GATE CSE 2021 Set 2 | Question: 54

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Consider a network using the pure $\text{ALOHA}$ medium access control protocol, where each frame is of length $1,000$ bits. The channel transmission rate is $1$ Mbps ($=10^6$ bits per second). The aggregate number of transmissions across all the nodes (including new frame transmissions and retransmitted frames due to collisions) is modelled as a Poisson process with a rate of $1,000$ frames per second. Throughput is defined as the average number of frames successfully transmitted per second. The throughput of the network (rounded to the nearest integer) is ______________

edited

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$L/B$ $→$ $1$ $ms$

$g*$ $e^{-2*g}$

$\text{pure aloha formula, if we put g=1, for one frame, we get 0.13533 }$
$\text{for 1000 frames, we get 135.33}$

$\text{Nearest integer 135}$

edited ago by
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What if we do like :

Efficiency in pure aloha = 18%.

Since Throughput = Efficiency*Bandwith

= 0.18 * 1 Mbps

Throughput in terms of frames = (0.18 * 10^6)/1000

=180
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efficiency in pure aloha is 18.4% when you have g as 0.5

that is the maxima

1
Thanks. Got the mistake.
1 vote
Frame Transmission Time : $T_{fr}= L/B$ = 1000/$10^{6}$ =1ms

System generates 1000 frames in 1 second.

Now, G is defined as average no of frames generated by the system in one frame transmission time.

in 1 second → 1000 frames

$\therefore$  in 1 ms → 1 frame

Therefore,  G= 1

average no of successful transmissions, $S=G*e^{-2G}$ = $1*e^{-2}=$$0.13534 i.e. 13.534% 13.5% of the frames get successfully transmitted. \therefore Throughput = \frac{13.534}{100}$$*1000$= 135.34 $\approx 135$
Read 1st two lines of the question & skipped it :/
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