Suppose that a 10-Mbps wireless station is transmitting 50-byte frames one immediately after the other.

i) How many frames is it transmitting per second?

ii) If the probability of a frame being damaged (having at least one wrong bit) is 0.004, approximately how many frames will be damaged in one hour?

8 bits = 1 byte
10 Mbps = 10/8 M bytes per second
1 frame = 50 bytes
Frame rate = 10 M / (8 x 50) = 10 M / 400 = 25,000 frames per second
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If each frame has an error probability of 0.004, then (0.004 x 25,000) = 100 frames per second are damaged.
1 hour = 3,600 seconds.
100 damaged frames per second = (100 x 3,600) = 360,000 damaged frames per hour.

Answered by AL2006 | 2024-06-10

There are 1'000'000 Bytes in a Megabyte. In 10 Megabytes there are 10'000'000 bytes. So if it were transmitting 1-byte frames then the answer would be 10'000'000 frames. However this is **not **the case, because it is transmitting 50-byte frames, so we must divide 10'000'000 by 50.
Therefore, the answer to part (i) is: 200'000 Frames /second
If there are 200'000 Frames/second, then there are 12'000'000 Frames/hour (we get that by multiplying 200'000 by 60)
If the probability of a damaged frame is 0.004, then (multiplying 0.004 by 12'000'000) we get (ii) 48'000 Damaged Frames/hour

Answered by brahmin1 | 2024-06-10

The wireless station transmits about 25,000 frames per second and approximately 360,000 frames will be damaged in one hour due to a damage probability of 0.004.
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Answered by AL2006 | 2024-12-26