The throughput of pure ALOHA is given by:

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The throughput of pure ALOHA is given by:

  1. A.

    \(S=G\)

  2. B.

    \(S=e^{2G}\)

  3. C.

    \(S=Ge^{2G}\)

  4. D.

    \(S=Ge^{-2G}\)

Show answer & explanation

Correct answer: D

Concept: Throughput S of a random-access protocol equals the offered load G multiplied by the probability that a transmitted frame survives without a collision — i.e. S = G × P(no collision). In pure ALOHA a station transmits the instant it is ready, with no time-slotting, so any two transmissions that overlap even partially collide.

  1. A frame with transmission time Tt sent at time t0 collides with any other frame whose transmission begins in the interval (t0 − Tt, t0 + Tt); this vulnerable period therefore has length 2Tt — twice the frame time (in slotted ALOHA, slot alignment shrinks the vulnerable period to just Tt).

  2. Frame arrivals are modelled as a Poisson process with average rate G frames per frame-time, so the number of arrivals k in an interval spanning τ frame-times has probability P(k) = (Gτ)ᵏ e−Gτ / k!.

  3. A frame survives only if zero other frames arrive during its own vulnerable period, τ = 2. Substituting k = 0, τ = 2 gives the success probability P(0) = e−2G.

  4. Multiplying the offered load by this success probability gives the throughput: S = G · e−2G.

Cross-check: Differentiating S = G·e−2G with respect to G gives dS/dG = e−2G(1 − 2G); setting this to zero gives G = 0.5, and substituting back gives Smax = 0.5 × e−1 ≈ 0.184 — the well-known 18.4% peak efficiency of pure ALOHA, confirming the derived formula. (The analogous derivation for slotted ALOHA drops the factor of 2 in the vulnerable-period length, giving S = G·e−G with a higher peak of 36.8% at G = 1 — consistent with slotting halving the vulnerable period.)

Explore the full course: Nta Ugc Net Paper 2

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