt per slot—double Pure ALOHA's optimal G* = 0.5.
Step 3: Verify it's a maximum
Second derivative test:
$$\frac{d^2S}{dG^2} = \frac{d}{dG}[e^{-G}(1-G)]$$ $$= -e^{-G}(1-G) + e^{-G}(-1)$$ $$= e^{-G}[-(1-G) - 1]$$ $$= e^{-G}(G - 2)$$
At G* = 1: $$\frac{d^2S}{dG^2}\bigg|_{G=1} = e^{-1}(1 - 2) = -e^{-1} < 0$$
The second derivative is negative, confirming G* = 1 is indeed a maximum.
Step 4: Calculate maximum throughput
$$S_{max} = S(1.0) = 1.0 \cdot e^{-1.0} = \frac{1}{e}$$
$$\boxed{S_{max} = \frac{1}{e} \approx 0.3679 \approx 36.8%}$$
Slotted ALOHA's maximum throughput is exactly 1/e ≈ 36.8%, achieved at offered load G=1.0. This is precisely double Pure ALOHA's 1/(2e) ≈ 18.4% at G=0.5—the direct consequence of halving the vulnerable period.
Let's create a comprehensive comparison of throughput across both ALOHA variants at various load levels.
| G | Pure ALOHA S=Ge^(-2G) | Slotted ALOHA S=Ge^(-G) | Ratio (Slotted/Pure) |
|---|---|---|---|
| 0.1 | 0.0819 (8.2%) | 0.0905 (9.0%) | 1.10× |
| 0.2 | 0.1341 (13.4%) | 0.1637 (16.4%) | 1.22× |
| 0.3 | 0.1646 (16.5%) | 0.2222 (22.2%) | 1.35× |
| 0.4 | 0.1797 (18.0%) | 0.2681 (26.8%) | 1.49× |
| 0.5 | 0.1839 (18.4%) | 0.3033 (30.3%) | 1.65× |
| 0.6 | 0.1807 (18.1%) | 0.3293 (32.9%) | 1.82× |
| 0.8 | 0.1615 (16.2%) | 0.3595 (36.0%) | 2.23× |
| 1.0 | 0.1353 (13.5%) | 0.3679 (36.8%) | 2.72× |
| 1.5 | 0.0747 (7.5%) | 0.3347 (33.5%) | 4.48× |
| 2.0 | 0.0366 (3.7%) | 0.2707 (27.1%) | 7.39× |
Key Observations:
Slotted ALOHA always outperforms Pure ALOHA at every load level
The ratio increases with load: At low G, the ratio is near 1 (both perform similarly). At high G, Slotted ALOHA's advantage becomes dramatic (7× at G=2.0)
Different optimal points: Pure ALOHA peaks at G=0.5; Slotted ALOHA peaks at G=1.0
Graceful degradation: Slotted ALOHA maintains >27% throughput even at G=2.0, while Pure ALOHA collapses to <4%
Practical implications: Slotted ALOHA can operate at higher loads while maintaining acceptable performance
The relationship 1/e = 2 × 1/(2e) is exact, not approximate. Slotted ALOHA's 36.79% is precisely double Pure ALOHA's 18.39%. This reflects the 2:1 vulnerable period ratio.
At Slotted ALOHA's optimal point G=1.0, some interesting properties emerge.
Slot Outcome Probabilities at G=1:
When G=1 (Poisson with λ=1):
| Outcome | Probability | Meaning |
|---|---|---|
| Idle slot (k=0) | e^{-1} = 36.8% | Channel wasted |
| Success (k=1) | 1·e^{-1} = 36.8% | Productive use |
| Collision (k≥2) | 1 - 2e^{-1} = 26.4% | Channel wasted |
Remarkable Observation:
At G=1, the probability of success (36.8%) equals the probability of an idle slot (36.8%)! This elegant symmetry arises from the Poisson distribution: P(0) = P(1) = e^{-1} when λ=1.
Why G=1 is Optimal:
The optimization balances two opposing forces:
At G=1, these forces are perfectly balanced:
Setting marginal benefit = marginal cost yields G=1.
The 36.8% Breakdown:
Of every 100 slots:
The 63.2% waste comes from the fundamental unpredictability of random access.
G=1.0 represents the 'sweet spot' where adding another transmission attempt would cost more in collisions than it would gain in success probability. Operating here maximizes channel utilization—but not per-frame success rate.
Let's apply Slotted ALOHA analysis to practical scenarios.
Example 1: Satellite Network Design
Scenario: A VSAT network uses Slotted ALOHA for uplink access. The satellite channel is 1 Mbps. Frame size is 1000 bits.
Find: Slot duration, maximum successful throughput, and optimal frame rate.
Solution:
Slot duration T = Frame time = 1000 bits / 1,000,000 bps = 1 ms
Maximum throughput = S_max × Channel capacity = (1/e) × 1 Mbps ≈ 368 Kbps
At optimal load G=1.0:
Conclusion: A 1 Mbps satellite channel using Slotted ALOHA can deliver ~368 Kbps of successful user data at maximum efficiency.
Example 2: Capacity Comparison
Scenario: Compare Pure ALOHA vs. Slotted ALOHA for a 100 Kbps channel.
Find: Maximum usable capacity for each protocol.
Solution:
| Metric | Pure ALOHA | Slotted ALOHA |
|---|---|---|
| S_max | 1/(2e) = 18.4% | 1/e = 36.8% |
| Max throughput | 18.4 Kbps | 36.8 Kbps |
| Improvement | Baseline | +100% |
Conclusion: Slotted ALOHA extracts exactly twice the useful capacity from the same channel.
| Operating Point G | Throughput S | Success Rate e^(-G) | Avg Attempts/Frame |
|---|---|---|---|
| 0.5 (conservative) | 30.3% | 60.7% | 1.65 |
| 0.8 (moderate) | 36.0% | 44.9% | 2.23 |
| 1.0 (optimal) | 36.8% | 36.8% | 2.72 |
| 1.2 (aggressive) | 36.1% | 30.1% | 3.32 |
| 1.5 (overloaded) | 33.5% | 22.3% | 4.48 |
Operating at maximum throughput (G=1.0) means 63% of frames need at least one retransmission. For latency-sensitive applications, operating at G=0.5 or lower provides much better per-frame success rates (61% vs. 37%) at the cost of only 18% less throughput (30.3% vs. 36.8%).
Beyond throughput, delay is crucial for system performance. Let's analyze Slotted ALOHA delay characteristics.
Average Number of Transmission Attempts:
Since each attempt succeeds with probability e^{-G}:
$$E[\text{attempts}] = \frac{1}{e^{-G}} = e^G$$
At optimal load (G=1): E[attempts] = e ≈ 2.72 attempts per successful frame.
Delay Components:
Total delay = Waiting delay + Transmission delay + Retransmission delay
Waiting delay: Average time until next slot = T/2 (uniform over [0, T])
Transmission delay: T per attempt
Backoff delay (after each collision): Average K×T/2 per collision, where K is the backoff range
Expected Delay Formula:
For backoff uniformly distributed over [0, K×T]:
$$E[\text{delay}] = \frac{T}{2} + T \cdot e^G + \frac{KT}{2}(e^G - 1)$$
Simplified: $$E[\text{delay}] = T\left[\frac{1}{2} + e^G\left(1 + \frac{K}{2}\right) - \frac{K}{2}\right]$$
Numerical Example:
With T = 1 ms, K = 10, G = 1.0:
$$E[\text{delay}] = 1\text{ms}\left[0.5 + 2.72(1 + 5) - 5\right]$$ $$= 1\text{ms}[0.5 + 16.32 - 5] = 11.82\text{ms}$$
A frame takes ~12 ms on average to successfully transmit, including wait time, transmission, and collision recovery.
| Load G | Avg Attempts | Expected Delay | Throughput S |
|---|---|---|---|
| 0.5 | 1.65 | 6.7 ms | 30.3% |
| 0.8 | 2.23 | 9.7 ms | 36.0% |
| 1.0 | 2.72 | 11.8 ms | 36.8% |
| 1.5 | 4.48 | 21.4 ms | 33.5% |
| 2.0 | 7.39 | 38.4 ms | 27.1% |
Maximum throughput (G=1.0) comes with ~76% higher delay than conservative operation (G=0.5). Network designers must balance these competing objectives based on application requirements.
Let's consolidate everything into a unified framework for ALOHA protocol analysis.
The General ALOHA Throughput Formula:
For any ALOHA-style protocol with vulnerable period αT:
$$S = G \cdot e^{-\alpha G}$$
Where:
Optimal Load:
$$G^* = \frac{1}{\alpha}$$
Maximum Throughput:
$$S_{max} = \frac{1}{\alpha e}$$
| Property | General Formula | Pure ALOHA (α=2) | Slotted ALOHA (α=1) |
|---|---|---|---|
| Vulnerable period | αT | 2T | T |
| Success probability | e^(-αG) | e^(-2G) | e^(-G) |
| Throughput S | Ge^(-αG) | Ge^(-2G) | Ge^(-G) |
| Optimal load G* | 1/α | 0.5 | 1.0 |
| S_max | 1/(αe) | 1/(2e) ≈ 18.4% | 1/e ≈ 36.8% |
| Avg attempts at G* | e^(αG*) = e | e ≈ 2.72 | e ≈ 2.72 |
| P(success) at G* | e^(-1) | 36.8% | 36.8% |
Key Insights from the Framework:
The vulnerable period is everything: It alone determines maximum efficiency
Success probability at optimum is invariant: Both protocols achieve 36.8% per-attempt success at their optimal points
Average attempts at optimum is invariant: Both require ~2.72 attempts per successful frame at optimal load
The slot boundary makes the difference: Synchronization halves α from 2 to 1
Further improvement requires reducing α below 1: This would require overlapping transmissions to succeed (e.g., CDMA spread spectrum approaches)
More advanced protocols (CSMA, CSMA/CD) reduce α further by avoiding transmissions during ongoing ones, approaching α→0 (vulnerable period → 0) in ideal conditions. However, they still have non-zero vulnerability due to propagation delay and sensing limitations.
We've completed our rigorous analysis of the ALOHA protocol family. Here's the complete picture:
The Master Equations:
| Protocol | Throughput | Optimal G | S_max |
|---|---|---|---|
| Pure ALOHA | S = Ge^{-2G} | 0.5 | 1/(2e) ≈ 18.4% |
| Slotted ALOHA | S = Ge^{-G} | 1.0 | 1/e ≈ 36.8% |
Module Complete:
You now have a comprehensive understanding of the ALOHA protocol family—from Pure ALOHA's pioneering 'transmit-at-will' philosophy, through the critical vulnerable period concept, to Slotted ALOHA's synchronized improvement. These protocols, though over 50 years old, remain foundational to modern wireless systems and provide the conceptual framework for understanding all contention-based multiple access.
Congratulations! You've mastered the ALOHA protocol family—Pure ALOHA's 18.4% and Slotted ALOHA's 36.8% maximum efficiencies, the mathematical derivations behind these limits, and the practical trade-offs involved. This knowledge is foundational for understanding CSMA, CSMA/CD, and all modern random access protocols.