Discussion
We have a LAN with 10 Mbps bandwidth and average 12000 frames per minutes and size of each frame is 10000 bits. Then its throughput will be ................... .
Answer: B
The throughput is 2 Mbps.
Throughput is the rate at which data actually passes, whatever the bandwidth:
\[ \begin{aligned} T &= \frac{12000 \times 10000\ \text{bits}}{60\ \text{s}} \\ &= 2{,}000{,}000\ \text{bits/s} = 2\ \text{Mbps} \end{aligned} \]
So the link carries only a fifth of its 10 Mbps bandwidth. 20 Mbps would exceed the bandwidth itself, 0.2 Mbps is off by a factor of ten, and 0.83 Mbps comes from no consistent use of the figures.
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