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

@Yaman Sahu range of answer should be 115-117 right? What if someone take 18.4Kbps = 18.4*1024 instead of 18.4*1000 then the answer will be 117 stations.

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@adad20 In CN, whenever given bandwidth/throughput we take K as 1000 and whenever we are given data then we take K as 1024.

ex:- 1Kbps = 1000 bps

1Kb = 1024 bit

Similarly for M, etc.

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Don't worry about the range. Last year GATE has given the range of 130-140 in the same Pure ALOHA question of set 2.

But what you are doing is conceptually incorrect. @Yaman Sahu has given the correct facts about when to take K to be 1024 and when 1000.

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@Vishal_kumar98 Yeah I got that from your previous comment.

Thanks @Yaman Sahu

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@Aditya_ @Yaman Sahu @Vishal_kumar98 Why are we not recalculating the value of $G$? Although max has been asked, still G (average number of frames generated by the system during one frame transmission time) must be taken out as :-

$1000/50 * (1000*8/(100*10^3)) = 8/5 $ 

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@Abhrajyoti00 Correct me if I’m wrong, but G is the number of frames generated by a station in one transmission time and what you are calculating is the number of bytes transmitted in one transmission time.

The question gives us no idea on the G, how many frames it is actually generating. It just gives us each station transmits one frame each 50 sec. Thus we are considering the highest efficiency we can achieve.

In simple language the question asks "Each station has the capability to transmit at a speed of 1000 Bytes per 50 Second while the channel can support up to 100 Kbps. Find maximum N we can have"

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@DebRC Thank you man! I got my mistake. 

One more thing throughput = 

(in Forouzan)

Then how can we relate throughput = efficiency (as done in the above answer)?

Computer Networks: GATE CSE 2021 Set 2 | Question: 54 (gateoverflow.in) This q can be easily solved with the concept given in Forouzan, but I can't relate it with the above q.

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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).

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

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You’re welcome, man. :)