Data Link Layer and Medium Access Sub Layer · Computer Networks

Pure ALOHA

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Pure ALOHA Protocol


1. Origin and Philosophy of ALOHA

The ALOHA protocol was devised in 1970 by Norman Abramson and his colleagues at the University of Hawaii to interconnect computers located on different Hawaiian islands using ground-based UHF radio broadcasting.

ALOHA represents the earliest and simplest Random Access Protocol (Contention-Based):

  • Core Rule: A station transmits a frame immediately whenever it has data ready to send.
  • There is no central controller and no carrier sensing ("talk before listening").
SVGVector Architecture Diagram (SVG)
Station 1: Frame 1.1 Frame 1.2 Station 2: Frame 2.1 Frame 2.2 Station 3: Frame 3.1 COLLISION! (Frames 1.1, 2.1, 3.1 overlap & destroy each other) t

2. Collision Mechanism & Vulnerable Time

Because stations transmit completely asynchronously without checking whether the channel is busy, frames frequently overlap in time. Any partial overlap, even by a single bit, corrupts both frames completely and is termed a Collision.

Vulnerable Time (VtV_t) Calculation:

Let all frames have a constant transmission duration of TfrT_{fr}.
Suppose Station A begins transmitting a frame at time t0t_0.

SVGVector Architecture Diagram (SVG)
t0 - Tfr t0 t0 + Tfr Station A Frame: Tfr Other stations start in (t0-Tfr, t0): Tfr Collides with beginning of A! Other stations start in (t0, t0+Tfr): Tfr Collides with end of A! Vulnerable Time = 2 * Tfr
  • If any other station transmits between t0−Tfrt_0 - T_{fr} and t0t_0, the tail of its frame will collide with the start of Station A's frame.
  • If any other station transmits between t0t_0 and t0+Tfrt_0 + T_{fr}, its beginning will collide with the tail of Station A's frame.
  • Vulnerable Time for Pure ALOHA:
Vt=2×TfrV_t = 2 \times T_{fr}

3. Mathematical Derivation of Throughput

Assuming frame generation follows a Poisson distribution with an average arrival rate of GG frames per frame-time TfrT_{fr} (where GG is the channel traffic load):

The probability that kk frames are generated during a time interval TT is:

P(k)=(G⋅TTfr)ke−G⋅TTfrk!P(k) = \frac{(G \cdot \frac{T}{T_{fr}})^k e^{-G \cdot \frac{T}{T_{fr}}}}{k!}

For a frame to be successfully transmitted without collision, zero other frames (k=0k = 0) must be generated during the entire vulnerable time interval Vt=2TfrV_t = 2 T_{fr}:

P(0)=(2G)0e−2G0!=e−2GP(0) = \frac{(2G)^0 e^{-2G}}{0!} = e^{-2G}

Throughput Equation (SS):

The channel throughput SS (average number of successful transmissions per frame-time) is:

S=G⋅P(0)=G⋅e−2GS = G \cdot P(0) = G \cdot e^{-2G}

Finding Maximum Throughput:

To find the load GG that maximizes throughput, differentiate SS with respect to GG and set to zero:

dSdG=e−2G−2Ge−2G=e−2G(1−2G)=0  ⟹  1−2G=0  ⟹  G=12=0.5\frac{dS}{dG} = e^{-2G} - 2G e^{-2G} = e^{-2G}(1 - 2G) = 0 \implies 1 - 2G = 0 \implies \mathbf{G = \frac{1}{2} = 0.5}

Substitute G=0.5G = 0.5 into the throughput equation:

Smax⁡=0.5⋅e−2(0.5)=0.5e=12e≈0.184(18.4%)S_{\max} = 0.5 \cdot e^{-2(0.5)} = \frac{0.5}{e} = \frac{1}{2e} \approx \mathbf{0.184 \quad (18.4\%)}

Key Takeaway: The maximum channel capacity of Pure ALOHA is only 18.4%18.4\%. Over 81.6%81.6\% of channel capacity is wasted due to collisions under heavy traffic!


4. Retransmission and Backoff in Pure ALOHA

MermaidDiagram
Rendering diagram…
  • When a collision occurs, stations must not retransmit immediately; doing so would guarantee another immediate collision.
  • Each station waits a random Backoff Time TB=R×TfrT_B = R \times T_{fr} (where RR is a random integer chosen from a range) before attempting to retransmit.

Frequently Asked Questions

QCalculate the throughput of a Pure ALOHA channel with an offered load of $G = 0.2$ frames per frame-time.

S=Ge−2G=0.2⋅e−2(0.2)=0.2⋅e−0.4=0.2×0.6703≈0.134(13.4%)S = G e^{-2G} = 0.2 \cdot e^{-2(0.2)} = 0.2 \cdot e^{-0.4} = 0.2 \times 0.6703 \approx \mathbf{0.134 \quad (13.4\%)}

QWhat is the vulnerable time of Pure ALOHA?

Vt=2×TfrV_t = 2 \times T_{fr}, spanning the duration of one frame-time prior to transmission up to the end of the current frame transmission.

QAt what traffic load $G$ does Pure ALOHA achieve its maximum throughput, and what is the value?

Maximum throughput occurs at G=0.5G = 0.5, yielding Smax⁡=12e≈18.4%S_{\max} = \frac{1}{2e} \approx 18.4\%.

QWhat does a Pure ALOHA station do after a collision?

It does not retransmit immediately (that would collide again). It waits a random backoff time TB=R×TfrT_B = R \times T_{fr} and then retransmits.

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