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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 ______________

But here they didn’t mention the unit so how will we get to know of required in bits/s or Kb/s or Mb/s   ?

Throughput is defined as the average number of frames successfully transmitted per second.

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Frame Transmission Time $:T_{fr}= L/B = 1000/10^{6} =1\text{ms}$

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.

• $1$ second $\to 1000$ frames
• $\therefore 1 \text{ms} \to 1$ frame

Therefore,  $G= 1$

Average number of successful transmissions, $S=G*e^{-2G}$ = $1*e^{-2}=0.13534$

Throughput is defined in question as the average number of frames successfully transmitted per second.

$\therefore$  Throughput $=0.13534*1000= 135.34 \approx 135.$

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$L/B$ $→$ $1$ $ms$

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

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

$\text{Nearest integer 135}$
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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
efficiency in pure aloha is 18.4% when you have g as 0.5

that is the maxima

Thanks. Got the mistake.
thanks ...i did the same mistake...now i got my doubt clear