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Module 2: Data Link Layer & MAC Sublayer

20217m

A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the throughput, if the system (all stations together) produces (i) 1000 frames per second and (ii) 500 frames per second?

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For pure ALOHA, throughput is given by:

S = G x e^(-2G)

where G is the offered load (average number of frames generated per frame-transmission-time).

Step 1: Frame transmission time

Frame size = 200 bits
Bandwidth  = 200 kbps = 200,000 bits/sec

Tfr = Frame size / Bandwidth = 200 / 200,000 = 0.001 sec = 1 ms

Case (i): 1000 frames/sec

G = (frames/sec) x Tfr = 1000 x 0.001 = 1

S = G x e^(-2G) = 1 x e^(-2) = 1 x 0.1353 = 0.1353  (13.53%)

Throughput = S x (frames/sec) = 0.1353 x 1000 ≈ 135 frames/sec successfully delivered

Case (ii): 500 frames/sec

G = 500 x 0.001 = 0.5

S = G x e^(-2G) = 0.5 x e^(-1) = 0.5 x 0.3679 = 0.1839  (18.39%)

Throughput = 0.1839 x 1000 ≈ 184 frames/sec successfully delivered

Observation: G = 0.5 gives the maximum possible throughput for pure ALOHA (~18.4%), which is exactly what case (ii) hits. Case (i)'s higher offered load (G=1) actually results in a lower successful throughput (13.53%) than case (ii), because more collisions occur as load increases beyond the optimal point — illustrating pure ALOHA's low ceiling on maximum achievable throughput.

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