Pure Aloha | Made Easy Full Syllabus Test

Question

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.

Answer 1

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

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

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

Comments

The throughput (or efficiency or $S$) defines the number of "successful" frames I can send in one transmission time.

In 2021 question they defined throughput as “number of successful frames I can send in one second”, so that’s why we multiplied $S$ with 1000 cause our transmission time was $1$ ms ($S$ in the question was calculated as the frames we can transmit in $1$ ms or 1 $T_t$).

Coming to this question, maximum efficiency tells us the maximum number of successful frames I can send in one $T_t$ multiplying it with the Channel Capacity gives us the total bits we can send in one $T_t$. For the opposite side we are doing the same where we are dividing the maximum bandwidth we can achieve to N number of channels each with a defined bandwidth (1000 Bytes each 50 seconds).

---

@DebRC Thank you. You’ve explained so nicely :) 

---

You’re welcome, man. :)