MadeEasy Test Series: Computer Networks - Lan Technologies

Question

The minimum frame length (in bytes) of the cable of length 240 meter for transmitting data at a rate of 500 Kbps in IEEE 802.3 LAN is __________. (Assume the signal speed in the cable to be, 2,00,000 m/s)

Comments

I think the min frame length in 802.3 i.e Ethernet is 46 Bytes. So other info is not required maybe. What is the ans?

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Yes, the answer is 46 Bytes.

But, are we supposed to remember that? Please solve this by using the values.

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There's no solving involved. If u remember the frame structure in LAN u can derive this val

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I agree that minimum data size in ethernet is $46 Byte$ but calculating using formula by using the given parameters

$150B$ is coming as answer.

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Actually min frame data is 46B and min frame length by adding checksum and addresses it would be 64B in ethernet! But for these values 150B is coming

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Sorry for stating the wrong answer previously. The answer is 150 bytes.

But how is this solved?

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lol  _jerry, why did u say yes? haha

In LAN the transmission delay should be at least twice the propagation delay(to detect collisions)

Therefore, Tt >= 2*Tp

i.e. L/B >= 2* Tp

L >= 2 * (dist/speed) * Bandwidth

L >= 2 * (240/200000) * 500000

L >= 1200 bits

L >= 1200/8 Bytes

Thus min length = 150 B

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If the options have both 150B and 46B, what should we choose?

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Ofcourse 150B.

Because min payload size is 46B and min frame length will be 64B

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@ braindead...Thanks.

Arey, I got confused yaar! There was some other question that I was doing at the same time for which the answer was 46 bytes. :p