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A group of N stations shares a 100 Kbps pure ALOHA channel. Each station outputs a 1000 byte frame on average once every 50 seconds. The maximum value of N is ________.

 

 

This is the solution they have provided, and i cannot make any sense out of it.

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Given.

$Bw = \ 100 \text{ Kbps}$

Let the total number of stations be $N$

Throughput of station $= \frac{\text{Total Data Transfer}}{\text{Time taken}} = \frac{1000 \ \times \ 8 \ bits}{50 \ sec}$  

$\eta_{\text{pure aloha}} =  0.184$ (max)

Throughput of link $ = \eta_{\text{pure aloha}} \times Bw = 0.184 \ \times \ 100 \ \text{ Kbps} = 18.4  \text{ Kbps}$

$\therefore $  Throughput of link $=$ Throughput of station $\times$ Total number of station

$18.4  \text{ Kbps} = \frac{1000 \ \times \ 8 \ bits}{50 \ sec} \times N$

$N = \frac{18.4 \times 10^3 \times 50}{1000 \times 8} = 115$
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